Mechanical Engineering - Hydraulic Machines - Discussion
Discussion Forum : Hydraulic Machines - Section 1 (Q.No. 7)
7.
A Pelton wheel develops 1750 kW under a head of 100 metres while running at 200 r.p.m. and discharging 2500 litres of water per second. The unit power of the wheel is
Discussion:
19 comments Page 1 of 2.
PRATIK said:
1 decade ago
Unit power is the the power required to raise the fluid up to 1m head.
So,
Unit power = total power / total head.
= 175/100.
= 1.75.
So,
Unit power = total power / total head.
= 175/100.
= 1.75.
Kapil said:
1 decade ago
Unit power= P/[H^(3/2)].
= 1750*10^3/100^(3/2).
= 1750 W.
= 1.75 KW.
= 1750*10^3/100^(3/2).
= 1750 W.
= 1.75 KW.
(1)
MAK said:
8 years ago
You are right @Kapil.
Vish said:
8 years ago
I agree with you @Kapil.
Vijay solanki said:
8 years ago
I agree with you @Pratik.
I.P.Dwivedi said:
8 years ago
Unit power = P/[H^(3/2)].
= 1750 * 10^3/100^(3/2).
= 1750 W.
= 1.75 KW.
= 1750 * 10^3/100^(3/2).
= 1750 W.
= 1.75 KW.
Piyush said:
8 years ago
Unit power = p/h^(3/2),
So, we get 1.75 KW.
So, we get 1.75 KW.
Lokesh said:
8 years ago
I agree with the explanation. Thank you all.
Nandlal yadav said:
7 years ago
I agree, thanks @Pratik.
MADHU said:
7 years ago
Thank you @KAPIL.
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