# Thread: A smokin' hot motor

1. Senior Member
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Originally Posted by gadfly56
This does not make physical sense. The pump will operate along the pump curve, and the output will be inversely proportional to the total head loss. The motor is never going to spin faster than 1,750 RPM. Think of it this way, if I had a tank 3 feet deep filled to the brim, and I set up the suction at 1 foot above the bottom of the tank and set up the discharge back into the tank at the same elevation, am I going to burn out the pump?

If I dead head the pump causing it to churn, the impeller only "slips" past the water. It doesn't draw more and more current in that case either.
Put it this way. Will the pump move more water as you have described or if you were raising the discharge 100 feet? More water means more work.

2. Originally Posted by gadfly56
This does not make physical sense. The pump will operate along the pump curve, and the output will be inversely proportional to the total head loss. The motor is never going to spin faster than 1,750 RPM. Think of it this way, if I had a tank 3 feet deep filled to the brim, and I set up the suction at 1 foot above the bottom of the tank and set up the discharge back into the tank at the same elevation, am I going to burn out the pump?
It's the gallons per minute the pump is trying to move that matters. Stand in your tank with a one gallon bucket, fill it, raise it up shoulder height and dump it, now switch to a five gallon bucket and go through the same motion the same number of times per minute.
If I dead head the pump causing it to churn, the impeller only "slips" past the water. It doesn't draw more and more current in that case either.
It will draw close to no load amps.

3. Originally Posted by gadfly56
This does not make physical sense. The pump will operate along the pump curve, and the output will be inversely proportional to the total head loss. The motor is never going to spin faster than 1,750 RPM. Think of it this way, if I had a tank 3 feet deep filled to the brim, and I set up the suction at 1 foot above the bottom of the tank and set up the discharge back into the tank at the same elevation, am I going to burn out the pump?

If I dead head the pump causing it to churn, the impeller only "slips" past the water. It doesn't draw more and more current in that case either.
Maybe, maybe not. It will depend on how much media is being moved and how much load the pump motor can deliver without being overloaded. That motor will be more loaded if it is moving 25 gallons a minute, even if it is going right back to the same place with little or no restriction that it was taken from, then if it is only moving 5 gallons a minute. Now close a valve in the line and it still has some load from the resistance the media exerts on it, but it is just swirling the same media around and around the impeller and won't be as much load as when it is actually moving some volume with the valve open. Excess energy will heat the media as well in that scenario.

4. Originally Posted by Coppersmith
OK, squishy wasn't a very exact word. Normally, a slight push of a breaker handle in the off direction would cause it turn off with a "snap". When I pressed on the handle it didn't snap, it resisted my pushing, felt spongy, and finally turned off when I pressed very hard. For a moment, i thought it was going to be locked-on and I would be in deep doodoo.
This is what a breaker does after it has tripped, as opposed to being turned off. There are three positions of a breaker handle, not just two; On, Off and Tripped. The "Tripped" position of a Q0 breaker is a mid-point and in order to reset it, you must push is all the way to the full Off position first and make it latch, only then you can snap it back on.
I was suggesting that the obstruction was on the output side of the pump. I'm no expert, but it seems to me that if the pump was attempting to push a column of water and it wasn't succeeding this would slow the impeller which would raise the current draw.
As Golddigger pointed out, it doesn't matter where the obstruction is. Flow = power, less flow = less power.

I'd also look for a throttling valve because as I said, that may be the simplest solution. I worked on several fountains and can assure you that the "artistic" types who commission having them built do not fully understand the engineering principles involved and on more than one occasion, I have found that they see a valve and assume they can open it up to get more flow etc., not realizing that it can overload the pump.

5. Pump hp = head (ft) x gpm / (eff x 3957)

look at a pump curve
as head decreases flow increases (as does hp)
so too low head will ol the motor for some pumps

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Last large line shaft turbine pump I installed was a 300 hp. At "deadhead" it reduces to a 150 hp load.

7. Senior Member
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Originally Posted by Dzboyce
Last large line shaft turbine pump I installed was a 300 hp. At "deadhead" it reduces to a 150 hp load.
That, and with what has been written, makes me think the OP has a motor that is extremely overloaded, that no amount of system throttling will allow a 7.5HP motor to work satisfactorily. That it's "the same size that burned up" makes me think a 20HP motor is more what's needed.

8. This thing will cost him like dollar an hour to run it even if you get the motor running within the rated load.

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The OP did not confirm it is a centrifugal pump. A positive displacement pump such as a piston pump will overload severely with obstruction on the output side.

Originally Posted by Coppersmith
I was suggesting that the obstruction was on the output side of the pump. It seems to me that if the pump was attempting to push a column of water and it wasn't succeeding this would raise the current draw.
Thanks.

10. Originally Posted by Sahib
The OP did not confirm it is a centrifugal pump. A positive displacement pump such as a piston pump will overload severely with obstruction on the output side.
The frame on the motor is a 215JP, it's tied onto an impeller pump.

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