Mechanical Engineering - Hydraulic Machines - Discussion
Discussion Forum : Hydraulic Machines - Section 2 (Q.No. 41)
41.
The power developed by a turbine is (where H = Head of water under which the turbine is working)
Discussion:
9 comments Page 1 of 1.
Shiva said:
5 years ago
You are right @Shivanand. Thanks.
Kajal Patil said:
5 years ago
Power develop by turbine given by.
P = η * Ï * g * h * Q.
P = η * Ï * g * h * Q.
Subrata ghorai said:
5 years ago
H proportional to H3/2.
Ref: RS Khurmi Gupta.
Ref: RS Khurmi Gupta.
Jitumoni said:
6 years ago
I think it's inversely proportional.
Shubham said:
7 years ago
P=pQgH.
Here q is proportional to h^1/2 as Q=av here v=2gh^1/2.
Here q is proportional to h^1/2 as Q=av here v=2gh^1/2.
Partha said:
7 years ago
The correct Answer is C.
It is given for centrifugal pumps.
(P1/P2)=(D1/D2)^3 and H1/H2=(D/D2)^2
Now equate the above equation.
It is given for centrifugal pumps.
(P1/P2)=(D1/D2)^3 and H1/H2=(D/D2)^2
Now equate the above equation.
Shivanand said:
8 years ago
Pu = P/H^3/2.
Hence, P = Pu* H^3/2.
Hence, P = Pu* H^3/2.
(1)
Daya said:
8 years ago
I think the Answer is h^5/4.
Siva Kumar G said:
9 years ago
I think the answer is Directly proportional to H^(5/2).
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