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
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
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
Solution to 10.6
H = 16 ft
BHP = 5.9 +/- 0.1 hp
Eff. = 80%
Problem 10.7
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
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
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