#1
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:

(Blurb about National Grid wiring)

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
#2
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.
#3
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.
#4
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Dude screw the exam, sell all this hash you're making.

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#5
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:

(Blurb about National Grid wiring)

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.
#6
Well aside from the power and current and voltage you don't get anything
Last edited by philipisabeast at Apr 6, 2008,
#7
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.
#8
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,
#9
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.
#11
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.
#12
Ah, electromagnetism always was the worst part of physics.
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#13
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
#15
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