Chapter 5, Problem 5.49P

(a)

Interpretation Introduction

Interpretation:

Calculate the lost work of given process in$\left(\text{kJ}\right){\left(\text{mole}\right)}^{-1}$ of the flue gas.

Concept Introduction:

The lost work of any process is calculated by following formula:

  $W_{\text{lost}}=T_{\sigma }\stackrel{•}{S_G}$    ....(1)

Answer to Problem 5.49P

  $W_{\text{lost}}=14.8208\frac{\text{kJ}}{\text{mol }}$

Explanation of Solution

Given information:

The process is given as a flue gas is cooled from${1100}^{\circ }\text{C}$ to${150}^{\circ }\text{C}$, heat is used to generate saturated steam at${100}^{\circ }\text{C}$ in a boiler.

The heat capacity of flue gas is given as,

  $\frac{C_P}{R}=3.83+0.000551T/\left(K\right)$

Latent heat of vaporization of water to generate steam is given as$\Delta H_V=2256.9\frac{\text{kJ}}{\text{kg}}$ and water enters the boiler at${100}^{\circ }\text{C}$ and surrounding temperature is given as${25}^{\circ }\text{C}$ .

  $T_0={1100}^{\circ }\text{C}+273.15=1373.15\text{K}$

  $T={150}^{\circ }\text{C}+273.15=423.15\text{K}$

  $T_{\sigma }={25}^{\circ }\text{C}+273.15=298.15\text{K}$

  $T_{\text{steam}}={100}^{\circ }\text{C}+273.15=373.15\text{K}$

For rate of entropy generation, the total entropy generation in boiler is the sum of the entropy generation during the heating of water in order to generate steam and entropy generation during the cooling of gas which are find out by entropy balance of the open system equation.

  $\stackrel{•}{S_G}=\stackrel{•}{S_{Gas}}+\stackrel{•}{S_{Steam}}$

Entropy generation during cooling of gas is

  $\left(\Delta S_{gas}\stackrel{•}{n}\right)=S_{Gas}$

The entropy change of gas is

  $\begin{array}{l}dS=\frac{C_P}{T}dT\\ \Rightarrow \Delta S_{gas}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{T}\frac{dT}{R}}\\ \Rightarrow \Delta S_{gas}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{3.83+0.000551T}{T}dT}\\ \Delta S_{gas}=R\left[3.83\times \ln \frac{423.15\text{K}}{1373.15\text{K}}+0.000551\left(423.15\text{K}-1373.15\text{K}\right)\right]\\ \Delta S_{gas}=8.314\frac{\text{J}}{\text{mol K}}\times -5.03\\ \Delta S_{gas}=-41.83\frac{\text{J}}{\text{mol K}}\end{array}$

And

Entropy generation during vaporization of water is

  $\left(\Delta S_{steam}\stackrel{•}{m}\right)=\stackrel{•}{S_{steam}}$

The entropy change of water is

  $\begin{array}{l}\Delta S_{steam}=\frac{\Delta H_V}{T_{steam}}\\ \Delta S_{steam}=\frac{2256.9\frac{\text{kJ}}{\text{kg}}}{373.15\text{K}}=6.048\frac{\text{kJ}}{\text{kg K}}\end{array}$

Now for the calculations of mass and molar flow rates, applying the steady state energy balance across the boiler

