Please check whether the heat balance of this distillation column is correct

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1. Given:
(1) n-Butanol: boiling point 117.25°C, relative density d = 0.8098; specific heat capacity at 20°C under constant pressure is 2.33 kJ/(kg·K).
(2) Water: boiling point 100°C, relative density d = 1; specific heat capacity is 4.2 kJ/(kg·°C) in the liquid phase and 2.1 kJ/(kg·°C) in the gas phase.
(3) 1,4-Butanediol (BDO): boiling point 230°C, relative density d = 1.0171; specific heat capacity at 20°C under constant pressure is 2.2 kJ/(kg·K).

2. The steam pressure in the reboiler is 1.8 Mpa, and the corresponding temperature is approximately 205°C.

3. It is assumed that t1 = 0°C, and Δt = t2 – t1.

4. Feed composition: 5% butanol, 30% BDO, 65% water; total flow rate = 10 m³/h, temperature = 30°C.
Effluent composition: 8% butanol, 92% water; total flow rate = 6.5 m³/h, top temperature = 110°C.
Bottom product composition: 96% BDO, 4% water; total flow rate = 3.5 m³/h, bottom temperature = 174°C.
Reflux composition: 8% butanol, 92% water; total flow rate = 0.3 m³/h, top temperature = 30°C.

5. Material balance: Total flow rate ≈ 10 m³/h = 3.5 + 6.5. Although the densities differ, the difference is not significant; approximate calculations can be used.

6. Heat balance: Q_reboiler + Q_feed + Q_reflux = Q_effluent + Q_bottom_product + Q_loss. (Here, Q_loss = 0.05 * Q_reboiler.)

7. Calculations:
(1) Q_feed = Q_butanol + Q_BDO + Q_water = c1m1Δt1 + c2m2Δt2 + c3m3Δt3 = 10.49*10^8 J/h.
(2) Q_top = Q_butanol + Q_water. Where Q_butanol = Q_sensible + Q_vaporization = 3.69*10^8 J/h, and Q_water = Q_sensible_liquid + Q_vaporization + Q_sensible_gas = 168.97*10^8 J/h. Thus, Q_top = 172.66*10^8 J/h.
(3) Q_bottom_product = Q_BDO + Q_water = Q_BDO + Q_sensible_liquid + Q_vaporization + Q_sensible_gas = 17.05*10^8 J/h.
(4) Q_reflux = Q_butanol + Q_water = 0.36*10^8 J/h.

8. From Q_reboiler + Q_feed + Q_reflux = Q_effluent + Q_bottom_product + Q_loss, it follows that Q_reboiler = 188.27*10^8 J/h.

9. For the reboiler, Q_reboiler = Km * A * Δtc. Here, Km represents the heat transfer coefficient, with values of 600, 800, and 1140 (typical values for heat transfer between organic solutions and steam). Δtc = 205°C – 174°C – 3°C = 28°C; the 3°C value represents the temperature resistance. A represents the heat transfer area. Therefore:
- When K = 600, A = 311 m².
- When K = 800, A = 233 m².
- When K = 1140, A = 163 m².

10. The actual heat transfer area of the reboiler is 112 m².
Question: I want to calculate the area of the reboiler. Why is there such a large difference? ? Could some experts help me check whether my algorithm is correct? ? I hope you can give me a lot of guidance~~