## Fluid Mechanics and Hydraulic Machinery Miscellaneous

#### Fluid Mechanics and Hydraulic Machinery

1. 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

1.  PM = Pp = Nm 3 Dm 5 Np Dp

 = 300 2000 3 1 5 = 2.34 kW 1000 5

##### Correct Option: A

 PM = Pp = Nm 3 Dm 5 Np Dp

 = 300 2000 3 1 5 = 2.34 kW 1000 5

1. The head loss for a laminar incompressible flow through a horizontal circular pipe is h1. 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 h2. The ratio h2 /h1. is

1.  Head loss, h ∝ uavg D2

 ∴ h2 = D1 2 uavg,2 = 22 = 2 h1 D2 uavg,1

 h2 = 8 h1

##### Correct Option: C

 Head loss, h ∝ uavg D2

 ∴ h2 = D1 2 uavg,2 = 22 = 2 h1 D2 uavg,1

 h2 = 8 h1

1. Three parallel pipes connected at the two ends have flow-rates Q1, Q2 and Q3 respectively, and the corresponding frictional head losses are hL1, hL2 and hL3 respectively. The correct expressions for total flow rate (Q) and frictional head loss across the two ends (hL) are

1. Total flow rate Q = Q1 + Q2 + Q3
Head loss h = hL1 = hL2 = hL3

##### Correct Option: B

Total flow rate Q = Q1 + Q2 + Q3
Head loss h = hL1 = hL2 = hL3

1. 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

1.  Head loss = fLv2 = fl(Q/A)2 2gd 2gd

 ⇒  Head loss ∝ L d5

 h1* ∝ L d5

 h2* ∝ 2.L = L × 64 (d/2)5 d5

 h1 = 1/64 ⇒ h2 = 64 h2 h1

##### Correct Option: D

 Head loss = fLv2 = fl(Q/A)2 2gd 2gd

 ⇒  Head loss ∝ L d5

 h1* ∝ L d5

 h2* ∝ 2.L = L × 64 (d/2)5 d5

 h1 = 1/64 ⇒ h2 = 64 h2 h1

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

1.  ∆ P = 32μUL ∆²

 Q = AV = π d²V 4

 V = 4Q πd²

 ∆P = 128μQL πD4

Above equation shows that
 ∆P ∝ 1 D4

 ∆P1 = C = C where D1 = D D41 D4

 & ∆P2 = C = C where D2 = 1.5D D42 (d42)

% change in pressure drop
 % = ∆P1 - ∆P2 × 100 ∆P1

= 80.24% ≃ 80%

##### Correct Option: D

 ∆ P = 32μUL ∆²

 Q = AV = π d²V 4

 V = 4Q πd²

 ∆P = 128μQL πD4

Above equation shows that
 ∆P ∝ 1 D4

 ∆P1 = C = C where D1 = D D41 D4

 & ∆P2 = C = C where D2 = 1.5D D42 (d42)

% change in pressure drop
 % = ∆P1 - ∆P2 × 100 ∆P1

= 80.24% ≃ 80%