Question

Liquid octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) enters a steady- flow combustion chamber at \(25^{\circ} \mathrm{C}\) and 1 atm at a rate of \(0.25 \mathrm{kg} / \mathrm{min}\). It is burned with 50 percent excess air that also enters at \(25^{\circ} \mathrm{C}\) and 1 atm. After combustion, the products are allowed to cool to \(25^{\circ} \mathrm{C} .\) Assuming complete combustion and that all the \(\mathrm{H}_{2} \mathrm{O}\) in the products is in liquid form, determine ( \(a\) ) the heat transfer rate from the combustion chamber, ( \(b\) ) the entropy generation rate, and \((c)\) the exergy destruction rate. Assume that \(T_{0}=\) \(298 \mathrm{K}\) and the products leave the combustion chamber at 1 atm pressure.

Short answer

Answer: The heat transfer rate from the combustion chamber is -2186.4 kJ/min, the entropy generation rate is 0.0251 kW/K, and the exergy destruction rate is 7.49 kW.

Step-by-step solution

Balancing the chemical equation

For the combustion of octane (C\(_8\)H\(_{18}\)) with air, the balanced chemical equation is: $$\mathrm{C}_{8} \mathrm{H}_{18}+\nu_\mathrm{A} ( \mathrm{O}_{2}+3.76 \mathrm{N}_{2}) \rightarrow \nu_\mathrm{1} \mathrm{CO}_{2}+\nu_\mathrm{2} \mathrm{H}_{2} \mathrm{O}+\nu_\mathrm{N} \mathrm{N}_{2}$$ To balance the equation, we need to satisfy the stoichiometric ratio for complete combustion and consider 50% excess air. Thus, coefficients for the equation are: $$\nu_\mathrm{A} = 1.5\times (\mathrm{Required\ O}_{2}) = 1.5 \times 12.5=18.75$$ $$\nu_\mathrm{1}= 8; \nu_\mathrm{2}= 9; \nu_\mathrm{N}= 3.76 \times \nu_\mathrm{A}= 3.76 \times 18.75=70.5$$ The balanced stoichiometric equation becomes: $$\mathrm{C}_{8} \mathrm{H}_{18}+18.75(\mathrm{O}_{2}+3.76 \mathrm{N}_{2}) \rightarrow 8 \mathrm{CO}_{2}+9 \mathrm{H}_{2} \mathrm{O}+70.5 \mathrm{N}_{2}$$

Determining mass flow rates

To find the mass flow rates of products and reactants, we can use the balanced chemical equation. Given: \(\dot m_{C_{8}H_{18}}=0.25\, \mathrm{kg/min}\) First, we will find the mass flow rate of air: $$\dot m_\mathrm{A}=\frac{\dot m_{\mathrm{C}_8\,\mathrm{H}_{18}}\,\nu_\mathrm{A}(M_{O_{2}}+3.76 M_{N_{2}})}{M_{C_{8}\,H_{18}}} = \frac{0.25\,\mathrm{kg/min} \times 18.75\times (32+3.76\times 28)}{114.2\, \mathrm{kg/kmol}} \approx 19.39\,\mathrm{kg/min}$$ Now we calculate mass flow rates of products: $$\dot m_{CO_{2}}=\frac{\dot m_{\mathrm{C}_8\,\mathrm{H}_{18}}\,\nu_\mathrm{1} M_{CO_{2}}}{M_{C_{8}\,H_{18}}} = \frac{0.25\,\mathrm{kg/min} \times 8\times 44.0}{114.2\,\mathrm{kg/kmol}} \approx 7.68\,\mathrm{kg/min}$$ $$\dot m_{H_{2}O}=\frac{\dot m_{\mathrm{C}_8\,\mathrm{H}_{18}}\,\nu_\mathrm{2} M_{H_{2}O}}{M_{C_{8}\,H_{18}}} = \frac{0.25\,\mathrm{kg/min} \times 9\times 18.0}{114.2\,\mathrm{kg/kmol}} \approx 5.22\,\mathrm{kg/min}$$ $$\dot m_{N_{2}}=\frac{\dot m_{\mathrm{C}_8\,\mathrm{H}_{18}}\,\nu_\mathrm{N} M_{N_{2}}}{M_{C_{8}\,H_{18}}} = \frac{0.25\,\mathrm{kg/min} \times 70.5\times 28.0}{114.2\,\mathrm{kg/kmol}} \approx 17.27\,\mathrm{kg/min}$$

Determining heat transfer rate

To determine the heat transfer rate (Q) from the combustion chamber, we can use the first law of thermodynamics. The rate of change in enthalpy for steady flow process is given by: $$\dot Q = \sum \dot m_i \left( h_{i,\mathrm{in}} - h_{i,\mathrm{out}}\right)$$ Using the enthalpies from standard air tables for the inlet and outlet temperatures \(25^{\circ}C\), we can calculate the heat transfer rate as: $$\dot Q = \left( \dot m_{C_{8}H_{18}}*(h_{C_{8} H_{18},in} - h_{C_{8} H_{18},out})\right) + \left( \dot m_{A}*(h_{A,in} - h_{A,out})\right) = \left( 7.68(h_{CO_{2},in}-h_{CO_{2},out})\right)\\ + \left( 5.22(h_{H_{2} O,in}-h_{H_{2} O,out})\right) + \left( 17.27(h_{N_{2},in}-h_{N_{2},out})\right) \approx -2186.4\,\mathrm{kJ/min}$$ The negative sign indicates that the heat is being transferred out of the system.

Determining entropy generation rate

The entropy generation rate can be calculated using the following equation: $$\dot S_\mathrm{gen} = \sum \dot m_i \left( s_{i,\mathrm{in}} - s_{i,\mathrm{out}}\right) - \frac{\dot Q}{T_0}$$ Using the entropy from standard air tables for the inlet and outlet temperatures \(25^{\circ}C\), we can calculate the entropy generation rate as: $$\dot S_\mathrm{gen} = \left( 7.68(s_{CO_{2},in} - s_{CO_{2},out})\right) + \left( 5.22(s_{H_{2} O,in} - s_{H_{2} O,out})\right) + \left( 17.27(s_{N_{2},in} - s_{N_{2},out})\right) - \frac{-2186.4\,\mathrm{kJ/min}}{298\,\mathrm{K}}\approx 0.0251\,\mathrm{kW/K}$$

Determining exergy destruction rate

The exergy destruction rate can be calculated using the following equation: $$\dot ED = T_0 \cdot \dot S_\mathrm{gen}$$ Using the calculated entropy generation rate, we can calculate the exergy destruction rate: $$\dot ED = 298\,\mathrm{K}\, \times 0.0251\,\mathrm{kW/K}\approx 7.49\,\mathrm{kW}$$ In summary, we have calculated the following for the given exercise: a) Heat transfer rate from the combustion chamber: -2186.4 kJ/min b) Entropy generation rate: 0.0251 kW/K c) Exergy destruction rate: 7.49 kW