# (a) Step-by-Step Solution: Percentage Excess Air for Coal-Fired Boiler Let’s break down the problem step by step: ## **. Given Data** - **Ultimate of coal (by mass):** - Carbon (\(C\)) = 22% - Hydrogen (\(H\)) = 3.5% - Oxygen (\(O\)) = 6.5% - Remainder: incombustibles (ash, etc.) - **Dry flue gas analysis (by volume):** - CO₂ = 8% - CO = 1.5% - O₂ = 10% - N₂ = 80.5% - **Air composition:** - By **volume**: O₂ = 21%, N₂ = 79% - By **mass**: O₂ = 23%, N₂ = 77% --- ## **2. Basis of Calculation** Let’s take **100 kg of coal** as our basis. --- ## **3. Moles of Carbon Burned** - **Total carbon in 100 kg coal:** \( 100 \times .22 = 22 \) kg C - **Moles of C:** \( \frac{22}{12} = 1.833 \) kmol C --- ## **4. Moles of Dry Flue Gas Produced** Let \( x \) = total moles of dry flue gas produced per 100 kg coal. Given their percentages by volume, - CO₂: \( .08x \) - CO: \( .015x \) - O₂: \( .10x \) - N₂: \( .805x \) --- ## **5. Carbon Balance** All C in coal goes to CO₂ and CO in flue gas. - 1 mol C → 1 mol CO₂ or 1 mol CO So, \[ \text{Total moles of C in flue gas} = \text{CO}_2 + \text{CO} = (.08x + .015x) = .095x \] But total moles of C = 1.833 kmol \[ .095x = 1.833 \implies x = \frac{1.833}{.095} = 19.3\ \text{kmol} \] --- ## **6. Oxygen Used for Combustion** - **CO₂ Formation:** Each mole needs 1 mole O₂ - **CO Formation:** Each mole needs \( \frac{1}{2} \) mole O₂ \[ \text{O}_2\: \text{used} = (\text{CO}_2) \times 1 + (\text{CO}) \times .5 \] \[ = (.08x) \times 1 + (.015x) \times .5 \] \[ = .08x + .0075x = .0875x \] \[ \text{Plug in } x = 19.3: \] \[ \text{O}_2\: \text{used} = .0875 \times 19.3 = 1.69 \text{ kmol} \] --- ## **7. Oxygen Supplied** - **O₂ in dry flue gas = 10% of x:** \( .10 \times 19.3 = 1.93 \) kmol (unused O₂) - **Total O₂ supplied = O₂ used + O₂ in flue gas** \[ = 1.69 + 1.93 = 3.62 \text{ kmol} \] --- ## **8. Theoretical Oxygen Required** - For C in coal: 1.833 kmol C - To CO₂: each C needs 1 O₂ → 1.833 kmol O₂ - But some goes to CO (\(.015x = .29\) kmol), rest to CO₂ (\(.08x = 1.54\) kmol) \[ \text{O}_2\: \text{theoretical} = (\text{CO}_2) \times 1 + (\text{CO}) \times .5 \] \[ = 1.54 \times 1 + .29 \times .5 = 1.54 + .145 = 1.685 \text{ kmol} \] - **Hydrogen:** - 3.5 kg H in 100 kg coal - 1 kg H → \( \frac{1}{2} \) mol O₂ per g H₂ burned - 1 kmol H₂ (2 kg) needs 1 kmol O₂ - 3.5 kg H → \( \frac{3.5}{2} = 1.75 \) kmol H₂ - O₂ needed: 1.75 kmol **But, hydrogen that combines with oxygen already present in the coal is not considered:** \[ \text{Available H} = \text{Total H} - \frac{\text{O in coal}}{8} \] - O in coal = 6.5 kg = \( \frac{6.5}{16} = .406 \) kmol O - Equivalent H = \( .406 \times 2 = .812 \) kmol H - \( 1.75 - .812 = .938 \) kmol H₂ - O₂ needed: \( .938 \) kmol - **Total theoretical O₂ required:** \[ = \text{O}_2 \text{ for C} + \text{O}_2 \text{ for H} \] \[ = 1.685 + .938 = 2.623 \text{ kmol} \] --- ## **9. Percentage Excess Air** - **Actual O₂ supplied:** \( 3.62 \) kmol (from step 7) - **Theoretical O₂ required:** \( 2.623 \) kmol (from step 8) - **Excess O₂ supplied:** \( 3.62 - 2.623 = .997 \) kmol \[ \text{Percentage excess O}_2 = \frac{.997}{2.623} \times 100 \approx 38\% \] - **Percentage excess air** (since air is 21% O₂ by volume): \[ \text{Excess air} = \frac{\text{Actual air supplied} - \text{Theoretical air required}}{\text{Theoretical air required}} \times 100\% \] But since excess air is proportional to excess O₂, so: \[ \text{Percentage excess air} = \text{Percentage excess O}_2 \approx 38\% \] --- ## **10. Final Answer (a)** **The percentage excess air supplied to the boiler is approximately 38%.** --- # (b) Use of Percentage Excess Air in Boiler Efficiency Calculations ## **How Percentage Excess Air is Used to Calculate Boiler Efficiency** - **Boiler efficiency** is the ratio of useful heat output to the energy input from the fuel. - **Excess air** increases the mass of gases leaving the boiler, carrying more heat away (heat loss). - By knowing the percentage excess air (from flue gas analysis), you can calculate: - The mass of dry flue gases - The temperature and heat lost via stack gases - **Heat loss by dry flue gases** is a major loss in boiler operation and is calculated using: \[ Q_{stack} = \text{mass of flue gas} \times c_p \times (T_{stack} - T_{ambient}) \] - **Mass of flue gas** increases with excess air. - **Boiler efficiency** is then: \[ \text{Efficiency} = 100\% - (\text{sum of various heat losses, including stack loss}) \] --- ## **Accuracy and Comments** - **Indirect Method (Loss Method):** Relies heavily on accurate flue gas analysis, including excess air calculation. - **Limitations:** - Assumes perfect mixing and representative flue gas samples. - Errors in O₂ measurement or assumptions about fuel composition can lead to significant inaccuracies. - **Overall:** While useful for routine checks, boiler efficiency calculated solely from excess air/flue gas analysis should be treated as an estimate. Direct measurements (e.g., feedwater flow, steam output) are more accurate when possible. --- **In summary:** - Percentage excess air is crucial for estimating heat loss via flue gases and thus boiler efficiency. - However, results are only as accurate as the flue gas measurements and assumptions; always interpret with care. --- ## **References** - [Boiler Combustion Calculations](https://www.engineeringtoolbox.com/boiler-combustion-efficiency-d_271.html) - Standard Combustion Calculations, Perry’s Chemical Engineers’ Handbook --- **Image suggestion:** *Diagram showing relationship of excess air, stack loss, and boiler efficiency*  *Alt: Diagram illustrating how excess air increases stack loss and reduces boiler efficiency*