Chapter 3.9, Problem 31P

(A)

Interpretation Introduction

Interpretation:

The rate at which heat is added to the toluene in the reboiler

Concept Introduction:

Thesteady state energy balance equation on reboiler (toluene).

$\frac{d}{dt}\left\{N\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{n}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{n}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{\text{S}}+W_{\text{EC}}+\dot{Q}\end{array}\right]$

Here, time taken is $t$, number of moles is $N$, specific internal energy is $\widehat{U}$, velocity is $V$, acceleration due to gravity is $g$, height is $h$, initial molar flow rate is ${\dot{n}}_{in}$, initial specific enthalpy is ${\widehat{H}}_{in}$, initial velocity is $V_{in}$, initial height of the gas is $h_{in}$, final molar flow rate is ${\dot{n}}_{out}$, final height of the gas is $h_{out}$, rate at which shaft work is added to the system is ${\dot{W}}_{\text{S}}$, rate at which work is added to the system through expansion or contraction of the system is ${\dot{W}}_{\text{EC}}$, and rate at which heat is added to the system is $\dot{Q}$.

Explanation of Solution

Given information:

Mass flow rate of toluene is $10\text{mol}/\text{s}$.

80% toluene is heated up to saturated vapor at pressure of 1.1 bar.

Mass flow rate of benzene is $10\text{mol}/\text{s}$.

25% benzene is heated up to saturated vapor at pressure of 0.9 bar.

Calculate the heat transfer using steady state energy balance equation on reboiler (toluene).

$\frac{d}{dt}\left\{N\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{n}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{n}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{\text{S}}+W_{\text{EC}}+\dot{Q}\end{array}\right]$        (1)

Rewrite the steady state Equation (1) by neglecting the work done, potential energy, and kinetic energy.

$\begin{array}{l}0={\dot{n}}_{in}{\underset{\_}{H}}_{in}-{\dot{n}}_{out}{\underset{\_}{H}}_{out}+\dot{Q}\\ \dot{Q}=-{\dot{n}}_{in}{\underset{\_}{H}}_{in}+{\dot{n}}_{out,L}{\underset{\_}{H}}_{out,L}+{\dot{n}}_{out,V}{\underset{\_}{H}}_{out,V}\end{array}$        (2)

Here, initial molar enthalpy is ${\underset{\_}{H}}_{in}$, final molar enthalpy is ${\underset{\_}{H}}_{out}$, final molar flow rate of liquid and vapor is ${\dot{n}}_{out,L},\text{and}{\dot{n}}_{out,V}$ respectively, and final molar enthalpy of liquid and vapor is ${\underset{\_}{H}}_{out,L},\text{and}{\underset{\_}{H}}_{out,V}$ respectively.

Refer the NIST webbook, obtain the enthalpy of saturated liquid and vapor toluene corresponding to pressure of $\text{1}\text{.1}\text{bar}$:

$\begin{array}{c}{\underset{\_}{H}}_{in}={\underset{\_}{H}}_{out,L}\\ =0.541\text{kJ}/\text{mol}\end{array}$

${\underset{\_}{H}}_{out,V}=33.60\text{kJ}/\text{mol}$

Calculate final molar flow rate of liquid $\left({\dot{n}}_{out,L}\right)$.

$\begin{array}{l}0=-{\dot{n}}_{in}+{\dot{n}}_{out,V}+{\dot{n}}_{out,L}\\ {\dot{n}}_{out,L}=-{\dot{n}}_{out,V}+{\dot{n}}_{in}\end{array}$        (3)

Substitute $0.8\left(\text{10}\text{mol}/\text{s}\right)$ for ${\dot{n}}_{out,V}$ and $10\text{mol}/\text{s}$ for ${\dot{n}}_{in}$ in Equation (3).

$\begin{array}{c}{\dot{n}}_{out,L}=-8\text{mol}/\text{s}+10\text{mol}/\text{s}\\ =2\text{mol}/\text{s}\end{array}$

Substitute $10\text{mol}/\text{s}$ for ${\dot{n}}_{in}$, $0.541\text{kJ}/\text{mol}$ for ${\underset{\_}{H}}_{in},\text{and}{\underset{\_}{H}}_{out,L}$, $8\text{mol}/\text{s}$ for ${\dot{n}}_{out,V}$, $2\text{mol}/\text{s}$ for ${\dot{n}}_{out,L}$, and $8\text{mol}/\text{s}$ for ${\dot{n}}_{out,V}$, and $33.60\text{kJ}/\text{mol}$ for ${\underset{\_}{H}}_{out,V}$ in Equation (2).

