A smokin' hot motor

[Coppersmith]

Here is the data plate for a motor I hooked up. (It powers a waterfall feature in an artificial mountain behind a large residence. Gotta love rich people.) I'm told it's the same size motor as one that burned up. The existing breaker was a 2-pole 40 amp @ 240v. The wiring is #6 THHN/THWN. The motor tripped the breaker after about 15 seconds. (Tried several times.) I looked up the calculations and if I am correct it should be a 70 amp breaker. Note that the data plate says "31 amps" and not "31 FLA". I assumed they were the same thing. I installed the 70, the motor ran about 2 minutes and then started to smoke. I flipped off the breaker. The breaker was very warm and the handle was squishy. After the breaker cooled for a few minutes, it appeared to operate with it's normal feel. There is no motor starter or heaters in the circuit.

1) Is 70 amps correct?

2) Why would the breaker handle get squishy? It's a SqD QO.

3) Is there an electrical issue here (beside the breaker size)? I personally think the flow is obstructed causing excessive motor loading. I suggested to the owner he call the plumber back (who installed the motor) to check for obstructions.

4) I also suggested that if there is no obstruction, a call to a motor repair specialist would be in order to check the motor for proper operation. Was this a good call?

5) Should there be a DOL motor starter with heaters in the circuit or it it OK without heaters?

 

[junkhound]

Load is too big for 7.5 HP.

re: check for obstructions or the valve is too FAR open, increased flow at constant head (waterfall) = higher load. Need to look at the pump flow vs. power curve vs. head curve. Plumber did not set things up correctly for 7.5 HP.

The statement: After the breaker cooled for a few minutes, it appeared to operate with it's normal feel is ambiguous - as is ran for a few minutes OK? or ran for a few hours OK? or ran fine with valve partially closed? or ran with ho head?, etc.....

What is the desired flow and how high is the 'waterfall'.

 

[Jraef]

Here is the data plate for a motor I hooked up. (It powers a waterfall feature in an artificial mountain behind a large residence. Gotta love rich people.) I'm told it's the same size motor as one that burned up. The existing breaker was a 2-pole 40 amp @ 240v. The wiring is #6 THHN/THWN. The motor tripped the breaker after about 15 seconds. (Tried several times.) I looked up the calculations and if I am correct it should be a 70 amp breaker. Note that the data plate says "31 amps" and not "31 FLA". I assumed they were the same thing. I installed the 70, the motor ran about 2 minutes and then started to smoke. I flipped off the breaker. The breaker was very warm and the handle was squishy. After the breaker cooled for a few minutes, it appeared to operate with it's normal feel. There is no motor starter or heaters in the circuit.

1) Is 70 amps correct?

No, not if there is no starter with overloads in the circuit. In fact if that's the case, even the 40A is not legal. Running overload protection must be maximum of 125%, no "next size up". 125% of 31A is 38.75A, a 40A breaker is not legal for this purpose per 430.32. 70A is so far off it's not even funny, but if you DO install a motor starter with OL heaters, then 70A works. I assume you determined 70A by virtue of 430.52, but that only counts AFTER you have satisfied 430.32

2) Why would the breaker handle get squishy? It's a SqD QO.

Squishy? You mean the breaker seems like it will trip easily? That could be that the breaker has tripped so many times now that it is wearing out.

3) Is there an electrical issue here (beside the breaker size)? I personally think the flow is obstructed causing excessive motor loading. I suggested to the owner he call the plumber back (who installed the motor) to check for obstructions.

If flow is obstructed in a centrifugal pump, the current would DROP. Flow = work performed, work performed = power. Less flow, less power.

4) I also suggested that if there is no obstruction, a call to a motor repair specialist would be in order to check the motor for proper operation. Was this a good call?

It's more likely that the pump was improperly sized for the desired flow and you are over loading it as a result. In other words someone may have sized it for a particular head (height) and pipe size, then when the fountain was built, the head height was lower and/or the pipe was larger, so there is MORE flow than the pump is rated for, leading to the motor overloading and since there is no starter w/ OL heaters, only the breaker takes it off line eventually. It's also possible that someone installed a throttling valve knowing that the pump flow needed modulation, then someone later opened that valve further because they wanted more flow, not understanding the effect this would have on the pump.

5) Should there be a DOL motor starter with heaters in the circuit or it it OK without heaters?

Absolutely. If that motor were the type that had internal thermal trips, it is REQUIRED to state that on the nameplate, which it does not. So when it does not say that, you must provide it externally.

So in conclusion:
1) You MUST add a motor starter with thermal protection set for 31A
2) most likely I'm willing to be that it will trip, because something in your fountain design is wrong and the pump is overloading due to excessive flow. You will either need to add a throttling valve and use a clamp-on ammeter to adjust the flow to where the motor current is 31A or less, or get a larger motor.

