Showing That the Energy and Entropy Balances Can Be Used to Determine Whether a Process Is Possible An engineer claims to have invented a steady-flow device that will take air at 4 bar and 20◦C and separate it into two streams of equal mass, one at 1 bar and −20◦C and the second at 1 bar and 60◦C.

Question

Showing That the Energy and Entropy Balances Can Be Used to Determine Whether a Process Is Possible An engineer claims to have invented a steady-flow device that will take air at 4 bar and 20°C and separate it into two streams of equal mass, one at 1 bar and −20°C and the second at 1 bar and 60°C. Furthermore, the inventor states that his device operates adiabatically and does not require (or produce) work. Is such a device possible? [Air can be assumed to be an ideal gas with a constant heat capacity of [latex]C_{ P }^{*}[/latex] = 29.3 J/(mol K)].

Accepted Answer

The three principles of thermodynamics—(1) conservation of mass, (2) conservation of energy, and (3) [latex]\dot{S}_{ ext {gen }} \geq 0[/latex]—must be satisfied for this or any other device. These principles can be used to test whether any device can meet the specifications given here.

The steady-state mass balance equation for the open system consisting of the device and its contents is [latex]d N / d t=0=\sum_{k} \dot{N}_{k}=\dot{N}_{1}+\dot{N}_{2}+\dot{N}_{3}[/latex] . Since, from the problem statement, [latex]\dot{N}_{2}=\dot{N}_{3}=-\frac{1}{2} \dot{N}_{1}[/latex] , mass is conserved. The steady-state energy balance for this device is

[latex]\begin{aligned}\frac{d U}{d t}=0 &=\sum_{k} \dot{N}_{k} \underline{H}_{k}=\dot{N}_{1} \underline{H}_{1}-\frac{1}{2} \dot{N}_{1} \underline{H}_{2}-\frac{1}{2} \dot{N}_{1} \underline{H}_{3} \&=\dot{N}_{1} C_{ P }^{*}\left(293.15 K -\frac{1}{2} imes 253.15 K -\frac{1}{2} imes 333.15 K ight)=0\end{aligned}[/latex]

so the energy balance is also satisfied. Finally, the steady-state entropy balance is

[latex]\begin{aligned}\frac{d S}{d t}=0 &=\sum_{K} \dot{N}_{k} \underline{S}_{k}+\dot{S}_{ ext {gen }}=\dot{N}_{1} \underline{S}_{1}-\frac{1}{2} \dot{N}_{1} \underline{S}_{2}-\frac{1}{2} \dot{N}_{1} \underline{S}_{3}+\dot{S}_{ ext {gen }} \&=\frac{1}{2} \dot{N}_{1}\left[\left(\underline{S}_{1}-\underline{S}_{2} ight)+\left(\underline{S}_{1}-\underline{S}_{3} ight) ight]+\dot{S}_{ ext {gen }}\end{aligned}[/latex]

Now using Eq. 4.4-3, we have

[latex]\underline{S}\left(T_{2}, P_{2} ight)-\underline{S}\left(T_{1}, P_{1} ight)=C_{ P }^{*} \ln \left(\frac{T_{2}}{T_{1}} ight)-R \ln \left(\frac{P_{2}}{P_{1}} ight)[/latex] (4.4-3)

[latex]\begin{aligned}\dot{S}_{ gen } &=-\frac{1}{2} \dot{N}_{1}\left(C_{ P }^{*} \ln \frac{T_{1}}{T_{2}}-R \ln \frac{P_{1}}{P_{2}}+C_{ P }^{*} \ln \frac{T_{1}}{T_{3}}-R \ln \frac{P_{1}}{P_{3}} ight) \&=-\frac{1}{2} \dot{N}_{1}\left(29.3 \ln \frac{293.15 imes 293.15}{253.15 imes 333.15}-8.314 \ln \frac{4 imes 4}{1 imes 1} ight)=11.25 \dot{N}_{1} \frac{ J }{ Ks }\end{aligned}[/latex]

Therefore, we conclude, on the basis of thermodynamics, that it is possible to construct a device with the specifications claimed by the inventor. Thermodynamics, however, gives us no insight into how to design such a device. That is an engineering problem.
Two possible devices are indicated in Fig. 4.5-5. The first device consists of an air-driven turbine that extracts work from the flowing gas. This work is then used to drive a heat pump (an air conditioner or refrigerator) to cool part of the gas and heat the rest. The second device, the Hilsch-Ranque vortex tube, is somewhat more interesting in that it accomplishes the desired change of state with only a valve and no moving parts. In this device the air expands as it enters the tube, thus gaining kinetic energy at the expense of internal energy (i.e., at the end of the expansion process we have high-velocity air of both lower pressure and lower temperature than the incoming air). Some of this cooled air is withdrawn from the center of the vortex tube. The rest of the air swirls down the tube, where, as a result of viscous dissipation, the kinetic energy is dissipated into heat, which increases the internal energy (temperature) of the air. Thus, the air being withdrawn at the valve is warmer than the incoming air.

Figure 4.5-5 Two devices to separate compressed air into two low-pressure air streams of different temperature.

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