Fluid mechanics and hydraulics miscellaneous
- Critical depth at a section of a rectangular channel is 1.5 m. The specific energy at that section is
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yc = 1.5 m
Ec = 3 yc 2 = 3 × 1.5 = 2.25 m 2 Correct Option: D
yc = 1.5 m
Ec = 3 yc 2 = 3 × 1.5 = 2.25 m 2
- A partially open sluice gate discharges water into a rectangular channel. The tail water depth
formed at a downstream of the sluice gate after the vena contracta of the jet coming out from the sluice gae, the sluice gate opening should be (coefficient of contraction Cc = 0.9)in the channel is 3 m and Froude number is 1 . If a free hydraulic jump is to be 2 √2
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y1 = y2 [-1 + √1 + 8Fr22] 2
= 0.62 m
Coeff. of contraction, CcCc = area of vena contract area of opening ⇒ 0.9 = 0.62 × b y × b ∴ y = 0.62 = 0.69 m 0.9
Correct Option: C
y1 = y2 [-1 + √1 + 8Fr22] 2
= 0.62 m
Coeff. of contraction, CcCc = area of vena contract area of opening ⇒ 0.9 = 0.62 × b y × b ∴ y = 0.62 = 0.69 m 0.9
- A stream function is given by Ψ = 2x2y + (x + 1)y2
The flow rate across a line joining points A(3, 0) and B(0,2) is
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Ψ = 2x2y + (x + 1)y2
At A (3, 0)
Ψ(3, 0) = 2(3)2 - (0) + (3 + 1)(0)2
= 0
At B(2, 0)
Ψ(2, 0) = 2(0)2(2) + (0 + 1)(2)2
= 0 + 4 = 4
Flow rate across line joining A and B = |ΨA - ΨB|
= |0 - 4| unitsCorrect Option: C
Ψ = 2x2y + (x + 1)y2
At A (3, 0)
Ψ(3, 0) = 2(3)2 - (0) + (3 + 1)(0)2
= 0
At B(2, 0)
Ψ(2, 0) = 2(0)2(2) + (0 + 1)(2)2
= 0 + 4 = 4
Flow rate across line joining A and B = |ΨA - ΨB|
= |0 - 4| units
- Cross-section of an object (having same section normal to the paper) submerged into a fluid consists of a square of sides 2 m and triangle as shown in the figure. The object is hinged at point P that is one meter below the fluid free surface. If the object is to be kept in the position as shown in the figure, the value ‘x’ should be
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Take moment of vertical component of hydrastatic force on two section
V10 + 2 = V2 × x 2 3 ρ gv1 × 1 = ρgv2 × x 3 ∴ v1 = = v2 x 3 (2 × z × w) = 1 x × 2 (w) × x 2 3
⇒ x = √12 = 2√3
Correct Option: A
Take moment of vertical component of hydrastatic force on two section
V10 + 2 = V2 × x 2 3 ρ gv1 × 1 = ρgv2 × x 3 ∴ v1 = = v2 x 3 (2 × z × w) = 1 x × 2 (w) × x 2 3
⇒ x = √12 = 2√3
- The circulation ‘Γ’ around a circle of radius 2 units for the velocity field u = 2x + 3y and v = 2y is
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→V = ui + vj
u = 2x + 3y , v = -2y∂u = 3 , ∂v = 0 ∂y ∂x circulation = ∂v - ∂u × Area ∂x ∂y
= (0 - 3) × πR2
= (-3)π × 22
= – 12π units
Correct Option: B
→V = ui + vj
u = 2x + 3y , v = -2y∂u = 3 , ∂v = 0 ∂y ∂x circulation = ∂v - ∂u × Area ∂x ∂y
= (0 - 3) × πR2
= (-3)π × 22
= – 12π units