 

[gadfly56]

The spec information is here. It looks like it shouldn't have heaters, it's direct on line start.

We'd need the pump specs, but in any event, you can't overload the motor if the head is too low. The pump output will vary as the sum of the static and velocity head, but it won't overload unless you grab the impeller and lock it. Something else must be at work here. Even if you dead head the pump, if it's centrifugal, it will simply churn the water in the case. You can heat the water to boiling this way, but it would take more than two minutes. Has anyone checked to see if there is excessive runout, offset, or angular misalignment?

 

[Jraef]

The spec information is here. It looks like it shouldn't have heaters, it's direct on line start.

We'd need the pump specs, but in any event, you can't overload the motor if the head is too low. The pump output will vary as the sum of the static and velocity head, but it won't overload unless you grab the impeller and lock it. Something else must be at work here. Even if you dead head the pump, if it's centrifugal, it will simply churn the water in the case. You can heat the water to boiling this way, but it would take more than two minutes. Has anyone checked to see if there is excessive runout, offset, or angular misalignment?

Being DOL start has nothing to do with needing OL protection or not. The "heater" in that spec sheet is referring to a motor space heater, i.e. to keep condensation out of it. The part of the spec where you can tell it needs OVERLOAD protection (aka "heaters") is where it says

Thermal Device - Winding None

That means, per 430.32, YOU must supply external motor running overload protection.

 

[ActionDave]

The spec information is here. It looks like it shouldn't have heaters, it's direct on line start.

We'd need the pump specs, but in any event, you can't overload the motor if the head is too low. The pump output will vary as the sum of the static and velocity head, but it won't overload unless you grab the impeller and lock it. Something else must be at work here. Even if you dead head the pump, if it's centrifugal, it will simply churn the water in the case. You can heat the water to boiling this way, but it would take more than two minutes. Has anyone checked to see if there is excessive runout, offset, or angular misalignment?

A motor has to have heaters, either internal or on a starter. This one has to start across the line because it's single phase and you can't soft start it.

If the pump is moving more water than the motor can handle then the motor will pull high amps. It needs some back pressure to work right.

 

[Coppersmith]

Squishy? You mean the breaker seems like it will trip easily? That could be that the breaker has tripped so many times now that it is wearing out.

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.

If flow is obstructed in a centrifugal pump, the current would DROP. Flow = work performed, work performed = power. Less flow, less power.

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.

It's more likely that the pump was improperly sized for the desired flow and you are over loading it as a result. In other words someone may have sized it for a particular head (height) and pipe size, then when the fountain was built, the head height was lower and/or the pipe was larger, so there is MORE flow than the pump is rated for, leading to the motor overloading and since there is no starter w/ OL heaters, only the breaker takes it off line eventually. It's also possible that someone installed a throttling valve knowing that the pump flow needed modulation, then someone later opened that valve further because they wanted more flow, not understanding the effect this would have on the pump.

So in conclusion:
1) You MUST add a motor starter with thermal protection set for 31A
2) most likely I'm willing to be that it will trip, because something in your fountain design is wrong and the pump is overloading due to excessive flow. You will either need to add a throttling valve and use a clamp-on ammeter to adjust the flow to where the motor current is 31A or less, or get a larger motor.

I'll relay this to the owner and suggest he have the design checked. I'll also inform him a starter with heaters must be installed. Thanks.

 

[Coppersmith]

Has anyone checked to see if there is excessive runout, offset, or angular misalignment?

I'll give these details to the owner. Thanks.

 

[GoldDigger]

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.

A centrifugal pump spins the water to that "centrifugal force" slings the water out the output fitting. If no water moves, no water is being accelerated and no energy is transferred into the water (except for a small amount of friction.)
Same thing happens with an impeller. Only a positive displacement pump will take more power as head increases (as long as the same volume of water moves) because the piston is moving water the same distance against a higher pressure.

It does not matter whether you restrict the input or the output, the physics of the moving (or not moving) water is the same. Water is not compressible.

You can see this effect with a vacuum cleaner. If you block the input or the output the motor speeds up.

 

[gadfly56]

A motor has to have heaters, either internal or on a starter. This one has to start across the line because it's single phase and you can't soft start it.

If the pump is moving more water than the motor can handle then the motor will pull high amps. It needs some back pressure to work right.

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.

 

[ptonsparky]

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.

 

[ActionDave]

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.

 

[kwired]

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.

 

[Jraef]

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.

 

[Ingenieur]

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

http://cdn.qleapahead.com/zoeller-63/132945.pdf?P185

 

[Dzboyce]

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

 

[JFletcher]

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.

 

[Electric-Light]

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

 

[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.

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.

 

[ActionDave]

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.