Chapter 2, Problem 13P

The pressure in an automobile tire depends on the temperature of air in the tire. When the air temperature is $25°C$ , the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m³, determine the pressure rise in the tire when the air temperature in the tire rises to 50°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa.

To determine

The pressure rise in the tire.

The amount of air bled off.

Answer to Problem 13P

The pressure rise is $26.01\text{kPa}$.

The amount of air bled off is $7.0152\times {10}^{-3}\text{kg}$.

Explanation of Solution

Given information:

The initial temperature in tire is $25°\text{C}$, Initial pressure in the tire is $210\text{kPa}$, Volume of the tire is $0.025{\text{m}}^{\text{3}}$, atmospheric pressure is $\text{100kPa}$ and the final temperature of the tire is $50°\text{C}$.

Write the expression for the initial absolute pressure.

  $P_1=P_{gauge}+P_{atm}$   ...... (I)

Here, initial absolute pressure is $P_1$, gauge pressure is $P_{gauge}$ and atmospheric pressure is $P_{atm}$.

Write the expression for the initial temperature conversion from Degree Celsius to Kelvin.

  $T_1=\left(273+t_1\right)\text{K}$   ...... (II)

Here, the initial temperature in Kelvin is $T_1$ and initial temperature in degree Celsius is $t_1$.

Write the expression for the final temperature conversion from Degree Celsius to Kelvin.

  $T_2=\left(273+t_2\right)\text{K}$   ...... (III)

Here, the final temperature in Kelvin is $T_2$ and final temperature in degree Celsius is $t_2$.

Write the expression for the final pressure in tire.

  $P_2=P_1\times \frac{T_2}{T_1}$   ...... (IV)

Here, the final pressure is $P_2$.

Write the expression for the pressure rise.

  $\Delta P_{rise}=P_2-P_1$   ...... (V)

Here, pressure rise is $\Delta P_{rise}$.

Write the expression for the amount of air bled off.

  $\Delta m=\frac{P_{1}V}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$   ...... (VI)

Here, amount of air bled off is $\Delta m$, gas constant is $R$ and volume is $V$.

Calculation:

Substitute $210\text{kPa}$ for $P_{gauge}$ and $100\text{kPa}$ for $P_{atm}$ in Equation (I).

  $\begin{array}{c}{P}_1=210\text{kPa}+100\text{kPa}\\ =310\text{kPa}\end{array}$

Substitute $25°\text{C}$ for $t_1$ in Equation (II).

  $\begin{array}{c}{T}_1=\left(25\text{°C}+273\right)\text{K}\\ \text{=298}\text{K}\end{array}$

Substitute $50°\text{C}$ for $t_2$ in Equation (III).

  $\begin{array}{c}{T}_2=\left(273+50\text{°C}\right)\text{K}\\ =323\text{K}\end{array}$

Substitute $310\text{kPa}$ for $P_1$, $\text{298}\text{K}$ for $T_1$ and $323\text{K}$ for $T_2$ in Equation (IV).

  $\begin{array}{c}{P}_2=310\text{kPa}\left(\frac{323\text{K}}{298\text{K}}\right)\\ =310\text{kPa}\times 1.0838\\ =336.01\text{kPa}\end{array}$

Substitute $310\text{kPa}$ for $P_1$ and $336.01\text{kPa}$ for $P_2$ in Equation (V).

  $\begin{array}{c}\Delta P_{rise}=336.01\text{kPa}-310\text{kPa}\\ =26.01\text{kPa}\end{array}$

Substitute $310\text{kPa}$ for $P_1$, $0.025{\text{m}}^{\text{3}}$ for $V$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $298\text{K}$ for $T_1$ and $323\text{K}$ for $T_2$ in Equation (VI).

  $\begin{array}{c}\Delta m=\frac{\left(310\text{kPa}\right)\left(0.025{\text{m}}^{\text{3}}\right)}{0.287\text{kJ}/\text{kg}\cdot \text{K}}\left(\frac{1}{298\text{K}}-\frac{1}{323\text{K}}\right)\\ =27.01\left(\text{kPa}\cdot {\text{m}}^{\text{3}}/\left(\text{kJ}/\text{kg}\cdot \text{K}\right)\right)\times \left(\left(2.597\times {10}^{-4}\right){\text{K}}^{-1}\right)\\ =7.0152\times {10}^{-3}\left(\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kJ}}\right)\left(\frac{\text{kg}\cdot \text{K}}{\text{K}}\right)\\ =7.0152\times {10}^{-3}\left(\frac{\text{kPa}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)\cdot {\text{m}}^{\text{3}}}{\text{kJ}\left(\frac{1\text{N}\cdot \text{m}}{1\text{J}}\right)}\right)\text{kg}\end{array}$

  $\Delta m=7.0152\times {10}^{-3}\text{kg}$

Conclusion:

Pressure rise in the tire is $26.01\text{kPa}$.

Amount of air bled off is $7.0152\times {10}^{-3}\text{kg}$.