Where does the power 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.