Calculating power consumption from knowing suction & disch. pressures 3

[luciom]

If the suction & discharge pressures are read-off from gauges located on a running centrifugal pump suction & discharge how is the power consumption calculated assuming the pump is pumping water & flow rate is known ?

 

[tstead]

A couple ways:

1. Take an amp reading on the motor, get the motor performance (efficiency & power factor) to back into the pump bhp that way.

2. Get a performance curve, go to flow and match that to the horsepower. All pump curves should have the performance and power requirements plotted.

You can compare the answers from both different approaches and come up with a good answer.

Of course, you could also put a torque coupling between the pump and motor and measure the torque directly and (since the speed is known) calculate the power fairly easily. Problem is, torque couplings tend to be just a tad on the expensive side and not too many people have them just laying around.

 

[luciom]

Thanks Tstead but I made a small mistake in my post. I said I knew the flow but that was wrong. Actually I would like to calculate the flow knowing the suction & discharge pressures.

 

[tstead]

In that case, you need to know the impeller diameter if you want to get to it from the performance curve point of view.

You don't need to know the flowrate - or pressure readings if you go about it from the motor power point of view - you do need, however, reliable info on the motor (e.g. power factor & motor efficiencies).

Another trick, if you don't know the impeller diameter and if you are running on 1750 RPM speeds, run the pump to shutoff head, measure the discharge pressure, convert this to TDH in feet and take the square root. You'll get approximately the impeller diameter in inches.

 

[luciom]

Is it not possible to calculate the TDH knowing both suction & disch. pressures ? Would the difference between the two not be the TDH ?

 

[ArtR]

TDH = (Disch. psi - Suct. psi) x 2.31 ft/psi

Horsepower is approximately (GPM X TDH) / 4000 (for water)

tstead is right about the impeller diameter. If your pump runs at 3500 rpm, the head will be 4 times what it was at 1750 rpm (head is proportional to speed squared)

 

[tstead]

You can calculate the TDH as ArtR said above, that is not a problem (you do need to know the specific gravity though - that is a very crucial part of the equation just like suction and discharge pressure).

The horsepower eq. he gave is water horsepower - not brake horsepower. To find the brake horsepower using just flow and head, you NEED to know the efficiency of the pump at the operaitng point. Since there is no gauge you can mount on to the pump and measure the efficiency, you need to go about it through measuring or calculating the horsepower directly. Using that, you can back into the pump's efficiency if you really want that piece of information at that point (it would be uselss at that point).

By the way, H.I. standards say to use 3960 not 4000. Small difference, I know. Also, if you were to run through the derivation of that equation (it's not difficult at all), you'd see that it is actually 3956, but I digress.

Tim

 

[electricpete]

One other suggestion in addition to the excellent answers already posted:

If you use a Tach to determine the speed of the motor, then you can compute shaft horsepower from the motor's torque-speed curve. Multiply SHP by estimated pump efficiency to get fluid power. Divide fluid power by differential pressure to get volumetric flow rate (carry your units through). Surprisingly, The SG of the fluid is NOT required using this approach (or when using motor input power and efficiency to estimate shp). The units of volume flow rate times differential power convert to units of power without any dependence on SG.