Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 6 (Q.No. 50)
50.
The Brake Horse power of the motor (efficiency 60%) required for a pump of capacity 0.075 cumec for a total lift of 12 m, is
Discussion:
5 comments Page 1 of 1.
Jamahali said:
3 years ago
Can anyone explain what 75 means?
P = QγH.
P = (0.075)(9.81)(12).
P = 8.829 kN^2m/s or kWatts.
Convert kW to hp to use the power efficiency formula;
P = 8.829kW (1hP/0.746kW).
P = 11.835 hP.
Efficiency=(wHP/bHP)*100.
bHP = (11.835hP/60)x100.
bHP = 19.725 => 20.
P = QγH.
P = (0.075)(9.81)(12).
P = 8.829 kN^2m/s or kWatts.
Convert kW to hp to use the power efficiency formula;
P = 8.829kW (1hP/0.746kW).
P = 11.835 hP.
Efficiency=(wHP/bHP)*100.
bHP = (11.835hP/60)x100.
bHP = 19.725 => 20.
Anirban Roy said:
5 years ago
Total work done = gain in PE = p*q*h÷75= 1000*0.075*12*100/(75*60) = 20 hp.
Raj said:
7 years ago
Thank you both of you for explaining the answer.
Sufazo said:
9 years ago
1200/75 = 12, 60% = 12 its mean 100% for 20.
BALU MARGE said:
1 decade ago
= ((W*Q*H)/(75*n))*100.
= ((9.81*.075*12)/(75*.6)*100.
= 19.62 = 20%.
= ((9.81*.075*12)/(75*.6)*100.
= 19.62 = 20%.
(4)
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