Where does the power go ?

[PinheadBE]

Hi there,

Here's a somewhat dumb question, but it haunts me for a couple of nights now while it probably shouldn't....

All our means of producing electrical power are running 24/7.
But, obviously, at some times, all of the power which is produced is not used.  For example, during the night.

So, if I simplify the whole grid to one producer (say a 1 GW nuclear power plant) and a bunch of consumers which at a certain time consumes only 500 MW, where do the other 500 MW go ? 

The decay of uranium in the reactor produces the same amount of heat which is transformed in the same amount of steam which in turn makes an alternator turn at the same speed, but the counter-eletromagnetic force exerced on it may be lowered (due to the lower consumption on the grid).
So, where does the remaining power go ?

(The same reasoning could be applied with other energy sources such as a wind farm or a coal power plant; the nuclear reactor is only used as an example)

Sorry if this is, indeed, dumb, but pls, help me to regain some restorative sleep.... 

Thanks, and good night 

[stmdude]

The decay of uranium in the reactor produces the same amount of heat

The decay is actively controlled by "control-rods", which can increase or decrease the output of the reactor.

Now, the demands of the grid can change faster than the control-rods can change the power-output of the reactor, which is (one of the) reason that the voltage in your outlet actually varies.
In my home, I've seen 210V -> 240V RMS.

[Someone]

Wikipedia has a good introduction to this:
https://en.wikipedia.org/wiki/Load_following_power_plant
There aren't many references to learn more details, so you'd need to hunt down some industry information to learn more about how each specific type of plant handles load changes. But for the very fast changes its often just pouring extra energy into the condensers, simply dumping that energy as heat into the atmosphere.

[IanB]

All our means of producing electrical power are running 24/7.

But, obviously, at some times, all of the power which is produced is not used.  For example, during the night.

So, if I simplify the whole grid to one producer (say a 1 GW nuclear power plant) and a bunch of consumers which at a certain time consumes only 500 MW, where do the other 500 MW go ?

Broadly speaking, the amount of power generated is continuously adjusted to match the demand. The grid managers use models and historical data to predict how the demand will change at various times of the day and so they can plan ahead for how to adjust generator output.

If there is excess power generation at any instant it can also be stored for later use, for example in hydroelectric facilities by pumping water up into a higher reservoir.

[Ian.M]

Also https://en.wikipedia.org/wiki/Demand_response
e.g an industrial consumer with a large electric smelter, furnace or kiln can get a preferential tariff for scheduling their peak power usage to suit the grid management authority, and can also load-shed on electronic request for short periods up to a few minutes without significantly affecting the process due to the large thermal inertia.

[T3sl4co1l]

Many power plants can't vary their output much (nuclear is a good example of a base load generator, in fact).  What they can't control directly, they shunt regulate: that is, the excess power is sent to the cooling towers, or cooling water, or burned in very large resistors (think power transmission towers, but strung up with stainless steel rods that run red hot under load!).

This continues to be a problem as nuclear fuel is changed: the removed fuel continues to "glow" with significant power for some time (decades).  It must be stored in a pool with circulating water.  (After some time, it's not that it's sweltering hot, in and of itself, but just that there's an industrial quantity of it in the pool, and if the water isn't kept circulating, the pool eventually boils off.  A fuel bundle, some years after removal, could be stored outside without being noticeably hot (thermally).  It might be a bit toxic with radiation though.   (In contrast, water blocks radiation very nicely; you could swim pretty far in such a spent fuel pool.)

Tim

[Jr460]

I don't know of any plants where they "dump" energy.

I'm most familiar with fossil fired plants.  So that is where I will start.

You would be surprised how quickly a unit can ramp up load.   I've seen the load charts in controls go from 250MW to over 500MW in about 10-15 minutes.   Why 250MW as a base?   Well that was about 50% load on the unit and below that the efficiency of the unit drops rapidly.

If you need faster response to overall system load, then smaller units or gas turbines (jet engine basically) that top out at about 90MW can be started and follow load quickly until you larger units in the system ramp up.  If you still can't keep up, then then the system frequency drops for a bit, but track that over time and make up for it later so over a day the looks that run on small synchronous motors are not off by more than a second.

Any way, as you add load to a generator, it wants to slow down.  The controls then open the steam control valves a bit more to keep it spinning right at 3600RPM.  Of course then the steam flow goes up which means the pressure goes down, and level in the steam drum goes down.  This causes the feed water pumps (run by a small turbine off the first stage extraction steam) to ramp up and put water faster into the boiler.  This causes the temp to start go down which causes the forced draft fans to push more air into the coal mills which increase the fuel flow.  Then of course other controls adjust induced draft fans to match the full flow.

Seems like a bunch of steps, but it is all interconnected with feedback and feed forward control systems.  So having a load of X schedules a basic fuel flow, and the controls fine tune everything.

The only times things are "dumped", and I've seen it....  is a unit trip.  Full load and the unit trips, the control cut off fuel and slam all kinds of valves closed.  The fans and dampers for boiler air to to max to push all the air they can into the boiler.  Even with nothing burning a a 14 story high boiler has a lot of mass that can't cool that quickly.   Thus the water/steam trapped in the system increase in pressure until a big safety pop-off values opens on the roof.   The whole build shakes.

The reverse happens if the load goes down, load or heat, nothing is dumped.  Everything is saved.   Just the cooling for the stator in the generator is routed to help pre-heat air into the boiler.

Now a nuke.  In a BWR design, the controls rods make gross adjustments.  Reactor power is control by adjusting the flow of the recirculation pumps.   Run them faster and you more cool and denser water into the core.  Denser water means more neutron moderation and reactivity and power levels go up.  I'll admit, I'm not up on PWR reactors and power control.

Bottom line.  Power doesn't go anywhere because they never generate it in the first place.  Yes, larger units react ti changing load a bit slower.  Most reasons for minimum loads or staying at fixed load is based on operational issues and the mostly running the unit at the most efficient point.

[IanB]

Any way, as you add load to a generator, it wants to slow down.  The controls then open the steam control valves a bit more to keep it spinning right at 3600RPM.  Of course then the steam flow goes up which means the pressure goes down, and level in the steam drum goes down.  This causes the feed water pumps (run by a small turbine off the first stage extraction steam) to ramp up and put water faster into the boiler.  This causes the temp to start go down which causes the forced draft fans to push more air into the coal mills which increase the fuel flow.  Then of course other controls adjust induced draft fans to match the full flow.

Seems like a bunch of steps, but it is all interconnected with feedback and feed forward control systems.  So having a load of X schedules a basic fuel flow, and the controls fine tune everything.

The control system is like a finely orchestrated ballet. One interesting point about the part highlighted above. When the pressure goes down, the steam bubbles in the tubes expand, causing the level to go up, initially. This effect (and the reverse) is known as "shrink and swell". It obviously has an impact on the response of the level controller, which is the kind of thing which complicates the design of control systems.

[Someone]

I don't know of any plants where they "dump" energy.

I'm most familiar with fossil fired plants.  So that is where I will start.

You would be surprised how quickly a unit can ramp up load.   I've seen the load charts in controls go from 250MW to over 500MW in about 10-15 minutes.   Why 250MW as a base?   Well that was about 50% load on the unit and below that the efficiency of the unit drops rapidly.

If you need faster response to overall system load, then smaller units or gas turbines (jet engine basically) that top out at about 90MW can be started and follow load quickly until you larger units in the system ramp up.  If you still can't keep up, then then the system frequency drops for a bit, but track that over time and make up for it later so over a day the looks that run on small synchronous motors are not off by more than a second.

Any way, as you add load to a generator, it wants to slow down.  The controls then open the steam control valves a bit more to keep it spinning right at 3600RPM.  Of course then the steam flow goes up which means the pressure goes down, and level in the steam drum goes down.  This causes the feed water pumps (run by a small turbine off the first stage extraction steam) to ramp up and put water faster into the boiler.  This causes the temp to start go down which causes the forced draft fans to push more air into the coal mills which increase the fuel flow.  Then of course other controls adjust induced draft fans to match the full flow.

Seems like a bunch of steps, but it is all interconnected with feedback and feed forward control systems.  So having a load of X schedules a basic fuel flow, and the controls fine tune everything.

The only times things are "dumped", and I've seen it....  is a unit trip.  Full load and the unit trips, the control cut off fuel and slam all kinds of valves closed.  The fans and dampers for boiler air to to max to push all the air they can into the boiler.  Even with nothing burning a a 14 story high boiler has a lot of mass that can't cool that quickly.   Thus the water/steam trapped in the system increase in pressure until a big safety pop-off values opens on the roof.   The whole build shakes.

The reverse happens if the load goes down, load or heat, nothing is dumped.  Everything is saved.   Just the cooling for the stator in the generator is routed to help pre-heat air into the boiler.

Now a nuke.  In a BWR design, the controls rods make gross adjustments.  Reactor power is control by adjusting the flow of the recirculation pumps.   Run them faster and you more cool and denser water into the core.  Denser water means more neutron moderation and reactivity and power levels go up.  I'll admit, I'm not up on PWR reactors and power control.

Bottom line.  Power doesn't go anywhere because they never generate it in the first place.  Yes, larger units react ti changing load a bit slower.  Most reasons for minimum loads or staying at fixed load is based on operational issues and the mostly running the unit at the most efficient point.

Have a look at the efficiency of the plant during slewing down, it drops substantially and that waste energy is dumped as heat. There are many cascaded control loops in such a large and complex system but since many of them run hard against the material limits to improve efficiency any perturbations (particularly down slew) creates a lot of waste heat while the system stabilises again. Efficiency/waste heat gets progressively worse as the slew rates increase right up to the catastrophic fast 100% down to 0% unplanned trip/scram.

[denverpilot]

Besides the excellent comments above, from hanging around family members in the power biz...

In addition to efficiency there's also designs of plants that simply aren't stable enough to operate outside of certain limits. Especially in nuclear power. But boilers have both low and high limits too. Turbines are a lot less sensitive overall, but have lubrication and speed limits as well.

Chernobyl being the prime example of doing that, even as a test, and doing it way wrong, of course.

[IanMacdonald]

Basically there is only limited storage capacity on the Grid, so generation has to be matched to demand as an ongoing process. In the UK anyway, the frequency is used as a means of signalling over or under supply. If it gets a bit below 50Hz then all contributing stations will tend to ramp their output up. Not sure if other Grids use this principle or not.

When heavier loads are anticipated, extra stations are asked to become 'spinning reserve' -That is, turbines revolving and synced-up bit not delivering any significant output. Thus they can quickly come onstream to cope with demand. If it turns out their services aren't required, the operators get paid 'constraint payments' for the fuel used to spin-up needlessly.

The controversial aspect is that renewables operators also get constraint payments, only in their case  it's for all energy which could have been delivered even if no-one could possibly have wanted it.

http://gridwatch.templar.co.uk/

Nuclear cannot vary its output rapidly, plus the main costs are the fixed ones, whilst the fuel is cheap. Thus it makes financial sense to run nuclear stations at full power as much as possible.

The main source of reserve in the UK is combined cycle gas turbine plant. Although this can come onstream very quickly it takes time (an hour or so) for the secondary steam cycle to start up, and that means accepting a significant reduction in efficiency under start/stop operating conditions. 

[Jr460]

Let me correct a few things I see.

The turbine speed.   It is held at 3600RPM.   It does not vary according to load.

A unit has lower efficiency when run below full load, that is a given.   I've had to measure it many times and produce the graphs of heat rate vs load.  Not easy, it is an all day task for several people to gather the data.  Then a few days of compiling and crunching all the numbers.  You can do it quick and fast, but it is not very accurate. 

However it was implied that to run at lower loads a unit dumps energy.  That is just not true.  You put less energy into the system to start with.   All kinds of odd things can happen as load shifts quickly,  but still no control system or part of the plant that is designed to "dump" energy.  One exception as listed is a unit trip.

Efficiency can be very confusing.  And example, for the same fuel input I can get 5% more load out of a unit, but it's efficiency goes in the toilet.

Any unit has diagrams showing a heat balance around the whole system, including what is required by either river water or cooling tower for the condenser.  These are shown for several different load conditions.   Break out your steam tables and look at one. 

Might not be the best example, but a quick Goggle search shows one with a lot of the details.  https://www.slideshare.net/jitendramechi/heat-balance-diagram

[petergebruers]

In Belgium, they store electricuty in "Coo", a pumped-storage hydroelectric power station, it can generate 1.2 GW:

https://en.wikipedia.org/wiki/Coo-Trois-Ponts_Hydroelectric_Power_Station

I do not think the maximum energy transfer is mentioned anywhere, but I seem to remember it can deliver that 1.2 GW for "a few hours" (2 hours?). My memory is rusty, I studied energy production in 1990 and at that time it didn't generate that amount of power, but it was smaller too, so run time might be the same.

They do mention: "It generates about 1 million MWh annually and consumes about 20 percent more in pumping mode."

[Someone]

However it was implied that to run at lower loads a unit dumps energy.  That is just not true.  You put less energy into the system to start with.   All kinds of odd things can happen as load shifts quickly,  but still no control system or part of the plant that is designed to "dump" energy.  One exception as listed is a unit trip.

Its during the slew that energy is dumped, faster down slew, more dumping. The slow thermal processes can't change quickly and they can't be left to heat up as the energy demand drops, there are some co/tri gen plants that can store that energy into their thermal outputs but the majority of plants are going to dump the excess in the cooling systems. We agree in steady state that an efficient equilibrium will be found minimising the thermal waste, but they are slow processes. I introduced the dumping specifically during slewing not as a steady state condition.

[madires]

Based on statistics about power usage you can estimate how much power has to be generated at a specific date and time. Power plants with a slow response time (nuclear) are used for the base load. For periods of higher power demand additional power is provided by plants with faster response times, like coal. For peak situations there are gas turbines and other fast acting power sources. Before PV and wind farms the whole controlling system required only a few manual corrections. There's pretty good software for calculating forecasts of the power generation by PV and wind farms but it's not 100% accurate. So we have now much more manual corrections to keep the power grid stable.

[Jr460]

However it was implied that to run at lower loads a unit dumps energy.  That is just not true.  You put less energy into the system to start with.   All kinds of odd things can happen as load shifts quickly,  but still no control system or part of the plant that is designed to "dump" energy.  One exception as listed is a unit trip.

Its during the slew that energy is dumped, faster down slew, more dumping. The slow thermal processes can't change quickly and they can't be left to heat up as the energy demand drops, there are some co/tri gen plants that can store that energy into their thermal outputs but the majority of plants are going to dump the excess in the cooling systems. We agree in steady state that an efficient equilibrium will be found minimising the thermal waste, but they are slow processes. I introduced the dumping specifically during slewing not as a steady state condition.

Yes I understand what you are saying.  However I'm skeptical.   I say this because even if the load drops quickly, you still are generating hundreds of MW.  So let's say load drops from 500 to 300 MW in a minute, and remember this one of many units on the grid, so each unit is trying to ramp back a similar amount.  300MW is still a good amount of power by anyone's standards, so I ramp the fuel back to what would be needed for 250MW rather than 300 and let the thermal flywheel keep me going at 300 for a bit.  Like I said early in this thread, you would be surprised how quickly you can change the load.  For example, a unit being spun up.  The generator is not connect to the grid and the turbine is spinning at 3600 RPM.  When the breaker closes in less than a second I saw it jump to 20 MW, before the operator put the controls in hold mode rather than load-follow.

I don't have access to the charts from a unit anymore, otherwise I find one of load and compare with water temp being sent to the cooling tower during a ramp down in load.  Because that is only place the waste process heat goes.  And you just can't dump anything you want into it, it has limits as to how much heat it can remove.  If you do, then the temp in the condenser goes up, the back pressure goes up and you are start hurting the last stage turbine buckets.  (note this is not a problem when the unit trips, turbine speeds down to a few RPM and put on the turning gear, and dump valve opens from the main steam lines past the stop and control valves directly into the condenser.  It prevents that trapped steam from over speeding the turbine, but is minor in the amount of heat in the whole boiler and unit piping.)

[Mr. Scram]

Many power plants can't vary their output much (nuclear is a good example of a base load generator, in fact).  What they can't control directly, they shunt regulate: that is, the excess power is sent to the cooling towers, or cooling water, or burned in very large resistors (think power transmission towers, but strung up with stainless steel rods that run red hot under load!).

This continues to be a problem as nuclear fuel is changed: the removed fuel continues to "glow" with significant power for some time (decades).  It must be stored in a pool with circulating water.  (After some time, it's not that it's sweltering hot, in and of itself, but just that there's an industrial quantity of it in the pool, and if the water isn't kept circulating, the pool eventually boils off.  A fuel bundle, some years after removal, could be stored outside without being noticeably hot (thermally).  It might be a bit toxic with radiation though.   (In contrast, water blocks radiation very nicely; you could swim pretty far in such a spent fuel pool.)

Tim

I've been told that swimming in a spent fuel pool actually exposes you to less radiation that being outside of one, due to the water also blocking background radiation. Only when you dive deep and come quite close to the rods, you get into trouble.

[rs20]

Have I been living under a misconception? I thought when supply dropped, the main consequence was:
-- Reduced current flowing through the generators at the plant,
-- This reduced current means there's less torque holding back the turbines, so the resulting power imbalance causes the rotor's speed start to increase (assuming no control system intervenes)
-- This higher rotor speed manifests as a higher frequency in our power sockets at home, not higher voltage (the voltage at the plant might maybe move as well, but there's so much voltage regulation/autotransformers between there and your house that there's not going to be any correlation left over.)

Thus, in a very real sense, all the generators in the grid are rotating in lockstep, as if they're connected by some sort of mystical invisible (albeit slightly stretchy) belt, with all the steam turbines trying to spin the generators faster and all the users extracting energy by trying to slow the rotor down (extracting power in the process). Hence, the problem of matching supply to demand reduces to a problem of regulating steam to a turbine in order to regulate its rotational speed (albeit actually very complicated with so many different plants etc). Of course, the grid operators have control over many of our hot water cylinders, which is an excellent place to dump pulses of excess load (if it helps, think of it as a "brake" for the turbines), in addition to pumped hydro and battery storage, etc.

Hence, the short-term imbalance storage capacity of the grid is in the combined mechanical inertia of all the generators in the grid, not (traditionally) capacitors. This manifests as the mains frequency shifting over time, as shown here. The cool part is that many grid operators guarantee that although the frequency might change in the short term, the total number of cycles per day or per week are tightly regulated to be correct; so you can use it as a pretty decent time source for a clock (and indeed, many clocks used to work this way.)

[IanB]

Have I been living under a misconception? I thought when supply dropped, the main consequence was:
-- Reduced current flowing through the generators at the plant,
-- This reduced current means there's less torque holding back the turbines, so the resulting power imbalance causes the rotor's speed start to increase (assuming no control system intervenes)
-- This higher rotor speed manifests as a higher frequency in our power sockets at home

This is true in principle if you had a single generator running the load. But when you have a whole grid with many generators connected, they all get synchronized to the same speed. There is so much inertia in the system that one generator cannot really affect the system frequency.

What happens instead is that the generator with too little load advances its phase angle slightly relative to the grid, while still keeping the same rotational speed. The increase in phase angle causes more power to be pulled from the generator, thus balancing the situation. Therefore the generator remains "locked" to the grid frequency even when there is a load imbalance.

