Question

08 % Problem 4 (25 points) 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 — 417-in dia.com NPSH, ft 15 %od %98 %88 10 300 38-in dia. %68 Total head, ft 250 — 35-in dia. 86% 84% 200 K12 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

ANSWER
G Hege Given 71=180 79=150 length & pipe L-a0km Friction factor =0.019 ke=0.5 Kenit-1. Now consider minor loss applying Bernoullis Exualian) 0 P2 Pg 29 + Zetkent us + Kenit +150 +05 29. +180-150 = ( since there is open surface in lank so R =B=e] +180 = +os - 2 30 =25 ano Now applying certinuity Efuation 29 +30 - 2.5 &= A Ng = A2V2 44 4&

2 simillasly Nga V242 Put these values in Elciaton then we get Q. 30=25%) ya 30-215 x 1602 2972 1602 71999 - 4092 1602 29² 2972 30 - 2402 29712 2402 2x9.8149.86 30 - 5303.5= 2402- 5803,5 24 Q2= 241-81 15.95m)s Flow rate in gipum (=246997-333pm. powest segured is given by the formula pa feabfco flu² ago hp =

fo hupa 0-VRO X103 (21-7) 2 2x9.81X0-465 1630 6:22 d = 38 in (Pup) 9649cm -)076 1076510] [ Ven 15.458 4-60-966.p = 9181613] 68579.096 he -- 18.7333 hp-=4734-89 put the value in E91 we get npt value. P-99849-81X15.65885734-82 p=873520510.9 watt pe 1140.939 kHp [1. Hp -- 746 untt] Net positive action head = ha-henit-henk. But for the value Sechon head from graph we need to Tolol head +3=5734-22 +(180-150) H=5734.82 +-30 5764.821 H - 1799446 PE (1 ft=0:31] (more than 25ft). :: SO, NPSH = 394_