💧

Multi-Effect Evaporator Design

Design forward-feed evaporator trains with boiling point elevation (BPE) correction.

Effect 1 Effect 2 Effect 3 Steam ($T_s$) To Condenser Product ($x_p$)

Evaporator Parameters

Multi-effect evaporators use boiling chambers in series to vaporize water efficiently:

  • Feed flow: Initial liquid flow entering Effect 1.
  • Solute concentrations: Increase from feed ($x_f$) to product ($x_p$).
  • BPE correction: Solute concentration raises boiling point, reducing available $\Delta T$.
  • Steam Economy: Water evaporated per unit steam feed, improving with effect count.

Sizing Inputs

Calculation Results

Results and visual trends will be displayed here upon completion of the computation.

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:

  1. McCabe, W. L., Smith, J. C., & Harriott, P. (2005). Unit Operations of Chemical Engineering (7th ed.). McGraw-Hill.
  2. 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: $Q = 1.359 \text{ kg/s} \cdot 2260 \text{ kJ/kg} = 3071 \text{ kW}$. $$A = \frac{3,071,000 \text{ W}}{2500 \text{ W/m²K} \cdot 22.65\text{ K}} \approx 54.2 \text{ m² (per effect)}$$ $$\text{Total Area } A_{tot} = 3 \cdot 54.2 \approx 162.6 \text{ m²}$$