Chapter 3.9, Problem 13P

(A)

Interpretation Introduction

Interpretation:

Determine the work and heat transfer for this process.

Concept Introduction:

An energy balance equation piston cylinder arrangement.

$\Delta \left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\displaystyle \sum _{j=1}^{j=J}\left\{m_{j,in}\left({\widehat{H}}_j+\frac{V_j^2}{2}+g{h}_j\right)\right\}}\\ -{\displaystyle \sum _{k=1}^{k=K}\left\{m_{k,out}\left({\widehat{H}}_k+\frac{V_k^2}{2}+g{h}_k\right)\right\}}+W_{EC}+W_S+Q\end{array}\right]$

Here, mass of the system is $M$, specific internal energy is $\widehat{U}$, velocity is $V$, acceleration due to gravity is $g$, height is $h$, individual quantities of mass added to and removed from the process is $m_{j,in}\text{and}{m}_{k,out}$ respectively, initial specific enthalpy is ${\widehat{H}}_j$, final specific enthalpy is ${\widehat{H}}_k$, initial velocity is $V_j$, final velocity is $V_k$, initial height is $h_j$, final height is $h_k$, shaft work addition is $W_{\text{S}}$, work addition through expansion or contraction of the system is $W_{\text{EC}}$, and heat addition is $Q$.

Here, the height, velocity, shaft work is zero.

Rewrite the above steady state equation.

$\begin{array}{l}\Delta \left\{M\widehat{U}\right\}=W_{EC}+Q\\ M\left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)=W_{EC}+Q\end{array}$

Here, initial specific internal energy is ${\widehat{U}}_{initial}$ and final specific internal energy is ${\widehat{U}}_{final}$.

The expression to calculate the change in enthalpy.

$d\widehat{H}=C_P^{\ast }dT$

Here, derivative of specific enthalpy is $d\widehat{H}$, constant pressure heat capacity is $C_P$, and change in temperature is $dT$

Write the expression to calculate the constant pressure heat capacity $\left(C_P\right)$.

$\begin{array}{l}\frac{C_P}{R}=A+BT+C{T}^2\\ C_P=R\left(A+BT+C{T}^2\right)\end{array}$

Here, gas constant is $R$, temperature is $T$, and constants are $A,B,C,D\text{and}E$ respectively.

Substitute $R\left(A+BT+C{T}^2\right)$ for $C_P$ in change in enthalpy equation.

$d\widehat{H}=R\left(A+BT+C{T}^2\right)dT$

Integrate the above Equation with respect to initial and final condition.

$\begin{array}{l}{\displaystyle \underset{initial}{\overset{final}{\int }}d\widehat{H}}={\displaystyle \underset{initial}{\overset{final}{\int }}R\left(A+BT+C{T}^2\right)dT}\\ {\widehat{H}}_{final}-{\widehat{H}}_{initial}=R\left[\begin{array}{l}A\left(T_{final}-T_{initial}\right)+\frac{B}{2}\left(T^2{}_{final}-T^2{}_{initial}\right)+\\ \frac{C}{3}\left(T^3{}_{final}-T^3{}_{initial}\right)\end{array}\right]\end{array}$

Here, final specific enthalpy is ${\widehat{H}}_{final}$, initial specific enthalpy is ${\widehat{H}}_{initial}$, final temperature is $T_{final}$, and initial temperature is $T_{initial}$.

Write the expression to calculate final specific volume $\left({\widehat{V}}_{final}\right)$.

${\widehat{V}}_{final}=\frac{1}{{\rho }_{final}}$

Here, final density is ${\rho }_{final}$.

Write the expression to calculate initial specific volume $\left({\widehat{V}}_{initial}\right)$.

${\widehat{V}}_{initial}=\frac{1}{{\rho }_{initial}}$

Here, initial density is ${\rho }_{initial}$.

Write the relation between the enthalpy and internal energy.

$\begin{array}{l}H=U+PV\\ d\widehat{H}=d\widehat{U}+Pd\widehat{V}\\ \left({\widehat{H}}_{final}-{\widehat{H}}_{final}\right)=\left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)+P\left({\widehat{V}}_{final}-{\widehat{V}}_{initial}\right)\end{array}$

Here, enthalpy is $H$, internal energy is $U$, pressure is $P$, volume is $V$, change in specific enthalpy is $d\widehat{H}$, change in internal energy is $d\widehat{U}$, change in volume is $d\widehat{V}$, final specific enthalpy is ${\widehat{H}}_{final}$, initial specific enthalpy is ${\widehat{H}}_{initial}$, initial specific internal energy is ${\widehat{U}}_{initial}$, final specific internal energy is ${\widehat{U}}_{final}$, change in initial specific volume is ${\widehat{V}}_{initial}$, and change in final specific volume is ${\widehat{V}}_{final}$.

