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
0.25 kW
0.75 kW
1.75 kW
3.75 kW
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 1 of 2.

Dharmesh Patel said:   6 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.

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.

Vishal said:   4 years ago
Unit power = P/[H^(3/2)].

= 1750 * 10^3/100^(3/2).

= 1750 W.

= 1.75 KW.

I.P.Dwivedi said:   7 years ago
Unit power = P/[H^(3/2)].

= 1750 * 10^3/100^(3/2).

= 1750 W.

= 1.75 KW.

Unit power= P/[H^(3/2)].

= 1750*10^3/100^(3/2).

= 1750 W.
= 1.75 KW.
(1)

Rajendra said:   6 years ago
Unit power of Pelton wheel = P/H^3/2 so 1750/100^3/2=1.75KW.

Sousend said:   5 years ago
Unit power= (Outpur power/H^(3/2)).

Hence 1750/1000=1.75.
(1)

Piyush said:   7 years ago
Unit power = p/h^(3/2),

So, we get 1.75 KW.

Lokesh said:   7 years ago
I agree with the explanation. Thank you all.

Sattibabu said:   6 years ago
Thank you for the explanation.