[djacobow]

Dumping electric power into resistors is very uncommon, but it has been done. I encourage anyone to Google the "Chief Joe Braking Resistor" operated by the Bonneville Power Administration for decades. At times it was cheaper and easier to dump excess hydropower this way then it was to throttle dams. Keep in mind that many dams are under legal requirements to flow a certain amount of water, so turning the flow down is operationally problematic.

Nowadays, controllers have access to a large array of resources in any balancing area and can combine small changes at many units to get the ramps they need. Typically this will not involve nukes and other "baseload" units, but more flexible generation like combustion turbines and combined cycle generators. Also, operationally, renewables are more like baseload, as they cannot be dispatched "up" and it's economically problematic to dispatch them "down."

[Someone]

Many power plants can't vary their output much (nuclear is a good example of a base load generator, in fact).  What they can't control directly, they shunt regulate: that is, the excess power is sent to the cooling towers, or cooling water, or burned in very large resistors (think power transmission towers, but strung up with stainless steel rods that run red hot under load!).

This continues to be a problem as nuclear fuel is changed: the removed fuel continues to "glow" with significant power for some time (decades).  It must be stored in a pool with circulating water.  (After some time, it's not that it's sweltering hot, in and of itself, but just that there's an industrial quantity of it in the pool, and if the water isn't kept circulating, the pool eventually boils off.  A fuel bundle, some years after removal, could be stored outside without being noticeably hot (thermally).  It might be a bit toxic with radiation though.   (In contrast, water blocks radiation very nicely; you could swim pretty far in such a spent fuel pool.)

Tim

I've been told that swimming in a spent fuel pool actually exposes you to less radiation that being outside of one, due to the water also blocking background radiation. Only when you dive deep and come quite close to the rods, you get into trouble.

This is from an old xkcd "what if":
https://what-if.xkcd.com/29/
Its focused on the fuel as a source and missing many of the other pathways that a human could be exposed to radiation in such a pool. The activation products in the water:
https://en.wikipedia.org/wiki/Activation_product
Which come from impurities and corrosion material would pose a significant safety hazard, so much that significant decontamination is required for objects coming out of the water. There are some numbers in here:
http://www-pub.iaea.org/MTCD/Publications/PDF/te_1081_prn.pdf

Swimming in such a mixed and unpredictable media is insane, I don't recall ever seeing a certain famous radiation "tourist" ever going swimming where neutron sources could be present even though they would happily eat an apple:

Understanding the nuclide chains, chemistry of the materials, and exposure pathways is very complicated. But we know that nuclear storage ponds are going to have some biologically active impurities which could enter your body if exposed to the water.

[Red Squirrel]

If you are Hydro One, what you do is sell all the excess power at a loss to the states, and then when you need extra power, you buy it from Quebec at 5x the market value.  When you realize you are losing money doing this, you just keep increasing everyone's hydro bill.   

On more serious note, I had a similar question once and from what I recall, turbines can sorta act like flywheels, based on demand a turbine that is in motion will either lag behind and act as a motor or try to push harder, causing the others to act as motors.  At least in simple terms, I think that is sorta what happens. 

[Mr. Scram]

In Belgium, they store 1.2 GW in "Coo", a pumped-storage hydroelectric power station:

https://en.wikipedia.org/wiki/Coo-Trois-Ponts_Hydroelectric_Power_Station

You mean 1.21 GW?

[dr.diesel]

Jr has correctly answered, and I wrote quite a bit about this (coal plant operation from the control system side) several years ago here.  But if there are further questions please ask and I'll elaborate.

[Zbig]

In Belgium, they store 1.2 GW in "Coo", a pumped-storage hydroelectric power station:

https://en.wikipedia.org/wiki/Coo-Trois-Ponts_Hydroelectric_Power_Station

[..]
If there is excess power generation at any instant it can also be stored for later use, for example in hydroelectric facilities by pumping water up into a higher reservoir.

I have stopped by the nearby power station of this type last week: https://en.wikipedia.org/wiki/%C5%BBarnowiec_Pumped_Storage_Power_Station Impressive stuff.
Rocking my "Negative feedback" T-Shirt for bonus points

[rs20]

In Belgium, they store 1.2 GW in "Coo", a pumped-storage hydroelectric power station:

https://en.wikipedia.org/wiki/Coo-Trois-Ponts_Hydroelectric_Power_Station

Looks like someone needs to brush up:

[petergebruers]

Ha! Power versus Energy! You are quite right. I do now the difference, but I did not pay attention. Stupid me, posting too fast 

I do not think the maximum energy transfer is mentioned anywhere, but I seem to remember it can deliver that 1.2 GW for "a few hours" (2 hours?). My memory is rusty, I studied energy production in 1990 and at that time it didn't generate that amount of power, but it was smaller too, so run time might be the same.

They do mention: "It generates about 1 million MWh annually and consumes about 20 percent more in pumping mode."

[jmelson]

In Belgium, they store electricuty in "Coo", a pumped-storage hydroelectric power station, it can generate 1.2 GW:

Ameren has a pumped energy storage facility in central Missouri (US).  They pump water up a mountain at night, and then it runs downhill during peak energy usage times.  The plant is called Taum Sauk, the upper reservoir is Proffit Mountain.

Jon

[Codebird]

No-one here has addressed the more fundamental question: Where does unused energy go? Or, more practically, why do grid operators go to such lengths to carefully balance power flow to ensure that there is no excess energy to go anywhere?

Whenever energy is drawn from the grid, it is drawn from the kinetic energy of the generators: Hundreds of tons of rapidly spinning metal gets a tiny bit slower. The governor mechanisms quickly detect this and a few steam valves open to push the turbines a tiny bit harder. Likewise if consumption drops, the metal spins faster and the governor systems close those valves.

If you could somehow convince everyone in the country to run around turning off their factories, appliances, and lights at once, then the resulting sudden drop in consumption could result in generators spinning so fast that turbine blades become shrapnel flying through the halls before the governor systems can react and get those valves closed. Fortunately this is not likely to happen.

[IanB]

If you could somehow convince everyone in the country to run around turning off their factories, appliances, and lights at once, then the resulting sudden drop in consumption could result in generators spinning so fast that turbine blades become shrapnel flying through the halls before the governor systems can react and get those valves closed. Fortunately this is not likely to happen.

I think a generator with suddenly no load is indeed likely to happen, and fortunately the damaging consequences do not actually occur. If generators started spinning so fast that turbine blades started flying around then the engineers who designed them would not be worth their job titles. So no, not gonna happen.

A generator can end up with no load if its output circuit trips open for any reason. The control system around the generator is designed to ensure this will not be a catastrophic event.

[Mr. Scram]

No-one here has addressed the more fundamental question: Where does unused energy go? Or, more practically, why do grid operators go to such lengths to carefully balance power flow to ensure that there is no excess energy to go anywhere?

Whenever energy is drawn from the grid, it is drawn from the kinetic energy of the generators: Hundreds of tons of rapidly spinning metal gets a tiny bit slower. The governor mechanisms quickly detect this and a few steam valves open to push the turbines a tiny bit harder. Likewise if consumption drops, the metal spins faster and the governor systems close those valves.

If you could somehow convince everyone in the country to run around turning off their factories, appliances, and lights at once, then the resulting sudden drop in consumption could result in generators spinning so fast that turbine blades become shrapnel flying through the halls before the governor systems can react and get those valves closed. Fortunately this is not likely to happen.

I have trouble believing that our technology wouldn't intervene before that happens, but I like the picture you painted

[Ian.M]

Isn't that scenario equivalent to the power station's grid interconnect tripping out - an event that the turbines and control systems must be designed to survive?  Due to the damage that a serous turbine over-speed would do, I suspect that it would be a requirement that it would take three faults to cause damage, as loss of load occasionally (rarely) happens and steam valves do occasionally jam.   

[Jr460]

No-one here has addressed the more fundamental question: Where does unused energy go? Or, more practically, why do grid operators go to such lengths to carefully balance power flow to ensure that there is no excess energy to go anywhere?

Answered early on in the thread.   The energy comes from the fuel.  If you don't need that load level, you don't put the fuel into the boiler to start with and don't generate excess energy.

If you had no grid operators, a system of generators with changing loads will follow the overall system load.  That is as long as your control system keeps each unit spinning at 3600RPM.  (This is why any movie or tin foil hat person that talks about hacking the power plant/system computers is an idiot.  Plants can and do run just fine without gird operators or computers of any kind.)

Grid operators are balancing several things.  One is you can't overload feeder lines.  Another big one is overall system efficiency.  You have a bunch of units, built at different times, and they all have a different curve for heat rate vs. load.  In fact it is not the operator that does this but a routine that runs abut once a second that computes the best load for each unit and then bumps one ups 1MW and another down the same amount.   Overall the total generated power is the same, but now it is more efficient overall.

The other thing they do is sell or buy power.  Some it short term, some longer term contracts. 

Say you have a city with several units generating power for it.  Outside of your city you have tie lines to other utility companies.  Just sum up the power flow on the tie lines with outbound being negative and in being positive.   When it sums to zero, then you know that what you are generating is the load of your city.  If someone is buying power from you, then all you do is bias the equation, rather than zero, the amount you are buying or selling.   Then about once every tens seconds you make up for any over or under that flowed on the tie lines.   

Now in terms of sudden load loss.   Yea that happens, like when a unit trips and the generator breaker opens.  The control system is very very fast, and the hydraulics that move the valves are to be respected.  Not much overspeed is tolerated by the controls.  The emergency  all mechanical trip is set at 110% percent of rated speed if I remember right.

Normally you have the main steam feed split into two "stop valves".  As the name implies they are meant to stop the flow, right now.  Either open or closed.  Pressure is required to keep them open, and large springs and the steam flow wants to push them closed.  Pull the top off of one and take the stuff out and you can sit inside one.  From that point it goes to 4 control valves.   These modulate the steam flow.   Once a unit trips these 6 along with the reheat control and stop valves slam closed

Or if you want to lose your job there is a big red button on the front of the turbine that is an emergency trip.   Just push it.

That button does get tested, also the turbine overspeed trip is tested, and then the backup mechanical overspeed trip is tested.  This happens after an outage, (once a year for 2 or more weeks) after the turbine is buttoned back up and just before it is put back online.  Of course during the test the generator breaker is open, so the trurbine is free to run at any speed.  The "load" is just the frictional losses.

[Codebird]

I did exaggerate a little to make the point. Flying turbine blades is very much the worst case scenario - while it is a distant possibility, there are indeed multiple safety mechanisms in place to prevent that from happening, all of which would have to fail at once.

[Jr460]

I did exaggerate a little to make the point. Flying turbine blades is very much the worst case scenario - while it is a distant possibility, there are indeed multiple safety mechanisms in place to prevent that from happening, all of which would have to fail at once.

With all the stuff I talked about, it can and has happened.  However a unit trip or sudden shedding of load is rarely the cause.  It requires a several things to go wrong.

I learned a lot about turbines from Bill.  I was young and just starting out.  Bill was old as dirt and worked for GE.  He died in 2006 at the age of 83.

If I had a question he took the time to explain why things were designed the way they were, and what could go wrong.   I think he had a few stories of commissioning new turbines in Israel and one of them due to a sudden imbalance tossing blades through the turbine casing.

[Old Don]

The electrical grid in the USA is divided into 4 sub-grids. Power in each grid is controlled by adding or removing power generating units. Some generators are able to adjust their output quicker than others (nuclear are slow to change) and are used shift power as loads vary. To shift power from one unit to another the operators first parallel the outputs to match frequency and phase. Once the two generators are in sync they are connected via breakers and the two are in parallel with each other and share the total load. To drive power to the new oncoming unit they try to raise the speed of that generator and/or lower the speed of the other unit and the causes the new generator to pick up part of the load. This can continue until all the power load is on the new generator or they can remain in parallel and split the loads. This is how power is shared in each grid and total load is shared by many generators at any one time. Should there be a sudden change of load (up or down) the change is shared by all. In the case of a major failure the generators are supposted to split, but there has been several times in the past where one plant takes down the whole grid. Babies occur 9 months later in these cases.  But other than screw ups like these the load continues to shift between the various power generators and there is no excess power generated. Hydro plants can shift generators and turn some into pumps and so at night when power needs drop they pump water back up into storage lakes and also at the same time maintain output on the operating generators. 

[G7PSK]

I don't think that a turbine and alternator weighing several hundred tons is likely to so suddenly increase speed so rapidly that it will burst before some one or something applies the brakes. In my experience with gen sets (up to half Mw) you switch the load of and the voltage only spikes by a few volts and the frequency (RPM) only increases a nominal amount before the regulators and governors get back in control.

[retrolefty]

To try and keep it on a simpler example, if a generator needs to connect to the grid they first synchronize the generator to the grid frequency before closing it's output breaker, then close the breaker. If they want to take on load they slowly increase it's phase angle Vs the grid and to decrease the load on the generator they slowly decrease it's phase output. Automatic (usually PID based) control keeps the generator at it's operator's desired setpoint value.

[SeanB]

Get the synchronising wrong ( it was possible with older synchroscopes to have it be exactly 180 degrees out of phase and no error shown on the display, though the addition of 3 incandescent lamps to the circuitry to show the much larger differential voltage was always done to tell the difference between right and wrong on the dial otherwise) and close the contactor and you will have several hundreds of tons of turbine and alternator exit the building is very large chunks. Been done at quite a few power plants, either by accident or with a faulty synchroniser. Generally this also results in fatalities, as the pieces are going to be sent in all directions, along with large chunks of structure as well.

With wind turbines there is the added issue of the brakes not really being able to dissipate the full load power as well, relying on the feathering system to provide power reduction so that the brakes only have to dissipate and hold a minimum power from having zero angle of attack. Been a few turbines that have burned when the feathering mechanism failed due to leaking seals and losing all the hydraulic fluid and thus control of both blade positioning and slewing of the nacelle. They do not put a brake capable of dissipating 1MW at the top of the tower, it would be as large as the alternator itself and a lot heavier as well.

[G7PSK]

A lot of modern wind turbine use inverters so the speed regulation is not so critical, I was watching one the other day and noticed that as two blades were in the rising side the rota ion speed dropped and as one went over the top dead center the speed picked up until there were again two blades rising.

[jmelson]

If you could somehow convince everyone in the country to run around turning off their factories, appliances, and lights at once, then the resulting sudden drop in consumption could result in generators spinning so fast that turbine blades become shrapnel flying through the halls before the governor systems can react and get those valves closed. Fortunately this is not likely to happen.

I have trouble believing that our technology wouldn't intervene before that happens, but I like the picture you painted

This is called a load dump, and is one of the conditions that the power plants are designed to handle.  If a circuit breaker on an HV transmission line trips, the plant can lose its load.
They have a fast-acting emergency steam valve that shuts off the steam input to the turbine in a fraction of a second.  Since the turbine-alternator set had truly MASSIVE inertia, it won't speed up much before the valve can be shut.

Jon

[jmelson]

A lot of modern wind turbine use inverters so the speed regulation is not so critical, I was watching one the other day and noticed that as two blades were in the rising side the rota ion speed dropped and as one went over the top dead center the speed picked up until there were again two blades rising.

I think this must be an optical illusion, easy for this to happen if the shaft was not pointing right at you.  The blade set is well-balanced, so there is never a heavy side and a light side.  The rotational inertia of these blades is hard to comprehend, they just don't permit the rotor to speed up and slow down quickly.

Yes, you'd think that with one blade sticking straight out to the right, and two angled up and down to the left, the left side would be heavier.  But, those blades on the left are angled, and therefore their centers of gravity are closer to the hub on the horizontal plane, and EXACTLY balance the weight of the single blade's center of gravity on the right, because it is farther out.  You can check this with any 3-bladed household fan, or work out the relatively simple trig yourself.

Jon

[dr.diesel]

Overspeed is protected by multiple redundant systems, and these systems are individually tested on regular intervals.  This testing is performed at the end of an outage or maintenance cycle as well to ensure no outage activities messed with the operation.  Now before anybody says it, yes these systems can fail, but when you look at historical events, these are extremely rare and usually caused by human error.

Even if a unit is at max power and the load instantly disappears (I have personally witnessed this many many times) there is no danger of overspeed.  The steam valves slam shut lightning fast, and scare you half to death when not expecting it!  (also first hand experience ha ha)

Another interesting bit, many generators are hydrogen gas cooled, so if one does happen to explode it will be spectacular.   

[Jr460]

Even if a unit is at max power and the load instantly disappears (I have personally witnessed this many many times) there is no danger of overspeed.  The steam valves slam shut lightning fast, and scare you half to death when not expecting it!  (also first hand experience ha ha)

Another interesting bit, many generators are hydrogen gas cooled, so if one does happen to explode it will be spectacular.   

Yep seen it happen.  This is what I remember as it was many many years ago.

The unit had three 480V 3 phases busses, A, B and C with most systems split across two busses.  So two pumps for a system would be on different buss.  This was the case the stator cooling water pumps. 

One day the 4160V to 480V transformer for buss B smoked.  Normally not a big problem, with loss of half of things, the controls would have run the unit back to no more than 50% load.  However, 3 months before the same thing happened on buss C.  The issue was quickly fixed by opening the breaks to the C buss transformer and closing a buss tie from B to C buss.  In this case the two pumps in questions were on buss B and C, so when the buss B failed, it took down B and C and both pumps....  Along with tons of other things.   No stator cooling and the controls ran the turbine back to 10% load in less than a second.

Off course the boiler controls followed and shut the coal feed down, but the thermal mass of boiler drove the pressure up until it hit the upper limit, with then caused a full unit trip.  Main steam pressure still when up, and the large pop off values on the roff opened shaking the whole 14 story high structure.

Funny thing was the replacement for the smoked transformer for C buss had just shown up the week before and plans were being finalized on putting it in place and going.

As to hydrogen cooling in the generator, even the oldest units I saw from the 1930s had that feature.  Keep it above 95% pure and it will not burn.  If I remember right about 10% to 90% and a spark can make it ignite.

[Ghydda]

A lot of modern wind turbine use inverters so the speed regulation is not so critical, I was watching one the other day and noticed that as two blades were in the rising side the rota ion speed dropped and as one went over the top dead center the speed picked up until there were again two blades rising.

I think this must be an optical illusion, easy for this to happen if the shaft was not pointing right at you.  The blade set is well-balanced, so there is never a heavy side and a light side.  The rotational inertia of these blades is hard to comprehend, they just don't permit the rotor to speed up and slow down quickly.

I beg to differ.
The rotor plane is weight balanced, that part is correct. And obvoius. The rotational inertia is very big. But the shaft torque is incomprehensible too.
And define quickly by the way.

Anyhow, wind rushes past a turbine at different rates at the top of the rotor plane and the bottom. Thus when two blades occupy the upper half the the rotor plane, the rotor speeds up, and when two blades occupy the lower half the rotor slows down. A little.

This lead to frequency variation on the generator and the power converter handles the difference. If not, then the stresses on the drivetrain (gearbox) would quickly eat said gearbox.

This speed change is indeed perceivable when standing inside the nacelle looking into the hub of the rotor.
From afar, outside, I find this is much less visible to the point I doubt it happens. There is something about the lowest blade rushing past the tower, that screws with my brain.

The amount of speed pickup and drop (3 times per rotor plane rotation) is most pronounced in big turbines with huge rotor planes on fairly short towers, as the ratio of wind difference between top and bottom gets big and the low angular rotor speed helps to make it perceivable.

On a side note, the amount of rocking back and forth of the nacelle/tower on a big turbine is really disconcerting. The brain insists the motion is somehow wrong and not meant to be. This is again a result of different wind speeds across the rotor plane.

