power rating

[Heidi]

Dear friends,

It reads in my text - "Assume that a generator is rated at 600V, 120 kVA. This means that it is capable of supplying I =
120 kVA/600V
= 200A."

I have a few questions about this statement:
#1 Does it mean that the generator can provide an open-circuit sinusoidal voltage such as v(t) = 600√2 cos(2∏ft)? If it does, why don't we need to specify the value of frequency?

#2 Is '120kVA' obtained under the assumption that the gnerator will generate 120kVA/600V=200A current from a voltage source of 600V? But since we don't know what kind of load is going to be connected to the generator, how can we be sure what current value it will supply?

Thank you!

 

[alec_t]

1) Yes it does. Frequency should be specified, since it is significant for inductive or capacitive loads.
2) If the source is AC the 600V is assumed to be the rms value. The 200A is the rms current for the rated power irrespective of load type.

 

[Heidi]

I feel even more confused looking at the generator.jpg below.

'When the generator supplying a load with P=120kW and Q=160kVAR, the generator current is 333.3A.' said in my text.

Since we don't know what the frequency of the exciteation and the inductance of the inductor are, how can we compute the current through the inductor with a 600V across it?

 

[alec_t]

how can we compute the current through the inductor?

Inductor current = 160kVA/600 = 266A

 

[johansen]

1)its generally assumed that the frequency is the same as the grid in the country you're located in.
2) a generator's "rating" is normally its nominal rating at which it can operate continuously at rated temperature rise (40C above ambient perhaps)

also, keep in mind that 333 amps at 600 volts probably means 192 amps per phase. 333/root(3)

 

[Heidi]

Inductor current = 160kVA/600 = 266A

Is the resistor current = 120kW/600V = 200A? But then 200A + 266A = 466A ≠ 333.3A?

The 200A is the rms current for the rated power irrespective of load type.

Does that mean the generator's rated power(120kVA) will not change no matter what load is connected to the generator and 600V voltage is still across that load?

The basic problem for me is that I totally don't understand what ' a generator is rated at 600V, 120 kVA' is supposed to mean. Could you please clarify it for me first?

 

[johansen]

Is the resistor current = 120kW/600V = 200A? But then 200A + 266A = 466A ≠ 333.3A?

Does that mean the generator's rated power(120kVA) will not change no matter what load is connected to the generator and 600V voltage is still across that load?

The basic problem for me is that I totally don't understand what ' a generator is rated at 600V, 120 kVA' is supposed to mean. Could you please clarify it for me first?

the resistor is consuming 120KW of non reactive load.
the inductor is consuming another 160KVAR of inductive reactance.

so the total KVA as seen by the generator is sqrt 120^2 plus 160^2 or 200KVAR.
only 120 of it is real but the generator still has 333 amps flowing through it.
if the generator is rated for 120KW then the copper is facing 2.7 times the nominal power loss, and so it takes probably 30 minutes to catch on fire.

 

[Ramussons]

The basic problem for me is that I totally don't understand what ' a generator is rated at 600V, 120 kVA' is supposed to mean. Could you please clarify it for me first?

It means that the Terminal Voltage of the generator is 600V RMS.

The Current it can supply is 120000/600 = 200 A RMS

It does NOT mention anything about the Power Factor or the Phase Difference between the Voltage and Current.

In other words, a Voltmeter across the output will show 600 V and an AC Ammeter at the output can read up to 200 A before the generator starts getting Overloaded.

As johansen mentioned, the Frequency is assumed to be that of that country, Normally 50 Hz or 60 Hz.

Ramesh

 

[Heidi]

Thanks friends. But I think I must have missed something. Please allow me to restart from the beginning and please check where I go wrong.

The scenario is this: Assume a generator is rated at 600V, 120kVA. The generator is supplying a load which consists of a resistor and inductor in parallel with P=120kW and Q=160kVAR
I want to know why the generator has a current of 333.3A.

I tried a different way than the one in textbook, however, step by step trying to figure out where exactly I was going wrong.
First I tried to find out the inductance of the inductor from Q=160kVAR, assuming that the frequency of line voltage is 50Hz. (My original idea is that if I could get the max currents through the resistor and the inductor, then the max current through the generator = sqrt (I_R^2 + I_L^2) )

From the definition of a inductor's reactive power Q=VI where V and I are the RMS voltage across and RMS current through the inductor respectively. Since V=600V, I=160000/600=266.67A. According to V/I=X_L=ωL, ω=2∏f=100∏, so L=V/(ωI)=7.162mH

Next from P=120kW=VI to find the resistor's resistance. I=120000/600=200A, so R=V/I=600/200=3Ω.

But once I get the values of the resistance and indutance, I put them into PSpice to stimulate the max current in the generator, it goes from 592A to -346A sinusoidally, where did I go wrong?

