Chapter 6, Problem 6.46P

(a)

Interpretation Introduction

Interpretation:

Calculate values of $G^l$ and $G^v$ for saturated liquid and vapor at $\text{150}\text{psia}$ . Comment whether these two have same value or not.

Concept Introduction:

The Gibbs free energy is calculated as:

  $G=H-TS$

Answer to Problem 6.46P

  $G^l=-89.9\frac{\text{btu}}{{\text{lb}}_{\text{m}}}$

  $G^v=-89.9\frac{\text{btu}}{{\text{lb}}_{\text{m}}}$

Explanation of Solution

Given information:

It is given that pressure is $P\text{=150}\text{psia}$ and steam is saturated.

From steam tables of saturated steam in Appendix E, table E.1

At pressure $P\text{=150}\text{psia}$

  $P\text{=150}\text{psia}\times \frac{6.89476\text{kPa}}{\text{1}\text{psia}}\text{=1034}\text{.214}\text{kPa}$

Since, pressure $P\text{=1034}\text{.214}\text{kPa}$ lies in between pressures $\text{1049}\text{.6}\text{kPa}$ and $\text{1002}\text{.7}\text{kPa}$ with temperatures $\text{455}\text{.15}\text{K}$ and $\text{453}\text{.15}\text{K}$, with enthalpies of saturated vapor are $H^{vap}=2778.0\frac{\text{J}}{\text{gm}}$ and $H^{vap}=2776.3\frac{\text{J}}{\text{gm}}$, enthalpies of saturated liquid $H^{liq}=772.0\frac{\text{J}}{\text{gm}}$ and $H^{liq}=763.1\frac{\text{J}}{\text{gm}}$, with entropies of saturated vapor are $S^{vap}=6.5660\frac{\text{J}}{\text{gm}\text{K}}$ and $S^{vap}=6.5819\frac{\text{J}}{\text{gm}\text{K}}$, entropies of saturated liquid are $S^{liq}=2.1587\frac{\text{J}}{\text{gm}\text{K}}$ and $S^{liq}=2.1393\frac{\text{J}}{\text{gm}\text{K}}$ .

From linear interpolation, if $M$ is the function of a single independent variable $X$ then the value of $M$ at $X$ is intermediate between two given values, $M_1$ at $X_1$ and $M_2$ at $X_2$

Temperature corresponding to $P\text{=150}\text{psia}$

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ T=\left(\frac{\text{1049}\text{.6}\text{kPa}-\text{1034}\text{.214}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times \text{453}\text{.15}\text{K}+\left(\frac{\text{1034}\text{.214}\text{kPa}-\text{1002}\text{.7}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times \text{455}\text{.15}\text{K}\\ \\ T=\text{454}\text{.49}\text{K}\end{array}$

Enthalpy of saturated liquid and entropy of saturated liquid and vapor is

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ H^{vap}=\left(\frac{\text{1049}\text{.6}\text{kPa}-\text{1034}\text{.214}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 2776.3\frac{\text{J}}{\text{gm}}+\left(\frac{\text{1034}\text{.214}\text{kPa}-\text{1002}\text{.7}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 2778.0\frac{\text{J}}{\text{gm}}\\ \\ H^{vap}=2777.44\frac{\text{J}}{\text{gm}}\end{array}$

And,

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ H^{liq}=\left(\frac{\text{1049}\text{.6}\text{kPa}-\text{1034}\text{.214}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 763.1\frac{\text{J}}{\text{gm}}+\left(\frac{\text{1034}\text{.214}\text{kPa}-\text{1002}\text{.7}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 772\frac{\text{J}}{\text{gm}}\\ \\ H^{liq}=769.08\frac{\text{J}}{\text{gm}}\end{array}$

Entropy of saturated liquid and entropy of saturated liquid and vapor is

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ S^{liq}=\left(\frac{\text{1049}\text{.6}\text{kPa}-\text{1034}\text{.214}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 2.1393\frac{\text{J}}{\text{gm}\text{K}}+\left(\frac{\text{1034}\text{.214}\text{kPa}-\text{1002}\text{.7}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 2.1587\frac{\text{J}}{\text{gm}\text{K}}\\ \\ S^{liq}=2.1523\frac{\text{J}}{\text{gm}\text{K}}\end{array}$

