Fluid Mechanics and Hydraulic Machinery Miscellaneous
 Two pipes of uniform section but different diameters carry water at the same flow rate. Water properties are the same in the two pipes. The Reynolds number, based on the pipe diameter

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Reynolds number, R_{e} = Vd υ Discharge, Q = AV = π d^{2}V 4 From above, = 4Q πd^{2} ∴ Reynolds No. = 4Q × d πd^{2} 2 Reynolds No. = 4Q ....(1) πd From (1), R_{e} ∝ 1 2
Reynolds number is large in the Narrow pipe.Correct Option: B
Reynolds number, R_{e} = Vd υ Discharge, Q = AV = π d^{2}V 4 From above, = 4Q πd^{2} ∴ Reynolds No. = 4Q × d πd^{2} 2 Reynolds No. = 4Q ....(1) πd From (1), R_{e} ∝ 1 2
Reynolds number is large in the Narrow pipe.
Direction: Consider a linear programming problem with two variable and two constraints. The objective function is maximize x_{1} + x_{2}. The corner points of the feasible region are (0, 0), (0,2) (2, 0) and (4/3, 4/3)
 Oil is being pumped through a straight pipe, the pipe length, diameter and volumetric flow rate are all doubled in a new arrangement. The pipe friction factor, however, remains constant. The ratio of pipe frictional losses in the new arrangement to that in the original configuration would be

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For original configuration
l_{1} = l
d_{1} = d
Q_{1} = Q
For new configuration
l_{2} = 21
d_{2} = 2d
Q_{2} = 2Q
Head loss due to friction,h_{f} = flV² 2gb
where f = friction factor
l = Length of pipe
v = Average velocity
d = Diameter pipeDischarge: Q = AV = π a²V 4 or V = 4Q πa² h_{f} = fl × 4Q ² = 8flQ² 2gd πa² πga^{5}
For new Pipe,h_{f2} = 8fl_{1}Q_{2}² = 8f × 2f × 4Q² = 2fQ² πgd_{a}^{5} π × 2^{5} × a^{5} πga^{5} h_{1} = 2flQ² × πga^{5} = 1 h_{2}² πga^{5} 2flQ² 4 Correct Option: A
For original configuration
l_{1} = l
d_{1} = d
Q_{1} = Q
For new configuration
l_{2} = 21
d_{2} = 2d
Q_{2} = 2Q
Head loss due to friction,h_{f} = flV² 2gb
where f = friction factor
l = Length of pipe
v = Average velocity
d = Diameter pipeDischarge: Q = AV = π a²V 4 or V = 4Q πa² h_{f} = fl × 4Q ² = 8flQ² 2gd πa² πga^{5}
For new Pipe,h_{f2} = 8fl_{1}Q_{2}² = 8f × 2f × 4Q² = 2fQ² πgd_{a}^{5} π × 2^{5} × a^{5} πga^{5} h_{1} = 2flQ² × πga^{5} = 1 h_{2}² πga^{5} 2flQ² 4

The ratio P_{A} − P_{B} . (1/2)ρu_{0}²
(where p_{A} and p_{B} are the pressures at section A and B, respectively, and ρ is the density of the fluid), is

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Applying Bernoulli’s equation at sections A and B, we get
P_{A} + u_{0}² + Z_{A} = P_{A} + V_{m}² + Z_{B} ρg 2g ρg 2g P_{A} + u_{0}² = P_{B} + V_{m}² ρg 2g ρg 2g or P_{A} + P_{B} = V_{m}² + u_{0}² P_{g} ρg 2g 2g
Now, the ratioP_{A} − P_{B}/(1/2)ρ_{0}² = ρV_{m}² − ρu_{0}² 2 2 1 ρu_{0}² 2 = V_{m}² − u_{0}² = V_{m} ² − 1 u_{0}² u_{0} Substituting the Value of V_{m} = 1 u_{0} 1 − (δ/H)
we getP_{A} − P_{B} = 1 − 1 (1/2)ρ_{0}² [1 − (δ/H)]² Correct Option: B
Applying Bernoulli’s equation at sections A and B, we get
P_{A} + u_{0}² + Z_{A} = P_{A} + V_{m}² + Z_{B} ρg 2g ρg 2g P_{A} + u_{0}² = P_{B} + V_{m}² ρg 2g ρg 2g or P_{A} + P_{B} = V_{m}² + u_{0}² P_{g} ρg 2g 2g
Now, the ratioP_{A} − P_{B}/(1/2)ρ_{0}² = ρV_{m}² − ρu_{0}² 2 2 1 ρu_{0}² 2 = V_{m}² − u_{0}² = V_{m} ² − 1 u_{0}² u_{0} Substituting the Value of V_{m} = 1 u_{0} 1 − (δ/H)
we getP_{A} − P_{B} = 1 − 1 (1/2)ρ_{0}² [1 − (δ/H)]²
Direction: Consider a steady incompressible flow through a channel as shown below:
The velocity profile is uniform with a value of u_{0} at the inlet section A. The velocity profile at section B downstream is
 The ratio V_{m}/u_{0} is

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Applying continuity equation at sections A and B, mass flow rate at inlet section A = mass rate at section B.
∴ u = V_{m}Y δ u_{0}H = V_{m}(H − 2δ) + 2 V_{m} Y² ^{δ} δ 2 _{0} u_{0}H = V_{m}(H − 2δ) + 2 V_{m} × δ² δ 2
u_{0}H = V_{m}(H − 2δ) + V_{m}δ
u_{0}H = V_{m}H − 2V_{m}δ + V_{m}δ
u_{0}H = V_{m}H − V_{m}δ
u_{0}H = V_{m}(H − δ)or V_{m} = H = 1 u_{0} H − δ 1 − (δ/H) Correct Option: C
Applying continuity equation at sections A and B, mass flow rate at inlet section A = mass rate at section B.
∴ u = V_{m}Y δ u_{0}H = V_{m}(H − 2δ) + 2 V_{m} Y² ^{δ} δ 2 _{0} u_{0}H = V_{m}(H − 2δ) + 2 V_{m} × δ² δ 2
u_{0}H = V_{m}(H − 2δ) + V_{m}δ
u_{0}H = V_{m}H − 2V_{m}δ + V_{m}δ
u_{0}H = V_{m}H − V_{m}δ
u_{0}H = V_{m}(H − δ)or V_{m} = H = 1 u_{0} H − δ 1 − (δ/H)
 Consider steady laminar incompressible axisymmetric fully developed viscous flow through a straight circular pipe of constant crosssectional area at a Reynolds number of 5. The ratio of inertia force to viscous force on a fluid particle is

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Reynold’s number = Inertia force = 5 Viscous force Correct Option: A
Reynold’s number = Inertia force = 5 Viscous force