  $\begin{array}{l}\stackrel{•}{n}{\displaystyle \underset{T_0}{\overset{T}{\int }}{C}_{P}dT}+\stackrel{•}{m}\Delta H_V=0\\ \Rightarrow \frac{\stackrel{•}{m}}{\stackrel{•}{n}}=-\frac{{\displaystyle \underset{T_0}{\overset{T}{\int }}{C}_{P}dT}}{\Delta H_V}=\frac{R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}dT}}{\Delta H_V}\\ \frac{\stackrel{•}{m}}{\stackrel{•}{n}}=-\frac{{\displaystyle \underset{T_0}{\overset{T}{\int }}{C}_{P}dT}}{\Delta H_V}=-\frac{8.314\frac{\text{J}}{\text{mol K}}\times {\displaystyle \underset{T_0}{\overset{T}{\int }}\left(3.83+0.000551T\right)dT}}{970.3\frac{\text{Btu}}{{\text{lb}}_{\text{m}}}}\\ \frac{\stackrel{•}{m}}{\stackrel{•}{n}}=-\frac{8.314\frac{\text{J}}{\text{mol K}}\times \left[3.83\times \left(T-T_0\right)+0.000551\times \frac{T^2}{2}-0.000551\times \frac{T_0{}^2}{2}\right]}{2256.9\frac{\text{kJ}}{\text{kg}}}\\ \frac{\stackrel{•}{m}}{\stackrel{•}{n}}=-\frac{8.314\frac{\text{J}}{\text{mol K}}\times \left[3.83\times \left(423.15\text{K}-1373.15\text{K}\right)+0.000551\times \frac{{423.15}^2}{2}-0.000551\times \frac{{1373.15}^2}{2}\right]}{2256.9\frac{\text{kJ}}{\text{kg}}}\\ \frac{\stackrel{•}{m}}{\stackrel{•}{n}}=-\frac{-34159.205\frac{\text{J}}{\text{mol}}\times \frac{1\text{kJ}}{1000\text{J}}}{2256.9\frac{\text{kJ}}{\text{kg}}\times \frac{1\text{kg}}{1000\text{g}}}\\ \frac{\stackrel{•}{m}}{\stackrel{•}{n}}=15.14\frac{\text{gm}}{\text{mol}}\end{array}$

Hence

  $\begin{array}{l}\stackrel{•}{S_G}=\stackrel{•}{S_{Gas}}+\stackrel{•}{S_{Steam}}\\ \stackrel{•}{S_G}=\left(\Delta S_{gas}\stackrel{•}{n}\right)+\left(\Delta S_{steam}\stackrel{•}{m}\right)\\ \frac{\stackrel{•}{S_G}}{\stackrel{•}{n}}=\frac{\stackrel{•}{m}}{\stackrel{•}{n}}\Delta S_{steam}+\Delta S_{gas}\\ \frac{\stackrel{•}{S_G}}{\stackrel{•}{n}}=15.14\frac{\text{gm}}{\text{mol}}\times 6.048\frac{\text{J}}{\text{gm K}}+\left(-41.83\frac{\text{J}}{\text{mol K}}\right)\\ \frac{\stackrel{•}{S_G}}{\stackrel{•}{n}}=49.709\frac{\text{J}}{\text{mol K}}\end{array}$

Hence the lost work in$\left(\text{kJ}\right){\left(\text{mole}\right)}^{-1}$ is

  $\begin{array}{l}{W}_{\text{lost}}=T_{\sigma }\stackrel{•}{S_G}\\ W_{\text{lost}}=298.15\text{K}\times 49.709\frac{\text{J}}{\text{mol K}}\\ W_{\text{lost}}=14820.805\frac{\text{J}}{\text{mol }}\times \frac{1\text{kJ}}{1000\text{J}}\\ W_{\text{lost}}=14.8208\frac{\text{kJ}}{\text{mol }}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Calculate the maximum work of given process in$\left(\text{kJ}\right){\left(\text{mole}\right)}^{-1}$ of the flue gas that can be accomplished by the saturated steam if it condenses only and does not subcool.

Concept Introduction:

The maximum work of any process is ideal work and calculated by following formula:

  $W_{ideal}=\Delta H-T_{\sigma }\Delta S$

    ....(1)

Answer to Problem 5.49P

  $W_{ideal}=-6.867\frac{\text{kJ}}{\text{kmol}}$

Explanation of Solution

Given information:

Latent heat of vaporization of water to generate steam is given as$\Delta H_V=2256.9\frac{\text{kJ}}{\text{kg}}$ and water enters the boiler at${100}^{\circ }\text{C}$ and surrounding temperature is given as${25}^{\circ }\text{C}$ .

