Fluid mechanics and hydraulics miscellaneous
- An aicraft is flying in level flight at a speed of 200 km/hr through air (density, r = 1.2 kg/m2, and viscosity m = 1.6 × 10–5 N-s/m2). The lift co-efficient at this speed is 0.4 and the drag co-efficient is 0.0065. The mass of the aircraft is 800 kg. The effective lift area of the aircraft is
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0.4 × 1.2 × 
200 
2 = 800 × 9.81 2 3.6
A = 10.594 m2
A = 10.594 m2 = 10.6 m2Correct Option: B
0.4 × 1.2 × 200 2 = 800 × 9.81 2 3.6
A = 10.594 m2
A = 10.594 m2 = 10.6 m2
- The x component of velocity in a two dimensional incompressible flow is given by u = 1.5 x. At the point (x, y) = (1, 0), the y component of velocity v = 0. The equation for the y component of velocity is
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Apply continuity equation (2D, incompressible flow)
∂u + ∂v + ∂w = 0 ∂x ∂y ∂z ⇒ ∂ (1.5x) + ∂v + 0 = 0 ∂x ∂y ⇒ 1.5 + ∂v = 0 ∂x
∴ ∂v = - 1.5 ∂y
Integrate both sides,
∫ dv = -1.5∫ dy
⇒ v = -1.5y + c
At (x, y), = (1, 0), y component is zero
⇒ 0 = 1.5 × 0 + c
∴ c = 0
∴ v = – 1.5yCorrect Option: D
Apply continuity equation (2D, incompressible flow)
∂u + ∂v + ∂w = 0 ∂x ∂y ∂z ⇒ ∂ (1.5x) + ∂v + 0 = 0 ∂x ∂y ⇒ 1.5 + ∂v = 0 ∂x
∴ ∂v = - 1.5 ∂y
Integrate both sides,
∫ dv = -1.5∫ dy
⇒ v = -1.5y + c
At (x, y), = (1, 0), y component is zero
⇒ 0 = 1.5 × 0 + c
∴ c = 0
∴ v = – 1.5y
- In the inclined manometer shown in the figure below, the reservoir is large. Its surface may be assumed to remain at a fixed elevation. A is connected to a gas pipeline and the deflection noted on the inclined glass tube is 100 mm. Assuming q = 30° and the manometric fluid as oil with specific gravity of 0.86, the pressure at A is
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Pressure at A = Poil. g. hoil
= Poil g. (l. × sinθ)
= (0.86 × 1000) × 9.81 × (0.1 × sin 30°)
= 421.83 N/m2
Also, Pressure at A = Pw .g.hw
⇒ 421.83 = 1000 × 9.81 × hw = 43 mm of waterCorrect Option: B
Pressure at A = Poil. g. hoil
= Poil g. (l. × sinθ)
= (0.86 × 1000) × 9.81 × (0.1 × sin 30°)
= 421.83 N/m2
Also, Pressure at A = Pw .g.hw
⇒ 421.83 = 1000 × 9.81 × hw = 43 mm of water
- Two pipelines, one carrying oil (mass density = 900 kg/m3) and the other water, are connected to a manometer as shown in the figure, By what amount the pressure in the water pipe should be increased so that the mercury levels in both the limbs of the manometer become equal? (Mass density of mercury = 13,550 kg/m3 and g = 9.81 m/s2)
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Initial case,
PA .Po .gho = PB + ρw g.hw + ρm g.hm
⇒ PA + 900 × 9.81 × 3 = PB × 1000 × 9.81 × 1.5 + 13550 × 9.81 × 0.2
⇒ PA – PB = 14.81 × 103 N/m2 = 14.81 kPa
To make mercury level equal, lower mercury by 0.1 m at side B
So that at side A it rises by 0.1 m and becomes equal.
Resultant figure
PA ' + 900 × 9.81 × 2.9 = PB ' + 1000 × 9.81 × 1.6
∴ PB ' – PA ' = 9.908 kPa
∴ PA ' – PB ' = –9.908 kPa
∴ Pressure to be increased = (PA. PB) – (PA ' – PB ')
= 14.81 – (– 9.908) = 24.73 kPaCorrect Option: A
Initial case,
PA .Po .gho = PB + ρw g.hw + ρm g.hm
⇒ PA + 900 × 9.81 × 3 = PB × 1000 × 9.81 × 1.5 + 13550 × 9.81 × 0.2
⇒ PA – PB = 14.81 × 103 N/m2 = 14.81 kPa
To make mercury level equal, lower mercury by 0.1 m at side B
So that at side A it rises by 0.1 m and becomes equal.
Resultant figure
PA ' + 900 × 9.81 × 2.9 = PB ' + 1000 × 9.81 × 1.6
∴ PB ' – PA ' = 9.908 kPa
∴ PA ' – PB ' = –9.908 kPa
∴ Pressure to be increased = (PA. PB) – (PA ' – PB ')
= 14.81 – (– 9.908) = 24.73 kPa
- A hydraulic jump takes place in a triangular channel of vertex angle 90°, as shown in figure. The discharge is 1m3 /s and the pre-jump depth is 0.5m. What will be the post-jump? (Take g = 9.81 m/s2)
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Taking specific force at 2 sections:
z = y 3 ⇒ 12 + y12 × y1 = 12 + y22 × y2 9.81 × y12 3 9.81 × y12 3
y1 = 0.5 m⇒ 0.449 = 1 + y22 9.81y23 3
⇒ y2 = 1.02 mCorrect Option: C
Taking specific force at 2 sections:
z = y 3 ⇒ 12 + y12 × y1 = 12 + y22 × y2 9.81 × y12 3 9.81 × y12 3
y1 = 0.5 m⇒ 0.449 = 1 + y22 9.81y23 3
⇒ y2 = 1.02 m
