GATE Mechanical (ME) Test: Properties of Fluids & Turbulent Flow in Pipes

Given:

T₁ = 20°C → 273 + 20 = 293 K
ν₁ = 1.6 × 10⁻⁵ m²/s
T₂ = 70°C → 273 + 70 = 343 K
Since kinematic viscosity (ν) is proportional to T³/², we use:

(ν₁ / T₁³/²) = (ν₂ / T₂³/²)

Rearranging for ν₂:

ν₂ = ν₁ × (T₂ / T₁)³/²

Substituting values:

ν₂ = 1.6 × 10⁻⁵ × (343 / 293)³/²

ν₂ = 2.02 × 10⁻⁵ m²/s

Final Answer:
ν₂ = 2.02 × 10⁻⁵ m²/s

Step 1: Understanding Each Fluid Type
Ideal Fluid:
An ideal fluid has zero viscosity, meaning no shear stress is present.
Correct match → (4) Shear stress is zero.

Newtonian Fluid:
A Newtonian fluid follows Newton’s law of viscosity, meaning shear stress is directly proportional to the rate of strain.
Correct match → (1) Shear stress varies linearly with the rate of strain.

Non-Newtonian Fluid:
A Non-Newtonian fluid does not obey Newton’s law of viscosity, meaning shear stress does not vary linearly with strain rate.
Correct match → (2) Shear stress does not vary linearly with the rate of strain.

Bingham Plastic:
A Bingham plastic behaves like a solid until a certain yield stress is exceeded. After that, it flows like a Newtonian fluid.
Correct match → (3) Fluid behaves like a solid until a minimum yield stress, beyond which it exhibits a linear relationship between shear stress and the rate of strain.

Step 2: Matching List-I with List-II

Thus, the correct order is:
A → 4, B → 1, C → 2, D → 3

Ans. (d)

• Area of plate, A = 0.6 m²
• Weight of plate, W = 280 N
• Velocity of plate, u = 0.36 m/s
• Thickness of film, t = dy = 1.8 mm = 1.8 × 10⁻³ m

Viscosity of the fluid, μ:
Component of W along the plate:
W sinθ = 280 sin 30° = 140 N

Shear force on the bottom surface of the plate:
F = 140 N

Shear stress:
τ = F / A = 140 / 0.6 = 233.33 N/m²

Where,

• du = change of velocity = u - 0 = 0.36 m/s
• dy = t = 1.8 × 10⁻³ m

Using the shear stress equation:

233.33 = μ × (0.36 / 1.8 × 10⁻³)

Solving for μ:

μ = (233.33 × 1.8 × 10⁻³) / 0.36

μ = 1.166 N·s/m² = 11.66 poise

Given Data:

• Diameter of tube, d = 2.5 mm = 2.5 × 10^-3 m
• Surface tension, σ = 4 N/m
• Contact angle, θ = 135°
• Density of liquid, ρ = 13600 kg/m³
• Acceleration due to gravity, g = 9.81 m/s²

Step 1: Capillary Rise (or Fall) Formula
The capillary rise or fall is given by the formula:
h = (4σ cosθ) / (w × d)
where:
w = ρ × g (Specific weight of the liquid)
d = Diameter of the tube
Step 2: Substituting the Values
w = ρ × g = 13600 × 9.81 = 133416 N/m³
h = (4 × 4 × cos 135°) / (133416 × 2.5 × 10^-3)
h = (16 × (-0.707)) / (133416 × 2.5 × 10^-3)
h = (-11.31) / (333.54)
h = -3.39 × 10^-3 m or -3.39 mm

Step 3: Interpretation
The negative sign indicates a capillary depression (fall in liquid level).
Final Answer: The capillary fall is 3.39 mm.

For water and glass, Contact Angle (θ)=0º and Surface tension is 0.0725 N/m
For mercury and glass, Contact Angle (θ)=128º and Surface tension is 0.52 N/m

Ans. (d) =


or without calculating this we may say that it must be less than one and option (d) is only choice.

Given Data:

• Diameter of first capillary tube = 1.0 mm = 0.001 m
• Diameter of second capillary tube = 1.5 mm = 0.0015 m
• Surface tension of water = 0.0075 kg/m (or N/m)
• Contact angle = 0° (assumed to be perfectly wetting)
• Density of water (ρ) = 1000 kg/m³
• Acceleration due to gravity (g) = 9.81 m/s²

We need to calculate the difference in the liquid levels due to capillarity.

Step 1: Calculate Capillary Rise in Each Tube
The capillary rise (h) in a tube is given by the formula:
h = (4σ) / (ρ g d)
where:
σ = Surface tension = 0.0075 N/m
ρ = Density of water = 1000 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
d = Diameter of the tube
For the First Capillary (d = 0.001 m):
h₁ = (4 × 0.0075) / (1000 × 9.81 × 0.001)
h₁ = (0.03) / (9.81)
h₁ = 3.06 mm
For the Second Capillary (d = 0.0015 m):
h₂ = (4 × 0.0075) / (1000 × 9.81 × 0.0015)
h₂ = (0.03) / (14.715)
h₂ = 2.04 mm

Step 2: Calculate the Difference in Liquid Levels
Difference in menisci levels due to capillarity:
Δh = h₁ - h₂
Δh = 3.06 - 2.04
Δh = 1 mm

Final Answer:
The difference in liquid levels due to capillarity is 1 mm.

