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Why does the frequency increase when generation exceeds load?(2)

il102 (Electrical) (OP) 
14 Mar 12 11:09 
I could rephrase the same question, why does the frequency decrease when load exceeds generation?
I know these are fundamental operating characteristics of a power system network but I'm not sure I can mathematically prove it. My best guess is that everything dissipates more power when the frequency increases, or less power when the frequency decreases. So if you have an imbalance of generation/load, the change in frequency effectively forces the load to dissipate more power or consume less power, right?
Which equation explicitly shows this? The AC instantaneous current formula?
I'm familiar with frequency control schemes, etc, I just want to make sure I understand the mathematical concepts behind it.
Thanks 

dpc (Electrical) 
14 Mar 12 12:42 
Why does you car engine speed up when you push in the clutch pedal? 

il102 (Electrical) (OP) 
14 Mar 12 13:06 

Simply put, the energy put into a closed system must equal the energy taken out of the closed system. Or Pin=Pout
Frequency is not the only thing that can change when there is an imbalance. And by imbalance I would include stored energy in parts of the system like fly wheel effect. The stored energy is why the system takes time to change. 

DTR2011 (Electrical) 
14 Mar 12 13:55 

il102 (Electrical) (OP) 
14 Mar 12 14:02 
Still haven't really answered my original question explicitly. I'm looking for a mathematical proof of what is happening. 

Is this a homework problem? 

il102 (Electrical) (OP) 
14 Mar 12 14:09 
No, I just want to mathematically understand the effect of power imbalance on the grid?
Everyone I ask tells me the frequency goes up or down and there are control systems in place to adjust the generation as needed. That's all well and good but I want to understand why exactly the frequency goes up? Is the desire to actually understand the science behind concepts so rare as opposed to just accepting a subjective answer at face value? 

DTR2011 (Electrical) 
14 Mar 12 14:50 
There were actually good answers supplied already. The frequency increases because the generators are huge (large mass) and take time to settle down after a sudden loss of load. There is effectively a gas pedal (governor or steam valve) to supply fuel, but nothing equivalent to a break pedal. Assuming you are talking about a generator connected to a utility grid, the math may not be so easy as there are many different scenarios of load and generation to consider. From a mechanical standpoint, consider the math in the following: http://en.wikipedia.org/wiki/Flywheel#Physics I would consider this link as a good starting point, as well as the Pin=Pout statement listed above. BTW a clutch is the third pedal in a stick shift car. "A clutch is a mechanical device that provides for the transmission of power (and therefore usually motion) from one component (the driving member) to another (the driven member). The opposite component of the clutch is the brake.: http://en.wikipedia.org/wiki/Clutch 

Check out the ACE equation. It interrelates generation, load, and system frequency. I think this is what you're looking for, or at least should be a good starting point. The formula for calculation of ACE follows: ACE = (NIA  NIS)  10b (FA  FS) Tob + IME Where, •NIA represents actual net interchange (MWs) •NIS represents scheduled net interchange (MWs) •b represents the control area's frequency bias setting (MW/0.1 Hz) •FA represents actual system frequency (Hz) •FS represents scheduled system frequency (60.0 Hz in America) •Tob represents scheduled interchange energy used to bilaterally correct inadvertent accumulations (MWs) •IME represents a manually entered amount to compensate for known equipment error (MWs) http://everything2.com/title/Area+Control+Error 

Find a good power system stability text. Lots of equations there.
Consider a generator and load in balance. You have P_{M}=P_{E}+P_{loss}. Now, make a step reduction in P_{E}. There can be no corresponding step reduction in P_{M} or step increase in P_{loss}. So, either certain conservation laws get violated or there has to be another term in the original equation giving P_{M}=P_{E}+P_{loss}+P_{accel}. During steady state operation P_{accel}=0. On a step decrease in P_{E} there is a equal step increase in P_{accel} (and vice versa). Given time to react, the governor system reduces P_{M} enough to drive P_{accel} negative and then brings it back up to regain the original balance. 

il102 (Electrical) (OP) 
14 Mar 12 15:03 
Thanks for the responses, that helps a bit.
I think I'm getting closer to understanding. Pin=Pout is really the sticking point of this issue. At steady state lets say Pin = 100MW and Pout = 100MW. Now 1MW of load is lost. Examining this transient period of time  Pin is still 100MW because your generator has not had time to react.
Pout is still 100MW because of the conservation of energy we just spoke of (Pin=Pout). So my question really was, where does this extra 1MW of power get dissipated during this transient period? Let's assume there is no energy storage or flywheels.
There are only two places I can see where this extra energy is dissipated. It would either be A) in the mechanical generating turbines themselves, or B) the overall connected load helps compensate, or C) some combination of A and B.
I was thinking  when the frequency increases, the inductive reactance of a connected load increases as well, right? So does the connected grid dissipate slightly more power at 61 Hz as opposed to 60 Hz. Or does all of the extra power get burned up as mechanical energy?
Do you get where I'm going with this, at least? 

Nope, P_{out} drops. You have to have a P_{internal} in the form of P_{accel} to absorb the lost P_{out}. 

il102 (Electrical) (OP) 
14 Mar 12 15:19 
K  if that's true I think that answers my question then. So the generating turbines ultimately compensate for system wide transient power imbalances by accelerating or decelerating therefore absorbing or producing excess power (within realistic limits until they could become damaged)? Since this is a mechanical speed change if causes the frequency change. 

ijl (Electrical) 
14 Mar 12 15:30 
Check any (elementary :) book on physics. The password is Newton. That is F = ma, or, because of rotational motion, T = J alpha, or alpha = T/J, where T is torque and J is the moment of inertia, and alpha is the angular acceleration. The torque is the total torque = torque of the prime mover minus the torque of the generator. One more hint: power = torque times angular velocity. You should now be able to work it out yourself (I hope :). 

As you incerase frequency system loss will increase, and so will current through resisitive load (so voltage also goes up).
Without a control system to dampen the system change, the system will come back into balance with Pin=Pout, where Pout will be increased losses, and load growth (load increases with frequency and voltage).
You will also increase energy stored in the rotating mass, as well as reactive energy. This storage of energy is why the system will not jump to a new steady state value, but will tend to ring. 

@cranky108 It's safe to say that resistive load (nonmotor load) doesn't depend on frequency deviation Without a control system the generator should trip, unless you want to see the rotor running around the power station if overspeeding or lose the blades on the turbine if running well below nominal frequency. May you grow up to be righteous, may you grow up to be true... 

I wasen't trying to guess at the generator type, or prime mover. I was going at this from a theory point of view. If the prime mover was say an otto cycle engine, it would have a greater range than some other types.
But you are correct that there maybe limits other than the electrical system.
As I said above, the frequency is not the only value that changes. 

waross (Electrical) 
15 Mar 12 20:26 
Many older systems and many contemporary systems use a proportional control system with about 3% proportional band. The speed and frequency would vary 3% from no load to full load and this variation from set point was used to control the energy supply to the prime mover. The speed and frequency would be 3% fast at no load and drop to the desired value at full load. Sudden load changes will cause a dip or surge in speed/frequency until the control system and prime mover respond. An isochronous control system will run at the set speed, but will respond to fast load changes as described above. If the control system fails to curtail the energy supplied to the prime mover in such a way that the energy supplied exceeds the energy demanded, the generator will accelerate until some factor balances the input energy and the output energy. This factor may be abrupt, dramatic, rapid and expensive. Bill  "Why not the best?" Jimmy Carter 