Emergency stops in a big wind turbine under full load is also not for the faint hearted. The tower sways so much you can't see from the top to the bottom. Yikes.

Cheers!

[Ice-Tea]

What happens instead is that the generator with too little load advances its phase angle slightly relative to the grid, while still keeping the same rotational speed. The increase in phase angle causes more power to be pulled from the generator, thus balancing the situation. Therefore the generator remains "locked" to the grid frequency even when there is a load imbalance.

AKA droop.

OT: Check out the "electricity map" app for some insights. Great stuff.

[IanMacdonald]

Here, the politicos are talking about aiming for 100% renewable energy by 2050. I just wonder if they've thought through what happens on a cloudy windless day. Or more importantly, through a month of cloudy windless days. 

The greens are putting-out claims about energy storage being available soon to solve all these issues, but I don't see any evidence of that being the case. If the politicos sign a massive wind contract on the strength of pure vapour 'energy storage' predictions, then they could be wasting a huge amount of public money. Did a few calculations on the basis of the Tesla Powerwall, and I reckon that it would cost a trillion UP Pounds to provide a week's backup that way. 

Point of fact a week wouldn't be enough anyway unless there were plans to un-mothball fossil fuel plant in such a contingency, because wind outages can last a month or two, not weeks. (and by Sod's Law it's a fair bet that some of the mothballed plant wouldn't start, compounding the problem!)   

http://iwrconsultancy.co.uk/renewables-backup

Anyone see a missteak in my reckoning. by all means say so.

[DenzilPenberthy]

I just wonder if they've thought through what happens on a cloudy windless day.

Yeah I think they've probably thought of that. 

Anyone see a missteak in my reckoning. by all means say so.

Well for a start you are analysing a technical solution to something 33 years in the future using current day technology and pricing.  At the moment today the grid is 22% wind powered and 13% solar powered. There is almost no coal used in the UK grid any more. This would have been unthinkable 33 years ago and proposing that level of renewable infrastucture would have got me laughed out of the room by people such as yourself. What makes you think that the grid in 33 years' time will be powered by £50 car batteries bought from the local Halfords? 

[Someone]

Anyone see a missteak in my reckoning. by all means say so.

Money is not the primary constraint as renewables are getting lifecycle energy production costs well below traditional generators. Read this:
https://www.withouthotair.com
The constraints for the UK are real but they aren't dominated by the economics.

[Old Don]

Where has all the power gone?
Long time passing
Where has all the power gone?
Long time ago
Where has all the power gone?
Computer nerds used it all
When will they ever learn?
When will they ever learn?

 

[cdev]

Why not use air compression for power storage? Excess capacity could be stored in compressed air.

[Ian.M]

Because adiabatic compression leaves the compressed gas significantly hotter than it was originally and adiabatic expansion cools the exhaust gas well below ambient, and that's all wasted energy.  For short term storage, its not so bad -just keep the compressed air hot with a well lagged tank but for long term storage, that's less than ideal.

[T3sl4co1l]

Also, it's been done, there are some underground salt domes being used for that.

Water is probably better: there are reservoirs up hills / mountains that are pumped for storage.  I wonder what the losses are like.

Tim

[Ian.M]

Wikepedia quotes the Dinorwig Pumped Storage Power Station as operating at 74–76% efficiency.  With separate pumps and turbines^*, and improved penstock design, the round trip efficiency could exceed 85%.

* Which could be on the same shaft, closely coupled to the motor/generator - just drain the pump when the turbine is generating and drain the turbine when the motor is pumping so the unused section is spinning dry. 

[max_torque]

One of the current important facts in grid control is that there are many consumers, and just a few generators.  The result is a massive "averaging" of the load, and a load that changes in an extremely repeatable, forecast able way!   "Big industry" that uses many MW or GW of power is the exception, and it generally has to ask permission before switching on/off it's load so the generation assets are ready to respond.  The grid operators have got very very good at "guessing" what events will cause the load to vary significantly, with seasonal, social and political events all being considered

[Jr460]

The grid operators have got very very good at "guessing" what events will cause the load to vary significantly, with seasonal, social and political events all being considered

It is not just guessing, it is based on reeds over the past few decades.

For example, a weekday and a Saturday and a Sunday all have different loads.  Factor in the date to get the expected outside temps and also sunrise and sunset and you can predict very load.

When I was young I used to use the term "phase of the moon" function to describe a program that was overblown and doing things it really doesn't need to do.   I used that term once to old hand in the electric production department and he gave me a funny look and said his program does have a phase of the moon function.  Oh course I had to understand why.

His program predicted load for 10-15 years out so they could enter into long term contracts to buy coal at better rates.  Seems the load profile changed if it was holiday, or a holiday weekend.  New Years, Christmas, and others on fixed dates were easy.   Easter was a problem.   If I remember right, the first Sunday after the first full moon after the vernal equinox.   Thus, it moves every year.  The soonest is March 22nd, the last is April 25th.  Here in the US midwest big outside temp changes between those times, means the load profile is very different.

[nes999]

I used to work for a power plant. An obscene amount is wasted due to not using/heat heat in the lines.

Sent from my VS988 using Tapatalk

[kastnerd]

New York has an interesting "Pumped Storage Power Station" Built in 1973
https://en.wikipedia.org/wiki/Blenheim-Gilboa_Hydroelectric_Power_Station

Fills the reservoirs when there is extra power on the grid.   Drains the reservoirs when demand is high.  Is this more efficient then battery's?

[T3sl4co1l]

New York has an interesting "Pumped Storage Power Station" Built in 1973
https://en.wikipedia.org/wiki/Blenheim-Gilboa_Hydroelectric_Power_Station

Fills the reservoirs when there is extra power on the grid.   Drains the reservoirs when demand is high.  Is this more efficient then battery's?

IIRC, pumped storage is roughly comparable to chemical (battery) storage.  A little bit better?  I haven't seen a comparison in a while, would be good to search on it.

Note that, while pumped gas storage can be used as well (e.g., underground caverns, depleted formations, mined-out salt domes), due to adiabatic temperature swing, it's much more lossy than water pumping!

Thermal storage is also used, for processes that are primarily thermal, like solar towers.  This melts/freezes a huge vat of molten salts (mainly sodium and potassium nitrate, AFAIK), which has pretty high heat capacity (comparable to water?), and a relatively low melting point (a few hundred C).

Tim

[rs20]

Note that, while pumped gas storage can be used as well (e.g., underground caverns, depleted formations, mined-out salt domes), due to adiabatic temperature swing, it's much more lossy than water pumping!

If I were a god, I'd be tempted to change the laws of physics to make this not be the case. But then again, if I did, refrigerators would be impossible...

[DougSpindler]

Jr has correctly answered, and I wrote quite a bit about this (coal plant operation from the control system side) several years ago here.  But if there are further questions please ask and I'll elaborate.

I have an unanswered question.  Let's say there's a power plant that's produces 100 MWatts of electricity to match a demand load of 100 MWatts.  As the demand tapers off to say MWatts and the generator continues to produce 100 MWatts what happens to the other 50 MWatts being produced by the generators?  There has to be a conservation of energy going on here.  I'm thinking reduced load (less current draw) would result in  much higher voltages and possibly higher Hz?  Where/how is that extra energy dissipated?

[dr.diesel]

I have an unanswered question.  Let's say there's a power plant that's produces 100 MWatts of electricity to match a demand load of 100 MWatts.  As the demand tapers off to say MWatts and the generator continues to produce 100 MWatts what happens to the other 50 MWatts being produced by the generators?  There has to be a conservation of energy going on here.  I'm thinking reduced load (less current draw) would result in  much higher voltages and possibly higher Hz?  Where/how is that extra energy dissipated?

Think of it like a giant lake, power plants are pouring into the lake and customers are drawing out of the lake.  If a few customers suddenly stop drawing, the lake height will start to rise, but at a miniscule rate, plenty of time for generation to ever so slightly taper off.

In the event of large scale load sheds the power plants will simply trip off completely, venting the stream out the roof of the building.  (Which will scare you half to death if you just happen to be on the roof)  This actually happens quite frequently, ie Mother Nature taking out a distribution line or a field breaker opens up, which instantly drops all load off the generator.  I've seen this happen hundreds of times, the multi-redundant control system is plenty fast enough to deal with the situation.


 

[Jr460]

Jr has correctly answered, and I wrote quite a bit about this (coal plant operation from the control system side) several years ago here.  But if there are further questions please ask and I'll elaborate.

I have an unanswered question.  Let's say there's a power plant that's produces 100 MWatts of electricity to match a demand load of 100 MWatts.  As the demand tapers off to say MWatts and the generator continues to produce 100 MWatts what happens to the other 50 MWatts being produced by the generators?  There has to be a conservation of energy going on here.  I'm thinking reduced load (less current draw) would result in  much higher voltages and possibly higher Hz?  Where/how is that extra energy dissipated?

This seems to be where people are confused.

As the load tappers off, the generator follows and does not continue to to produce 100 MW.

As the load drops, and with the turbine still pushing as hard, the generator will want to speed up since the force holding the generator back is now less the force pushing it faster.

However.......

The controls notice the very very slight increase in speed and close the steam valves a bit.  As the load continues to drop the control system keeps closing the valves and and the coal/gas/oil feed to boiler.

Thus nothing to dissipate.   The generator and turbine follow the load.

The generator always runs at a speed that gives 50/60 Hz depending on your system.  Voltage is not controlled by speed, it is controlled by alter-ex/gener-ex/exciter which provides DC to slip rings on the rotor that provides the magnetic field.  Stronger field gives higher voltage.

Load is a function of how hard you have to "push" the generator to stay at the right speed.

[IanB]

I have an unanswered question.  Let's say there's a power plant that's produces 100 MWatts of electricity to match a demand load of 100 MWatts.  As the demand tapers off to say MWatts and the generator continues to produce 100 MWatts what happens to the other 50 MWatts being produced by the generators?  There has to be a conservation of energy going on here.  I'm thinking reduced load (less current draw) would result in  much higher voltages and possibly higher Hz?  Where/how is that extra energy dissipated?

Think of it like a giant lake, power plants are pouring into the lake and customers are drawing out of the lake.  If a few customers suddenly stop drawing, the lake height will start to rise, but at a miniscule rate, plenty of time for generation to ever so slightly taper off.

Continuing the lake analogy there is also the operating strategy of the power plant. Some plants (base load) might be operated so they pump 100 MW into the lake no matter what. Other plants may be operated to handle load peaks, so they will adjust the amount of power they produce as the water level goes up and down.

There will be a grid control center where each power plant is directed how to operate according to a system of contracts, a bidding system, and current overall demand requirements.

[DougSpindler]

I have an unanswered question.  Let's say there's a power plant that's produces 100 MWatts of electricity to match a demand load of 100 MWatts.  As the demand tapers off to say MWatts and the generator continues to produce 100 MWatts what happens to the other 50 MWatts being produced by the generators?  There has to be a conservation of energy going on here.  I'm thinking reduced load (less current draw) would result in  much higher voltages and possibly higher Hz?  Where/how is that extra energy dissipated?

Think of it like a giant lake, power plants are pouring into the lake and customers are drawing out of the lake.  If a few customers suddenly stop drawing, the lake height will start to rise, but at a miniscule rate, plenty of time for generation to ever so slightly taper off.

Continuing the lake analogy there is also the operating strategy of the power plant. Some plants (base load) might be operated so they pump 100 MW into the lake no matter what. Other plants may be operated to handle load peaks, so they will adjust the amount of power they produce as the water level goes up and down.

There will be a grid control center where each power plant is directed how to operate according to a system of contracts, a bidding system, and current overall demand requirements.

Guess I'm not asking the question in the right way.  My question is not what mechanisms reduce the power, but what happens to the energy if there is nothing to consume it.

I know when gas powered generators are running there RPM is related to load or watts being consumed.  When the load decreases/less watts are being used and the engine RPMs decrease.  But what would happen if the RPMs did not decrease when the load was removed?  That non-consumed energy in the form of electricity has to go somewhere.  I would think the voltage would increase and with the engine RPM increased the hz would also increase.  And now that I think about it a bit more, I would think some of this excess electricity would be converted to heat.

[IanB]

Guess I'm not asking the question in the right way.  My question is not what mechanisms reduce the power, but what happens to the energy if there is nothing to consume it.

It doesn't work that way though. In simple terms you can imagine the grid to be an infinite sink. Whatever you push into it, it will absorb.

I know when gas powered generators are running there RPM is related to load or watts being consumed.  When the load decreases/less watts are being used and the engine RPMs decrease.  But what would happen if the RPMs did not decrease when the load was removed?  That non-consumed energy in the form of electricity has to go somewhere.  I would think the voltage would increase and with the engine RPM increased the hz would also increase.  And now that I think about it a bit more, I would think some of this excess electricity would be converted to heat.

Gas powered generators have a local load, so there is a direct connection between the power consumed and the power generated.

The grid doesn't work this way. It connects many generators with many loads, and it has a huge amount of inertia. Yes, you could try to change the grid voltage or frequency with your generator, but the effect you could have would be infinitesimal.

[DougSpindler]

Guess I'm not asking the question in the right way.  My question is not what mechanisms reduce the power, but what happens to the energy if there is nothing to consume it.

It doesn't work that way though. In simple terms you can imagine the grid to be an infinite sink. Whatever you push into it, it will absorb.

I know when gas powered generators are running there RPM is related to load or watts being consumed.  When the load decreases/less watts are being used and the engine RPMs decrease.  But what would happen if the RPMs did not decrease when the load was removed?  That non-consumed energy in the form of electricity has to go somewhere.  I would think the voltage would increase and with the engine RPM increased the hz would also increase.  And now that I think about it a bit more, I would think some of this excess electricity would be converted to heat.

Gas powered generators have a local load, so there is a direct connection between the power consumed and the power generated.

The grid doesn't work this way. It connects many generators with many loads, and it has a huge amount of inertia. Yes, you could try to change the grid voltage or frequency with your generator, but the effect you could have would be infinitesimal.

Like I said I don't think we are communicating.  Yes the grid has many generators which are much larger than a little power generator, but the principals are still the same.  Maybe your power company never has brownouts and blackouts.  Where I live in California we have them from time to time.  But those are low voltage conditions.  But then again we have over voltage conditions which when transformers blow-up from over voltage and too much current.  Only about 50 miles away close to 6,000 homes were burned to the ground and close to 50 people died as a result of an over current condition by the local power company.

After the over current situation the power company attempted to shed the load for tens of thousands of customers.  There is no way the power generators could have reacted so quickly.  So again my question is what happens to all of that extra electricity (power) that's on the grid?  It can't be destroyed, so where does it go?

[denverpilot]

Blackouts are extremely low voltage scenarios. LOL... yes. Zero usually. :-)

50 houses don’t burn to the ground from “over current” because there’s no reason for them to draw more current. Something else in Ohm’s equation must have changed.

Could you link to a technical article on the “destructive over current” scenario you’re describing so we aren’t reading utter nonsense (that sounds like it’s from a typical California TV news station) and can have a chance of explaining it rationally?

[DougSpindler]

Blackouts are extremely low voltage scenarios. LOL... yes. Zero usually. :-)

50 houses don’t burn to the ground from “over current” because there’s no reason for them to draw more current. Something else in Ohm’s equation must have changed.

Could you link to a technical article on the “destructive over current” scenario you’re describing so we aren’t reading utter nonsense (that sounds like it’s from a typical California TV news station) and can have a chance of explaining it rationally?

Are you trying to say over current conditions do not start fires?  A few years ago the city was installing fourth of July decorations.  Two men were in a cherry picker and which hit the HV power line.  It took 20 minutes for the FD and power company to cut the power to that section of the city while the men in the cherry picker were being electrocuted.  (You do understand how they were in the current path getting electrocuted while in the metal cherry picker, right?)

You tube is full of videos resulting in fires from over current conditions.  Here are just a few.

[IanB]

After the over current situation the power company attempted to shed the load for tens of thousands of customers.  There is no way the power generators could have reacted so quickly.

Of course they could and did react so quickly. The system is designed by engineers to handle such scenarios. A few years ago there was a major blackout in the San Diego area due to a power line fault that took out a whole section of the grid in San Diego county. Millions of consumers were plunged into darkness for hours. The grid handled that, as it had to. There was no alternative. Power plants don't blow up in these situations.

So again my question is what happens to all of that extra electricity (power) that's on the grid?  It can't be destroyed, so where does it go?

It gets consumed, or wasted. Here's a thought experiment for you: suppose you connect an array of 50 or so 12 V 1 W bulbs to a bench power supply set at 12 V. You will see the power supply indicating a load of about 50 W. Now suppose you simulate turning on a new generator by increasing the voltage to 13 V. The power supply will now indicate something like 55 W. It is pumping an extra 5 W into the "grid". Where is the extra 5 W going?

A similar situation happens in the real power grid, only on a much larger scale. If you pump extra power into the grid it will be dissipated somewhere, somehow, over the millions of square miles that the grid covers.

[denverpilot]

Are you trying to say over current conditions do not start fires?

No, I said exactly what I intended to say. In this example of yours in which you’ve posted no technical detail, and sounds like the way a TV news reporter would have described it, the phrase “over current” is gibberish.

Please provide a link to this multiple house fire event that has any better level of detail of the fault, and we’ll speak engineering here on an engineering board. Not gibberish.

[IanB]

You tube is full of videos resulting in fires from over current conditions.  Here are just a few.

But those are not over current conditions, those are electrical faults. If there is a real over current condition it will cause a protective device to trip and cut off the power. It is when the current is within the normal range that fun happens, because then the breaker doesn't trip and the current keeps flowing, producing the fireworks.

[DougSpindler]

After the over current situation the power company attempted to shed the load for tens of thousands of customers.  There is no way the power generators could have reacted so quickly.

Of course they could and did react so quickly. The system is designed by engineers to handle such scenarios. A few years ago there was a major blackout in the San Diego area due to a power line fault that took out a whole section of the grid in San Diego county. Millions of consumers were plunged into darkness for hours. The grid handled that, as it had to. There was no alternative. Power plants don't blow up in these situations.

So again my question is what happens to all of that extra electricity (power) that's on the grid?  It can't be destroyed, so where does it go?

It gets consumed, or wasted. Here's a thought experiment for you: suppose you connect an array of 50 or so 12 V 1 W bulbs to a bench power supply set at 12 V. You will see the power supply indicating a load of about 50 W. Now suppose you simulate turning on a new generator by increasing the voltage to 13 V. The power supply will not indicate something like 55 W. It is pumping an extra 5 W into the "grid". Where is the extra 5 W going?

A similar situation happens in the real power grid, only on a much larger scale. If you pump extra power into the grid it will be dissipated somewhere, somehow, over the millions of square miles that the grid covers.

Now we are getting around to my original question.  What happens to the unused energy?  Where dose it go? 

[denverpilot]

Now we are getting around to my original question.  What happens to the unused energy?  Where dose it go?

Where does the energy go in the wires in your house when you turn off a light switch? Do the wires spring a leak?

[rs20]

A similar situation happens in the real power grid, only on a much larger scale. If you pump extra power into the grid it will be dissipated somewhere, somehow, over the millions of square miles that the grid covers.

Not to mention that all the spinning generators will see less load, and so less of the mechanical energy being put into them will get converted into electricity and some will go to increasing the rotational speed (and therefore the frequency) of the grid. Control systems will detect this and reduce the amount of mechanical energy being put in to bring this under control on the timescale of, oh I dunno, seconds.

