A question about maximum available current.

Thread Starter

DEWFPO

Joined Feb 17, 2009
6
I need some expert advice.

I have tried to research this on the internet but the more I research, the more frustrated I have become.

I just need a straight answer please (if that is possible).

Setup : I will be buying an electrically driven water pump to pump water out of a creek up a hill (a horizontal distance of about 400 feet and a vertical distance of about 50 feet.). I need to provide about 9 gpm at about 40 psi at the highest, furthest point. I will be running 250 feet of 12/3 H.D. construction extension cord to the location of the pump with a dedicated GFCI in the cord. The cord will be plugged into an outside outlet on the house that will have nothing else running on that circuit at the time the pump is running.

Question : I need to know how much current (the maximum) available for me to use at the physical pump location. This will determine how big a pump (1.0 or 1.5 HP) I can install.

I would like to go with the biggest pump I can safely run for hours on end.

Your advise is greatly appreciated.

DEWFPO
 

beenthere

Joined Apr 20, 2004
15,819
Well, all that 12/3 is going to drop some voltage (and so the current draw of the pump will go up). The NEC limit for continuous draw is 80% of the breaker rating, or 16 amps in your case. That will be a bit much for that length of cord, but it's still enough to supply 2.5 HP. 16 amps X 120 VAC = 1920 watts. 1920 / 768 (watts per horsepower) = 2.5 HP.
 

TanTJ

Joined Mar 6, 2008
21
Beenthere, you beat me to it!

I was just going to say, current through a series circuit is the same everywhere so the maximum current is that of what the breaker is rated at. I figured there would be a "safe" continuous rating as well, just wasn't sure what it was.

I've got couple questions for you guys though. Taking into account the resistance of the extension cords might give us a different answer for calculating which pump to get. Read on and please comment on where I messed up.

I got really bored and figured out the resistance of 500 feet of extension cord (he said 250, but the distance from pump to outlet was 400, I figured I'd base the calculations on 500 ft assuming there might be excess extension cord.)

The total resistance (considering there are 2 conductors carrying the current for a total length of 1000ft) was 1.545 Ω. This is based on 12 gauge copper wire.

Since I was still bored I figured total resistance in the circuit would be 7.5 Ω if the voltage is 120v and a current draw of 16 A. Considering the extension cord eats up 1.545 Ω that leaves 5.955 Ω for the pump motor.

Now here's an interesting question for you guys. Knowing all this, a motor that has 5.955 Ω of resistance will actually draw roughly 20 Amps without the 500 ft of extension cord. 20 Amps times 120 volts gets us 2400 Watts, or 3.2 Horsepower. So should he not buy a 3 or 3 1/4 Horsepower pump knowing that 500 ft. of extension cord will provide enough resistance to drop the current to 16Amps?

Is this an extreme over analysis of a simple situation or do I have some merit here? I'm leaning towards the former, but what do you think?
 

beenthere

Joined Apr 20, 2004
15,819
I could never recommend using an extension cord as a limiting resistor. The 1 1/2 horse pump leaves some overhead. One of the problems with voltage drops is that the load ends up pulling more current because of it - loads use power. As soon as you start to heat your conductors, the resistance goes up, and you enter a death spiral when the load tries to pull yet more current - making the conductor that much hotter.

I've seen a 10 ga wire burn through because a lug got a bit resistive and the wire got hot.
 

TanTJ

Joined Mar 6, 2008
21
Thanks for the response Beenthere! That actually makes perfect sense now that I think about it a little bit more. To put it into other words, if you think of the extension cord simply as a resistor, they will dissipate power in the form of heat. As it gets hotter the resistance will go up, and the power dissipated will go up, and the spiral you talked about keeps going until a wire starts to burn.
 

Thread Starter

DEWFPO

Joined Feb 17, 2009
6
I just checked the specs on the 1.5HP pump and it lists the "motor amps" as 19.9 at 115VAC. I'm a bit confused. I doesn't say if that's the startup voltage but I will research a bit more.

Thanks, DEWFPO
 

Thread Starter

DEWFPO

Joined Feb 17, 2009
6
O.K., the online manual for this 1.5 HP pump lists the following specs.(for 115VAC) :

Max. Load Amp = 19.2 (different from the spec sheet I referenced above @ 19.9).
Branch Fuse Rating Amps = 25 amps
AWG Min. Wire Size = 10
Distance in ft. from motor to supply (201-300 ft.), AWG wire size 6!

I guess I better check out the specs for the 1HP motor.

I wonder why it would be drawing so many more amps than expected?

DEWFPO
 

Thread Starter

DEWFPO

Joined Feb 17, 2009
6
Looks like I'll be stuck with a 1 HP motor on that pump. These are common Simer/Flotec pumps with AO Smith motors.

The specs for the 1 HP Pump (FP5162) are as follows :

Max. Load Amps = 14.8 amps
Branch fuse rating amps = 2- amps
AWG Min. Wire size = 12

But again, it says I should use a 6 gauge wire for a 250" distance from motor to supply. I couldn't find a starting load anywhere in the manual.

DEWFPO
 

beenthere

Joined Apr 20, 2004
15,819
The "excess" current must represent the difference between the usable output of the pump and the inefficiency of the motor.

For insurance, if nothing else, you had better stay within those specs. I really hate to think about the cost of 6 ga.
 

italo

Joined Nov 20, 2005
205
I think the following info will sufice.The #AWG to limit the voltage drrop to 2% on a 110v ac circuit is as folows.
16 amps only
awg #12-25' #6 100' #2 300' #0 400' #00 at 500'
 
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