Chapter 5, Problem 5.46P

Interpretation Introduction

Interpretation :

An air stream with conditions ${20}^{\circ }\text{C}\text{and 5 bar}$ enters in a Hilsch vortex tube which splits air stream into two streams at ${27}^{\circ }\text{C and }-{22}^{\circ }\text{C},\text{both at 1 atm}$ . Comment on these results possible for this process.

Concept Introduction :

A thermodynamic process is possible if and only if it follows laws of thermodynamics, first and second law of thermodynamics.

According to the first law of thermodynamics the energy or enthalpy of the universe is constant or conserved and it can convert from one form into another form and hence $\Delta H$ is essentially zero or and according to the first law of thermodynamics, for a steady state flow process after omission of kinetic and potential terms, the enthalpy of the process for ideal gas is given by

  $\Delta H=Q+W_S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}dT}$

According to the second law of thermodynamics process is process if and only if the total entropy (system surrounding) is greater than or equal to zero for any change of state of the system. Hence,

  $S_G\ge 0$

And formula for entropy generation in steady state flow process with one input stream is

  $\Delta S-{\displaystyle \sum _j\frac{\stackrel{•}{Q_j}}{T_{\sigma ,j}}}=S_G\ge 0$

And for the Hilsch vortex tube, considering adiabatic vortex tube $\stackrel{•}{Q_j}=0$

  $\Delta S=S_G\ge 0$

And for an ideal gas

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}-R\ln \frac{P_2}{P_1}$

Answer to Problem 5.46P

Process is valid with values

  $\Delta H=2.1\times {10}^{-3}\frac{\text{kJ}}{\text{mol}}$

  $S_G=0.0132\frac{\text{kJ}}{\text{mol K}}$

Explanation of Solution

Given information:

It is given that Hilsch vortex tube operates with no mechanical parts splits air stream entering at ${20}^{\circ }\text{C}\text{and 5 bar}$ into two streams: one warmer and other cooler than entering stream at ${27}^{\circ }\text{C and -}{22}^{\circ }\text{C},\text{both at 1 atm}$ .

Hence

  $T_1={20}^{\circ }\text{C}+273.15=293.15\text{K}$

  $T_H={27}^{\circ }\text{C}+273.15=300.15\text{K}$

  $T_C=-{22}^{\circ }\text{C}+273.15=251.15\text{K}$

  $P_1=5\text{bar}$

  $P_2=1\text{atm=1}\text{.01325 bar}$

The mass flow rate of warm air leaving is six times that of the cool air mass flow rate.

It is given that air is assumed to be an ideal gas at the conditions given.

Basis: $\stackrel{•}{m_C}=1\frac{\text{kg}}{\text{s}}$ of cooler air leaving from the tube

Hence

Mass flow rate of warm air leaving is six times that of the cool air mass flow rate i.e.

  $\stackrel{•}{m_H}=6\frac{\text{kg}}{\text{s}}$

Therefore, the mass fraction of the cooler air in the output stream of air is

  $\begin{array}{l}{x}_C=\frac{\stackrel{•}{m_C}}{\stackrel{•}{m_C}+\stackrel{•}{m_H}}\\ x_C=\frac{1}{1+6}=\frac{1}{7}\end{array}$

And mass fraction of warm air stream in the output is

  $x_H=1-\frac{1}{7}=\frac{6}{7}$

Hence

Total enthalpy change of the output stream is:

  $\Delta H=x_1\Delta H_1+x_2\Delta H_2$

And we know that for an ideal gas,

  $\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}dT}$

Thus,

  $\begin{array}{l}\Delta H=x_H\Delta H_H+x_C\Delta H_C\\ \Delta H=x_H\Delta H_H+x_C\Delta H_C\\ \Delta H=\frac{6}{7}\times R\times {\displaystyle \underset{T_1}{\overset{T_H}{\int }}\frac{C_P}{R}dT}+\frac{1}{7}\times R\times {\displaystyle \underset{T_1}{\overset{T_C}{\int }}\frac{C_P}{R}dT}\end{array}$

Where ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$ is calculated by following formula

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(1)

And

  $\tau =\frac{T}{T_0}$

For warmer air stream

Values of above constants for air in equation (1) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{air}3.3550.5750-0.016\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T_H}{T_1}\\ \tau =\frac{300.15\text{K}}{293.15\text{K}}=1.024\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_H}{\int }}\frac{\Delta C_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_1}{\overset{T_H}{\int }}\frac{\Delta C_P}{R}dT}=3.355\times 293.15\times (1.024-1)+\frac{0.575\times {10}^{-3}}{2}\times {293.15}^2\times ({1.024}^2-1)\\ +\frac{0}{3}\times {293.15}^3({1.024}^3-1)+\frac{-0.016\times {10}^5}{293.15}(\frac{1.024-1}{1.024})\\ {\displaystyle \underset{T_1}{\overset{T_H}{\int }}\frac{\Delta C_P}{R}dT}=24.677\text{K}\end{array}$