Now we are getting around to my original question.  What happens to the unused energy?  Where dose it go? 

He specifically answered your question with the thought experiment. If the voltage of the grid rises, all the light burns just a teensy bit brighter. All the motors get a teensy bit warmer. And, as I added, any remaining deficit goes into the mechanical rotational inertia of all the generators; which then leads to less steam being let into the turbines and ultimately less coal being burned or less water being let out of the dam.

[DougSpindler]

After the over current situation the power company attempted to shed the load for tens of thousands of customers.  There is no way the power generators could have reacted so quickly.

Of course they could and did react so quickly. The system is designed by engineers to handle such scenarios. A few years ago there was a major blackout in the San Diego area due to a power line fault that took out a whole section of the grid in San Diego county. Millions of consumers were plunged into darkness for hours. The grid handled that, as it had to. There was no alternative. Power plants don't blow up in these situations.

So again my question is what happens to all of that extra electricity (power) that's on the grid?  It can't be destroyed, so where does it go?

It gets consumed, or wasted. Here's a thought experiment for you: suppose you connect an array of 50 or so 12 V 1 W bulbs to a bench power supply set at 12 V. You will see the power supply indicating a load of about 50 W. Now suppose you simulate turning on a new generator by increasing the voltage to 13 V. The power supply will not indicate something like 55 W. It is pumping an extra 5 W into the "grid". Where is the extra 5 W going?

A similar situation happens in the real power grid, only on a much larger scale. If you pump extra power into the grid it will be dissipated somewhere, somehow, over the millions of square miles that the grid covers.

Now we are getting around to my original question.  What happens to the unused energy?  Where dose it go?

Maybe we have different definitions for the same word.  I was taught an electrical fault is an over current condition.  Since you think it's not an over current condition what's your definition? 

[DougSpindler]

A similar situation happens in the real power grid, only on a much larger scale. If you pump extra power into the grid it will be dissipated somewhere, somehow, over the millions of square miles that the grid covers.

Not to mention that all the spinning generators will see less load, and so less of the mechanical energy being put into them will get converted into electricity and some will go to increasing the rotational speed (and therefore the frequency) of the grid. Control systems will detect this and reduce the amount of mechanical energy being put in to bring this under control on the timescale of, oh I dunno, seconds.

Now we are getting around to my original question.  What happens to the unused energy?  Where dose it go? 

He specifically answered your question with the thought experiment. If the voltage of the grid rises, all the light burns just a teensy bit brighter. All the motors get a teensy bit warmer. And, as I added, any remaining deficit goes into the mechanical rotational inertia of all the generators; which then leads to less steam being let into the turbines and ultimately less coal being burned or less water being let out of the dam.

Look back at my original post.  I said I thought the voltage and frequency would increase.  But now as I think about this I think I have my answer.  Say one has a 12 watt/12 volt light bulb.  At 12 v the current draw would be 1 amp.  But when the same light is connected to 120 volts the  wattage and current draw remains the same, but since the voltage is 10 times higher the light can not dissipate the heat fast enough and the light burns out for excess heat.

And now that I think about it same think happens with the charging circuit of a car.  If the voltage regulator shorts the generator/alternator in the car will continue to generate electricity.  That excess electricity will "cook" the battery due to over voltage.

Same thing must happen on the power grid should breakers fail.  The load on the steam/water turbines decreases so they spin faster.  The voltage and frequency will increase.  This excess energy (from the higher voltage) will be heat the wires and be dissipated as heat energy.  If the heat energy can not be dissipated fast enough it results in an oxidizing chemical reaction, (fire).  This is what's happening in the videos.

Thanks for making me think through this.

[rs20]

Same thing must happen on the power grid should breakers fail.  The load on the steam/water turbines decreases so they spin faster.  The voltage and frequency will increase.  This excess energy (from the higher voltage) will be heat the wires and be dissipated as heat energy.  If the heat energy can not be dissipated fast enough it results in an oxidizing chemical reaction, (fire).  This is what's happening in the videos.

Those explosions happen when there are short-circuit/overcurrent conditions (either downstream from the transformer or within the transform complex itself) that aren't caught by circuit breakers, leading to large I^2 R power losses in the windings, excess heat, in turn leading to the insulating/cooling oil boiling, venting, and the resulting vaporised/atomised flammable oil cloud being ignited by the sparks. You can frequently see the puffy white oil cloud, shortly before the explosion. These explosion are not caused by people turning off their lights and there therefore being "excess power that has to go somewhere"; there's no reason why the excess power would be concentrated on a single transformer.

Fast-spinning generators have absolutely nothing to do with the transformer explosions.

[DougSpindler]

Same thing must happen on the power grid should breakers fail.  The load on the steam/water turbines decreases so they spin faster.  The voltage and frequency will increase.  This excess energy (from the higher voltage) will be heat the wires and be dissipated as heat energy.  If the heat energy can not be dissipated fast enough it results in an oxidizing chemical reaction, (fire).  This is what's happening in the videos.

Those explosions happen when there are short-circuit/overcurrent conditions (either downstream from the transformer or within the transform complex itself) that aren't caught by circuit breakers, leading to large I^2 R power losses in the windings, excess heat, in turn leading to the insulating/cooling oil boiling, venting, and the resulting vaporised/atomised flammable oil cloud being ignited by the sparks. You can frequently see the puffy white oil cloud, shortly before the explosion. These explosion are not caused by people turning off their lights and there therefore being "excess power that has to go somewhere"; there's no reason why the excess power would be concentrated on a single transformer.

Fast-spinning generators have absolutely nothing to do with the transformer explosions.

I really appreciate your replies, but I think we have a language issue which is preventing us from communicating.  There was a post inviting further questions which I responded to with a a very specific question.  While I appreciate all of the answers they were confusing and did not answer the question being asked.  After talking to some other knowledgeable about this topic I found everything you were saying was correct, but was misapplied to the question I asked.  I'm an American so I'm thinking we are having language/cultural differences in our communications.

[IanB]

I really appreciate your replies, but I think we have a language issue which is preventing us from communicating.  There was a post inviting further questions which I responded to with a a very specific question.  While I appreciate all of the answers they were confusing and did not answer the question being asked.  After talking to some other knowledgeable about this topic I found everything you were saying was correct, but was misapplied to the question I asked.  I'm an American so I'm thinking we are having language/cultural differences in our communications.

The trouble is, we answered your question but you are not hearing the answer.

I gave you an example with the bench power supply and an array of bulbs. If you make the power supply deliver more power into the grid of bulbs then all the bulbs glow a little bit brighter until the extra power is accounted for.

What you fail to appreciate is that the grid is big. Really big.

When a river flows into the ocean, where does the water go? It spreads out everywhere and the sea level rises imperceptibly.

The grid is like an ocean. If you put more power into the grid the same thing happens. It spreads out throughout the grid and the voltage rises imperceptibly. At the same time all the loads on the grid consume a little more power with the rise in voltage, and all the other generators cut back on their power output a to compensate for the excess power being provided. Before very long everything balances out again.

[PinheadBE]

Don't be messed up     I'm french-speaking, and sometimes, it's hard for me too to get all the subtilities between British-English, Australian-English, US-english....   

Anyway, I understand, from an engineering POV what is being said, but I would like to rephrase my original question (the subject of this topic): Considering "the" power mix that feeds my house (some percent of nuclear, some percent of fossils, some percent of renewables), when all those units are producing what they can or what they designed to produce, considering I have, lets' say, a TV-set consuming 1 Wh while in stand-by, what happens if I turn it completely off ?  Will that 1 Wh be registred somewhere else than my own power meter,and be effectively contributing to a lesser consumption of ressources or not ?   
If not, let's just consider that a million people do the same, thus saving 1 Wh times 1 million.  Will it be contributing to a lesser consumption of ressources or not ?     
In any case, where's the threshold, then?

(Mind blowing, isn't it ?  )

[DougSpindler]

I really appreciate your replies, but I think we have a language issue which is preventing us from communicating.  There was a post inviting further questions which I responded to with a a very specific question.  While I appreciate all of the answers they were confusing and did not answer the question being asked.  After talking to some other knowledgeable about this topic I found everything you were saying was correct, but was misapplied to the question I asked.  I'm an American so I'm thinking we are having language/cultural differences in our communications.

The trouble is, we answered your question but you are not hearing the answer.

I gave you an example with the bench power supply and an array of bulbs. If you make the power supply deliver more power into the grid of bulbs then all the bulbs glow a little bit brighter until the extra power is accounted for.

What you fail to appreciate is that the grid is big. Really big.

When a river flows into the ocean, where does the water go? It spreads out everywhere and the sea level rises imperceptibly.

The grid is like an ocean. If you put more power into the grid the same thing happens. It spreads out throughout the grid and the voltage rises imperceptibly. At the same time all the loads on the grid consume a little more power with the rise in voltage, and all the other generators cut back on their power output a to compensate for the excess power being provided. Before very long everything balances out again.

Again you are not answering my question.  Let me try again.
What happens if the lights are powered by a turbine which spins at twice the normal speed?  Won't the voltage double?  And won't the frequency double as well?
P = I *E    Now that the voltage has doubled it also means the power/wattage has doubled as well.  The current would remain the same, so where's all of that extra power/energy going?

Yes the grid is large.  But again where does all of the energy go in over voltage conditions?   Conservation of energy requires it all be accounted for. 

[IanB]

Again you are not answering my question.  Let me try again.
What happens if the lights are powered by a turbine which spins at twice the normal speed?  Won't the voltage double?  And won't the frequency double as well?
P = I *E    Now that the voltage has doubled it also means the power/wattage has doubled as well.  The current would remain the same, so where's all of that extra power/energy going?

If the generator is connected directly to the bulbs, then yes the voltage may go up, and yes the frequency may go up. If the voltage goes up the current goes up too--that's how bulbs and many other things work. Power is voltage x current, so if both voltage and current go up, the power goes up. More power comes out of the generator, and more power goes into the bulbs. So it balances.

Now when the generator is connected to the grid, the generator cannot change speed. It is locked to the grid frequency at 60 Hz. The reasons why need more than a few words to explain. You can study the theory in appropriate textbooks. But regardless, if you push more power into the grid then the consumers will consume more and the other generators will generate less. After a small fluctuation it will balance out.

Yes the grid is large.  But again where does all of the energy go in over voltage conditions?   Conservation of energy requires it all be accounted for.

As everyone has told you many times it gets consumed by consumers (or absorbed temporarily by other generators). Why is this so hard to understand?

[DougSpindler]

Don't be messed up     I'm french-speaking, and sometimes, it's hard for me too to get all the subtilities between British-English, Australian-English, US-english....   

Anyway, I understand, from an engineering POV what is being said, but I would like to rephrase my original question (the subject of this topic): Considering "the" power mix that feeds my house (some percent of nuclear, some percent of fossils, some percent of renewables), when all those units are producing what they can or what they designed to produce, considering I have, lets' say, a TV-set consuming 1 Wh while in stand-by, what happens if I turn it completely off ?  Will that 1 Wh be registred somewhere else than my own power meter,and be effectively contributing to a lesser consumption of ressources or not ?   
If not, let's just consider that a million people do the same, thus saving 1 Wh times 1 million.  Will it be contributing to a lesser consumption of ressources or not ?     
In any case, where's the threshold, then?

(Mind blowing, isn't it ?  )

I understand you completely.  The local power company has an excellent presentation on the power industry.   
https://www.pge.com/includes/docs/pdfs/shared/solar/solareducation/pv_basics.pdf

When you turn of a light, appliance, heather etc. the electricity stops flowing to your device.  The power company is turbines/generators now have slightly less resistance so they will spin a bit faster which will increase the voltage and frequency of the power grid.  The extra power would be dissipated throughout the grid and in your neighbors homes as heat and motors will spin a bit faster.  Now in reality your TV draws so little power compared to the size of the grid it could not be measured.

The way the grid works in the US is there are I think 5.  They are independent of each other and are regional.  West, Texas, south, northeast etc.  The grid on the west coast has 100's of power companies connected to it.  (Consumers)  They buy there power for companies who generate power, producers.  (Coal, nuclear, hydro, wind, biomass, solar etc.)  The power companies place orders for the number of kWhr they will consume minute by minute 24 hrs. in advance so the producers know who much to produce.  Should the orders not cover the demand there's a brown out.  Remember it takes an hour our more to get the grid to open vales to produce more electricity.

I live in California where we have a lot of solar and wind produced electricity.  Since know one knows how windy or if a cloud will cover one of the solar farms we have we can have brown/black-outs on clear summer days.  This has become a real problem for the power companies.

The history of electricity and power companies is quite interesting.  Over 100 years ago thee was no grid and we had local power companies.  As cities grew so did the demand for electricity and larger generation facilities like Hover dam.  To distribute the power grids were created. 

But a grid means a mistake/accident at one power company can domino and bring down and entire grid.  It's interesting if you like this kind of stuff.


 

[DougSpindler]

Again you are not answering my question.  Let me try again.
What happens if the lights are powered by a turbine which spins at twice the normal speed?  Won't the voltage double?  And won't the frequency double as well?
P = I *E    Now that the voltage has doubled it also means the power/wattage has doubled as well.  The current would remain the same, so where's all of that extra power/energy going?

If the generator is connected directly to the bulbs, then yes the voltage may go up, and yes the frequency may go up. If the voltage goes up the current goes up too--that's how bulbs and many other things work. Power is voltage x current, so if both voltage and current go up, the power goes up. More power comes out of the generator, and more power goes into the bulbs. So it balances.

Now when the generator is connected to the grid, the generator cannot change speed. It is locked to the grid frequency at 60 Hz. The reasons why need more than a few words to explain. You can study the theory in appropriate textbooks. But regardless, if you push more power into the grid then the consumers will consume more and the other generators will generate less. After a small fluctuation it will balance out.

Yes the grid is large.  But again where does all of the energy go in over voltage conditions?   Conservation of energy requires it all be accounted for.

As everyone has told you many times it gets consumed by consumers (or absorbed temporarily by other generators). Why is this so hard to understand?

When you say absorbed what exactly do you mean?  In high school physics I learned energy can not be created or destroyed transferred or transformed.

 

[Ice-Tea]

I know when gas powered generators are running there RPM is related to load or watts being consumed.  When the load decreases/less watts are being used and the engine RPMs decrease.  But what would happen if the RPMs did not decrease when the load was removed?  That non-consumed energy in the form of electricity has to go somewhere.  I would think the voltage would increase and with the engine RPM increased the hz would also increase.  And now that I think about it a bit more, I would think some of this excess electricity would be converted to heat.

Perhaps this is where you are wrong. RPM does not relate to power. Normally a fixed frequency is chosen, which results in the frequency of the electricity being matched to 50 or 60Hz. The generator will always run at that RPM. After that, the 'throttle' is changed in function of the load while keeping the RPM constant.

Analogy: cruise control. You pick a speed and after that the cruise control manipulates the throttle to keep that speed. If the load is decreased (ie you are going downhill) the throttle will be released a bit and that's it. No drama.

[IanB]

When you say absorbed what exactly do you mean?  In high school physics I learned energy can not be created or destroyed transferred or transformed.

Energy can be stored or accumulated. A grid may have some storage elements, such as large batteries, or pumped storage hydroelectric facilities. Energy can also be stored as the rotational kinetic energy of generators. This may not seem like much, but for small imbalances averaged over many generators it can be enough to smooth out fluctuations and prevent sudden changes.

[rs20]

What happens if the lights are powered by a turbine which spins at twice the normal speed?  Won't the voltage double?  And won't the frequency double as well?

A turbine spinning at twice the speed has four times the kinetic energy. Now recall, that power (as measure in Watts or Joules per second) is the flow of energy (measured in Joules) over time. How did this generator, along with all the other generators and AC motors that are virtually mechanically linked, end up with double the energy?

This is why your hypothetical question is silly, it presupposes a situation that will never happen except in the most extreme case of mismanagement. In reality, an excess of power (or the portion thereof that doesn't get dissipated as heat in all the lights in the country as correctly explained by IanB) gets converted to kinetic energy in the turbines. The turbines end up spinning negligibly faster, but not for long because the device metering the steam/water into the turbine (analogy: cruise control as correctly provided by Ice-Tea) notices the slight overspeed and holds back the hydroelectric water flow or steam flow (and, further back down the chain of other control systems, less coal or gas being burned).  So when you turn off your TV, negligibly less coal/gas is burned or less water is let through the dam.

Short term (ms): Lights in your neighborhood burn brighter, with the vast rotational inertia of all motors and generators in the country hold frequency nearly steady
Medium term (100ms - multiple seconds): Turbine flow control in power stations is adjusted
Long term (integrated/infinity): Less coal burned or more water still in dam.

[DougSpindler]

What happens if the lights are powered by a turbine which spins at twice the normal speed?  Won't the voltage double?  And won't the frequency double as well?

A turbine spinning at twice the speed has four times the kinetic energy. Now recall, that power (as measure in Watts or Joules per second) is the flow of energy (measured in Joules) over time. How did this generator, along with all the other generators and AC motors that are virtually mechanically linked, end up with double the energy?

This is why your hypothetical question is silly, it presupposes a situation that will never happen except in the most extreme case of mismanagement. In reality, an excess of power (or the portion thereof that doesn't get dissipated as heat in all the lights in the country as correctly explained by IanB) gets converted to kinetic energy in the turbines. The turbines end up spinning negligibly faster, but not for long because the device metering the steam/water into the turbine (analogy: cruise control as correctly provided by Ice-Tea) notices the slight overspeed and holds back the hydroelectric water flow or steam flow (and, further back down the chain of other control systems, less coal or gas being burned).  So when you turn off your TV, negligibly less coal/gas is burned or less water is let through the dam.

Short term (ms): Lights in your neighborhood burn brighter, with the vast rotational inertia of all motors and generators in the country hold frequency nearly steady
Medium term (100ms - multiple seconds): Turbine flow control in power stations is adjusted
Long term (integrated/infinity): Less coal burned or more water still in dam.

Your statement, "This is why your hypothetical question is silly, it presupposes a situation that will never happen except in the most extreme case of mismanagement" made me laugh out loud.  So aren't you really saying something that will never happen can happen?  So when what's supposed to never happen, happens, what happens?

So a generator that has more KE than another, as in it's spinning faster, will speed up slower ones and be slowed by the slower ones.  Now if one has 4x the KE as in spinning twice as fast I would suspect the electron flow and magnetic fields would heat things quite a bit with reduced R resistance or I should say reactive resistance.  If the R is low enough that excess energy could result in a massive exothermal oxidative reaction or fire.  Similar to what we are seeing in the transformer fires.
 

[ruffy91]

As load increases or power generation decreases the frequency of the grid will decrease.When regulation failsand the frequency decreases more than 1% load shedding occurs (1/3 of all consumers will be shed of the grid, which consumers is rotated on a weekly base, so that it remains fair). On the opposite if load is decreasing or power generation increases generating capacity is shed of the grid and loads are switched on. I know of cases where you would get 700$/MWh of used energy for a few seconds-minutes. Some power companies even have huge resistor arrays which can burn a few MW for a few minutes.
The regulation mechanisms are called primary regulation (simple proportional regulator which uses the difference between the set point (normally 50/60Hz) and the actual frequency to regulate power generation). Then you have secondary regulation which varies the set point so you have exctly 50/60Hz in average on a day. And finally tertiary regulation which is like wall street where energy is sold and bought.

