[Help] Power Loss in power cable

[Jamie Bristow]

A small factory has decided to go Carbon neutral and has installed its own wind turbine. The wind turbine is installed on land belonging to the factory 1 km away. The factory owner needs 2km of copper wire of diameter 3mm to connect the turbine to the factory. The resistance of the cable is 1.2 Ω.

a) What is the resistivity of the wire?

The turbine generates 12V dc electricity and needs to supply the factory with 1kW of power. The factory owner has the choice inverting the supply to 230V or transmitting it at 12V. By calculating:

b) The current that passes along the cables if they are to deliver 1kW of power to the factory

c) The power that would be lost in the connecting wires

d) The total power that must be generated

e) The efficiency of the transmission

For both voltages decide which is a more efficient mode of transmission?
For: Resistivity
p= RA/L
I got 4.242x10^-9 ohmmeters to 4 s.f.

not sure how to implement the Power loss Formulas:
P = IV = I^2R = V^2/R

Would I use V=IR to find current and Voltage drop and the add to the power loss formulas or use I=p/v for current and use the initial voltages?

How do I go about finding the total power generated to deliver the 1KW of power to the destination? would it be a case of simple addition or is there a formula for this?

 

[gneill]

In order to deliver 1 kW of power to the factory, the factory end of the cable must have a combination of voltage and current such that ##I V = 1kW##.

The thing is, for any given current there will be a voltage drop along the cable due to its resistance. In the case of the 12 V source the factory won't see that 12 V, it'll see 12 V less whatever drop occurred on the cables. See if you can set up an equation that reflects this. Write an expression that will give you the voltage ##V_{end}## at the factory end given that some current ##I## is being drawn. Find the current that will make ##I V_{end} = 1 kW##. (Hint: it may not be possible!)

 

[Jamie Bristow]

I was thinking this because this poor factory owner has built a windmill that generates 12 volts... surely this Imaginary businessman's not that stupid.
I tried V=IR and P=VI but any combination just results in silly numbers.
Thank you for the response by the way.

 

[Jamie Bristow]

For 12 volts I got

I=100012=83.3
V=2503(1.2)=100

So lost 88 more volts than started with. no use in continuing with 12v?For 230 Volts:
I=1000/230=4.347826087
V=4.35(1.2)=5.217391304
Power loss = 22.68431002W

dont know if i got this right?

also not sure about the formula for finding the corrected power input to get 1KW output.

 

[CWatters]

also not sure about the formula for finding the corrected power input to get 1KW output.

You haven't calculated the current correctly yet. You assumed 1000W was sent rather than 1000W was received. You will need to write two equations before you can solve them.

One equation is for the delivered Power at the factory end in terms of the voltage at the factory V_f and the current eg..

P_f = V_f * I

The other one you need is an equation for the voltage at the factory end V_f in terms of the source voltage V_s (=230V) and the voltage loss in the cable. I'll let you write that one.

Solve for I and V_f

 

[Jamie Bristow]

Power generated=? Power Received = 1KW
Voltage generated 12 or 230V Voltage received = ?
Current = ?
Resistance of wire= 1.2 ohms

So if I did V=IR and found the current along the line by using
I=12v/1.2ohms = 10
or
I=230v/1.2ohms = 191.6

then used the current for P=IV at the factory then
V=1000w/10=100 volts
or
V = 1000w/191.6 = 5.219 volts

so the voltage drop was
12 - 100 = -88
and
230 - 5.219 = 224.781

so the new p.d. at factory is either zero or 224v.

which makes some sense but if I add this back to P=IV or P=V²/R then I get silly numbers.

I can't see any other way of doing it

 

[gneill]

Here's a diagram we can work to:

Write KVL for the loop using ##V_g## and ##V_f## for the potentials at each end of the cable.

Consider how you might change or what you might do to the equation so that there will be a term for the factory power in it.

 

[Jamie Bristow]

So I'm not that great with KVL since I only just read about it.
Are you Implying that Vg(Pw)-Vf(Pf)=0?

also is that +- on the resistor correct?
I thought this was going to be simple.

 

[gneill]

So I'm not that great with KVL since I only just read about it.
Are you Implying that Vg(Pw)-Vf(Pf)=0?

No, I'm not implying that.

