Mechanical Engineering - Hydraulic Machines - Discussion
Discussion Forum : Hydraulic Machines - Section 1 (Q.No. 1)
1.
Power required to drive a centrifugal pump is directly proportional to __________ of its impeller.
Discussion:
35 comments Page 1 of 4.
Nani said:
4 years ago
As per affinity law, break horse power is directly proportional to cube of diameter or cube of rpm.
(8)
Sandeep said:
9 years ago
d^4 will be the correct answer. Since the question is asked as directly proportional the tangent flow of an impulse turbine always takes 4th power to diameter.
(2)
Mahi ss said:
10 years ago
The right answer is P proportional to D^5.
(1)
Sharan said:
9 years ago
Ans: C is correct.
i.e d^3--p=TW=(Tou*pi*d^3/16*T)*(v/r).
i.e d^3--p=TW=(Tou*pi*d^3/16*T)*(v/r).
(1)
Pushpender said:
9 years ago
Affinity laws applied to axial and radial flows pumps and turbines. Its a tangent flow ; impulse turbine so I think 4th power is correct.
Dipanjan said:
9 years ago
I agree @Sreeejith.
Indrajeet said:
9 years ago
According to affinity law of pump.
Power is directly proportional to the fifth Power of diameter.
Power is directly proportional to the fifth Power of diameter.
Jigar k patel said:
9 years ago
Correct Ans is C.
Law 2c Power is proportional to the cube of impeller diameter (assuming constant shaft speed):
{\displaystyle {P_{1} \over P_{2}}={\left({D_{1} \over D_{2}}\right)^{3}}} {P_{1} \over P_{2}}={\left({D_{1} \over D_{2}}\right)^{3}}.
Law 2c Power is proportional to the cube of impeller diameter (assuming constant shaft speed):
{\displaystyle {P_{1} \over P_{2}}={\left({D_{1} \over D_{2}}\right)^{3}}} {P_{1} \over P_{2}}={\left({D_{1} \over D_{2}}\right)^{3}}.
Vinit said:
9 years ago
The Correct answer will be d^5.
Pabitra said:
9 years ago
Fifth power of diameter will be the correct answer.
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