Pumping Calculations - Basic Mathematics for Water and Wastewater Operators - page 211

Solution:

whp
bhp

18 hp
20 hp

Pump efficiency = × 100 = × 100 = 90%

■ Example 14.39
Problem: What is the overall efficiency if an electric power equivalent to 30 hp is
supplied to the motor and 18 hp of work is accomplished?

Solution:

whp
bhp

18 hp
30 hp

Overallefficiency = × 100 = × 1000 0 = %

■ Example 14.40
Problem: Suppose that 30-kW power is supplied to a motor. If the brake horsepower
is 18 bhp, what is the efficiency of the motor?

Solution:

1hp
0.746 kW

30 kW × = 40.2 hp

= bhp
mhp

18 bhp
40.2 hp

Motorefficiency ×= ×= 100 100 44 7.%

■ Example 14.41
Problem: Suppose that 12-kW power is supplied to a motor. If the brake horsepower
is 10 bhp, what is the efficiency of the motor?

Solution:

1hp
0.746 kW

12 kW × = 16.08 hp

bhp
mhp

10 bhp
16.08 hp

Motorefficiency = × 100 = × 100 = 62 .%

■ Example 14.42
Problem: The motor horsepower required for a particular pumping job is 40 hp. If
your power cost is $0.09/kWh, what is the cost of operating the motor for 1 hr?

Solution:

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

= 40 hp × 0.746 kW/hp × $0.09/kWh

= $269 /hr