/sci/ - Science & Math

[Anonymous No.4849443]

(Advanced) Physics problem?
An electric power plant generates electricity @ 22 kV
and 100 A. The voltage is stepped up to 240 kV before
being transmitted to a city 15 km away over a copper
wire that has a radius of 2 mm. Energy is lost in heating
the wire during transmission. If the transformer is 95%
efficient, and the price of electricity is 15¢/kWh, How
much money is saved per day by stepping the voltage up
instead of transmitting it at the original voltage?

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I understand most of the equations, but exactly how would you solve this and in what manner? If you could also state the given value for each variable beforehand, that would be great too. (e.g. I/current=?, etc.) Thanks in advance.

[Anonymous No.4849493]

forever arone. bump

[Anonymous No.4849502]

http://www.youtube.com/watch?v=UEaKX9YYHiQ

[Anonymous No.4849537]

Wouldn't you need 2 copper wires and 2 transformers?

[Anonymous No.4849591]

>>4849537
http://www.youtube.com/watch?v=31g0YE61PLQ

[Anonymous No.4849614]

I hope the following method may give good results.

The out put current Is in the transformer secondary is given by ..

Vs Is = Vp Ip x 0.95
Is = 22 x 100 x 0.95 / 240 = 8.708 A

The power loss during transmission without step up and with step up are ...
P1 = Ip^2 R
P2 = Is^2 R
R is the resistance of the copper wire of length 2l = L = 30x10^3 m
Specific resistance of copper = pi d^2 R /4 L , where d = 4x10-3 m
R =
So if I know the specific resistance I get the value of R
(P1 - P2) gives the difference in power loss in watts.
This can be converted into Kwh and hence the saving
___________________
Sp.resistance of copper = 1.7x120^-4 ohm-m
R = 4.058x10^5 ohms
P1 = 405.8 x10^7 watts
P2 = 3.078x10^7 watts
P1 - P2 = 402.7 x10^7 watts
Difference in Power consumed per day
= (P1 - P2)x10^-3 x3600 x 24 Kw
Difference in the no. of units of power consumed is
(P1 - P2)x10^-3 x 24 / 3600 Kwh = 9.665 x10^7 Kwh
This multiplied by the currency gives the money saved

[Anonymous No.4849620]

>>4849614
holy shit thanks bro.

[Anonymous No.4849625]

>>4849614
give this man praise bump

[Anonymous No.4849753]

100A over 15km of 2mm wires?
Someone must be kidding here.

A=πr^2=3.14mm^2
J=I/A=100A/3.14mm^2=31.8A/mm^2

Way too high. This will not work.

Rloop=2*rho*l/A=2*0.018(Ohm*mm^2/m)*30e3m/3.14mm^2=344Ω

Ishort=22KV/344Ω=64A

Ploadmax=U^2/4Rloop=350KW (another 350KW heat the wires)

22KV*100A=2.2MW
No way.

240KV: Ploadmax=42MW
May work, still lossy.

Suggest 2cm wire diameter :)

[Anonymous No.4849810]

Error. Radius is 2mm, not diameter..

Rloop=86Ω
Ploadmax=1.4MW
Still bad.
240V:
Ploadmax=167MW
This works.