KWH Calculation for Pumps 1

[Naggs]

Dear Electrical Guru's

I am a Chemical Engineer . I do technical consulting for Industrial wastewater treatment. We often run into estimating the KWH/day for the pumps in our treatment plants. The procedure we follow to arrive at the KWH/day is as follows

Pump BHP = (Flow*psig)/(1714*Pump.efficiency)

Line Current, I = (BHP*746)/(1.732*P.F*E*MotorEfficiency)

where 'E" is the Line voltage (120,460,2300,4160 etc.. assumed based on the Motor HP)

where&quot;P.F&quot; is the power factor (0.9 for motors > 50 H.P and 0.75 for motors < 50 H.P)

KVA = (1.732*E*I)/1000

K.W = (KVA*PF)

KWH/day = (K.W * No of hours of Operaton per day)

With above procedure, we notice that the effect of power factor is zero . The reason is that we are dividing the Line current by P.F and then multiplying the KVA by P.F. So the net effect is zero , as we multiply and divide with the same factor, which kinda of says that, there is something wrong in the above formula's that i am following.

Can anyone advice on where i am going wrong. I definetely appreciate a response on the same

Thanx

Naggs

 

[meihm]

KW = BHP*746/EFF

No need to go through other steps since you don't care about line voltage,line current or KVA

 

[gigahertz]

Power is power. So the power required by the pump is the power supplied by the motor. The power supplied to the motor is the motor output power divided by the efficiency, and that is the power used. But, if the motor is not operated near full load, it will operate at a low power factor (.75 or even lower) and probably result in payment of a power factor penalty. Typically, the adjusted kW-Hrs for billing is (Measured kW-Hrs)*(0.95/PF) where PF is the average power factor over a billing cycle. Note that this applies to all of the energy consumed on this account over the billing cycle - not just the motor. This penalty usually applies when the average power factor over the billing period is below 0.95, and you don't get any credit for being above 0.95.

 

[lyledunn]

Power taken by motor is termed useful,ie used to drive the motor, the energy required to set up the magnetic field due to the reactive components is returned to the circuit and therefore not &quot;consumed&quot;. Only the useful component is charged for. However, utilities will require additional capacity for their plant, cables etc to deal with the &quot;additional&quot; current required for the wattless component. Thus the penalty mentioned by Sailfishc. Regards,

Lyledunn