Velocity Gradient Computation

Problem Statement:

Given the velocity profile for laminar flow in a pipe:

$$ u(y) = u_{\text{max}} \left(1 - \frac{y^2}{R^2}\right) $$

where:

$$ u_{\text{max}} $$ is the maximum velocity at the center of the pipe.
$$ R $$ is the radius of the pipe.
$$ y $$ is the distance from the center of the pipe.

Find the velocity gradient $$ \frac{\partial u}{\partial y} $$ with respect to $$ y $$.

Step-by-Step Computation:

1. Write the Velocity Profile:

We start with the given velocity profile:

$$ u(y) = u_{\text{max}} \left(1 - \frac{y^2}{R^2}\right) $$

2. Differentiate the Velocity Profile:

We need to find the derivative of $$ u(y) $$ with respect to $$ y $$:

$$ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \left[ u_{\text{max}} \left(1 - \frac{y^2}{R^2}\right) \right] $$

3. Apply the Chain Rule:

The chain rule states that the derivative of a product is the derivative of the first function times the second function plus the first function times the derivative of the second function. Here, $$ u_{\text{max}} $$ is a constant, so we can factor it out:

$$ \frac{\partial u}{\partial y} = u_{\text{max}} \cdot \frac{\partial}{\partial y} \left(1 - \frac{y^2}{R^2}\right) $$

4. Differentiate the Inner Expression:

Now, we need to differentiate $$ 1 - \frac{y^2}{R^2} $$ with respect to $$ y $$:

$$ \frac{\partial}{\partial y} \left(1 - \frac{y^2}{R^2}\right) = \frac{\partial}{\partial y} (1) - \frac{\partial}{\partial y} \left(\frac{y^2}{R^2}\right) $$

Since the derivative of a constant (1) is zero:

$$ \frac{\partial}{\partial y} (1) = 0 $$

Next, we differentiate $$ \frac{y^2}{R^2} $$:

$$ \frac{\partial}{\partial y} \left(\frac{y^2}{R^2}\right) = \frac{1}{R^2} \cdot \frac{\partial}{\partial y} (y^2) $$

5. Explanation of $$ y^2 $$ Becoming $$ 2y $$:

The derivative of $$ y^2 $$ with respect to $$ y $$ is found using the power rule. The power rule states that if you have a term $$ y^n $$, its derivative with respect to $$ y $$ is $$ n \cdot y^{n-1} $$. For $$ y^2 $$:

$$ \frac{\partial}{\partial y} (y^2) = 2 \cdot y^{2-1} = 2y $$

So, we have:

$$ \frac{\partial}{\partial y} \left(\frac{y^2}{R^2}\right) = \frac{1}{R^2} \cdot 2y = \frac{2y}{R^2} $$

6. Combine the Results:

Substitute the result back into the original differentiation:

$$ \frac{\partial}{\partial y} \left(1 - \frac{y^2}{R^2}\right) = 0 - \frac{2y}{R^2} = -\frac{2y}{R^2} $$

7. Multiply by the Constant:

Now, multiply by the constant $$ u_{\text{max}} $$:

$$ \frac{\partial u}{\partial y} = u_{\text{max}} \cdot \left(-\frac{2y}{R^2}\right) $$

Final Result:

The velocity gradient $$ \frac{\partial u}{\partial y} $$ is:

$$ \frac{\partial u}{\partial y} = -\frac{2 u_{\text{max}} y}{R^2} $$

This shows how the velocity changes with respect to the distance $$ y $$ from the center of the pipe. The negative sign indicates that the velocity decreases as you move away from the center towards the pipe walls.

Velocity Gradient Computation