Fluid Mechanics and Hydraulic Machinery Miscellaneous
 A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m. For initial testing, a 1: 4 scale model of the turbine operates under a head of 10 m. The power generated by the model (in kW)will be

View Hint View Answer Discuss in Forum
P_{M} = P_{p} = N_{m} ^{3} D_{m} ^{5} N_{p} D_{p} = 300 2000 ^{3} 1 ^{5} = 2.34 kW 1000 5
Correct Option: A
P_{M} = P_{p} = N_{m} ^{3} D_{m} ^{5} N_{p} D_{p} = 300 2000 ^{3} 1 ^{5} = 2.34 kW 1000 5
 The head loss for a laminar incompressible flow through a horizontal circular pipe is h_{1}. Pipe length and fluid remaining the same, if the average flow velocity doubles and the pipe diameter reduces to half its previous value, the head loss is h_{2}. The ratio h_{2} /h_{1}. is

View Hint View Answer Discuss in Forum
Head loss, h ∝ u_{avg} D^{2} ∴ h_{2} = D_{1} ^{2} u_{avg,2} = 2^{2} = 2 h_{1} D_{2} u_{avg,1} h_{2} = 8 h_{1} Correct Option: C
Head loss, h ∝ u_{avg} D^{2} ∴ h_{2} = D_{1} ^{2} u_{avg,2} = 2^{2} = 2 h_{1} D_{2} u_{avg,1} h_{2} = 8 h_{1}
 Three parallel pipes connected at the two ends have flowrates Q_{1}, Q_{2} and Q_{3} respectively, and the corresponding frictional head losses are h_{L1}, h_{L2} and h_{L3} respectively. The correct expressions for total flow rate (Q) and frictional head loss across the two ends (h_{L}) are

View Hint View Answer Discuss in Forum
Total flow rate Q = Q_{1} + Q_{2} + Q_{3}
Head loss h = h_{L1} = h_{L2} = h_{L3}Correct Option: B
Total flow rate Q = Q_{1} + Q_{2} + Q_{3}
Head loss h = h_{L1} = h_{L2} = h_{L3}
 Consider fully developed flow in a circular pipe with negligible entrance length effects. Assuming the mass flow rate, density and friction factor to be constant, if the length of the pipe is doubled and the diameter is halved, the head loss due to friction will increase by a factor of

View Hint View Answer Discuss in Forum
Head loss = fLv^{2} = fl(Q/A)^{2} 2gd 2gd ⇒ Head loss ∝ L d^{5} h_{1}^{*} ∝ L d^{5} h_{2}^{*} ∝ 2.L = L × 64 (d/2)^{5} d^{5} h_{1} = 1/64 ⇒ h_{2} = 64 h_{2} h_{1} Correct Option: D
Head loss = fLv^{2} = fl(Q/A)^{2} 2gd 2gd ⇒ Head loss ∝ L d^{5} h_{1}^{*} ∝ L d^{5} h_{2}^{*} ∝ 2.L = L × 64 (d/2)^{5} d^{5} h_{1} = 1/64 ⇒ h_{2} = 64 h_{2} h_{1}
 Water is flowing through a horizontal pipe of constant diameter and the flow is laminar. If the diameter of the pipe is increased by 50% keeping the volume flow rate constant, then the pressure drop in the pipe due to friction will decrease by

View Hint View Answer Discuss in Forum
∆ P = 32μUL ∆² Q = AV = π d²V 4 V = 4Q πd² ∆P = 128μQL πD^{4}
Above equation shows that∆P ∝ 1 D^{4} ∆P_{1} = C = C where D_{1} = D D^{4}_{1} D^{4} & ∆P_{2} = C = C where D_{2} = 1.5D D^{4}_{2} (d^{4}_{2})
% change in pressure drop% = ∆P_{1}  ∆P_{2} × 100 ∆P_{1}
= 80.24% ≃ 80%Correct Option: D
∆ P = 32μUL ∆² Q = AV = π d²V 4 V = 4Q πd² ∆P = 128μQL πD^{4}
Above equation shows that∆P ∝ 1 D^{4} ∆P_{1} = C = C where D_{1} = D D^{4}_{1} D^{4} & ∆P_{2} = C = C where D_{2} = 1.5D D^{4}_{2} (d^{4}_{2})
% change in pressure drop% = ∆P_{1}  ∆P_{2} × 100 ∆P_{1}
= 80.24% ≃ 80%