And,

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ S^{Vap}=\left(\frac{\text{1049}\text{.6}\text{kPa}-\text{1034}\text{.214}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 6.5819\frac{\text{J}}{\text{gm}\text{K}}+\left(\frac{\text{1034}\text{.214}\text{kPa}-\text{1002}\text{.7}\text{kPa}}{\text{1049}\text{.6}\text{kPa}-\text{1002}\text{.7}\text{kPa}}\right)\times 6.5660\frac{\text{J}}{\text{gm}\text{K}}\\ \\ S^{Vap}=6.571\frac{\text{J}}{\text{gm}\text{K}}\end{array}$

So,

  $\begin{array}{l}{G}^l=H^l-T{S}^l\\ G^l=769.08\frac{\text{J}}{\text{gm}}-\text{454}\text{.49}\text{K}\times 2.1523\frac{\text{J}}{\text{gm}\text{K}}\\ G^l=-209.12\frac{\text{J}}{\text{gm}}=-89.9\frac{\text{btu}}{{\text{lb}}_{\text{m}}}\end{array}$

And

  $\begin{array}{l}{G}^{vap}=H^{vap}-T{S}^{vap}\\ G^v=2777.44\frac{\text{J}}{\text{gm}}-\text{454}\text{.49}\text{K}\times 6.571\frac{\text{J}}{\text{gm}\text{K}}\\ G^v=-209.01\frac{\text{J}}{\text{gm}}=-89.9\frac{\text{btu}}{{\text{lb}}_{\text{m}}}\end{array}$

Both values of $G^l$ and $G^v$ for saturated liquid and vapor at $P\text{=150}\text{psia}$ are same.

(b)

Interpretation Introduction

Interpretation:

Calculate values for $\frac{\Delta H^{lv}}{T}$ and $\Delta S^{lv}$ at $P\text{=150}\text{psia}$ . Comment whether these two have same value or not.

Concept Introduction:

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

  $\Delta H^{lv}=H^v-H^l$

And

  $\Delta S^{lv}=S^v-S^l$

Answer to Problem 6.46P

  $\frac{\Delta H^{lv}}{T}=1.055\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}$

  $\Delta S^{lv}=1.055\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}$

Explanation of Solution

From subpart (a), values of enthalpies of saturated vapor and liquid as well as values of entropies of saturated vapor and liquid are:

$H^l=769.08\frac{\text{J}}{\text{gm}}$, $S^l=2.1523\frac{\text{J}}{\text{gm}\text{K}}$, $H^{vap}=2777.44\frac{\text{J}}{\text{gm}}$ and $S^{vap}=6.571\frac{\text{J}}{\text{gm}\text{K}}$

And temperature is:

  $T=\text{454}\text{.49}\text{K}$

So,

  $\begin{array}{l}\frac{\Delta H^{lv}}{T}=\frac{H^v-H^l}{T}\\ \frac{\Delta H^{lv}}{T}=\frac{2777.44\frac{\text{J}}{\text{gm}}-769.08\frac{\text{J}}{\text{gm}}}{\text{454}\text{.49}\text{K}}\\ \frac{\Delta H^{lv}}{T}=4.419\frac{\text{J}}{\text{gm}\text{K}}=1.055\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}\end{array}$

And

  $\begin{array}{l}\Delta S^{lv}=S^v-S^l\\ \Delta S^{lv}=6.571\frac{\text{J}}{\text{gm}\text{K}}-2.1523\frac{\text{J}}{\text{gm}\text{K}}\\ \Delta S^{lv}=4.4187\frac{\text{J}}{\text{gm}\text{K}}=1.055\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}\end{array}$

Both values of $\frac{\Delta H^{lv}}{T}$ and $\Delta S^{lv}$ for saturated liquid and vapor at $P\text{=150}\text{psia}$ are same.

(c)

Interpretation Introduction

Interpretation:

Calculate values for $V^R$, $H^R$ and $S^R$ for saturated vapor at $P\text{=150}\text{psia}$ and compare the values with suitable generalized correlation.

Concept Introduction:

The residual properties $V^R$ is calculated as:

  $V^R=V-V^{ig}=V-\frac{RT}{P}$

The residual properties $H^R$ is calculated as:

  $H^R=H-H^{ig}$

The residual properties $S^R$ is calculated as:

  $S^R=S-S^{ig}$

Answer to Problem 6.46P

  $V^R=-0.23\frac{{\text{ft}}^{\text{3}}}{{\text{lb}}_{\text{m}}}$

  $H^R=-27.39\frac{\text{btu}}{{\text{lb}}_{\text{m}}}$

  $S^R=-0.025\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}$

Explanation of Solution

At $P\text{=150}\text{psia}$, for saturated vapor

$H=2777.44\frac{\text{J}}{\text{gm}}$ and $S=6.571\frac{\text{J}}{\text{gm}\text{K}}$, $T=\text{454}\text{.49}\text{K}$

$V\text{at 1002}\text{.7}\text{kPa}=193.8\frac{{\text{cm}}^{\text{3}}}{\text{gm}}$, $V\text{at 1049}\text{.6}=185.5\frac{{\text{cm}}^{\text{3}}}{\text{gm}}$

So,

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ V=\left(\frac{\text{1002}\text{.7}\text{kPa}-1034.214\text{kPa}}{\text{1002}\text{.7}\text{kPa}-\text{957}\text{.36}\text{kPa}}\right)\times 193.8\frac{{\text{cm}}^{\text{3}}}{\text{gm}}+\left(\frac{1034.214\text{kPa}-\text{957}\text{.36}\text{kPa}}{\text{1002}\text{.7}\text{kPa}-\text{957}\text{.36}\text{kPa}}\right)\times 185.5\frac{{\text{cm}}^{\text{3}}}{\text{gm}}\\ \\ V=188.22\frac{{\text{cm}}^{\text{3}}}{\text{gm}}\end{array}$

Now, for the hypothetical ideal gas values of volume, enthalpy and entropy at same temperature and pressure, it cannot find using steam table because steam tables only give values of real gases not ideal gases. However, we can make an approximation of low pressure in real gases at the same temperature to convert real gas into ideal one. So, we are considering low pressure or $\text{1}\text{kPa}\text{and 454}\text{.49}\text{K}$ . Now from steam tables of superheated steam in Appendix E, table E.2, at $\text{1}\text{kPa}\text{and 454}\text{.49}\text{K}$

Since,

$\text{1}\text{kPa}\text{and 454}\text{.49}\text{K}$ lies in between temperatures $\text{448}\text{.15}\text{K}$ and $\text{473}\text{.15}\text{K}$, with volumes $206810\frac{{\text{cm}}^3}{\text{gm}}$ and $218350\frac{{\text{cm}}^3}{\text{gm}}$, enthalpies $2831.7\frac{\text{J}}{\text{gm}}$ and $2880.1\frac{\text{J}}{\text{gm}}$, entropies $9.8629\frac{\text{J}}{\text{gm}\text{K}}$ and $9.9679\frac{\text{J}}{\text{gm}\text{K}}$ .

Hence from linear interpolation, at $\text{1}\text{kPa}\text{and 454}\text{.49}\text{K}$

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ V^{ig}=\left(\frac{\text{473}\text{.15}\text{K}-\text{453}\text{.03}\text{K}}{\text{473}\text{.15}\text{K}-\text{448}\text{.15}\text{K}}\right)\times 206810\frac{{\text{cm}}^3}{\text{gm}}+\left(\frac{\text{453}\text{.03}\text{K}-\text{448}\text{.15}\text{K}}{\text{473}\text{.15}\text{K}-\text{448}\text{.15}\text{K}}\right)\times 218350\frac{{\text{cm}}^3}{\text{gm}}\\ \\ V^{ig}=209062\frac{{\text{cm}}^3}{\text{gm}}\end{array}$

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ H^{ig}=\left(\frac{\text{473}\text{.15}\text{K}-\text{453}\text{.03}\text{K}}{\text{473}\text{.15}\text{K}-\text{448}\text{.15}\text{K}}\right)\times 2831.7\frac{\text{J}}{\text{gm}}+\left(\frac{\text{453}\text{.03}\text{K}-\text{448}\text{.15}\text{K}}{\text{473}\text{.15}\text{K}-\text{448}\text{.15}\text{K}}\right)\times 2880.1\frac{\text{J}}{\text{gm}}\\ \\ H^{ig}=2841.15\frac{\text{J}}{\text{gm}}\end{array}$

