Civil Engineering - Water Supply Engineering - Discussion
Discussion Forum : Water Supply Engineering - Section 8 (Q.No. 13)
13.
For 3.25 x 10-2 cumecs discharging from a well, a pump is installed to lift water against a total head of 30 m. The minimum required horse power, is
Discussion:
10 comments Page 1 of 1.
Arnold Schwarzenegger said:
4 years ago
Here, P = WQH/75n.
(3)
Muhammad Junaid Michaelson said:
5 years ago
@Gizachew Dirsha.
The efficiency of a Pump can be taken or assumed as 65%.
The efficiency of a Pump can be taken or assumed as 65%.
(1)
Gizachew dirsha said:
6 years ago
What is the limitation of assuming efficiency? Why not use 75%?
Mani said:
7 years ago
P = WQH/75.
W= Density of Fluid = 1000 kg/m^3.
P=1000 * 3.25 * 10^-2 * 30/75 = 13,
Assume efficiency of 65%,
Required Power = 13/0.65 = 20 HP.
W= Density of Fluid = 1000 kg/m^3.
P=1000 * 3.25 * 10^-2 * 30/75 = 13,
Assume efficiency of 65%,
Required Power = 13/0.65 = 20 HP.
(1)
Starlord said:
7 years ago
W= unit weight of water take 1000 kg per cubic meter.
Starlord said:
7 years ago
@Bablu.
W is unit weight of water.
W is unit weight of water.
Bablu basak said:
8 years ago
What is W, why taken of its value 10?
Priya said:
8 years ago
I can't understand getting the value of W=10.
Please explain.
Please explain.
Sadashiva said:
8 years ago
P=WQH/75 * efficiency of the pump.
Assume maximum efficiency of the pump=65%.
Then,p=10 * 3.25 * 10^-2 * 30/75 * 0.65.
P=0.2 * 100.
P=20HP.
Assume maximum efficiency of the pump=65%.
Then,p=10 * 3.25 * 10^-2 * 30/75 * 0.65.
P=0.2 * 100.
P=20HP.
Vivek said:
8 years ago
Here, P = wqH I am getting 10.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers