Chapter 6, Problem 6.37P

(a)

Interpretation Introduction

Interpretation:

Calculate the change in enthalpy $\Delta H$ and entropy $\Delta S$ of steam if it undergoes a change from an initial state to a final state using steam table data.

Concept Introduction:

During the heat transferred in the process, total mass and volume remains constant. So, the total volume after the heat addedis same as total volume at $\text{101}\text{.325}\text{kPa}$ . So, find the total volume and mass at $\text{101}\text{.325}\text{kPa}$ using the saturated steam table and use them to find $\Delta H$ and entropy $\Delta S$ .

Answer to Problem 6.37P

  $\Delta H=-611.17\frac{\text{J}}{\text{gm}}$

  $\Delta S=0.03\frac{\text{J}}{\text{gm}\text{K}}$

Explanation of Solution

Given information:

It is given that at initial state

$P_1\text{=3000}\text{kPa}$, $T_1{\text{=450}}^{\circ }\text{C+273}\text{.15=723}\text{.15}\text{K}$

And at final state,

$P_2\text{=235}\text{kPa}$, $T_1{\text{=140}}^{\circ }\text{C+273}\text{.15=413}\text{.15}\text{K}$

From steam tables of superheated steam in Appendix E, table E.2

At pressure $P_1\text{=3000}\text{kPa}$, $T_1{\text{=450}}^{\circ }\text{C+273}\text{.15=723}\text{.15}\text{K}$,

$V_1=107.79\frac{{\text{cm}}^{\text{3}}}{\text{gm}}$ and $H_1=3344.6\frac{\text{J}}{\text{gm}}$, $S_1=7.0854\frac{\text{J}}{\text{gm K}}$

At pressure $P_2\text{=235}\text{kPa}$, $T_1{\text{=140}}^{\circ }\text{C+273}\text{.15=413}\text{.15}\text{K}$,

Since, pressure $\text{235}\text{kPa}$ lies in between pressures $\text{225}\text{kPa}$ and $\text{250}\text{kPa}$ and temperature $\text{413}\text{.15}\text{K}$ lies in between temperatures $\text{398}\text{.15}\text{K}$ and $\text{423}\text{.15}\text{K}$ . So, there are two independent variables and hence double interpolation must be applicable on it. So, corresponding values of volume, enthalpyand entropy are:

At $\text{225}\text{kPa}$ and $\text{398}\text{.15}\text{K}$,

  $V=795.25\frac{{\text{cm}}^{\text{3}}}{\text{gm}}$

$H=2713.8\frac{\text{J}}{\text{gm}}$ and $S=7.0928\frac{\text{J}}{\text{gm K}}$

At $\text{225}\text{kPa}$ and $\text{423}\text{.15}\text{K}$,

  $V=850.97\frac{{\text{cm}}^{\text{3}}}{\text{gm}}$

$H=2766.5\frac{\text{J}}{\text{gm}}$ and $S=7.2213\frac{\text{J}}{\text{gm K}}$

At $\text{250}\text{kPa}$ and $\text{398}\text{.15}\text{K}$,

  $V=713.698\frac{{\text{cm}}^{\text{3}}}{\text{gm}}$

$H=2711.12\frac{\text{J}}{\text{gm}}$ and $S=7.0404\frac{\text{J}}{\text{gm K}}$

At $\text{250}\text{kPa}$ and $\text{423}\text{.15}\text{K}$,

  $V=764.09\frac{{\text{cm}}^{\text{3}}}{\text{gm}}$

$H=2764.5\frac{\text{J}}{\text{gm}}$ and $S=7.1689\frac{\text{J}}{\text{gm K}}$

Now, From linear double interpolation, if $M$ is the function of two independent variable $X$ and $Y$ then the value of quantity $M$ at two independent variables $X$ and $Y$ adjacent to the given values are represented as follows:

  $\begin{array}{l}{X}_{1}X{X}_2\\ Y_1{M}_{1,1}{M}_{1,2}\\ YM=?\\ Y_2{M}_{2,1}{M}_{2,2}\end{array}$

So,

Similarly, enthalpy is

And entropy

Hence,

Change in entropy and enthalpy is,

  $\begin{array}{l}\Delta H=H_2-H_1\\ \\ \Delta H=2733.43\frac{{\text{cm}}^{\text{3}}}{\text{gm}}-3344.6\frac{\text{J}}{\text{gm}}\\ \\ \Delta H=-611.17\frac{\text{J}}{\text{gm}}\end{array}$

And

  $\begin{array}{l}\Delta S=S_2-S_1\\ \\ \Delta S=7.11276\frac{\text{J}}{\text{gm}\text{K}}-7.0854\frac{\text{J}}{\text{gm K}}\\ \\ \Delta S=0.03\frac{\text{J}}{\text{gm}\text{K}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Calculate the change in enthalpy $\Delta H$ and entropy $\Delta S$ of steam if it undergoes a change from an initial state to a final state using equation of an ideal gas.