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ S^{ig}=\left(\frac{\text{473}\text{.15}\text{K}-\text{453}\text{.03}\text{K}}{\text{473}\text{.15}\text{K}-\text{448}\text{.15}\text{K}}\right)\times 9.8629\frac{\text{J}}{\text{gm}\text{K}}+\left(\frac{\text{453}\text{.03}\text{K}-\text{448}\text{.15}\text{K}}{\text{473}\text{.15}\text{K}-\text{448}\text{.15}\text{K}}\right)\times 9.9679\frac{\text{J}}{\text{gm}\text{K}}\\ \\ S^{ig}=9.883\frac{\text{J}}{\text{gm}\text{K}}\end{array}$

But we want values of volume, enthalpy and entropy of ideal gas at $1034.214\text{kPa}\text{and 454}\text{.49}\text{K}$, not $\text{1}\text{kPa}\text{and 454}\text{.49}\text{K}$ .Since temperature is not changing ( $\text{454}\text{.49}\text{K}$ ) and hence it is constant so, it is isothermal process and hence

  $P_1{V}_1=P_2{V}_2$ is applicable

  $\begin{array}{l}1034.214\text{kPa}\times V_1=1\times 209062\frac{{\text{cm}}^3}{\text{gm}}\\ \\ V^{ig}=202.146\frac{{\text{cm}}^{\text{3}}}{\text{gm}}\end{array}$

And

Enthalpy of ideal gas is defined as

  $\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P{}^{ig}}{R}dT}$

Which is not the function of pressure, so enthalpy of ideal gas at $1034.214\text{kPa}\text{and 454}\text{.49}\text{K}$ is same as enthalpy of ideal gas at $\text{1}\text{kPa}\text{and 454}\text{.49}\text{K}$ and hence,

  $H^{ig}=2841.15\frac{\text{J}}{\text{gm}}$

Now,

For entropy we know that

  $d{S}^{ig}=C_P{}^{ig}\frac{dT}{T}-R\frac{dP}{P}$

Since temperature s constant so first term at right hand side will be zero and hence,

  $\begin{array}{l}d{S}^{ig}=-R\frac{dP}{P}\\ \Delta S^{ig}=-R{\displaystyle \underset{P_1}{\overset{P_2}{\int }}\frac{dP}{P}}=-R\ln \frac{P_2}{P_1}\\ \Delta S^{ig}=-8.314\frac{\text{J}}{\text{mol K}}\times \frac{1\text{mol}}{18\text{gm}}\times \ln \frac{1034.214}{1}\\ \Delta S^{ig}=-3.206\frac{\text{J}}{\text{gm}\text{K}}\\ S_2{}^{ig}-S_1{}^{ig}=-3.206\frac{\text{J}}{\text{gm}\text{K}}\end{array}$

  $S_2{}^{ig}=S_1{}^{ig}-3.206\frac{\text{J}}{\text{gm}\text{K}}$

  $S_2{}^{ig}=9.883\frac{\text{J}}{\text{gm}\text{K}}-3.206\frac{\text{J}}{\text{gm}\text{K}}=6.677\frac{\text{J}}{\text{gm}\text{K}}$

So,

Residual properties are:

  $V^R=V-V^{ig}$

  $\begin{array}{l}{V}^R=188.22\frac{{\text{cm}}^{\text{3}}}{\text{gm}}-202.146\frac{{\text{cm}}^{\text{3}}}{\text{gm}}\\ V^R=-13.926\frac{{\text{cm}}^{\text{3}}}{\text{gm}}\times \frac{0.016\frac{{\text{ft}}^{\text{3}}}{{\text{lb}}_{\text{m}}}}{1\frac{{\text{cm}}^{\text{3}}}{\text{gm}}}=-0.23\frac{{\text{ft}}^{\text{3}}}{{\text{lb}}_{\text{m}}}\end{array}$

And

  $\begin{array}{l}{H}^R=H-H^{ig}\\ H^R=2777.44\frac{\text{J}}{\text{gm}}-2841.15\frac{\text{J}}{\text{gm}}\\ H^R=-63.71\frac{\text{J}}{\text{gm}}=-27.39\frac{\text{btu}}{{\text{lb}}_{\text{m}}}\end{array}$

And

  $\begin{array}{l}{S}^R=S-S^{ig}\\ S^R=6.571\frac{\text{J}}{\text{gm}\text{K}}-6.677\frac{\text{J}}{\text{gm}\text{K}}\\ S^R=-0.106\frac{\text{J}}{\text{gm}\text{K}}=-0.025\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}\end{array}$

Now calculations of residual properties of volume, enthalpy and entropy for saturated vapor from generalized correlations are given below,

$V^R=\frac{RT}{P}\left(Z-1\right)$

                      ....(1a)

Where, $Z=1+B^0\frac{P_r}{T_r}+\omega B^1\frac{P_r}{T_r}$

                      ....(1b)

$\frac{H^R}{R{T}_C}=HRB\left(T_r,P_r,\omega \right)$

                      ....(2a)

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

                      ....(2b)

And $\frac{S_2{}^R}{R}=SRB\left(T_r,P_r,\omega \right)$

                      ....(3a)

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

                      ....(3b)

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

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

  $\begin{array}{l}{T}_r=\frac{T}{T_c}\\ T_r=\frac{\text{454}\text{.49}\text{K}}{647.1\text{K}}\text{=0}\text{.7}\end{array}$,

  $\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{1034.214\text{kPa}\times \frac{1\text{bar}}{100\text{kPa}}}{220.55\text{bar}}\text{=0}\text{.0469}\end{array}$

So,

  $\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ \\ B^0=0.083-\frac{0.422}{{0.7}^{1.6}}\\ \\ B^0=-0.664\end{array}$

And

  $\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ \\ B^1=0.139-\frac{0.172}{{0.7}^{4.2}}\\ \\ B^1=-0.63\end{array}$

For differentiative terms in equation (2b) and (3b),

  $\begin{array}{l}{B}^0=0.083-\frac{0.422}{T_r{}^{1.6}}\\ \\ \frac{d{B}^0}{d{T}_r}=\frac{1.6\times 0.422}{T_r{}^{2.6}}=\frac{1.6\times 0.422}{\text{0}{\text{.7}}^{2.6}}=1.707\end{array}$

And,

  $\begin{array}{l}{B}^1=0.139-\frac{0.172}{T_r{}^{4.2}}\\ \\ \frac{d{B}^1}{d{T}_r}=\frac{4.2\times 0.172}{T_r{}^{5.2}}=\frac{4.2\times 0.172}{\text{0}{\text{.7}}^{5.2}}=4.616\end{array}$

For residual volume calculations, From equation (1b)

  $\begin{array}{l}Z=1+B^0\frac{P_r}{T_r}+\omega B^1\frac{P_r}{T_r}\\ Z=1+\left(-0.664\right)\times \frac{\text{0}\text{.0469}}{\text{0}\text{.7}}+0.345\times \left(-0.63\right)\times \frac{\text{0}\text{.0469}}{\text{0}\text{.7}}\\ Z=0.941\end{array}$

So, from equation (1a)

  $\begin{array}{l}{V}^R=\frac{RT}{P}\left(Z-1\right)\\ \\ V^R=\frac{8.314\frac{\text{kPa l}}{\text{mol K}}\times 454.49\text{K}}{1034.214\text{kPa}}\times \frac{1000{\text{cm}}^3}{1\text{l}}\times \frac{1\text{mol}}{18\text{gm}}\left(0.941-1\right)\\ \\ V^R=-11.976\frac{{\text{cm}}^3}{\text{gm}}\times \frac{0.016\frac{{\text{ft}}^{\text{3}}}{{\text{lb}}_{\text{m}}}}{1\frac{{\text{cm}}^{\text{3}}}{\text{gm}}}=-0.192\frac{{\text{ft}}^{\text{3}}}{{\text{lb}}_{\text{m}}}\end{array}$

For residual enthalpy calculations, From equation (2b)

