Problem Statement
A piston of weight 90N slides in a lubricated pipe. The clearance between piston and pipe is 0.025mm. If the piston decelerates at 0.6m/s² when the speed is 0.5m/s, what is the viscosity of the oil?Given Data
Weight of piston (W) = 90N
Clearance (dy) = 0.025mm = 0.000025m
Deceleration (a) = 0.6m/s²
Change in velocity (du) = 0.5m/s
Diameter of piston (D) = 125mm = 0.125m
Length of piston (L) = 130mm = 0.13m
Viscosity of oil (μ) = ?
Clearance (dy) = 0.025mm = 0.000025m
Deceleration (a) = 0.6m/s²
Change in velocity (du) = 0.5m/s
Diameter of piston (D) = 125mm = 0.125m
Length of piston (L) = 130mm = 0.13m
Viscosity of oil (μ) = ?
Solution
Step 1: Surface Area Calculation
A = πDL
A = π × 0.125 × 0.13
A = 0.0511 m²
A = π × 0.125 × 0.13
A = 0.0511 m²
Step 2: Force Analysis
W = F + ma
90 = F + (90/9.81) × 0.6
90 = F + 5.5
F = 84.5 N
90 = F + (90/9.81) × 0.6
90 = F + 5.5
F = 84.5 N
Step 3: Shear Stress Calculation
τ = F/A
τ = 84.5/0.0511
τ = 1653.6 N/m²
τ = 84.5/0.0511
τ = 1653.6 N/m²
Step 4: Viscosity Determination
τ = μ × (du/dy)
μ = τ × (dy/du)
μ = 1653.6 × (0.000025/0.5)
μ = 0.094 NS/m²
μ = τ × (dy/du)
μ = 1653.6 × (0.000025/0.5)
μ = 0.094 NS/m²
Final Result: The viscosity of the oil (μ) = 0.094 NS/m² (Pascal-seconds)
Technical Explanation
1. Force Balance Analysis
• Total weight (90N) is balanced by viscous force and inertial force
• Inertial force = ma = (90/9.81) × 0.6 = 5.5N
• Resulting viscous force = 84.5N drives fluid deformation
• Inertial force = ma = (90/9.81) × 0.6 = 5.5N
• Resulting viscous force = 84.5N drives fluid deformation
2. Shear Stress Development
• Shear stress is calculated from force per unit area
• Surface area considers the cylindrical contact surface
• Stress magnitude indicates fluid resistance to motion
• Surface area considers the cylindrical contact surface
• Stress magnitude indicates fluid resistance to motion
3. Viscosity Analysis
• Based on Newton’s law of viscosity
• Velocity gradient = 0.5/0.000025 = 20,000 s⁻¹
• Final viscosity (0.094 NS/m²) indicates medium-weight oil
• Suitable for typical machinery applications
• Velocity gradient = 0.5/0.000025 = 20,000 s⁻¹
• Final viscosity (0.094 NS/m²) indicates medium-weight oil
• Suitable for typical machinery applications
