Chapter 16.6, Problem 91RP

a)

To determine

The amount of heat required for the process.

Answer to Problem 91RP

The amount of heat required for the process is $324,033\text{kJ}$.

Explanation of Solution

Write the energy balance equation for the reported process.

$E_{in}-E_{out}=\Delta E_{system}$ (I)

Here, input energy is $E_{in}$, output energy is $E_{out}$, and the change in system energy is $\Delta E_{system}$.

Write the expression to obtain the amount of heat required for the process $\left(Q_{in}\right)$.

$Q_{in}=N\left({\overline{u}}_2-{\overline{u}}_1\right)$ (II)

Here, number of moles is N, internal energy of the system at state 1 is ${\overline{u}}_1$, and internal energy of the system at state 2 is ${\overline{u}}_2$.

Write the expression to obtain the internal energy of the system at state 1 $\left({\overline{u}}_1\right)$.

${\overline{u}}_1=h_1-R_u{T}_1$ (III)

Here, enthalpy of the system at state 1 is $h_1$, universal gas constant is $R_u$, and temperature of the system at state 1 is $T_1$.

Write the expression to obtain the internal energy of the system at state 2 $\left({\overline{u}}_2\right)$.

${\overline{u}}_2=h_2-R_u{T}_2$ (IV)

Here, enthalpy of the system at state 2 is $h_2$, and temperature of the system at state 2 is $T_2$.

Write the expression to obtain the change in enthalpy of the system $\left(h_2-h_1\right)$.

$h_2-h_1={\displaystyle \underset{1}{\overset{2}{\int }}{c}_{p}dT}$ (V)

Conclusion:

Substitute $\left(h_1-R_u{T}_1\right)$ for ${\overline{u}}_1$, and $\left(h_2-R_u{T}_2\right)$ for ${\overline{u}}_2$ in Equation (II).

$\begin{array}{c}{Q}_{in}=N\left({\overline{u}}_2-{\overline{u}}_1\right)\\ =N\left(h_2-R_u{T}_2-\left(h_1-R_u{T}_1\right)\right)\\ =N\left(h_2-h_1-R_u\left(T_2-T_1\right)\right)\end{array}$ (VI)

Refer Table A-2c, “Ideal-gas specific heats of various common gases”, obtain the specific heat relation as $a+bT+c{T}^2+d{T}^3$.

Substitute $a+bT+c{T}^2+d{T}^3$ for $c_p$ in Equation (V) and then integrate.

$\begin{array}{c}{h}_2-h_1={\displaystyle \underset{1}{\overset{2}{\int }}{c}_{p}dT}\\ ={\displaystyle \underset{1}{\overset{2}{\int }}\left(a+bT+c{T}^2+d{T}^3\right)dT}\\ =a{\left[T\right]}_1^2+b{\left[\frac{T^2}{2}\right]}_1^2+c{\left[\frac{T^3}{3}\right]}_1^2+d{\left[\frac{T^4}{4}\right]}_1^2\\ =a\left[T_2-T_1\right]+\frac{b}{2}\left[T_2^2-T_1^2\right]+\frac{c}{3}\left[T_2^3-T_1^3\right]+\frac{d}{4}\left[T_2^4-T_1^4\right]\end{array}$ (VII)

Here, constants are a, b, c and d.

Refer Table A-2c, “Ideal-gas specific heats of various common gases”, obtain the values of constants a, b, c and d for methane as 19.89, $\text{5}\text{.024}\times {10}^{-2}$, $1.269\times {10}^{-5}$, and $-11.01\times {10}^{-9}$ respectively.

Substitute 19.89 for a, $\text{5}\text{.024}\times {10}^{-2}$ for b, $1.269\times {10}^{-5}$ for c, $-11.01\times {10}^{-9}$ for d, 1,000 K for $T_2$, and 298 K for $T_1$ in Equation (VII).

$\begin{array}{c}{h}_2-h_1=a\left[T_2-T_1\right]+\frac{b}{2}\left[T_2^2-T_1^2\right]+\frac{c}{3}\left[T_2^3-T_1^3\right]+\frac{d}{4}\left[T_2^4-T_1^4\right]\\ =\left\{\begin{array}{l}19.89\left(1,000\text{K}-298\text{K}\right)\\ +\left(\frac{\text{5}\text{.024}\times {10}^{-2}}{2}\right)\left({\left(1,000\text{K}\right)}^2-{\left(298\text{K}\right)}^2\right)\\ +\left(\frac{1.269\times {10}^{-5}}{3}\right)\left({\left(1,000\text{K}\right)}^3-{\left(298\text{K}\right)}^3\right)\\ +\left(\frac{-11.01\times {10}^{-9}}{4}\right)\left({\left(1,000\text{K}\right)}^4-{\left(298\text{K}\right)}^4\right)\end{array}\right\}\\ =13,962.78+22,889+4,118-2,730\\ =38,239\text{kJ}/\text{kmol}\end{array}$

Substitute $38,239\text{kJ}/\text{kmol}$ for $\left(h_2-h_1\right)$, $8.314\text{kJ}/\text{kmol}\cdot \text{K}$ for $R_u$, 1,000 K for $T_2$, and 298 K for $T_1$ in Equation (VI).

