wut, remember, 1 million watts in an MW and 1000 watts in a kW
Energy loss in superconducting wire
= 3kW x 100
= 300000 W
= 100 MW
therefore efficiency of C is 99.7% while B is 98.4%, therefore C is the best. The question didnt ask for price only calculations
dont read the question, picked superconductor cuz is da lastest tech yo, i mean they say they need to use da superconductor for da future transmission, then this question ask this BULLsht.i really cbfd reading what everyone wrote so im just gonna say how i did it (so sorry if someone else has said it)
the question was a bit sketchy, cos it said for design B that 40ohms was the 'total resistance', but ohms is resistance per metre. so i did 40 times 100km (which is 4000000 metres or something) and got a current of 0.127A (using R = V/I) or something similar which i thought made sense cos it was high voltage and low current.
and power loss for the design A was 20MW (you could get that from the difference in output from the power station and input from the substation in the question). and design C used 3x10^6 watts in cooling the conductor for the whole thing.
so i just did the power loss equation with 0.127^2 x 4000000 (I^2.R) and i got like 65000 watts or something. someone please tell me they did the same thing hahah
but yeh i got design B as the most efficient
...lol key word - futuredont read the question, picked superconductor cuz is da lastest tech yo, i mean they say they need to use da superconductor for da future transmission, then this question ask this BULLsht.
If I remember correctly, the source was 508kV, and the input was 500kV with resistance at 40 ohms, so thats a loss of 8kV.Can someone please explain the design B's power loss? I got something really low so I'm guessing it's probably wrong. Had no idea what I was doing there...
Whoops... think I subbed in P=IR^2 instead...If I remember correctly, the source was 508kV, and the input was 500kV with resistance at 40 ohms, so thats a loss of 8kV.
use V=IR to find I, so 8x10^3/40, I=200A
then sub I into P=I^2R to find power loss which was 1.6MW
I Think you'll lose a mark for incorrectly using the equation, and if you didn't say the best one was B, then thats probably another mark gone. Im just guessing btwWhoops... think I subbed in P=IR^2 instead...
any ideas if its just the 1 mark lost for that if I had the equations right but plugged it into the calc wrong?