Gesendet von meinem BLA-L29 mit Tapatalk

[timb]

Same thing must happen on the power grid should breakers fail.  The load on the steam/water turbines decreases so they spin faster.  The voltage and frequency will increase.  This excess energy (from the higher voltage) will be heat the wires and be dissipated as heat energy.  If the heat energy can not be dissipated fast enough it results in an oxidizing chemical reaction, (fire).  This is what's happening in the videos.

Those explosions happen when there are short-circuit/overcurrent conditions (either downstream from the transformer or within the transform complex itself) that aren't caught by circuit breakers, leading to large I^2 R power losses in the windings, excess heat, in turn leading to the insulating/cooling oil boiling, venting, and the resulting vaporised/atomised flammable oil cloud being ignited by the sparks. You can frequently see the puffy white oil cloud, shortly before the explosion. These explosion are not caused by people turning off their lights and there therefore being "excess power that has to go somewhere"; there's no reason why the excess power would be concentrated on a single transformer.

Fast-spinning generators have absolutely nothing to do with the transformer explosions.

I really appreciate your replies, but I think we have a language issue which is preventing us from communicating.  There was a post inviting further questions which I responded to with a a very specific question.  While I appreciate all of the answers they were confusing and did not answer the question being asked.  After talking to some other knowledgeable about this topic I found everything you were saying was correct, but was misapplied to the question I asked.  I'm an American so I'm thinking we are having language/cultural differences in our communications.

FWIW, I’m an American and understood the replies to your question perfectly. Honestly, I think the communication problem is on your side; I.e., you’re misunderstanding certain terms or concepts.

A good example of this would be referring to short circuit events as “over current” events. If you touch a live wire and are electrocuted, it’s a short circuit. Over current would imply a specific type of event, with one of the many possible causes being a short circuit; as a result of that event, a breaker would trip or current limiting would kick in. In your example of the linemen being electrocuted in the cherry picker, obviously there wasn’t an over current event, or upstream breakers would have kicked in instantly; the power lines were obviously able to handle the current it took to electrocute them, as it took the power company 20 minutes to shut down that part of the grid.

(Also, I wonder how they were electrocuted to begin with, as the buckets on a cherry picker are generally made from fiberglass or composite plastic and therefor non-conductive. Even if the metal arm of the picker touched the line, the men in the bucket should have been OK as the electricity wouldn’t have a route through them.)

Anyway, your question has been answered repeatedly. Read back through the answers and maybe share what you think some of the terminology means. That way we can all get on the same page. [emoji4]

[IanB]

So a generator that has more KE than another, as in it's spinning faster, will speed up slower ones and be slowed by the slower ones.

Yes, good so far.

Now if one has 4x the KE as in spinning twice as fast I would suspect the electron flow and magnetic fields would heat things quite a bit with reduced R resistance or I should say reactive resistance.  If the R is low enough that excess energy could result in a massive exothermal oxidative reaction or fire.  Similar to what we are seeing in the transformer fires.

But here what you have said does not conform with proper scientific or engineering understanding. It is almost gobbledygook. Therefore there is no way to give any response to this statement.

[Jr460]

Short term (ms): Lights in your neighborhood burn brighter, with the vast rotational inertia of all motors and generators in the country hold frequency nearly steady
Medium term (100ms - multiple seconds): Turbine flow control in power stations is adjusted
Long term (integrated/infinity): Less coal burned or more water still in dam.

I think you are way off. 
Short term (100 us)
Medium term (ms) Turbine controls are adjusted - and because of feed forward in the control systems, fuel is adjusted.

Myself and others have explained back about 3 pages.  There is no "extra" or excess energy or power that needs to go anywhere.

Let me ask this.

Say I have a lab switching power supply power supply and I set it 12V and connect it to a set of 12V lamps.  The power supply keeps the voltage at 12V as you add or remove load/lamps.  Since it is a switcher you don't have power being dissipated in a large pass transistor.  The current changes and thus the power provided by the supply.

Now I suddenly remove all the lamps, were does all that extra engird from the power supply go?

The answer is nowhere, the control loop sees the voltage start going up because the load is now gone, and reduces the PWM quickly.  Sure you might see on a scope some slight (better supply with better control will have less by amount also time) over/under shoot of the 12V as the load quickly changes, but nothing radical.

Did the supply switch in resistors to handle the loss of the load?   No.
Did your circuit of lamps for a storage element in it?  No.

[DougSpindler]

Same thing must happen on the power grid should breakers fail.  The load on the steam/water turbines decreases so they spin faster.  The voltage and frequency will increase.  This excess energy (from the higher voltage) will be heat the wires and be dissipated as heat energy.  If the heat energy can not be dissipated fast enough it results in an oxidizing chemical reaction, (fire).  This is what's happening in the videos.

Those explosions happen when there are short-circuit/overcurrent conditions (either downstream from the transformer or within the transform complex itself) that aren't caught by circuit breakers, leading to large I^2 R power losses in the windings, excess heat, in turn leading to the insulating/cooling oil boiling, venting, and the resulting vaporised/atomised flammable oil cloud being ignited by the sparks. You can frequently see the puffy white oil cloud, shortly before the explosion. These explosion are not caused by people turning off their lights and there therefore being "excess power that has to go somewhere"; there's no reason why the excess power would be concentrated on a single transformer.

Fast-spinning generators have absolutely nothing to do with the transformer explosions.

I really appreciate your replies, but I think we have a language issue which is preventing us from communicating.  There was a post inviting further questions which I responded to with a a very specific question.  While I appreciate all of the answers they were confusing and did not answer the question being asked.  After talking to some other knowledgeable about this topic I found everything you were saying was correct, but was misapplied to the question I asked.  I'm an American so I'm thinking we are having language/cultural differences in our communications.

FWIW, I’m an American and understood the replies to your question perfectly. Honestly, I think the communication problem is on your side; I.e., you’re misunderstanding certain terms or concepts.

A good example of this would be referring to short circuit events as “over current” events. If you touch a live wire and are electrocuted, it’s a short circuit. Over current would imply a specific type of event, with one of the many possible causes being a short circuit; as a result of that event, a breaker would trip or current limiting would kick in. In your example of the linemen being electrocuted in the cherry picker, obviously there wasn’t an over current event, or upstream breakers would have kicked in instantly; the power lines were obviously able to handle the current it took to electrocute them, as it took the power company 20 minutes to shut down that part of the grid.

(Also, I wonder how they were electrocuted to begin with, as the buckets on a cherry picker are generally made from fiberglass or composite plastic and therefor non-conductive. Even if the metal arm of the picker touched the line, the men in the bucket should have been OK as the electricity wouldn’t have a route through them.)

Anyway, your question has been answered repeatedly. Read back through the answers and maybe share what you think some of the terminology means. That way we can all get on the same page. [emoji4]

Not sure we agree on the definitions or use of words in the same way.
Are you saying a short does not cause or result in an an over current event?
A short circuit is one where electrons/current flow through an unintended path with very no or very little resistance.
The body is not a perfect conductor, it's a resistive load.  It can be from 1.000 to 100.000 ohms.  At higher voltages after a bit of cooking the skin resistance is about 500 ohms.  Not sure I would consider that a short.   If we were perfect conductors as I think you are implying we could not live.  Your heart could not beat, you could not think or move.  And I guess if you thought hard enough you could get a shock from all of that electrical activity.
I was taught electrocution is what happens when a person becomes a conductor either intentionally or accidentally accompanied with an external flow of electrons/current through the body.  Not sure why you would consider that a short.  There are many medical procedures which use electricity.  The procedure would not work if the body were a perfect conductor.  Intensely there are people who can not get electrocuted if they are on certain drugs and have a sweating disorder.

Not sure what the cherry picker was made out of.  I'm surprised if you understand electricity why you don't understand why the two guys in the cherry picker were being electrocuted.  This is something firemen are taught with live wires.  If a live wire falls on the ground firemen are taught to keep both feet in contact with the ground and shuffle away from the live wire.  If they take a stride the resistance of the ground can be high enough current will flow up one leg and down the other electrocuting the fireman.  I remember hearing of three police horses being killed in Florida because of a live wire in a field.  The resistance of the ground between the legs of the horses was enough that the current flowed through the horses killing them.

This is what I was taught.  Not say I'm right, but then again I've been looking up the definition of the words to make sure I'm using them as properly. 

I do want to thank you for questioning me.  It's required me to do a fair amount of researcher to confirm what I saying is correct.

[DougSpindler]

So a generator that has more KE than another, as in it's spinning faster, will speed up slower ones and be slowed by the slower ones.

Yes, good so far.

Now if one has 4x the KE as in spinning twice as fast I would suspect the electron flow and magnetic fields would heat things quite a bit with reduced R resistance or I should say reactive resistance.  If the R is low enough that excess energy could result in a massive exothermal oxidative reaction or fire.  Similar to what we are seeing in the transformer fires.

But here what you have said does not conform with proper scientific or engineering understanding. It is almost gobbledygook. Therefore there is no way to give any response to this statement.

I thought we agreed if the generator is spinning faster voltage would increase.  And if spinning at twice the speed there were by 4 X the kinetic energy.  My question is where is all of that extra energy going?  I'm re-reading what I wrote and it seems very clear to me.  I would suspect the electron flow and magnetic fields would heat things quite a bit with reduced R resistance or I should say reactive resistance.  If the R is low enough that excess energy could result in a massive exothermal oxidative reaction or fire.  Similar to what we are seeing in the transformer fires.

What is it you are not understanding?

[DougSpindler]

Short term (ms): Lights in your neighborhood burn brighter, with the vast rotational inertia of all motors and generators in the country hold frequency nearly steady
Medium term (100ms - multiple seconds): Turbine flow control in power stations is adjusted
Long term (integrated/infinity): Less coal burned or more water still in dam.

I think you are way off. 
Short term (100 us)
Medium term (ms) Turbine controls are adjusted - and because of feed forward in the control systems, fuel is adjusted.

Myself and others have explained back about 3 pages.  There is no "extra" or excess energy or power that needs to go anywhere.

Let me ask this.

Say I have a lab switching power supply power supply and I set it 12V and connect it to a set of 12V lamps.  The power supply keeps the voltage at 12V as you add or remove load/lamps.  Since it is a switcher you don't have power being dissipated in a large pass transistor.  The current changes and thus the power provided by the supply.

Now I suddenly remove all the lamps, were does all that extra engird from the power supply go?

The answer is nowhere, the control loop sees the voltage start going up because the load is now gone, and reduces the PWM quickly.  Sure you might see on a scope some slight (better supply with better control will have less by amount also time) over/under shoot of the 12V as the load quickly changes, but nothing radical.

Did the supply switch in resistors to handle the loss of the load?   No.
Did your circuit of lamps for a storage element in it?  No.

I agree with everything you have stated.  But that's not the scenario we are taking about.
Let's say you have a generator located at Hover Dam.  You are located 1,000 miles away.  The power lines are directly connected to several motors and light blubs at your location.  All of the motors are running and the lights are lit.  More or less same scenario as your power supply but the distance between ps and load in much greater.

You then turn off half of the motors.  The generator is still spinning at the same speed producing the same amount of energy because it does not know the motors are turned off and you need less energy.  For that split second the wires are conducting far more energy than what you need.  What happens to that excess energy?  Where does it go?  Do we agree the speed of electrons in copper wire is about 1/100 the speed of light?  Same would happen in your power supply example if you added 1,000 miles of wire.

After that split second the load on the generator is less so it will spin faster and the voltage will increase and with the increase in voltage the current should decrease.  (And since the generator is spinning faster the hz will increase.) 

   

[IanB]

Not sure we agree on the definitions or use of words in the same way.
Are you saying a short does not cause or result in an an over current event?
A short circuit is one where electrons/current flow through an unintended path with very no or very little resistance.

A short circuit is when the current takes a shorter path than it is intended to take. Hence "short" circuit. A short may be high resistance or low resistance, but it is true that a low resistance is commonly assumed in casual speech.

An over current situation is again what the words describe. It is a situation when the current is over what it is intended to be, or over the maximum designed for or allowed for. Hence "over" current. Usually an over current situation will be protected against by tripping a fuse or circuit breaker.

I was taught electrocution is what happens when a person becomes a conductor either intentionally or accidentally accompanied with an external flow of electrons/current through the body.

Electrocution is to be killed by electricity passing through the body. If you don't die it is not electrocution.

Not sure what the cherry picker was made out of.  I'm surprised if you understand electricity why you don't understand why the two guys in the cherry picker were being electrocuted.

If the cherry picker was made out of an insulator like GRP, then no current would flow at all and the people riding on it would be safe. This is the case with any cherry pickers that might be used near overhead wires.

If the cherry picker was made out of steel then its resistance would be very low and as soon as it contacted the wires it would cause a short circuit and an over current situation (see above) and the breaker would trip, cutting power to the wires. In the unlikely event that the breaker didn't trip, then the low resistance of the steel would prevent any differential voltage being high enough to cause harm. (Unless the unfortunate person was between the high voltage wires and the cherry picker. In that case they would not only be electrocuted, they would be instantly incinerated.)

If a live wire falls on the ground firemen are taught to keep both feet in contact with the ground and shuffle away from the live wire.  If they take a stride the resistance of the ground can be high enough current will flow up one leg and down the other electrocuting the fireman.

Just a quick footnote: it is quite unlikely a human can be electrocuted in this way, because the electrical path up one leg and down the other does not pass through a human's heart. In addition to which, the fireman is probably wearing heavy insulated boots. It is sensible to take the proper precautions and avoid unnecessary risks, but the actual risk is not that great.

[IanB]

You then turn off half of the motors.  The generator is still spinning at the same speed producing the same amount of energy because it does not know the motors are turned off and you need less energy.  For that split second the wires are conducting far more energy than what you need.  What happens to that excess energy?  Where does it go?

You have to think of it like this: the generator is "pushing" electricity into the grid.

Imagine you are pushing someone along on a bicycle, but they have the brakes on. You are pushing really hard, but they are only moving slowly. Suddenly they let off the brakes. You are still pushing really hard, but the resistance has suddenly dropped. You are pushing much harder than you really need to. What happens to all your excess energy?

[rs20]

I thought we agreed if the generator is spinning faster voltage would increase.

Yep.

And if spinning at twice the speed there were by 4 X the kinetic energy.

If the turbine spins at twice the speed, the turbine will have 4x the kinetic energy, yes.

My question is where is all of that extra energy going?

The energy is not going anywhere. A faster spinning turbine has more kinetic energy, just like a fully charged battery has more chemical energy in it. Energy doesn't have to go places, you can just have a system with lots of gravitational potential energy (fully filled dam), lots of chemical potential energy (charged batteries, a pile of coal), or lots of kinetic energy (a rapidly spinning heavy thing), and that's just fine. The energy can stay where it is, it's not a flow (like power) and so it doesn't have to flow, let alone flow to any particular place.

I think you're confusing energy and power? Energy is measured in Joules, and power is measured in Joules per second (also known as Watts). There's a recent EEVBlog video that goes into some detail on this, you really need to have this concept down before you can discuss subtle things like an imbalance in the power going into/out of a turbine leading to a change in the kinetic energy of the turbine (and that being the end of the story and just fine) -- just like an imbalance in the water flow going into/out of a bucket leading to a change in the quantity of water in the bucket with no need to "explain" where all that "extra water" "goes".

[T3sl4co1l]

You then turn off half of the motors.  The generator is still spinning at the same speed producing the same amount of energy because it does not know the motors are turned off and you need less energy.  For that split second the wires are conducting far more energy than what you need.  What happens to that excess energy?  Where does it go?

Directly into the spark(s) that follow switching such a load.

Power is going to ~zero every cycle, too.  It doesn't need to go anywhere at all.  A "lucky" switch can open during the zero crossing.

(This is the case for single phase.  Three phase always has power going somewhere; you'd need an extremely lucky contactor to open two lines in the correct phase so that no current is being switched.)

Do we agree the speed of electrons in copper wire is about 1/100 the speed of light?  Same would happen in your power supply example if you added 1,000 miles of wire.

Slower than that. But electron drift velocity (which averages zero in mains, BTW) is completely irrelevant to power transmission -- the contactor opening or closing is communicated at light speed [in the conductors], so the force on the rotor changes almost immediately (10s-100s microseconds).

After that, all the mechanical dynamics and control systems play a role.  The instant effect is the generator accelerating incrementally.  Its frequency does not change, in that it makes much less than one full rotation with respect to the grid.  No, its angle change draws power from neighboring generators (others within the plant, and from the grid at large), and that angle oscillates back and forth, like a torsional spring, until the transient energy is dissipated.  Dissipation occurs through losses in the system (generators contain shorting slats*, to reduce harmonic oscillations and dampen these vibrations), transmission losses, and the controller.

*More specifically, synchronous machines do.  Shorting slats are simply shorted turns on the armature, which act to slow any change in its magnetic field (which is an electromagnet, so the field is supposed to be static anyway).  Note: I haven't studied modern generators per se -- their design may vary.  Anyway, even if this isn't the exact solution used, the same effect is required, however it's achieved.


The amount of energy stored in the grid is extremely small.  Only rotating machinery has any inertia at all, and it only accounts for about 10% of the total available power (IIRC).  PFC capacitors might account for 30% of that in turn.

When the grid goes down, whether it's a small neighborhood or the entire northeast, it simply disappears, poof.  Likewise, when it's restarted, the load is there, ~instantaneously, no windup, no springiness.  (In practice, there's a lot of inrush associated with restart.  But they can deal with that, as, of course, they must.)

Tim

[DougSpindler]

You then turn off half of the motors.  The generator is still spinning at the same speed producing the same amount of energy because it does not know the motors are turned off and you need less energy.  For that split second the wires are conducting far more energy than what you need.  What happens to that excess energy?  Where does it go?

You have to think of it like this: the generator is "pushing" electricity into the grid.

Imagine you are pushing someone along on a bicycle, but they have the brakes on. You are pushing really hard, but they are only moving slowly. Suddenly they let off the brakes. You are still pushing really hard, but the resistance has suddenly dropped. You are pushing much harder than you really need to. What happens to all your excess energy?

Exactly  At the moment the brakes are let of the amount of energy you are pushing with has to go somewhere so you fall forward and hit the ground.  That extra KE you aren't using to push is instantaneously converted to PE (since you have no resistance) than as you begin to fall it's converted to PE (as you are falling) then it's converted to heat and sound as you hit the ground.

This is exactly what I've been saying would happen with the generator. 

 

[DougSpindler]

I thought we agreed if the generator is spinning faster voltage would increase.

Yep.

And if spinning at twice the speed there were by 4 X the kinetic energy.

If the turbine spins at twice the speed, the turbine will have 4x the kinetic energy, yes.

My question is where is all of that extra energy going?

The energy is not going anywhere. A faster spinning turbine has more kinetic energy, just like a fully charged battery has more chemical energy in it. Energy doesn't have to go places, you can just have a system with lots of gravitational potential energy (fully filled dam), lots of chemical potential energy (charged batteries, a pile of coal), or lots of kinetic energy (a rapidly spinning heavy thing), and that's just fine. The energy can stay where it is, it's not a flow (like power) and so it doesn't have to flow, let alone flow to any particular place.

I think you're confusing energy and power? Energy is measured in Joules, and power is measured in Joules per second (also known as Watts). There's a recent EEVBlog video that goes into some detail on this, you really need to have this concept down before you can discuss subtle things like an imbalance in the power going into/out of a turbine leading to a change in the kinetic energy of the turbine (and that being the end of the story and just fine) -- just like an imbalance in the water flow going into/out of a bucket leading to a change in the quantity of water in the bucket with no need to "explain" where all that "extra water" "goes".

