A 15 cm diameter vertical cylinder rotates concentrically inside another cylinder of diameter 15.10 cm. Both cylinders are 25 cm high. The space between the cylinders is filled with a liquid of unknown viscosity. If a torque of 12.0 Nm is required to rotate the inner cylinder at 100 rpm, determine the viscosity of the fluid.

Viscosity Determination in Concentric Cylinders

Problem Statement

A 15 cm diameter vertical cylinder rotates concentrically inside another cylinder of diameter 15.10 cm. Both cylinders are 25 cm high. The space between the cylinders is filled with a liquid whose viscosity is unknown. If a torque of 12.0 Nm is required to rotate the inner cylinder at 100 r.p.m., determine the viscosity of the fluid.

Given Data

Diameter of inner cylinder, $D_i = 15 \, \text{cm} = 0.15 \, \text{m}$
Diameter of outer cylinder, $D_o = 15.10 \, \text{cm} = 0.151 \, \text{m}$
Length of cylinders, $L = 25 \, \text{cm} = 0.25 \, \text{m}$
Torque, $T = 12.0 \, \text{Nm}$
Speed of inner cylinder, $N = 100 \, \text{r.p.m.}$

Solution

1. Calculate Tangential Velocity of Inner Cylinder

The tangential velocity $u$ of the inner cylinder is given by:

$$ u = \frac{\pi D_i N}{60} $$ $$ u = \frac{\pi \times 0.15 \, \text{m} \times 100 \, \text{r.p.m.}}{60} $$ $$ u \approx 0.7854 \, \text{m/s} $$

2. Determine the Gap Thickness (dy)

The space between the cylinders, $dy$, is half the difference in their diameters:

$$ dy = \frac{D_o - D_i}{2} $$ $$ dy = \frac{0.151 \, \text{m} - 0.150 \, \text{m}}{2} $$ $$ dy = \frac{0.001 \, \text{m}}{2} = 0.0005 \, \text{m} $$

3. Calculate the Surface Area of the Inner Cylinder

The surface area $A$ of the inner cylinder in contact with the fluid is:

$$ A = \pi D_i L $$ $$ A = \pi \times 0.15 \, \text{m} \times 0.25 \, \text{m} $$ $$ A \approx 0.1178 \, \text{m}^2 $$

4. Apply Newton's Law of Viscosity to find Shear Stress

Newton's law of viscosity states that shear stress $\tau$ is proportional to the velocity gradient $\frac{du}{dy}$:

$$ \tau = \mu \frac{du}{dy} $$

Here, $du = u - 0 = u = 0.7854 \, \text{m/s}$ (assuming the outer cylinder is stationary).

$$ \tau = \mu \frac{0.7854 \, \text{m/s}}{0.0005 \, \text{m}} $$

5. Calculate Shear Force (F)

The shear force $F$ acting on the inner cylinder is the shear stress multiplied by the surface area:

$$ F = \tau \times A $$ $$ F = \left( \mu \frac{0.7854}{0.0005} \right) \times 0.1178 $$

6. Relate Torque to Shear Force and Determine Viscosity

The torque $T$ required to rotate the inner cylinder is the shear force multiplied by the radius of the inner cylinder ($D_i/2$):

$$ T = F \times \frac{D_i}{2} $$ $$ 12.0 \, \text{Nm} = \left( \mu \frac{0.7854}{0.0005} \times 0.1178 \right) \times \frac{0.15}{2} $$

Now, solve for $\mu$:

$$ \mu = \frac{12.0 \times 0.0005 \times 2}{0.7854 \times 0.1178 \times 0.15} $$ $$ \mu \approx 0.864 \, \text{N s/m}^2 $$

Final Result:

The viscosity of the fluid is $ \mu = 0.864 \, \text{N s/m}^2 $.

Converting to Poise (1 N s/m$^2$ = 10 Poise):

$ \mu = 0.864 \times 10 = 8.64 \, \text{Poise} $

Explanation

1. Velocity Calculation:
The tangential velocity of the inner cylinder is calculated based on its diameter and rotational speed. This velocity represents the relative motion between the fluid layer adhering to the inner cylinder and the fluid layer near the outer cylinder.

2. Gap and Area:
The small radial gap between the cylinders is crucial for determining the velocity gradient. The surface area of the inner cylinder in contact with the fluid is used to calculate the total shear force.

3. Newton's Law of Viscosity:
Assuming a linear velocity profile across the small gap (which is a reasonable approximation for small gaps), Newton's law of viscosity is applied. This law directly relates shear stress to the fluid's viscosity and the velocity gradient.

4. Torque Relationship:
The applied torque is the product of the shear force acting on the inner cylinder's surface and the radius of the inner cylinder. By equating the given torque to this expression, the unknown viscosity can be determined.

Physical Meaning

1. Viscosity as Resistance to Flow:
The calculated viscosity ($\mu$) quantifies the fluid's internal resistance to flow. A higher viscosity means the fluid offers more resistance to deformation, requiring a greater torque to maintain the same rotational speed.

2. Shear Stress in Annular Flow:
In this setup, the fluid layers near the inner rotating cylinder experience a high shear stress, which gradually decreases towards the stationary outer cylinder. This stress is responsible for transmitting the torque.

3. Practical Applications:
This type of problem is fundamental in understanding the behavior of lubricants in bearings, the operation of viscometers (devices used to measure viscosity), and the design of fluid-filled mechanical systems where rotational motion occurs in confined spaces.

4. Importance of Small Gap:
The very small gap between the cylinders allows for the assumption of a linear velocity profile, simplifying the calculation. In real-world applications, this small gap ensures efficient power transmission or lubrication with minimal fluid loss.