Chapter 11, Problem 55EP

To determine

The drag force acting on the top surface and side surfaces.

The power requires to overcome the drag.

Answer to Problem 55EP

The drag force acting on the top surface and side surfaces is $17.023\text{lbf}$.

The power requires to overcome the drag is $2.37\text{kW}$.

Explanation of Solution

Given information:

The velocity of the truck is $70\text{mi}/\text{h}$, the length of the rectangular refrigerated compartment is $20\text{ft}$, the height is $8\text{ft}$, the width is $9\text{ft}$, the air temperature is $80°\text{F,}$ and the air pressure is $1\text{atm}$.

Write the expression for the area of the top and side surfaces of the truck.

   $A=\left(2\times l\times h\right)+\left(l\times w\right)$ …… (I)

Here, the length of the compartment is $l$, the height of the compartment is $h,$ and the width of the compartment is $w$.

Write the expression for the Reynolds number for the length of the truck.

   ${\mathrm{Re}}_{\text{L}}=\frac{Vl}{v}$ …… (II)

Here, the velocity of the truck is $V$, the length of the compartment is $l$ and the kinematic viscosity is $v$.

Write the expression for the friction coefficient.

   $C_f=\frac{0.074}{{\left({\mathrm{Re}}_{\text{L}}\right)}^{1/5}}$ …… (III)

Write the expression for the drag force.

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

Here, the density of the air is $\rho$.

Write the expression for power required to overcome the drag.

   $W=F_D\times V$ …… (V)

Calculation:

Substitute $20\text{ft}$ for $l$, $8\text{ft}$ for $h$ and $9\text{ft}$ for $w$ in Equation (I).

   $\begin{array}{c}A=\left(2\left(20\text{ft}\right)\left(8\text{ft}\right)\right)+\left(\left(20\text{ft}\right)\left(9\text{ft}\right)\right)\\ =320\text{ft+}180\text{ft}\\ \text{=500}{\text{ft}}^2\end{array}$

Refer to table A-22E, “Properties of air at $1\text{atm}$ ’ to obtain the values of $\rho$ as $0.07350\text{lbm}/{\text{ft}}^{\text{3}}$ and $v$ as $1.697\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$ at $80°\text{F}$.

Substitute $1.697\times {10}^{-4}{\text{m}}^{\text{2}}/\text{s}$ for $v$, $20\text{ft}$ for $l$ and $70\text{mi}/\text{h}$ for $V$ in Equation (II).

   $\begin{array}{c}{\mathrm{Re}}_{\text{L}}=\frac{\left(70\text{mi}/\text{h}\right)\left(20\text{ft}\right)}{1.697\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =\frac{\left(70\text{mi}/\text{h}\left(\frac{1.4667\text{ft}/\text{s}}{1\text{mi}/\text{h}}\right)\right)\left(20\text{ft}\right)}{1.697\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =\frac{\left(102.67\text{ft}/\text{s}\right)\left(20\text{ft}\right)}{1.697\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =1.210\times {10}^7\end{array}$

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

Substitute $1.210\times {10}^7$ for ${\mathrm{Re}}_{\text{L}}$ in Equation (III).

   $\begin{array}{c}{C}_f=\frac{0.074}{{\left(1.210\times {10}^7\right)}^{1/5}}\\ =\frac{0.074}{26.097}\\ =2.83\times {10}^{-3}\end{array}$

Substitute $2.83\times {10}^{-3}$ for $C_f$, $\text{500}{\text{ft}}^2$ for $A$, $0.07350\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho ,$ and $70\text{mi}/\text{h}$ for $V$ in Equation (IV).

   $\begin{array}{c}{F}_D=\frac{\left(2.83\times {10}^{-3}\right)\left(0.07350\text{lbm}/{\text{ft}}^{\text{3}}\right){\left(70\text{mi}/\text{h}\right)}^2\left(\text{500}{\text{ft}}^2\right)}{2}\\ =\frac{\left(2.83\times {10}^{-3}\right)\left(0.07350\text{lbm}/{\text{ft}}^{\text{3}}\right){\left(70\text{mi}/\text{h}\left(\frac{1.4667\text{ft}/\text{s}}{1\text{mi}/\text{h}}\right)\right)}^2\left(\text{500}{\text{ft}}^2\right)}{2}\\ =548.15\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{32.2\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{2}}}\right)\\ =17.023\text{lbf}\end{array}$

Substitute $17.023\text{lbf}$ for $F_D$ and $70\text{mi}/\text{h}$ for $V$ in Equation (V).

   $\begin{array}{l}W=17.023\text{lbf}\times 70\text{mi}/\text{h}\\ =17.023\text{lbf}\times \left(70\text{mi}/\text{h}\left(\frac{1.4667\text{ft}/\text{s}}{1\text{mi}/\text{h}}\right)\right)\\ =\left(17.023\text{lbf}\right)\left(102.67\text{ft}/\text{s}\right)\\ =1747.778\text{lbf}\cdot \text{ft}/\text{s}\end{array}$

   $\begin{array}{c}W=1747.778\text{lbf}\cdot \text{ft}/\text{s}\left(\frac{1\text{W}}{0.737\text{lbf}\cdot \text{ft}/\text{s}}\right)\\ =2371.48\text{W}\left(\frac{1\text{kW}}{1000\text{W}}\right)\\ =2.37\text{kW}\end{array}$

Conclusion:

The drag force acting on the top surface and side surfaces is $17.023\text{lbf}$.

The power requires to overcome the drag is $2.37\text{kW}$.