motor drawing excess current
motor drawing excess current
(OP)
We have a deep-well submersible pump driven by a 75hp 460v vertical induction motor, with nameplate FLA=87A.
The pump curve indicates maximum BHP is 63HP at 525gpm. BHP is lower if you increase or decrease flow from that point.
At 525 gpm AND 130psig discharge pressure we twice measured the currents. Both times we measured 90, 92, 94A. Voltage was 490v balanced within accuracy of our voltmeter.
We wondered how the pump could be overloaded above 75hp when pump is incapable of drawing more than 62hp. We did realize that the motor measured speed is 1775rpm while pump curve is based on 1760 rpm. Assuming a cubed-law variation of power we still conclude pump should not draw more than 64hp. [I have to remember to double-check the nameplate speed... I think it's around 1775rpm).
Based on the fact that the HP had to be going somewhere we suspected a rub. We had the maintenance dept adjust the pump lift. They did an adjustment and reported no signs of a rub (rotated smoothly). After the lift adjustment the current had decreased approx 2A to 88A, 89A, 92A (still 490V, still 130 psig/525gpm).
We varied flow by pumping to bypass. We confirmed that for any discharge pressure higher or lower than 130A, the current readings went down (this is in agreement with the shape of the BHP curve).
We cannot precisely determine the head of the pump since the suction pressure (far underground is unknown), but we believe the flow info is sufficient to locate us on the curve. We are perplexed that the motor can deliver more current than the pump appears capable of consuming on paper.
I don't think it is a problem with the motor itself since the motor has been replaced and the previous motor had similar high current draw. In fact this motor has been replaced about 3 times in last 10 years for various reasons I am not totally familiar with. It is a tough outdoor environment. With 1.15sf it should be able to withstand this level of overload but I would like to try to understand it better.
Any suggestions what might be going on or how to troubleshoot?
The pump curve indicates maximum BHP is 63HP at 525gpm. BHP is lower if you increase or decrease flow from that point.
At 525 gpm AND 130psig discharge pressure we twice measured the currents. Both times we measured 90, 92, 94A. Voltage was 490v balanced within accuracy of our voltmeter.
We wondered how the pump could be overloaded above 75hp when pump is incapable of drawing more than 62hp. We did realize that the motor measured speed is 1775rpm while pump curve is based on 1760 rpm. Assuming a cubed-law variation of power we still conclude pump should not draw more than 64hp. [I have to remember to double-check the nameplate speed... I think it's around 1775rpm).
Based on the fact that the HP had to be going somewhere we suspected a rub. We had the maintenance dept adjust the pump lift. They did an adjustment and reported no signs of a rub (rotated smoothly). After the lift adjustment the current had decreased approx 2A to 88A, 89A, 92A (still 490V, still 130 psig/525gpm).
We varied flow by pumping to bypass. We confirmed that for any discharge pressure higher or lower than 130A, the current readings went down (this is in agreement with the shape of the BHP curve).
We cannot precisely determine the head of the pump since the suction pressure (far underground is unknown), but we believe the flow info is sufficient to locate us on the curve. We are perplexed that the motor can deliver more current than the pump appears capable of consuming on paper.
I don't think it is a problem with the motor itself since the motor has been replaced and the previous motor had similar high current draw. In fact this motor has been replaced about 3 times in last 10 years for various reasons I am not totally familiar with. It is a tough outdoor environment. With 1.15sf it should be able to withstand this level of overload but I would like to try to understand it better.
Any suggestions what might be going on or how to troubleshoot?





RE: motor drawing excess current
RE: motor drawing excess current
Has the pump ever been rebuilt in this ten years?
RE: motor drawing excess current
The water table is a plane under the surface. When you start pumping the profile around the pump drops ( you pump makes a "hole" in the water table). You can see this when you have monitoring wells around a production well. The "hole" your maiking could be deeper than normal for some reason (drought, well plugging, etc.)
The work your pump does is measured from the surface of the water its pumping to the discharge. Weight X height /Time.
The dynamic level of the water is probably lower than the static level. You may think the pump is incapable of drawing only 62 HP but in fact it may be trying to do more.
RE: motor drawing excess current
For your intial readings of 92A ave, 490V, assumed 3 phase, and a power factor of 60%, implies 63HP being drawn. Admittedly, 60% PF is rather low, but I just got done measuring a compressor motor at 64,72,72 PF on each of 3 legs.
