Well Pump/Plant Efficiencies

[lsceeng]

After performing an in-situ pump test for a vertical turbine well pump, I have a few remaining questions.

First let me explain my understanding of the fundamentals to well pump efficiency.

The formula for Water Horsepower (WHP), or power imparted to the liquid by the pump, is WHP=(Q*H)/(3960)

Input power must be measured off the electrical panel through measurement of voltages (V), amperages (I), and power factors (PF), the formula for Input Horsepower is (I*E*1.73*PF)/1000 and for HP its (I*E*2.32*PF)/1000

To find the Overall Plant Efficiency, the formula is Water Horsepower divided by Input Horsepower

At this point, the only two remaining variables left are pump efficiency and brake horsepower to the pump

I have found the formula for brake horsepower (BHP) to be BHP=(Q*H)/(3960*e) and the formula for pump efficiency to be e=WHP/BHP

Substitution of variables doesn’t get anywhere as all variables cancel out.

Here are my questions:

1) How do you formulaically solve for pump efficiency? Am I able to use the relationship of e_ope=e_pump * e_motor? Or are there other efficiencies Im missing?

2) At zero flow (dead head) the water horsepower, and efficiencies are equal to zero, therefore BHP is indeterminate (0 divided by 0). However this cant be true as the motor is still doing work (maintaining dead head pressure) How am I able to solve for BHP?

I have one more question not related to formulas

How is the velocity head, water stage elevation, and total suction head related to the total dynamic head? I have always concluded the TDH is the sum of the elevation head, discharge pressure, and minor losses (fittings and column friction losses)

Thanks for your help in advance

 

[BigInch]

Substitution of variables doesn't get anywhere as all variables cancel out.

You don't need to go anywhere else.
You already have formulas for both WHP and BHP.
And "Ep" = pump efficiency, Q and H are given values, "Ep" is typically obtained from reading the pump curve.

If you can't read Pe from the pump curve, you must measure the electrical power consumed at the motor and you must include the motor efficiency "Em", read from the motor efficiency curve, and calculate the power delivered to the pump shaft, BHP. Now with the known values for BHP, Q and H, you can calculate WHP, and using Ep = WHP/BHP solve for Ep.

Only the pump efficiency and motor efficiency are needed to describe pump driven directly by a motor. If you have a belt or gearbox, you must include Eb or Eg. If you want to to calculate power at the electrical meter, you must also include electrical line losses between the motor and pump, with an Electrical Line Efficiency term "Ee". If you have a VFD, you must include the efficiency of the VFD, Evfd.

Be aware that motor efficiency is not constant and varies nonlinerly with the ratio Motor_load / motors_rated_load.
VFD efficiency varies with the VFD_load / VFD_rated_load.

If you're not at usual temperatures and atmospheric pressures, motors and VFD could require additonal correction factors. VFD installations may need additonal cooling using a cooling fan. Cooling fan power draw may not be included, in which case that power must also be added to the input power.

Technically everything in the sketch has an efficiency. Sometimes they are small enough to ignore, sometimes not. If your electrical cables are short, fat, cold and ugly, efficiency might be ignored. If they are tall, slender, hot and pretty, as we all know, its going to cost more money.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch

http://virtualpipeline.spaces.live.com/

 

[ccfowler]

lsceeng,

As always, BigInch has provided excellent information for you, but I am a bit confused between the title of your posting and the content of the posting. What is the problem that you are trying to resolve? Are you trying to evaluate the performance of an old pump, a new pump, a pump & motor combination, the pumping station as a whole, etc.?

The more clearly you are able to state your problem, tne better the chances are for you to get the advice that you really need.

Regarding your question #2, the efficiency is indeed zero because no useful work is being done (no flow). Motor efficiency is not zero because it is effectively powering what has become a shaft driven water heater.