JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year FLUID MECHANICS AND HYDRAULICS Previous Question Paper (October - 2020 - R18)

To determine the dynamic and kinematic viscosity of the oil, we apply the principles of Newton's Law of Viscosity.

Given Data:

• Side of square plate (L) = 60 cm = 0.6 m
• Area of plate (A) = 0.6 × 0.6 = 0.36 m²
• Thickness of oil film (dy) = 12.5 mm = 0.0125 m
• Velocity of upper plate (du) = 2.5 m/s
• Force required (F) = 98.1 N
• Specific gravity of oil (S) = 0.95

1. Calculation of Dynamic Viscosity (μ):

According to Newton's Law of Viscosity, the shear stress (τ) is given by:

τ = μ (du / dy)

We know that shear stress τ = F / A:

τ = 98.1 / 0.36 = 272.5 N/m²

Substituting the values into the viscosity equation:

272.5 = μ (2.5 / 0.0125)

272.5 = μ (200)

μ = 272.5 / 200 = 1.3625 N·s/m²

Since 1 N·s/m² = 10 poise:

μ = 13.625 poise

2. Calculation of Kinematic Viscosity (ν):

Kinematic viscosity is defined as the ratio of dynamic viscosity to mass density (ρ):

ν = μ / ρ

Density of oil (ρ) = S × ρ_water = 0.95 × 1000 = 950 kg/m³.

ν = 1.3625 / 950 = 0.001434 m²/s

Since 1 m²/s = 10,000 stokes:

ν = 0.001434 × 10,000 = 14.34 stokes

Final Results:

• Dynamic Viscosity = 13.625 poise
• Kinematic Viscosity = 14.34 stokes

To calculate the total pressure and the center of pressure on one side of the tank, we consider the two distinct liquid layers acting on the vertical side wall of width w = 2 m.

• Layer 1 (Water): Height h₁ = 0.5 m, Density ρ₁ = 1000 kg/m³.
• Layer 2 (Immiscible liquid): Height h₂ = 1.0 m, Specific gravity S₂ = 0.8, Density ρ₂ = 800 kg/m³.

1. Total Pressure Force (F): The total force is the sum of forces exerted by the two liquids.

• Force due to immiscible liquid (F₁): The pressure at the interface is P_i = ρ₂gh₂ = 800 × 9.81 × 1 = 7848 N/m². The force is the area of the pressure diagram (trapezoid): F₁ = (P_i × h₂ × w) = 7848 × 1 × 2 = 15696 N.
• Force due to water (F₂): The pressure at the base is P_b = P_i + ρ₁gh₁ = 7848 + (1000 × 9.81 × 0.5) = 7848 + 4905 = 12753 N/m². The force is the area of the trapezoid: F₂ = ((P_i + P_b)/2) × h₁ × w = ((7848 + 12753)/2) × 0.5 × 2 = 20601 N.
• Total Force F = F₁ + F₂ = 15696 + 20601 = 36297 N.

2. Center of Pressure (h̅): The center of pressure from the base is calculated by taking moments of the forces about the base.

• Moment of F₁ about base: M₁ = F₁ × (h₁ + h₂/3) = 15696 × (0.5 + 1/3) = 13080 Nm.
• Moment of F₂ about base: M₂ = (Area_rect × h₁/2) + (Area_tri × h₁/3) = (7848 × 0.5 × 2) × 0.25 + (0.5 × 4905 × 0.5 × 2) × (0.5/3) = 1962 + 408.75 = 2370.75 Nm.
• Total distance from base h̅ = (M₁ + M₂) / F = (13080 + 2370.75) / 36297 ≈ 0.4257 m from the base.