dead head pump 4

[ivoryhorseman]

how to calculate the output power of a dead head pump; since we donot have a flow at dead head condition and the head is maximum. I do have the input power which is close to 2.7KW.
what considerations of the loss should i take(is it the impeller losses and the motor losses along with the LRP)
sombody kindly help......

 

[hydtools]

All the input power is being wasted. There is no useful output.

Ted

 

[1gibson]

Efficiency is 0% with no flow. I think by "output power" you mean "water horsepower."

http://www.engineeringtoolbox.com/pumping-water-horsepower-d_753.html

Pwhp = q h sg / 3960 (1)

where

Pwhp = water horsepower (hp)

q = flow (gal/min)

h = head (ft)

sg = specific gravity

 

[Artisi]

As the OP is talking in kW's the calculation for water power in metric terms is

l/s x h (metres)x sg (1 for H2O)/ 102 = kW.

However, the OP has asked for output power at CV which is a meaningless question as it is presented, the enquiry needs to be clarified.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[Pumpsonly]

There is no hydraulic power output at shut valve condition.
Power in put will be converted into heat energy but not all the 2.7kw of electrical power to the motor.
Need to minus the losses in the motor and mechanical looses in the pump bearings.