Calculating Real Power in a 3 Phase Circuit

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Discussion Overview

The discussion revolves around calculating the real power in a balanced three-phase circuit with a star-connected load, specifically focusing on the effects of line resistance and voltage drop. Participants are exploring the application of formulas and the implications of circuit parameters on power calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for real power using the formula P=V.I.Cos(angle) but arrives at a different answer than expected, prompting requests for assistance.
  • Another participant questions the source voltage used in the calculations, pointing out that the voltage drop across the supply cable must be considered.
  • A participant acknowledges the need to adjust the voltage due to internal resistance and seeks clarification on how to account for this in their calculations.
  • Discussion includes the significance of the cable resistance relative to the load resistance, suggesting that it cannot be ignored in this context.
  • Participants discuss the need to model the circuit accurately, including the placement of line resistance in simulations to reflect the real power delivered to the load.
  • There is a query about whether a specific circuit simulation accurately represents the original problem, particularly regarding the exclusion of internal resistance in power calculations.

Areas of Agreement / Disagreement

Participants generally agree on the importance of considering line resistance in power calculations, but there is no consensus on the specific approach to incorporate it or the implications for the calculated power values.

Contextual Notes

Participants express uncertainty regarding the correct voltage to use in calculations and how to account for the internal resistance of the cable. There are unresolved questions about the impact of these factors on the overall power calculation.

orla22
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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 can't 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
 
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orla22 said:
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 can't 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...
 
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?
 
orla22 said:
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.
 
xcvxcvvc said:
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.
 
orla22 said:
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...
 
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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orla22 said:
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...
 
where do i put that then?
 
  • #10
berkeman said:
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?
 
  • #11
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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