$\begin{array}{c}\dot{Q}=-{\dot{n}}_{in}{\underset{\_}{H}}_{in}+{\dot{n}}_{out,L}{\underset{\_}{H}}_{out,L}+{\dot{n}}_{out,V}{\underset{\_}{H}}_{out,V}\\ =-\left(10\text{mol}/\text{s}\right)\left(0.541\text{kJ}/\text{mol}\right)+\left(2\text{mol}/\text{s}\right)\left(0.541\text{kJ}/\text{mol}\right)+\left(8\text{mol}/\text{s}\right)\left(33.60\text{kJ}/\text{mol}\right)\\ =-5.41\text{kJ}/\text{s}+1.082\text{kJ}/\text{s}+268.8\text{kJ}/\text{s}\\ =264.46\text{kJ}/\text{s}\end{array}$

Thus, the rate at which heat is added to the toluene in the reboiler is $264.46\text{kJ}/\text{s}$.

(B)

Interpretation Introduction

Interpretation:

The flow rate of steam entering the reboiler at pressure of $\text{1}\text{bar}$, $3\text{bar}$, and $\text{5}\text{bar}$.

Concept Introduction:

Write the steady state energy balance equation around the water passing through the reboiler.

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{\text{S}}+W_{\text{EC}}+\dot{Q}\end{array}\right]$

Here, mass is $M$, initial mass flow rate is ${\dot{m}}_{in}$, and final mass flow rate is ${\dot{m}}_{out}$.

Explanation of Solution

Given information:

For steam at 1 bar:

For steam at 3 bar:

For steam at 5 bar:

Calculate the heat transfer using steady state energy balance equation on reboiler (toluene).

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{\text{S}}+W_{\text{EC}}+\dot{Q}\end{array}\right]$        (4)

Rewrite the steady state Equation (4) by neglecting the work done, potential energy, and kinetic energy.

$\begin{array}{l}0={\dot{m}}_{in}{\widehat{H}}_{in}-{\dot{m}}_{out}{\widehat{H}}_{out}+\dot{Q}\\ \dot{m}=\frac{\dot{Q}}{{\widehat{H}}_{out}-{\widehat{H}}_{in}}\end{array}$        (5)

Refer the Appendix table A-1, “Saturated steam-Pressure increments”, obtain the value of initial and final specific enthalpy corresponding to pressure of $\text{1}\text{bar}$.

$\begin{array}{l}{\widehat{H}}_{in}=2674.9\text{kJ}/\text{kg}\\ {\widehat{H}}_{out}=417.5\text{kJ}/\text{kg}\end{array}$

Substitute $-264.46\text{kJ}/\text{s}$ for $\dot{Q}$, $2674.9\text{kJ}/\text{kg}$ for ${\widehat{H}}_{in}$, and $417.5\text{kJ}/\text{kg}$ for ${\widehat{H}}_{out}$ in Equation (5).

$\begin{array}{c}\dot{m}=\frac{\dot{Q}}{{\widehat{H}}_{out}-{\widehat{H}}_{in}}\\ =\frac{-264.46\text{kJ}/\text{s}}{417.5\text{kJ}/\text{kg}-2,674.9\text{kJ}/\text{kg}}\\ =0.117\text{kg}/\text{s}\end{array}$

Thus, the flow rate of steam entering the reboiler at pressure of $\text{1}\text{bar}$ is $0.117\text{kg}/\text{s}$.

Refer the Appendix table A-1, “Saturated steam-Pressure increments”, obtain the value of initial and final specific enthalpy corresponding to pressure of $\text{3}\text{bar}$.

$\begin{array}{l}{\widehat{H}}_{in}=2724.9\text{kJ}/\text{kg}\\ {\widehat{H}}_{out}=561.4\text{kJ}/\text{kg}\end{array}$

Substitute $-264.46\text{kJ}/\text{s}$ for $\dot{Q}$, $2724.9\text{kJ}/\text{kg}$ for ${\widehat{H}}_{in}$, and $561.4\text{kJ}/\text{kg}$ for ${\widehat{H}}_{out}$ in Equation (5).

