Use of Mass and Energy Balances with an Ideal Gas A compressed-air tank is to be repressurized to 40 bar by being connected to a high-pressure line containing air at 50 bar and 20°C. The repressurization of the tank occurs so quickly that the process can be assumed to be adiabatic; also, there is n

Question

Use of Mass and Energy Balances with an Ideal Gas

A compressed-air tank is to be repressurized to 40 bar by being connected to a high-pressure line containing air at 50 bar and 20°C. The repressurization of the tank occurs so quickly that the process can be assumed to be adiabatic; also, there is no heat transfer from the air to the tank. For this illustration, assume air to be an ideal gas with [latex]C_{ V }^{*}[/latex] = 21 J/(mol K).

a. If the tank initially contains air at 1 bar and 20°C, what will be the temperature of the air in the tank at the end of the filling process?

b. After a sufficiently long period of time, the gas in the tank is found to be at room temperature (20°C) because of heat exchange with the tank and the atmosphere. What is the new pressure of air in the tank?

Accepted Answer

(Each of these problems contains only a single unknown thermodynamic property, so solutions should be possible.)
a. We will take the contents of the tank to be the system. The difference form of the mass (or rather mole) and energy balances for this open system are

[latex]N_{2}-N_{1}=\Delta N[/latex] (a)

[latex]N_{2} \underline{U}_{2}-N_{1} \underline{U}_{1}=(\Delta N) \underline{H}_{ in }[/latex] (b)

In writing the energy balance we have made the following observations:
1. The kinetic and potential energy terms are small and can be neglected.

2. Since the tank is connected to a source of gas at constant temperature and pressure, [latex]\underline{H}_{ ext {in }}[/latex] is constant.

3. The initial process is adiabatic, so Q = 0, and the system (the contents of the tank) is of constant volume, so ΔV = 0

Substituting Eq. a in Eq. b, and recognizing that for the ideal gas [latex]\underline{H}(T)=C_{ P }^{*}\left(T-T_{R} ight)[/latex] and [latex]\underline{U}(T)=C_{ V }^{*}\left(T-T_{R} ight)-R T_{R}[/latex] , yields

[latex]N_{2}\left\{C_{ V }^{*}\left(T_{2}-T_{R} ight)-R T_{R} ight\}-N_{1}\left\{C_{ V }^{*}\left(T_{1}-T_{R} ight)-R T_{R} ight\}=\left(N_{2}-N_{1} ight) C_{ P }^{*}\left(T_{ in }-T_{R} ight)[/latex]

or

[latex]N_{2} C_{ V }^{*} T_{2}-N_{1} C_{ V }^{*} T_{1}=\left(N_{2}-N_{1} ight) C_{ P }^{*} T_{ in }[/latex]

(Note that the reference temperature [latex]T_{R}[/latex] cancels out of the equation, as it must, since the final result cannot depend on the arbitrarily chosen reference temperature.) Finally, using the ideal gas equation of state to eliminate [latex]N_{1} ext { and } N_{2}, ext { and recognizing that } V_{1}=V_{2}[/latex], yields

[latex]\frac{P_{2}}{T_{2}}=\frac{P_{1}}{T_{1}}+\frac{C_{ V }^{*}}{C_{ P }^{*}}\left(\frac{P_{2}-P_{1}}{T_{ in }} ight) \quad ext { or } \quad T_{2}=\frac{P_{2}}{\frac{P_{1}}{T_{1}}+\frac{C_{ V }^{*}}{C_{ P }^{*}}\left(\frac{P_{2}-P_{1}}{T_{ in }} ight)}[/latex]

The only unknown in this equation is [latex]T_{2}[/latex], so, formally, the problem is solved. The answer is T = 405.2 K = 132.05°C.
Before proceeding to the second part of the problem, it is interesting to consider the case in which the tank is initially evacuated. Here [latex]P_{1}=0[/latex], and so

[latex]T_{2}=\frac{C_{ P }^{*}}{C_{ V }^{*}} T_{ in }[/latex]

independent of the final pressure. Since [latex]C_{ P } ext { is always greater than } C_{ V }[/latex], the temperature of the gas in the tank at the end of the filling process will be greater than the temperature of gas in the line. Why is this so?
b. To find the pressure in the tank after the heat transfer process, we use the mass balance and the equation of state. Again, choosing the contents of the tank as the system, the mass (mole) balance is [latex]N_{2}=N_{1}[/latex], since there is no transfer of mass into or out of the system during the heat transfer process (unless, of course, the tank is leaking; we do not consider this complication here). Now using the ideal gas equation of state, we have

[latex]\frac{P_{2}}{T_{2}}=\frac{P_{1}}{T_{1}} \quad ext { or } \quad P_{2}=P_{1} \frac{T_{2}}{T_{1}}[/latex]

Thus, [latex]P_{2}=28.94 ext { bar. }[/latex]

Question 3.4-3: Use of Mass and Energy Balances with an Ideal Gas A compress...

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