  $\begin{array}{l}HRB\left(T_r,P_r,\omega \right)=\frac{H^R}{R{T}_C}=P_r\times \left[B^0-T_r\frac{d{B}^0}{d{T}_r}+\omega \left(B^1-T_r\frac{d{B}^1}{d{T}_r}\right)\right]\\ \\ \frac{H^R}{R{T}_C}=\text{0}\text{.0469}\times \left[-0.664-\text{0}\text{.7}\times 1.707+0.345\times \left(-0.63-\text{0}\text{.7}\times 4.616\right)\right]\\ \\ HRB\left(T_r,P_r,\omega \right)=\frac{H^R}{R{T}_C}=-0.1497\end{array}$

So, from equation (2a)

  $\begin{array}{l}\frac{H^R}{R{T}_C}=HRB\left(T_r,P_r,\omega \right)\\ H^R=R{T}_C\times HRB\left(T_r,P_r,\omega \right)\\ \\ H^R=8.314\frac{\text{J}}{\text{mol K}}\times 647.1\text{K}\times \frac{1\text{mol}}{18\text{gm}}\times -0.1497\\ \\ H^R=-44.74\frac{\text{J}}{\text{gm}}=19.23\frac{\text{btu}}{{\text{lb}}_{\text{m}}}\end{array}$

For residual entropy calculations, from equation (3b)

  $\begin{array}{l}SRB\left(T_r,P_r,\omega \right)=\frac{S^R}{R}=-P_r\left[\frac{d{B}^0}{d{T}_r}+\omega \frac{d{B}^1}{d{T}_r}\right]\\ \\ \frac{S^R}{R}=-\text{0}\text{.0469}\times \left[1.707+0.345\times \left(4.616\right)\right]\\ \\ SRB\left(T_r,P_r,\omega \right)=\frac{S^R}{T_C}=-0.1547\end{array}$

So, from equation (3a)

  $\begin{array}{l}\frac{S^R}{R}=SRB\left(T_r,P_r,\omega \right)\\ S^R=R\times SRB\left(T_r,P_r,\omega \right)\\ \\ S^R=8.314\frac{\text{J}}{\text{mol K}}\times \frac{1\text{mol}}{18\text{gm}}\times -0.1547\\ \\ S^R=-0.0714\frac{\text{J}}{\text{gm}\text{K}}=-0.0170\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}\end{array}$

Results do not agree but approximately they do agree.

(d)

Interpretation Introduction

Interpretation:

Calculate values for $\frac{d{P}^{sat}}{dT}$ at $P\text{=150}\text{psia}$ and apply the Clapeyron equation to evaluate $\Delta S^{lv}$ at $P\text{=150}\text{psia}$ . Comment whether results follow steam table values or not.

Concept Introduction:

First draw a graph between $\ln P$ and $\frac{1}{T}$, find its slope then calculate the $\frac{d{P}^{sat}}{dT}$ from slope.

The Clapeyron equation is:

  $\frac{d{P}^{sat}}{dT}=\frac{\Delta H^{lv}}{T\Delta V^{lv}}$

Answer to Problem 6.46P

  $\Delta S^{lv}=0.976\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}$

Explanation of Solution

From saturated steam table in Appendix E, table E.2

  $\begin{array}{l}{P}^{sat}(\text{kPa)}T(\text{K)}\\ 957.36451.15\\ 1000453.03\\ 1049.6455.15\end{array}$

Now, draw a graph between $\ln P$ and $\frac{1}{T}$

Slope of the graph from equation of graph is $-4.3627$

Or,

  $\text{slope=}\frac{d\ln P^{sat}}{d(1/T)}=-4.3627\times \frac{1}{{10}^{-3}}=\frac{1}{P^{sat}}\times -\frac{T^2}{1}\times \frac{d{P}^{sat}}{dT}$

So,

  $\begin{array}{l}\frac{d{P}^{sat}}{dT}=\text{slope}\times \frac{-P^{sat}}{T^2}\\ \frac{d{P}^{sat}}{dT}=-4362.7\times \frac{\text{-1034}\text{.214}\text{kPa}}{{454.49}^2\text{K}}\\ \frac{d{P}^{sat}}{dT}=21.84\end{array}$