$\begin{array}{c}\dot{m}=\frac{\dot{Q}}{{\widehat{H}}_{out}-{\widehat{H}}_{in}}\\ =\frac{-264.46\text{kJ}/\text{s}}{561.4\text{kJ}/\text{kg}-2724.9\text{kJ}/\text{kg}}\\ =0.122\text{kg}/\text{s}\end{array}$

Thus, the flow rate of steam entering the reboiler at pressure of $\text{3}\text{bar}$ is $0.122\text{kg}/\text{s}$.

Refer the Appendix table A-1, “Saturated steam-Pressure increments”, obtain the value of initial and final specific enthalpy corresponding to pressure of $\text{5}\text{bar}$.

$\begin{array}{l}{\widehat{H}}_{in}=2748.1\text{kJ}/\text{kg}\\ {\widehat{H}}_{out}=640.1\text{kJ}/\text{kg}\end{array}$

Substitute $-264.46\text{kJ}/\text{s}$ for $\dot{Q}$, $2748.1\text{kJ}/\text{kg}$ for ${\widehat{H}}_{in}$, and $640.1\text{kJ}/\text{kg}$ for ${\widehat{H}}_{out}$ in Equation (5).

$\begin{array}{c}\dot{m}=\frac{\dot{Q}}{{\widehat{H}}_{out}-{\widehat{H}}_{in}}\\ =\frac{-264.46\text{kJ}/\text{s}}{640.1\text{kJ}/\text{kg}-2748.1\text{kJ}/\text{kg}}\\ =0.125\text{kg}/\text{s}\end{array}$

Thus, the flow rate of steam entering the reboiler at pressure of $\text{5}\text{bar}$ is $0.125\text{kg}/\text{s}$.

(C)

Interpretation Introduction

Interpretation:

Comments on the result of part B, explanation about high pressure steam turbine.

Explanation of Solution

Here, the mass flow rate is almost equal at different pressures. Though, these estimations do not explain the temperature driving force needed in a heat exchanger.

Referred data from NIST Webbook, the boiling temperature for toluene at pressure of $\text{1}\text{.1}\text{bar}$ is $\text{386}\text{.66}\text{K}\text{or}\text{ 113}\text{.41°C}$. Therefore, it is warmer than saturated water at pressure of 1 bar; heat will not transfer to the toluene from the steam. It is observe that, however we know practically that heat cannot transfer to a hotter fluid from a cold fluid, such a process obeys the second first law of thermodynamics; we were capable to do energy balances and acquire answers for this impossible process.

The high pressure steam has a higher temperature of $\text{151°C}$ at a pressure of $\text{5}\text{bar}$, and temperature of $\text{133°C}$ at a pressure of $\text{3}\text{bar}$, which means any of these could be used in the reboiler.

While the higher-pressure steam is high cost, the calculations in part (B) do not description for the equipment cost. A higher the temperature dissimilarity between the toluene and steam means faster heat exchange, which means a smaller heat exchanger is required. The exchange between equipment costs and operating costs are typically explained in design courses in the chemical engineering program.

(D)

Interpretation Introduction

Interpretation:

Flow rate of water if the water is used as coolant.

Concept Introduction:

Write the steady state energy balance equation for benzene entering and leaving the condenser.

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{\text{S}}+W_{\text{EC}}+\dot{Q}\end{array}\right]$

Write the expression to determine the heat capacity of liquid water from Appendix D:

${\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}{C}_{P}dT}$

Here, final and initial specific enthalpy is ${\widehat{H}}_{out},\text{and}{\widehat{H}}_{in}$ respectively, constant pressure heat capacity is $C_P$, initial temperature is $T_{in}$, final temperature is $T_{out}$, and change in temperature is $dT$.

Write the formula 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 constant are $A,B,\text{and}C$ respectively.

Explanation of Solution

Calculate the heat transfer using steady state energy balance equation benzene for entering and leaving the condenser.

