Figure 1 Problem 2 (35 points) If a pump with the characteristic curve shown in
ID: 1765751 • Letter: F
Question
Figure 1
Problem 2 (35 points) If a pump with the characteristic curve shown in the accompanying graph (Figure 2) is used to pump water from the lower to the upper tank (water flows from Tank 2 to Tank 1 in this question) shown in Figure 1. Use information provided in Figure 1 and assume that diameter of pipe AB is 600 mm (DAB 600 mm) and diameter of pipe BC is 300 mm DBc 300 mm). Find the operation point of this system Calculate the power to operate the pump, if pump efficiency is 70% a. b. h (m) 45 30- N1100 rpm 15 0 0.15 0.30 0.45Explanation / Answer
Concept : Using the definition of operating point of the pump ( point where characteristic curve of pump and system curve collide)
System head hs = dh +hL
dh = h2-h1 that is the static elevation difference between inlet and outlet of the system.
dh = 100-85 =15
hL= major and minor head loss in the system
= .5(vBC)2/2g(head loss at entry to the pipe) + 4fLAB(vAB)2/2gdAB (head loss due to friction in pipe AB) + 4fLBC(vBC)2/2gdBC (head loss due to friction in pipe AB) +(vAB)2/2g ((head loss at exit of the pipe)
Using the continuity equation we can say that
ABCvBC= AABvAB
Hence ?/4*3002*vBC = ?/4*6002*vAB Hence vBC = 4vAB
From the figure it can be seen that the head in the pump is 15m Hence corresponding to head of 15m discharge is 0.42m3/s.
Using Q = Av we can obtain vAB and vBC
Hence vAB = 1.485 m/s and vBC = 5.942 m/s.
Putting the values of velocities obtained in the head loss equation to obtain hL = 1.083
Hence hs = 15+1.083 =16.083
a) Operating point of the system is the point where the characteristic curve of the pump and system curve coincide. Hence operating point discharge will be obtained corresponding to an operating head of 16.083. Hence discharge obtained is nearly .43m3/s.
Hence the operating point is (.43, 16.083)
b) Given pump efficiency = 70 %.
We know that efficiency is Water Horsepower/Brake Horsepower where BHP(Brake Horsepower) is the power required to operate the pump.
WHP = Q*H/3960 = .43*16.083/3960 = 1.75*10-3 HP
BHP = WHP/efficiency = 1.75*10-3*100/70 = 2.5*10-3 HP.
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