Concept Introduction:

The change in enthalpy and entropy in ideal gas is defined as:

  $\begin{array}{l}d{H}^{ig}=C_P{}^{ig}dT\\ \text{Or,}\\ \Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P{}^{ig}}{R}dT}\end{array}$

...(1a)

And

  $\begin{array}{l}d{S}^{ig}=C_P{}^{ig}\frac{dT}{T}-R\frac{dP}{P}\\ \\ \Delta S=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}-R\ln \frac{P_2}{P_1}\end{array}$

....(1b)

Where,

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

.....(2a)

Such that $\tau =\frac{T}{T_0}$

And from Appendix D,

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

.....(2b)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 6.37P

  $\Delta H=-621.05\frac{\text{J}}{\text{gm}}$

  $\Delta S=0.0594\frac{\text{J}}{\text{gm}\text{K}}$

Explanation of Solution

Given information:

It is given that at initial state

$P_1\text{=3000}\text{kPa}$, $T_1{\text{=450}}^{\circ }\text{C+273}\text{.15=723}\text{.15}\text{K}$

And at final state,

$P_2\text{=235}\text{kPa}$, $T_2{\text{=140}}^{\circ }\text{C+273}\text{.15=413}\text{.15}\text{K}$

  $\tau =\frac{T}{T_0}=\frac{413.15}{723.15}=0.571$

Values of above constants are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ steam3.471.4500.121\end{array}$

So, from equation (2a)

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

Put the values

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{ig}{}_P}{R}dT}=3.47\times 723.15\times (0.571-1)+\frac{1.45\times {10}^{-3}}{2}{723.15}^2({0.571}^2-1)\\ +\frac{0}{3}{723.15}^3({0.571}^3-1)+\frac{0.121\times {10}^5}{723.15}(\frac{0.571-1}{0.571})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=-1344.6\text{K}\end{array}$

Now put the value in equation (1a),

  $\begin{array}{l}\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P{}^{ig}}{R}dT}\\ \Delta H=8.314\frac{\text{J}}{\text{K mol}}\times -1344.6\text{K}\\ \Delta H=-11178.97\frac{\text{J}}{\text{mol}}\\ \Delta H=-11178.97\frac{\text{J}}{\text{mol}}\times \frac{1\text{mol}}{18\text{gm}}\\ \Delta H=-621.05\frac{\text{J}}{\text{gm}}\end{array}$

Now, from equation (2b)

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

Put the values,

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{ig}{}_P}{R}\frac{dT}{T}}=\left[3.47+\left\{1.45\times {10}^{-3}\times \text{723}\text{.15}\text{K}+\left(0\times \text{723}\text{.15}{\text{K}}^2+\frac{0.121\times {10}^5}{{0.571}^2\times \text{723}{\text{.15}}^2{\text{K}}^2}\right)\left(\frac{0.571+1}{2}\right)\right\}\left(\frac{0.571-1}{\ln 0.571}\right)\right]\\ \times \ln 0.571\\ \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{ig}{}_P}{R}\frac{dT}{T}}=-2.418\end{array}$

Now put the value in equation (1b),

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}-R\ln \frac{P_2}{P_1}\\ \text{Or,}\end{array}$

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}+R\ln \frac{P_1}{P_2}\\ \Delta S=8.314\frac{\text{J}}{\text{K mol}}\times -2.418+8.314\frac{\text{J}}{\text{K mol}}\times \ln \frac{\text{3000}\text{kPa}}{\text{235}\text{kPa}}\\ \Delta S=1.0689\frac{\text{J}}{\text{mol}\text{K}}\end{array}$

  $\begin{array}{l}\Delta S=1.0689\frac{\text{J}}{\text{mol}\text{K}}\times \frac{1\text{mol}}{18\text{gm}}\\ \Delta S=0.0594\frac{\text{J}}{\text{gm}\text{K}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Calculate the change in enthalpy $\Delta H$ and entropy $\Delta S$ of steam if it undergoes a change from an initial state to a final state using appropriate generalized correlations.

