Fluid mechanics and hydraulics miscellaneous
- Match the following
Group 1 Group 2 P. Release valve 1. Reduce high inlet pressrue & lower outlet pressure Q. Check valve 2. Limit the flow of water to single direction R. Gate valve 3. Remove air from the pipeline S. Pilot valve 4. Stopping the flow of water in the pipeline
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NA
Correct Option: A
NA
- The thickness of the laminar boundary layer on a flat plate at a point A is 2 cm and at a point B, 1 m downstream of A, is 3 cm. What is the distance of A from the leading edge of the plate?
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As per Blasius equation,
→V = ui + vj
u = 2x + 3y , v = -2yδ = 5 , Re = PVx x √Re μ
∴ δ ∝ √x
x – distance from leading edge.
At x, δ1 = 2 cm
At x + 1, δ2 = 3 cmδ1 = √x = 2 δ2 √x + 1 3
∴ x = 0.8 m.
Correct Option: B
As per Blasius equation,
→V = ui + vj
u = 2x + 3y , v = -2yδ = 5 , Re = PVx x √Re μ
∴ δ ∝ √x
x – distance from leading edge.
At x, δ1 = 2 cm
At x + 1, δ2 = 3 cmδ1 = √x = 2 δ2 √x + 1 3
∴ x = 0.8 m.
- A hydraulic jump occurs in a rectangular, horizontal, frictionless channel. What would be the pre-jump depth if the discharge per unit width 2 m3 /s/m and the energy loss is 1 m?
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yc = 
q2 
1 / 3 = 22 1 / 3 = 0.74 g 9.81 2q2 = y1y2(y1 + y2) g = 2 × 22 = 0.815 9.81 ∆E = (y2 - y1)3 = 1 m 4y1 y2
By trial and error, we get 0.3 m.
Correct Option: B
yc = q2 1 / 3 = 22 1 / 3 = 0.74 g 9.81 2q2 = y1y2(y1 + y2) g = 2 × 22 = 0.815 9.81 ∆E = (y2 - y1)3 = 1 m 4y1 y2
By trial and error, we get 0.3 m.
- A very wide rectangular channel is designed to carry a discharge of 5 m3 /s per meter width. The design is based on the Manning’s equation with the roughness coefficient obtained from the grain size using Strickler’s equation and results in a normal depth of 1.0 m. By mistake, however the engineer used the grain diameter in mm in the Stickler’s equation instead of in meter. What should be the correct normal depth?
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R = A = B × y P (B + 2y) ∵ B >> y , R ≈ B × y ≈ y B n = d1 / 6 ( Stickler's equation ) 24 qold = 1 Rold2 / 3 S1 / 2 = 1 × yold5 / 3 . S1 / 2..........(1) nold nold qnew = 1 × ynew5 / 3 . S1 / 2..........(2) nnew
(1) ÷ (2) ,nold = yold5 / 3 nnew ynew5 / 3 ynew = nnew 3 / 5 × yold nold = 1 3 / 5 = 0.5 m 10001 / 6
Correct Option: B
R = A = B × y P (B + 2y) ∵ B >> y , R ≈ B × y ≈ y B n = d1 / 6 ( Stickler's equation ) 24 qold = 1 Rold2 / 3 S1 / 2 = 1 × yold5 / 3 . S1 / 2..........(1) nold nold qnew = 1 × ynew5 / 3 . S1 / 2..........(2) nnew
(1) ÷ (2) ,nold = yold5 / 3 nnew ynew5 / 3 ynew = nnew 3 / 5 × yold nold = 1 3 / 5 = 0.5 m 10001 / 6
- The flow of glycerin (kinematic viscosity, v = 5 × 10-4 m2 /s) in an open channel is to be modeled in a laboratory flume using water (v = 10-6 m2 /s) as the flowing fluid. If both gravity and viscosity are important, what should be the length scale (i.e. ratio of prototype to model dimensions) for maintaining dynamic similarity?
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Viscous force → Reynold’s number ρVd μ Gravity force → Froude’s number V √gy
Here, ‘d’ and ‘y’ are same as viscosity and gravity are equally important.PVd = Vd = VLr μ v v VLr = V vr √Lr
(Lr)3 / 2 = vrL = (vr)2 / 3 = Vm 2 / 3 vo = Vm 2 / 3 vo = 10-6 2 / 3 = 0.0158 5 × 10-4 Lr = Lm Lp ∴ Lp = 1 = 1 ≈ 63 Lm Lr 0.0158
Correct Option: C
Viscous force → Reynold’s number ρVd μ Gravity force → Froude’s number V √gy
Here, ‘d’ and ‘y’ are same as viscosity and gravity are equally important.PVd = Vd = VLr μ v v VLr = V vr √Lr
(Lr)3 / 2 = vrL = (vr)2 / 3 = Vm 2 / 3 vo = Vm 2 / 3 vo = 10-6 2 / 3 = 0.0158 5 × 10-4 Lr = Lm Lp ∴ Lp = 1 = 1 ≈ 63 Lm Lr 0.0158
