# Finding the time taken for pressurised tank to reach 1 atm (Exam Prep)

A gas tank has an internal volume of 0.3m3. It is filled with helium pressurized to 32 bar (absolute). The gas tank is in thermal equilibrium with its surroundings and so the gas is at a temperature of 20°C. During transportation, the tank is accidentally punctured. The hole has a cross sectional area of 0.08 cm2. Assuming isentropic flow, estimate how long it will take for the tank to reach atmospheric pressure?

(For helium, the ratio of specific heats is 1.66 and the gas constant is 2077 J/kgK)

I am trying to do this question for exam preparation but cannot derive a solution. I calculates that the mass flow rate as follows:

I cannot figure out how to then use integration to derive the time, I keep getting negative values. Any help is appreciated.

I usually don't do long answers, but I am seeking shelter from tropical storm Hilary and have time!

The isentropic flow equation relates pressure and density for an ideal gas:

$$P_1 V_1^\gamma = P_2 V_2^\gamma$$

Where:

• $$P_1$$ = Initial pressure
• $$V_1$$= Initial volume
• $$\gamma$$ = Specific heat ratio

The initial temperature of the gas is equal to the surroundings, i.e., $$T_1 = T_2 = 20°C = 293.15 K .$$

Using the ideal gas equation $$P V = m R T \quad,$$ where m is the mass of the gas.

$$V = \frac{m R T}{P}$$

Substituting this expression for V into the isentropic flow equation and solving for m :

$$P_1 \left(\frac{m R T_1}{P_1}\right)^\gamma = P_2 \left(\frac{m R T_2}{P_2}\right)^\gamma$$

$$m = \frac{P_1 V_1}{R T_1} \left(\frac{P_2}{P_1}\right)^{1/\gamma}$$

Now, the mass flow rate $$\dot{m}$$ through the vent hole can be expressed as:

$$\dot{m} = \rho A v$$

Where:

• $$\rho$$ = Density of the gas
• A = Cross-sectional area of the vent hole
• v = Velocity of the gas through the hole

Using the equation of state for an ideal gas $$\rho = \frac{P}{R T} ,$$ we can express $$\dot{m}$$ in terms of velocity:

$$\dot{m} = \frac{P A v}{R T}$$

$$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$

Since $$P_2$$ is much smaller than $$P_1$$ and $$v_1$$ is negligible compared to $$v_2$$, we can approximate the equation as:

$$\frac{1}{2} \rho v^2 = P_2$$

Solving for $$v$$:

$$v = \sqrt{\frac{2 P_2}{\rho}}$$

Substitute the expression for $$\rho$$ and solve for $$v$$:

$$v = \sqrt{\frac{2 P_2 R T}{P}}$$

Now, we can substitute the expression for $$\dot{m}$$ and $$v$$ into the equation for time $$t$$ it takes to reach atmospheric pressure:

$$t = \frac{m}{\dot{m}} = \frac{P_1 V_1}{R T_1} \left(\frac{P_2}{P_1}\right)^{1/\gamma} \cdot \frac{R T}{P A \sqrt{\frac{2 P_2 R T}{P}}}$$

Now plug in the values:

• $$P_1$$ = 3200 \times 10^3 ) Pa
• $$V_1$$= 0.3 m³
• R = 2077 J/kgK
•  T_1 = 293.15 K
• $P_2$ = 101300 Pa
• $\gamma$ = 1.66
• A = 0.08 * 10^-4 m²

Simplify the equation:

$$t = \frac{P_1 V_1}{R T_1} \left(\frac{P_2}{P_1}\right)^{1/\gamma} \cdot \frac{R T}{P A \sqrt{\frac{2 P_2 R T}{P}}}$$

$$t = \frac{V_1}{A} \sqrt{\frac{P}{2 P_2}} \left(\frac{P_2}{P_1}\right)^{1/\gamma} \sqrt{\frac{R T}{R T_1}}$$

$$t = \frac{V_1}{A} \sqrt{\frac{P}{2 P_2}} \left(\frac{P_2}{P_1}\right)^{1/\gamma} \sqrt{\frac{T}{T_1}}$$

Substitute the values and calculate $$t$$:

$$t = \frac{0.3}{0.08 \times 10^{-4}} \sqrt{\frac{3200 \times 10^3}{2 \cdot 101300}} \left(\frac{101300}{3200 \times 10^3}\right)^{1/1.66} \sqrt{\frac{293.15}{293.15}}$$

I let you do the rest.