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Electricity demand for the pumps
2

Electricity demand for the pumps

Electricity demand for the pumps

(OP)
Hi Folks,
I am trying to calculate the electrical power demand for a process plant. Where pumps and agitator motors are main consumers for the electricity.
Is there some rule of thumb to calculate the electricity demand for the pumps (based on flowrate) and motors.
Thanks in davnce for your input on the issue.
Best Regards
Zahid

RE: Electricity demand for the pumps

Yes

Power = flowrate * density * pressure differential / efficiency of 0.7

Now divide by motor efficiency = 0.9
and add another 10% for electric losses

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

**********************
"Being GREEN isn't easy" ..Kermit
http://www.youtube.com/watch?v=hpiIWMWWVco

http://vir

RE: Electricity demand for the pumps

Be careful in using consistent units.

 

RE: Electricity demand for the pumps

Actually I don't think you need the density.

Power = Volumetric flowrate * pressure differential / efficiency

A quick unit check.
If volume flow rate in m^3/sec and dp in N/m^2 then the product will be in Joule per second as required.  If you put density in there you will have something different than power.
 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: Electricity demand for the pumps

And efficiency would be product of efficiency of pump and motor and drive as mentioned above.  The efficiency (losses) are the hardest to figure.. all of them will change with operating point.

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Eng-tips forums: The best place on the web for engineering discussions.

RE: Electricity demand for the pumps

Two points:

(a) I repeat, one must be careful with the units' consistency.
To estimate the power input to pumps P = g.ρ.Q.H/η .
For example, given
  Q (flow rate) in m3/s,
  ρ (density) in kg/m3,
  H (head) in m,
  η (fractional efficiency) dimensionless, and assuming
  g = 9.8 m/s2,
P, kW =ρ.Q.H/102.η

(b) Regarding efficiencies, zamu2010 asked for a ROT, and BigInch submitted one.

RE: Electricity demand for the pumps

As the major unknown is the pump efficiency, give the problem to a good pump guy who would be able to estimate the power requirement for each unit as he / she would have a good feel for the pump eff. based on its size, type and application etc, otherwise use the ROT BigInch offered which is very conservative.  

RE: Electricity demand for the pumps

25362 - What is the purpose of "I repeat"... do you disagree with my comments in some way ?

I have provided consistent units including differential pressure.  You introduced head and density as alternative to differential pressure - also very valid.   My only reason for mentioning units was to demonstrate that the previously posted "flowrate * density * pressure differential / efficiency" is not correct (I assumed flow rate represents volumetric.... but also not correct if flow rate was mass, since in that case density would have to go in the denominator).

BigInch knows more about pumps than I'll ever know. I was just pointing out what seems to be a typo.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: Electricity demand for the pumps

Opps.  I must have had excess "pressure" on the brain and forgot my "head".

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

**********************
"Being GREEN isn't easy" ..Kermit
http://www.youtube.com/watch?v=hpiIWMWWVco

http://vir

RE: Electricity demand for the pumps

Before I make any rash statements that I might have to eat, what about the power factor at the plant and power factor correction or lack thereof?

rmw

RE: Electricity demand for the pumps

Well, as the sparkies explain it to me, (and if I remember correctly) PF is a separate phenomenon that doesn't actually affect the watts registered at the meter.  Its the vector sum of the  active and reactive power components in the motor, which can upset the grid when severe, which is why power companies charge extra for PFs over a value of 0.00005 (or something like that).  But the PF bill component is a penalty for screwing up the grid's karma and is very much separate from the actual bill for kWh, which PF does not affect at all.
  
OK.  Sparkies - fire at will.

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

**********************
"Being GREEN isn't easy" ..Kermit
http://www.youtube.com/watch?v=hpiIWMWWVco

http://vir

RE: Electricity demand for the pumps

It may not affect the watts at the meter but it does affect what the pump motors do with the power they receive, especially since the motors need the reactive power component of that vector diagram for the magnetizing current needed to create the rotaing magnetic field that runs the motors.  I think reactive power monkeys with the voltage (in a way that I have NEVER understood) and that can affect a motor's current demand which then affects I^2R losses in the electrical system from wherever the vars are produced to where they are used.

Bring it on distinguished Sparkies.  We are like sponges here.

I may go over to the electrical motor forum and start a thread that says rmw and biginch are all yours over in the pump engineering forum-have at 'em.

rmw

RE: Electricity demand for the pumps

Not bad BigInch, not bad at all.

Think of it as the kW demand is determined by the load and the efficiency of the motor and the kVAR demand (the reactive bit) is something inherent in the motor caused by the design of the magnetic circuit in the machine. The reactive power doesn't cross the boundary from the electrical world into the mechanical world so you guys don't have to worry about it other than to pay the 'leccy bill. At an ultra-simplistic level it means that the current drawn by the motor is higher than you would expect based purely on the kW demanded by the load, and we have to size conductors and equipment to accomodate this higher current.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: Electricity demand for the pumps

So you're saying PF does register on the meter as a higher current draw. I certainly didn't have that clear before.  I thought it was more like an extra billing component that was "assessed" rather than actually registered on the meter.  Now that I remember better, I think somebody told me it did't cause higher power, probably meaning at the pump only, then I assumed he meant at the meter too.  I was thinking Watts = V*I and when amps went higher, V dipped and Watts stayed the same, so there would be no change at the meter.  Oh well. Thanks for clearing that up!

Actually these sparky guys never give you the same story twice.  Something to do with job security I think.  It keeps me coming back asking the same dumb questions over and over Opps. I mean, Thank You Scotty!!

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

**********************
"Being GREEN isn't easy" ..Kermit
http://www.youtube.com/watch?v=hpiIWMWWVco

http://vir

RE: Electricity demand for the pumps


To electricpete, my message intended to reinforce yours.

RE: Electricity demand for the pumps

Yes, a non-unity PF does register on an ammeter as increased current draw compared with the same power delivered at unity PF. It does not register on a kWh meter because these are specifically designed to respond only to the in-phase component of current and to ignore the quadrature component. Similarly a kVARh meter ignores the in-phase component of current and only responds to the quadrature component. In order for either type of meter to work it requires a voltage reference against which it resolves the in-phase and quadrature components. A simple ammeter doesn't have this capability so it just reports the composite current without any distinction between in-phase and quadrature. That's the reason why an ammeter is a lousy tool to measure power or energy consumption.
 
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: Electricity demand for the pumps

"It does not register on a kWh meter".  So, PF does not cause any change to the basic electricy "power" bill, but might be responsible for additonal surcharges.

Well ...  That's what I was thinking all along.  I said it's, "a phenomenon that doesn't actually affect the watts registered at the meter".  Why did I get confused this time?

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

**********************
"Being GREEN isn't easy" ..Kermit
http://www.youtube.com/watch?v=hpiIWMWWVco

http://vir

RE: Electricity demand for the pumps

Ok, purists will argue that there are some additional losses in the wiring and system components downstream of the meter due to the increased current at a non-unity PF - and they would be right - but to a first approximation PF does not affect the active power recorded at the meter. For industrial customers a non-unity PF does normally incur other charges as you say.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: Electricity demand for the pumps

YES!

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying."  Tony Hayward CEO, BP

**********************
"Being GREEN isn't easy" ..Kermit
http://www.youtube.com/watch?v=hpiIWMWWVco

http://vir

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