Different current values for motor pums 2

[parlitu]

In one instalation of two identical centrifugal pumps that feed the same collector, I observe that if bouth pumps are running the current is around 13.5A, but if only one is running and the other one is in stand-by, the current is 11.8A. The nominal current of motor is 12.5A. Can someone explain why this diference? Thank you!

 

[dpc]

13.5 A for each motor or total for the two motors?

"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg

 

[jraef]

Assuming you mean 13.5A each, I'd bet it's probably a hydraulics issue more than anything. In a centrifugal pump, motor load is related to flow. With all other things equal, flow is related to head pressure and restrictions. If for instance with both pumps running there is a siphon action that is created by having greater pressure to overcome the head, then flow will increase and both pumps will pull more current. That's just one scenario, there are an infinite number of possibilities. You should run this by a pump expert who can see the whole system.


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[electricpete]

Dpc's question is relevant.
jraef's comment is undoubtedly valid - the answer lies in studing the fluid system and pump behavior.

A tangentially related subject is variation of horsepower with flow for centrifugal pumps. Seems to be a recurring topic so I have added a FAQ: faq237-1543

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[parlitu]

13.5A each. The instalation is design to work with one pump in stand-by and only one pump running and we usualy do this way, but 2 weeks ago, bouth pumps where on, and I observe this diference between currents. Probably the current is bigger when bouth pumps are running due to a bigger pressure. Since I coudn't explaine myself why this is happening, I asked, and I am very curios to find out an anser. Thank you again!

 

[7anoter4]

I agree with jraef and electricpete but I'll try to create a certain scenario. The intention is to demonstrate that the pressure loss on a common pipe will decrease in the case of a single pump functioning so the required power will be less.
This will be an example:
The water supply system is provided with 2 pipe of 2" from outlet of each pump up to
a common pipe [collector] of 4".The length of 2" pipe will be 20 m and the common 4" will be 7 m [up to the tank].
The inner dia of 2" pipe is 52 mm and 4" pipe 102 mm.
If two pumps are functioning the total flow will be- let's say- 0.036 cub.m/sec.
The power required for a centrifugal pump [bkw] = Q[cub.m/sec]*H[m]*ro/102/etha
in metric units. Where Q = water flow [cub.m/sec] etha=pump efficiency let's say =0.8
H[head] =Total Static Head+velocityhead+friction+tank pressure
Total Static Head=3 m [let's say].
Friction head [according to Lang [see Siemens] for the common pipe =
Hfriction=(0.02+0.0018/SQRT(commspeed*(commcondia/1000)))*commspeed^2/2/9.81*conmonlength/(commcondia/1000)/1.25[m]
Commspeed=Q/common conduit area=0.036/(.102^2*pi()/4)[m/sec]=17 if both pumps are functioning and will be a half[8.5] in the single pump functioning case.
Then Hfriction in common pipe will be 17.2 m in the first case and 0.323 m in the second case.
The Hfriction for the each individual pump will be:
Hfrictionsingle=(0.02+0.0018/SQRT(singlespeed*(singlcondia/1000)))*singlespeed^2/2/9.81*singlength/(singlcondia/1000)/1.25
Where singlespeed will be 0.018/((singlecondia/1000)^2/4*pi())=8.48 m/sec
The individual pipe Hfrictionsingle=25.59 m
As the flow is direct proportional with motor rpm and as an induction motor speed is approximately constant the flow per a pump will be the same[in both cases] and in return the individual pipe Hfrictionsingle=25.59 m.
For the first case the total head will be [neglecting velocity head and tank pressure!]
3+17.2+25.59/2=33 m
The required power will be: 0.036*33/102/.8/2=7.27 kW per each pump.
If Voltage=400 V pf=0.85 motor efficiency=91.5%. then I=13.5 A.
In the second case H=3+.323+25.59=28.91 m.
The required power will be :0.018*28.91/102/.8= 6.377 kw
Then I=6.77*1000/400/SQRT(3)/0.85/.915=11.8 A

 

[electricpete]

Your example is a little contrived (as you stated it was) to exagerate the friction loss effects in the individual pump pipe legs.... and since we double the effective length of individual pump pipe legs with two pumps with two pumps it makes a big difference in your example.

Some factors that seem a little unrealistic:
1 - double the diameter in the common collector pipe section as compared to the individual pump legs... that gives four times the area for the common pipe leg as compared to the individual. 4 times the area for at most double the flow doesn't make sense. Especially considering the length of the individual pump legs is longer than the common header.

2 - All of the pressure loss in this system is associated iwth piping. In most applications at our plant the biggest component of pressure loss is some kind of heat exchanger load (would play the same role as the common header in your example to diminish the relative importance in those individual pipe legs).

I am sure none of what I said is news to 7anoter4 since he intentionally set up the problem to illustrate a point. I just wanted to mention in most cases this is far from realistic and the effect of losses in those individual pump legs tends to be a pretty small effect.

Even though it is not realistic for most systems, I still vote lps to 7anoter4 for bringing up another interesting aspect to consider.

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[7anoter4]

Thank you, electricpete, and I agree with you!

 

[7anoter4]

Since I agree with electripete I learned I have to change something. So the new hydraulic system will be:
10 m common pipe of 60 mm inner dia. 20 m individual pipe of 52 mm 3m Hstat and the results will be:
First case:
Q=0.0335 cub.m/sec [2 pumps together]
Common friction loss H= 21.11 m
Total H= 35.24 m and required Power [for both pumps] = 14.47 KW [7.235 KW per one]
I= 13.425 A
Q=0.01675 cub.m/sec
Common friction loss H= 5.49 m
Total H= 30.74 m and required Power =6.31
I= 11.71
Thank you again electricpete!

 

[rockman7892]

7anoter4

From the equation for pump hp above it appears that hp is a function of flow times the head pressure. On pump curves I have seen hp increases linerally with decreasing head pressure as flow increases. As head pressure decreases and flow increases I would expect the hp to stay roughly the same since one variable is increasing as the other is decreasing. Why then does the hp continue to increase as head decreases and flow increases?

 

[electricpete]

Fluid power is definitely proportional to product of DP and volumetric flow rate. However the hp curve you see on the pump curve is not fluid power, it is bhp which includs fluid power plus pump losses.

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[rockman7892]

electricpete

So you are saying that it is because of pump losses that we see the bhp increase linearly as one of the other two variables increase and decrease?

 

[electricpete]

I think a more direct answer to your first question is:
Fluid power = Volumetric Flowrate * DP

We know that if we increased flow all the way to DP=0 on pump curve, fluid power goes to 0. So the product of DP and flow is not necessarily constant even though flowrate is increasing and DP is decreasing.

That in itself answers your first question I believe.

But the pump curve does not show fluid power, it shows brakehorsepower. To come up with brake horsepower we need to multiply fluid power times efficiency.
Brakehorsepower = Efficiency * FluidPower = Efficieincy * Flow * DP

Efficiency is high near BEP and decreases as we move farther above or below BEP

Far above BEP, we have DP decreasing and efficiency decreasing as flow increaseses.

The final result of multiplying all three is the BHP curve. Varies depending on the flavor of "centrifugal" pump you are talking about as per FAQ linked above.


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