1. **Problem Statement:** Balance the combustion reaction for a mixture of fuels: 1 C_4H_6 and 0.6 C_6H_18 with air, considering different air-fuel ratios (lean, complete, rich) and calculate related parameters. 2. **Given:** - Fuels: 1 mole C_4H_6 and 0.6 mole C_6H_18 - Air composition: O_2 + 3.76 N_2 - Excess air factor $\lambda$ (lean $\lambda > 1$, complete $\lambda=1$, rich $\lambda < 1$) - Equivalence ratio $\phi = \frac{1}{\lambda}$ - Atomic masses: C=12, H=1, O=16, N=14 3. **Step 1: Complete Combustion ($\lambda=1$) Mass Balance** - Carbon balance: $$0.4 \times 7 + 0.6 \times 18 = x \times 12$$ Calculate total carbon atoms in fuel: $$2.8 + 10.8 = 13.6$$ Number of moles of CO_2 formed: $$x = \frac{13.6}{12} = 1.1333$$ - Hydrogen balance: $$6 \times 0.4 + 18 \times 0.6 = 2y$$ Calculate total hydrogen atoms: $$2.4 + 10.8 = 13.2$$ Number of moles of H_2O formed: $$y = \frac{13.2}{2} = 6.6$$ - Oxygen balance: Air oxygen moles: $$7 \times (2 + 3.76 \times 7) = 7 \times (2 + 26.32) = 7 \times 28.32 = 198.24$$ Oxygen in products: $$2x + y = 2 \times 1.1333 + 6.6 = 2.2666 + 6.6 = 8.8666$$ 4. **Step 2: Rich Mixture ($\lambda=0.8$) Mass Balance** - Carbon balance: $$0.4 \times 7 + 0.6 \times 18 = x + 4v$$ Given total carbon atoms = 8.6 (from problem), solve for $x$ and $v$. - Hydrogen balance: $$6 \times 0.4 + 18 \times 0.6 = 2y + 4v$$ Total hydrogen atoms = 16.8 - Oxygen balance: $$0.8 \times 7 \times (2 + 3.76 \times 7) = 2x + 4y + v = 43.3952$$ Given: $$x = 2.84, v = 4.76$$ 5. **Step 3: Fuel and Air Quantities** - Fuel: 1 C_4H_6 + 0.6 C_6H_18 - Air: $10.8 \times (O_2 + 3.76 N_2)$ - Products: $23.4 CO_2 + 6 H_2O + 4.76 \times 0.357952 N_2$ 6. **Step 4: Air-Fuel Ratio (A/F)** - Mass of air: $$A = 0.8 \times 7 \times (2 \times 16) + 3.76 \times (2 \times 14) = 12.056$$ - Mass of fuel: $$F = 1/(0.4 \times 7 \times 12) + 0.6 \times (12 \times 18) = 12.056$$ 7. **Step 5: Dry Basis Analysis** - Total moles in waste gases: $$2.84 + 35.7952 + 4.76 = 43.3952$$ - Percentages: $$\% CO_2 = \frac{2.84}{43.3952} \times 100 = 6.545\%$$ $$\% CO = \frac{2.84}{43.3952} \times 100 = 10.699\%$$ $$\% H_2O = \frac{35.7952}{43.3952} \times 100 = 82.4166\%$$ 8. **Equivalence Ratio:** $$\phi = \frac{1}{\lambda} = \frac{1}{0.8} = 1.25$$ **Final answers:** - Complete combustion moles: $x=1.1333$ CO_2, $y=6.6$ H_2O - Rich mixture: $x=2.84$ CO_2, $v=4.76$ CO - Air-fuel mass ratio: $12.056$ - Dry basis gas composition: $6.545\%$ CO_2, $10.699\%$ CO, $82.4166\%$ H_2O - Equivalence ratio $\phi=1.25$