Electricity in transmission lines (Simple V^2/R =P)

[bonbon22]

Homework Statement

The National Grid uses high-voltage transmission lines to carry electrical power
around the UK. A particular transmission line delivers 800 MW of power at
132 kV to the user. It loses 1% of the transmitted power as heat.
What is the resistance of the transmission line?
[1 mark]
A 0.2 Ω
B 6 Ω
C 20 Ω
D 2000 Ω

Relevant Equations

V^2/R = P

I understand how to get the answer but i don't understand how the answer 0.2 ohms is equal to the resistance of the transmission lines
If V^2/P = R then (132kV)^2/800MW = the resistance of the transmission lines, why is the 1% tansmitted as heat relevant at all as this is just one form of energy released from the resistance ?

 

[gneill]

Problem Statement: The National Grid uses high-voltage transmission lines to carry electrical power
around the UK. A particular transmission line delivers 800 MW of power at
132 kV to the user. It loses 1% of the transmitted power as heat.
What is the resistance of the transmission line?
[1 mark]
A 0.2 Ω
B 6 Ω
C 20 Ω
D 2000 Ω
Relevant Equations: V^2/R = P

If V^2/P = R then (132kV)^2/800MW = the resistance of the transmission lines, why is the 1% tansmitted as heat relevant at all as this is just one form of energy released from the resistance ?

The transmission line is not the load connected to the end of the transmission line. Clearly, if all 132 kV were dropped across the transmission line then no power would be available for the actual load.

If 800 MW is actually delivered to the load, then there must be both voltage and current involved. That current is what leads to ohmic losses in the resistance of the transmission line.

 

[sojsail]

why is the 1% tansmitted as heat relevant at all as this is just one form of energy released from the resistance ?

Remember that the transmission lines deliver 800 MW of power to the user. What is lost as heat is lost. The 800 MW delivered is 99% of the power that the generating plant put into the transmission line. Another relevant equation for power is P = V*I. The power lost as heat is I^2*R.

 

[bonbon22]

Remember that the transmission lines deliver 800 MW of power to the user. What is lost as heat is lost. The 800 MW delivered is 99% of the power that the generating plant put into the transmission line. Another relevant equation for power is P = V*I. The power lost as heat is I^2*R.

Remember that the transmission lines deliver 800 MW of power to the user. What is lost as heat is lost. The 800 MW delivered is 99% of the power that the generating plant put into the transmission line. Another relevant equation for power is P = V*I. The power lost as heat is I^2*R.

ok. I think i understand it a bit better, when doing the calculation for resistance i got a value of 21.78ohms what does this value represent the answer was 0.01 percent of this value.

 

[gneill]

ok. I think i understand it a bit better, when doing the calculation for resistance i got a value of 21.78ohms what does this value represent the answer was 0.01 percent of this value.

You'll have to show your work so that we can see what you've calculated.

 

[bonbon22]

You'll have to show your work so that we can see what you've calculated.

132kV ^2/ 800mw

You'll have to show your work so that we can see what you've calculated.

(132*10^3)^2/80010^6 = 21.78

 

[gneill]

(132*10^3)^2/80010^6 = 21.78

Once again you're taking the voltage drop on the transmission line to be the entire voltage delivered to the load. That is incorrect. Only a small fraction of the voltage produced by the generating station is "lost" traveling the transmission line.

 

[sojsail]

The power lost via P = I^2*R heating is just 1% of the total power. Now, if you only knew the current ...

 

[bonbon22]

The power lost via P = I^2*R heating is just 1% of the total power. Now, if you only knew the current ...

0.01 of (800 * 10^6) = 8*10^6
(132*10^3)^2/ 8*10^6 = 2178

Once again you're taking the voltage drop on the transmission line to be the entire voltage delivered to the load. That is incorrect. Only a small fraction of the voltage produced by the generating station is "lost" traveling the transmission line.

so i take 1 percent of both power and voltage , i got the right answer now cheers.

 

[sojsail]

I was trying to steer you to a method of solving that I thought would demonstrate a more clear way to look at it:

You could calculate the current using P = V*I.
And then find the resistance, using as data the above current and the power that is lost to heat (P_lost=1% of P) while that current flows thru the R.
This formula, P_lost = I^2*R, rearranged as R = P_lost / I^2, would then have given you the same result.

 

[bonbon22]

I was trying to steer you to a method of solving that I thought would demonstrate a more clear way to look at it:

You could calculate the current using P = V*I.
And then find the resistance, using as data the above current and the power that is lost to heat (P_lost=1% of P) while that current flows thru the R.
This formula, P_lost = I^2*R, rearranged as R = P_lost / I^2, would then have given you the same result.

sojsail i appreciate your help, slightly confused are you saying that only taking the power at 1 percent but not the voltage would give me the same result ? is it bad practice to use V^2 /R so its better to use I ^2 * R?
thanks again.

 

[gneill]

Consider the phrasing of the original problem statement:

A particular transmission line delivers 800 MW of power at
132 kV to the user. It loses 1% of the transmitted power as heat.

800 MW at 132 kV is delivered to the user. That's what ends up across the eventual load. If 1% of the generated power is being lost in transmission, the generating station must actually be generating 808 MW at its end of the line.

You know what the user is receiving. From that information you can determine the current in the load and hence the transmission line since they are in series.

The transmission line is dissipating ΔP = 8 MW of power. So $\Delta P = I^2 R_T$ and solve for $R_T$.

 

[bonbon22]

Consider the phrasing of the original problem statement:

800 MW at 132 kV is delivered to the user. That's what ends up across the eventual load. If 1% of the generated power is being lost in transmission, the generating station must actually be generating 808 MW at its end of the line.

You know what the user is receiving. From that information you can determine the current in the load and hence the transmission line since they are in series.

View attachment 243259

The transmission line is dissipating ΔP = 8 MW of power. So $\Delta P = I^2 R_T$ and solve for $R_T$.

ahhh gotcha makes perfect sense now , cheers m8