# Adiabatic Combustion of Methane: Final Temperature Calculation Let's solve the problem step by step: ## **Step 1: Write the Combustion Reaction** For complete combustion of methane (\(\mathrm{CH_4}\)): \[ \mathrm{CH_4} + 2\ \mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\ \mathrm{H_2O} \] Air contains about 21% \( \mathrm{O_2} \) by volume (23.3% by mass), and the rest is mainly \( \mathrm{N_2} \). ## **Step 2: Calculate Oxygen Supplied by Air** **Air supplied:** 50.0 lb/min **Oxygen in air:** \( 23.3\% \) by mass \[ \text{O}_2 \text{ supplied} = 50.0\ \text{lb/min} \times 0.233 = 11.65\ \text{lb/min} \] ## **Step 3: Calculate Methane Combustion Requirements** **Moles of methane per minute:** - Methane molar mass: \( 16.04\ \mathrm{lb/lbmol} \) - \( 1.00\ \mathrm{lb/min} \div 16.04\ \mathrm{lb/lbmol} = 0.0624\ \mathrm{lbmol/min} \) **Oxygen required per mole of methane:** From the reaction: 1 mole \(\mathrm{CH_4}\) needs 2 moles \(\mathrm{O_2}\). - \( 0.0624\ \mathrm{lbmol/min} \times 2 = 0.1248\ \mathrm{lbmol/min} \) O\(_2\) required **Mass of O\(_2\) required:** - O\(_2\) molar mass = \( 32.00\ \mathrm{lb/lbmol} \) - \( 0.1248\ \mathrm{lbmol/min} \times 32.00\ \mathrm{lb/lbmol} = 3.99\ \mathrm{lb/min} \) **Excess O\(_2\) supplied:** - O\(_2\) supplied: 11.65 lb/min - O\(_2\) required: 3.99 lb/min - **Excess O\(_2\):** \( 11.65 - 3.99 = 7.66\ \mathrm{lb/min} \) So, there will be excess O\(_2\), and the combustion is with excess air. ## **Step 4: Calculate Moles of Other Components in Air** **Nitrogen supplied:** - Nitrogen is about \( 76.7\% \) by mass in air. - \( 50.0\ \mathrm{lb/min} \times 0.767 = 38.35\ \mathrm{lb/min} \) - Molar mass N\(_2\) = 28.01 lb/lbmol - \( 38.35\ \mathrm{lb/min} \div 28.01\ \mathrm{lb/lbmol} = 1.37\ \mathrm{lbmol/min} \) ## **Step 5: Write the Actual Combustion Equation** - \( \mathrm{CH_4} = 0.0624\ \mathrm{lbmol/min} \) - Total O\(_2\) supplied: \( 11.65\ \mathrm{lb/min} \div 32.00\ \mathrm{lb/lbmol} = 0.364\ \mathrm{lbmol/min} \) - O\(_2\) required: \( 0.1248\ \mathrm{lbmol/min} \) - O\(_2\) excess: \( 0.364 - 0.1248 = 0.239\ \mathrm{lbmol/min} \) - N\(_2\) supplied: \( 1.37\ \mathrm{lbmol/min} \) **Overall reaction:** \[ \mathrm{CH_4} + 2\ \mathrm{O_2} + 7.63\ \mathrm{N_2} + \text{excess O}_2 \rightarrow \mathrm{CO_2} + 2\ \mathrm{H_2O} + \text{excess O}_2 + \text{all N}_2 \] But since we're working with mass and moles, let's write the products per minute: - CO\(_2\): 0.0624 lbmol/min - H\(_2\)O: \( 2 \times 0.0624 = 0.1248 \) lbmol/min - O\(_2\) (excess): 0.239 lbmol/min - N\(_2\): 1.37 lbmol/min ## **Step 6: Set up Energy Balance** ### **Enthalpy Reference** We use \( 77^\circ F\) (\(25^\circ C\), \(298\) K) as the reference state for enthalpies (\(h^\circ\)). Let \(T_0 = 77^\circ F\) (rounded from \(70^\circ F\) for standard enthalpy tables). #### **Energy Balance (Adiabatic)** \[ \text{Total enthalpy in (reactants)} = \text{Total enthalpy out (products at } T_\text{final} \text{)} \] \[ \sum n_{R} \left[ h_{R}(T_{R}) - h^\circ (T_0) \right] + \Delta H_\text{comb} = \sum n_{P} \left[ h_{P}(T_{final}) - h^\circ (T_0) \right] \] But for adiabatic combustion, the heat of reaction is released as sensible heat to raise the temperature of the products: \[ \text{Total sensible enthalpy of products at } T_f = \text{Total sensible enthalpy of reactants at initial T} + \Delta H_{rxn, std} \] #### **Sensible Enthalpy Change** - For each component: \( n \times C_p \times (T - T_0) \) - We'll use average \( C_p \) (specific heat) values. ### **Substance Data (approximate)** | Substance | \( C_p \) (Btu/lbmol·°F) | Molar Mass (lb/lbmol) | |-----------|--------------------------|-----------------------| | CH\(_4\) (g) | 8.0 | 16.04 | | O\(_2\) (g) | 7.0 | 32.00 | | N\(_2\) (g) | 6.97 | 28.01 | | CO\(_2\) (g) | 8.9 | 44.01 | | H\(_2\)O (g) | 8.0 | 18.02 | ### **Heats of Formation (\( \Delta H_f^\circ \), 77°F, 1 atm)** | Substance | \( \Delta H_f^\circ \) (Btu/lbmol) | |-----------|--------------------------| | CH\(_4\) (g) | -17,889 | | CO\(_2\) (g) | -94,052 | | H\(_2\)O (g) | -44,012 | | O\(_2\) (g), N\(_2\) (g) | 0 | ### **Heat of Reaction per lbmol CH\(_4\)** \[ \Delta H_{rxn}^\circ = \left[ \Delta H_f^\circ(\text{CO}_2) + 2 \times \Delta H_f^\circ(\text{H}_2O) \right] - \left[ \Delta H_f^\circ(\text{CH}_4) + 2 \times \Delta H_f^\circ(\text{O}_2) \right] \] \[ = [-94,052 + 2 \times (-44,012)] - [-17,889 + 0] = [-94,052 - 88,024] + 17,889 = -182,076 + 17,889 = -164,187\ \text{Btu/lbmol} \] ### **Total Heat Released per min** \[ \text{Moles CH}_4 = 0.0624\ \text{lbmol/min} \] \[ \text{Heat released per min} = 0.0624 \times 164,187 = 10,245\ \text{Btu/min} \] ## **Step 7: Compute Enthalpy of Reactants** ### **Reactants** - \( \mathrm{CH_4} \): \(0.0624\ \mathrm{lbmol} \), \(70^\circ F\) - Air: \(50.0\ \mathrm{lb/min}\), \(200^\circ F\) - Air: \(0.364\) lbmol O\(_2\), \(1.37\) lbmol N\(_2\) **Enthalpy change (relative to 77°F, per lbmol):** - For CH\(_4\): \( (T_{in} - 77)\ \times C_p \), \( T_{in} = 70^\circ F \Rightarrow T_{in} - 77 = -7^\circ F \) - For air: \( T_{in} = 200^\circ F \Rightarrow 200 - 77 = 123^\circ F \) #### **CH\(_4\):** \[ \Delta H_{\mathrm{CH_4}} = 0.0624 \times 8.0 \times (-7) = -3.5\ \text{Btu/min} \] #### **O\(_2\):** \[ \Delta H_{\mathrm{O_2}} = 0.364 \times 7.0 \times 123 = 313\ \text{Btu/min} \] #### **N\(_2\):** \[ \Delta H_{\mathrm{N_2}} = 1.37 \times 6.97 \times 123 = 1,174\ \text{Btu/min} \] #### **Total Sensible Enthalpy of Reactants:** \[ \text{Total} = -3.5 + 313 + 1,174 = 1,483.5\ \text{Btu/min} \] ## **Step 8: Compute Enthalpy of Products** - CO\(_2\): \(0.0624 \) lbmol/min - H\(_2\)O: \(0.1248 \) lbmol/min - O\(_2\) (excess): \(0.239\) lbmol/min - N\(_2\): \(1.37\) lbmol/min Let \( T_f \) be the final temperature in \( ^\circ F \). All products start at 77°F. \[ \Delta H_{\text{products}} = \sum n_i \times C_{p, i} \times (T_f - 77) \] \[ = 0.0624 \times 8.9 \times (T_f - 77) + 0.1248 \times 8.0 \times (T_f - 77) + 0.239 \times 7.0 \times (T_f - 77) + 1.37 \times 6.97 \times (T_f - 77) \] \[ = \left[0.555 + 1.0 + 1.673 + 9.55 \right] \times (T_f - 77) \] \[ = 12.78 \times (T_f - 77) \] ## **Step 9: Set up and Solve Energy Balance Equation** \[ \text{Sensible enthalpy of products} = \text{Sensible enthalpy of reactants} + \text{Heat of reaction} \] \[ 12.78 (T_f - 77) = 1,483.5 + 10,245 \] \[ 12.78 (T_f - 77) = 11,728.5 \] \[ T_f - 77 = \frac{11,728.5}{12.78} = 917.8 \] \[ T_f = 77 + 917.8 = 994.8^\circ F \] **Final answer (adiabatic):** \[ \boxed{995^\circ F} \] --- ## **Step 10: Repeat with 10% Heat Loss** 10% heat lost, so only 90% of the released heat raises product temperature: \[ \text{Usable heat} = 0.9 \times 10,245 = 9,220\ \text{Btu/min} \] \[ 12.78 (T_f - 77) = 1,483.5 + 9,220 = 10,703.5 \] \[ T_f - 77 = \frac{10,703.5}{12.78} = 837.7 \] \[ T_f = 77 + 837.7 = 914.7^\circ F \] **Final answer (10% heat loss):** \[ \boxed{915^\circ F} \] --- # **Summary Table** | Condition | Final Exhaust Temperature (\(^\circ F\)) | |----------------|-----------------------------------------| | Adiabatic