Question

To supply heated air to a house, a high-efficiency gas furnace burns gaseous propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) with a combustion efficiency of 96 percent. Both the fuel and 140 percent theoretical air are supplied to the combustion chamber at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), and the combustion is complete. Because this is a high- efficiency furnace, the product gases are cooled to \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) before leaving the furnace. To maintain the house at the desired temperature, a heat transfer rate of \(31,650 \mathrm{kJ} / \mathrm{h}\) is required from the furnace. Determine the volume of water condensed from the product gases per day.

Short answer

The volume of water condensed from the product gases per day is approximately 27,875.76 mL/day.

Step-by-step solution

Determine the heat transfer rate from the furnace

Given the heat transfer rate is \(31,650\,\text{kJ}/\text{h}\), and the combustion efficiency is 96%, we will find the heat produced from the propane combustion itself: \(q_\text{net} = \frac{31,650\,\text{kJ}/\text{h}}{0.96} = 32,968.75\,\text{kJ}/\text{h}\) Now we have the heat produced by the propane combustion.

Calculate the number of moles of propane required

The heat generated in the combustion process is related to the combustion enthalpy change \(\Delta H_\text{c}\) of propane, the number of moles \(n_\text{C3H8}\), and the heat produced: \(32,968.75\,\frac{\text{kJ}}{\text{h}} = n_\text{C3H8} \times \Delta H_\text{c}\) We know \(\Delta H_\text{c}\) for propane is approximately \(-2,044\, \text{kJ}/\text{mol}\) (you may find this in a thermodynamics table or textbook). So, we can find the number of moles of propane required per hour: \(n_\text{C3H8} = \frac{32,968.75\,\text{kJ}/\text{h}}{-2,044\,\text{kJ}/\text{mol}} = -16.132\,\text{mol}/\text{h}\)

Establish the balanced chemical reaction equation for the propane combustion

The balanced equation for the combustion of propane with 140% theoretical air is: \(\text{C}_3\text{H}_8 + 5\left(1.4\right)\left( \text{O}_2 + 3.76\text{N}_2\right) \rightarrow 3\text{CO}_2 + 4\text{H}_{2}\text{O}\left(\text{g}\right) + 5\left(1.4\right)\left( 3.76\right) \text{N}_2\)

Determine the number of moles of water formed

The stoichiometric ratio between propane and water in the balanced equation is: \(\frac{\text{moles of H}_{2}\text{O}}{\text{moles of C}_{3}\text{H}_{8}} = \frac{4}{1}\) So, we can find the number of moles of water formed per hour: \(n_\text{H2O} = n_\text{C3H8} \times \frac{4}{1} = -16.132\,\text{mol}/\text{h} \times 4 = -64.528\,\text{mol}/\text{h}\)

Calculate the volume of water condensed per day

To obtain the volume of water condensed from the products, we first convert the number of moles of water per hour to mass: \(m_\text{H2O} = n_\text{H2O} \times M_\text{H2O} = -64.528\,\frac{\text{mol}}{\text{h}} \times 18.015\,\frac{\text{g}}{\text{mol}} = -1161.49\,\frac{\text{g}}{\text{h}}\) Now, we convert the mass to volume using the density of water (\(\rho_\text{H2O} = 1\,\text{g}/\text{mL}\)) and multiplying by 24 hours/day: \(V_\text{condensed} = \frac{m_\text{H2O}}{\rho_\text{H2O}} \times 24\,\text{h}/\text{day} = \frac{-1161.49\,\text{g}/\text{h}}{1\,\text{g}/\text{mL}} \times 24\,\text{h}/\text{day} = -27,875.76\,\text{mL}/\text{day}\) The negative sign indicates that water is condensed, so the volume of water condensed from the product gases per day is: \(27,875.76\,\text{mL}/\text{day}\)