You should be able to write KVL for a loop if you've been given this problem as an assignment. I would think that Kirchhoff's laws would be one of the first things covered after Ohm's law.

KVL states that the sum of all potential changes around a closed path is zero (that is, if you set off from one spot in a circuit and "walk" around a path that ends up back where you started, then all the changes in electric potential that you crossed along the way will sum to zero and you'll be back at the same potential).

See if you can write that for this loop.

also is that +- on the resistor correct?

Yes, it represents the change in potential across the resistor due to the current flowing through it. Think Ohm's law.

 

[Jamie Bristow]

The course I am studying is mostly home study, we get one lesson a week on physics and one month to study for an exam. I have kind of neglected KVL since I know that most of the Exam is going to be using maxwells laws and drift velocities of elctrons and so on.
I can understand this is a simple task but I just wrote 27 pages on everything else to do with electricity and my brain is in overload right now.

Ok so KVL makes sense because you can't get something from nothing right? except our universe but that's off topic.

so if current starts at windmill with 0v and gain voltage, gets pushed anticlockwise and looses voltage toward factory then It must lose voltage as as it returns since the resistivity is constant through the wire?
Im just not getting something and its bugging me since I usually get this stuff really easily.

 

[gneill]

Going around the loop clockwise starting from the bottom left corner of the circuit:

Gain of Vg due to windmill generator
Drop across 1.2 Ohms
Drop of Vf across factory
Return to bottom left corner.

Write it as an expression.

 

[Jamie Bristow]

for 230V -

The voltage drop in the wire is 5.22V am i correct?

so starting from the bottom-left

gain of 230V
minus 5.22V across the first wire
minus 219.56V across the factory
and minus 5.22V across the second wire

 

[CWatters]

for 230V -

The voltage drop in the wire is 5.22V am i correct?

No that assumes you know what the current it. Don't substitute values yet. The voltage drop in the resistor is I*R.

I is unknown but don't worry about that for the moment just write the equation.

 

[Jamie Bristow]

Ok so V₀+V_g-V₁-V_p=V₀?

or I could do V²/R to find powerfrom windmill?

 

[Jamie Bristow]

I don't get it? I'll just Find another way of doing it.
I'm sorry you guys are probably frustrated with me, I have a learning disability and I get that a lot.

cant you just give me the order of the formulas I need to use?

because that is what is confusing me.

 

[gneill]

You need to be able to write a KVL equation using the given voltages, component values (resistors) and current variables. That's a basic necessity.

There's only one current (unknown, variable: ##I##)
There's one resistor (known value: R = 1.2 Ω)
One voltage source (known value(s): ##V_g## = 12 V or 230 V)
One load voltage (unknown, variable: ##V_f##)

There is one constraint given: Load power is to be 1kW.

Ok so V₀+V_g-V₁-V_p=V₀?

I don't recognize some of those variables from the diagram. You'll have to define what they are for the equation to make sense. Also, V₀, whatever that is, appears on both sides of the equation, so should vanish (cancel).

I'm making a guess that your V₁ is meant to represent the voltage drop across the cable resistance? If so, replace it with the voltage drop due to current through the resistor (Ohm's law).

or I could do V²/R to find powerfrom windmill?

No, that would be true if the factory end of the cable was shorted out (no load, the wire just loops back to the generator). In that case all the generator would be accomplishing is heating the cable!

Note that the power lost on the cable ##I^2 R## is not the power that the factory sees. The factory sees ##P_f = I V_f##.

Can you take another stab at writing the equation for the loop? I think you're getting very close.

 

[Jamie Bristow]

Ok so ill try this
V_g-V_d-V_p=0?
Just got a nice book out on KVL and KCL ill give this a read for an hour or so.

 

[gneill]

Ok so ill try this
V_g-V_d-V_p=0?

I don't know what ##V_d## or ##V_p## are: They haven't been defined. You'll want to use the current and resistor values to express potential drops when possible.

Just got a nice book out on KVL and KCL ill give this a read for an hour or so.

Okay.

 

[Jamie Bristow]

I don't know what ##V_d## or ##V_p## are: They haven't been defined. You'll want to use the current and resistor values to express potential drops when possible.

Okay.

Sorry, I meant f, not p

So: V_g - 1.2Ω - V_f = 0?