Concept Introduction:

The change in enthalpy and entropy according to generalized correlationsare defined as:

  $\begin{array}{l}\Delta H=\Delta H^{ig}+H_2{}^R-H_1{}^R\\ \text{where,}\\ \frac{H_2{}^R}{R{T}_C}=HRB\left(T_r,P_r,\omega \right)\\ \\ \text{Hence}\\ \Delta H=\Delta H^{ig}+R{T}_C\left\{HRB\left(T_{r2},P_{r2},\omega \right)-HRB\left(T_{r1},P_{r1},\omega \right)\right\}\end{array}$

...(1a)

And

  $\begin{array}{l}\Delta S=\Delta S^{ig}+S_2{}^R-S_1{}^R\\ \text{where,}\\ \frac{S_2{}^R}{R}=SRB\left(T_r,P_r,\omega \right)\\ \\ \text{Hence}\\ \Delta S=\Delta S^{ig}+R\left\{SRB\left(T_{r2},P_{r2},\omega \right)-SRB\left(T_{r1},P_{r1},\omega \right)\right\}\end{array}$

....(1b)

Where,

  $HRB\left(T_r,P_r,\omega \right)=\frac{H^R}{R{T}_C}=P_r\left[B^0-T_r\frac{d{B}^0}{d{T}_r}+\omega \left(B^1-T_r\frac{d{B}^1}{d{T}_r}\right)\right]$

.....(2a)

And

  $SRB\left(T_r,P_r,\omega \right)=\frac{S^R}{R}=-P_r\left[\frac{d{B}^0}{d{T}_r}+\omega \frac{d{B}^1}{d{T}_r}\right]$

.....(2b)

Answer to Problem 6.37P

  $\Delta H=-602.46\frac{\text{J}}{\text{gm}}$

  $\Delta S=0.07\frac{\text{J}}{\text{gm}\text{K}}$

Explanation of Solution

Given information:

It is given that at initial state

$P_1\text{=3000}\text{kPa}$, $T_1{\text{=450}}^{\circ }\text{C+273}\text{.15=723}\text{.15}\text{K}$

And at final state,

$P_2\text{=235}\text{kPa}$, $T_2{\text{=140}}^{\circ }\text{C+273}\text{.15=413}\text{.15}\text{K}$

From equation (2a),

  $HRB\left(T_r,P_r,\omega \right)=\frac{H^R}{R{T}_C}=P_r\left[B^0-T_r\frac{d{B}^0}{d{T}_r}+\omega \left(B^1-T_r\frac{d{B}^1}{d{T}_r}\right)\right]$

Here $B^0=0.083-\frac{0.422}{T_r{}^{1.6}}$ and $B^1=0.139-\frac{0.172}{T_r{}^{4.2}}$

Properties of pure species of steam are given in Table B.1 Appendix B as water,

  $\omega =0.345,T_C=647.1\text{K,}{P}_C=220.55\text{bar}$

For initial state

  $\begin{array}{l}{T}_{r1}=\frac{T_1}{T_c}\\ T_{r1}=\frac{723.15\text{K}}{647.1\text{K}}\text{=1}\text{.117}\end{array}$,

  $\begin{array}{l}{P}_{r1}=\frac{P_1}{P_c}\\ P_{r1}=\frac{3000\text{kPa}\times \frac{1\text{bar}}{100\text{kPa}}}{220.55\text{bar}}\text{=0}\text{.136}\end{array}$

So,

  $\begin{array}{l}{B}^0{}_1=0.083-\frac{0.422}{T_{r1}{}^{1.6}}\\ \\ B^0{}_1=0.083-\frac{0.422}{{1.117}^{1.6}}\\ \\ B^0{}_1=-0.27\end{array}$

And

  $\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ \\ B^1=0.139-\frac{0.172}{{1.117}^{4.2}}\\ \\ B^1{}_1=0.031\end{array}$

For differentiative terms in equation (2a),

  $\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ \\ \frac{d{B}_1{}^0}{d{T}_r}=\frac{0.422}{T_r{}^{2.6}}=\frac{0.422}{\text{1}{\text{.117}}^{2.6}}=0.3165\end{array}$

  $\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ \\ \frac{d{B}_1{}^1}{d{T}_r}=\frac{0.172}{T_r{}^{5.2}}=\frac{0.172}{\text{1}{\text{.117}}^{5.2}}=0.09675\end{array}$

So,

From equation (2a),

  $\begin{array}{l}HRB\left(T_r,P_r,\omega \right)=\frac{H^R}{R{T}_C}=P_{r1}\times \left[B^0{}_1-T_{r1}\frac{d{B}^0}{d{T}_{r1}}+\omega \left(B^1{}_1-T_{r1}\frac{d{B}^1{}_1}{d{T}_{r1}}\right)\right]\\ \\ \frac{H^R}{R{T}_C}=\text{0}\text{.136}\times \left[-0.27-\text{1}\text{.117}\times 0.3165+0.345\times \left(0.031-\text{1}\text{.117}\times 0.09675\right)\right]\\ \\ HRB\left(T_{r1},P_{r1},\omega \right)=\frac{H^R{}_1}{R{T}_C}=-0.0884\end{array}$

Similarly, for final state

  $\begin{array}{l}{T}_{r2}=\frac{T_2}{T_c}\\ T_{r2}=\frac{413.15\text{K}}{647.1\text{K}}\text{=0}\text{.638}\end{array}$