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{\text{S}}+W_{\text{EC}}+\dot{Q}\end{array}\right]$        (6)

Rewrite the steady state Equation (6) by neglecting the work done, potential energy, and kinetic energy.

$\begin{array}{l}0={\dot{n}}_{in}{\underset{\_}{H}}_{in}-{\dot{n}}_{out}{\underset{\_}{H}}_{out}+\dot{Q}\\ \dot{Q}=-{\dot{n}}_{in}{\underset{\_}{H}}_{in}+{\dot{n}}_{out}{\underset{\_}{H}}_{out}\end{array}$        (7)

Refer the NIST webbook, obtain the value of ${\underset{\_}{H}}_{out}$ corresponding to saturated liquid and vapor benzene at pressure of $\text{0}\text{.9}\text{bar}$:

${\underset{\_}{H}}_{out}=-0.559\text{kJ}/\text{mol}$

${\underset{\_}{H}}_{in}=30.483\text{kJ}/\text{mol}$

Substitute $\text{8}\text{mol}/\text{s}$ for ${\dot{n}}_{in},\text{and}{\dot{n}}_{out}$, $-0.559\text{kJ}/\text{mol}$ for ${\underset{\_}{H}}_{out}$, and $30.483\text{kJ}/\text{mol}$ for ${\underset{\_}{H}}_{in}$ in Equation (7).

$\begin{array}{c}\dot{Q}=-{\dot{n}}_{in}{\underset{\_}{H}}_{in}+{\dot{n}}_{out}{\underset{\_}{H}}_{out}\\ =-\left(\text{8}\text{mol}/\text{s}\right)\left(30.483\text{kJ}/\text{mol}\right)+\left(\text{8}\text{mol}/\text{s}\right)\left(-0.559\text{kJ}/\text{mol}\right)\\ =-248.3\text{kJ}/\text{s}\end{array}$

Rewrite the Equation (6).

$\begin{array}{l}0={\dot{m}}_{in}{\widehat{H}}_{in}-{\dot{m}}_{out}{\widehat{H}}_{out}+\dot{Q}\\ \dot{m}=\frac{\dot{Q}}{{\widehat{H}}_{out}-{\widehat{H}}_{in}}\end{array}$        (8)

Therefore the coolant enters the condenser at temperature of $\text{25°C}\left(\text{298}\text{.15}\text{K}\right)$ and leaves the condenser at 339.44 K or 66.29°C. This is true regardless of which coolant is used.

Determine the heat capacity of liquid water from Appendix D:

${\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}{C}_{P}dT}$        (9)

Calculate the constant pressure heat capacity.

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

Substitute $R\left(A+BT+C{T}^2\right)$ for $C_P$ in Equation (9).

$\begin{array}{c}{\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}R\left(A+BT+C{T}^2\right)dT}\\ =R\left(MW\right){\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}R\left(A+BT+C{T}^2\right)dT}\\ =R\left(MW\right)\left[A\left(T_{out}-T_{in}\right)+\frac{B\left(T^2{}_{out}-T^2{}_{in}\right)}{2}+\frac{C\left(T^3{}_{out}-T^3{}_{in}\right)}{3}\right]\end{array}$        (10)

Here, molecular weight is $MW$, and gas constant is $R$.

Refer the Appendix table D, “Heat Capacity”, obtain the constant values of constant $A,B,\text{and}C$ as $8.712,1.25\times {10}^{-3},\text{and}-0.18\times {10}^{-6}$ respectively.

Substitute $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$, $\text{18}\text{.02}\text{g}/\text{mol}$ for $MW$, $8.712,1.25\times {10}^{-3},\text{and}-0.18\times {10}^{-6}$ for $A,B,\text{and}C$ respectively, $\text{298}\text{.15}\text{K}$ for $T_{in}$, and $\text{339}\text{.44}\text{K}$ for $T_{out}$ in Equation (10).

$\begin{array}{c}{\widehat{H}}_{out}-{\widehat{H}}_{in}=\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\frac{1\text{mol}}{\text{18}\text{.02}\text{g}}\left[\begin{array}{l}8.712\left(\text{339}\text{.44}\text{K}-\text{298}\text{.15}\text{K}\right)+\\ \frac{1.25\times {10}^{-3}\left[{\left(\text{339}\text{.44}\text{K}\right)}^2-{\left(\text{298}\text{.15}\text{K}\right)}^2\right]}{2}+\\ \frac{-0.18\times {10}^{-6}\left[{\left(\text{339}\text{.44}\text{K}\right)}^3-{\left(\text{298}\text{.15}\text{K}\right)}^3\right]}{3}\end{array}\right]\\ =173.2\text{J}/\text{g}\\ \simeq 173.2\text{kJ}/\text{kg}\end{array}$

Substitute $173.2\text{kJ}/\text{kg}$ for ${\widehat{H}}_{out}-{\widehat{H}}_{in}$, and $248.3\text{kJ}/\text{s}$ for $\dot{Q}$ in Equation (8).

$\begin{array}{c}\dot{m}=\frac{\dot{Q}}{{\widehat{H}}_{out}-{\widehat{H}}_{in}}\\ =\frac{248.3\text{kJ}/\text{s}}{173.2\text{kJ}/\text{kg}}\\ =1.43\text{kg}/\text{s}\end{array}$

Thus, the flow rate of wateris $1.43\text{kg}/\text{s}$.

(E)

Interpretation Introduction

Interpretation:

The flow rate of air

Concept Introduction:

Write the expression to calculate the change in specific enthalpy $\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)$.

$\begin{array}{c}{\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}{C}_{P,air}^{\ast }}dT\\ =R\left(MW\right){\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}0.79C_{P,N_2}^{\ast }+0.21C_{P,O_2}^{\ast }}dT\\ =R\left(MW\right){\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}{C}_{P,air(0.79N_2+0.21O_2)}^{\ast }}dT\end{array}$

Here, constant pressure heat capacity of nitrogen for ideal gas is $C_{P,N_2}^{\ast }$, constant pressure heat capacity of oxygen for ideal gas is $C_{P,O_2}^{\ast }$, and mixture of 0.79% of nitrogen and 0.21% of oxygen of constant pressure heat capacity is $C_{P,air(0.79N_2+0.21O_2)}^{\ast }$.

Write the expression to calculate the mixture of 0.79% of nitrogen and 0.21% of oxygen of constant pressure heat capacity $\left(C_{P,air(0.79N_2+0.21O_2)}^{\ast }\right)$.

$\begin{array}{l}\frac{C_{P,air(0.79N_2+0.21O_2)}^{\ast }}{R}=A+BT+C{T}^2+D{T}^3+E{T}^4\\ C_{P,air(0.79N_2+0.21O_2)}^{\ast }=R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)\end{array}$

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

Explanation of Solution

Given information:

Pressure is 1 bar and specific heat capacity is $\left(C_{P,air(0.79N_2+0.21O_2)}^{\ast }\right)$.

Calculate the mixture of 0.79% of nitrogen and 0.21% of oxygen of constant pressure heat capacity $\left(C_{P,air(0.79N_2+0.21O_2)}^{\ast }\right)$.

$C_{P,air(0.79N_2+0.21O_2)}^{\ast }=R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)$

Calculate the change in specific enthalpy $\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)$.

$\begin{array}{c}{\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}{C}_{P,air}^{\ast }}dT\\ =R\left(MW\right){\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}0.79C_{P,N_2}^{\ast }+0.21C_{P,O_2}^{\ast }}dT\\ =R\left(MW\right){\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}{C}_{P,air(0.79N_2+0.21O_2)}^{\ast }}dT\end{array}$        (10)

Substitute $R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)$ for $C_{P,air(0.79N_2+0.21O_2)}^{\ast }$ in Equation (10).

$\begin{array}{c}{\widehat{H}}_{out}-{\widehat{H}}_{in}=R\left(MW\right){\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}{C}_{P,air(0.79N_2+0.21O_2)}^{\ast }}dT\\ =R\left(MW\right){\displaystyle \underset{T_{in}}{\overset{T_{out}}{\int }}R\left(A+BT+C{T}^2+D{T}^3+E{T}^4\right)}dT\\ =R\left(MW\right)\left[\begin{array}{c}A\left(T_{out}-T_{in}\right)+\frac{B}{2}\left(T^2{}_{out}-T^2{}_{in}\right)+\\ \frac{C}{3}\left(T^3{}_{out}-T^3{}_{in}\right)+\frac{D}{4}\left(T^4{}_{out}-T^4{}_{in}\right)+\frac{E}{5}\left(T^5{}_{out}-T^5{}_{in}\right)\end{array}\right]\end{array}$        (11)

From Appendix D.1, “Ideal Gas Heat Capacity”, obtain and write the constant values of nitrogen and oxygen as in Table (1).

