Previous in Forum: Fluid Couplings and VVVFs   Next in Forum: resonance
Close
Close
Close
Rate Comments: Nested
Participant

Join Date: May 2009
Posts: 1

Real Power Transferred to Receiving End

03/25/2010 12:52 AM

Hi

I have a question that I am having trouble doing. I will tell you how I have attempted it to help you know what I'm thinking. Don't solve it, just help me understand what I have to do or what it's asking. Thanks

It is part d) to a lossless long transmission line question and goes "what is the maximum amount of real power that can be transferred to the load at unity power factor if the required load (receiving end) voltage should always be greater than 0.9 p.u. For this case assume |Vs| = 500 kV.

First I thought about the theoretical maximum real power transfer equation:
Prmax = (VrVs/X`) - (AVr^2/x`)cos(θB - θA) but this seems independent of power factor and so I'm not sure if this is the right way to go considering they have specified the unity power factor in the question.

From an earlier part I am assuming that the |Vr| = 500 kV x 0.9. Also that the load is 800MW.

So then I used the equation Pr = (VsVr/B) sin(δ) to find the load angle.
then I plugged it into this equation, Pr = (VsVr/B)cos(θB - δ) - (AVr^2/B)cos(θB - θA) and the second term will become zero because it's a lossless line. And then I just go around in circle and find that it's 800MW, which is what I started with.

I'm missing some fundamental understanding here and how the 0.9 pu fits into this.

Can anybody help me?

David.

Register to Reply
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Anonymous Poster
#1

Re: Real Power Transferred to Receiving End

03/25/2010 12:39 PM

This is just a bit too complicated .

Loss less Power Line means R=0, Though Z≠ 0 (since there is a drop in the voltage). Though this is supposed to be capacitive (as per theory), but here it is inductive.- Vrec = 0.9Vsup

Upto now OK, but then the problem comes.

The maximum power transfer equation for DC says the

Rth = RL

ie the source thevenin equivalent resistance equals the load resistance.

For AC however it is conjugate for maximum power transfer

ie Load impedance should be conjugate of source reactance.

Source being pure inductive (Loss less so R=0), Load should be purely capacitive - then the tubned circuit gives totally undefined power at load (with I→ ∞ )

So the maximum power transfer equation does not work.

The only way is to make an equivalent circuit with R and L,

Find the current as a function of R, L.

Put the constraint of 0.9V at the Load point.

You will get an equation

Use this current to get the power I2R

Use calculus (maxima/minima)

see what happens.

Register to Reply
Register to Reply

Previous in Forum: Fluid Couplings and VVVFs   Next in Forum: resonance

Advertisement