Chapter 15.7, Problem 99RP

(a)

To determine

The theoretical air fuel ratio required for the combustion reaction.

Answer to Problem 99RP

The theoretical air fuel ratio required for the combustion reaction is $14.83\text{kg}\text{air}/\text{kg fuel}$.

Explanation of Solution

Write the chemical reaction for the combustion of liquid gas fuel mixture of 90% octane and 10% alcohol with 200% excess theoretical amount of air.

$\begin{array}{l}\text{0}{\text{.9C}}_8{\text{H}}_{18}\left(l\right)+\text{0}{\text{.1C}}_2{\text{H}}_5\text{OH}+2a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to x{\text{CO}}_2+y{\text{H}}_2\text{O}+z{\text{O}}_2+w{\text{N}}_2\end{array}$ (I)

Write the expression to calculate the theoretical air-fuel ratio $\left({\text{AF}}_{\text{th}}\right)$.

$\begin{array}{c}{\text{AF}}_{\text{th}}=\frac{m_{\text{air}}}{m_{\text{fuel}}}\\ {\text{AF}}_{\text{th}}=\frac{\left(N_{\text{air}}{M}_{\text{air}}\right)}{\left(N_{{\text{C}}_8{\text{H}}_{18}}{M}_{{\text{C}}_8{\text{H}}_{18}}\right)+\left(N_{{\text{C}}_2{\text{H}}_5\text{OH}}{M}_{{\text{C}}_2{\text{H}}_5\text{OH}}\right)}\end{array}$ (II)

Here, mass of the air is $m_{\text{air}}$, mass of the fuel is $m_{\text{fuel}}$, number of moles of air is $N_{\text{air}}$, molar mass of air is $M_{\text{air}}$, number of moles of octane is $N_{{\text{C}}_8{\text{H}}_{18}}$, molar mass of octane is $M_{{\text{C}}_8{\text{H}}_{18}}$, number of moles of alcohol is $N_{{\text{C}}_2{\text{H}}_5\text{OH}}$, and molar mass of alcohol is $M_{{\text{C}}_2{\text{H}}_5\text{OH}}$.

Conclusion:

Balance for the Carbon from Combustion reaction Equation (I).

$\begin{array}{l}x=8\times 0.9+2\times 0.1\\ x=7.4\end{array}$

Balance for the Hydrogen from Combustion reaction Equation (I).

$\begin{array}{l}18\times 0.9+6\times 0.1=2y\\ y=8.4\end{array}$

Balance for the excess Oxygen from Combustion reaction Equation (I).

$a_{\text{th}}=z$

Balance for the Oxygen from Combustion reaction Equation (I).

$0.1\left(1\right)+2\times 2a_{\text{th}}=2x+y+2z$

Substitute $7.4$ for $x$, $8.4$ for $y$, and $a_{\text{th}}$ for $z$.

$\begin{array}{l}0.1\left(1\right)+2\times 2a_{\text{th}}=2\left(7.4\right)+8.4+2\left(a_{\text{th}}\right)\\ a_{\text{th}}=11.55\end{array}$

Balance for the Nitrogen from Combustion reaction Equation (I).

$3.76\times 2a_{\text{th}}=w$

Substitute $11.55$ for $a_{\text{th}}$.

$\begin{array}{l}3.76\times 2\left(11.55\right)=w\\ w=86.86\end{array}$

Rewrite the complete balanced chemical reaction for combustion as follows:

$\begin{array}{l}\text{0}{\text{.9C}}_8{\text{H}}_{18}\left(l\right)+\text{0}{\text{.1C}}_2{\text{H}}_5\text{OH}+2\left(11.55\right)\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to 7.4{\text{CO}}_2+8.4{\text{H}}_2\text{O}+11.55{\text{O}}_2+86.86{\text{N}}_2\end{array}$