When the boys pulled the pump, did they happen to measure the impeller diameter. Knowledge of the impeller diameter is critical to use of pump curves. Perhaps you could still infer the diameter by measuring shutoff head on the supply side (i.e. up top), this implies impeller diameter by comparing to pump curve at 0 GPM.
So you are pulling high amps, and if low power factor is the culprit, it could be worthwhile to install some PF correction capacitors at the motor control center...This bring the V phase more in line with the A phase, thus droppping the A, and all of the I2R losses throughout the system. This could reduce your electric bill, as well as provide longer life on insulation and transformers due to the lower A and associated heat generated.
PacificSteve
RE: motor drawing excess current
Buzzp - I am not assuming that all the HP going into the motor is going into the pump. The pump is rated for 75HP OUTPUT (shaft horsepower) at nameplate conditions 87A. Since we are greater than 87A I am assuming that greater than or comparable to 75HP output is being delivered. Motor power factor and efficiency at nameplate conditions should already be accounted for in the nameplate FLA. I agree that our slight increase in voltage might tend to drive current in either direction, but not that much. Well is 225 feet deep.
I am not entirely sure if "submersible" is the correct term. Motor is at ground level, pump is several hundred feet down. Is that called submersible? The pump has been rebuilt twice. I will check on assumed temperature in the BHP curve.
BJC - I agree it is complicated problem to compute the fluid power. We have no accurate way to compute the DP. But we CAN accurately measure the flow rate which should locate us on pump curve. We deliberately throttled to locate the flow-rate at the point of max BHP on the pump curve. A little experimentation (changing throttling configuration) confirmed that amps decreased if we increased or decreased throttling from this point. From where I'm standing, we have properly accounted for the unkown dp. (I think).
Steve- wow good idea about measuring the shutoff head to confirm the pump curve. I will have to remember that one for future use. Unfortunately I'm not sure of the exact depth of the pump which would throw uncertainty into my calculation. I'm a little skeptical that poor power factor is the culprit since #1 - poor power factor would show up as higher FLA on the nameplate (I'm only comparing current to nameplate, not making any explicit assumption about power factor); and #2 - at least two motors have drawn high current, suggesting the problem is not associated with the motor. I will have to look at the history to see if anyone has recorded the impeller diameter or changes to the pump which may not be reflected in our (old) design documents.
RE: motor drawing excess current
So this is shaft driven? Could the long shaft be part of the power drag, is this what you meant about "checking for a rub"? Could this be some vibrational type of energy dissapation. You know, the system creates the horsepower demand, the trusty ol' motor just tries to serve it. The solution could also be a larger motor, which would be a quick fix for early life failure concerns.
It only takes 5 minutes to measure true kW or PF if you have the right meter. Similar motors in same facility tend to run about the same PF, just by experience.
I would like to hear how this one turns out!
PS
PacificSteve
RE: motor drawing excess current
RE: motor drawing excess current
RE: motor drawing excess current
The power factor of the source will be determined by the loads connected, specifically meaning that the sum of all reactive and real power comsumed by the loads will determine the power factor of the system.
My 'guess' is that the pump curve you are using is not correct either due to changes in the pump or due to conditions in the well.
RE: motor drawing excess current
I'll stop beating this dead horse after this last comment. kW=HP*.746. kW=V*A*1.732*PF. If you are only measuring amps and voltage, PF remains an important factor, and you really have no idea what the kW is. Thus you have no idea what the HP is.
So my suggestion is to define the magnitude of the suspected problem. This is useful in chasing down the cause.
In this case, using presently available measured data
HP=(V*A*1.732*PF)/.746=(490*92*1.732*PF)/.746
HP=104.66*PF.
PacificSteve
RE: motor drawing excess current
2)Si el problema es de la parte hidraulica, por cálculo deficiente, impulsor mal dimensionado, ....etc., se produce un aumento en los amperios y de un orden similar en los kilowatios.
Midiendo amperios y potencia consumida en Kilowatios, puede conocerse si el problema es en 1) o 2)
RE: motor drawing excess current
I would have a look at the power factor and see what it actually is, also have a look at the open shaft current. Is this an installation that has been operating correctly and is now going off the rails? or is it a new installation?
