# 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
400
450
500
580
600
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Asif dar(AD) said:   2 years ago
Discharge = 1 m3/sec (converted),
dia = 1.2* (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,
Thus H=45m total.
Now, WHP for pump= wQH/735.
= (1000x9.81) x1 x 45/735,
= 600m exactly.

Note: Here, The 735 is the conversion factor between watt and HP as 1HP =735 watts.
(2)

Sunil said:   2 years ago

Santosh Kumar said:   2 years ago
Discharge = 1 m3/sec (converted),
dia = 1.2 √ (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
Thus H=45m total.
Now, WHP for pump= wQH/735.
= (1000x9.81) x1 x 45/735,
= 600m exactly.
Note: Here, he 735 is the conversion factor between watt and HP as 1HP =735 watt.

Endale Abebe said:   3 years ago
(1)

Nomore said:   3 years ago
@Dhanu.

How did you come up with the value of 75 on H.P of pump= WQH/75?

Deepti said:   5 years ago
Nice explanation thank you.

Hariom CHAURASIA said:   5 years ago
Thanks for explaining @Dhanu.

Srv said:   5 years ago
Thanks @Dhanu.

Pawan said:   5 years ago
Thanks @Dhanu.

Pawan said:   5 years ago
Thanks @Dhanu.