$\begin{array}{l}\text{0}{\text{.9C}}_8{\text{H}}_{18}\left(l\right)+\text{0}{\text{.1C}}_2{\text{H}}_5\text{OH}+23.1\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to 7.4{\text{CO}}_2+8.4{\text{H}}_2\text{O}+11.55{\text{O}}_2+86.86{\text{N}}_2\end{array}$ (III)

From the table A-2 of “Molar mass, gas constants, and critical-point properties”, select the molar masses of carbon dioxide, water, air, oxygen, nitrogen, octane and alcohol as,

$\begin{array}{l}{M}_{{\text{CO}}_2}=44\text{kg}/\text{kmol}\\ M_{\text{air}}=29\text{kg}/\text{kmol}\\ M_{{\text{H}}_2\text{O}}=18\text{kg}/\text{kmol}\\ M_{{\text{O}}_{\text{2}}}=32\text{kg}/\text{kmol}\end{array}$

$\begin{array}{l}{M}_{{\text{N}}_{\text{2}}}=28\text{kg}/\text{kmol}\\ M_{{\text{C}}_8{\text{H}}_{18}}=114\text{kg}/\text{kmol}\\ M_{{\text{C}}_2{\text{H}}_5\text{OH}}=46\text{kg}/\text{kmol}\end{array}$

Substitute $\left(11.55\times 4.76\text{kmol}\right)$ for $N_{\text{air}}$, $29\text{kg}/\text{kmol}$ for $M_{\text{air}}$, $0.9\text{kmol}$ for $N_{{\text{C}}_8{\text{H}}_{18}}$, $114\text{kg}/\text{kmol}$ for $M_{{\text{C}}_8{\text{H}}_{18}}$, $0.1\text{kmol}$ for $N_{{\text{C}}_2{\text{H}}_5\text{OH}}$, and $46\text{kg}/\text{kmol}$ for $M_{{\text{C}}_2{\text{H}}_5\text{OH}}$ in Equation (II).

$\begin{array}{c}{\text{AF}}_{\text{th}}=\frac{\left(11.55\times 4.76\text{kmol}\right)\left(29\text{kg}/\text{kmol}\right)}{\left(0.9\text{kmol}\right)\left(114\text{kg}/\text{kmol}\right)+\left(0.1\text{kmol}\right)\left(46\text{kg}/\text{kmol}\right)}\\ =14.83\text{kg}\text{air}/\text{kg fuel}\end{array}$

Thus, the theoretical air fuel ratio required for the combustion reaction is $14.83\text{kg}\text{air}/\text{kg fuel}$.

(b)

To determine

The ratio of products of the combustion to the fuel.

Answer to Problem 99RP

The ratio of products of the combustion to the fuel is $30.54\text{kg product}/\text{kg fuel}$.

Explanation of Solution

Write the formula to calculate the molar mass of the product gases $\left(M_{\text{prod}}\right)$ .

$M_{\text{prod}}=\frac{N_{\text{CO}{2}_{}}{M}_{\text{CO}{2}_{}}+N_{{\text{H}}_{\text{2}}\text{O}}{M}_{{\text{H}}_{\text{2}}\text{O}}+N_{\text{O}{2}_{}}{M}_{\text{O}{2}_{}}+N_{{\text{N}}_{\text{2}}}{M}_{{\text{N}}_{\text{2}}}}{N_{\text{CO}{2}_{}}+N_{{\text{H}}_{\text{2}}\text{O}}+N_{\text{O}{2}_{}}+N_{{\text{N}}_{\text{2}}}}$ (IV)

Here, number of moles of carbon dioxide is $N_{\text{CO}{2}_{}}$, molar mass of carbon dioxide is $M_{\text{CO}{2}_{}}$, number of moles of water is $N_{{\text{H}}_{\text{2}}\text{O}}$, molar mass of hydrogen is $M_{{\text{H}}_{\text{2}}\text{O}}$, number of moles of oxygen is $N_{\text{O}{2}_{}}$, molar mass of oxygen is $M_{\text{O}{2}_{}}$, number of moles of nitrogen is $N_{{\text{N}}_{\text{2}}}$, and molar mass of hydrogen is $M_{{\text{N}}_{\text{2}}}$.

