Question

Liquid propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}(\ell)\right)\) enters a combustion chamber at \(25^{\circ} \mathrm{C}\) and 1 atm at a rate of \(0.4 \mathrm{kg} / \mathrm{min}\) where it is mixed and burned with 150 percent excess air that enters the combustion chamber at \(25^{\circ} \mathrm{C}\). The heat transfer from the combustion process is \(53 \mathrm{kW}\). Write the balanced combustion equation and determine \((a)\) the mass flow rate of air; \((b)\) the average molar mass (molecular weight) of the product gases; \((c)\) the average specific heat at constant pressure of the product gases; and ( \(d\) ) the temperature of the products of combustion.

Short answer

Answer: The four main properties are: (a) Mass flow rate of air = 5 kg/min (b) Average molar mass (Molecular weight) of the product gases = 26.72 g (c) Average specific heat at constant pressure of the product gases = 1.289 kJ/kgK (d) Temperature of the products of combustion = 442.35^{\circ}C

Step-by-step solution

For complete combustion of propane, an equation can be written as: $$C_3H_8(l) + O_2(g) \rightarrow CO_2(g) + H_2O(g)$$ By balancing the equation, we get: $$C_3H_8(l) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)$$ **Step 2: Calculate the mass flow rate of oxygen and air**

Since there is 150% excess air, that means 250% of the theoretical air requirement is supplied. To calculate the mass flow rate of oxygen, we use stoichiometry: Mass flow rate of propane = 0.4 kg/min Using molar ratios from the balanced equation: Mass flow rate of \(O_2\) = (0.4 kg/min) * (5 mol \(O_2\) / 1 mol \(C_3H_8\)) * (32 g/mol \(O_2\) / 1) * (1 kg / 1000 g) = 2 kg/min = 2 kg/min Now, the mass flow rate of the supplied air is 250% of the theoretical amount. Mass flow rate of supplied air = 2.5 * Mass flow rate of \(O_2\) = 2.5 * 2 kg/min = 5 kg/min So, (a) Mass flow rate of air = 5 kg/min **Step 3: Calculate the molar flow rates of the products**

Using stoichiometry, we can calculate the molar flow rates of the products: Molar flow rate of \(CO_2\) = (0.4 kg/min) * (3 mol \(CO_2\) / 1 mol \(C_3H_8\)) * (1 mol / 44 g) = 0.0273 mol/s Molar flow rate of \(H_2O\) = (0.4 kg/min) * (4 mol \(H_2O\) / 1 mol \(C_3H_8\)) * (1 mol / 18 g) = 0.0736 mol/s **Step 4: Calculate the average molar mass of the product gases**

To find the average molar mass of the product gases, we should first identify all the gases involved in the combustion process. We have \(CO_2\), \(H_2O\), and the nitrogen from the air which did not participate in the combustion reaction. Mass flow rate of nitrogen in supplied air = Mass flow rate of supplied air - Mass flow rate of oxygen = 5 kg/min - 2 kg/min = 3 kg/min Molar flow rate of nitrogen, $$N_2 = (3 * kg / min * 1 mol / 28 g) = 1.7857 mol/s$$ Total molar flow rate of the product gases = Molar flow rate of \(CO_2\) + Molar flow rate of \(H_2O\) + Molar flow rate of \(N_2\) Total molar flow rate = 0.0273 + 0.0736 + 1.7857 = 1.8866 mol/s Sum of mass flow rates by product gas components = (0.0273 * 44) + (0.0736 * 18) + (1.7857 * 28) = 50.4 g/s Hence, (b) Average molar mass (Molecular weight) of the product gases = 50.4 g/s * 1 mol/s / 1.8866 mol = 26.72 g **Step 5: Determine the average specific heat at constant pressure of the product gases**

To find the average specific heat at constant pressure of the product gases, we have to find the mass ratio. For each product gas component, use the individual specific heat at constant pressure values. $$c_p(CO_2) = 0.842 kJ/kgK; c_p(H_2O) = 1.996 kJ/kgK; c_p(N_2) = 1.039 kJ/kgK$$ $$\text{Mass fraction of } CO_2 = \frac{0.0273 * 44}{50.4} = 0.2978$$ $$\text{Mass fraction of } H_2O = \frac{0.0736 * 18}{50.4} = 0.2625$$ $$\text{Mass fraction of } N_2 = \frac{1.7857 * 28}{50.4} = 0.4396$$ Now, we can find the average specific heat at constant pressure: $$c_p(avg)= c_p(CO_2)* \text{mass fraction of } CO_2 + c_p(H_2O) * \text{mass fraction of } H_2O + c_p(N_2) * \text{mass fraction of } N_2$$ $$c_p(avg)= (0.842)(0.2978) + (1.996)(0.2625) + (1.039)(0.4396)$$ Thus, (c) Average specific heat at constant pressure of the product gases = 1.289 kJ/kgK **Step 6: Determine the temperature of the products of combustion**

To find the temperature of the products of combustion, we will use an energy balance equation: $$Q_{in} - Q_{out} - Q_{transfer} = m \cdot c_p(avg) \cdot (T_{out} - T_{in})$$ Here, \(T_{in} = 25^\circ C\), \(Q_{transfer} = 53 kW\), \(m = \text{Total mass flow rate of the reactants}\), and \(T_{out}\) is the temperature of the products of combustion. $$m = \text{Mass flow rate of } C_3H_8 + \text{Mass flow rate of air} = 0.4 + 5 = 5.4 kg/min$$ Now, plugging the values, we will get: $$53 = (5.4 kg / min) * (1.289 kJ/kgK) * (T_{out} - 25)$$ T_{out} = 442.35^{\circ}C$$ So, (d) Temperature of the products of combustion = 442.35^{\circ}C In conclusion, we have found: (a) Mass flow rate of air = 5 kg/min (b) Average molar mass (Molecular weight) of the product gases = 26.72 g (c) Average specific heat at constant pressure of the product gases = 1.289 kJ/kgK (d) Temperature of the products of combustion = 442.35^{\circ}C