Question

A 38-in pump is installed for the system below. The pipe (100-cm diameter) has a friction factor of 0.012 and a total length of 20 km. Given that Ke = 0.5 (pipe entrance) and Ke = 1.0 (pipe exit). For water at 80 degrees Fahrenheit, estimate the flow rate (in gpm), the power required (in horsepower, HP), and the net positive suction head required (in ft) to avoid cavitation for the system when the pump running. Elevation = 180 m Elevation = 150 m 20 km Pump 100 cm The measured performance curves of the pump are shown below: n = 710 r/min 400 25 NPSH 20 350 -41%-in dia. NPSH, ft 15 72% 80% %09 -84% %98- 88% 10 300 38-in dia. 89% Total head, ft 250 35-in dia. -88% 86% 84% 200 1500 bhp 150 1000 bhp 1250 bhp 100 0 4 8 28 12 16 20 24 U.S. gallons per minute x 1000

Answer 1

Civan data Z, = 180 Z2 = 150m Length of pipe t = 20 km friction factor f- 0.012 dia of tipe (d) = 100 cm = 1m Kexit Kent = 0.5 0 & (2) ag * 9 Nowo cessides Moues loss of flying Bernecellis equatia b +V12 +2 = P + V2 ² + 2q + Kent V & Kenit e Pg 29 since there is open surface in both tank so P = 22 =0 Pg 29 - V ² V2 + 180 29 + 150 + 0.5&² + 1 ₂2 ag 29 23 2,5 V₂2 42 +30 29 २१ 30 = 2.54² ~42 29 ag Now applying continuity equation Q=AV, =A₂ų 2 Similarly Uz = Q IW Vi = 7/༡༧ 4 ų = 4Q t 40 T U equation o put these value in

2 *49)** **009) 30€ 2.5X 1602 1602 Tipeg agn² 4002 - 1602 30 dgn? agree 2 30 24 Q2 12. Q = 151558 m3/ flow rate in gem IQ = 246599.339pm 29 d =96.52cm = 0.965m power required is given by the formula p= fg a hf hf = fL² d = 38 in (pump) = 0.0 12x20x103 (21-27² VE B 2x9.818 0.965 A m = 151558 hf = 5734.82 would € 10.96572 this 5734.82 In HP u=21-27m/s put this value of he in equar (6) hf p = 998 x 9.81X15:558 X 5734:52 p = 87352051019 Walt P= 1170.939 K HP [ <HP +746 watt] In Horse Power

of Net positive suction head - ha herit - hart But for this value suction head from graph we need to Total head H = 5734.82 + (180-150) 5764 82m 0.3m [ s&t=0 1729:44 6pt = 39 ft [More than 25 85ft So NPSH