I will disagree with you.  The moment the load is removed from the circuit the generator "doesn't know" and will still be sending the same amount of energy/watts in the wires.  Similar concept in plumbing with pipes hammering.  The hammering is how the excess energy is being dissipated.  What's the equivalent to pipe hammering with the generator.

[DougSpindler]

Not sure we agree on the definitions or use of words in the same way.
Are you saying a short does not cause or result in an an over current event?
A short circuit is one where electrons/current flow through an unintended path with very no or very little resistance.

A short circuit is when the current takes a shorter path than it is intended to take. Hence "short" circuit. A short may be high resistance or low resistance, but it is true that a low resistance is commonly assumed in casual speech.

An over current situation is again what the words describe. It is a situation when the current is over what it is intended to be, or over the maximum designed for or allowed for. Hence "over" current. Usually an over current situation will be protected against by tripping a fuse or circuit breaker.

I was taught electrocution is what happens when a person becomes a conductor either intentionally or accidentally accompanied with an external flow of electrons/current through the body.

Electrocution is to be killed by electricity passing through the body. If you don't die it is not electrocution.

Not sure what the cherry picker was made out of.  I'm surprised if you understand electricity why you don't understand why the two guys in the cherry picker were being electrocuted.

If the cherry picker was made out of an insulator like GRP, then no current would flow at all and the people riding on it would be safe. This is the case with any cherry pickers that might be used near overhead wires.

If the cherry picker was made out of steel then its resistance would be very low and as soon as it contacted the wires it would cause a short circuit and an over current situation (see above) and the breaker would trip, cutting power to the wires. In the unlikely event that the breaker didn't trip, then the low resistance of the steel would prevent any differential voltage being high enough to cause harm. (Unless the unfortunate person was between the high voltage wires and the cherry picker. In that case they would not only be electrocuted, they would be instantly incinerated.)

If a live wire falls on the ground firemen are taught to keep both feet in contact with the ground and shuffle away from the live wire.  If they take a stride the resistance of the ground can be high enough current will flow up one leg and down the other electrocuting the fireman.

Just a quick footnote: it is quite unlikely a human can be electrocuted in this way, because the electrical path up one leg and down the other does not pass through a human's heart. In addition to which, the fireman is probably wearing heavy insulated boots. It is sensible to take the proper precautions and avoid unnecessary risks, but the actual risk is not that great.

Can you give me an example of a short circuit with high resistance?

[timb]

Not sure what the cherry picker was made out of.  I'm surprised if you understand electricity why you don't understand why the two guys in the cherry picker were being electrocuted.  This is something firemen are taught with live wires.  If a live wire falls on the ground firemen are taught to keep both feet in contact with the ground and shuffle away from the live wire.  If they take a stride the resistance of the ground can be high enough current will flow up one leg and down the other electrocuting the fireman.  I remember hearing of three police horses being killed in Florida because of a live wire in a field.  The resistance of the ground between the legs of the horses was enough that the current flowed through the horses killing them.

This is what I was taught.  Not say I'm right, but then again I've been looking up the definition of the words to make sure I'm using them as properly. 

I do want to thank you for questioning me.  It's required me to do a fair amount of researcher to confirm what I saying is correct.

I’m surprised why you think they *would* be electrocuted.

If the metal arm of the cherry picker touched the wire, there would be no path through the men in the bucket to ground, right? Hence they would be safe.

If just the (fiberglass or composite) bucket touched the live wire, there would be no current flow because the bucket is an insulator, and again they would be safe.

If the bucket was metal and touched the wire, they *may* be safe so long as they didn’t touch the sides of the bucket.

The only case in which they’d be harmed is if they touched the wire in a metal bucket, attached to an uninsulated truck (the type that uses hydraulic pistons to lower metal feet into the ground, for stability). In that case the current would go through the person, base of the bucket, down the arm, over the metal skin of the truck and down the steel feet into the ground.

This is all assuming they touched a “low voltage” wire, like the type attached to wooden power poles. The high voltage transport wires attached to the tall metal towers are a different story, but those are generally out of the reach of cherry pickers anyway. (Those are serviced via helicopters! The technicians fly right up to them and use an insulated bar to attach a cable between the copter and wire, to equalize the potentials.)

As for live cables on the ground, I’ve also heard it’s best to shuffle or hop with both feet together when moving away from a fallen cable. However, that’s a completely different scenario to the cherry picker one. (The potential difference surrounding a live wire  that’s fallen on the ground drops off rapidly as you move away from the wire, so walking with a large stride could cause current to flow through you.) There’s a PDF out there with diagrams and images that talks about all this. It’s been posted on the forum before, but I can’t seem to find it at the moment.

[timb]

I thought we agreed if the generator is spinning faster voltage would increase.

Yep.

And if spinning at twice the speed there were by 4 X the kinetic energy.

If the turbine spins at twice the speed, the turbine will have 4x the kinetic energy, yes.

My question is where is all of that extra energy going?

The energy is not going anywhere. A faster spinning turbine has more kinetic energy, just like a fully charged battery has more chemical energy in it. Energy doesn't have to go places, you can just have a system with lots of gravitational potential energy (fully filled dam), lots of chemical potential energy (charged batteries, a pile of coal), or lots of kinetic energy (a rapidly spinning heavy thing), and that's just fine. The energy can stay where it is, it's not a flow (like power) and so it doesn't have to flow, let alone flow to any particular place.

I think you're confusing energy and power? Energy is measured in Joules, and power is measured in Joules per second (also known as Watts). There's a recent EEVBlog video that goes into some detail on this, you really need to have this concept down before you can discuss subtle things like an imbalance in the power going into/out of a turbine leading to a change in the kinetic energy of the turbine (and that being the end of the story and just fine) -- just like an imbalance in the water flow going into/out of a bucket leading to a change in the quantity of water in the bucket with no need to "explain" where all that "extra water" "goes".

I will disagree with you.  The moment the load is removed from the circuit the generator "doesn't know" and will still be sending the same amount of energy/watts in the wires.  Similar concept in plumbing with pipes hammering.  The hammering is how the excess energy is being dissipated.  What's the equivalent to pipe hammering with the generator.

We’ve given you the answer numerous times.

The generator is powered by steam. There’s a controller that operates the valves that control the flow of steam. They monitor the generator’s output. When the load is removed from the generator the output will start to rise, the controller will sense this and close the steam valves a bit, causing the generator to slow and the output to come back down. This all happens in the blink of an eye.

That’s a control loop. It works no different to an Op-Amp.

Think about it. I have a simple op-amp follower circuit (no gain), and I apply 5V to the + input and attach a 100 Ohm load to the output. Now the output is at 5V. If I remove the 100 Ohm Load, the output voltage will briefly spike by a few millivolts before returning to 5V. Where did the power go? Nowhere. The op-amp used negative feedback to close the internal “steam valve” (a transistor) in order to keep the voltage at the setpoint.

Is this making sense yet? We really can’t make it much simpler.

[timb]

Can you give me an example of a short circuit with high resistance?

Let’s say I have a circuit only capable of 1uA output current. If I place a 1MOhm resistor on it’s output, it’s essentially a short circuit.

[rs20]

I will disagree with you.  The moment the load is removed from the circuit the generator "doesn't know" and will still be sending the same amount of energy/watts in the wires.

All I can say is, you're asking a vast number of interesting questions, but you need to focus your attention a bit. There are all sorts of different time scales on which generators and loads balance out:
- Milliseconds: Speed-of-light causality limit, which you seem to have brought up for the first time
- Short term (ms): Lights in your neighborhood burn brighter, with the vast rotational inertia of all motors and generators in the country hold frequency nearly steady
- Medium term (100ms - multiple seconds): Turbine flow control in power stations is adjusted
- Long term (integrated/infinity): Less coal burned or more water still in dam.
- Even longer term (years): Power stations getting built and decomissioned.

Now, when you're talking about instants of time so short that the speed of light does not allow a power source and load to even communicate let alone match current, you're getting into the realm of transmission lines. You'll note that the linked webpage there notes that the study transmission is (typically) restricted to "radio frequency, that is, currents with a frequency high enough that their wave nature must be taken into account". This wouldn't normally refer to 50 Hz power, but the huge size of transmission networks, and the newly introduced question of a device being turned off instantaneously (a hard edge with unbounded frequency content) makes transmission line theory necessary to answer your question with the sort of rigor that you seem to be demanding.

The rest of us on this thread are simply assuming that the load and generators are communicating and matching currents practically instantaneously, an assumption which a) simplifies the thought process massively and b) does not introduce any practically relevant errors when answering the question "what happens when I turn my TV off". I'd strongly encourage you to allow yourself to make this assumption, understand things within this simplified framework, and then start studying transmission lines if you're still bothered by the "spooky action at a distance" aspect. I mean, you can start developing formulae for the positions of individual electrons if you want, but none of us are going to hold your hand through that process! We're electronics people, not physicists.

Similar concept in plumbing with pipes hammering.  The hammering is how the excess energy is being dissipated.  What's the equivalent to pipe hammering with the generator.

That's because the water has inertia. The corresponding concept in electronics is inductance. Inductance of power lines is totally irrelevant when answering the question "what happens when I turn my TV off", except that power lines have both distributed inductance and distributed capacitance, which means they form... a transmission line. But as mentioned previously, you shouldn't think about this aspect unless you really want to. Allow yourself to assume that the load and the generator are connected by ideal, inductance-free, superconducting, short wires first.

[rs20]

Just for sake of a analogy, if you were explaining to someone how a dam "knows" to stop "sending" water when someone turns off a tap, and they asked what happens at the instant the tap is turned off, the dam doesn't know yet, where does the water go... the fully rigorous answer is that the pressure builds up behind the turned-off tap, and that high pressure then travels, as a sound wave, all the way back to the dam. After a few reflections back and forth, the system eventually settles into a new equilibrium. Does this explanation fully satisfy all the rules of causality and no spooky-action-at-a-distance? Why yes. Is it a useful or practical explanation? Nope. Would it even be used by professional water pipe planning people? Not even. Much simpler, and practically identical to reality, to assume that the dam instantaneously stops sending the extra water.

Btw, transmission lines vs wires is like sound waves in pipes vs standard hydraulics.

[IanB]

Exactly  At the moment the brakes are let of the amount of energy you are pushing with has to go somewhere so you fall forward and hit the ground.  That extra KE you aren't using to push is instantaneously converted to PE (since you have no resistance) than as you begin to fall it's converted to PE (as you are falling) then it's converted to heat and sound as you hit the ground.

Lol! I'm rather assuming you wouldn't be that clumsy 

What should happen is that as the bicycle runs away from you you put a foot out in front to catch yourself and you take an extra pace or to control your speed.

[timb]

You know, thinking about this question a bit more, there’s two critical things I think @DougSpindler fails to grasp: A generator can run freely without any load and power isn’t what he thinks it is.

Think about this, I have a propane based whole home generator with automatic switchover. When it starts up, the engine is spinning the generator windings just fine even though they aren’t connected to an electrical load. At this point the generator is producing 120VAC@60Hz and putting out 0A. Suddenly, a solenoid throws an A/B switch and my entire house is connected as the load. Now, you can hear the engine connected to the generator briefly whine and slow down a bit. At this point it might be producing 110VAC@58Hz and putting out 30A. Within a second or so, the control loop inside the unit increases gas flow to the engine, to compensate for this newly added load. Now we’re back to 120VAC@60Hz. After a few minutes, say my Air Con kicks off, decreasing the load on the generator, so now the voltage jumps to 125VAC@61Hz until the control loop slows the engine.

Now, I know what you’re thinking, at the end there the “power” jumped up, right? Well, no. The voltage may have gone up five volts, however the current would have gone down an equivalent amount, keeping the total power draw the same.

So, while voltage may fluctuate as I add and shed loads to my generator, the overall wattage used will stay the same (since an appliance will use less current at a higher voltage and vice versa).

Keep in mind this is different for the typical power distribution network, as there are so many generators spread out over such a large area that as loads are added and shed it’s basically imperceptible to the home user.

[jmelson]

*More specifically, synchronous machines do.  Shorting slats are simply shorted turns on the armature, which act to slow any change in its magnetic field (which is an electromagnet, so the field is supposed to be static anyway).  Note: I haven't studied modern generators per se -- their design may vary.  Anyway, even if this isn't the exact solution used, the same effect is required, however it's achieved.

I've had the opportunity to study a few GE power house alternators in a repair shop and before installation.  Very interesting stuff.  So, I'll describe the alternator that went into the Calloway County nuclear plant in Missouri.  The rotor was a single piece of steel about 15 feet long.  The active rotor section was, as I recall, about 18" diameter and about 6 feet long.  It fits within the stator iron laminations with just a few thousandths of an inch gap.  The rotor is a solid piece of steel, no laminations, and has a pair of rectangular-spiral grooves machined into it to form the poles.  The magnetic field is produced by a winding of about a dozen turns of a copper bar fitted into the groove.  The bar is about 1/4" thick and 2" wide.  The groove plunges deep under the winding to allow one end of the bar to pass from the center to the outside, where it is connected to the exciter. The bar is insulated with some kind of paper or fiberglass material inserted into the slot first.  The rotor winding is fed something like 10,000 Amps, with about  100 V drop across it.  Beyond the active rotor section, the shafts were somewhere between 9 - 12" diameter, with big flanges on one end where the turbine was attached.

These alternators are sealed as best as they can be (with a rotating shaft passing through) and filled with hydrogen gas.  There is a water-hydrogen heat exchanger to cool the hydrogen.  The stator laminations have channels in them for the hydrogen to pass through and pick up heat.

Jon

[DougSpindler]

Can you give me an example of a short circuit with high resistance?

Let’s say I have a circuit only capable of 1uA output current. If I place a 1MOhm resistor on it’s output, it’s essentially a short circuit.

How is that a short circuit?  Can you provide a reference source which describes the scenario you described as a short circuit?

Here's what I was taught.
A short circuit is an abnormal connection between two nodes of an electric circuit intended to be at different voltages. This results in an electric current limited only by the Thévenin equivalent resistance of the rest of the network which can cause circuit damage, overheating, fire or explosion. Although usually the result of a fault, there are cases where short circuits are caused intentionally, for example, for the purpose of voltage-sensing crowbar circuit protectors.

In circuit analysis, a short circuit is defined as a connection between two nodes that forces them to be at the same voltage. In an 'ideal' short circuit, this means there is no resistance and thus no voltage drop across the connection.

[IanB]

Let’s say I have a circuit only capable of 1uA output current. If I place a 1MOhm resistor on it’s output, it’s essentially a short circuit.

How is that a short circuit?  Can you provide a reference source which describes the scenario you described as a short circuit?

Well, you provided your own reference yourself below:

In circuit analysis, a short circuit is defined as a connection between two nodes that forces them to be at the same voltage. In an 'ideal' short circuit, this means there is no resistance and thus no voltage drop across the connection.

Let's imagine the 1 µA flowing through the 1 meg-ohm resistor. By Ohm's law, the voltage difference across the resistor will be 1 V. Now in practical terms, if the two nodes are only one volt apart, they are effectively "at the same voltage". Especially since a circuit capable of only 1 µA output is likely to produce thousands of volts in the normal mode of operation. Compare 1 volt to thousands of volts and the two nodes may as well be at the same voltage. So it's a short circuit.

[T3sl4co1l]

I've had the opportunity to study a few GE power house alternators in a repair shop and before installation.  Very interesting stuff.

[snip]

Cheers!

So, that solid iron rotor will have pretty good eddy currents induced in it by harmonics, so that should count for something.

Tim

[T3sl4co1l]

Let's imagine the 1 µA flowing through the 1 meg-ohm resistor. By Ohm's law, the voltage difference across the resistor will be 1 V. Now in practical terms, if the two nodes are only one volt apart, they are effectively "at the same voltage". Especially since a circuit capable of only 1 µA output is likely to produce thousands of volts in the normal mode of operation. Compare 1 volt to thousands of volts and the two nodes may as well be at the same voltage. So it's a short circuit.

Could you please read a textbook?

Thanks,

Tim

[IanB]

Could you please read a textbook?

OK 

Where do you think I made an error?

[T3sl4co1l]

Could you please read a textbook?

OK 

Where do you think I made an error?

Thanks for asking --

Let's imagine the 1 µA flowing through the 1 meg-ohm resistor. By Ohm's law, the voltage difference across the resistor will be 1 V. Now in practical terms, if the two nodes are only one volt apart, they are effectively "at the same voltage".

[citation needed]

Especially since a circuit capable of only 1 µA output is likely to produce thousands of volts in the normal mode of operation.

Making assumptions about a completely unspecified source.

Compare 1 volt to thousands of volts and the two nodes may as well be at the same voltage. So it's a short circuit.

TIL the voltage sense divider on a high voltage supply is definitely a short already, so it won't matter if I jam a screwdriver into it.

Another example: I was working on a circuit a few days ago with uA bias currents on BJTs.  Vbe is a mere 0.5V at this level.

I guess that whole circuit was accidentally a double short, then...
</sardonic>

But seriously, 1V isn't even so low that you can't do anything with it.

Tim

[IanB]

OK, I guess context is important.

We could imagine an electrostatic machine with an output current of 1 µA that in normal operation would charge up an output terminal to a high voltage. If we connected this output terminal to ground with a 1 meg resistor, then the output wouldn't generate any voltage at all. In that context it would effectively be shorted.

I accept that this example needs to be stated for it to make sense otherwise everything is open to interpretation.

Besides that there are cases of sensitive electronic circuits where contamination on the circuit board can lead to stray currents that compromise operation of the device. Maybe that's not a "short" circuit, but it is an unintended current path.

[T3sl4co1l]

OK, I guess context is important.

We could imagine an electrostatic machine with an output current of 1 µA that in normal operation would charge up an output terminal to a high voltage. If we connected this output terminal to ground with a 1 meg resistor, then the output wouldn't generate any voltage at all. In that context it would effectively be shorted.

This is actually a very sneaky, and -- probably unintentionally -- contrived situation!

Most static generators work on positive feedback.  Kelvin dropper, Wimshurst machine, those kinds: they operate on induction from the existing charge.

If you "short out" the output, the gain also goes to zero, and the output -- well and truly -- is shorted.  But it's also not 1uA (say) anymore.  Which makes it kind of disingenuous to say so!

A triboelectric or otherwise powered generator (like a van De Graaf generator -- often, static is "sprayed" onto the belt with a separate, lower (still "high") voltage, electronic supply), delivers constant charge (roughly speaking) and so cannot be shorted out with any load other than zero ohms.

A short is simply zero ohms -- to imply otherwise, whether in context or not, is simply not good usage!

Tim

[rs20]

I'm not sure this thread is too healthy -- we've got one person perpetually asking for more information at twice the rate that he is able to process it; another person taking on the challenge of providing a bulletproof definition for the ultimately fuzzy and colloquial concept of a "short-circuit", and another person poking holes in the previous attempts by piling on the interesting specifics of real-world static generators. I'm not sure if anyone is winning here.

Regardless, I, for one, have no quibble with the claim that a nominally CV power supply running at 1% or even 10% of its nominal output voltage due to a overcurrent condition is a power supply that (at the very least, colloquially) is being "shorted". Regardless of any oxidative reactions that might be happening in the vicinity.

[timb]

A short is simply zero ohms -- to imply otherwise, whether in context or not, is simply not good usage!

