electricity generation

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Shandre

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when generating electricity at a PF of 1 a certain amount of steam is required.
My question is as follows. With that same amount of steam, generating at the same fixed voltage and the same current flow but now at a different phase angle I produce less Megawatts.Is this statement correct?
Does mechanical input produce MVA or Mw's ?
 
Shandre said:
when generating electricity at a PF of 1 a certain amount of steam is required.
My question is as follows. With that same amount of steam, generating at the same fixed voltage and the same current flow but now at a different phase angle I produce less Megawatts.Is this statement correct?
Does mechanical input produce MVA or Mw's ?
Click to expand...

Yes, the real power in MW falls as the current lead or lag increases, and the steam power decreases proportionally--in an ideal system.

In general, real and apparent power are related as shown below,

Pr (MW) = Pa (MVA) x Power factor (cos(lead or lag))

The problem with low PF is the copper loss resulting from increased current required to deliver the real power through the distribution system.
 
Power factor (PF) is a function of the load, not the source. A given amount of steam produces a given amount of power, the electrical version of which is wattage, which is volts x amps (assuming unity PF).

A load with a poor power factor causes the current to peak (rise, peak, and fall, actually) before or after the voltage does during each half-cycle. (Let's call this poor 'efficiency' for this simple discussion.)

Because of that, a bit less of the electrical energy is converted into work, yet the electrical system must carry enough voltage and current (at offset times) to still produce the desired power (at one time).

Normally, amps x volts = watts, but if a bit more current (at a given voltage) is needed to supply a given amount of power delivered to the load, the 'excess' current must be generated, but doesn't get used by the load.

See here for more: http://www.faqs.org/docs/electric/AC/AC_11.html
 
mdshunk said:
Is there a formula for pounds per hour of steam at some pressure to convert to watts?
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It could be calculated. We know the energy required to produce a pound of steam from a pound of water plus the energy to provide superheated steam. We would also need to know the efficiency of the turbine plus all the conversion factors. In my student days, I had to take a course called, "Heat Power" otherwise known as thermodynamics, but all that steam has rusted my brain, so don't ask anything specific.
 
If you really want to the math - go ahead.....

But might I suggest a little history with the homework:
http://www.history.rochester.edu/steam/marshall/
http://en.wikipedia.org/wiki/Watt
http://en.wikipedia.org/wiki/Horsepower#PS
http://en.wikipedia.org/wiki/James_Prescott_Joule
http://en.wikipedia.org/wiki/Joule
http://en.wikipedia.org/wiki/Joule-Thomson_effect

And OT - in relation to refrigeration PBS recently (last week) had some slight mention of the Joule Thomson effect:
http://www.pbs.org/wgbh/nova/zero/ (Fascinating show....)

I mention Joule and Watt as they share a lot in common - Breweries and steam.
 
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Isn't there an adjustement on the exciter current or voltage that changes the amount of VARs that the generator can supply?
Don
 
Shandre said:
With that same amount of steam, generating at the same fixed voltage and the same current flow but now at a different phase angle I produce less Megawatts. Is this statement correct?
Click to expand...
No. You are trying to hold too many things constant.

Looking only at the ?same fixed voltage and the same current,? those two will imply that the MVA is constant. If you can alter something within the system, and have as a result that the power factor is now less than one, it will mean that the MW produced by the generator will be lower. That part of your statement is correct. But I don?t think you can do this and also say that the amount of steam needed to drive the turbine has also remained constant.

Shandre said:
Does mechanical input produce MVA or Mw's ?
Click to expand...
The ?real power? rating of a machine, the MW rating, is based on the capacity of the prime mover (in this case, the mechanical input power of steam driving a turbine). The ?apparent power? rating, the MVA rating, is based on the ability of the generator to reject heat from its windings.
 
This is a good question and something to think about.

Let's compare two theoretical situations:

Situation 1: A theoretically perfect generator is applying 100 volts to a theoretically perfect 10 ohm resistive circuit. Power factor is 100%. All the energy is converted to watts in the resistive circuit.

I x E x Pf = 10 amps x 100 volts x 1 = 1000 watts of power

Considering the perfect generator, it requires 1000/746 equals 1.34 horsepower to turn it to supply the load.

