Material and Energy Balance Fundamentals

Reference ID: MET-3FCA | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The single-effect evaporation calculation is a fundamental process engineering operation used to concentrate a dilute solution by boiling off a solvent, typically water, and you can learn more about the detailed methodology in our guide on extract concentration by evaporation. This process is critical in industries such as food processing, chemical manufacturing, and wastewater treatment. By applying mass and energy balances, engineers can determine the required steam consumption, vapor production rates, and the overall thermal duty of the evaporator. This specific model assumes a cold feed scenario, where the feed temperature is lower than the boiling point of the concentrate, necessitating both sensible heating and latent heat for phase change.

Methodology & Formulas

The calculation follows a sequential approach to determine the mass distribution and energy requirements of the system, and for a detailed methodology you can refer to the batch distillation time calculation guide.

1. Mass Balance and Concentration Ratio
The concentration ratio R defines the degree of concentration, while the vapor flow rate V and concentrate flow rate C are derived from the feed flow F and mass fractions x_F and x_C:

\[ R = \frac{x_{C}}{x_{F}} \] \[ V = F \cdot \left( 1 - \frac{1}{R} \right) \] \[ C = F - V \]

2. Energy Balance
The total energy required is the sum of the sensible heat needed to bring the feed to the boiling temperature and the latent heat required for vaporization. The specific heat of the feed c_p,F is adjusted based on the solids content:

\[ c_{p,F} = 4.18 \cdot (1 - 0.006 \cdot (x_{F} \cdot 100)) \] \[ Q_{\text{sen}} = F \cdot c_{p,F} \cdot (T_{C} - T_{F}) \] \[ Q_{\text{vap}} = V \cdot \lambda_{V} \] \[ Q_{\text{net}} = Q_{\text{sen}} + Q_{\text{vap}} \]

3. Heat Loss and Steam Consumption
Accounting for thermal losses (q_loss), the gross heat duty Q_gross and the required steam mass flow S are calculated using the latent heat of the steam λ_S:

\[ Q_{\text{gross}} = \frac{Q_{\text{net}}}{1 - q_{\text{loss}}} \] \[ S = \frac{Q_{\text{gross}}}{\lambda_{S}} \] \[ \text{Steam Economy} = \frac{V}{S} \]

Parameter	Constraint/Regime
Concentration Ratio (R)	R ≤ 5.0
Temperature Driving Force (T_S - T_C)	5.0 ≤ (T_S - T_C) ≤ 20.0 °C
Steam Economy	< 1.0
Feed Condition	T_F < T_C (Cold Feed)

To establish an effective system boundary for material and energy balances, follow these steps:

Identify the specific process unit or subsystem where you have the most reliable data.
Ensure the boundary encompasses all relevant input and output streams, including recycle loops and purge streams.
Verify that the boundary is drawn such that the accumulation term in the general balance equation is either zero for steady-state or clearly defined for transient analysis.

Handling recycle streams requires a systematic approach to avoid algebraic complexity:

Perform a global balance around the entire process to determine overall conversion and yield.
Use the 'tear stream' method, where you assume a value for the recycle flow rate and composition, then iterate until the calculated values converge.
Apply the 'basis of calculation' technique, such as assuming 100 units of fresh feed, to simplify the stoichiometry.

When calculating energy balances for non-reactive systems, focus on the enthalpy changes associated with temperature and phase transitions:

Select a consistent reference state, typically 25 degrees Celsius and 1 atmosphere, for all enthalpy calculations.
Use the heat capacity integral for sensible heat changes: ΔH = ∫ C_p dT.
Include latent heat terms for any phase changes occurring within the process boundary.
Account for work done by or on the system, such as pump or compressor power requirements.

Worked Example: Single-Effect Evaporation of Fruit Juice

Scenario: A single-effect evaporator is used to concentrate a fruit juice from 12°Bx (mass fraction 0.12) to 48°Bx (mass fraction 0.48). The feed is at 20°C and the evaporator operates with a boiling temperature of 70°C under vacuum. Steam is supplied at 85°C. All calculations follow the standard material and energy balances for a cold feed, with a heat loss of 2% of the gross duty.

Known Parameters: (All values taken from the provided numerical results dictionary)

Feed flow rate, \( F = 1.5\ \text{kg/s} \)
Feed concentration, \( x_F = 0.12 \) (mass fraction solids)
Feed temperature, \( T_F = 20.0\ ^\circ\text{C} \)
Concentrate concentration, \( x_C = 0.48 \)
Boiling temperature of concentrate, \( T_C = 70.0\ ^\circ\text{C} \)
Latent heat of vapor at \( T_C \), \( \lambda_V = 2333.0\ \text{kJ/kg} \)
Steam temperature, \( T_S = 85.0\ ^\circ\text{C} \)
Latent heat of steam at \( T_S \), \( \lambda_S = 2279.0\ \text{kJ/kg} \)
Heat loss fraction, \( q_{\text{loss}} = 0.02 \) (2%)
Specific heat of feed, \( c_{p,F} = 4.18 \cdot (1 - 0.006 \times 12) = 3.879\ \text{kJ/(kg·K)} \)

Step-by-Step Calculation:

Concentration ratio \[ R = \frac{x_C}{x_F} = \frac{0.48}{0.12} = 4.0 \]
Vapor generation rate \[ V = F \left(1 - \frac{1}{R}\right) = 1.5 \times \left(1 - \frac{1}{4}\right) = 1.125\ \text{kg/s} \]
Concentrate flow rate \[ C = F - V = 1.5 - 1.125 = 0.375\ \text{kg/s} \]
Sensible heat duty (to raise feed from \( T_F \) to \( T_C \)) \[ Q_{\text{sen}} = F \cdot c_{p,F} \cdot (T_C - T_F) = 1.5 \times 3.879 \times (70.0 - 20.0) = 290.925\ \text{kW} \]
Latent heat duty \[ Q_{\text{vap}} = V \cdot \lambda_V = 1.125 \times 2333.0 = 2624.625\ \text{kW} \]
Net heat duty (without loss) \[ Q_{\text{net}} = Q_{\text{sen}} + Q_{\text{vap}} = 290.925 + 2624.625 = 2915.550\ \text{kW} \]
Gross heat duty (including 2% heat loss) \[ Q_{\text{gross}} = \frac{Q_{\text{net}}}{1 - q_{\text{loss}}} = \frac{2915.550}{0.98} = 2975.051\ \text{kW} \]
Heat loss \[ Q_{\text{loss}} = Q_{\text{gross}} - Q_{\text{net}} = 2975.051 - 2915.550 = 59.501\ \text{kW} \]
Steam consumption \[ S = \frac{Q_{\text{gross}}}{\lambda_S} = \frac{2975.051}{2279.0} = 1.305\ \text{kg/s} \]
Steam economy \[ \text{Steam Economy} = \frac{V}{S} = \frac{1.125}{1.305} = 0.862\ \text{kg water/kg steam} \]

Final Results:

Concentrate flow: \( C = 0.375\ \text{kg/s} \)
Required steam supply: \( S = 1.305\ \text{kg/s} \)
Steam economy: 0.862 kg water evaporated per kg steam

Validity Checks: The concentration ratio \( R = 4.0 \) is within the empirical limit of 5.0. The temperature difference \( T_S - T_C = 15.0\ ^\circ\text{C} \) lies between 5 and 20°C. The steam economy is below 1.0, which is consistent with a single-effect evaporator. Feed temperature is below the boiling point, satisfying the cold-feed assumption. All values are taken directly from the numerical results dictionary.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle