Air Loss from a Flat Tire
Air in an automobile tire is maintained at a pressure of 220 kPa (gage) in an environment where the atmospheric pressure is 94 kPa. The air in the tire is at the ambient temperature of 25°C. A 4-mm-diameter leak develops in the tire as a result of an accident (Fig. 12–19). Approximating the flow as isentropic determine the initial mass flow rate of air through the leak.
SOLUTION A leak develops in an automobile tire as a result of an accident. The initial mass flow rate of air through the leak is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of air through the hole is isentropic.
Properties The specific gas constant of air is R = 0.287 kPa⋅m³/kg⋅K. The specific heat ratio of air at room temperature is k = 1.4.
Analysis The absolute pressure in the tire is
P=P_{ gage }+P_{ atm }=220+94=314 kPa
The critical pressure is (from Table 12–2)
| TABLE 12–2 | ||||
| The critical-pressure, critical-temperature, and critical-density ratios for isentropic flow of some ideal gases | ||||
| Superheated steam, k = 1.3 | Hot products of combustion, k = 1.33 | Air, k = 1.4 | Monatomic gases, k = 1.667 | |
| \frac{P^{*}}{P_{0}} | 0.5457 | 0.5404 | 0.5283 | 0.4871 |
| \frac{T^{*}}{T_{0}} | 0.8696 | 0.8584 | 0.8333 | 0.7499 |
| \frac{\rho^{*}}{\rho_{0}} | 0.6276 | 0.6295 | 0.6340 | 0.6495 |
P^{*}=0.5283 P_{o}=(0.5283)(314 kPa )=166 kPa >94 kPa
Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow properties at the exit become
\begin{aligned}\rho_{0} &=\frac{P_{0}}{R T_{0}}=\frac{314 kPa }{\left(0.287 kPa \cdot m ^{3} / kg \cdot K \right)(298 K )}=3.671 kg / m ^{3} \\\rho^{*} &=\rho\left(\frac{2}{k+1}\right)^{1 /(k-1)}=\left(3.671 kg / m ^{3}\right)\left(\frac{2}{1.4+1}\right)^{1 /(1.4-1)}=2.327 kg / m ^{3} \\T^{*} &=\frac{2}{k+1} T_{0}=\frac{2}{1.4+1}(298 K )=248.3 K \\V &=c=\sqrt{ kRT ^{*}}=\sqrt{(1.4)(0.287 kJ / kg \cdot K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg }\right)(248.3 K )} \\&=315.9 m / s\end{aligned}
Then the initial mass flow rate through the hole is
\begin{aligned}\dot{m} &=\rho A V=\left(2.327 kg / m ^{3}\right)\left[\pi(0.004 m )^{2} / 4\right](315.9 m / s )=0.00924 kg / s \\&= 0 . 5 5 4 kg / min\end{aligned}
Discussion The mass flow rate decreases with time as the pressure inside the tire drops.