Situation 2: Same as above, except supplying a theoretically perfect capacitive circuit with 10 ohms capacitive reactance. Power factor is 0. None of the energy is converted to watts.

I x E x Pf = 10 amps x 100 volts x 0 = 0 watts

Now, what about the energy required to turn the generator? Does the generator turn freely without the addition of any horsepower? Or would it still take 1.34 hp to turn it?
 
If you really mean "ideal," in the sense of having no resistance in the wire, no friction internal to the spinning generator, no inductive or capacitive coupling between any conductor in the system and the outside world, all of that sort of stuff, then yes, the generator will spin freely, without any additional input of mechanical power. Keep in mind that it will take some horsepower to get the generator turning in the first place, and to allow it to start sending current down the line. But once you have equilibrium established, all that will happen thereafter is that there will be an exchange of energy between the electric field of the capacitor and the magnetic field of the generator windings. That can continue forever, even if you somehow are able to unhook the generator from its prime mover (e.g., the diesel engine or the steam turbine).
 
Charlie b, thank you for the input.

Some other thoughts:

I assume that during the half-cycle where the capacitor is charging, it takes input into the generator shaft to charge the capacitor. Then, during the half-cycle when the capacitor is discharging back through the generator windings, the generator would actually be acting like a motor, and imparting enough momentum to the generator to rotate through the next half-cycle to charge the capacitor?

Or is the above "way out there" and "crazy talk"?
 
Unless they've changed the laws of physics recently, energy can be neither created nor destroyed. It can however, be changed from one form to another. Rotating motion to electrical (generator). Electrical to rotating motion (motor).

I believe Charlies example to be true, as no energy is being changed to anything, except electrical; back and forth. None is lost to heat or anything else.

I have a question though. In the above example, when the gen is disconnected from the prime mover, would its speed change (to reach capacitive-inductive equilibrium), or remain constant. I suspect it would remain constant, as the rotor contains stored energy (in the form of inertia), and cannot change unless acted upon by an outside force.

Then again, I could be way off! Interesting mental exercise.
 
Say the generator is producing DC voltage. Horsepower must be applied to the generator shaft to charge the capacitor, which stores the energy.

The same thing must be true of the AC generator during the portion of the cycle where the capacitor is charging. It does take energy to charge the capacitor. So, something must be making the shaft rotate to charge the capacitor. Is it the inertia left over from the generator acting like a motor when the capacitor discharges back through the windings?
 
It is not the generator shaft that gives and receives energy. It is the generator windings. Energy is stored in the windings, in the form of a magnetic field. The intensity of that field goes up and down in a sinusoidal pattern. At the same time, energy that is stored in the capacitor (in the form of an electric field) is going down and up in a sinusoidal pattern. If there are no losses, there will be no need to replace those losses by adding mechanical power to the shaft.
 
So basically the generator doesn't even need to spin once the capacitor is charged.... the inductance of the windings and the capacitance will be a tank circuit which circulates the energy?
 
charlie b said:
But once you have equilibrium established, all that will happen thereafter is that there will be an exchange of energy between the electric field of the capacitor and the magnetic field of the generator windings.
Click to expand...
And to add, in order for this to happen, the conductors between the generator and the capacitors will still have to carry this unproductive charging/discharging current, so also must be of zero impedance.

In the real world, we would want the system to be useful, so any current that actually is used at the load end would be in addition to the unproductive current described above; the system must carry both.

When correction means (capacitors for inductive loads) are added to compensate for loads with poor power factors, the unproductive current only has to traverse the conductors between the caps and the load.

If the compensation is exactly matched to the load, the service and supply conductors only have to be sized to carry the real power. Power companies are in the same boat, and add a surcharge to low-PF customers.

A watt-hour meter is basically an electric motor whose speed depends on the strength of both the voltage and the current. It measures volts and amps (VA, apparent power) but it reads out in watts (real power) over time.

If the load is purely resistive, the voltage and current peaks occur simultaneously, there is no phase shift and the PF is 1. If they don't occur at the same time, the motor spins slower, and isn't as accurate.

That's because the system must still be capable of carrying both the productive and unproductive currents, yet can only make use of the latter. Volts is volts, and amps is amps, but they ain't always watts.
 
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