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 K = 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. 180 m Elevation V Elevation = 150 m 20 km Pump 100 cm The measured performance, curves of the pump are shown below: 710 r/min 400 NPSH 25 20 15 NPSH, At 350 - 41.india. 10 300 38-in dia 99% Total head, ft 889 86% 250-35-in dia 84% 200 1250 bhp 1500 bhp 150 1000 bhp 100 0 28 4 12 16 20 24 U.S. gallons per minute x 1000

Answer 1

Crin data 2, = 180 72 = 150m Length of tipe £ = 20 km friction factor fa o old diacy fibe (d) = lorem = 1m Kexit Kent = 0.5 Now consider Manor class af flying 2 0&(0) Selain Bernaulli's equatice b +V2 +2 = P2 + V2² + 2q + Kent &? + Kenit u Kentu since there is open surface in both tank so Pi = R2 =0 ag - 1² V, 2 + 180 ag + 150 + 015 V² + 1 2² ལ། ༢༡ of cols 2150₂² *** 29 ag CÙ 30 = 2.54² ~42 ag ag Now applying continuity equation Q=A, V = Azų Q U₂ = B Similarly 02 1/02 4 ų = 40 - 40 s Equation ( put these value in

2 XL L x64 20 = 2.54 (4,2) ) ** 160² 30 2.5x 1602 291² 400² - 1602 པ་ཙེལ། 30 = agrz 30 agna 24 02 2911² Q = 151558772/ flow rate in gpm IQ = 246599.339pm ) (pump) power required is given by the formule p= fg g hf hf = flu? d = isin =9 6.52cm agd -0.965m B =0.012 4208103(21-27 2x91818 0.965 hf = 5734.82 wahom -15.558 chte 4 10.96572 5734.82 In HP U 52-37ml put this value of hf in equal (1) p = 998 x 9.81X15.558 X 5734.82 P = 87352051019 walt p= 1170.939 K HP (1 HP =746 watt In Horse Power

Net positive suction head a ha- hexit- haat But for this value of suction head from graph we need to H = 5734.82+ (180-150) Total head = 57648am [szka = 0.3m = 1729-446pt [More than 85 ft - So NPSH 39 ft