# Mechanical Engineering - Hydraulic Machines - Discussion

### Discussion :: 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

 [A]. 0.25 kW [B]. 0.75 kW [C]. 1.75 kW [D]. 3.75 kW

Explanation:

No answer description available for this question.

 Pratik said: (Oct 6, 2013) 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.

 Kapil said: (Oct 21, 2013) Unit power= P/[H^(3/2)]. = 1750*10^3/100^(3/2). = 1750 W. = 1.75 KW.

 Mak said: (Jul 7, 2016) You are right @Kapil.

 Vish said: (Jul 29, 2016) I agree with you @Kapil.

 Vijay Solanki said: (Aug 20, 2016) I agree with you @Pratik.

 I.P.Dwivedi said: (Sep 7, 2016) Unit power = P/[H^(3/2)]. = 1750 * 10^3/100^(3/2). = 1750 W. = 1.75 KW.

 Piyush said: (Sep 29, 2016) Unit power = p/h^(3/2), So, we get 1.75 KW.

 Lokesh said: (Dec 11, 2016) I agree with the explanation. Thank you all.

 Nandlal Yadav said: (Jul 20, 2017) I agree, thanks @Pratik.

 Madhu said: (Jul 23, 2017) Thank you @KAPIL.

 Sattibabu said: (Jan 7, 2018) Thank you for the explanation.

 Rajendra said: (Jan 31, 2018) Unit power of Pelton wheel = P/H^3/2 so 1750/100^3/2=1.75KW.

 Dharmesh Patel said: (Feb 1, 2018) 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.

 Krupal said: (May 2, 2019) I agree, thanks @Kapil.

 Dev said: (Jun 7, 2019) Thanks all of you.

 Sousend said: (Jun 9, 2019) Unit power= (Outpur power/H^(3/2)). Hence 1750/1000=1.75.

 Vishal said: (Oct 21, 2019) Unit power = P/[H^(3/2)]. = 1750 * 10^3/100^(3/2). = 1750 W. = 1.75 KW.

 Vihan said: (May 15, 2020) Agree @Kapil.

 George Mwenya said: (Oct 10, 2020) Thanks all for explaining.