12RJHO 3 stage vertical turbine pump questions

[Doggie]

Hello all: I am new to this forum. I am no pump expert and am just beginning a project involving nothing but pumps. I have so many questions and would greatly appreciate some good advice. I am trying to figure out what our pumps in the cold well are actually putting out (GPM). Sounds easy but there are no flow meters, just pressure at the outlet. I have the pump curves for these vertical turbine (centrifugal) pumps and many equations set up in my math solving app, we've measured currect on each leg of the 75 hp motors that are coupled with our pumps, and I've been trying to back out head and flow with the data. What would be the most straight forward approach for solving this problem? I have other data to share that may or may not help in this problem. Also is BHP the same thing as Power(pump,shaft)? Or in these equations which is BHP ->

W(pump,shaft) = W(pump,u))/eff(pump)

W(elect) = W(pump,u)/ ( eff(pump)*eff(motor) ) ,

I realize these are power and not work formulas please excuse the fact that there are no dots over the "W" terms. Thanks for your help,

Jason (Doggie)

 

[fredb]

Hello Jason,

Just approximately now,

The hydraulic or water horsepower delivered by the pump is:

wHp = Q x H x (sp. gr.)/3960

where: Q = gals./min., H = head in feet

The shaft power is: P (watts) = 9797 x Q x H x (sp. gr.)

The brake horsepower is:

BHp = Q x H x (sp. gr.)/3960 x Pump Efficiency

and pump efficiency is: wHp/BHp

you can get an approximation of flow by converting your gage reading to head, add the suction lift in feet and using this number, read the flow from the appropriate pump capacity curve.

If the pumps are turbine type rather than centrifugal there are other things that might have to be considered.

 

[Doggie]

fredb -

Thank You for responding and helping me out! I guess I was just confused on the equations you gave me b/c I was using some I found in my old thermo book that are as follows:

Wpump,shaft = Wpump,u / eff(pump) = rho*V*g*h(pump,u) / eff(pump)

and

Welect = Wpump,u / (eff(pump)*eff(motor)) but with these I didn't see the Brake HP.

I am just trying to solve a system of equations b/c I do have online pump curves (3 stages), don't have a suction pressure meter. So I meterd the current going to the motor for one of the pumps and calculated the "Welect" term knowing the eff of the motor, power factor, etc.

Then I setup these equations: Assuming S.G. = 1
Nm = eff motor
Np = eff pump
h = TDH feet
Pd = gauge pressure at discharge + 14.7 PSI atmosphere, since the pump curves are all in Absolute and 2.31 to convert PSI to Feet
Ps = unknown suction side of pump

P[elect]:=evalf(sqrt(3)*V[L]*Iline*pf):
---- metered value of current = 61 amps, known pf, V[L]
> eq1:=Wpump=P[elect]*Nm*Np/1000:
---Divide by 1000, so the values are in Kw----
> eq2:=3*h=(Pd-3*Ps)*2.31:
---- 3 stage pump, TDH eqn-------------------
> eq3:=BHp=(V*h)/(3960*Np):
------brake hp equation given to me-----units still Ok?
> eq4:=2840=1760*sqrt(V)/(h^(3/4)):
---relating known specific speed of 2840 provided in online pump curve and N =1760 = pump speed, RPM
> eq5:=Np=Wpump/BHp:

So, here's my problem: 5 eqns and 6 unknowns
What assumption do I make about the system? If I assume a pump effeciency these equations solve fine but when I go and plug into the online pump curves (inputs can be either V (flow) or H (head) then the values for BHp, Flow, and Head are all out of sink with what my equations are telling me. What am I doing wrong? Bad equations / setup? Any help or suggestions are greatly appreciated!!

Thanks,

Jason (doggie)

 

[vanstoja]

A possible explanation for your mismatch is that your pump is not operating on the system resistance curve that produces best efficiency conditions for the pump. In using the specific speed equation without knowing either flow or head you are assuming that you're operating at best efficiency conditions with impellers designed for 1760 or 1800 RPM synchronous speed. Actual system piping design and other factors may have you operating well away from the best efficiency system resistance condition which shows up in your motor load current values. Another problem with specific speed usage is that a modified impeller, if originally designed by specific speed methods (eg. Stepanoff's method), may produce a different pseudo specific speed when retested values of flow and head are used for the new specific speed rating. It is helful to know the origins of impeller designs eg., 1st, 2nd ,3rd ..etc. generation for some of the oldtimers in service. Vanstoja

 

[fredb]

Hello Doggie

Sorry for not getting back to you sooner, but I was away for extended holiday weekend.

Back to your problem:
I don't think you can make an assumption for pump efficiency, but you could make one for motor eff. Use 87% for old motor use 93% for new one.

In you first equation you have two unknowns Wpump and Pump Eff.

In your second equation: TDH = hd + hs (for a suction lift)
and:
hd = gage reading + (velocity head, I would neglect this)
hs = static suction lift - velocity head + friction head in
suction line
if you again neglect the velocity head, then you can solve for TDH.

In you third equation you have three unknowns: BHp, Flow, and Pump Eff.

In your fourth equation you have one unknown: Flow.

In your fifth equation you have three unknowns: Pump Eff., Wpump, and BHp.

However if you count up the unique unknowns, there are only four: Pump Eff., BHp, Flow and Wpump.

 

[Doggie]

fredb -

Once again, Thanks for your response. After I posted my last note I noticed this problem with my equations. I was however assuming around .9 eff motor and we did back out about .84 - .88 power factor at the pumps. But I do have some good news on this situation. About 1 foot or so out of the pump there was an additonal pressure gauge! They were painted over but after the paint was removed they could be read. So, I have a control boundary problem. I can use the good old energy balance (bernoulli) to back out flow b/c I can read at the pump a P1 and P2 and model the system in between these two gauges. I can do this problem but just need to find some k values (total my losses) for a 60 % open 8 inch butterfly valve, 8 inch check valve, 8 x 6 90 degree reducing elbow. I can email (if you care to help me out) a sketch of my "system" in an autocad .dwg file b/c me description isn't very descriptive. I'll figure out the kind of pipe, exact dimensions, heights, flanges, etc. in order to do a decent calculation. Once again thanks for the good feedback and help!

- Jason (doggie)