Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 1 (Q.No. 13)
13.
A hydraulic turbine has a discharge of 5 m3/s, when operating under a head of 20 m with a speed of 500 rpm. If it is to operate under a head of 15 m, for the same discharge, the rotational speed in rpm will approximately be
Discussion:
13 comments Page 1 of 2.
Shoma barbhuiya said:
5 years ago
Specific speed of pump = 403rpm.
(1)
Pratiksha said:
5 years ago
N1 * H2^(1/2) = N2*H1^(1/2).
500*(15)^(1/2) = N2*(20)^(1/2),
[500*(15)^(1/2)]/[20^(1/2)] = N2.
N2 = 433.01.
500*(15)^(1/2) = N2*(20)^(1/2),
[500*(15)^(1/2)]/[20^(1/2)] = N2.
N2 = 433.01.
(3)
Pravin Prakash Kadam said:
5 years ago
Thanks @Mohan.
Riddhan said:
6 years ago
I am also getting an answer 402. 963 which is equal to 403. So according to me, option B is correct.
(1)
Ankush said:
7 years ago
I am also getting 403, So according to me, it is option B.
Bhau said:
8 years ago
There should pump in question instead of turbine. So we will get answer 403.
Ismail said:
8 years ago
No, the right answer is 403.
Nq= (N*Q^0.5)/ H^0.75,
Nq1= Nq2,
(500*5^0.5)/20^0.75 =(N2 * 5^0.5)/ 15^0.75,
N2 =403.
Nq= (N*Q^0.5)/ H^0.75,
Nq1= Nq2,
(500*5^0.5)/20^0.75 =(N2 * 5^0.5)/ 15^0.75,
N2 =403.
(1)
Rohit said:
9 years ago
N directly proportional to h^3/4. So answer is 403.
Pankaj soni said:
9 years ago
n1/h1^1/2 = n2/h2^1/2.
DIWAKAR said:
10 years ago
N2 = N1*H2^(1/2)/H1(1/2).
= 500*15^0.5/20^0.5 = 433.
= 500*15^0.5/20^0.5 = 433.
(1)
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