What is the design voltage of the motor?
The additional current could be due to shaft load, but it could also be due to increased losses in the motor. If the motor was wound for 440V and operated on 490, the flux in the iron could be saturating and hence an increase in iron loss and magnetising current. This would show up as excessive open shaft current and low PF.
Mark Empson
http://www.lmphotonics.com
RE: motor drawing excess current
I still think it could be a hydraulics problem.
Suggestions:
1.) Next time the well is down and has had time to rise ( asusming my theory about drawing the water table down is correct ) put a recording ammeter on the pump. If the water table is being drawn down over time the trend might show up as an increasing current. This assumes the pump cannot instantly draw the water down but needs a little time to draw it down. Did you make all you test right after a long shut down?
The recording ammeter might find some unusal usage pattern like does someon fill a large tank, vat or swimming pool at 1 am every day.
2.) If the work the pump is doing is not going into lifting the water then it's got to be heating the water. At 220 ft my guess is your well water is somewhere between 55 and 65 deg F. I don't now if you could measure the temperature difference ( assuming you know the water temperature down hole).
3>) Checking impeller diameter was mentioned but that should have impeller diameterS (Not just one impeller), I would guess this pump has five or six stages.
Pumps like this are usually installed well below the surface of the water. Is it possile that drawing down the water level is lowering the pressure and causing caviation? Cavitation is funny stuff it can break impellers or just wear them out.
Just some thoughts
BJC
RE: motor drawing excess current
RE: motor drawing excess current
Submersible pupms are generally located at far off locations or in other words the length of power cable is very much. As I have seen voltmeters are installed in Substations while Ammeters are installed in field. The terminals of submersible motors are also mostly inaccessible for direct measurement of voltage at its terminals. The voltage you have mentioned 490 V is measured at the motor terminals (or in a junction box close to the motor) or was it in the substation.
What I want to say is that the power cable of the motor may be undersize and giving voltage drop at motor terminals, in running condition. Try to measure the voltage as close to the motor terminals as possible.
What is the size and length of power cable ? Is it adequate.
Some Submersible motors like of KSB, have water circulating between the stator and rotor for cooling. You can also test the motor in workshop by dipping it completely in a tank of water. Try finding its no-load current in this way, which should be approximately 30 - 40 % of full load current. Generally the lower is the NL current as compared to the rated, the better is the design of the motor.
You can also megger the motor and its power cable at the same time from the substation end, at 500 V.
In general if all these parameters are OK then the pump is the main reason for increase in amps of the motor, because motor is the driver. Its current increases indefinitely in response to the demand of the load. Of course various protections are there to prevent current increase above a certain value.
RE: motor drawing excess current
Voltage was measured at the motor term box.
I was incorrect in calling it a submersible. The motor is above ground, the pump is ~225' under.
RE: motor drawing excess current
kW=HP*.746. kW=V*A*1.732*PF*Eff
HP=(V*A*1.732*PF)/.746=(490*92*1.732*PF)/.746
HP=104.66*PF*Eff
Hope this is of some help for you!
RE: motor drawing excess current
I wonder if pump manufacturers use the motor efficiency factor when they draw up their pump curves?
Steve
RE: motor drawing excess current
RE: motor drawing excess current
Steve - I'm not sure if I understood your comment about pump curves accounting for motor efficiency. I assume you are referring to the curve of BHP vs flow-rate. BHP is by definition the motor output and independent of motor efficiency.
RE: motor drawing excess current
I jumped one step of logic in my comment. It is kind of like BHP versus SHP, brake horsepower in a car at the wheels, versus shaft horsepower, right off the crankshaft.
"Shaft horsepower" for the pump would be only what the pump itself requires.
But there are other power demands of the system, such as the HP that the motor needs to turn itself and the drive mechanism to the pump. This total would be the "brake horsepower" in my automotive analogy.
When you try to infer horsepower from electrical readings, if you do your measurements accurately, you will get the "brake horsepower".
Wiring feed losses also can't be ignored. I just got done measuring a pump kW using a true kW meter, and found a difference of almost 30% kW between the kW at the pump motor electrical connection, and the kW at the electrical panel that feeds it. On this 3way valve chillwater system, I would expect any significant changes in head of the system between one measurement and the next (5 minutes apart) so I assume it must be wiring loss.