Name	Formula	A	$B\times {10}^3$	$C\times {10}^5$	$D\times {10}^8$	$E\times {10}^{11}$
Nitrogen	${\text{N}}_{\text{2}}$	3.539	$-0.261$	0.007	0.157	$-0.099$
oxygen	${\text{O}}_{\text{2}}$	3.630	$-1.794$	0.658	$-0.601$	$0.179$
Air	$\text{0}\text{.79}\left({\text{N}}_{\text{2}}\right)+\text{0}\text{.21}\left({\text{O}}_{\text{2}}\right)$	3.5581	$-0.5829$	0.1437	$-0.0021$	$-0.0406$

Substitute 3.5581 for A, $-0.5829\times {10}^{-3}$ for B, $0.1437\times {10}^{-5}$ for C, $-0.0021\times {10}^{-8}$ for D, $-0.0406\times {10}^{-11}$ for E, $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for R, $\text{298}\text{.15}\text{K}$ for $T_{in}$, $\text{28}\text{.84}\text{g}/\text{mol}$ for $MW$, and $\text{339}\text{.44}\text{K}$ for $T_{out}$ in Equation (11).

$\begin{array}{c}{\widehat{H}}_{out}-{\widehat{H}}_{in}=\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\frac{1\text{mol}}{28.84\text{g}}\left[\begin{array}{l}3.5581\left[\left(\text{339}\text{.44}\text{K}\right)-\left(\text{298}\text{.15}\text{K}\right)\right]\\ +\frac{-0.5829\times {10}^{-3}}{2}\left[{\left(\text{339}\text{.44}\text{K}\right)}^2-{\left(\text{298}\text{.15}\text{K}\right)}^2\right]+\\ \frac{\begin{array}{c}0.1437\end{array}\times {10}^{-5}}{3}\left[{\left(\text{339}\text{.44}\text{K}\right)}^3-{\left(\text{298}\text{.15}\text{K}\right)}^3\right]+\\ \frac{-0.0021\times {10}^{-8}}{4}\left[{\left(\text{339}\text{.44}\text{K}\right)}^4-{\left(\text{298}\text{.15}\text{K}\right)}^4\right]+\\ \frac{-0.0406\times {10}^{-11}}{5}\left[{\left(\text{339}\text{.44}\text{K}\right)}^5-{\left(\text{298}\text{.15}\text{K}\right)}^5\right]\end{array}\right]\\ =41.8\text{J}/\text{g}\\ \simeq 41.8\text{kJ}/\text{kg}\end{array}$

Substitute $41.8\text{kJ}/\text{kg}$ for ${\widehat{H}}_{out}-{\widehat{H}}_{in}$, and $248.3\text{kJ}/\text{s}$ for $\dot{Q}$ in Equation (8).

$\begin{array}{c}\dot{m}=\frac{\dot{Q}}{{\widehat{H}}_{out}-{\widehat{H}}_{in}}\\ =\frac{248.3\text{kJ}/\text{s}}{41.8\text{kJ}/\text{kg}}\\ =5.94\text{kg}/\text{s}\end{array}$

Thus, the flow rate of air is $5.94\text{kg}/\text{s}$.

(F)

Interpretation Introduction

Interpretation:

Suggestion for using water instead of air.

Explanation of Solution

The mass flow rate of the air is 4 times greater than the mass flow rate of the water, for the equal amount of cooling. Again, these estimations do not account the cost of equipment.

The volume of 5.94 kg of air is orders of magnitude larger than the volume of 1.43 kg of liquid water at ambient pressure, which has implications for the equipment size.

Transfer of heat to liquids is much more efficient compared to heat transfer to low-pressure gases that also effects the size, and therefore the equipment cost of the heat transfer.

Thus, liquid water has the coolant has many benefits regardless of the fact that air free but water isn’t.