motor drawing excess current 8

[electricpete]

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?

 

[electricpete]

good comments. Since I have compared my current readings to nameplate, I have already acconted for both efficiency and power factor.

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.

 

[PacificSteve]

EP-

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

 

[AESMC]

Expanding on the ideas and facts of previously posted notes, I thought it might be interesting to experiment with the numbers. I hope this will cause some overlooked items to become obvious by better illustrating the problem.

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.

 

[Wirenut]

PUT A 100 HP MOTOR ON IT. MOVE ON TO OTHER PROBLEMS.... Wirenut

 

[electricpete]

I am pursuing upgrading the motor, based on large number of historical failures and the apparent overload condition. 100HP motor does not come in this frame size 365TP16 vertical hollow-shaft motor. A distributor told me there is no standard conversion kit for mounting the larger motor. It can certainly be done but it's a matter of how much work. Any suggestions on how to mount a larger motor are welcome... although I'm not expecting much help from folks who haven't seen the installation. I did hope that someone might have some ideas on WHY the motor can apparently draw more power than pump is capable of absorbing... based on comments from this thread I'm working toward the conclusion that our pump does not match our paperwork.

 

[jbartos]

Suggestion to the previous posting marked ///\\\:
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.\\\

 

[marktb]

The power absorbed as reported by the pump mnufacturer on its rating plate would normally exclude the shaft losses from motor to pump. It is also unlikely that the rating would include the efficiency of the motor. This means that power absorbed by the pump is nett of all external loses.

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.