Pump electric motor load in process plant
Pump electric motor load in process plant
(OP)
Hello,
There is a process plant with some pumps pumping fluid out of the plant
I use 2545*P*((1-eff)/eff).FUM.FLM classic pump motor heat dissipation formula from ASHRAE fundamentals handbook for these.
Now there are a lot of other pumps in this plant which only carry fluid between H/Xs, evaporators, indoor tanks etc. and fluid they pump is just transferred to the next process within the plant. Both pump and motor is indoors.
Do I use 2545*(P/eff)*FUM*FLM formula in this case?
This equation is giving me HUGE load, but hey- if it is the correct equation so be it.
Thanks!
eff=pump motor's efficiency
FUM=motor use factor
FLM=motor load factor
There is a process plant with some pumps pumping fluid out of the plant
I use 2545*P*((1-eff)/eff).FUM.FLM classic pump motor heat dissipation formula from ASHRAE fundamentals handbook for these.
Now there are a lot of other pumps in this plant which only carry fluid between H/Xs, evaporators, indoor tanks etc. and fluid they pump is just transferred to the next process within the plant. Both pump and motor is indoors.
Do I use 2545*(P/eff)*FUM*FLM formula in this case?
This equation is giving me HUGE load, but hey- if it is the correct equation so be it.
Thanks!
eff=pump motor's efficiency
FUM=motor use factor
FLM=motor load factor





RE: Pump electric motor load in process plant
I'm more concerned on how you get the efficiency since that depends on operation.
RE: Pump electric motor load in process plant
It seems like this question has been answered before:
http://www.eng-tips.com/viewthread.cfm?qid=301417
http://www.eng-tips.com/viewthread.cfm?qid=59911
The correct equation to use in my case is
2545*(P/eff)*FUM*FLM
P=rated horsepower of pump's motor
eff=pump motor's efficiency
FUM=motor use factor
FLM=motor load factor