Tim

So, you’re saying if I have a 1uA current source with a, say, 10V compliance voltage and I place a 10, 100 or even 1kOhm resistance across it, it wouldn’t be considered shorted? Who knew!

[Ice-Tea]

A short is simply zero ohms -- to imply otherwise, whether in context or not, is simply not good usage!

Tim

So, you’re saying if I have a 1uA current source with a, say, 10V compliance voltage and I place a 10, 100 or even 1kOhm resistance across it, it wouldn’t be considered shorted? Who knew!

I would consider it heavily loaded

[DougSpindler]

Interesting in how in over a hundred years of using electricity we still can't agree on the use of words and terms such as short-circuit.

[dr.diesel]

Interesting in how in over a hundred years of using electricity we still can't agree on the use of words and terms such as short-circuit.

IMO a short circuit is simply an unintended path or a bypass, it certainly doesn't require zero ohms to happen.

[DougSpindler]

You know, thinking about this question a bit more, there’s two critical things I think @DougSpindler fails to grasp: A generator can run freely without any load and power isn’t what he thinks it is.

Think about this, I have a propane based whole home generator with automatic switchover. When it starts up, the engine is spinning the generator windings just fine even though they aren’t connected to an electrical load. At this point the generator is producing 120VAC@60Hz and putting out 0A. Suddenly, a solenoid throws an A/B switch and my entire house is connected as the load. Now, you can hear the engine connected to the generator briefly whine and slow down a bit. At this point it might be producing 110VAC@58Hz and putting out 30A. Within a second or so, the control loop inside the unit increases gas flow to the engine, to compensate for this newly added load. Now we’re back to 120VAC@60Hz. After a few minutes, say my Air Con kicks off, decreasing the load on the generator, so now the voltage jumps to 125VAC@61Hz until the control loop slows the engine.

Now, I know what you’re thinking, at the end there the “power” jumped up, right? Well, no. The voltage may have gone up five volts, however the current would have gone down an equivalent amount, keeping the total power draw the same.

So, while voltage may fluctuate as I add and shed loads to my generator, the overall wattage used will stay the same (since an appliance will use less current at a higher voltage and vice versa).

Keep in mind this is different for the typical power distribution network, as there are so many generators spread out over such a large area that as loads are added and shed it’s basically imperceptible to the home user.

@Timb, you are following me. 

Let's take your scenario but add a second generator.  What happens if the second generator is 108 degrees out of phase with the other generator.  I'm guessing here.....   But I would assume there would be almost no current flowing to you home and all of the current would be flowing between the generators effectively resulting it a short and probably a fire (exothermic oxidative reaction) from over heating.

Second question.
What if the generators used permanent magnets and the speed of the generator was constant.  IF the generator was connected to a load let's say 1,000 watts are being generated and consumed.  When the switch is opened and current flow stops does that mean the generator has stopped producing 1,000 watts?  I would not think so.  The generator is still spinning at the same speed, the magnetic field is the same.  So where does that 1,000 watts go?

[timb]

You know, thinking about this question a bit more, there’s two critical things I think @DougSpindler fails to grasp: A generator can run freely without any load and power isn’t what he thinks it is.

Think about this, I have a propane based whole home generator with automatic switchover. When it starts up, the engine is spinning the generator windings just fine even though they aren’t connected to an electrical load. At this point the generator is producing 120VAC@60Hz and putting out 0A. Suddenly, a solenoid throws an A/B switch and my entire house is connected as the load. Now, you can hear the engine connected to the generator briefly whine and slow down a bit. At this point it might be producing 110VAC@58Hz and putting out 30A. Within a second or so, the control loop inside the unit increases gas flow to the engine, to compensate for this newly added load. Now we’re back to 120VAC@60Hz. After a few minutes, say my Air Con kicks off, decreasing the load on the generator, so now the voltage jumps to 125VAC@61Hz until the control loop slows the engine.

Now, I know what you’re thinking, at the end there the “power” jumped up, right? Well, no. The voltage may have gone up five volts, however the current would have gone down an equivalent amount, keeping the total power draw the same.

So, while voltage may fluctuate as I add and shed loads to my generator, the overall wattage used will stay the same (since an appliance will use less current at a higher voltage and vice versa).

Keep in mind this is different for the typical power distribution network, as there are so many generators spread out over such a large area that as loads are added and shed it’s basically imperceptible to the home user.

@Timb, you are following me. 

Second question.
What if the generators used permanent magnets and the speed of the generator was constant.  IF the generator was connected to a load let's say 1,000 watts are being generated and consumed.  When the switch is opened and current flow stops does that mean the generator has stopped producing 1,000 watts?  I would not think so.  The generator is still spinning at the same speed, the magnetic field is the same.  So where does that 1,000 watts go?

It goes nowhere! The generator is no longer producing 1000W! It’s simply spinning unloaded. It’s producing xxxxV@60HZ with 0A output. Remember, power is V*I. If zero amps are flowing (no load) then zero watts are being produced.

Think about this: Say I take a wall adapter capable of 10W output (a USB charger perhaps) but instead of hooking my tablet to it (which can draw 10W) I plug in my phone (which can only draw 5W). Where does that extra 5W go? Nowhere!

Another thought experiment: You’re driving along in your car, the motor is at 3000RPM and you shift into neutral. Where does the “power” from the engine go? In this analogy power is mechanical force in the form of torque applied to the gearbox, which is applied to the drive shaft, which is applied to the wheels. Answer: Nowhere! Now the only difference is it takes less fuel flow to spin the crankshaft at the same RPM, since it has no load. So, as you engaged the clutch, you’d back off the throttle. In this scenario, *you’re* part of the control loop.

I think the critical flaw in your thinking is that generator is somehow “producing current” that must be “consumed” by a load. That is incorrect. A load *draws* current from a source. The current draw at a particular voltage is what makes up the power output of the generator. If the load is drawing no current, then the generator is sourcing zero watts and spinning freely.

Another experiment: Take a small motor (like the kind used in a toy) and connect it to a multimeter. Try to spin the shaft with your fingers. It should spin easily and you should see a voltage produced. (At this point the motor has a 10Mohm load.)

Now, connect a 1 ohm resistor across the terminals of the motor and try to spin the shaft. It should be a lot harder to spin. That’s because the current draw on the output is loading it down.

In the unloaded case, very little mechanical energy is required to spin the shaft.

In the loaded case, much more mechanical energy is required to spin the shaft.

Basically, it works like this: A source of heat (burning coal, natural gas or a fissile material) is used to boil water, which results in steam. The steam is run through narrow pipes to increase the pressure. This high pressure steam pushes a turbine that turns a generator. Heat is a source of energy. So, you can directly equate the amount of heat required to xxW of energy produced by the generator. If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is heat into mechanical energy into electrical energy.)

Is this making any sense?

Let's take your scenario but add a second generator.  What happens if the second generator is 108 degrees out of phase with the other generator.  I'm guessing here.....   But I would assume there would be almost no current flowing to you home and all of the current would be flowing between the generators effectively resulting it a short and probably a fire (exothermic oxidative reaction) from over heating.

That scenario doesn’t reflect how generators in power distribution work. The generators at each plant would be locked in phase with the rest of the generators on the grid. I don’t know the technical details, but I imagine it involves monitoring the frequency of the grid and adjusting according.

[IanB]

That scenario doesn’t reflect how generators in power distribution work. The generators at each plant would be locked in phase with the rest of the generators on the grid. I don’t know the technical details, but I imagine it involves monitoring the frequency of the grid and adjusting according.

Yes, if you want to connect a generator to the grid you have to synchronize it first, by getting the speed and phase angle to match closely to the grid. In the old days people might have done this by hand, but these days it is done automatically. If you happen to close the switch when the generator is out of phase with the grid then there is indeed a catastrophically large current surge, a mechanical eruption when the turning generator meets an immovable force, and generally nothing good comes of it. In short, don't do that.

[dr.diesel]

In the old days people might have done this by hand

Still done by hand on large units!

[DougSpindler]

You know, thinking about this question a bit more, there’s two critical things I think @DougSpindler fails to grasp: A generator can run freely without any load and power isn’t what he thinks it is.

Think about this, I have a propane based whole home generator with automatic switchover. When it starts up, the engine is spinning the generator windings just fine even though they aren’t connected to an electrical load. At this point the generator is producing 120VAC@60Hz and putting out 0A. Suddenly, a solenoid throws an A/B switch and my entire house is connected as the load. Now, you can hear the engine connected to the generator briefly whine and slow down a bit. At this point it might be producing 110VAC@58Hz and putting out 30A. Within a second or so, the control loop inside the unit increases gas flow to the engine, to compensate for this newly added load. Now we’re back to 120VAC@60Hz. After a few minutes, say my Air Con kicks off, decreasing the load on the generator, so now the voltage jumps to 125VAC@61Hz until the control loop slows the engine.

Now, I know what you’re thinking, at the end there the “power” jumped up, right? Well, no. The voltage may have gone up five volts, however the current would have gone down an equivalent amount, keeping the total power draw the same.

So, while voltage may fluctuate as I add and shed loads to my generator, the overall wattage used will stay the same (since an appliance will use less current at a higher voltage and vice versa).

Keep in mind this is different for the typical power distribution network, as there are so many generators spread out over such a large area that as loads are added and shed it’s basically imperceptible to the home user.

@Timb, you are following me. 

Second question.
What if the generators used permanent magnets and the speed of the generator was constant.  IF the generator was connected to a load let's say 1,000 watts are being generated and consumed.  When the switch is opened and current flow stops does that mean the generator has stopped producing 1,000 watts?  I would not think so.  The generator is still spinning at the same speed, the magnetic field is the same.  So where does that 1,000 watts go?

It goes nowhere! The generator is no longer producing 1000W! It’s simply spinning unloaded. It’s producing xxxxV@60HZ with 0A output. Remember, power is V*I. If zero amps are flowing (no load) then zero watts are being produced.

Think about this: Say I take a wall adapter capable of 10W output (a USB charger perhaps) but instead of hooking my tablet to it (which can draw 10W) I plug in my phone (which can only draw 5W). Where does that extra 5W go? Nowhere!

Another thought experiment: You’re driving along in your car, the motor is at 3000RPM and you shift into neutral. Where does the “power” from the engine go? In this analogy power is mechanical force in the form of torque applied to the gearbox, which is applied to the drive shaft, which is applied to the wheels. Answer: Nowhere! Now the only difference is it takes less fuel flow to spin the crankshaft at the same RPM, since it has no load. So, as you engaged the clutch, you’d back off the throttle. In this scenario, *you’re* part of the control loop.

I think the critical flaw in your thinking is that generator is somehow “producing current” that must be “consumed” by a load. That is incorrect. A load *draws* current from a source. The current draw at a particular voltage is what makes up the power output of the generator. If the load is drawing no current, then the generator is sourcing zero watts and spinning freely.

Another experiment: Take a small motor (like the kind used in a toy) and connect it to a multimeter. Try to spin the shaft with your fingers. It should spin easily and you should see a voltage produced. (At this point the motor has a 10Mohm load.)

Now, connect a 1 ohm resistor across the terminals of the motor and try to spin the shaft. It should be a lot harder to spin. That’s because the current draw on the output is loading it down.

In the unloaded case, very little mechanical energy is required to spin the shaft.

In the loaded case, much more mechanical energy is required to spin the shaft.

Basically, it works like this: A source of heat (burning coal, natural gas or a fissile material) is used to boil water, which results in steam. The steam is run through narrow pipes to increase the pressure. This high pressure steam pushes a turbine that turns a generator. Heat is a source of energy. So, you can directly equate the amount of heat required to xxW of energy produced by the generator. If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is heat into mechanical energy into electrical energy.)

Is this making any sense?

Let's take your scenario but add a second generator.  What happens if the second generator is 108 degrees out of phase with the other generator.  I'm guessing here.....   But I would assume there would be almost no current flowing to you home and all of the current would be flowing between the generators effectively resulting it a short and probably a fire (exothermic oxidative reaction) from over heating.

That scenario doesn’t reflect how generators in power distribution work. The generators at each plant would be locked in phase with the rest of the generators on the grid. I don’t know the technical details, but I imagine it involves monitoring the frequency of the grid and adjusting according.

@timb

 Agree with everything you are saying, but you are forgetting it's still taking energy to spin the generator without the load.  Energy is being "consumed" when the circuity is energized and when it's not.  Difference is when there's an electrical load the there's additional resistance as a result of the EMF in the motor.  When the circuit is open there no current flow, not electron flow and no EMF to place resist the spinning of the generator.  So the energy loss of energy is to friction/heat.

[timb]

You know, thinking about this question a bit more, there’s two critical things I think @DougSpindler fails to grasp: A generator can run freely without any load and power isn’t what he thinks it is.

Think about this, I have a propane based whole home generator with automatic switchover. When it starts up, the engine is spinning the generator windings just fine even though they aren’t connected to an electrical load. At this point the generator is producing 120VAC@60Hz and putting out 0A. Suddenly, a solenoid throws an A/B switch and my entire house is connected as the load. Now, you can hear the engine connected to the generator briefly whine and slow down a bit. At this point it might be producing 110VAC@58Hz and putting out 30A. Within a second or so, the control loop inside the unit increases gas flow to the engine, to compensate for this newly added load. Now we’re back to 120VAC@60Hz. After a few minutes, say my Air Con kicks off, decreasing the load on the generator, so now the voltage jumps to 125VAC@61Hz until the control loop slows the engine.

Now, I know what you’re thinking, at the end there the “power” jumped up, right? Well, no. The voltage may have gone up five volts, however the current would have gone down an equivalent amount, keeping the total power draw the same.

So, while voltage may fluctuate as I add and shed loads to my generator, the overall wattage used will stay the same (since an appliance will use less current at a higher voltage and vice versa).

Keep in mind this is different for the typical power distribution network, as there are so many generators spread out over such a large area that as loads are added and shed it’s basically imperceptible to the home user.

@Timb, you are following me. 

Second question.
What if the generators used permanent magnets and the speed of the generator was constant.  IF the generator was connected to a load let's say 1,000 watts are being generated and consumed.  When the switch is opened and current flow stops does that mean the generator has stopped producing 1,000 watts?  I would not think so.  The generator is still spinning at the same speed, the magnetic field is the same.  So where does that 1,000 watts go?

It goes nowhere! The generator is no longer producing 1000W! It’s simply spinning unloaded. It’s producing xxxxV@60HZ with 0A output. Remember, power is V*I. If zero amps are flowing (no load) then zero watts are being produced.

Think about this: Say I take a wall adapter capable of 10W output (a USB charger perhaps) but instead of hooking my tablet to it (which can draw 10W) I plug in my phone (which can only draw 5W). Where does that extra 5W go? Nowhere!

Another thought experiment: You’re driving along in your car, the motor is at 3000RPM and you shift into neutral. Where does the “power” from the engine go? In this analogy power is mechanical force in the form of torque applied to the gearbox, which is applied to the drive shaft, which is applied to the wheels. Answer: Nowhere! Now the only difference is it takes less fuel flow to spin the crankshaft at the same RPM, since it has no load. So, as you engaged the clutch, you’d back off the throttle. In this scenario, *you’re* part of the control loop.

I think the critical flaw in your thinking is that generator is somehow “producing current” that must be “consumed” by a load. That is incorrect. A load *draws* current from a source. The current draw at a particular voltage is what makes up the power output of the generator. If the load is drawing no current, then the generator is sourcing zero watts and spinning freely.

Another experiment: Take a small motor (like the kind used in a toy) and connect it to a multimeter. Try to spin the shaft with your fingers. It should spin easily and you should see a voltage produced. (At this point the motor has a 10Mohm load.)

Now, connect a 1 ohm resistor across the terminals of the motor and try to spin the shaft. It should be a lot harder to spin. That’s because the current draw on the output is loading it down.

In the unloaded case, very little mechanical energy is required to spin the shaft.

In the loaded case, much more mechanical energy is required to spin the shaft.

Basically, it works like this: A source of heat (burning coal, natural gas or a fissile material) is used to boil water, which results in steam. The steam is run through narrow pipes to increase the pressure. This high pressure steam pushes a turbine that turns a generator. Heat is a source of energy. So, you can directly equate the amount of heat required to xxW of energy produced by the generator. If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is heat into mechanical energy into electrical energy.)

Is this making any sense?

Let's take your scenario but add a second generator.  What happens if the second generator is 108 degrees out of phase with the other generator.  I'm guessing here.....   But I would assume there would be almost no current flowing to you home and all of the current would be flowing between the generators effectively resulting it a short and probably a fire (exothermic oxidative reaction) from over heating.

That scenario doesn’t reflect how generators in power distribution work. The generators at each plant would be locked in phase with the rest of the generators on the grid. I don’t know the technical details, but I imagine it involves monitoring the frequency of the grid and adjusting according.

@timb

 Agree with everything you are saying, but you are forgetting it's still taking energy to spin the generator without the load.  Energy is being "consumed" when the circuity is energized and when it's not.  Difference is when there's an electrical load the there's additional resistance as a result of the EMF in the motor.  When the circuit is open there no current flow, not electron flow and no EMF to place resist the spinning of the generator.  So the energy loss of energy is to friction/heat.

Re-read my post carefully. I basically said this at the end of my reply to your second question:

“If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is, heat into mechanical energy into electrical energy.)”

Part of the mechanical energy losses would be friction, obviously.

The point I was making is that it requires orders of magnitude more energy to spin a generator under full load than it does to spin one under no load. If a generator under full load suddenly has a load removed, it will obviously speed up slightly until the control loop responds and slows it down again. Generally there’s a system in place that can respond in the hundreds of millisecond time range, that will divert the steam from the turbine and dump it somewhere in case of a full load removal. So, that’s where the energy goes!

[IanMacdonald]

Which is basically what constraint payments are about. The operator gets paid for the fuel used to spin the generator pointlessly.

"Any sufficiently advanced magic is indistinguishable from technology"
-For example, the palantir is basically a round iPhone. Sooner or later they'll figure out that a thinner rectangular one is more convenient.

[DougSpindler]

You know, thinking about this question a bit more, there’s two critical things I think @DougSpindler fails to grasp: A generator can run freely without any load and power isn’t what he thinks it is.

Think about this, I have a propane based whole home generator with automatic switchover. When it starts up, the engine is spinning the generator windings just fine even though they aren’t connected to an electrical load. At this point the generator is producing 120VAC@60Hz and putting out 0A. Suddenly, a solenoid throws an A/B switch and my entire house is connected as the load. Now, you can hear the engine connected to the generator briefly whine and slow down a bit. At this point it might be producing 110VAC@58Hz and putting out 30A. Within a second or so, the control loop inside the unit increases gas flow to the engine, to compensate for this newly added load. Now we’re back to 120VAC@60Hz. After a few minutes, say my Air Con kicks off, decreasing the load on the generator, so now the voltage jumps to 125VAC@61Hz until the control loop slows the engine.

Now, I know what you’re thinking, at the end there the “power” jumped up, right? Well, no. The voltage may have gone up five volts, however the current would have gone down an equivalent amount, keeping the total power draw the same.

So, while voltage may fluctuate as I add and shed loads to my generator, the overall wattage used will stay the same (since an appliance will use less current at a higher voltage and vice versa).

Keep in mind this is different for the typical power distribution network, as there are so many generators spread out over such a large area that as loads are added and shed it’s basically imperceptible to the home user.

@Timb, you are following me. 

