Civil Engineering - Water Supply Engineering - Discussion
Discussion Forum : Water Supply Engineering - Section 6 (Q.No. 20)
20.
To pump water from a water reservoir 3 m deep and maximum water level at 135 m, a pump is installed to lift water up to R.L. 175 m at a constant rate of 36, 00, 000 litres per hour. If the length of the pipe is 1506 m and f = 0.01, ignoring other minor losses and assuming the economical diameter from Lea's formula D = 1.2 Q, the water horse power of the pump is
Discussion:
13 comments Page 2 of 2.
Dhanu said:
8 years ago
Discharge = 1 m3/sec (converted),
dia = 1.2 root (1) = 1.2 m,
Q = area * velocity.
velocity = 0.884 m/s.
Lift of pump = R.L of high level - R.L of low level + depth of sump well
175-135+3 = 43m,
Frictional loss of head in the rising main = 4flv^2/2gd.
4*0.01*1506*0.884^2 / (2*9.81*1.2) = 1.999m,
assume loss in bends = 0.3m,
TOTAL HEAD = 43+0.3+1.999 = 45.30 m.
H.P of pump = WQH/75.
1000*1*45.30/75 = 604 = 600 given.
dia = 1.2 root (1) = 1.2 m,
Q = area * velocity.
velocity = 0.884 m/s.
Lift of pump = R.L of high level - R.L of low level + depth of sump well
175-135+3 = 43m,
Frictional loss of head in the rising main = 4flv^2/2gd.
4*0.01*1506*0.884^2 / (2*9.81*1.2) = 1.999m,
assume loss in bends = 0.3m,
TOTAL HEAD = 43+0.3+1.999 = 45.30 m.
H.P of pump = WQH/75.
1000*1*45.30/75 = 604 = 600 given.
(1)
Sujoy basak said:
8 years ago
Give the workout of this problem.
Mayank said:
9 years ago
The answer is 580.
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