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Chapter 8
Pumping, Storage, and Dual Water Systems
In theSIormetricsystem:
8.2.1 Power Requirements and Efficiencies
of Pumps
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MP = (Q,m ∕s)(1,000 kg∕m )(H,m)(kW∕101.97 m-kg∕s)
Work can be expressed as lifting a weight a certain vertical
MP = 9.8066(Q)(H)
(8.7b)
distance. It is usually defined in terms of ft-lb of work. Power
is work per unit of time. These statements are expressed
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where Q = water flow, m /s; H = head or lift, m; and MP =
mathematically by the following relationships:
power in metric unit, kW. Here 1 kW = 101.97 m-kg/s =
1,000 W; 1 W = 0.102 m-kg/s = 1 J/s; and 1 hp = 0.746 kW =
(8.5)
Work = WH
550 ft-lb/s.
Power = Work∕t
(8.6)
The total power input can be determined using the fol-
lowing equation:
where
INHP = input power in US customary unit = (V)(A)∕746
Work = work, ft-lb (m-kg)
(8.8a)
W = weight, lb (kg)
INMP = input power in metric unit = 0.001(V)(A)
H = height, ft (m)
(8.8b)
Power = power, ft-lb/s (m-kg/s)
t = time, s where V = voltage, V; and A = current, amp.
The wire-to-water efficiency is the efficiency of the total
If the water flow from a pump is converted to weight power input to produce water horsepower:
of water and multiplied by the vertical distance it is lifted,
E ww = (E pump )(E motor )(E ) (8.9)
o
the amount of work or power done can be represented by the
following equation: E ww = water horsepower∕input horsepower (8.10)
Power = Work∕t = WH∕t where
HP = (Q, gpm)(8.34 lb∕gal)(H,ft)(min ∕60 s)
E ww = wire-to-water efficiency
× (horsepower∕550 ft-lb∕s)
E pump = efficiency of the pump
HP = (Q)(H)∕3,957 (8.7a) E motor = efficiency of the motor driving the pump
where Q = water flow, gpm; H = head or lift, ft; and HP = E = efficiency of any other parts in the entire motor-
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power in US customary unit, hp. pump-wiring system
EXAMPLE 8.2 DETERMINATION OF PUMP POWER
Determine the water horsepower, break horsepower, and motor horsepower for a pump operating under the following conditions:
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water flow of 620 gpm (0.039 m ∕s = 39 L∕s) is to be pumped against a total head of 135 ft (41.15 m); the pump efficiency is 80%;
and the motor driving the pump has an efficiency of 90%.
Solution 1 (US Customary System):
WHP = water horsepower in US customary unit = (Q)(H)∕3,957
= (620) (135)∕3,957 = 21.2 hp (8.11)
(= 8.7a)
BHP = break horsepower in US customary unit = WHP∕E pump
= 21.2∕0.80 = 26.5 hp (8.12)
MHP = motor horsepower in US customary unit = BHP∕E motor
= 26.5∕0.90 = 29.4 hp (8.13)
where Q = water flow, gpm; and H = head or lift, ft.