 

[alec_t]

You don't need to know the peak current or frequency or R or L.
Resistor rms current = P/V = 200A.
Inductor rms current = Q/V = 266A.
Now draw a vector diagram triangle with one side 200 units long and a 266 units long side at right angles to it. What is the length of the hypotenuse (which represents the total current supplied by the generator)?

 

[Heidi]

Yes, sqrt(200^2 +266^2) = RMS current through the generator. But if I want to stimulate the circuit, I need to know the frequency, R, and L.

I sued 600*(square root of 2) sin (2*∏*50*t) as the voltage source, R and L as evaluted in my last post, why didn't PSpice show the expected generator max current of 333.3A? Where did I go wrong?

 

[MrAl]

Thanks friends. But I think I must have missed something. Please allow me to restart from the beginning and please check where I go wrong.

The scenario is this: Assume a generator is rated at 600V, 120kVA. The generator is supplying a load which consists of a resistor and inductor in parallel with P=120kW and Q=160kVAR View attachment 73677
I want to know why the generator has a current of 333.3A.

I tried a different way than the one in textbook, however, step by step trying to figure out where exactly I was going wrong.
First I tried to find out the inductance of the inductor from Q=160kVAR, assuming that the frequency of line voltage is 50Hz. (My original idea is that if I could get the max currents through the resistor and the inductor, then the max current through the generator = sqrt (I_R^2 + I_L^2) )

From the definition of a inductor's reactive power Q=VI where V and I are the RMS voltage across and RMS current through the inductor respectively. Since V=600V, I=160000/600=266.67A. According to V/I=X_L=ωL, ω=2∏f=100∏, so L=V/(ωI)=7.162mH

Next from P=120kW=VI to find the resistor's resistance. I=120000/600=200A, so R=V/I=600/200=3Ω.

But once I get the values of the resistance and indutance, I put them into PSpice to stimulate the max current in the generator, it goes from 592A to -346A sinusoidally, where did I go wrong?
View attachment 73679

Hello,

You seem to have the inductor value L and resistor value R calculated correctly. Now all you need to do is calculate the total current.

The total current is the square root of the sum of squares of the resistor current and the inductor current (one is real and the other is considered imaginary because it is always 90 degrees out of phase with the other):

Current in resistor: iR=E/R=600/3=200 amps
Current in inductor: iL=E/(w*L)=600/(2*pi*50*0.007162)=266.666 amps

Now take the square root of the sum of squares of those two currents:
I=sqrt(200^2+266.666^2) = sqrt(111110.755556) = 333.333 amps approximate

Once you know the real current and the imaginary currents you use the 'norm' or 'abs' or 'resulting vector length' which is sqrt(Real^2+Imag^2).

The exercise shows the generator burning up because the current exceeds the max of the generator. The max current is the max the generator can handle and that is 120000/600= 200 amps. That should be the total current we calculated above as "I", but ours was much higher so the generator burns up.

As you can see this is an important aspect of generators and transformers...things that are made of wire...because the total current even though out of phase with the mains voltage still generates real power in the wire itself that the device is made up of. Wires have their own rating that can not be exceeded, and that rating will be reflected in the rating of the generator. Another way of looking at it is that at some point the losses in the wires in the generator become too great so it starts to deteriorate. So when we see a VA rating we know that it is really a rating of the wires internal to the device, but we dont have to bother to think about all that, just make sure we dont exceed the stamped VA rating.

 

[Heidi]

Thanks, MrAl.

Now I would like to go one step further. Are there any ways we can use in PSpice to stimulate the whole generator-with-load situation to get the total current of 333.3A? I tried but I failed.

First, is this circuit fine to get started?

 

[MrAl]

Hi again Heidi,

As you know from past experience with inductors and resistors in circuits, you sometimes have to solve for the initial current in the inductor. In this circuit with the small 1u Ohm series resistor, the time constant is something like 7162 seconds which means it takes over 35000 seconds to reach the AC steady state response. That's almost 10 hours. If you want to do a simulation that is that lengthy, you can do that, but you'll have to wait for at least 35000 seconds before you see the current waveform center itself around zero when the exponential part of the response goes away. It takes that long. If you use a bigger resistor larger than 1u Ohm you'll see this happen faster, but it probably would take a 1m Ohm resistor to see this happen within say 50 cycles, but at least then you would see it START to stabilize. Of course 1m Ohm would interfere with the circuit so you dont want to keep that value except for this test if you care to do it.

So see if you can calculate the initial inductor current and use that in the simulation. Once you get the right value, you'll see the waveform centered vertically about zero. You can even use trial and error if you feel like fooling with it. Just keep changing the initial current in an organized way until you start to see the more expected results. You may have to use a negative value for the initial current.

Also, if you are saying that 600v is the peak that is different than if it is 600v rms im sure you know. But if you want to work in rms values then just use 600 for the line voltage and then read the graphical peaks as "volts rms" instead of "volts peak" because everything scales accordingly. If you dont feel comfortable with this though you dont have to do it that way i just thought i would mention it because it often makes reading the simulation output graphs a tiny bit easier.