Write the formula to calculate the mass of the product gases per unit mass of fuel $\left(m_{\text{prod}}\right)$.

$m_{\text{prod}}=\frac{N_{\text{CO}{2}_{}}+N_{{\text{H}}_{\text{2}}\text{O}}+N_{\text{O}{2}_{}}+N_{{\text{N}}_{\text{2}}}}{\left(N_{{\text{C}}_8{\text{H}}_{18}}{M}_{{\text{C}}_8{\text{H}}_{18}}\right)+\left(N_{{\text{C}}_2{\text{H}}_5\text{OH}}{M}_{{\text{C}}_2{\text{H}}_5\text{OH}}\right)}$ (V)

Conclusion:

Substitute $7.4\text{kmol}$ for $N_{\text{CO}{2}_{}}$, $44\text{kg}/\text{kmol}$ for $M_{\text{CO}{2}_{}}$, $8.4\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $18\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$, $11.55\text{kmol}$ for $N_{\text{O}{2}_{}}$, $32\text{kg}/\text{kmol}$ for $M_{\text{O}{2}_{}}$, $86.86\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (IV).

$\begin{array}{c}{M}_{\text{prod}}=\frac{\left\{\begin{array}{l}\left(7.4\text{kmol}\right)\left(44\text{kg}/\text{kmol}\right)+\left(8.4\text{kmol}\right)\left(18\text{kg}/\text{kmol}\right)+\\ \left(11.55\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)+\left(86.86\text{kmol}\right)\left(28\text{kg}/\text{kmol}\right)\end{array}\right\}}{7.4\text{kmol}+8.4\text{kmol}+11.55\text{kmol}+86.86\text{kmol}}\\ =28.72\text{kg}/\text{kmol}\end{array}$

Substitute $7.4\text{kmol}$ for $N_{\text{CO}{2}_{}}$, $8.4\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $11.55\text{kmol}$ for $N_{\text{O}{2}_{}}$, $86.86\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.9\text{kmol}$ for $N_{{\text{C}}_8{\text{H}}_{18}}$, $114\text{kg}/\text{kmol}$ for $M_{{\text{C}}_8{\text{H}}_{18}}$, $0.1\text{kmol}$ for $N_{{\text{C}}_2{\text{H}}_5\text{OH}}$, and $46\text{kg}/\text{kmol}$ for $M_{{\text{C}}_2{\text{H}}_5\text{OH}}$ in Equation (V).

$\begin{array}{c}{m}_{\text{prod}}=\frac{7.4\text{kmol}+8.4\text{kmol}+11.55\text{kmol}+86.86\text{kmol}}{\left(0.9\text{kmol}\right)\left(114\text{kg}/\text{kmol}\right)+\left(0.1\text{kmol}\right)\left(46\text{kg}/\text{kmol}\right)}\\ =30.54\text{kg product}/\text{kg fuel}\end{array}$

Thus, the ratio of products of the combustion to the fuel is $30.54\text{kg product}/\text{kg fuel}$.

(c)

To determine

The mass flow rate of the air required for the combustion.

Answer to Problem 99RP

The mass flow rate of the air required for the combustion is $148.3\text{kg}/\text{s}$.

Explanation of Solution

Write the formula to calculate the actual air-fuel ratio $\left({\text{AF}}_{\text{act}}\right)$.

${\text{AF}}_{\text{act}}=2\left({\text{AF}}_{\text{th}}\right)$ (VI)

Write the formula to calculate the mass flow rate of air $\left({\dot{m}}_{\text{air}}\right)$.