Write the expression to calculate the initial volume $\left(V_{inital}\right)$.

$V_{initial}={\widehat{V}}_{initial}M$

Write the expression to calculate the work addition $\left(W_{\text{EC}}\right)$.

$\begin{array}{c}{W}_{\text{EC}}=-PdV\\ =-P\left(V_{final}-V_{initial}\right)\end{array}$

Here, final volume is $V_{final}$ and initial volume is $V_{initial}$.

Write the expression to calculate the final volume $\left(V_{final}\right)$.

$V_{final}={\widehat{V}}_{final}M$

Here, mass of liquid water is $M$.

Explanation of Solution

Calculate the change in enthalpy.

${\widehat{H}}_{final}-{\widehat{H}}_{initial}=R\left[\begin{array}{l}A\left(T_{final}-T_{initial}\right)+\frac{B}{2}\left(T^2{}_{final}-T^2{}_{initial}\right)+\\ \frac{C}{3}\left(T^3{}_{final}-T^3{}_{initial}\right)\end{array}\right]$        (1)

Refer the appendix table D.1, “Ideal Gas Heat Capacity”, obtain the constant values of water.

Name	Formula	A	$B\times {10}^3$	$C\times {10}^6$
Water	${\text{H}}_{\text{2}}\text{O}$	$8.712$	$1.25$	$-0.18$

Substitute 8.712 for A, $1.25\times {10}^{-3}$ for B, $-0.18\times {10}^{-6}$ for C, $8.314\text{kJ}/\text{kmol}\cdot \text{K}$ for R, $\text{95°C}$ for $T_{final}$, and $\text{5°C}$ for $T_{initial}$ in Equation (1).

$\begin{array}{c}{\widehat{H}}_{final}-{\widehat{H}}_{initial}=R\left[\begin{array}{l}A\left(T_{final}-T_{initial}\right)+\frac{B}{2}{\left(T^2{}_{final}-T^2{}_{initial}\right)}^2+\\ \frac{C}{3}\left(T^3{}_{final}-T^3{}_{initial}\right)\end{array}\right]\\ =\left[\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(\frac{\text{kmol}}{18.01\text{kg}}\right)\right]\left[\begin{array}{l}\left(8.712\right)\left(368-278\right)+\\ \frac{\left(1.25\times {10}^{-3}\right)}{2}\left({368}^2-{278}^2\right)+\\ \frac{\left(-0.18\times {10}^{-6}\right)}{3}\left({368}^3-{278}^3\right)\end{array}\right]\\ =0.46163\left[784.08+36.34-1.7011\right]\\ \simeq 378\text{kJ}/\text{kg}\end{array}$

Calculate the change in final specific volume $\left({\widehat{V}}_{final}\right)$.

${\widehat{V}}_{final}=\frac{1}{{\rho }_{final}}$        (2)

Substitute $\text{962}\text{.3}\text{g}/\text{L}$ for ${\rho }_{final}$ in Equation (2).

$\begin{array}{c}{\widehat{V}}_{final}=\frac{1}{\text{962}\text{.3}\text{g}/\text{L}}\\ =0.00103917\text{L}/\text{g}\left(\frac{1,000\text{g}}{\text{kg}}\right)\\ =1.03917\text{L}/\text{kg}\end{array}$

Calculate the change in initial specific volume $\left({\widehat{V}}_{initial}\right)$.

${\widehat{V}}_{initial}=\frac{1}{{\rho }_{initial}}$        (3)

Substitute $\text{1,000}\text{.4}\text{g}/\text{L}$ for ${\rho }_{final}$ in Equation (3).

$\begin{array}{c}{\widehat{V}}_{initial}=\frac{1}{\text{1,000}\text{.4}\text{g}/\text{L}}\\ =0.00099960\text{L}/\text{g}\left(\frac{1,000\text{g}}{\text{kg}}\right)\\ =0.99960\text{L}/\text{kg}\end{array}$

Write the relation between the enthalpy and internal energy.

$\left({\widehat{H}}_{final}-{\widehat{H}}_{final}\right)=\left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)+P\left({\widehat{V}}_{final}-{\widehat{V}}_{initial}\right)$        (4)

Substitute $378.34\text{kJ}/\text{kg}$ for ${\widehat{H}}_{final}-{\widehat{H}}_{initial}$, $\text{1}\text{bar}$ for $P$, $1.03917\text{L}/\text{kg}$ for ${\widehat{V}}_{final}$, and $0.99960\text{L}/\text{kg}$ for ${\widehat{V}}_{initial}$ in Equation (4).

