Inductor current = 160kVA/600 = 266Ahow can we compute the current through the inductor?
Is the resistor current = 120kW/600V = 200A? But then 200A + 266A = 466A ≠ 333.3A?Inductor current = 160kVA/600 = 266A
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 200A is the rms current for the rated power irrespective of load type.
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 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?
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
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.
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 use a bigger resistor larger than 1u Ohm you'll see this happen faster
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.
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