Problem Statement
A cylinder of 12cm radius rotates concentrically inside a fixed cylinder of 12.6cm radius. Both cylinders are 0.3m long. Determine the viscosity of the liquid that fills the space between the cylinders if a torque of 0.9 Nm is required to maintain an angular velocity of 60rpm.
Solution
Given:
- Radius of outer cylinder (r₁) = 12.6cm = 0.126m
- Radius of inner cylinder (r₂) = 12cm = 0.12m
- Clearance (dr) = dy = r₁ – r₂ = 0.126 – 0.12 = 0.006m
- Length of cylinder (L) = 0.3m
- Torque (T) = 0.9 Nm
- Speed (N) = 60rpm
Calculations:
1. Angular velocity (ω):
ω = (2 × π × N) / 60 = (2 × π × 60) / 60 = 6.28 rad/s
2. Tangential velocity of inner cylinder (u):
u = r₂ × ω = 0.12 × 6.28 = 0.75 m/s
3. Torque (T) and shear stress (τ):
T = F × r₂ = μ × (du/dy) × (2πr₂L) × r₂
Substitute the values:
0.9 = μ × (0.75 / 0.006) × (2π × 0.12 × 0.3) × 0.12
4. Simplifying for viscosity (μ):
μ = 0.265 NS/m²
Result:
Viscosity of the liquid (μ): 0.265 NS/m²
Explanation
This problem involves determining the viscosity of a liquid in the gap between two concentric cylinders, one of which is fixed while the other rotates. Here’s the explanation:
- Angular velocity (ω): The angular velocity of the rotating cylinder is calculated from the given speed (rpm).
- Tangential velocity (u): The tangential velocity of the inner cylinder’s surface is determined by multiplying its radius by the angular velocity.
- Shear stress and torque: The resisting torque (T) due to the liquid’s viscosity is related to the shear stress, velocity gradient (du/dy), and the contact surface area. This relationship allows us to solve for the liquid’s viscosity (μ).
- Final viscosity calculation: Substituting all known values into the equation and solving gives the liquid’s viscosity as 0.265 NS/m².
This solution demonstrates the application of fluid dynamics principles in determining the viscosity of a liquid based on the resistance it offers to relative motion between two cylinders.
