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.
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)
Sousend said:
5 years ago
Unit power= (Outpur power/H^(3/2)).
Hence 1750/1000=1.75.
Hence 1750/1000=1.75.
(1)
George Mwenya said:
4 years ago
Thanks all for explaining.
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.
Vihan said:
4 years ago
Agree @Kapil.
Vishal said:
5 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.
Dev said:
5 years ago
Thanks all of you.
Krupal said:
5 years ago
I agree, thanks @Kapil.
Dharmesh Patel said:
7 years ago
Unit power is power by water can raise head for 1 meter.
Then,
We can apply the formula to find the Unit power as below:
Unit Power = p/H^3/2.
=1750*1000/100^3,
=1750000/1000000,
=1750 W,
=1.75 KW.
Then,
We can apply the formula to find the Unit power as below:
Unit Power = p/H^3/2.
=1750*1000/100^3,
=1750000/1000000,
=1750 W,
=1.75 KW.
Rajendra said:
7 years ago
Unit power of Pelton wheel = P/H^3/2 so 1750/100^3/2=1.75KW.
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