The velocity distribution for flow over a plate is given by u= (3/2)y-y^3/2 where u is the point velocity in metre per second at a distance of y metre above the plate. Determine the shear stress at y= 9cm . Assume the dynamic viscosity as 8 poise
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To determine the shear stress at \( y = 9 \) cm, we first need to calculate the velocity gradient at that point. The velocity distribution is given by:

\[ u = \frac{3}{2} y - y^{3/2} \]

The shear stress (\(\tau\)) in a fluid flow is given by the equation:

\[ \tau = \mu \frac{du}{dy} \]

where:
- \(\mu\) is the dynamic viscosity
- \(\frac{du}{dy}\) is the velocity gradient

Given:
- Dynamic viscosity, \(\mu = 8\) poise = \( 8 \times 0.1 \) Pa·s (since 1 poise = 0.1 Pa·s)
- \( y = 9 \) cm = 0.09 m

First, let's find the velocity gradient (\(\frac{du}{dy}\)):

\[ u = \frac{3}{2} y - y^{3/2} \]

Taking the derivative of \(u\) with respect to \(y\):

\[ \frac{du}{dy} = \frac{d}{dy} \left( \frac{3}{2} y - y^{3/2} \right) \]

\[ \frac{du}{dy} = \frac{3}{2} - \frac{3}{2} y^{1/2} \]

Now, substituting \( y = 0.09 \) m into the velocity gradient:

\[ \frac{du}{dy} \bigg|_{y=0.09} = \frac{3}{2} - \frac{3}{2} (0.09)^{1/2} \]

\[ \frac{du}{dy} \bigg|_{y=0.09} = \frac{3}{2} - \frac{3}{2} \sqrt{0.09} \]

\[ \frac{du}{dy} \bigg|_{y=0.09} = \frac{3}{2} - \frac{3}{2} \times 0.3 \]

\[ \frac{du}{dy} \bigg|_{y=0.09} = \frac{3}{2} - \frac{3}{2} \times 0.3 \]

\[ \frac{du}{dy} \bigg|_{y=0.09} = \frac{3}{2} - 0.45 \]

\[ \frac{du}{dy} \bigg|_{y=0.09} = 1.5 - 0.45 \]

\[ \frac{du}{dy} \bigg|_{y=0.09} = 1.05 \, \text{m/s} \]

Finally, we calculate the shear stress (\(\tau\)):

\[ \tau = \mu \frac{du}{dy} \]

\[ \tau = 0.8 \times 1.05 \]

\[ \tau = 0.84 \, \text{Pa} \]

Therefore, the shear stress at \( y = 9 \) cm is \( 0.84 \) Pa.

Would you like more details or have any questions?

Here are 5 numbered questions you might want to ask:

1. How do you convert dynamic viscosity from poise to Pascal-seconds?
2. Can you explain the concept of shear stress in fluid mechanics?
3. What is the significance of the velocity gradient in calculating shear stress?
4. How would the shear stress change if the dynamic viscosity were different?
5. Can you provide a step-by-step explanation of taking the derivative of the velocity equation?

Tip: Always ensure units are consistent when performing calculations involving physical quantities.

Learn how to calculate shear stress in fluid mechanics using the velocity distribution formula. This example problem involves determining shear stress at a specified distance from a plate, given dynamic viscosity. Suitable for undergraduate-level studies.

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