Chapter 7, Problem 124RQ

To determine

The time required to cool the truck from $25°\text{C}$ to $5°\text{C}$.

Explanation of Solution

Given:

The volume of the truck $\left({\nu }_{\text{truck}}\right)$ is,

The length of the truck $\left(l\right)$ is 12 m

The breath of the truck $\left(b\right)$ is 2.3 m.

The height of the truck $\left(h\right)$ is 3.5 m.

The initial temperature of the truck $\left(T_1\right)$ is $25°\text{C}$.

The final temperature of the truck $\left(T_2\right)$ is $5°\text{C}$.

The rate of transmission heat gain occurs $\left(UA\right)$ is $120\text{W}/\text{°C}$.

The refrigeration capacity $\left(Q_{ref}\right)$ is 11 kW.

Calculation:

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, for the average temperature $15°\text{C}$.

The value of density of air $\left({\rho }_{\text{air}}\right)$ is $1.2\text{kg}/{\text{m}}^{\text{3}}$

The specific heat $\left(c_p\right)$ air is $1.0\text{kJ}/\text{kg}\cdot °\text{C}$.

Calculate the mass of the air $\left(m_{\text{air}}\right)$ in the truck.

  $\begin{array}{c}{m}_{\text{air}}={\rho }_{\text{air}}\times {\nu }_{\text{truck}}\\ ={\rho }_{\text{air}}\times {\left(l\times b\times h\right)}_{\text{truck}}\end{array}$

  $\begin{array}{c}{m}_{\text{air}}=\left(1.2\text{kg}/{\text{m}}^{\text{3}}\right)\times {\left(12\text{m}\times 2.3\text{m}\text{}\times 3.5\text{m}\right)}_{\text{truck}}\\ =115.92\text{kg}\\ \cong \text{116}\text{kg}\end{array}$

Calculate the amount of heat removed from the truck.

  $\begin{array}{c}{Q}_{\text{cooling,air}}=\left(m_{\text{air}}{c}_p\Delta T\right)\\ =m_{\text{air}}{c}_p\left(T_1-T_2\right)\end{array}$

  $\begin{array}{c}{Q}_{\text{cooling,air}}=\left(116\text{kg}\right)\times \left(1.0\text{kJ}/\text{kg}\cdot °\text{C}\right)\times \left(25-5\right)°\text{C}\\ =2,320\text{kJ}\end{array}$

Calculate the average temperature in the truck during precooling.

  $T_{\text{avg}}=\frac{T_1+T_2}{2}$

  $\begin{array}{c}{T}_{\text{avg}}=\frac{\left(25+5\right)°\text{C}}{2}\\ =15°\text{C}\end{array}$

Calculate the average rate of heat gain by transmission.

  $\begin{array}{c}{\dot{Q}}_{transmission,avg}=UA\Delta T\\ =UA\left(T_1-T_{\text{avg}}\right)\end{array}$

  $\begin{array}{c}{\dot{Q}}_{transmission,avg}=\left(120\text{W}/\text{°C}\right)\times \left(25°\text{C}-15°\text{C}\right)\\ =\left(120\text{W}/\text{°C}\right)\times \left(10°\text{C}\right)\\ =1200\text{W}\times \left(\frac{{10}^{-3}\text{kJ}/\text{s}}{1\text{W}}\right)\\ =1.2\text{ kJ/s}\end{array}$

Calculate the time required to cool the truck.

  $\begin{array}{l}{\dot{Q}}_{refrig}\times \Delta t=Q_{cooling,air}+{\dot{Q}}_{transmission}\times \Delta t\\ \Delta t=\frac{Q_{cooling,air}}{{\dot{Q}}_{refrig}-{\dot{Q}}_{transmission}}\end{array}$

  $\begin{array}{c}\Delta t=\frac{2320\text{kJ}}{\left(11-1.2\right)\text{kJ}/\text{s}}\\ =\frac{2320\text{kJ}}{9.8\text{kJ}/\text{s}}\\ =236.73\text{s}\times \left(\frac{1\min }{60\sec }\right)\\ \cong 3.95\text{ min}\end{array}$

Thus, the time required to cool the truck from $25°\text{C}$ to $5°\text{C}$ is $\underset{\_}{\text{3}\text{.95}\text{min}}$.