 

[gneill]

Sorry, I meant f, not p

So: V_g - 1.2Ω - V_f = 0?

Multiply the resistance by ##I## and you've got it. Ohm's law: ##V = I R##. ##I## is the (as yet unknown) current passing through the resistor, so the (as yet unknown) potential drop across the resistor is ##I~1.2 Ω##. All the terms of an equation must have the same units. In this case, volts. So the lone resistance value kind of stuck out.

Now we don't yet know what the factory voltage ##V_f## is, nor do we know the current ##I## that it draws. However! We do know that the factory wants to use 1 kW of power. Power is given by ##I V##. So how might you modify your KVL equation to incorporate that constraint? Here's the equation in its current form:

##V_g - I R - V_f = 0##

Note the ##V_f## variable sitting there. It's an unknown...now if only it were the load power instead...

 

[Jamie Bristow]

##V_g - I R - V_f = 0##

Ah of course. This is suprisingly difficult to improvise having never seen it be done before.

I'm tempted to say V_g - IR - IV_f = 0

with IV_f equalling 1kW

 

[gneill]

If you want to multiply one term in an equation by something, you must multiply every term by the same thing. So, what does that yield?

 

[Jamie Bristow]

If you want to multiply one term in an equation by something, you must multiply every term by the same thing. So, what does that yield?

IV_g - I²R - IV_f = 0

They're all looking a lot like power equations now

 

[gneill]

IV_g - I²R - IV_f = 0

Excellent. Note that each term represents a power, either generated or dissipated by one of the loads (the cable resistance counts as a load as far as the generator is concerned). You are given the desired factory load power, so you have a number to insert in place of that that ##I V_f## term. The only unknown left in the equation is the current ##I##. Solve for it.

The equation is quadratic in ##I## so you'll have the possibility of two solutions to choose from (imaginary solutions would rule out a physically realizable solution; note the results when you examine the solutions for the 12 V generator voltage).

What do you find?

 

[Jamie Bristow]

Oh my. I didn't anticipate this.

So arranging for the quadratic formula I get
-1.2(I)² + 12(I) - 1000 = 0
and
-1.2(I)² + 230(I) - 1000 = 0

for 12V, the discriminant was a negative root, so 12V is impossible?

for 230V, i got that I could equal 4.45...A, and 187.215...A

 

[gneill]

Oh my. I didn't anticipate this.

So arranging for the quadratic formula I get
-1.2I² + 12I - 1000 = 0
and
-1.2I² + 230I - 1000 = 0

for 12V, the discriminant was a negative root, so 12V is impossible?

for 230V, i got that I could equal 4.45...A, and 187.215...A

Huzzah! Yes.

So the "12 V solution" that the factory owner dreamt about was purely wishful thinking. Can't be done with 12 V.

The two real solutions that turn up for the 230 V case bear some investigation. Why two realizable solutions? Plug the current values into the individual power terms in your equation and see if you can spot a problem with one of them.

 

[Jamie Bristow]

The two real solutions that turn up for the 230 V case bear some investigation. Why two realizable solutions? Plug the current values into the individual power terms in your equation and see if you can spot a problem with one of them.

Would I be right to move the IV_f term over to the right-hand side of the equation, and then divide the left-hand side by this I to find V_f?

 

[gneill]

Would I be right to move the IV_f term over to the right-hand side of the equation, and then divide the left-hand side by this I to find V_f?

Yes, you could do that. Although it may be more instructive to determine the power values for the two solutions first. How is the generated power being used in each case?

 

[Jamie Bristow]

Yes, you could do that. Although it may be more instructive to determine the power values for the two solutions first. How is the generated power being used in each case?

Ahh, I think I see. With 4.45A, 1023W is generated, and 23W is lost in the wires.
and with 187A, 43,010W is being generated, with only 41962W being lost in the wires?

So 4.45A is the more sensible choice?

 

[gneill]

I think you get the picture

Now that you've identified the practical choice, you should have all the information required to complete the problem sections.

 

[Jamie Bristow]

I think you get the picture

Now that you've identified the practical choice, you should have all the information required to complete the problem sections.

Thank you so much.
I feel like this would have been next to impossible without your guidance

 

[gneill]

Glad I could help.