High voltage input question?

Percentage of input power available...Urgent!?

• 5 points to best answer. I need help with iv, but you have to use the figures from previous questions to answer it. Is it okay to post this in this section? Hmm, I wasn't sure. Its a simple question really. The input voltage to each cable is 400kV. The cable carries a current of 40A. Calculate i) the input power. P=IV= 1.8x10^8 ii) number of cable required to transmit the power from a 2000MW power station= 11.36..well we cant have 11.36 bits of cable to does it become 11 or 12? iii)the power lost per km of cable( is that the same as output power?)P=I^2r= 1.0x 10^4 ( where r is given as 0.052 ohms) iv) the percentage of the input power that is available from and distance of 100km...O.o what? how would could I a) calculate the efficiency out of curiosity...and b) the percentage of input power available as asked. Please explain in simple words and show the calculations involved.

• Answer:

  Assuming 3 phase supply. At a voltage of 400k (line voltage VL) and a current of 40 amp (line current IL) then power drawn would, in watts be :- Sq Rt 3 x VL x IL x power factor of load, = 1.732 x 400,000 x 40 x Pf.(power factor) which you don't mention. VA would be same but without Pf. Divide both by 1000 for kW and kVA resp. The number of cables required would depend on size of cable used and distance (length). 1 cable may be able to carry 40 amp, but voltage drop, (mV/amp/metre), may mean that the voltage at the far end of cable is too low. You may need to use a much larger cable, or two smaller cables Power lost in the cable is it's AC resistance (Impedance) over it's total length x the current carried. As said it depends on size of cable used. (Total resistance) Power lost from generation due to usage is dependant on the power factor of the consumer. If a consumer has a low power factor, then the generator has to generate more current than that used. This requires more fuel input to prime mover of generator, which is not being charged for. Sorry I have to go now. Back again. When you were told insufficient information was given, 'Billruss' was right. You either haven't posted all information, or your lecturer hasn't given you sufficient. Is the question about 3 phase, single phase or DC? If the resistance of the cable is 0.052 ohms per kM. then power lost is current squared x 0.052/kM. If so then the resistance over 100kM would be 5.2 ohms. Percentage Power available, is power available / power generated x 100