  $T_{\sigma }={25}^{\circ }\text{C}+273.15=298.15\text{K}$

  $T_{\text{steam}}={100}^{\circ }\text{C}+273.15=373.15\text{K}$

The maximum work of flue gas if it condenses only and does not subcoolis same as ideal work done by water to vaporize. Hence

  $W_{ideal}=\Delta H_{steam}-T_{\sigma }\Delta S_{steam}$

And$\Delta H_{steam}=\Delta H_V=-2256.9\frac{\text{kJ}}{\text{kg}}$

From the part (a), we found that

  $\begin{array}{l}\Delta S_{steam}=\frac{\Delta H_V}{T_{steam}}\\ \Delta S_{steam}=\frac{-2256.9\frac{\text{kJ}}{\text{kg}}}{373.15\text{K}}=-6.048\frac{\text{kJ}}{\text{kg K}}\end{array}$

Hence

  $\begin{array}{l}{W}_{ideal}=\Delta H_V-T_{\sigma }\Delta S_{steam}\\ W_{ideal}=-2256.9\frac{\text{J}}{\text{gm}}-298.15\text{K}\times -6.048\frac{\text{kJ}}{\text{kg K}}\\ W_{ideal}=-453.62\frac{\text{kJ}}{\text{kg}}\end{array}$

The maximum work in terms of$\left(\text{kJ}\right){\left(\text{mole}\right)}^{-1}$ is

From the part (a), we found that

  $\frac{\stackrel{•}{m}}{\stackrel{•}{n}}=15.14\frac{\text{gm}}{\text{mol}}$

Hence

  $\begin{array}{l}{W}_{ideal}\times \frac{\stackrel{•}{m}}{\stackrel{•}{n}}=-453.62\frac{\text{kJ}}{\text{kg}}\times 15.14\frac{\text{gm}}{\text{mol}}\times \frac{1\text{kg}}{1000\text{gm}}\\ W_{ideal}=-6.867\frac{\text{kJ}}{\text{kmol}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Compare the results of part (b) with the theoretical value of maximum work done by flue gas itself.

Concept Introduction:

The maximum work of any process is ideal work and calculated by following formula:

  $W_{ideal}=\Delta H-T_{\sigma }\Delta S$

    ....(1)

Answer to Problem 5.49P

  $W_{ideal}=-21.687\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

The maximum work of flue gas is same as ideal work done by flue gas. Hence

  $W_{ideal}=\Delta H_{gas}-T_{\sigma }\Delta S_{gas}$

  $\begin{array}{l}\Delta H_{gas}={\displaystyle \underset{T_0}{\overset{T}{\int }}{C}_{P}dT}\\ \approx \Delta H_{gas}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{C_P}{R}dT}\\ \Delta H_{gas}=8.314\frac{\text{J}}{\text{mol K}}\times {\displaystyle \underset{T_0}{\overset{T}{\int }}\left(3.83+0.000551T\right)dT}\\ \Delta H_{gas}=8.314\frac{\text{J}}{\text{mol K}}\times \left[3.83\times \left(423.15\text{K}-1373.15\text{K}\right)+0.000551\times \frac{{423.15}^2}{2}-0.000551\times \frac{{1373.15}^2}{2}\right]\\ \Delta H_{gas}=-34159.205\frac{\text{J}}{\text{mol}}\times \frac{1\text{kJ}}{1000\text{J}}\\ \Delta H_{gas}=-34.159\frac{\text{kJ}}{\text{mol}}\end{array}$

From the part (a), we found that

  $\Delta S_{gas}=-41.83\frac{\text{J}}{\text{mol K}}$

Hence

  $\begin{array}{l}{W}_{ideal}=\Delta H_{gas}-T_{\sigma }\Delta S_{gas}\\ W_{ideal}=-34159.205\frac{\text{J}}{\text{mol}}-298.15\text{K}\times -41.83\frac{\text{J}}{\text{mol K}}\\ W_{ideal}=-21687.59\frac{\text{J}}{\text{mol}}\times \frac{1\text{kJ}}{1000\text{J}}\\ W_{ideal}=-21.687\frac{\text{kJ}}{\text{mol}}\end{array}$