And

  $\begin{array}{l}{P}_{r2}=\frac{P_2}{P_c}\\ P_{r2}=\frac{235\text{kPa}\times \frac{1\text{bar}}{100\text{kPa}}}{220.55\text{bar}}\text{=0}\text{.011}\end{array}$

So,

  $\begin{array}{l}{B}^0{}_2=0.083-\frac{0.422}{T_{r1}{}^{1.6}}\\ \\ B^0{}_2=0.083-\frac{0.422}{{0.638}^{1.6}}\\ \\ B^0{}_2=-0.783\end{array}$

And

  $\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ \\ B^1{}_2=0.139-\frac{0.172}{{0.638}^{4.2}}\\ \\ B^1{}_2=-0.997\end{array}$

For differentiative terms in equation (2a),

  $\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ \\ \frac{d{B}_2{}^0}{d{T}_{r2}}=\frac{0.422}{T_{r2}{}^{2.6}}=\frac{0.422}{\text{0}{\text{.638}}^{2.6}}=1.358\end{array}$

  $\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ \\ \frac{d{B}_2{}^1}{d{T}_{r2}}=\frac{0.172}{T_{r2}{}^{5.2}}=\frac{0.172}{\text{0}{\text{.638}}^{5.2}}=1.78\end{array}$

So,

From equation (2a),

  $\begin{array}{l}HRB\left(T_r,P_r,\omega \right)=\frac{H^R}{R{T}_C}=P_{r2}\times \left[B^0{}_2-T_{r2}\frac{d{B}^0{}_2}{d{T}_{r2}}+\omega \left(B^1{}_2-T_{r2}\frac{d{B}^1{}_2}{d{T}_{r2}}\right)\right]\\ \\ \frac{H^R}{R{T}_C}=\text{0}\text{.011}\times \left[-0.783-\text{0}\text{.638}\times 1.358+0.345\times \left(-0.997-\text{0}\text{.638}\times 1.78\right)\right]\\ \\ HRB\left(T_{r2},P_{r2},\omega \right)=\frac{H^R{}_2}{R{T}_C}=-0.0262\end{array}$

Now, change in enthalpy will be according to generalized correlations,

  $\begin{array}{l}\Delta H=\Delta H^{ig}+R{T}_C\left\{HRB\left(T_{r2},P_{r2},\omega \right)-HRB\left(T_{r1},P_{r1},\omega \right)\right\}\\ \\ \Delta H=-621.05\frac{\text{J}}{\text{gm}}+8.314\frac{\text{J}}{\text{mol}\text{K}}\times 647.1\text{K}\times \frac{1\text{mol}}{18\text{gm}}\times \left\{\left(-0.0262\right)-\left(-0.0884\right)\right\}\\ \\ \Delta H=-602.46\frac{\text{J}}{\text{gm}}\end{array}$

Now from equation (2b),

  $\begin{array}{l}SRB\left(T_{r1},P_{r1},\omega \right)=\frac{S^R{}_1}{R}=-P_{r1}\left[\frac{d{B}^0{}_1}{d{T}_{r1}}+\omega \frac{d{B}^1{}_1}{d{T}_{r1}}\right]\\ \\ SRB\left(T_{r1},P_{r1},\omega \right)=\frac{S^R{}_1}{R}=-\text{0}\text{.136}\times \left[0.3165+0.345\times 0.09675\right]\\ \\ SRB\left(T_{r1},P_{r1},\omega \right)=-0.04758\end{array}$

Similarly

  $\begin{array}{l}SRB\left(T_{r2},P_{r2},\omega \right)=\frac{S^R{}_2}{R}=-P_{r2}\left[\frac{d{B}^0{}_2}{d{T}_{r2}}+\omega \frac{d{B}^1{}_2}{d{T}_{r2}}\right]\\ \\ SRB\left(T_{r2},P_{r2},\omega \right)=\frac{S^R{}_2}{R}=-\text{0}\text{.011}\times \left[1.358+0.345\times 1.78\right]\\ \\ SRB\left(T_{r2},P_{r2},\omega \right)=-0.02169\end{array}$

Hence,

  $\begin{array}{l}\Delta S=\Delta S^{ig}+R\left\{SRB\left(T_{r2},P_{r2},\omega \right)-SRB\left(T_{r1},P_{r1},\omega \right)\right\}\\ \\ \Delta S=0.0594\frac{\text{J}}{\text{gm}\text{K}}+8.314\frac{\text{J}}{\text{mol}\text{K}}\times \frac{1\text{mol}}{18\text{gm}}\times \left\{\left(-0.02169\right)-\left(-0.04758\right)\right\}\\ \\ \Delta S=0.07\frac{\text{J}}{\text{gm}\text{K}}\end{array}$