Ok, not exactly physics, but I am stuck.

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?

Well I don't know what the pu is, but if it is a factor that when multiplied by the input voltage is what should be the least voltage dropped across the load resistance and you have a power dissipation from the load as 800MW, then the current is 1600 A. The power factor being 1 is a good thing as it simplifies things greatly. Typically the real power across a load can be given by just V*I*cos(theta) where cos(theta) is your power factor. If the voltage and current are 90 degrees out of phase with each other then no real power gets dissipated, but rather all of it is stored and released by reactive elements. So if the 0.9pu is some factor times the input voltage that is the minimum voltage that should be dropped across the load resistor then the answer should just be:

500kV*1600A*0.9*1 = 720 MW

Again, this answer is only correct if the 0.9pu is a factor times the input voltage that is the minimum voltage that should be dropped across the load resistor and that your value of 800 MW is correct for the load power dissipation.

The reason why the maximum power transfer function appears to be independent of the power factor is because it is. That function assumes that your load and source resistances are real and matched for maximum real power transfer, so there is no phase difference between the voltage and current because everything is real and there are no reactive elements that can displace the current in phase with respect to the voltage signal.

If you only know the load resistance value and not that of the source then you don't know if the two are matched, and hence shouldn't assume so using that function, as it does assume the load and source resistances are matched for a maximum power transfer to occur.

That's the best I can do for you without knowing more about the problem, like the value of the load resistance, the input power, the driving current, the input frequency, what pu is, basically anything that you might know but didn't include in the description of the question; but, if the pu is just a multiplication factor and the current flow is 1600 A, the power factor being 1, then it should just be the 500kV*1600A*0.9pu*1 = 72 MW.

Come to think of it, if you say the load power is 800MW then you have your answer, which leads me to believe you meant the load is 800MOhms. If so we'll do the same thing also using the 0.9pu as some kind of factor times the input voltage that should be the minimum voltage across the load, so:

Power = (V^2/R)*0.9*1 = 281.25 Watts, which makes sense because the load resistance is huge and so the current flow is only 500kV/800MOhms = 625 micro-amps. So that should give you the same answer as VIcos(theta) = 500kV*625 micro-amps*0.9*1 = 281.25 Watts, not that it proves that answer correct as the current is derived from the voltage and resistance (which I'm guessing you meant to be 800MOhms), and I still don't know what the pu is, but if it is just a factor times the input voltage that should be the load voltage and the resistance of the load is 800MOhms, with a power factor of 1, means it should be 281.25 Watts. If any of these assumptions are wrong, so is that answer.

Lastly, if the load resistance is 800MOhms, then there is a very good chance the source resistance is not, as a source ideally has no resistance, but typical function generators or signal generators have an internal resistance of about 50 Ohms, so the two are definitely not matched and so the maximum power transfer function will not work for you.

Many Smiles,
Craig

Craig

That's an awesome response considering I haven't told you everything and you don't even know what P.U. is. YOu are right about it being a factor. It's something that power engineers use in their analysis of circuits.
Thanks for putting that effort in, and I really appreciate it.
It was alittle bit helpful, but I will keep you posted as to how I do on this question if I get some more out.

regards
David.