I would assume that pump manufacturers list brake horsepower, by making assumptions about the motor and mechanical power transmission equipment that will be used. Maybe not, and I am hoping there is someone out there who knows for sure. Perhaps they think that it is OK to graph the "shaft horsepower", and that the motor service factor has enough slack to keep them out of trouble and cover the full brake horsepower. Most times, with that extra 15% slack factor, they would be right.
I've measured pumps in the field operating at nearly twice the motors amperage rating...I'm sure they won't last 10 years though, and with the super high temperatures within the motor, the motor itself will have large electrical losses.
PacificSteve
RE: motor drawing excess current
Motor Nameplate:
75 HP
460v
FLA = 87
Speed = 1775
Motor mounted above ground
Pump is below ground with ~225 Ft of shaft between the motor and pump
Pump Curve Data:
BHP = 63 HP at 525 gpm and Speed of 1760
BHP = ~ 64 HP at Speed of 1775
Assuming these loads don't take into account the length and weight of the shaft
Applied Voltage = 490
Current readings = 90, 92, 94 amps
If we assume a 100% PF, we can estimate motor efficiency using the nameplate.
KW = HP * 746
KW = 75 * 746
KW = 55950
Eff = KW/(V * I * 1.73)
Eff = 55950/(460 * 87 * 1.73)
Eff = .80812 or 80.8%
Using an estimated efficiency of 80.8% and assuming 100% PF, we can estimate the approximate nameplate FLA at 490v.
FLA = KW/(E * 1.73 * Eff)
FLA = 55950/(490 * 1.73 * .80812)
FLA = 81.67
If the motor is turning at its rated speed of 1775 rpm and the true power consumed remains constant at 55950 KW, we can estimate our PF from the current readings.
PF = KW/(E * 1.73 * I * Eff)
PF = 55950/(490 * 1.73 * 92 * .80812)
PF = ~. 887757 or 88.78%
If we subtract our current reading from the expected FLA, we get a difference of 10.33 amps. (92 - 81.67 = 10.33)
We can theorize that a long shaft will add additional burden on the motor resulting in excess power drawn from the source. (Approximately 225 Ft of shaft) Although as previously mentioned by others, the shaft can be one of many factors adding load. To quantify this added burden, the excess load can be expressed in HP.
HP = (Eapp * Idiff * 1.73 * Eff * PF)/746
HP = (490 * 10.33 * 1.73 * .80812 * .887757)/746
HP = 8.42
Adding the excess load to the pump load, we have the following:
64 HP + 8.42 HP = 72.42 HP (load drawn by the pump and shaft on the motor)
I'm not quite sure how to calculate the load created by the shaft alone. This could be
significant telling us whether we should look elsewhere to identify a large portion of the excess load. (Ex: voltage drop, viscosity of the liquid, bad bearings at the pump impeller, harmonics, bent shaft, high-speed vibrations, etc.)
At an estimated PF of 88.78%, we can approximate the maximum HP we could expect to receive from the motor without overloading as the manufacturer designed it.
HP = (E * I * 1.73 * Eff * PF)/746
HP = (460 * 87 * 1.73 * .80812 * .887757)/746
HP = 66.58
In light of the obvious, comparing 66.58 with 72.42 HP, the motor appears to be overloaded. (Changed three times in 10 years) It appears there is a need to reduce the load or increase the available horsepower.
It might be interesting to know the manufacture's expected life of this motor at various
intervals such as when overloaded approximately 8.77%.
The next common size motor may be 100 HP, but to increase the size of the motor may not be cost effective. It may also be physically difficult.
The problem may not be worth solving.
RE: motor drawing excess current
Wirenut
RE: motor drawing excess current
RE: motor drawing excess current
I did hope that someone might have some ideas on WHY the motor can apparently draw more power than pump is capable of absorbing...
///Please notice that common mode current can accomplish this. Namely, they can increase imput power to the motor noticeably higher than the pump needs.\\\
based on comments from this thread I'm working toward the conclusion that our pump does not match our paperwork.
///Highly commendable.\\\
RE: motor drawing excess current
Since over 12kW are going somewhere and you haven't snapped a shaft yet I would suggest that the hydraulic end of the pump is not what is stated on the rating plate.