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Engineering 3 phase circuit - real power

  • Thread starter orla22
  • Start date
22
0
If the line current is 240V, 400Hz, 3 phase generator supplies a balanced 3 phase star connected load through a long 3 core cable. load phase is a 0.25 ohms resistor in series with a 130 (nano)H inductor. 3 core cable has negligable resistance and a line resistance of 0.1 ohms.

How do hand calculate the real power.

I can work it out but cant get the right answer can someone help....
I know the formula is P=V.I.Cos(ange)

i know to find angle:
Z=R+jX
X=XL-XC
i get X to equal 3.267e-4 ohms.
therefore Z = 0.25 + j3.267e-4
so the angle is 0.0749

i know how to find current:
V=IZ
115/0.25+j3.267e-4
I=460

so real power = 115 x 460 x cos (0.0749) = 52900 W but the answer is 59113 W
 

berkeman

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5,515
If the line current is 240V, 400Hz, 3 phase generator supplies a balanced 3 phase star connected load through a long 3 core cable. load phase is a 0.25 ohms resistor in series with a 130 (nano)H inductor. 3 core cable has negligable resistance and a line resistance of 0.1 ohms.

How do hand calculate the real power.

I can work it out but cant get the right answer can someone help....
I know the formula is P=V.I.Cos(ange)

i know to find angle:
Z=R+jX
X=XL-XC
i get X to equal 3.267e-4 ohms.
therefore Z = 0.25 + j3.267e-4
so the angle is 0.0749

i know how to find current:
V=IZ
115/0.25+j3.267e-4
I=460

so real power = 115 x 460 x cos (0.0749) = 52900 W but the answer is 59113 W
Welcome to the PF. Where did the 115 come from? The problem says the voltage is 240Vrms. And remember that 240Vrms is the source voltage -- you are dropping some voltage across the supply cable, which doesn't make it to the load...
 
22
0
sorry yes your right 240/root 3 = 138v then you say i have to take voltage off this because of internal resistance which is 0.1 ohms? how would i do that?
 
sorry yes your right 240/root 3 = 138v then you say i have to take voltage off this because of internal resistance which is 0.1 ohms? how would i do that?
check your PMs. I sent you a question regarding this problem.
 

berkeman

Mentor
55,339
5,515
check your PMs. I sent you a question regarding this problem.
Turns out to be a good side question. We're working on it. Thanks.
 

berkeman

Mentor
55,339
5,515
sorry yes your right 240/root 3 = 138v then you say i have to take voltage off this because of internal resistance which is 0.1 ohms? how would i do that?
Well, normally 0.1 Ohms would be pretty negligible as cable resistance. But with the load being only 0.25 Ohms itself, the 0.1 Ohms of cable/source resistance is non-negligible. So if they are asking about power delivered to the load, you will have a voltage divider between the cable and the load resistance (or impedance). You need to include the cable resistance in your calculations of the voltages and currents, in order to get an accurate power delivery calculation....
 
22
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is this the circuit that represents the question above? the 16.144Kw being the total real power ( the power excluding any internal resistance?)
 

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berkeman

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is this the circuit that represents the question above? the 16.144Kw being the total real power ( the power excluding any internal resistance?)
I don't see the 0.1 Ohms of line resistance in that simulation...
 
22
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where do i put that then?
 

berkeman

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I don't see the 0.1 Ohms of line resistance in that simulation...
You tell us. Describe what each thing is in your simulation schematic, and how that corresponds to the things in the original problem statement. Then tell us where would be a logical place to put the cable resistance, in order to model it and the whole system accurately. Then please also show us how the answer changes when you include the cable resistance... How big of a change is it?
 
22
0
is this the circuit that represents the question above? the 16.144Kw being the total real power ( the power excluding any internal resistance?) notice i did say "the power excluding any internal resistance?" i .e. without the 0.1 ohm resistance
 

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