$\begin{array}{c}{Q}_{in}=N\left(h_2-h_1-R_u\left(T_2-T_1\right)\right)\\ =\left(10\text{mol}\right)\left(\begin{array}{l}38,239\text{kJ}/\text{kmol}\\ -\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(1,000\text{K}-298\text{K}\right)\end{array}\right)\\ =324,033\text{kJ}\end{array}$

Thus, the amount of heat required for the process is $324,033\text{kJ}$.

b)

To determine

The amount of heat required for the process.

Answer to Problem 91RP

The amount of heat required for the process is $245,200\text{kJ}$.

Explanation of Solution

Write the stoichiometric reaction for the dissociation process.

${\text{CH}}_{\text{4}}\rightleftharpoons \text{C}+2{\text{H}}_{\text{2}}$

From the stoichiometric reaction, infer that the stoichiometric coefficient for methane $\left(v_{{\text{CH}}_{\text{4}}}\right)$ is 1, for hydrogen $\left(v_{{\text{H}}_{\text{2}}}\right)$ is 2, and for carbon $\left(v_{\text{C}}\right)$ is 1.

Write the expression to obtain the actual reaction for the dissociation process.

${\text{CH}}_{\text{4}}\to x{\text{CH}}_{\text{4}}+y\text{C}+z{\text{H}}_{\text{2}}$ (VIII)

From the actual reaction, infer that the equilibrium composition contains x amount of methane $\left(N_{{\text{CH}}_{\text{4}}}\right)$, y amount of carbon $\left(N_{\text{C}}\right)$, and z amount of hydrogen $\left(N_{{\text{H}}_{\text{2}}}\right)$.

Write the expression to obtain the total number of moles $\left(N_{\text{total}}\right)$.

$N_{\text{total}}=N_{{\text{CH}}_{\text{4}}}+N_{\text{C}}+N_{{\text{H}}_{\text{2}}}$ (IX)

Here, number of moles of ${\text{CH}}_{\text{4}}$ is $N_{{\text{CH}}_{\text{4}}}$, number of moles of $\text{C}$ is $N_{\text{C}}$, and number of moles of ${\text{H}}_2$ is $N_{{\text{H}}_2}$.

Write the expression to obtain the equilibrium constant $\left(K_p\right)$ for the dissociation process.

$K_p=\frac{N_{\text{C}}^{v_{\text{C}}}{N}_{{\text{H}}_2}^{v_{{\text{H}}_2}}}{N_{{\text{CH}}_4}^{v_{{\text{CH}}_4}}}{\left(\frac{P}{N_{\text{total}}}\right)}^{\left(v_{\text{C}}+v_{{\text{H}}_2}-v_{{\text{CH}}_4}\right)}$ (X)

Here, pressure is P.

Write the expression to obtain the mole fraction of Methane $\left(y_{{\text{CH}}_{\text{4}}}\right)$.

$y_{{\text{CH}}_{\text{4}}}=\frac{N_{{\text{CH}}_{\text{4}}}}{N_{\text{total}}}$ (XI)

Write the expression to obtain the mole fraction of carbon $\left(y_{\text{C}}\right)$.

$y_{\text{C}}=\frac{N_{\text{C}}}{N_{\text{total}}}$ (XII)

Write the expression to obtain the mole fraction of hydrogen $\left(y_{{\text{H}}_2}\right)$.

$y_{{\text{H}}_2}=\frac{N_{{\text{H}}_2}}{N_{total}}$ (XIII)

Write the expression to obtain the amount of heat required for the process $\left(Q_{in}\right)$.

$Q_{in}=N\left(y_{{\text{CH}}_{\text{4}}}{c}_{V,{\text{ CH}}_4}{T}_2+y_{{\text{H}}_{\text{2}}}{c}_{V,{\text{ H}}_{\text{2}}}{T}_2+y_{\text{C}}{c}_{V,\text{ C}}{T}_2\right)-N{c}_{V,{\text{ CH}}_4}{T}_1$ (XIV)

Here, specific heat of methane is $c_{V,{\text{ CH}}_4}$, specific heat of hydrogen is $c_{V,{\text{ H}}_{\text{2}}}$, and specific heat of carbon is $c_{V,\text{ C}}$.

Conclusion:

Write the carbon balance equation from Equation (VIII).

$\begin{array}{l}1=x+y\\ y=1-x\end{array}$ (XV)

Write the hydrogen balance equation from Equation (VIII).

