Discussion :: Water Supply Engineering - Section 4 (Q.No.39)
| Shahzeb said: (Jun 2, 2016) | |
| The correct formula is WQH/75. Plese tell me if I'm wrong. |
|
| Sagar said: (Feb 27, 2017) | |
| Agree @Shahzeb. It is Wqh/75. |
|
| Farhan said: (Oct 3, 2017) | |
| Yes, it should be 75 in place of 15. | |
| Mrinal Kanti Biswas said: (Nov 14, 2017) | |
| If w = unit wt of water in KN/cum. Q = Discharge in cum/sec. H = total head. Then, required H.P. = wQH/0.735. |
|
| Gubendhiran said: (Jan 3, 2018) | |
| Hers, H.P = (wQH)/(0.735*efficiency of pump). | |
| Joe said: (Jan 25, 2020) | |
| Water horse power= w*Q*H/75. Break horse power= w*Q*H/ efficiency. |
|
| Pavan said: (Apr 24, 2020) | |
| Yeah, it's WQH/75. Not 15. |
|
| Rosie said: (Sep 30, 2021) | |
| No it Should be 75 instead of 15. | |
Post your comments here:
Name *:
Email : (optional)
» Your comments will be displayed only after manual approval.