# Civil Engineering - Water Supply Engineering - Discussion

### Discussion :: 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

 [A]. 400 [B]. 450 [C]. 500 [D]. 580 [E]. 600

Explanation:

No answer description available for this question.

 Mayank said: (Nov 30, 2016) The answer is 580.

 Sujoy Basak said: (Jun 24, 2017) Give the workout of this problem.

 Dhanu said: (Sep 17, 2017) 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.

 Pawan said: (Mar 19, 2019) Thanks @Dhanu.

 Pawan said: (May 24, 2019) Thanks @Dhanu.

 Srv said: (Jun 20, 2019) Thanks @Dhanu.

 Hariom Chaurasia said: (Jul 22, 2019) Thanks for explaining @Dhanu.

 Deepti said: (Dec 4, 2019) Nice explanation thank you.

 Nomore said: (Apr 11, 2021) @Dhanu. How did you come up with the value of 75 on H.P of pump= WQH/75?

 Endale Abebe said: (Sep 23, 2021) Please explain the formula P=WQH/75.