$\begin{array}{c}378.34\text{kJ}/\text{kg}=\left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)+\left(1\text{bar}\right)\left(1.03917\text{L}/\text{kg}-0.99960\text{L}/\text{kg}\right)\\ \left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)=378.34\text{kJ}/\text{kg}-\left(0.03957\frac{\text{Lbar}}{\text{kg}}\right)\left\{\left(\frac{{10}^5\text{Pa}}{1\text{bar}}\right)\times \left(\frac{1{\text{m}}^{\text{3}}}{1,000\text{L}}\right)\right\}\\ =378.34\text{kJ}/\text{kg}-\left(3.957\frac{{\text{Pa m}}^3}{\text{kg}}\right)\left(\frac{1\text{kJ}}{1000\text{Pa}\cdot {\text{m}}^{\text{3}}}\right)\\ =378.34\text{kJ}/\text{kg}-0.003957\text{kJ}/\text{kg}\\ =378.336\text{kJ}/\text{kg}\end{array}$

Calculate the final volume $\left(V_{final}\right)$.

$V_{final}={\widehat{V}}_{final}M$        (5)

Substitute $1.03917\text{L}/\text{kg}$ for ${\widehat{V}}_{final}$, and $\text{1}\text{kg}$ for $M$ in Equation (5).

$\begin{array}{c}{V}_{final}=\left(1.03917\text{L}/\text{kg}\right)\left(\text{1}\text{kg}\right)\\ =1.03917\text{L}\end{array}$

Calculate the initial volume $\left(V_{inital}\right)$.

$V_{initial}={\widehat{V}}_{initial}M$        (6)

Substitute $0.99960\text{L}/\text{kg}$ for ${\widehat{V}}_{initial}$ and $\text{1}\text{kg}$ for $M$ in in Equation (6).

$\begin{array}{c}{V}_{initial}=\left(0.99960\text{L}/\text{kg}\right)\left(\text{1}\text{kg}\right)\\ =0.99960\text{L}\end{array}$

Calculate the work addition.

$W_{\text{EC}}=-P\left(V_{final}-V_{initial}\right)$        (7)

Substitute $\text{1}\text{bar}$ for $P$, $0.99960\text{L}$ for $V_{initial}$, and $1.03917\text{L}$ for $V_{final}$ in Equation (7).

$\begin{array}{c}{W}_{\text{EC}}=-P\left(V_{final}-V_{initial}\right)\\ =-\left(\text{1}\text{bar}\right)\left(1.03917\text{L}-0.99960\text{L}\right)\left[\left(\frac{{10}^5\text{Pa}}{1\text{bar}}\right)\left(\frac{1\text{J}}{1\text{Pa}\cdot {\text{m}}^{\text{3}}}\right)\left(\frac{1{\text{m}}^{\text{3}}}{1,000\text{L}}\right)\left(\frac{1\text{kJ}}{1,000\text{J}}\right)\right]\\ =-0.003957\text{kJ}\end{array}$

Thus, the work $\left(W_{\text{EC}}\right)$ addition is $-0.003957\text{kJ}$.

Write the modified steady state equation.

$M\left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)=W_{EC}+Q$        (8)

Substitute $\text{1}\text{kg}$ for $M$, $378.336\text{kJ}/\text{kg}$ for ${\widehat{U}}_{final}-{\widehat{U}}_{initial}$, and $-0.003957\text{kJ}$ for $W_{\text{EC}}$ in Equation (8).

$\begin{array}{l}M\left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)=W_{EC}+Q\\ \left(\text{1}\text{kg}\right)\left(378.336\text{kJ}/\text{kg}\right)=-0.003957\text{kJ}+Q\\ Q=378.34\text{kJ}\end{array}$

Thus, the heat $\left(Q\right)$ for the process is $378.34\text{kJ}$.

(B)

Interpretation Introduction

Interpretation:

Determine the heat transfer of the process.

Concept Introduction:

Write the expression to calculate the net specific enthalpy.

$\begin{array}{l}d\widehat{H}=C_{P}dT\\ {\widehat{H}}_{final}-{\widehat{H}}_{initial}=C_P\left(T_{final}-T_{initial}\right)\end{array}$

Write the expression to calculate final specific volume $\left({\widehat{V}}_{final}\right)$.

${\widehat{V}}_{final}=\frac{1}{{\rho }_{final}}$

Write the expression to calculate initial specific volume $\left({\widehat{V}}_{initial}\right)$.

${\widehat{V}}_{initial}=\frac{1}{{\rho }_{initial}}$

Here, initial density is ${\rho }_{initial}$.

Write the expression to calculate the final volume $\left(V_{final}\right)$.

$V_{final}={\widehat{V}}_{final}M$

Write the expression to calculate the initial volume $\left(V_{inital}\right)$.

$V_{initial}={\widehat{V}}_{initial}M$

Explanation of Solution

Given information:

Density = 1000 g/L.

$C_P=4.19\text{J/g}\cdot \text{K}$

Calculate the net specific enthalpy.

