Answer
$T_{dp}=113.2^{∘}F $
Work Step by Step
Analysis ( $a$ ) The theoretical combustion equation in this case can be written as $$
\mathrm{C}_4 \mathrm{H}_{10}+a_{\mathrm{sh}} \mathrm{O}_2+3.76 \mathrm{~N}_2 \longrightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}+3.76 a_{t h} \mathrm{~N}_2
$$ where $a_{\text {th}}$ is the stoichiometric coefficient for air. It is determined from $\mathrm{O}_2$ balance: $\quad a_{\mathrm{th}}=4+25 \longrightarrow a_{\text {th }}=6.5$
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, $$
\mathrm{AF}_{\text {th }}=\frac{m_{\text {air,th }}}{m_{\text {fuel }}}=\frac{(6.5 \times 4.76 \mathrm{lbmol})(29 \mathrm{lbm} / \mathrm{lbmol})}{(4 \mathrm{lbmol})(12 \mathrm{lbm} / \mathrm{lbmol})+(5 \mathrm{lbmol})(2 \mathrm{lbm} / \mathrm{lbmol})}=15.5\ \mathrm{lbm} \text { air/lbm fiel }
$$ The actual air-fuel ratio used is $$
\mathrm{AF}_{\mathrm{act}}=\frac{m_{\text {air }}}{m_{\text {fuel }}}=\frac{25 \mathrm{lbm}}{1 \mathrm{lbm}}=25 \mathrm{lbm} \text { air/lbm fiel }
$$ Then the percent theoretical air used can be determined from $$
\text { Percent theoretical air }=\frac{\mathrm{AF}_{\text {act }}}{\mathrm{AF}_{\text {th }}}=\frac{25 \mathrm{lbm} \text { air } / \mathrm{lbm} \text { fuel }}{15.5 \mathrm{lbm} \text { air } / \mathrm{lbm} \text { fuel }}=161 \%
$$ (b) The combustion is complete, and thus products will contain only $\mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{O}_2$ and $\mathrm{N}_2$. The air-fuel ratio for this combustion process on a mole basis is
$$
\overline{\mathrm{AF}}=\frac{N_{\text {air }}}{N_{\text {fuel }}}=\frac{m_{\text {air }} / M_{\text {air }}}{m_{\text {fuel }} / M_{\text {fuel }}}=\frac{(25 \mathrm{lbm})(29 \mathrm{lbm} / \mathrm{lbmol})}{(1 \mathrm{lbm}) /(58 \mathrm{lbm} / \mathrm{lbmol})}=50 \mathrm{lbmol} \text { air } / \mathrm{lbmol} \text { fuel }
$$ Thus the combustion equation in this case can be written as
$$
\mathrm{C}_4 \mathrm{H}_{10}+(50 / 4.76)\left[\mathrm{O}_2+3.76 \mathrm{~N}_2\right] \longrightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}+4 \mathrm{O}_2+39.5 \mathrm{~N}_2
$$ The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, $$
P_v=\left(\frac{N_v}{N_{\text {prod }}}\right) P_{\text {prod }}=\left(\frac{5 \mathrm{lbmol}}{52.5 \mathrm{lbmol}}\right)(14.7 \mathrm{psia})=1.4 \mathrm{psia}
$$ Thus, $$
T_{\mathrm{dp}}=T_{\mathrm{sat} g 1.4 \mathrm{psia}}=113.2^{\circ} \mathrm{F}
$$
