Question

One lbm of butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) is burned with 25 lbm of air that is at \(90^{\circ} \mathrm{F}\) and 14.7 psia. Assuming that the combustion is complete and the pressure of the products is 14.7 psia, determine \((a)\) the percentage of theoretical air used and \((b)\) the dew-point temperature of the products.

Short answer

Answer: The percentage of theoretical air used is approximately 827.48%, and the dew-point temperature of the products is approximately 46.4°F.

Step-by-step solution

1. Calculate the stoichiometric air-fuel ratio for the complete combustion of butane

Write the balanced equation for the combustion of butane with the required amount of air (stoichiometric air): \(C_4H_{10} + 6.5 (O_2 + 3.76N_2) -> 4CO_2 + 5H_2O + 24.44N_2\) The stoichiometric air-fuel ratio can be calculated from the balanced equation. First, calculate the molar mass of butane, \(C_4H_{10}\): \(M_{C_4H_{10}} = (4\times12.01) + (10\times1.01) = 58.12 \, lb_{mole}/lbm\) Now, calculate the molar mass of air required for stoichiometric combustion: \(M_{air} = 6.5 \times (32 + 3.76 \times 28) = 175.72 \, lb_{mole}/lbm\) Finally, the stoichiometric air-fuel ratio is: \(AFR_{st} = \frac{M_{air}}{M_{C_4H_{10}}} = \frac{175.72}{58.12} = 3.02\)

2. Determine the actual air-fuel ratio and calculate the percentage of theoretical air used

Given that 1 lbm of butane is burned with 25 lbm of air, the actual air-fuel ratio (AFR) is: \(AFR = \frac{25}{1} = 25\) Now, calculate the percentage of theoretical air used: \(\%\, of \, theoretical \, air = \frac{AFR}{AFR_{st}} \times 100 = \frac{25}{3.02} \times 100 \approx 827.48\%\)

3. Calculate the dew-point temperature of the products

First, find the moles of water vapor formed as a result of the combustion process. From the balanced equation, we have: \(5\,moles\,H_2O : 1\,mole\,C_4H_{10}\) So, for 1 lbm of butane, we have: \(\frac{5}{58.12}\, moles\, of\, H_2O\, per\, lbm\, of\, C_4H_{10}\) Now, the process occurs on a mass basis, so we multiply the mass of butane burnt by the moles of water vapor formed in the combustion process: \(n_{H_2O} = 1 \times \frac{5}{58.12} \approx 0.086\, lb_{mole}\) Next, we have to find the partial pressure of water vapor in the products. The total pressure of the gas mixture is given as 14.7 psia. The mole fraction of water vapor, \(y_{H_2O}\), can be calculated as: \(y_{H_2O} = \frac{n_{H_2O}}{\text{Total moles of combustion products}}\) The total moles of combustion products can be found from the balanced equation: \(1+6.5+24.44 = 31.94\, moles\) Hence, the mole fraction of water vapor is: \(y_{H_2O} = \frac{0.086}{31.94} \approx 0.0027\) Now, we can calculate the partial pressure of water vapor in the products: \(P_{H_2O} = y_{H_2O} \times P_{total} = 0.0027 \times 14.7 \, psia \approx 0.0397 \, psia\) Finally, using steam tables or the Antoine equation, we find the dew-point temperature of the products. This dew point temperature corresponds to the saturation temperature of water at 0.0397 psia. As per steam tables or the Antoine equation, we find that the dew-point temperature is approximately \(46.4^{\circ}\,F\).