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# Octane gas $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$ at $25^{\circ} \mathrm{C}$ is burned steadily with 80 percent excess air at $25^{\circ} \mathrm{C},$ 1 atm, and 40 percent relative humidity. Assuming combustion is complete and the products leave the combustion chamber at 1000 K, determine the heat transfer for this process per unit mass of octane.

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First the general reaction for the excess amount of air is written as follows:

\begin{align*} \text{C}_{8}\text{H}_{18}+1.8a(\text{O}_{2}+3.76\text{N}_{2})\:->b\text{CO}_{2}+c\text{H}_{2}\text{O}+0.8a\text{O}_{2}+d\text{N}_{2} \end{align*}

An analysis of the reaction gives the following values for the coefficients:

\begin{align*} &a=12.5\\ &b=8\\ &c=9\\ &d=84.6 \end{align*}

The reaction is then:

\begin{align*} \text{C}_{8}\text{H}_{18}+22.5(\text{O}_{2}+3.76\text{N}_{2})\:->8\text{CO}_{2}+9\text{H}_{2}\text{O}+10\text{O}_{2}+84.6\text{N}_{2} \end{align*}

The additional amount of water is obtained by first determining the pressure of the water vapor in the products. The saturation pressure at the given temperature is taken from A-4:

\begin{align*} P_{\text{H}_{2}\text{O}}&=\phi P_{\text{sat, 25}}\\ &=0.4\cdot3.1698\:\text{kPa}\\ &=1.268\:\text{kPa} \end{align*}

The amount of the additional water vapor is then:

\begin{align*} &N_{\text{H}_{2}\text{O}}=\dfrac{P_{\text{H}_{2}\text{O}}}{P}N\\ &N_{\text{H}_{2}\text{O}}=\dfrac{P_{\text{H}_{2}\text{O}}}{P}(N_{\text{dry}}+N_{\text{H}_{2}\text{O}}\\ &N_{\text{H}_{2}\text{O}}=\dfrac{1.268}{101.325}(4.76\cdot22.5+N_{\text{H}_{2}\text{O}}\\ &N_{\text{H}_{2}\text{O}}=1.36\:\text{kmol} \end{align*}

The final form of the reaction then is:

\begin{align*} \text{C}_{8}\text{H}_{18}+22.5(\text{O}_{2}+3.76\text{N}_{2})+1.36\text{H}_{2}\text{O}\:->8\text{CO}_{2}+10.36\text{H}_{2}\text{O}+10\text{O}_{2}+84.6\text{N}_{2} \end{align*}

The heat transfer is determined from the enthalpies of formation and the molar enthalpies which are obtained from A-18 to A-26 for the given temperatures:

\begin{align*} \overline{Q}&=(N(\overline{h_{f}^{\text{\textdegree}{°}}}+\overline{h}-\overline{h^{\text{\textdegree}{°}}}))_{\text{reac}}-(N(\overline{h_{f}^{\text{\textdegree}{°}}}+\overline{h}-\overline{h^{\text{\textdegree}{°}}}))_{\text{prod}}\\ &=(1(-208450)+1.36(-241820)-8(-393520+42769-9364)-10.36(-241820+35882-9904)\\ &-10(0+31389-8682)-84.6(0+30129-8669))\:\dfrac{\text{kJ}}{\text{kmol}}\\ &=2537132\:\dfrac{\text{kJ}}{\text{kmol}} \end{align*}

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