💧 Multi-Effect Evaporator Design
Sizing of forward-feed multi-effect evaporators (1 to 6 effects) with boiling point elevation (BPE) correction.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
Heat Transfer
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Sizing Inputs
Calculation Results
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Calculation Methodology
Mathematical Model & Theory
Multi-effect evaporators reduce energy consumption by utilizing vapor generated in one effect as the heating steam in the next effect. Material balances across the train are resolved as:
$$\dot{m}_f \cdot x_f = \dot{m}_{prod} \cdot x_p \implies \dot{m}_{evap} = \dot{m}_f - \dot{m}_{prod}$$
Boiling Point Elevation (BPE) occurs because dissolved solutes lower the solvent vapor pressure. BPE is linearized based on concentration:
$$\text{BPE}_i = \text{BPE}_c \cdot x_i$$
The total temperature drop available across the train is reduced by the sum of BPEs:
$$\Delta T_{avail} = T_{steam} - T_{condenser} - \sum_{i=1}^{N} \text{BPE}_i$$
Sizing uses the heat load ($Q_i$) and overall heat transfer coefficient ($U_i$) to determine area ($A_i$):
$$Q_i = U_i \cdot A_i \cdot \Delta T_{eff}, \quad \Delta T_{eff} = \frac{\Delta T_{avail}}{N}$$
Academic References:
- McCabe, W. L., Smith, J. C., & Harriott, P. (2005). Unit Operations of Chemical Engineering (7th ed.). McGraw-Hill.
- Geankoplis, C. J. (2003). Transport Processes and Separation Process Principles (4th ed.). Prentice Hall.
Worked Engineering Example
Problem Statement:
Size a triple-effect forward-feed evaporator concentrating sugar juice from $12\%$ to $65\%$ mass fraction. Juice flow is $5$ kg/s. Heating steam is supplied at $120$°C, and the condenser operates at $50$°C. Overall heat transfer coefficient $U = 2500$ W/m²K, and BPE coefficient $= 1.78$. Latent heat of vaporization $h_{fg} = 2260$ kJ/kg.
Step-by-Step Solution:
1. Calculate product and evaporation flows:
$$\dot{m}_{prod} = 5 \cdot 0.12 / 0.65 \approx 0.923 \text{ kg/s}$$ $$\dot{m}_{evap} = 5 - 0.923 \approx 4.077 \text{ kg/s}$$
2. Assess Boiling Point Elevation (BPE):
Average concentration $\bar{x} \approx (0.12 + 0.65)/2 = 0.385$. $$\text{BPE}_{tot} = 1.78 \cdot 0.385 \cdot 3 \approx 2.06^\circ\text{C}$$
3. Calculate Temp Differences:
$$\Delta T_{avail} = 120 - 50 - 2.06 = 67.94^\circ\text{C}$$ $$\Delta T_{eff} = 67.94 / 3 \approx 22.65^\circ\text{C}$$
4. Size Heat transfer Area:
Steam consumption for 3 effects: $\dot{m}_{steam} \approx \dot{m}_{evap}/3 = 1.359$ kg/s. $$\text{Economy} = 4.077 / 1.359 \approx 3.0$$ Heat duty per effect: $Q = 1.359 \text{ kg/s} \cdot 2260 \text{ kJ/kg} = 3071 \text{ kW}$. $$A = \frac{3,071,000 \text{ W}}{2500 \text{ W/m²K} \cdot 22.65\text{ K}} \approx 54.2 \text{ m² (per effect)}$$ $$\text{Total Area } A_{tot} = 3 \cdot 54.2 \approx 162.6 \text{ m²}$$
Size a triple-effect forward-feed evaporator concentrating sugar juice from $12\%$ to $65\%$ mass fraction. Juice flow is $5$ kg/s. Heating steam is supplied at $120$°C, and the condenser operates at $50$°C. Overall heat transfer coefficient $U = 2500$ W/m²K, and BPE coefficient $= 1.78$. Latent heat of vaporization $h_{fg} = 2260$ kJ/kg.
Step-by-Step Solution:
1. Calculate product and evaporation flows:
$$\dot{m}_{prod} = 5 \cdot 0.12 / 0.65 \approx 0.923 \text{ kg/s}$$ $$\dot{m}_{evap} = 5 - 0.923 \approx 4.077 \text{ kg/s}$$
2. Assess Boiling Point Elevation (BPE):
Average concentration $\bar{x} \approx (0.12 + 0.65)/2 = 0.385$. $$\text{BPE}_{tot} = 1.78 \cdot 0.385 \cdot 3 \approx 2.06^\circ\text{C}$$
3. Calculate Temp Differences:
$$\Delta T_{avail} = 120 - 50 - 2.06 = 67.94^\circ\text{C}$$ $$\Delta T_{eff} = 67.94 / 3 \approx 22.65^\circ\text{C}$$
4. Size Heat transfer Area:
Steam consumption for 3 effects: $\dot{m}_{steam} \approx \dot{m}_{evap}/3 = 1.359$ kg/s. $$\text{Economy} = 4.077 / 1.359 \approx 3.0$$ Heat duty per effect: $Q = 1.359 \text{ kg/s} \cdot 2260 \text{ kJ/kg} = 3071 \text{ kW}$. $$A = \frac{3,071,000 \text{ W}}{2500 \text{ W/m²K} \cdot 22.65\text{ K}} \approx 54.2 \text{ m² (per effect)}$$ $$\text{Total Area } A_{tot} = 3 \cdot 54.2 \approx 162.6 \text{ m²}$$