Pumping Calculations - Basic Mathematics for Water and Wastewater Operators - page 207

■ Example 14.20
Problem: Given that 25-kW power is supplied to a motor and the brake horsepower
is 30 hp, what is the efficiency of the motor?
Solution:

25 kW
0.746 kW/hp

= 33.5 hp

Brake horsepower
Motor horsepower

30 hp
33.5 hp

Percentmotor effic iency = ×100 = × 100 = 89.%

■ Example 14.21
Problem: A pump is discharging 1000 gpm against a head of 50 ft. The wire-to-
water (overall) efficiency is 75%. If the cost of power is $0.036/kWh, what is the cost
of the power consumed during a run of 100 hr?
Solution:

bhp
Motorefficiency

Flow (gpm)Head (ft × )

mhp = =

3960 × Pump efficiency × Motorefficiency

100 0 gpm × 50ft

= = 16.8 hp

3960 × 0.75

Cost ($/hr) =× × mhp 0.746 kW/hpCost($/kWh)

= 16 .. 8hp × 0.746 kW/hp × $0.036/kWh

= $0.45/hr

Total cost = $0.45/hr × 100 hr = $45.00

■ Example 14.22
Problem: What is the horsepower for a motor that is rated at 60 amperes and 440 volts?
Solution:

Volts Amperes
746 watts/hp

× 440 ×

746 watts/hp

50 amperes

hp = = = 29.5 hp

■ Example 14.23
Problem: Determine the power factor for a system that used 4867 watts and pulls 11
amperes at 440 volts.
Solution:

Watts
VoltsAmperes

4867 watts

Powerfactor = = = 1

× 4 40 volts × 11 amperes