Second question.
What if the generators used permanent magnets and the speed of the generator was constant.  IF the generator was connected to a load let's say 1,000 watts are being generated and consumed.  When the switch is opened and current flow stops does that mean the generator has stopped producing 1,000 watts?  I would not think so.  The generator is still spinning at the same speed, the magnetic field is the same.  So where does that 1,000 watts go?

It goes nowhere! The generator is no longer producing 1000W! It’s simply spinning unloaded. It’s producing xxxxV@60HZ with 0A output. Remember, power is V*I. If zero amps are flowing (no load) then zero watts are being produced.

Think about this: Say I take a wall adapter capable of 10W output (a USB charger perhaps) but instead of hooking my tablet to it (which can draw 10W) I plug in my phone (which can only draw 5W). Where does that extra 5W go? Nowhere!

Another thought experiment: You’re driving along in your car, the motor is at 3000RPM and you shift into neutral. Where does the “power” from the engine go? In this analogy power is mechanical force in the form of torque applied to the gearbox, which is applied to the drive shaft, which is applied to the wheels. Answer: Nowhere! Now the only difference is it takes less fuel flow to spin the crankshaft at the same RPM, since it has no load. So, as you engaged the clutch, you’d back off the throttle. In this scenario, *you’re* part of the control loop.

I think the critical flaw in your thinking is that generator is somehow “producing current” that must be “consumed” by a load. That is incorrect. A load *draws* current from a source. The current draw at a particular voltage is what makes up the power output of the generator. If the load is drawing no current, then the generator is sourcing zero watts and spinning freely.

Another experiment: Take a small motor (like the kind used in a toy) and connect it to a multimeter. Try to spin the shaft with your fingers. It should spin easily and you should see a voltage produced. (At this point the motor has a 10Mohm load.)

Now, connect a 1 ohm resistor across the terminals of the motor and try to spin the shaft. It should be a lot harder to spin. That’s because the current draw on the output is loading it down.

In the unloaded case, very little mechanical energy is required to spin the shaft.

In the loaded case, much more mechanical energy is required to spin the shaft.

Basically, it works like this: A source of heat (burning coal, natural gas or a fissile material) is used to boil water, which results in steam. The steam is run through narrow pipes to increase the pressure. This high pressure steam pushes a turbine that turns a generator. Heat is a source of energy. So, you can directly equate the amount of heat required to xxW of energy produced by the generator. If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is heat into mechanical energy into electrical energy.)

Is this making any sense?

Let's take your scenario but add a second generator.  What happens if the second generator is 108 degrees out of phase with the other generator.  I'm guessing here.....   But I would assume there would be almost no current flowing to you home and all of the current would be flowing between the generators effectively resulting it a short and probably a fire (exothermic oxidative reaction) from over heating.

That scenario doesn’t reflect how generators in power distribution work. The generators at each plant would be locked in phase with the rest of the generators on the grid. I don’t know the technical details, but I imagine it involves monitoring the frequency of the grid and adjusting according.

@timb

 Agree with everything you are saying, but you are forgetting it's still taking energy to spin the generator without the load.  Energy is being "consumed" when the circuity is energized and when it's not.  Difference is when there's an electrical load the there's additional resistance as a result of the EMF in the motor.  When the circuit is open there no current flow, not electron flow and no EMF to place resist the spinning of the generator.  So the energy loss of energy is to friction/heat.

Re-read my post carefully. I basically said this at the end of my reply to your second question:

“If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is, heat into mechanical energy into electrical energy.)”

Part of the mechanical energy losses would be friction, obviously.

The point I was making is that it requires orders of magnitude more energy to spin a generator under full load than it does to spin one under no load. If a generator under full load suddenly has a load removed, it will obviously speed up slightly until the control loop responds and slows it down again. Generally there’s a system in place that can respond in the hundreds of millisecond time range, that will divert the steam from the turbine and dump it somewhere in case of a full load removal. So, that’s where the energy goes!

Question I've been asking is where does the energy go in those 100 milliseconds.  The time between when the load is removed and the time the generator slows to a no load condition.  There's energy there that needs to be accounted for.  And let's say the load on the generator is quite large say tens of thousands of watts.  And in the same circuit is a 3 watt indicator light.  Seems to me during those 100 milliseconds the generator is still going to be producing tens of thousands of watts of energy with a load of just 3 watts.  That extra energy has to be accounted for somewhere.

[DougSpindler]

Here's a video which briefly covers the power industry.  (Production, transmission, consumption and storage.)

Very interesting.

[DougSpindler]

Where does the energy go with solar panels and no load?

If a solar panel is in the sun and has a load it will produce electrical energy.  What happens to the energy when the load is removed?
I'm thinking the panel would get a little warmer from the photons which are not being captured and covered to electricity. 

Does that sound correct?

Thanks

This has been a nice discussion. Thank you all.  You are really making me think.  (Thank you.)
 

[Jeroen3]

Question I've been asking is where does the energy go in those 100 milliseconds.  The time between when the load is removed and the time the generator slows to a no load condition.  There's energy there that needs to be accounted for.  And let's say the load on the generator is quite large say tens of thousands of watts.  And in the same circuit is a 3 watt indicator light.  Seems to me during those 100 milliseconds the generator is still going to be producing tens of thousands of watts of energy with a load of just 3 watts.  That extra energy has to be accounted for somewhere.

Go one step back. The energy is there because the engine is moving the rotor. And the generator rotor has magnetic flux. Coupled to the stator windings that output a certain voltage.

When you remove load, less flux is needed to stay at nominal voltage. When you remove a large load, the voltage will rise since it takes time for the regulation to act and the flux to collapse.
When you remove power, less flux is needed to stay at nominal voltage, and less mechanical force is needed to move the rotor. Thus the speed will increase, until the speed governor adjusts.
To answer, where does the power go? Well, it dissipates in raising both rotor speed and voltage.
Why can't you see that? Because the speed governor and voltage regulator have been optimised to stay within safe operation limits.

Notice the distinct load and power, since if you have an ideal inductive load there is no power, thus no mechanical force for the engine.

If you were instead to use a generator as VA source, as you seem to think, you'd have at no load a gazillion volts and infinite rotor speed. However, there is some real world stuff that prevents us from making any ideal source.

Where does the energy go with solar panels and no load?

Solar panels are like batteries. When not connected, they do not output an infinite high voltage. Instead, they settle at their maximum voltage.

[rs20]

Where does the energy go with solar panels and no load?

Solar panels are like batteries. When not connected, they do not output an infinite high voltage. Instead, they settle at their maximum voltage.

Correct, but you didn't answer Doug's question at all. The answer (easily googleable) is that the electron/hole pairs simply recombine (producing reradiation/heat) because the open-circuit voltage prevents further separation of charge. This means that the power goes towards making the solar panel hotter (or maybe re-radiating IR?), so the incoming power (again, Doug is conflating power and energy) is dissipated as heat or radiation.

[SeanB]

In a solar panel the energy is lost in the intrinsic forward biased diode that is the panel junction. The open circuit voltage is around 0.45V per cell at the set luminance level( equal roughly to the equator at noon in the Sahara desert), and this voltage is not going to increase, the panel loss will increase as the diode starts to conduct.

As to the generator thought the extra energy is going to be used to start to speed up multiple tons of rotating equipment, and the time it takes for this to increase in speed to any extent ( more than 1%, which is the typical trip speed limit that the control allows before it does an emergency stop of the lot, though in a Hydro dam this can be insufficient, as found out in the USSR recently) there is normally enough time for a bypass valve to be rapidly opened to dump pressure into the condenser, raising the pressure in there considerably above it's normal sub atmospheric pressure level. Then the control also has dump valves to dump steam to ambient if there is too much condenser pressure, all to keep the generator speed well below it's limiting speed, where it is rated to run without destruction. Dump the load and have one of those safety systems fail and the generator runs up in speed absorbing energy till something breaks. That is best viewed from a distance, around 5km is a minimum, and well upwind and upstream as well, because there will be large parts of very heavy equipment and building making a change in location.

My father saw a little 500kVA turboset get synchronised 180 degrees out of phase, it exited the turbine hall via the one wall in thousands of pieces, and very luckily nobody was in the way of it as well. Took them a good few years to replace that unit and rebuild the turbine hall as well. Synchronisation of those uses a phase meter for rough phase, using 2 needles, and fine adjust before closing the final connecting breaker to get the power flowing using a 3 lamp system to show phase difference between the large grid load ( which is the leader, simply as this is the bigger generation set) and the alternator phases, so you see the 3 lamps brighten and dim in sequence as the phase approaches equal. These days this is often done automatically using a computer that does the same thing.

[timb]

You know, thinking about this question a bit more, there’s two critical things I think @DougSpindler fails to grasp: A generator can run freely without any load and power isn’t what he thinks it is.

Think about this, I have a propane based whole home generator with automatic switchover. When it starts up, the engine is spinning the generator windings just fine even though they aren’t connected to an electrical load. At this point the generator is producing 120VAC@60Hz and putting out 0A. Suddenly, a solenoid throws an A/B switch and my entire house is connected as the load. Now, you can hear the engine connected to the generator briefly whine and slow down a bit. At this point it might be producing 110VAC@58Hz and putting out 30A. Within a second or so, the control loop inside the unit increases gas flow to the engine, to compensate for this newly added load. Now we’re back to 120VAC@60Hz. After a few minutes, say my Air Con kicks off, decreasing the load on the generator, so now the voltage jumps to 125VAC@61Hz until the control loop slows the engine.

Now, I know what you’re thinking, at the end there the “power” jumped up, right? Well, no. The voltage may have gone up five volts, however the current would have gone down an equivalent amount, keeping the total power draw the same.

So, while voltage may fluctuate as I add and shed loads to my generator, the overall wattage used will stay the same (since an appliance will use less current at a higher voltage and vice versa).

Keep in mind this is different for the typical power distribution network, as there are so many generators spread out over such a large area that as loads are added and shed it’s basically imperceptible to the home user.

@Timb, you are following me. 

Second question.
What if the generators used permanent magnets and the speed of the generator was constant.  IF the generator was connected to a load let's say 1,000 watts are being generated and consumed.  When the switch is opened and current flow stops does that mean the generator has stopped producing 1,000 watts?  I would not think so.  The generator is still spinning at the same speed, the magnetic field is the same.  So where does that 1,000 watts go?

It goes nowhere! The generator is no longer producing 1000W! It’s simply spinning unloaded. It’s producing xxxxV@60HZ with 0A output. Remember, power is V*I. If zero amps are flowing (no load) then zero watts are being produced.

Think about this: Say I take a wall adapter capable of 10W output (a USB charger perhaps) but instead of hooking my tablet to it (which can draw 10W) I plug in my phone (which can only draw 5W). Where does that extra 5W go? Nowhere!

Another thought experiment: You’re driving along in your car, the motor is at 3000RPM and you shift into neutral. Where does the “power” from the engine go? In this analogy power is mechanical force in the form of torque applied to the gearbox, which is applied to the drive shaft, which is applied to the wheels. Answer: Nowhere! Now the only difference is it takes less fuel flow to spin the crankshaft at the same RPM, since it has no load. So, as you engaged the clutch, you’d back off the throttle. In this scenario, *you’re* part of the control loop.

I think the critical flaw in your thinking is that generator is somehow “producing current” that must be “consumed” by a load. That is incorrect. A load *draws* current from a source. The current draw at a particular voltage is what makes up the power output of the generator. If the load is drawing no current, then the generator is sourcing zero watts and spinning freely.

Another experiment: Take a small motor (like the kind used in a toy) and connect it to a multimeter. Try to spin the shaft with your fingers. It should spin easily and you should see a voltage produced. (At this point the motor has a 10Mohm load.)

Now, connect a 1 ohm resistor across the terminals of the motor and try to spin the shaft. It should be a lot harder to spin. That’s because the current draw on the output is loading it down.

In the unloaded case, very little mechanical energy is required to spin the shaft.

In the loaded case, much more mechanical energy is required to spin the shaft.

Basically, it works like this: A source of heat (burning coal, natural gas or a fissile material) is used to boil water, which results in steam. The steam is run through narrow pipes to increase the pressure. This high pressure steam pushes a turbine that turns a generator. Heat is a source of energy. So, you can directly equate the amount of heat required to xxW of energy produced by the generator. If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is heat into mechanical energy into electrical energy.)

Is this making any sense?

Let's take your scenario but add a second generator.  What happens if the second generator is 108 degrees out of phase with the other generator.  I'm guessing here.....   But I would assume there would be almost no current flowing to you home and all of the current would be flowing between the generators effectively resulting it a short and probably a fire (exothermic oxidative reaction) from over heating.

That scenario doesn’t reflect how generators in power distribution work. The generators at each plant would be locked in phase with the rest of the generators on the grid. I don’t know the technical details, but I imagine it involves monitoring the frequency of the grid and adjusting according.

@timb

 Agree with everything you are saying, but you are forgetting it's still taking energy to spin the generator without the load.  Energy is being "consumed" when the circuity is energized and when it's not.  Difference is when there's an electrical load the there's additional resistance as a result of the EMF in the motor.  When the circuit is open there no current flow, not electron flow and no EMF to place resist the spinning of the generator.  So the energy loss of energy is to friction/heat.

Re-read my post carefully. I basically said this at the end of my reply to your second question:

“If the generator is unloaded, very little heat and steam are required to turn the generator. (What is required are part of the losses from turning one form of energy into another; that is, heat into mechanical energy into electrical energy.)”

Part of the mechanical energy losses would be friction, obviously.

The point I was making is that it requires orders of magnitude more energy to spin a generator under full load than it does to spin one under no load. If a generator under full load suddenly has a load removed, it will obviously speed up slightly until the control loop responds and slows it down again. Generally there’s a system in place that can respond in the hundreds of millisecond time range, that will divert the steam from the turbine and dump it somewhere in case of a full load removal. So, that’s where the energy goes!

Question I've been asking is where does the energy go in those 100 milliseconds.  The time between when the load is removed and the time the generator slows to a no load condition.  There's energy there that needs to be accounted for.  And let's say the load on the generator is quite large say tens of thousands of watts.  And in the same circuit is a 3 watt indicator light.  Seems to me during those 100 milliseconds the generator is still going to be producing tens of thousands of watts of energy with a load of just 3 watts.  That extra energy has to be accounted for somewhere.

I think a big part of the confusion here is the imprecise meaning of words. When you say “Where goes the energy/power go?” a lot of us are assuming you mean *electrical* power.

What you are *really* asking is, where does the energy from the boiler/steam turbines go in that 100 milliseconds. As SeanB stated, it goes into speeding up the generator. These generators are very, very, large and it takes a lot of energy and (on the order of) seconds for them to speed up by a significant amount. More than enough time for the control loop to respond *or* emergency systems to activate and shut things down.

If that fails, the mechanical systems speed until they explode.

As for solar panels, that’s a completely different thing. Simply put, without a load to allow current flow, no energy is produced.

Obviously the sun produces a lot of energy. Where does it go when those photos strike anything else? It’s converted into heat energy. (Plants being the exception, as they can convert it into chemical energy.)

[Jr460]

What you are *really* asking is, where does the energy from the boiler/steam turbines go in that 100 milliseconds. As SeanB stated, it goes into speeding up the generator. These generators are very, very, large and it takes a lot of energy and (on the order of) seconds for them to speed up by a significant amount. More than enough time for the control loop to respond *or* emergency systems to activate and shut things down.

If that fails, the mechanical systems speed until they explode.

The turbine speed changes maybe a small amount over a very short time and the steam value (very high power hydraulic system) quickly move driven by the control system to speed in check.  This in turns cuts the flow rate of steam.  The steam is kept at constant temp and pressure so another way to measure the load on a unit is the main steam flow rate.  Flow goes down and that will cause the pressure and temp to start to go up slightly.  The boiler controls will then back off on the feed water pumps and fuel feed.

If you go from full load of say 500MW to half that or nothing in a second, the turbine will still not over speed, but the boiler pressure goes up faster than it can control.  High main steam pressure causes a unit trip.  Everything is shut off, all values in the steam and feed water path close, pumps stop.   Heat is still trapped in the system, so the air dampers on the inlet and outlet side of the boiler go to full open, and the forced draft and induced draft fans go to full speed to remove heat from the boiler.  The condenser water pumps and valves go full open to remove any heat trapped in the condensate and lower parts of the feed water system to the cooling tower or river water.  In most cases this doesn't happen as quickly as one would like and the boiler pressure continues to rise as the thermal mass of a 14 story high boiler of steel tubes imparts more energy into the steam/water trapped in it.

It doesn't take too long and safety pressure relief valve(s) on the roof of the boiler building pop open with a loud roar that shakes the building and sends a plume of steam high into the air.

Once things calm down, depending on the valve type either it closes and again, or someone cranks on a large handle to reset it, and the operations staff goes to restart the unit after fixing the reason for the unit trip in the first places.

If something explodes, then the all the safety systems didn't work as designed.

[jmelson]

I've had the opportunity to study a few GE power house alternators in a repair shop and before installation.  Very interesting stuff.

[snip]

Cheers!

So, that solid iron rotor will have pretty good eddy currents induced in it by harmonics, so that should count for something.

Tim

Well, that's an interesting thought.  But, with 10KA flowing in the rotor winding, it is probably driven close to saturation, at least in the middle of the coil.  So, I don't know what the harmonics do to the rotor.

But, supposedly, the power station guys will tell you, years ago, you could not tell whether the alternator was on the grid or not while standing next to it.  Now, they will tell you you can feel the harmonic effect vibrating the floor when it is on line.

I didn't quite know what to listen/feel for, so I didn't detect it.

Jon

[T3sl4co1l]

Steel is quite lossy due to hysteresis, which will be reduced near saturation; but eddy currents remain, and in such a thick section, that will do.

Tim

[John Heath]

Where has all the power gone?
Long time passing
Where has all the power gone?
Long time ago
Where has all the power gone?
Computer nerds used it all
When will they ever learn?
When will they ever learn?

 

Nicely done. I would be remiss if not kicking the can again.

Where has all the energy gone.
long time passing
up in smoke all of it
long time ago

Hot fire and cold house
long time passing?
Warm fire warm house
Is where time goes
Warm fire warm house
Is where energy goes

With enthlapy in mind how do you charge a cell phone if you are on the surface of the sun? Everything is hot on the sun so there is no temperature difference for energy to charge a cell phone. How to solve this problem?

Maybe a parabolic dish facing the cold black sky. This should cause the focal point of the parabolic dish to be colder than the sun's surface. Now the cell phone can be charged. Would this work ??
   

[DougSpindler]

Where has all the power gone?
Long time passing
Where has all the power gone?
Long time ago
Where has all the power gone?
Computer nerds used it all
When will they ever learn?
When will they ever learn?

 

Nicely done. I would be remiss if not kicking the can again.

Where has all the energy gone.
long time passing
up in smoke all of it
long time ago

Hot fire and cold house
long time passing?
Warm fire warm house
Is where time goes
Warm fire warm house
Is where energy goes

With enthlapy in mind how do you charge a cell phone if you are on the surface of the sun? Everything is hot on the sun so there is no temperature difference for energy to charge a cell phone. How to solve this problem?

Maybe a parabolic dish facing the cold black sky. This should cause the focal point of the parabolic dish to be colder than the sun's surface. Now the cell phone can be charged. Would this work ??
 

Ummm couple of corrections.     
We can’t break the laws of physics here.   Energy is not gone, (can’t disappear) butgets transformed.
Sun does have a temperature gradient.  So yes it will be possible to charge your phone.....  But would your phone work?  Don’t think so with all of the charged highly particles and the stron magnetic fields I think the chips in your phone would be fried.