Also,

Volume of saturated liquid and entropy of saturated liquid and vapor at pressure $P\text{=1034}\text{.214}\text{kPa}$ from steam tables of saturated steam in Appendix E, table E.1

Since, pressure $P\text{=1034}\text{.214}\text{kPa}$ lies in between pressures $\text{1049}\text{.6}\text{kPa}$ and $\text{1002}\text{.7}\text{kPa}$ with volume of saturated vapor are $V{\text{}}^{vap}=185.5\frac{{\text{cm}}^3}{\text{gm}}$ and $V^{vap}=193.8\frac{{\text{cm}}^3}{\text{gm}}$, and volume of saturated liquid are $V{\text{}}^{liq}=1.130\frac{{\text{cm}}^3}{\text{gm}}$ and $V^{liq}=1.128\frac{{\text{cm}}^3}{\text{gm}}$, Hence from linear interpolation

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ V^{vap}=\left(\frac{\text{1002}\text{.7}\text{kPa}-1034.214\text{kPa}}{\text{1002}\text{.7}\text{kPa}-\text{957}\text{.36}\text{kPa}}\right)\times 193.8\frac{{\text{cm}}^{\text{3}}}{\text{gm}}+\left(\frac{1034.214\text{kPa}-\text{957}\text{.36}\text{kPa}}{\text{1002}\text{.7}\text{kPa}-\text{957}\text{.36}\text{kPa}}\right)\times 185.5\frac{{\text{cm}}^{\text{3}}}{\text{gm}}\\ \\ V^{vap}=188.22\frac{{\text{cm}}^{\text{3}}}{\text{gm}}\end{array}$

And,

  $\begin{array}{l}M=\left(\frac{X_2-X}{X_2-X_1}\right)M_1+\left(\frac{X-X_1}{X_2-X_1}\right)M_2\\ \\ V^{liq}=\left(\frac{\text{1002}\text{.7}\text{kPa}-1000\text{kPa}}{\text{1002}\text{.7}\text{kPa}-\text{957}\text{.36}\text{kPa}}\right)\times 1.128\frac{{\text{cm}}^3}{\text{gm}}+\left(\frac{1000\text{kPa}-\text{957}\text{.36}\text{kPa}}{\text{1002}\text{.7}\text{kPa}-\text{957}\text{.36}\text{kPa}}\right)\times 1.130\frac{{\text{cm}}^3}{\text{gm}}\\ \\ V^{liq}=1.129\frac{{\text{cm}}^3}{\text{gm}}\end{array}$

So,

  $\begin{array}{l}\Delta V^{lv}=V^{vap}-V^{liq}\\ \Delta V^{lv}=188.22\frac{{\text{cm}}^3}{\text{gm}}-1.129\frac{{\text{cm}}^3}{\text{gm}}\\ \Delta V^{lv}=187.091\frac{{\text{cm}}^3}{\text{gm}}\end{array}$

Now from Clapeyron equation,

  $\begin{array}{l}\frac{d{P}^{sat}}{dT}=\frac{\Delta H^{lv}}{T\Delta V^{lv}}\\ \text{Since, }\Delta S^{lv}=\frac{\Delta H^{lv}}{T}\\ \Rightarrow \frac{d{P}^{sat}}{dT}=\frac{\Delta S^{lv}}{\Delta V^{lv}}\\ \Rightarrow \Delta S^{lv}=\Delta V^{lv}\frac{d{P}^{sat}}{dT}\\ \Delta S^{lv}=187.091\frac{{\text{cm}}^3}{\text{gm}}\times 21.84\frac{\text{kPa}}{\text{K}}\times \frac{\text{1}{\text{m}}^3}{{\text{100}}^3{\text{cm}}^3}\times \frac{1000\text{Pa}}{\text{1}\text{kPa}}\\ \Delta S^{lv}=4.086\frac{\text{J}}{\text{gm}\text{K}}=0.976\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}\end{array}$

The value of $\Delta S^{lv}$ at $\text{1034}\text{.214}\text{kPa}$ from steam table is, take reference of subpart (b) for the value of $\Delta S^{lv}$ is,

  $\Delta S^{lv}=1.055\frac{\text{btu}}{{\text{lb}}_{\text{m}}\text{R}}$

So, the results approximately match with each other.