Given Data:

• Plate dimensions = 1.25 m × 1.25 m × 6 mm
• Weight of plate = 90 N
• Gap between vertical plates = 24 mm = 0.024 m
• The plate is placed midway, so each gap = 0.012 m
• Oil dynamic viscosity = 3.0 N·s/m²
• Velocity of plate = 0.15 m/s
• Specific gravity of oil = 0.85 (not needed in this calculation)

We need to determine the force required to lift the plate with a constant velocity.

Step 1: Calculate Shear Stress
Shear stress (τ) is given by:
τ = (μ × V) / h
where:
μ = dynamic viscosity = 3.0 N·s/m²
V = velocity of plate = 0.15 m/s
h = gap on one side = 0.012 m
τ = (3.0 × 0.15) / 0.012
τ = 0.45 / 0.012
τ = 37.5 N/m²

Step 2: Calculate Force Due to Viscosity
Force (F) is given by:
F = τ × Area
Area of the plate:
Area = 1.25 × 1.25 = 1.5625 m²
F = 37.5 × 1.5625
F = 58.59 N
Since there is oil on both sides of the plate, the total force required is:
Total force = 58.59 × 2 = 117.18 N

Step 3: Calculate Total Force Required to Lift the Plate
The total force required is the sum of viscous force and the weight of the plate:
Total Force = 117.18 + 90
Total Force = 168.08 N

Final Answer:
Force required to lift the plate = 168.08 N

Given Data:

• Diameter of bubble = 3.0 mm = 0.003 m
• Surface tension of water at 20°C = 0.0735 N/m
• Surrounding water pressure = 100.3 kN/m² = 100300 N/m²

We need to determine:
The excess air pressure required at the nozzle
The absolute pressure inside the bubble
Step 1: Calculate the Excess Pressure Required
The pressure inside a bubble due to surface tension is given by Laplace’s equation:
P_inside = P_surrounding + (4 × σ) / d

Substituting the values:
P_inside = 100300 + (4 × 0.0735) / 0.003
P_inside = 100300 + (0.294) / 0.003
P_inside = 100300 + 98
P_inside = 100398 N/m²
P_inside = 100.398 kN/m²

Step 2: Calculate the Excess Pressure at the Nozzle
Excess pressure required at the nozzle:
P_excess = P_inside - P_surrounding
P_excess = 100398 - 100300
P_excess = 98 N/m²
P_excess = 0.098 kN/m²

Final Answer:
Absolute pressure inside the bubble = 100.398 kN/m²
Excess pressure required at the nozzle = 0.098 kN/m²

Given Data:

• Diameter of shaft = 400 mm = 0.4 m
• Rotational speed = 200 rpm
• Bearing length = 120 mm = 0.12 m
• Oil film thickness = 1.5 mm = 0.0015 m
• Dynamic viscosity = 0.7 Ns/m²

Step 1: Calculate Tangential Velocity
U = (π × D × N) / 60
U = (3.1416 × 0.4 × 200) / 60
U = 4.188 m/s

Step 2: Calculate Shear Rate
Shear rate = U / h
Shear rate = 4.188 / 0.0015
Shear rate = 2792 s⁻¹

Step 3: Calculate Shear Stress
τ = μ × Shear rate
τ = 0.7 × 2792
τ = 1954.4 N/m²

Step 4: Calculate Viscous Force
F = τ × L
F = 1954.4 × 0.12
F = 234.53 N/m

Step 5: Calculate Torque
T = F × (D / 2)
T = 234.53 × (0.4 / 2)
T = 58.97 Nm

Step 6: Calculate Power Utilization
Angular velocity (ω) = (2π × N) / 60
ω = (2 × 3.1416 × 200) / 60
ω = 20.94 rad/s

Power (P) = T × ω
P = 58.97 × 20.94
P = 1.235 kW

Final Answer:
Torque required = 58.97 Nm
Power utilized = 1.235 kW

1. Absolute Viscosity (A)

• Absolute viscosity (also called dynamic viscosity) is measured in Poise.
• It represents the internal resistance to fluid flow.
• Correct match → (3) Poise

2. Kinematic Viscosity (B)

• Kinematic viscosity is the ratio of absolute viscosity to density.
• It is measured in Stokes (St).
• Correct match → (5) Stokes

3. Newtonian Fluid (C)

• A Newtonian fluid follows Newton's Law of Viscosity, which states that shear stress is proportional to strain rate.
• This means stress/strain rate is constant.
• Correct match → (4) Stress/Strain is constant

4. Surface Tension (D)

• Surface tension is a force per unit length and is measured in Newton per meter.
• Correct match → (2) Newton per meter

Ans. (a)

where f(t) is increasing

• A gas is often considered incompressible when the Mach number is less than 0.2. This is because the changes in density are usually negligible at this speed.
• A Newtonian fluid is characterised by a constant viscosity regardless of the applied stress. However, it is not necessarily incompressible or non-viscous.
• An ideal fluid is defined as having no viscosity and no surface tension.

The correct choice is 1 and 3, since:

• Statement 1 is correct: Gases are considered incompressible at low Mach numbers.
• Statement 3 is correct: An ideal fluid typically has negligible surface tension.

At the interface between a liquid and a gas at rest, the pressure difference is influenced by the shape of the interface.

• The pressure is higher on the concave side of the interface. This is due to the curvature affecting pressure distribution.
• This phenomenon is explained by the Young-Laplace equation, which relates surface tension and curvature to pressure difference.
• The pressure difference is crucial in understanding processes like bubble formation and capillary action.

In simple terms, when you have a curved interface, the side that bends inward (concave) experiences more pressure than the side that bulges outward (convex).