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.
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)
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)
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)
Mohan said:
1 decade ago
Velocity = sqrt 2*g*h & pi*d*N*.
So N is proportionate to Sqrt h,
N1/sqrt h1= N2/sqrt h2
So N is proportionate to Sqrt h,
N1/sqrt h1= N2/sqrt h2
(1)
M.ANBARASAN M.E said:
1 decade ago
N1/H(1/2) = N2/H(1/2).
500/20.5 = N2/H2.5.
500/20.5*H.5 = N2.
500/20.5*15.5 = 433.012.
500/20.5 = N2/H2.5.
500/20.5*H.5 = N2.
500/20.5*15.5 = 433.012.
Bhau said:
8 years ago
There should pump in question instead of turbine. So we will get answer 403.
Ankush said:
7 years ago
I am also getting 403, So according to me, it is option B.
PILU said:
1 decade ago
Ns = (NQ^1/2)/H^(3/4).
N directly proportional to H^3/4.
N directly proportional to H^3/4.
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)
Rohit said:
9 years ago
N directly proportional to h^3/4. So answer is 403.
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