Let's break down the problem and solve step by step. --- ## **Step 1: Basis of Calculation** Let’s take **100 kg of coal** as the basis. - C: 73 kg - H: 11.5 kg - S: 3.7 kg - Ash: 11.8 kg --- ## **Step 2: Air Supplied** 15% excess dry air is used at a flow rate of 23,000 kg/h. Let’s denote: - \( m_{coal} \) = mass of coal burned per hour - \( m_{air} \) = mass of dry air supplied per hour = 23,000 kg/h Let’s find how much coal is burned. --- ## **Step 3: Solid Waste Analysis** Solid waste (ash + unburned C + unburned S): - 28.7% C - 1.6% S - 69.7% ash Let \( x \) = mass of solid waste produced per 100 kg coal. Ash in solid waste = \( x \times .697 \) But this must equal initial ash input (11.8 kg per 100 kg coal): \[ x \times .697 = 11.8 \implies x = \frac{11.8}{.697} = 16.94 \, \text{kg} \] So, per 100 kg coal: - Solid waste = **16.94 kg** Unburned C = \( 16.94 \times .287 = 4.86 \) kg Unburned S = \( 16.94 \times .016 = .27 \) kg Burned: - Carbon burned = \( 73 - 4.86 = 68.14 \) kg - Sulfur burned = \( 3.7 - .27 = 3.43 \) kg --- ## **Step 4: Combustion Reactions** \[ \begin{align*} \text{C} + \text{O}_2 &\rightarrow \text{CO}_2 \\ \text{S} + \text{O}_2 &\rightarrow \text{SO}_2 \\ \text{H}_2 + \frac{1}{2}\text{O}_2 &\rightarrow \text{H}_2\text{O} \end{align*} \] O₂ required: - For C: \( 68.14 \times \frac{32}{12} = 181.04 \) kg O₂ - For S: \( 3.43 \times \frac{32}{32} = 3.43 \) kg O₂ - For H₂: \( 11.5 \times \frac{16}{4} = 46.00 \) kg O₂ Total theoretical O₂ = \( 181.04 + 3.43 + 46.00 = 230.47 \) kg O₂ --- ## **Step 5: Theoretical Air Required** Air contains 23% O₂ by mass. \[ \text{Theoretical air} = \frac{230.47}{.23} = 1002.04 \text{ kg air} \] With 15% excess: \[ \text{Actual air supplied} = 1002.04 \times 1.15 = 1152.35 \text{ kg air} \] --- ## **Step 6: Scaling to Plant Operation** Given that air supplied per hour = 23,000 kg So, the coal burned per hour (\( M_{coal} \)): \[ \frac{M_{air}}{\text{air per 100 kg coal}} = \frac{23,000}{1152.35} = 19.96 \] So, **coal burned per hour**: \[ M_{coal} = 100 \times 19.96 = 1996 \ \text{kg/h} \] --- ## **Step 7: Heat Released by Combustion** ### **Heats of Combustion (at 25°C):** - C: \( \Delta H_c = -393.5 \) kJ/mol - H₂: \( \Delta H_c = -285.8 \) kJ/mol - S: \( \Delta H_c = -296.8 \) kJ/mol Moles of each per hour: - C: \( 68.14 \times 19.96 = 136.3 \) kg/h; \( \frac{136.3}{12} = 113.36 \) kmol/h - H₂: \( 11.5 \times 19.96 = 229.5 \) kg/h; \( \frac{229.5}{2} = 114.75 \) kmol/h - S: \( 3.43 \times 19.96 = 68.47 \) kg/h; \( \frac{68.47}{32} = 2.14 \) kmol/h Total heat (per hour): \[ Q_{total} = (113.36 \times 393.5) + (114.75 \times 285.8) + (2.14 \times 296.8) \] \[ = 44,622 + 32,813 + 635 = 78,070 \ \text{kJ/h} \] --- ## **Step 8: Heat Available for Steam Generation** 90% is used: \[ Q_{use} = .9 \times 78,070 = 70,263 \ \text{kJ/h} \] --- ## **Step 9: Steam Generation Calculation** Water enters at 20°C, steam at 10 bar (saturated). - At 10 bar, \( T_{sat} = 179.9^\circ C \) - Enthalpy of saturated steam (\( h_g \)) at 10 bar = 2778 kJ/kg - Enthalpy of water at 20°C (\( h_{20} \)) ≈ 83.9 kJ/kg \[ \Delta h = 2778 - 83.9 = 2694.1 \ \text{kJ/kg} \] Steam produced per hour: \[ \text{Steam production rate} = \frac{Q_{use}}{\Delta h} = \frac{70,263}{2694.1} = 26.07 \ \text{kg/h} \] --- ## **Step 10: Final Answer** ### **Production rate of saturated steam generated from water at 20°C to 10 bar (abs) is:** \[ \boxed{26.1 \ \text{kg/h}} \] --- ### **Summary Table** | Parameter | Value / Formula | |------------------------|------------------------| | Coal burned per hour | 1996 kg/h | | Air supplied per hour | 23,000 kg/h | | Heat for steam | 70,263 kJ/h | | Enthalpy increase | 2,694.1 kJ/kg | | Steam production rate | 26.1 kg/h |