Chapter 11, Problem 59P

To determine

The drag force acting on the wall.

The drag force when the velocity is doubled.

Answer to Problem 59P

The drag force acting on the wall when velocity is $55\text{km}/\text{h}$ is $16.3\text{N}$.

The drag force acting on the wall when velocity is $110\text{km}/\text{h}$ is $57\text{N}$.

Explanation of Solution

Given information:

The wind velocity is $55\text{km}/\text{h}$, the temperature of the wind is $5°\text{C}$, pressure is $1\text{atm}$, the height of wall of house is $4\text{m,}$ and the length of the wall is $10\text{m}$.

Write the expression for the Reynolds number.

   ${\mathrm{Re}}_{\text{L}}=\frac{VL}{v}$ ...... (I)

Here, the velocity of flow is $V$, length of the wall is $L,$ and the kinematic viscosity of fluid is $v$.

Write the expression for the frictional coefficient.

   $C_f=\frac{0.074}{{\left({\mathrm{Re}}_{\text{L}}\right)}^{1/5}}-\frac{1742}{{\mathrm{Re}}_{\text{L}}}$ ...... (II)

Write the expression for the area of the wall.

   $A=L\times b$ ...... (III)

Here, the length of the wall is $L$ and the width height of the wall is $b$.

Write the expression for the drag force when the pressure drag is zero.

   $F_D=\frac{C_f\rho A{V}^2}{2}$ ...... (IV)

Here, the density of air is $\rho$ and the area of the wall is $A$.

Calculation:

Refer to Table A-9, “Properties of air” to obtain the values of $\rho$ as $1.269\text{kg}/{\text{m}}^{\text{3}}$ at

   $5°\text{C}$ and $v$ as $1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$ at $5°\text{C}$.

Substitute $1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$ for $v$, $10\text{m}$ for $L$ and $55\text{km}/\text{h}$ for $V$ in Equation (I).

   $\begin{array}{c}{\mathrm{Re}}_{\text{L}}=\frac{\left(55\text{km}/\text{h}\right)\left(10\text{m}\right)}{1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{\left(55\text{km}/\text{h}\left(\frac{5/18\text{m}/\text{s}}{1\text{km}/\text{h}}\right)\right)\left(10\text{m}\right)}{1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{\left(15.27\text{m}/\text{s}\right)\left(10\text{m}\right)}{1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =1.1\times {10}^7\end{array}$

The flow is laminar and turbulent as the obtained value of Reynolds number is greater than the critical Reynolds number.

Substitute $1.1\times {10}^7$ for ${\mathrm{Re}}_{\text{L}}$ in Equation (II).

   $\begin{array}{c}{C}_f=\frac{0.074}{{\left(1.1\times {10}^7\right)}^{1/5}}-\frac{1742}{1.1\times {10}^7}\\ =0.0029-0.00016\\ =0.00274\end{array}$

Substitute $10\text{m}$ for $L$ and $4\text{m}$ for $b$ in Equation (III).

   $\begin{array}{c}A=\left(10\text{m}\right)\left(4\text{m}\right)\\ =40{\text{m}}^2\end{array}$

Substitute $0.00274$ for $C_f$, $1.269\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $40{\text{m}}^{\text{2}}$ for $A$ and $55\text{km}/\text{h}$ for $V$ in Equation (IV).

   $\begin{array}{c}{F}_D=\frac{\left(0.00274\right)\left(1.269\text{kg}/{\text{m}}^{\text{3}}\right)\left(40{\text{m}}^{\text{2}}\right){\left(55\text{km}/\text{h}\right)}^2}{2}\\ =\frac{\left(0.00274\right)\left(1.269\text{kg}/{\text{m}}^{\text{3}}\right)\left(40{\text{m}}^{\text{2}}\right){\left(55\text{km}/\text{h}\left(\frac{5/18\text{m}/\text{s}}{1\text{km}/\text{h}}\right)\right)}^2}{2}\\ =16.3\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =16.3\text{N}\end{array}$

The velocity is doubled as stated.

   $V=110\text{km}/\text{h}$

Substitute $1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}$ for $v$, $10\text{m}$ for $L$ and $110\text{km}/\text{h}$ for $V$ in Equation (I).

   $\begin{array}{c}{\mathrm{Re}}_{\text{L}}=\frac{\left(110\text{km}/\text{h}\right)\left(10\text{m}\right)}{1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{\left(110\text{km}/\text{h}\left(\frac{5/18\text{m}/\text{s}}{1\text{km}/\text{h}}\right)\right)\left(10\text{m}\right)}{1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{\left(30.6\text{m}/\text{s}\right)\left(10\text{m}\right)}{1.382\times {10}^{-5}{\text{m}}^{\text{2}}/\text{s}}\\ =2.1\times {10}^7\end{array}$

Substitute $2.1\times {10}^7$ for ${\mathrm{Re}}_{\text{L}}$ in Equation (II).

   $\begin{array}{c}{C}_f=\frac{0.074}{{\left(2.1\times {10}^7\right)}^{1/5}}-\frac{1742}{2.1\times {10}^7}\\ =0.0025-0.0000788\\ =0.0024\end{array}$

Substitute $0.00274$ for $C_f$, $1.269\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $40{\text{m}}^{\text{2}}$ for $A$ and $110\text{km}/\text{h}$ for $V$ in Equation (IV).

   $\begin{array}{c}{F}_D=\frac{\left(0.00274\right)\left(1.269\text{kg}/{\text{m}}^{\text{3}}\right)\left(40{\text{m}}^{\text{2}}\right){\left(110\text{km}/\text{h}\right)}^2}{2}\\ =\frac{\left(0.00274\right)\left(1.269\text{kg}/{\text{m}}^{\text{3}}\right)\left(40{\text{m}}^{\text{2}}\right){\left(110\text{km}/\text{h}\left(\frac{5/18\text{m}/\text{s}}{1\text{km}/\text{h}}\right)\right)}^2}{2}\\ =57\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =57\text{N}\end{array}$

Conclusion:

The drag force acting on the wall when velocity is $55\text{km}/\text{h}$ is $16.3\text{N}$.

The drag force acting on the wall when velocity is $110\text{km}/\text{h}$ is $57\text{N}$.