STUPIDSID
Answer any one question from Q1 and Q2
1 (a)
Derive an expression for total pressure and centre of pressure for inclined plane submerged in liquid and hence derive the expression for centre of pressure for vertical plane.
6 M
1 (b)
The velocity vector in a fluid flow is given \[ V=(2x^3)\widehat i - (5x^2y)\widehat j + (2tz)\widehat k \] Obtain expression for velocity vector and acceleration vector at a point (2,1,3) at time t=1s. Also calculate the value of velocity and acceleration at the given point.
6 M
2 (a)
The space between two square flat parallel plates is filled with oil. Each side of the plate is 720 mm. The thickness of the oil film is 15 mm. The upper plate, which moves at 3m/s requires a force of 120 N to maintain the speed. Determine:
i) The dynamic viscosity of the oil.
ii) The kinematic viscosity of oil if the specific gravity of oil is 0.95.
i) The dynamic viscosity of the oil.
ii) The kinematic viscosity of oil if the specific gravity of oil is 0.95.
6 M
2 (b)
Write a short note on:
i) Velocity Potential
ii) Stream Function
iii) Vorticity
i) Velocity Potential
ii) Stream Function
iii) Vorticity
6 M
Answer any one question from Q3 and Q4
3 (a)
Derive an expression of conservation of momentum in differential form.
7 M
3 (b)
An oil of viscosity 10 poise and sp. gravity 0.8 flows through a horizontal pipe of 50 mm dia. If the pressure drops at the rate of 20kN/m2 per meter length, determine,
i) Rate of flow of oil in 1pm.
ii) The centre line velocity.
iii) Total frictional drag over 1 km of the pipe.
i) Rate of flow of oil in 1pm.
ii) The centre line velocity.
iii) Total frictional drag over 1 km of the pipe.
6 M
4 (a)
A 300 mm × 150 mm venturimeter is provided in a vertical pipeline carrying oil of specific gravity 0.9, flow being upward. The difference in elevation of the throat section and entrance section of the venturimeter is 300 mm.
The differential U-tube mercury manometer shows a gauge deflection of 250 mm. Calculate:
i) The discharge of oil, and
ii) The pressure difference between the entrance section and the throat section. Take Cd =0.98 and specific gravity of mercury as 13.6.
i) The discharge of oil, and
ii) The pressure difference between the entrance section and the throat section. Take Cd =0.98 and specific gravity of mercury as 13.6.
7 M
4 (b)
Derive an expression of velocity & shear stress distribution for laminar flow between fixed parallel plates.
6 M
Answer any one question from Q5 and Q6
5 (a)
What are repeating variables? What points are important while selecting repeating variables?
6 M
5 (b)
Two reservoirs, having a difference in elevation of 15 m, are connected by a 200 mm diameter siphon. The length of the siphon is 400 m and the summit is 3 m above the water level in the upper reservoir. The length of the pipe from upper reservoir to the summit is 120 m. If the coefficient
of friction is 0.02, determine:
i) Discharge through the siphon, and
ii) Pressure at the summit.
i) Discharge through the siphon, and
ii) Pressure at the summit.
6 M
6 (a)
Derive an expression for the power transmission through the pipes. Find also the condition for maximum transmission of power.
6 M
6 (b)
Show that the thrust on a propeller is given by: \[ F= pD^2V^2 \phi \left [ \dfrac {\omega D}{V}, \dfrac {pVD}{\mu} \right ] \] Where the terms carry usual meaning.
6 M
Answer any one question from Q7 and Q8
7 (a)
Derive an expression of the following boundary layer thickness:
i) Displacement thickness
ii) Momentum thickness
iii) Energy thickness
i) Displacement thickness
ii) Momentum thickness
iii) Energy thickness
9 M
7 (b)
After polishing the hull of a boat it was noticed that coefficient of drag has reduced by 20%.If the same driving power is used, what would be the % increase in speed.
4 M
8 (a)
Discuss the boundary layer formation over a flat plate.
7 M
8 (b)
Following are parametric forms of three velocity profiles over a stationary surface. Check whether the flow adheres to or detaches from or on the verge of separation from the surface. \[ \dfrac {u}{U} = \dfrac {3}{2} \eta - \dfrac {1}{2}\eta^3 \\ \dfrac {u}{U}= - \dfrac {3}{2}\eta + \dfrac {1}{2}\eta^3 + \eta^4 \\ \dfrac {u}{U}= 2\eta^2 + \eta^3 - 2 \eta^4 \\ where \ \eta = y/\delta. \]
6 M
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