Okay, this is a question that came up on my exam and I made a hash of it and can't really figure out where I need to go to get the answer; so if anyone would be happy to help, here's the Q:

i) Each cable operates at 400kv and the current is 100A
Find power: 4x10^7 W

iii) An acceptable power loss is 1.4kW per Km in a cable of current 100A.

Show that the conductance G of a 1.0kM length of cable must be greater than about 7 Siemens so that it does not lose more power than 1.4KW per kM

That last bit is what loses me; I know the equations for working out G but i'm not sure how to apply them to this case
Was this on the AS Exam in January? Cos if it is you need to do something like combine 2 formulas. Don't ask which ones cos I didn't really follow it.
You can make hash out of questions?

How much could you make out of that one?

Or that one?

Dude screw the exam, sell all this hash you're making.
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Dude screw the exam, sell all this hash you're making.

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Quote by philipisabeast
Okay, this is a question that came up on my exam and I made a hash of it and can't really figure out where I need to go to get the answer; so if anyone would be happy to help, here's the Q:

i) Each cable operates at 400kv and the current is 100A
Find power: 4x10^7 W

iii) An acceptable power loss is 1.4kW per Km in a cable of current 100A.

Show that the conductance G of a 1.0kM length of cable must be greater than about 7 Siemens so that it does not lose more power than 1.4KW per kM

That last bit is what loses me; I know the equations for working out G but i'm not sure how to apply them to this case

Did not give some info about the cables? I remember the question but I can't remember how to do it.
Well aside from the power and current and voltage you don't get anything
Last edited by philipisabeast at Apr 6, 2008,
Quote by philipisabeast

That last bit is what loses me; I know the equations for working out G but i'm not sure how to apply them to this case

Its been awhile since ive done this, if you remind me what the equation for G is i might be able to help.
G=(Conductivity x Area)/ L

G=1/R

G=I/V

Are equations for it, although neither directly apply, and using the second one with the derived R from V/I doesn't give 7S
Last edited by philipisabeast at Apr 6, 2008,
There must be some infomation missing.....you sure it doesnt give a factor which how much of the total power is lost in the cable? like "10% of the overall power of the cable is lost as heat" sort of thing?

Because it feels like we need to relate the power, to the power loss.
Well you get total power, and then the acceptable powerloss
On a slightly different note; does any one know a formula to get Percentage error? My coursework is due tomorrow and I need this formula to finish it.
Ah, electromagnetism always was the worst part of physics.
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SYK messaged me the answer which is a paragon of badass of him, if anyone wants it just let me know and ill copy it here

% Error = (Precision of instrument/ Measurement) x100
Power (acceptable loss) = 1.4 kW
Current = 100 A

P = IV
(transpose)
V = P/I
V = 1400 W / 100 A
V = 14 W/A = 14 V

V = IR
(transpose)
R = V/I
R = 14 V / 100 A
R = 0.14 V/A = 0.14 ohms

G = 1 / R
G = 1 / 0.14 ohms = 7.12 Siemens