${\dot{m}}_{\text{air}}={\text{AF}}_{\text{act}}\left({\dot{m}}_{\text{fuel}}\right)$ (VII)

Here, mass flow rate of fuel is ${\dot{m}}_{\text{fuel}}$.

Conclusion:

Substitute $14.83\text{kg}\text{air}/\text{kg fuel}$ for ${\text{AF}}_{\text{th}}$ in Equation (VI).

$\begin{array}{c}{\text{AF}}_{\text{act}}=2\left(14.83\text{kg}\text{air}/\text{kg fuel}\right)\\ =29.65\text{kg}\text{air}/\text{kg fuel}\end{array}$

Substitute $29.65\text{kg}\text{air}/\text{kg fuel}$ for ${\text{AF}}_{\text{act}}$, and $5\text{kg}/\text{s}$ for ${\dot{m}}_{\text{fuel}}$ in Equation (VII).

$\begin{array}{c}{\dot{m}}_{\text{air}}=\left(29.65\text{kg}\text{air}/\text{kg fuel}\right)\left(5\text{kg}/\text{s}\right)\\ =148.3\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the air required for the combustion is $148.3\text{kg}/\text{s}$.

(d)

To determine

The lower heating value of the fuel mixture.

Answer to Problem 99RP

The lower heating value of the fuel mixture is $43,760\text{kJ}/\text{kg}$.

Explanation of Solution

Write the expression to calculate the equation for steady-flow energy balance.

$H_R={\overline{q}}_{\text{LHV}}+H_P$ (VIII)

Here, enthalpy of product is $H_P$, lower heating value is ${\overline{q}}_{\text{LHV}}$, and enthalpy of reactant is $H_R$.

Write the expression to calculate the enthalpy of reactant $\left(H_R\right)$.

$H_R=\left\{\begin{array}{l}0.9{\left({\overline{h}}_f^o-{\overline{h}}_{298\text{ K}}\right)}_{{\text{C}}_8{\text{H}}_{18}}+0.1{\left({\overline{h}}_f^o-{\overline{h}}_{298\text{ K}}\right)}_{{\text{C}}_2{\text{H}}_5\text{OH}}\\ +23.1{\left({\overline{h}}_f^o\right)}_{{\text{O}}_2}+86.86{\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}\end{array}\right\}$ (IX)

Here, enthalpy of vaporization is ${\overline{h}}_f^o$ and the enthalpy of formation is ${\overline{h}}_{298\text{ K}}$.

Write the expression to calculate the enthalpy of product $\left(H_P\right)$.

$H_P=7.4{\left({\overline{h}}_f^o\right)}_{{\text{CO}}_{\text{2}}}+8.4{\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O}}+11.55{\left({\overline{h}}_f^o\right)}_{{\text{O}}_2}+86.86{\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$ (X)

Calculate the lower heating value on mass basis $\left(q_{\text{LHV}}\right)$.

$q_{\text{LHV}}=\frac{{\overline{q}}_{\text{LHV}}}{\left(N_{{\text{C}}_8{\text{H}}_{18}}{M}_{{\text{C}}_8{\text{H}}_{18}}\right)+\left(N_{{\text{C}}_2{\text{H}}_5\text{OH}}{M}_{{\text{C}}_2{\text{H}}_5\text{OH}}\right)}$ (XI)

Write the formula to calculate the lower heating value on mass basis $\left(q_{\text{LHV}}\right)$.

$q_{\text{LHV}}=\frac{{\overline{q}}_{\text{LHV}}}{\left(N_{{\text{C}}_8{\text{H}}_{18}}{M}_{{\text{C}}_8{\text{H}}_{18}}\right)+\left(N_{{\text{C}}_2{\text{H}}_5\text{OH}}{M}_{{\text{C}}_2{\text{H}}_5\text{OH}}\right)}$ (XII)

Conclusion:

From the Table A-18 through Table A-26, obtain the enthalpies of vaporization and enthalpies of formation for different substances as in Table 1.

Substance	Enthalpy of vaporization ${\overline{h}}_f^o\left(\text{kJ}/\text{kmol}\right)$	Enthalpy of formation at 298 K ${\overline{h}}_{298\text{K}}\left(\text{kJ}/\text{kmol}\right)$
${\text{C}}_8{\text{H}}_{18}$	$-130,847$	41,465
${\text{C}}_2{\text{H}}_5\text{OH}$	$-235,310$	42,340
${\text{O}}_2$	0	8682
${\text{N}}_2$	0	8669
${\text{CO}}_2$	$-393,520$	9364
${\text{H}}_2\text{O}\left(g\right)$	$-241,820$	9904

Substitute $-130,847\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{C}}_8{\text{H}}_{18}}$, $41,465\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{\text{298}\text{K}}\right)}_{{\text{C}}_8{\text{H}}_{18}}$, $-235,310\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{C}}_2{\text{H}}_5\text{OH}}$, $42,340\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{\text{298}\text{K}}\right)}_{{\text{C}}_2{\text{H}}_5\text{OH}}$, 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{O}}_2}$, and 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$ in Equation (IX).

$\begin{array}{c}{H}_R=\left\{\begin{array}{l}0.9\left(-130,847\text{kJ}/\text{kmol}-41,465\text{kJ}/\text{kmol}\right)+\\ 0.1\left(-235,310\text{kJ}/\text{kmol}-42,340\text{kJ}/\text{kmol}\right)+\\ 23.1\left(0\right)+86.86\left(0\right)\end{array}\right\}\\ =-252,697\text{kJ}/\text{kmol}\end{array}$

Substitute $-393,520\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}$, $-241,820\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O}}$, 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{O}}_2}$, and 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$ in Equation (X).

$\begin{array}{c}{H}_P=\left\{\begin{array}{c}7.4\left(-393,520\text{kJ}/\text{kmol}\right)+8.4\left(-241,820\text{kJ}/\text{kmol}\right)+\\ 11.55\left(0\right)+86.86\left(0\right)\end{array}\right\}\\ =-4.943\times {10}^6\text{kJ}/\text{kmol}\end{array}$

Substitute $-252,697\text{kJ}/\text{kmol}$ for $H_R$, and $-4.943\times {10}^6\text{kJ}/\text{kmol}$ for $H_P$ in Equation (XI).

$-252,697\text{kJ}/\text{kmol}={\overline{q}}_{\text{LHV}}+\left(-4.943\times {10}^6\text{kJ}/\text{kmol}\right)$

$\begin{array}{c}{\overline{q}}_{\text{LHV}}=4.943\times {10}^6\text{kJ}/\text{kmol}-252,697\text{kJ}/\text{kmol}\\ =4.691\times {10}^6\text{kJ}/\text{kmol}\end{array}$

Substitute $4.691\times {10}^6\text{kJ}/\text{kmol}$ for ${\overline{q}}_{\text{LHV}}$, $0.9\text{kmol}$ for $N_{{\text{C}}_8{\text{H}}_{18}}$, $114\text{kg}/\text{kmol}$ for $M_{{\text{C}}_8{\text{H}}_{18}}$, $0.1\text{kmol}$ for $N_{{\text{C}}_2{\text{H}}_5\text{OH}}$, and $46\text{kg}/\text{kmol}$ for $M_{{\text{C}}_2{\text{H}}_5\text{OH}}$ in Equation (XII).

$\begin{array}{c}{q}_{\text{LHV}}=\frac{4.691\times {10}^6\text{kJ}/\text{kmol}}{\left(0.9\text{kmol}\right)\left(114\text{kg}/\text{kmol}\right)+\left(0.1\text{kmol}\right)\left(46\text{kg}/\text{kmol}\right)}\\ =43,760\text{kJ}/\text{kg}\end{array}$

Thus, the lower heating value of the fuel mixture is $43,760\text{kJ}/\text{kg}$.