Calculating power absorbed in transmission lines

Click For Summary

Discussion Overview

The discussion revolves around calculating the power absorbed by a transmission line on open circuit, with a focus on the parameters provided and the appropriate equations to use. Participants explore the complexities involved in the problem, including the implications of open-circuit conditions and the characteristics of transmission lines.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The original poster (OP) presents parameters and equations but expresses uncertainty about how to start solving the problem.
  • One participant questions the clarity of the parameters provided (A1, A2, etc.) and suggests that the OP consult their instructor for further guidance.
  • The OP later clarifies that they have an open-circuit received voltage and provides additional formulas related to the transmission line, indicating a potential misunderstanding about the role of current in this scenario.
  • Another participant identifies the parameters as the ABCD 2-port parameters for the transmission line and suggests that if the OP's calculated values for voltage and current are correct, computing the power should be straightforward.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to approach the problem, with some expressing confusion over the parameters and others suggesting that the calculations could be straightforward if the initial values are correct. The discussion remains unresolved regarding the correct method to calculate the power absorbed.

Contextual Notes

There are limitations due to missing definitions for the parameters A1, A2, etc., and the OP's uncertainty about the application of the formulas provided. The discussion also highlights the potential need for additional context regarding the behavior of transmission lines in open-circuit conditions.

Help_required
Messages
3
Reaction score
0

Homework Statement


Calculate the power (Pso) absorbed from the supply by the transmission line on open circuit.

Given parameters are:

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

I have calculated Vs and Is earlier in the question

Vs= 77325 + j3149V
Is= j119.93A

Homework Equations

The Attempt at a Solution



I don't know where to start with this one to be honest and haven't got any real direction from my course notes. Can i use equation P=I^2*R? I suspect it's more complicated than that!
 

Attachments

Physics news on Phys.org
Ok, you clearly have not given enough here to solve the problem. What do these constants A1 etc. refer to? They don't appear on your diagram.

A transmission line can act like a big antenna. That means it has inductance and capacitance. And it can radiate power, and have current flowing even though there is no completed circuit. If you have not covered these things in class, and your text does not describe how to deal with it, you need to go to your instructor and get help.

There should be a formula that tells you how to calculate this. It will include such things as the frequency of the applied voltage.
 
Thanks DEvans,

Sorry, i should have included this information on the original post.

I have been given the open-circuit received voltage of 88.9kV and the only formulas i have for this question are:

Vs= Vr (A1+jA2) + Ir (B1+jB2)
Is= Vr (C1 + jC2) +Ir (D1 + jD2)

'S' and 'R' are for sending end and receiving end. These are what i used to calculate Vs and Is earlier in the question, although i used zero as 'Ir' due to it being open circuit (thinking no current could flow) but from what you've said above i may need to look at those answers again.

With regards A1 etc. i have no further information, this is all that is given on the assignment. Looks like i'll have to contact the tutor for this module.
 
I believe this problem has appeared earlier on PF homework help.

@DEvens, these are the ABCD 2-port parameters for the xmsn line.

If the OP got the right Vs and Is (post 1) then computing the power is straightforward.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
6
Views
6K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 28 ·
Replies
28
Views
14K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K