Fluid Mechanics and Hydraulic Machinery Miscellaneous
 A model of a hydraulic turbine is tested at a head of 1/4th of that under which the full scale turbine works. The diameter of the model is half of that of the full scale turbine. If N is the RPM of the full scale turbine, then the RPM of the model will be

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Given data:
Model: H_{m} = H_{p} ;D_{m} = D_{p} 4 9
N_{m} = ?
Prototype (Full scale turbine)
N_{p} = NC_{H} = H (ND)^{2}
(C_{H})_{model} = (C_{H})_{prototype}H_{m} = H_{p} N_{m}²D_{m}² N_{p}²D_{p}² H_{p}/4 = H_{p} N_{m}² ×(D_{p}²/4) N_{p}²D_{p}²
N_{m} = NCorrect Option: C
Given data:
Model: H_{m} = H_{p} ;D_{m} = D_{p} 4 9
N_{m} = ?
Prototype (Full scale turbine)
N_{p} = NC_{H} = H (ND)^{2}
(C_{H})_{model} = (C_{H})_{prototype}H_{m} = H_{p} N_{m}²D_{m}² N_{p}²D_{p}² H_{p}/4 = H_{p} N_{m}² ×(D_{p}²/4) N_{p}²D_{p}²
N_{m} = N
 The inlet angle of runner blades of a Francis turbine is 90°. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant through out the blade passage and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is

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Given data:
B_{i} = 90°
V_{wo} = 0V_{fi} = V_{fo} = U_{i} 2
Blade efficiency of the runner,η_{H} = V_{wi}u_{i} = U_{1}² ∴ V_{wi} = u_{i} gH gH
By energy balance equation,ρQgH = ρQV_{wi}U_{i} + ρQV_{0}² 2 gH = U_{i}² + V_{fo}² U_{i}² + U_{i}² 2 8 gH = 9 U_{i}² 8 ∴ η_{H} = U_{i}² = 8 = 0.8888 (9/8)U_{i}² 9
= 88.88% ≈ 98%Correct Option: D
Given data:
B_{i} = 90°
V_{wo} = 0V_{fi} = V_{fo} = U_{i} 2
Blade efficiency of the runner,η_{H} = V_{wi}u_{i} = U_{1}² ∴ V_{wi} = u_{i} gH gH
By energy balance equation,ρQgH = ρQV_{wi}U_{i} + ρQV_{0}² 2 gH = U_{i}² + V_{fo}² U_{i}² + U_{i}² 2 8 gH = 9 U_{i}² 8 ∴ η_{H} = U_{i}² = 8 = 0.8888 (9/8)U_{i}² 9
= 88.88% ≈ 98%
 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

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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

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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

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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}