And for cooler air stream

  $\begin{array}{l}\tau =\frac{T_C}{T_1}\\ \tau =\frac{251.15\text{K}}{293.15\text{K}}=0.857\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_C}{\int }}\frac{\Delta C_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_1}{\overset{T_C}{\int }}\frac{\Delta C_P}{R}dT}=3.355\times 293.15\times (0.857-1)+\frac{0.575\times {10}^{-3}}{2}\times {293.15}^2\times ({0.857}^2-1)\\ +\frac{0}{3}\times {293.15}^3({0.857}^3-1)+\frac{-0.016\times {10}^5}{293.15}(\frac{0.857-1}{0.857})\\ {\displaystyle \underset{T_1}{\overset{T_C}{\int }}\frac{\Delta C_P}{R}dT}=-146.29\text{K}\end{array}$

Hence Total energy change during the process is

  $\begin{array}{l}\Delta H=\frac{6}{7}\times R\times {\displaystyle \underset{T_1}{\overset{T_H}{\int }}\frac{C_P}{R}dT}+\frac{1}{7}\times R\times {\displaystyle \underset{T_1}{\overset{T_C}{\int }}\frac{C_P}{R}dT}\\ \Delta H=\frac{6}{7}\times 8.314\frac{\text{J}}{\text{mol K}}\times \left(24.677\text{K}\right)+\frac{1}{7}\times 8.314\frac{\text{J}}{\text{mol K}}\times \left(-146.29\text{K}\right)\\ \Delta H=2.1\times {10}^{-3}\frac{\text{kJ}}{\text{mol}}\end{array}$

Hence it is nearly zero and has finite value. So, first law of thermodynamics is valid.

Now, Total entropy change of the output stream is:

  $\Delta S=x_1\Delta S_1+x_2\Delta S_2$

And we know that for an ideal gas,

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}-R\ln \frac{P_2}{P_1}$

Thus,

  $\begin{array}{l}\Delta S=x_H\Delta S_H+x_C\Delta S_C\\ \Delta S=\frac{6}{7}\times \left[R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}-R\ln \frac{P_2}{P_1}\right]+\frac{1}{7}\times \left[R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}-R\ln \frac{P_2}{P_1}\right]\end{array}$

Where $R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$ is calculated by following formula

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(1)

And

  $\tau =\frac{T}{T_0}$

For warmer air stream

Values of above constants for air in equation (1) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{air}3.3550.5750-0.016\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T_H}{T_1}\\ \tau =\frac{300.15\text{K}}{293.15\text{K}}=1.024\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[3.355+\left[0.575\times {10}^{-3}\times 293.15+\left(0\times {293.15}^2+\frac{-0.016\times {10}^5}{{1.024}^2\times {293.15}^2}\right)\left(\frac{1.024+1}{2}\right)\right]\left(\frac{1.024-1}{\ln 1.024}\right)\right]\ln 1.024\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=0.0832\end{array}$

And for cooler air stream

  $\begin{array}{l}\tau =\frac{T_C}{T_1}\\ \tau =\frac{251.15\text{K}}{293.15\text{K}}=0.857\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[3.355+\left[0.575\times {10}^{-3}\times 293.15+\left(0\times {293.15}^2+\frac{-0.016\times {10}^5}{{0.857}^2\times {293.15}^2}\right)\left(\frac{0.857+1}{2}\right)\right]\left(\frac{0.857-1}{\ln 0.857}\right)\right]\ln 0.857\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=-0.5385\end{array}$

Hence Total entropy change during the process is

  $\begin{array}{l}\Delta S=x_H\Delta S_H+x_C\Delta S_C\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\left[\frac{6}{7}\times \left[0.0832-\ln \frac{1.01325}{5}\right]+\frac{1}{7}\times \left[-0.5385-\ln \frac{1.01325}{5}\right]\right]\\ \Delta S=13.225\frac{\text{J}}{\text{mol K}}=0.0132\frac{\text{kJ}}{\text{mol K}}\end{array}$

Hence

  $\Delta S=S_G=0.0132\frac{\text{kJ}}{\text{mol K}}\ge 0$

Hence total entropy (system surrounding) is greater than or equal to zero for any change of state of the system. So, second law of thermodynamics is valid.

So, we can say that both first and second law of thermodynamics are valid for the given process hence this process is possible.