# 7.34 Pump Power and Operating Cost

Given a pump provides 428 feet of head, pumps 700 gpm of 60 deg F water. The efficiencies of the pump, motor, and VSD are as follows: 70.7%, 95%, 96%.

Find the brake horsepower, electrical horsepower, and the operating cost (on an 8000 hour/year basis) given a power cost of $0.12/kWhr.

    [1]:

    from fluids.units import *
    rho = 62.364*u.lb/u.ft**3
    head = 428*u.foot
    Q = 700*u.gal/u.min
    pump_efficiency = 0.707
    total_efficiency = .707*.95*.96
    dP = head*rho*u.gravity

    # unfortunately pint does not support currency, so we keep it dimensionless
    rate = 0.12/(u.kW*u.hour)

    bhp = Q*dP/pump_efficiency
    print('The brake power is %s' %(bhp.to(u.hp)))
    power = Q*dP/total_efficiency
    print('The power is %s' %(power.to(u.hp)))
    cost = power*rate
    cost = (cost*8000*u.hour).to_base_units()
    print('The cost is  $%s dollars/year' %(cost))

    The brake power is 107.05591309130916 horsepower
    The power is 117.38586961766357 horsepower
    The cost is  $84033.24278348622 dimensionless dollars/year

The value given in Crane is $83970.