Chapter 1, Problem 1.51P

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(a)

To determine

By estimating the wall shear stress t_win Pa.

Answer to Problem 1.51P

The wall shear stress is $=0.011\text{Pa}$

Explanation of Solution

Given:

An approximation for the boundary layer shape

$u\left(y\right)\approx U\sin \left(\frac{\pi y}{2\delta }\right),0\le y\le \delta$

Where

$\begin{array}{l}\text{U}=\text{stream}\text{velocity}\text{far}\text{from}\text{wall}=10.8\text{m/s}\\ \text{δ}=\text{boundary}\text{layer}\text{thickness}=\text{3}\text{cm}\end{array}$

Concept Used:

Calculate the shear stain using the velocity of the boundary layer

$\begin{array}{c}\frac{du\left(y\right)}{dy}=\frac{d}{dx}\left[U\sin \left(\pi y/2\delta \right)\right]\\ =\frac{U\pi }{2\delta }\left[\cos \left(\pi y/2\delta \right)\right]\end{array}$

Calculation:

Calculate the shear strain at the wall

$\begin{array}{c}{\frac{du}{dy}|}_{y=o}=\frac{U\pi }{2\delta }\cos \left(\pi \left(0\right)/2\delta \right)\\ =\frac{U\pi }{2\delta }\end{array}$

Calculate the shear stress at the wall

$\begin{array}{c}{\tau }_w=\mu {\frac{du}{dy}|}_{y=0}\\ =\mu \left(U\pi /2\delta \right)\end{array}$

Here, the viscosity of helium at $200°\text{C}$ and 1 atm is $\mu$

Putting

$\begin{array}{l}\mu =1.97x{10}^{-5}\text{kg}\text{.m/s}\\ \delta =0.03\text{m}\\ U=10.8\text{m/s}\end{array}$

In the above equation

$\begin{array}{c}{\tau }_w=\left[\left(1.091\text{x}{10}^{-5}\right)\frac{10.8\pi }{2\text{x}0.03}\right]\\ =0.011{\text{N/m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the wall shear stress is

$\begin{array}{c}=0.011{\text{N/m}}^{\text{2}}\\ or\\ =0.011\text{Pa}\end{array}$

<

(b)

To determine

 By finding the position in the boundary layer where t is one half of t_w.

Answer to Problem 1.51P

The position in the boundary layer where t in one half of t_wis $=0.02\text{m}$

Explanation of Solution

Given:

An approximation for the boundary layer shape

$u\left(y\right)\approx U\sin \left(\frac{\pi y}{2\delta }\right),0\le y\le \delta$

Where

$\begin{array}{l}\text{U}=\text{stream}\text{velocity}\text{far}\text{from}\text{wall}=10.8\text{m/s}\\ \text{δ}=\text{boundary}\text{layer}\text{thickness}=\text{3}\text{cm}\end{array}$

Concept Used:

Calculate the shear stain using the velocity of the boundary layer

$\begin{array}{c}\frac{du\left(y\right)}{dy}=\frac{d}{dx}\left[U\sin \left(\pi y/2\delta \right)\right]\\ =\frac{U\pi }{2\delta }\left[\cos \left(\pi y/2\delta \right)\right]\end{array}$

Calculation:

Calculate the shear stress with the help of the ${\tau }_w$

$\begin{array}{c}\tau =\frac{{\tau }_w}{2}\\ =\frac{0.011}{2}\\ =5.57\text{x}{10}^{-3}{\text{N/m}}^{\text{2}}\end{array}$

Find the position of the boundary layer

$\begin{array}{c}\tau =\frac{du}{dy}\\ =\mu \frac{U\pi }{2\delta }\cos \left(\frac{\pi y}{2\delta }\right)\\ 5.57x{10}^{-3}=\left({1.9710}^{-5}\right)\left[\frac{10.8\pi }{2\text{x}0.03}\right]\cos \left(\pi y/2\text{x}0.03\right)\end{array}$

$\begin{array}{c}\cos \left(\pi y/2x0.03\right)=0.5\\ \frac{\pi y}{0.06}={\cos }^{-1}0.5\\ \frac{\pi y}{0.06}={\cos }^{-1}0.5\\ \frac{\pi y}{0.06}=\frac{\pi }{3}\\ y=0.02\text{m}\end{array}$

Conclusion:

Hence, the position in the boundary layer where t in one half of t_wis $=0.02\text{m}$