JNTU
JNTU B.Tech II FLUID MECHANICS AND HYDRAULICS Previous Question Paper (March - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
A vertical gap 2.2 cm wide of infinite extent contains a fluid of viscosity 2.0 N s/m2 and specific gravity 9.0. A metallic plate 1.2 m × 1.2 m × 0.2 cm is to be lifted up with a constant velocity of 0.15 m/s, through the gap. If the plate is in the middle of the gap, find the force required. The weight of the plate is 40 N.
Answer
To find the force required to lift the plate, we must account for the weight of the plate and the viscous shear force acting on both sides of the plate.
Given data:
- Gap width (W) = 2.2 cm = 0.022 m
- Plate thickness (t) = 0.2 cm = 0.002 m
- Plate dimensions = 1.2 m × 1.2 m
- Velocity (v) = 0.15 m/s
- Viscosity (μ) = 2.0 N s/m2
- Specific gravity (S) = 9.0
- Weight of plate (Wp) = 40 N
The clearance on each side of the plate (y) is:
The shear stress (τ) on each side of the plate is calculated using Newton's law of viscosity:
The total shear force (Fs) acting on both sides of the plate (Area A = 1.2 × 1.2 = 1.44 m2) is:
The buoyant force (Fb) acts upwards on the plate. The density of the fluid is ρ = 9.0 × 1000 = 9000 kg/m3. The volume of the plate is V = 1.2 × 1.2 × 0.002 = 0.00288 m3.
The force required (F) to lift the plate at constant velocity is:
The negative sign indicates that the plate is naturally buoyant and would require a downward force of 128.07 N to maintain the constant upward velocity of 0.15 m/s.
A tank contains water upto a height of 0.5 m above the base. An immiscible liquid of sp.gr. 0.8 is filled on top of the water upto 1 m height. Calculate total pressure on one side of the tank and position of center of pressure for one side of the tank, which is 2 m wide.
Answer
To calculate the total pressure and the center of pressure, we consider the two layers of liquids acting on the vertical side of the tank. The side is 2 m wide (b = 2 m). Let the water depth be h1 = 0.5 m (density ρ1 = 1000 kg/m³) and the oil depth be h2 = 1.0 m (density ρ2 = 800 kg/m³).
The pressure at the interface (depth h2 from the top) is P1 = ρ2 * g * h2 = 800 * 9.81 * 1.0 = 7848 N/m².
The pressure at the base (depth h1 + h2 = 1.5 m) is P2 = P1 + ρ1 * g * h1 = 7848 + (1000 * 9.81 * 0.5) = 7848 + 4905 = 12753 N/m².
The total pressure force (F) is the sum of the forces exerted by the two layers. For the oil layer (a rectangle and a triangle): F1 = (Area of pressure diagram) * width = [ (P1 * h2) + (0.5 * P1 * h2) ] * b = [ 7848 * 1 + 0.5 * 7848 * 1 ] * 2 = 23544 N. For the water layer: F2 = [ (P1 * h1) + (0.5 * (P2 - P1) * h1) ] * b = [ 7848 * 0.5 + 0.5 * (12753 - 7848) * 0.5 ] * 2 = 7848 + 2452.5 = 10300.5 N. Total Force F = F1 + F2 = 33844.5 N.
The center of pressure (h_cp) is found by taking moments about the free surface: h_cp = [ (F1 * y1) + (F2 * y2) ] / F, where y1 is the centroid of the oil pressure distribution and y2 is the centroid of the water pressure distribution. For the oil layer, the force acts at y1 = (h2/3) * (P1 + 2*0) / (P1 + 0) = 0.333 m from the interface. For the water layer, the centroid is at h2 + (h1/3) * (P1 + 2*P2) / (P1 + P2) = 1.0 + (0.5/3) * (7848 + 25506) / (7848 + 12753) = 1.0 + 0.278 = 1.278 m from the top. Calculating the weighted average yields the final center of pressure.
An open circular cylinder of 15 cm diameter and 100 cm long contains water upto a height of 80 cm. Find the maximum speed at which the cylinder is to be rotated about its vertical axis so that no water spills.
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A 25 cm diameter pipe carries oil of sp.gr. 0.9 at a velocity of 3 m/s. At another section the diameter is 20 cm. Find the velocity at this section and also mass rate of flow of oil.
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A stream function is given by
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A pipe of 300 mm diameter conveying 0.3 m3/s of water has a right-angled bend in a horizontal plane. Find the resultant force exerted on the bend if the pressure at inlet and outlet of the bend are 24.252 N/cm2 and 23.544 N/cm2.
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A horizontal venturimeter with inlet and throat diameters 30 cm and 15 cm respectively is used to measure the flow of water. The reading of differential manometer connected to the inlet and throat is 20 cm of mercury. Determine the rate of flow. Take Cd = 0.98.
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A pipeline of 0.6 m diameter is 1.5 km long. To increase the discharge, another pipeline of same diameter is introduced parallel to the first in the second half of the length. Neglecting minor losses, find the increase in discharge if 4f = 0.04. The head at inlet is 300 mm.
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With neat sketch explain principle and working of a centrifugal pump.
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What is reciprocating pump? Describe the principle and working of a reciprocating pump with a neat sketch. Why is a reciprocating pump not coupled directly to the motor? Discuss the reason in detail.
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