${\widehat{H}}_{final}-{\widehat{H}}_{initial}=C_P\left(T_{final}-T_{initial}\right)$        (9)

Substitute $\text{4}\text{.19}\text{J}/\text{g}\cdot \text{K}$ for $C_P$, $\text{368}\text{K}$ for $T_{final}$, and $\text{278}\text{K}$ for $T_{initial}$ in Equation (9).

$\begin{array}{c}{\widehat{H}}_{final}-{\widehat{H}}_{initial}=\text{4}\text{.19}\text{J}/\text{g}\cdot \text{K}\left(\text{368}\text{K}-\text{278}\text{K}\right)\\ =377.1\text{kJ}/\text{kg}\end{array}$

Assume that the net specific internal energy is equal to net specific enthalpy.

${\widehat{H}}_{final}-{\widehat{H}}_{initial}={\widehat{U}}_{final}-{\widehat{U}}_{initial}$        (10)

Substitute $377.1\text{kJ}/\text{kg}$ for ${\widehat{H}}_{final}-{\widehat{H}}_{initial}$ in Equation (10).

${\widehat{U}}_{final}-{\widehat{U}}_{initial}=377.1\text{kJ}/\text{kg}$

Calculate the final specific volume $\left({\widehat{V}}_{final}\right)$.

${\widehat{V}}_{final}=\frac{1}{{\rho }_{final}}$        (11)

Substitute $\text{1,000}\text{g}/\text{L}$ for ${\rho }_{final}$ in Equation (11).

$\begin{array}{c}{\widehat{V}}_{final}=\frac{1}{\text{1000}\text{g}/\text{L}}\\ =0.001\text{L}/\text{g}\left(\frac{1000\text{g}}{\text{kg}}\right)\\ =1\text{L}/\text{kg}\end{array}$

Calculate the initial specific volume $\left({\widehat{V}}_{initial}\right)$.

${\widehat{V}}_{initial}=\frac{1}{{\rho }_{initial}}$        (12)

Substitute $\text{1000}\text{g}/\text{L}$ for ${\rho }_{final}$ in Equation (12).

$\begin{array}{c}{\widehat{V}}_{initial}=\frac{1}{\text{1000}\text{g}/\text{L}}\\ =0.001\text{L}/\text{g}\left(\frac{1000\text{g}}{\text{kg}}\right)\\ =1\text{L}/\text{kg}\end{array}$

Calculate the final volume $\left(V_{final}\right)$.

$V_{final}={\widehat{V}}_{final}M$        (13)

Substitute $1\text{L}/\text{kg}$ for ${\widehat{V}}_{final}$, and $\text{1}\text{kg}$ for $M$ in Equation (13).

$\begin{array}{c}{V}_{final}=\left(1\text{L}/\text{kg}\right)\left(\text{1}\text{kg}\right)\\ =1\text{L}\end{array}$

Calculate the initial volume $\left(V_{inital}\right)$.

$V_{initial}={\widehat{V}}_{initial}M$        (14)

Substitute $1\text{L}/\text{kg}$ for ${\widehat{V}}_{initial}$, and $\text{1}\text{kg}$ for $M$ in Equation (14).

$\begin{array}{c}{V}_{initial}=\left(1\text{L}/\text{kg}\right)\left(\text{1}\text{kg}\right)\\ =1\text{L}\end{array}$

Substitute $\text{1}\text{bar}$ for $P$, $1\text{L}$ for $V_{initial}$, and $1\text{L}$ for $V_{final}$ in Equation (7).

$\begin{array}{c}{W}_{\text{EC}}=-P\left(V_{final}-V_{initial}\right)\\ =-\left(\text{1}\text{bar}\right)\left(1\text{L}-1\text{L}\right)\\ =0\end{array}$

Substitute $\text{1}\text{kg}$ for $M$, $377.1\text{kJ}/\text{kg}$ for ${\widehat{U}}_{final}-{\widehat{U}}_{initial}$, and $0$ for $W_{\text{EC}}$ in Equation (8).

$\begin{array}{l}M\left({\widehat{U}}_{final}-{\widehat{U}}_{initial}\right)=W_{EC}+Q\\ \left(\text{1}\text{kg}\right)\left(377.1\text{kJ}/\text{kg}\right)=0+Q\\ Q=377.1\text{kJ}/\text{kg}\end{array}$

Thus, the heat $\left(Q\right)$ for the process is $377.1\text{kJ}$.

(C)

Interpretation Introduction

Interpretation:

Compare the heat transfer values from part A and part B and verify the given process is legitimate and practical.

Explanation of Solution

The amount of heat transfer in part (A) is $\text{378}\text{.34}\text{kJ}$ and the amount of heat transfer in part (B) is $\text{377}\text{.1}\text{kJ}$ so both the answer nearly equal so the approximation seems to be legitimate and practical.