Chapter 10, Problem 10.38P

Interpretation Introduction

(a)

Interpretation:

Rate of air and oxygen flow in mol/s unit should be calculated.

Concept Introduction:

Let us assume $x$ represents rate of air flow (mol/s) and $y$ represents rate of pure oxygen flow rate (mol/s) into the process.

Component material balance equation will be used to determine rate of air flow and oxygen flow.

Interpretation Introduction

(b)

Interpretation:

Rate of heat transfer for this mixing process should be calculated.

Concept Introduction:

It is considered ideal gas law is followed for the entire mixing process, $PV=nRT$

This is an isothermal mixing process at constant pressure. Since no moving parts the system, so external work done is zero.

First law of thermodynamics equation, $dU=dQ+dW$

For isothermal process, $dU=0$

Therefore, $dQ=-dW$

Integrating for overall process, $Q=-W$

Here $W$ is the total reversible work done in the mixing process to be calculated as follows:

  $\begin{array}{l}W=W_A+W_B\\ =-{\displaystyle {\int }_{V_A}^{V_A+V_B}PdV}-{\displaystyle {\int }_{V_B}^{V_A+V_B}PdV}\\ =-{\displaystyle {\int }_{V_A}^{V_A+V_B}{n}_{A}RTdV/V-}{\displaystyle {\int }_{V_B}^{V_A+V_B}{n}_{B}RTdV}/V\\ =-n_{A}RT\ln V_A+V_B/V_A-n_{B}RT\ln V_A+V_B/V_B\\ =-n_{A}RT\ln n_A+n_B/n_A-n_{B}RT\ln n_A+n_B/n_B\\ =n_{A}RT\ln x_A+n_{B}RT\ln x_B\end{array}$ (applying ideal gas law equation)

Here $n_A$, $V_A$, $x_{{}_A}$ refers to mole of air, volume of air and final mole fraction of air. Similarly, $n_B$, $V_B$, $x_{{}_B}$ refers to mole of oxygen, volume of oxygen and final mole fraction of oxygen.

Rate of heat transfer, $Q=-n_{A}RT\ln x_A-n_{B}RT\ln x_B$

Interpretation Introduction

(c)

Interpretation:

Rate of entropy generation (WK-1) should be calculated.

Concept Introduction:

Entropy of mixing can be calculated using following formula:

  $\begin{array}{c}\Delta S=Q/T\\ =-n_{A}R\ln x_A-n_{B}R\ln x_B\end{array}$

Here, n is number of moles, R is Universal gas constant and x is mole fraction.