$\begin{array}{l}4=4x+2z\\ z=2-2x\end{array}$ (XVI)

Substitute x for $N_{{\text{CH}}_{\text{4}}}$, y for $N_{\text{C}}$, $z$ for $N_{{\text{H}}_2}$, $1-x$ for y, and $2-2x$ for z in Equation (IX).

$\begin{array}{c}{N}_{\text{total}}=x+y+z\\ =x+1-x+2-2x\\ =3-2x\end{array}$ (XVII)

Substitute $e^{-2.328}$ for $K_p$, x for $N_{{\text{CH}}_{\text{4}}}$, y for $N_{\text{C}}$, $z$ for $N_{{\text{H}}_2}$, $1-x$ for y, $2-2x$ for z.

$3-2x$ for $N_{\text{total}}$,1 atm for P, 2 for $v_{{\text{H}}_{\text{2}}}$, and 1 for both $v_{\text{C}}$ and $v_{{\text{CH}}_{\text{4}}}$ in Equation (X).

$\begin{array}{l}{e}^{-2.328}=\frac{\left(1-x\right){\left(2-2x\right)}^2}{x}{\left(\frac{1\text{atm}}{3-2x}\right)}^{\left(1+2-1\right)}\\ 0.0975x=\left(1-x\right){\left(2-2x\right)}^2\left(\frac{1}{{\left(3-2x\right)}^2}\right)\\ 0.0975x\left[9+4x^2-12x\right]=\left(1-x\right)\left(4+4x^2-8x\right)\\ 0.8775x+0.39x^3-1.17x^2=\left(4+4x^2-8x-4x-4x^3+8x^2\right)\end{array}$

$\begin{array}{l}3.61x^3-13.17x^2+12.8775x-4=0\\ x=0.641\end{array}$

Substitute 0.641 for x in equation (XV).

$\begin{array}{c}y=1-x\\ =1-0.641\\ =0.359\end{array}$

Substitute 0.641 for x in equation (XVI).

$\begin{array}{c}z=2-2x\\ =2-2\left(0.641\right)\\ =0.718\end{array}$

Substitute 0.641 for x in equation (XVII).

$\begin{array}{c}{N}_{\text{total}}=3-2x\\ =3-2\left(0.641\right)\\ =1.718\end{array}$

Substitute 0.641 for x, 0.359 for y, and 0.718 for z in Equation (VIII).

$\begin{array}{l}{\text{CH}}_{\text{4}}\to x{\text{CH}}_{\text{4}}+y\text{C}+z{\text{H}}_{\text{2}}\\ {\text{CH}}_{\text{4}}\to 0.641{\text{CH}}_{\text{4}}+0.359\text{C}+0.718{\text{H}}_{\text{2}}\end{array}$

Substitute 0.641 for x, and 1.718 for $N_{\text{total}}$ in Equation (XI).

$\begin{array}{c}{y}_{{\text{CH}}_{\text{4}}}=\frac{0.641}{1.718}\\ =0.37\end{array}$

Substitute 0.359 for x, and 1.718 for $N_{\text{total}}$ in Equation (XII).

$\begin{array}{c}{y}_{\text{C}}=\frac{0.359}{1.718}\\ =0.2\end{array}$

Substitute 0.718 for x, and 1.718 for $N_{\text{total}}$ in Equation (XIII).

$\begin{array}{c}{y}_{{\text{H}}_2}=\frac{0.718}{1.718}\\ =0.41\end{array}$

Substitute 10 kmol for N, 0.37 for $y_{{\text{CH}}_{\text{4}}}$, $63.3\text{kJ}/\text{kmol}\cdot \text{K}$ for $c_{V,{\text{ CH}}_4}$, 0.41 for $y_{{\text{H}}_{\text{2}}}$, $21.7\text{kJ}/\text{kmol}\cdot \text{K}$ for $c_{V,{\text{ H}}_{\text{2}}}$, 0.2 for $y_{\text{C}}$, $0.711\text{kJ}/\text{kmol}\cdot \text{K}$ for $c_{V,\text{ C}}$, 1,000 K for $T_2$, and 298 K for $T_1$ in Equation (XIV).

$\begin{array}{c}{Q}_{in}=\left\{\begin{array}{l}\left(10\text{kmol}\right)\left(\begin{array}{l}\left(0.37\right)\left(63.3\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(1,000\text{K}\right)\\ +\left(0.41\right)\left(21.7\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(1,000\text{K}\right)\\ +\left(0.2\right)\left(0.711\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(1,000\text{K}\right)\end{array}\right)\\ -\left(10\text{kmol}\right)\left(27.8\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(298\text{K}\right)\end{array}\right\}\\ =245,200\text{kJ}\end{array}$

Thus, the amount of heat required for the process is $245,200\text{kJ}$.