The velocity profile of a viscous fluid over a plate is parabolic with its vertex 20 cm from the plate, where the velocity is 120 cm/s. Calculate the velocity gradient and shear stress at distances of 0, 5, and 15 cm from the plate, given the viscosity of the fluid is 6 poise.

Shear Stress in Parabolic Flow

Problem Statement

Given Data

Vertex Distance, $y_{vertex} = 20 \, \text{cm} = 0.2 \, \text{m}$
Vertex Velocity, $u_{vertex} = 120 \, \text{cm/s} = 1.2 \, \text{m/s}$
Dynamic Viscosity, $\mu = 6 \, \text{poise}$
Distances for calculation: $y = 0, 5, 15 \, \text{cm}$

Solution

1. Establish the Velocity Profile Equation

A parabolic profile can be described by the equation $u = ay^2 + by + c$. We use the given boundary conditions to find the constants $a$, $b$, and $c$.

Condition 1: At the plate, y = 0, u = 0 (no-slip).

$$ u(0) = a(0)^2 + b(0) + c $$ $$ 0 = c $$ $$ \text{So, } u = ay^2 + by $$

Condition 2: At the vertex, the velocity gradient is zero.

$$ \frac{du}{dy} = 2ay + b $$ $$ \text{At } y = 0.2 \, \text{m}, \frac{du}{dy} = 0 $$ $$ 0 = 2a(0.2) + b $$ $$ b = -0.4a $$

Condition 3: At the vertex, y = 0.2 m, u = 1.2 m/s.

$$ u(0.2) = a(0.2)^2 + b(0.2) $$ $$ 1.2 = 0.04a + 0.2b $$

Substitute $b = -0.4a$ into the equation:

$$ 1.2 = 0.04a + 0.2(-0.4a) $$ $$ 1.2 = 0.04a - 0.08a $$ $$ 1.2 = -0.04a $$ $$ a = \frac{1.2}{-0.04} = -30 $$

Now find b:

$$ b = -0.4a $$ $$ b = -0.4(-30) = 12 $$

The final velocity profile equation is:

$$ u = -30y^2 + 12y $$

2. Convert Viscosity to SI Units

$$ \mu = 6 \, \text{poise} \times \frac{1 \, \text{N s/m}^2}{10 \, \text{poise}} $$ $$ \mu = 0.6 \, \text{N s/m}^2 $$

3. Calculations at y = 0 cm (0 m)

Velocity Gradient:

$$ \frac{du}{dy} = -60y + 12 $$ $$ \left. \frac{du}{dy} \right|_{y=0} = -60(0) + 12 $$ $$ \left. \frac{du}{dy} \right|_{y=0} = 12 \, \text{s}^{-1} $$

Shear Stress:

$$ \tau = \mu \frac{du}{dy} $$ $$ \tau_0 = 0.6 \, \text{N s/m}^2 \times 12 \, \text{s}^{-1} $$ $$ \tau_0 = 7.2 \, \text{N/m}^2 $$

4. Calculations at y = 5 cm (0.05 m)

Velocity Gradient:

$$ \left. \frac{du}{dy} \right|_{y=0.05} = -60(0.05) + 12 $$ $$ \left. \frac{du}{dy} \right|_{y=0.05} = -3 + 12 $$ $$ \left. \frac{du}{dy} \right|_{y=0.05} = 9 \, \text{s}^{-1} $$

Shear Stress:

$$ \tau = \mu \frac{du}{dy} $$ $$ \tau_5 = 0.6 \, \text{N s/m}^2 \times 9 \, \text{s}^{-1} $$ $$ \tau_5 = 5.4 \, \text{N/m}^2 $$

5. Calculations at y = 15 cm (0.15 m)

Velocity Gradient:

$$ \left. \frac{du}{dy} \right|_{y=0.15} = -60(0.15) + 12 $$ $$ \left. \frac{du}{dy} \right|_{y=0.15} = -9 + 12 $$ $$ \left. \frac{du}{dy} \right|_{y=0.15} = 3 \, \text{s}^{-1} $$

Shear Stress:

$$ \tau = \mu \frac{du}{dy} $$ $$ \tau_{15} = 0.6 \, \text{N s/m}^2 \times 3 \, \text{s}^{-1} $$ $$ \tau_{15} = 1.8 \, \text{N/m}^2 $$

Final Results:

At y = 0 cm: Velocity Gradient = 12 s⁻¹, Shear Stress = 7.2 N/m²

At y = 5 cm: Velocity Gradient = 9 s⁻¹, Shear Stress = 5.4 N/m²

At y = 15 cm: Velocity Gradient = 3 s⁻¹, Shear Stress = 1.8 N/m²

Explanation of the Method

1. Deriving the Profile:
The most critical step is to translate the problem's description into a mathematical equation. By using the standard form of a parabola ($u = ay^2 + by + c$) and applying the known physical constraints (velocity at the wall, position and velocity at the vertex), we can solve for the unknown constants to create a complete model of the flow.

2. Velocity Gradient and Shear Stress:
Once the velocity profile $u(y)$ is known, we can find the velocity gradient ($du/dy$) by differentiation. This gradient tells us how rapidly the fluid velocity changes with distance from the plate. According to Newton's law of viscosity, this gradient, when multiplied by the fluid's dynamic viscosity ($\mu$), gives the shear stress ($\tau$) at any point.

Physical Meaning

The results show that both the velocity gradient and the shear stress are highest at the plate (y=0) and decrease as we move away from it, reaching zero at the point of maximum velocity (the vertex at y=20 cm).

This makes physical sense:

The greatest change in velocity (the steepest gradient) occurs right next to the stationary plate, where the fluid goes from zero velocity to moving relatively quickly over a short distance.
This high rate of shearing results in the maximum internal friction, or shear stress, at the wall. This is the point where the fluid exerts the most drag force on the plate.
As you move towards the center of the flow (the vertex), the fluid layers are moving at more similar speeds, reducing the velocity gradient and thus lowering the shear stress.

The velocity profile of a viscous fluid over a plate is parabolic with its vertex 20 cm from the plate, where the velocity is 120 cm/s. Calculate the velocity gradient and shear stress at distances of 0, 5, and 15 cm from the plate, given the viscosity of the fluid is 6 poise.

Problem Statement

Given Data

Solution

1. Establish the Velocity Profile Equation

2. Convert Viscosity to SI Units

3. Calculations at y = 0 cm (0 m)

4. Calculations at y = 5 cm (0.05 m)

5. Calculations at y = 15 cm (0.15 m)

Explanation of the Method

Physical Meaning

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