Where does the power go ?

[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.