Maximize Efficiency with Transmission Line Help - 99.5%, 577.4A, 692.88V

[aliserwan]

Homework Statement

A smaller factory uses 400 kW of power from the electrical grid. The total resistance of the pair of transmission lines carrying electricity to the factory is 1.2 Ω. The heat produced by the pair of transmission lines due to the motion of charges constitutes a loss in the process. The loss is 2 kW.
1-What is the efficiency of electricity transmission?
2- What current is carried by the supplying transmission line?
3- What voltage are the transmission lines connected to?

Relevant Equations

P=RI^2
V=RI or P/I

a) 99.5%
B) 577.4 A
c) 692.88 V

 

[haruspex]

Homework Statement:: A smaller factory uses 400 kW of power from the electrical grid. The total resistance of the pair of transmission lines carrying electricity to the factory is 1.2 Ω. The heat produced by the pair of transmission lines due to the motion of charges constitutes a loss in the process. The loss is 2 kW.
1-What is the efficiency of electricity transmission?
2- What current is carried by the supplying transmission line?
3- What voltage are the transmission lines connected to?
Relevant Equations:: P=RI^2
V=RI or P/I

a) 99.5%
B) 577.4 A
c) 692.88 V

B and C are wrong. Please post your working.
Hint: typical voltage for such would be many kV.

 

[aliserwan]

P=R*I^2 so I=(400000/1.2)^0.5=577.4 A
V=R*I so V= 577.4*1.2 =692.88 V

 

[haruspex]

P=R*I^2 so I=(400000/1.2)^0.5=577.4 A
V=R*I so V= 577.4*1.2 =692.88 V

You are confusing the power consumed by the factory with that lost in the wires.

 

[aliserwan]

standard unit ? wat/aohm=Amber
v kvolt=kw/amber
??

 

[aliserwan]

You are confusing the power consumed by the factory with that lost in the wires.

if i calculate:
I=(40/1.2)^0.5=18.25
v=I*R=18.25*1.2=21.9
but what about Units ? watt or Kw >> v or kv

 

[ghaith7710]

efficiency=p-ploss /p =(400-2)/400 = 0.995
Ploss = I^2 * R then I = 40.8 A
P = V * I then V= 9.8 KV

 

[haruspex]

efficiency=p-ploss /p =(400-2)/400 = 0.995
Ploss = I^2 * R then I = 40.8 A
P = V * I then V= 9.8 KV

This is a homework thread.
Please read the forum guidelines. We provide hints, point out errors, but do not provide solutions.

 

[haruspex]

if i calculate:
I=(40/1.2)^0.5=18.25
v=I*R=18.25*1.2=21.9
but what about Units ? watt or Kw >> v or kv

You have not understood my post. All you have done there is lose the "k" in kW.

We do not know the resistance of the load in the factory. We know the resistance of the wires, and we know the power lost in the wires, so what is the current through the wires?

 

[bhobba]

Just a note in my capacity as moderator. I just want to reinforce what has been said - hints are provided - not answers. Just as a further hint - we are using KW ie thousands of watts. When calculating remember results will be in thousands as well. Now you know the resistance and power loss hence can calculate the current in thousands of amps. Knowing the resistance and amps, ohms law gives the voltage in thousands of volts across the wires, knowing the current in the device its connected to and power consumed you can calculate the voltage across it. Add them up.

Thanks
Bill