 

[JoeJester]

Heidi,

Do something like this ....

 

[ericgibbs]

Thanks, MrAl.

Now I would like to go one step further. Are there any ways we can use in PSpice to stimulate the whole generator-with-load situation to get the total current of 333.3A? I tried but I failed.

hi heidi,

This is the SIM in LTSpice, showing the required run time for the inductor current to stabilise, as pointed out by MrAl.

I have allowed 30seconds.

E.

 

[MrAl]

Hello there Eric,

Thanks for posting a simulation. I have to ask though, what is the internal series resistance for the voltage source you used (it may have a high default value if you did not set it). I ask because the analytical time constant is 7162 seconds which means the current envelope should reach about 37 percent of it's initial value in 7162 seconds, yet your simulation has reached that point in well under 50 seconds.

This could happen for at least two or three reasons:
1. The internal resistance of the source is higher than 1 microohm.
2. The minimum step time for the simulation is greater than 1ms (very approximate).
3. Both of the above.

You could check this out and see what happened if you like.

The analytical exponential offset coefficient is:
e^(-t*R1*R2/(L*(R1+R2)))

and so the time constant is L*(R1+R2)/(R1*R2)

and this is where R1 is the series resistance (stated to be 1u ohm)
and R2 is the 3 ohm resistor.

Check that out and see if you can get a similar result. I'll post the current i found using analytical techniques and graphed. In my graph shown below, the peak is shown as an RMS value, so when we see 600 amps that's means 600 amps RMS not 600 amps peak.

 

[JoeJester]

I set Ri to 1u for the generator, and cheated the time by using essentially a cosine wave by shifting the sine wave -90 degrees. The graph was adjusted by changing the timeline to start at the generator's zero crossing. That is why it starts at 5mS and ends at 45mS

 

[RCinFLA]

That is why generators and transformers are spec'd in KVA not KW. It is the current that is the dominate determining factor in their output capability.

 

[Heidi]

Hello, MrAl, I should have thanked you earlier. It took me quite a while thinking, doing calculations and even typing! I've tried the 'hard way' to find the complete solution to the generator current equation through which I realized why I had failed the simulation.

In this circuit with the small 1u Ohm series resistor, the time constant is something like 7162 seconds which means it takes over 35000 seconds to reach the AC steady state response. That's almost 10 hours.

First I changed the voltage source V1 and the resistor R1 into the Norton's equivalent (fig1), the 'unsteady' inductor current would be iL(t) = Ke^-(Rt/L), where L=7.162mH, R=R1*R2/(R1+R2)=10^(-6)Ω is the equivalent resistance of the two parallel resistors R1, R2, K is a constant, so the time constant would be L/R=7162 seconds, five times the time constant is nearky 10 hours, just as you predicted!

Next (fig0) I tried to find the 'steady' AC generator phasor current in frequency domain I=Vs/Z, where Vs = 600(-90°) is the generator phasor voltage (or vS(t)=600sin(wt) ) and Z = R1 + (R2)*(jwL)/(R2+jwL) = 1.8(53.13°)Ω is input impedance of the circuit, the result is I = 333.33(-143.13°) or i(t)=333.33cos(wt-143.13°), the same as evaluated by the 'complex power triangle' which I had a hard time understanding.

If you use a bigger resistor larger than 1u Ohm you'll see this happen faster

As you suggested, if I change the resistor R1 to 1m, time constant = L/R = L/(R1*R2/(R1+R2)) = 7.162 seconds, decay time 5*7.162=35.81 seconds is much acceptable!

With R1=1m, R2=3, L=7.162m, vS(t)=600sin(2*pi*50*t), step size=1m
(fig2)During 0~40 seconds, the generator current is gradually steady.
(fig3)During 0~1 second, the generator current was jumping around -100~600 amps RMS.
(fig4)During 36~37 seconds, it jumped around -329~332 amps RMS.
(fig5)During the last cycle, 39.98~40 seconds, the current centered at 0, with amplitude 332 amps RMS approximately.

But if you want to work in rms values then just use 600 for the line voltage and then read the graphical peaks as "volts rms" instead of "volts peak" because everything scales accordingly. If you dont feel comfortable with this though you dont have to do it that way i just thought i would mention it because it often makes reading the simulation output graphs a tiny bit easier.

I didn't notice that. You are very thoughtful. Thank you!

I also want to thank RCinFLA, JoeJester, ericgibbs, alec_t, Ramussons and johansen for all your help and your precious time!

I have one more question.

The description "The generator is supplying a load which consists of a resistor and inductor in parallel with P=120kW and Q=160kVAR" is strange enough to me. Don't we have to know the inductor's RMS voltage and the RMS current before we can evaluate its reactive power Q? In real life, P or Q can be known 'before' knowing corresponding voltage and current?