Chapter 7, Problem 7.43P

To determine

(a)

Stream velocity $U$.

Answer to Problem 7.43P

The stream velocity is 34 m/s.

Explanation of Solution

Given Information:

We have following figure:

And at $x=2$ m, shear stress is 2.1 Pa. The given flow is turbulent. Also, we know that air has density, $\rho =1.2$ kg/m³ and viscosity, $\mu =1.8\times {10}^{-5}$ kg/m·s.

Calculation:

Let’s calculate Reynolds’s number,

${\mathrm{Re}}_x=\frac{\rho Ux}{\mu }=\frac{1.2\times U\times 2}{1.8\times {10}^{-5}}$

And the coefficient of friction:

$C_f=\frac{0.027}{{\mathrm{Re}}_x{}^{1/7}}=\frac{0.027}{{[\rho Ux/\mu ]}^{1/7}}$

1. The shear stress is given as:

$\begin{array}{l}{\tau }_w=\frac{1}{2}{C}_f\rho U^2\\ \Rightarrow 2.1=\frac{1}{2}\left(\frac{0.027}{{[\rho Ux/\mu ]}^{1/7}}\right)\rho U^2\\ \Rightarrow 2.1=\frac{1}{2}\left(\frac{0.027}{{[1.2\times U\times 2/(1.8\times {10}^{-5})]}^{1/7}}\right)\times 1.2U^2\\ \Rightarrow U\approx 34\text{ m/s}\end{array}$

So, the stream velocity is 34 m/s.

Conclusion:

The stream velocity is 34 m/s.

To determine

(b)

The boundary layer thickness, $\delta$ at the element.

Answer to Problem 7.43P

The boundary layer thickness is 36 mm.

Explanation of Solution

Given Information:

We have following figure

And at $x=2$ m, shear stress is 2.1 Pa. The given flow is turbulent. Also, we know that air has density, $\rho =1.2$ kg/m³ and viscosity, $\mu =1.8\times {10}^{-5}$ kg/m·s.

Calculation:

Let’s calculate Reynolds’s number,

${\mathrm{Re}}_x=\frac{\rho Ux}{\mu }=\frac{1.2\times U\times 2}{1.8\times {10}^{-5}}$

And the coefficient of friction:

$C_f=\frac{0.027}{{\mathrm{Re}}_x{}^{1/7}}=\frac{0.027}{{[\rho Ux/\mu ]}^{1/7}}$

The shear stress is given as:

$\begin{array}{l}{\tau }_w=\frac{1}{2}{C}_f\rho U^2\\ \Rightarrow 2.1=\frac{1}{2}\left(\frac{0.027}{{[\rho Ux/\mu ]}^{1/7}}\right)\rho U^2\\ \Rightarrow 2.1=\frac{1}{2}\left(\frac{0.027}{{[1.2\times U\times 2/(1.8\times {10}^{-5})]}^{1/7}}\right)\times 1.2U^2\\ \Rightarrow U\approx 34\text{ m/s}\end{array}$

So, the stream velocity is 34 m/s.

Using this velocity, we can now, evaluate Reynolds’s number as below:

${\mathrm{Re}}_x=\frac{\rho Ux}{\mu }=\frac{1.2\times 34\times 2}{1.8\times {10}^{-5}}=4.54\times {10}^6$

Let’s calculate the boundary layer thickness, $\delta$ as below,

$\delta =\frac{0.16x}{{\mathrm{Re}}_x{}^{1/7}}=\frac{0.16\times 2}{{\left(4.54\times {10}^6\right)}^{1/7}}\approx 0.036\text{ m}=36\text{ mm}$

So, the boundary layer thickness is 36 mm.

Conclusion:

The boundary layer thickness is 36 mm.

To determine

(c)

Velocity, $u$ in m/s, at 5 cm above the element.

Answer to Problem 7.43P

The velocity at 5 cm above the element is 34 m/s.

Explanation of Solution

Given Information:

We have following figure

And at $x=2$ m, shear stress is 2.1 Pa. The given flow is turbulent. Also, we know that air has density, $\rho =1.2$ kg/m³ and viscosity, $\mu =1.8\times {10}^{-5}$ kg/m·s.

Calculation:

Let’s calculate Reynolds’s number,

${\mathrm{Re}}_x=\frac{\rho Ux}{\mu }=\frac{1.2\times U\times 2}{1.8\times {10}^{-5}}$

And the coefficient of friction,

$C_f=\frac{0.027}{{\mathrm{Re}}_x{}^{1/7}}=\frac{0.027}{{[\rho Ux/\mu ]}^{1/7}}$

The shear stress is given as,

$\begin{array}{l}{\tau }_w=\frac{1}{2}{C}_f\rho U^2\\ \Rightarrow 2.1=\frac{1}{2}\left(\frac{0.027}{{[\rho Ux/\mu ]}^{1/7}}\right)\rho U^2\\ \Rightarrow 2.1=\frac{1}{2}\left(\frac{0.027}{{[1.2\times U\times 2/(1.8\times {10}^{-5})]}^{1/7}}\right)\times 1.2U^2\\ \Rightarrow U\approx 34\text{ m/s}\end{array}$

So the stream velocity is 34 m/s.

Using this velocity, we can now, evaluate Reynolds’s number as below,

${\mathrm{Re}}_x=\frac{\rho Ux}{\mu }=\frac{1.2\times 34\times 2}{1.8\times {10}^{-5}}=4.54\times {10}^6$

Let’s calculate the boundary layer thickness, $\delta$ as below,

$\delta =\frac{0.16x}{{\mathrm{Re}}_x{}^{1/7}}=\frac{0.16\times 2}{{\left(4.54\times {10}^6\right)}^{1/7}}\approx 0.036\text{ m}=36\text{ mm}$

So the boundary layer thickness is 36 mm.

Boundary layer velocity at 5 cm above the element will be equal to stream velocity which is 34 m/s because this point is well above the boundary layer thickness.

Conclusion:

The velocity at 5 cm above the element is 34 m/s.