fun recreatinoal voltage drop quiz

[electrofelon]

Consider a 120V source, 500 feet from an infinitely variable resistive load. The circuit conductors are #12 Cu. What resistance results in the maximum power dissipated by the variable resistor and what is the power dissipated.

 

[gar]

160202-2417 EST

Maximum power in a variable load with a constant voltage source with a constant internal resistance of R occurs when the load resistance is R. Use calculus to prove it or trial and error with various values.

The load power is 1/2 of the total input power, or P = Vs*I/2 = (Vs*Vs/2*R)/2 = Vs^2/4*R.

.

 

[adamscb]

The resistance of the load to maximize power transfer would have to have the same resistance value of the wire. The resistance of 12 awg, 500 feet is .794 ohms. If the load is .794 ohms, then the total resistance is 1.588 ohms. This results in a current draw of 75.58 amps.

The power dissipated across the load is (I^2)*R, or (75.58^2)*.794, which is equal to 4.535 kW.

That is, if the #12 Cu doesn't burn out by then.

 

[electrofelon]

160202-2417 EST

Maximum power in a variable load with a constant voltage source with a constant internal resistance of R occurs when the load resistance is R. Use calculus to prove it or trial and error with various values.

The load power is 1/2 of the total input power, or P = Vs*I/2 = (Vs*Vs/2*R)/2 = Vs^2/4*R.

.

The resistance of the load to maximize power transfer would have to have the same resistance value of the wire.

Correct. I didnt realize that at first, although it seemed obvious in hindsight. I wrote an equation for the power in terms of resistance, and graphed it in maple with some different values. I was about to take the derivative to find the max, when the solution occured to me. Here is a graph of resistance vs power, its kinda cool to see.

P.S. adamscb, I think your wire resistance value was wrong.

 

[K8MHZ]

The resistance of the load to maximize power transfer would have to have the same resistance value of the wire. The resistance of 12 awg, 500 feet is .794 ohms. If the load is .794 ohms, then the total resistance is 1.588 ohms. This results in a current draw of 75.58 amps.

The power dissipated across the load is (I^2)*R, or (75.58^2)*.794, which is equal to 4.535 kW.

That is, if the #12 Cu doesn't burn out by then.

Maybe you factored this in and I missed it, but wouldn't a load 500 feet from the source require 1000 feet of conductor?

 

[K8MHZ]

We could look at this in a different way.

First, we are trying for a 'conjugate match' for maximum power transfer. For that to happen, the impedance of the load must = the impedance of the supply.

The theoretical question implies a constant voltage source with no internal impedance, therefore the only impedance seen by the load would be that of the conductors.

I found that 12 AWG has 1.588 Ohms per 1000 feet. That means maximum power transfer would be into a load of 1.588 Ohms.

I didn't figure out the power the resistor would dissipate, but it's not as simple as it seems. We first have to figure the voltage drop seen at the resistor and then apply Ohm's Law and I'm too lazy right now.

On edit, taking the lazy way out, the resistor and the conductors would have to dissipate the same amount of power to be matched. My SWAG would be 2267 watts using my dollar store calculator.

 

[winnie]

Approximately 2 ohms and 1800W.

The DC resistance of 12 AWG wire is given in table 8 as between 1.93 and 2.05 ohms per 1000 feet at 75C, depending on stranding and coating. Reasonable ambient temperature variations might mean that the resistance is 20% lower than these figures. So I just picked 2 ohms wire resistance as a reasonable value for 500 feet out and back.

Maximum power is delivered when load resistance = source resistance.

Total resistance is the sum of load and source, so we have 4 ohms across 120V, or 30A of current.

You have 1800W dissipated in the wire and 1800W dissipated in the load.

-Jon

 

[K8MHZ]

Approximately 2 ohms and 1800W.

The DC resistance of 12 AWG wire is given in table 8 as between 1.93 and 2.05 ohms per 1000 feet at 75C, depending on stranding and coating. Reasonable ambient temperature variations might mean that the resistance is 20% lower than these figures. So I just picked 2 ohms wire resistance as a reasonable value for 500 feet out and back.

Maximum power is delivered when load resistance = source resistance.

Total resistance is the sum of load and source, so we have 4 ohms across 120V, or 30A of current.

You have 1800W dissipated in the wire and 1800W dissipated in the load.

-Jon

Better answer than mine. I got my resistance value from a chart I found on line. Your source is better.

 

[Carultch]

The resistance of the load to maximize power transfer would have to have the same resistance value of the wire. The resistance of 12 awg, 500 feet is .794 ohms. If the load is .794 ohms, then the total resistance is 1.588 ohms. This results in a current draw of 75.58 amps.

The power dissipated across the load is (I^2)*R, or (75.58^2)*.794, which is equal to 4.535 kW.

That is, if the #12 Cu doesn't burn out by then.

Wire resistance is a function of the wire operating temperature, rather than ambient temperature. The NEC values are all at 75 Celsius operating temperature. In this particular example, your current is far greater than the NEC allowable current in #12 wire at 75C, (which means it is probably unsafe), so expect the resistance to be more than the NEC ohms/kft would indicate.

We have a model for how resistance varies with temperature, and we have the Neher-McGrath formula that I've never used, which would model the thermal behavior of the wire. I'd imagine you'd need to somehow reverse engineer the NMG formula, to calculate wire temperature as a function of current. This would have to be iterative, given the non-linearity of all factors combined.

 

[electrofelon]

Right my figures assume wire resistance of 2 ohms per thousand. I got thinking about this because I was first thinking about how we dont usually factor in the effect of voltage drop on the load, so for resistive loads, the voltage drop is less than it actually is (that is probably not worded very well). To prove the question in the OP, or to find the result if we didnt realize that it would be when source impedance matches load impedance (As I didnt at first), is not so simple. Finding the derivative of the function is not a trivial task to do by hand, although of course a computer algebra program like Maple has no trouble. See my attachment, the derivative is after the "diff" command.

 

[Carultch]

Right my figures assume wire resistance of 2 ohms per thousand. I got thinking about this because I was first thinking about how we dont usually factor in the effect of voltage drop on the load, so for resistive loads, the voltage drop is less than it actually is (that is probably not worded very well). To prove the question in the OP, or to find the result if we didnt realize that it would be when source impedance matches load impedance (As I didnt at first), is not so simple. Finding the derivative of the function is not a trivial task to do by hand, although of course a computer algebra program like Maple has no trouble. See my attachment, the derivative is after the "diff" command.

At least with finding a derivative, it's possible to do it in closed form for all elementary functions no matter how they are combined. Integrals on the otherhand, not so easy, as many do not have a closed form solution.

 

[electrofelon]

At least with finding a derivative, it's possible to do it in closed form for all elementary functions no matter how they are combined. Integrals on the otherhand, not so easy, as many do not have a closed form solution.

Yes the chain rule is a wonderful thing. Integrating by hand can be a real PIA, or as you say not possible at all beyond a numerical approximation. IIRC, calc 2 was my worst math grade.