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50 Hz High-Voltage Transmission Line

  • Engineering
  • Thread starter Joe85
  • Start date
  • #1
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Homework Statement:

Figure shows a 50 Hz, high-voltage, transmission line. The relationships between the sending and receiving end voltages and currents are given by the complex ABCD equations:

Homework Equations:

.
VS=VR(A1+jA2)+IR(B1+jB2)

IS=VR(C1+jC2)+IR(D1+jD2)

Given the parameter values in TABLE C and an open-circuit received voltage measured as 88.9 kV, calculate the values of VS and IS and hence the power (PSO) absorbed from the supply by the transmission line on open circuit.

VR= 88.9×103

Table C Values:
A1 = 0.8698
A2 = 0.03542
B1 = 47.94Ω
B2 = 180.8Ω
C1 = 0 S
C2 = 0.001349 S
D1 = 0.8698
D2 = 0.03542

Transmission Line is Open Circuit, thus:

IR = 0


Hi all,

I seem to be coming up with the correct answer but i'm getting a negative value for PSO. Just wanted to be sure i'm not making an error with my numbers.

IR = 0

VS=VR(A1+jA2)+IR(B1+jB2)
∴ VS=VR(A1+jA2)
∴ VS= 88.9×103(0.8698 +J 0.03542)
VS = 77325.22 + J3148.838 Volts

And:

IS=VR(C1+jC2)+IR(D1+jD2)
∴ IS=VR(C1+jC2)
∴ IS= 88.9×103(0 + J0.001349)
IS= J119.9261 Amps

PSO = Re{VSIS*}
∴ PSO = Re{(77325.22 + J3148.838)(J119.9261)
= J9273312.066 + J2377627.8609
= (-1)*377627.8609 + J9273312.066
= -377627.8609 Watts

I've seen that other answers using a similar method are positive, does this mean i disregard it being a negative value as it's a 'Total Power absorbed' so must a positive quantity?

Any help much appreciated.

Joe.
 

Answers and Replies

  • #2
gneill
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I note that you didn't incorporate the received voltage in your calculation of the power lost. Are you really supposing that the entire source voltage is being "lost" along the trip to the receiving end?

Secondly, when you calculated the power you indicated that the complex conjugate of the current should be used. I didn't see that happen in the next line.
 
  • #3
gneill
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I've seen that other answers using a similar method are positive, does this mean i disregard it being a negative value as it's a 'Total Power absorbed' so must a positive quantity?
No, you can't ignore the sign. Power absorbed implies a positive value absorbed (how you adapt your analysis method to arrive at that value is up to you).

In your analysis you didn't take the complex conjugate of the current when you solved for the (complex) power. I think you'll find that that will account for your sign discrepancy.
 
  • #4
421
118
CONJUGATE(IS)=-J119.9261

PSO = Re{√3(77325.22 + J3148.838)(-J119.9261)
 
  • #5
29
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I note that you didn't incorporate the received voltage in your calculation of the power lost. Are you really supposing that the entire source voltage is being "lost" along the trip to the receiving end?

Secondly, when you calculated the power you indicated that the complex conjugate of the current should be used. I didn't see that happen in the next line.

Doh, sorry Gneill. That's what i get for copy and pasting!

PS= Re{VSIS*}

I didn't get as PSO (which i assume will be the result of PR - PS? S & R being Sending and receiving?)

And would PR = Re{VRIS*} ?
 
  • #6
gneill
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You're looking for the power absorbed by the transmission line, so you want to work with the voltage dropped across the transmission line. Think of it like a very long resistor. You want the power absorbed by that resistor, so you don't just take the potential at one end.

Not that it's critical here, but note that long transmission lines can have not only resistance, but inductance and capacitance too. So don't be surprised if the open-ended received voltage is actually higher than the transmitted voltage! Think about what can happen with a resonant circuit.
 
  • #7
29
3
You're looking for the power absorbed by the transmission line, so you want to work with the voltage dropped across the transmission line. Think of it like a very long resistor. You want the power absorbed by that resistor, so you don't just take the potential at one end.

Not that it's critical here, but note that long transmission lines can have not only resistance, but inductance and capacitance too. So don't be surprised if the open-ended received voltage is actually higher than the transmitted voltage! Think about what can happen with a resonant circuit.
There’s so little in the learning materials to go on here.

isn’t A1+jA2 the reverse voltage gain. So

(A1+jA2)2/ ( VS2/P) would equal the power absorbed?
 
  • #8
gneill
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isn’t A1+jA2 the reverse voltage gain. So

(A1+jA2)2/ ( VS2/P) would equal the power absorbed?
What's P in that equation?
 
  • #9
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Sorry it was PS from post 5. Or 378KW.
 
  • #10
gneill
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Let's take a look at the units in your equation. The A's are unitless, Vs would be volts and P is VA. So:

##\frac{1^2}{\frac{V^2}{V A}} = \frac{A}{V} = \frac{1}{\Omega}##

That's not a power.

Consider this. The voltage change across the transmission line is:

##\Delta V = V_S - V_R##

and in this case since ##I_R = 0## you can write ##V_S = A V_R## where ##A = A_1 + j A_2##. Then:

##\Delta V = \left( A - 1 \right) V_r##

The complex power lost becomes:

##p = \Delta V I_s^* = \left( A - 1 \right) V_r I_s^*##
 
  • #11
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Thanks gneill, i'm eternally grateful thus far, but why cant we just carry out the subtraction for VS-VR since we know what both VS and VR are? And then insert this into the power absorbed equation?
 
  • #12
gneill
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Thanks gneill, i'm eternally grateful thus far, but why cant we just carry out the subtraction for VS-VR since we knoiw what both VS and VR are? And then insert this into the power absorbed equation?
You certainly can! That's how I originally did it myself when I worked your problem through. I just thought that your idea of using the A term was interesting and sought to show how it might work.

The thing you were missing in your original attempt was that you used the entire sending voltage in your power loss calculation. Not all of the sending voltage is dropped on the transmission line. It's the end to end ##\Delta V## across the line that represents the power lost.
 
  • #13
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So having quickly whacked the figures into Symbolab Calc on my phone, the PSO = 377627.86 Watts or 378KW. So the answer was correct but the method wrong, hence what you were saying about the entire source being lost?
 
  • #14
gneill
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So having quickly whacked the figures into Symbolab Calc on my phone, the PSO = 377627.86 Watts or 378KW. So the answer was correct but the method wrong, hence what you were saying about the entire source being lost?
Yup.
 
  • #15
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Yup.
Thank you gneill. You’ve been a real gent.
 
  • #16
gneill
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I'm happy that I could help!
 

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