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transmission question (1 Viewer)

v1

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u only get 1 mark for stating B is the most efficient... the other 5 marks come from evaluating each design in terms of physics and calculations
 

EinstenICEBERG

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doesnt make sense, that SUPER CONDUCTOR transmission lines would waste more energy than design B. the logic. SUPER conductor is the new shizz.
 

traiwit

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doesnt make sense, that SUPER CONDUCTOR transmission lines would waste more energy than design B. the logic. SUPER conductor is the new shizz.
But cooling system ..,
Btw doesnt question ask to do calculation to support the answer ?
 

cookeemonstah

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wut, remember, 1 million watts in an MW and 1000 watts in a kW

Energy loss in superconducting wire
= 3kW x 100
= 300kW
= 300000 W

Energy input
= 100 MW
= 100000000 W

Efficiency of C
= (1 - (300000/100000000)) x 100%
= 99.7%

therefore efficiency of C is 99.7% while B is 98.4%, therefore C is the best. The question didnt ask for theory only calculations.
 
Last edited:

mickstarify

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wut, remember, 1 million watts in an MW and 1000 watts in a kW

Energy loss in superconducting wire
= 3kW x 100
= 300kW
= 300000 W

Energy input
= 100 MW
= 100000000

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


It was 30kW lol
 

markGee

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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
 

EinstenICEBERG

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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
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.
 

markGee

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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.
...lol key word - future
the question is implying that we cant use superconductors yet cos its impractical as it wastes high amounts of power in cooling it
 

nucgaek

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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...
 

mickstarify

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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...
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
 

nucgaek

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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
Whoops... 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?
 

mickstarify

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Whoops... 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?
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 btw
 

someth1ng

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Yeah, you must end your response with a short paragraph (judgement) on which is best and which is the worst. Without it, you would get 1/6.
 

Poops

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Did you have to calculate everything using the values given? O___O

I just talked about the formulas P: I^2xR and V=IR ...
 

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