Problems and Solutions to Chapter 10: Water Supply

Chapter 10 is worth 35 points (5 points per problem)

Problem 10.3
For the centrifugal pump shown in Fig. 10.13 and an impeller diameter of 6-1/8 in., determine the head, efficiency, and the brake horsepower for a discharge of 400 gpm. Should this pump be recommended for these flow conditions?

Solution to 10.3

h = 130 ft +/- 5 ft
efficiency = 80.5%
BHP = 15.5 hp

Yes, because efficiency >70% (see page 178, 70-80% efficiency recommended)

 

Problem 10.4
For the turbine pump in Fig. 10.14 and an impeller speed of 2400 rpm, determine the recommended operating range of head and discharge.

Solution to 10.4

range of head = 105 - 110 ft (75 - 110 ft)
range of discharge = 1900 - 2400 gpm (1800 - 3250 gpm)

Note: don't go to the right on curves because net positive suction head (NPSH) becomes too large and you don't want the NPSH to be half the total head.

 

Problem 10.5
Determine the WHP and BHP for a pump if the flow rate is 500 gpm, the total lift is 100 ft, and the pump efficiency is 75 percent.

Solution to 10.5

Given:
Q = 500 gpm
H = 100 ft
Ep = 0.75

WHP = Q(gpm)*H(ft) / 3960 = (500 gpm)(100 ft) / 3960 = 12.6 hp
BHP = WHP / Ep = 12.6 hp / 0.75 = 16.8 hp

 

Problem 10.6
For the propeller pump in Fig. 10.15 and impeller B, determine the head, efficiency, and power for a discharge of 1200 gpm.

Solution to 10.6

H = 16 ft
BHP = 5.9 +/- 0.1 hp
Eff. = 80%

 

Problem 10.7
An irrigator is pumping 300 gpm with a total lift of 60 ft. The pump is driven by an electric motor with an efficiency of 90 percent, and the pump is 70 percent efficient. How many kilowatts are required if 1 hp = 0.746 kw. If electrical energy costs $0.10 per kw-h, how much will the energy cost for a 24-h period?

Solution to 10.7

Given:
Q = 300 gpm
H = 60 ft
Em = 0.90
Ep = 0.70
1 hp = 0.746 kW
electrical energy costs = $0.10/kWh
adjustment factor for continuous operation of electric motor = 1.0 (see Table 10.4, p. 183)

WHP = Q(gpm)*H / 3960 = (300 gpm)(60 ft) / 3960 = 4.55 hp
BHP = WHP / Ep = 4.55 hp / 0.70 = 6.5 hp
Input Power to motor = BHP / Em = 6.5 hp / 0.9 = 7.2 hp* 1.0 = 7.2 hp
Convert from hp to kW: 7.2 hp x 0.746 kW / hp = 5.4 kW
Cost = (5.4 kW)($0.10 / kWh)(24h) = $12.92

(NOTE: To estimate the maximum horsepower rating of the unit to purchase, must adjust engine horsepower requirement for continuous operation. Use Table 10.4 to determine adjustment factor. For continuously running electric motor, adjustment factor is 1.0. Therefore, BHP * Adj Factor = adjusted HP = 6.5 hp * 1 = 6.5 hp)

 

Problem 10.8
The irrigator in Problem 10.7 decides to add a sprinkler system that requires a pressure of 50 psi at the pump. Determine the energy required and cost for a 24-h period.

Solution to 10.8

Given (from Problem 10.7):
Q = 300 gpm
H = 60 ft
Em = 0.90
Ep = 0.70
1 hp = 0.746 kW
electrical energy costs = $0.10/kWh
adjustment factor for continuous operation of electric motor = 1.0 (see Table 10.4, p. 183)

Original head = 60 ft
New head = 60 ft + (50 psi x 2.31 ft/psi) = 175.5 ft

WHP = Q(gpm)*H / 3960 = (300 gpm)(175.5 ft) / 3960 = 13.3 hp
BHP = WHP / Ep = 13.3 hp / 0.70 = 19.0 hp * 1.0 = 19.0 hp
Input power to motor = BHP / Em = 19.0 hp / 0.9 = 21.1 hp x 0.746 kW / hp = 15.7 kW

Cost = (15.7 kW)($0.10/kWh)(24 h) = $37.78

(NOTE: To estimate the maximum horsepower rating of the unit to purchase, must adjust engine horsepower requirement for continuous operation. Use Table 10.4 to determine adjustment factor. For continuously running electric motor, adjustment factor is 1.0. Therefore, BHP * Adj Factor = adjusted HP = 19.0 hp * 1 = 19.0 hp)

 

Problem 10.9
Using the data from Problem 10.7, estimate the cost of pumping for 24 h with a diesel engine. Assume diesel engines average 15 hp-h per gallon of fuel and fuel costs $1.00 per gallon.

Solution to 10.9

Given (from Problem 10.7):
Q = 300 gpm
H = 60 ft
Ep = 0.70
1 hp = 0.746 kW
diesel fuel costs = $1.00/gallon
adjustment factor for continuous operation of diesel engine = 1.25 (see Table 10.4, p. 183)

WHP = 4.55 hp
BHP = 6.5 hp
Adjust for continuously operating diesel engine: BHP * adjustment factor = 6.5 hp * 1.25 = 8.1 hp

Cost = (8.1 hp)($1.00 / gal)(gal / 15 hp-h)(24 hr) = $12.96 or $13.00