Energy Balances in Bioprocess Engineering

Importance of Energy Effects

Energy effects are critical in bioprocesses due to the sensitivity of biological catalysts (such as enzymes and cells) to temperature changes. Proper energy management ensures optimal process conditions and prevents damage to biological systems.

• Heat generation during large-scale reactions can cause cell death or enzyme denaturation if not controlled.

• Energy balances are also essential in processes like steam sterilization.

Basic Energy Concepts

The law of conservation of energy forms the basis for energy accounting in chemical and bioprocess systems. Various forms of energy must be considered:

• Kinetic Energy (Ek): Energy due to motion.

• Potential Energy (Ep): Energy due to position in a field (e.g., gravitational, electromagnetic).

• Internal Energy (U): Sum of all molecular, atomic, and subatomic energies within matter.

Energy Transfer Mechanisms

• Heat (Q): Energy transferred due to a temperature difference between system and surroundings.

• Work (W): Energy transferred by means other than temperature difference.

• Shaft Work (Ws): Work done by moving parts within the system (e.g., impellers).

• Flow Work (Wf): Energy required to push matter into or out of the system.

Flow work is given by:

where p is pressure and V is volume.

Units of Energy

• SI Unit: Joule (J), where 1 J = 1 kg·m2/s2 = 1 N·m

• Calorie (cal): Energy to raise 1 g of water by 1°C at 1 atm. 1 cal = 4.184 J (thermochemical), 1 cal = 4.1868 J (steam table).

• British Thermal Unit (Btu): Energy to raise 1 lb of water by 1°F at 1 atm.

Enthalpy

Enthalpy (H) is a measure of total energy in a thermodynamic system, combining internal energy and the energy required to establish system volume and pressure.

• Specific enthalpy (h): Enthalpy per unit mass:

• Enthalpy is a state function (depends only on the state, not the path).

• Only changes in enthalpy can be measured, not absolute values.

General Energy Balance Equations

Law of Conservation of Energy

The total energy entering a system equals the total energy leaving plus the accumulation within the system. This principle is foundational for energy balance calculations in flow systems.

• Mass enters and exits the system, carrying energy as internal, kinetic, and potential energy.

• Flow work is performed to move fluids in and out.

• Energy can also be exchanged as heat (Q) or shaft work (Ws).

General Energy Balance Equation (Steady-State, No Shaft Work)

For a control volume under steady-state conditions (no accumulation):

Where:

• = mass flow rate

• = specific enthalpy

• = velocity

• = acceleration due to gravity

• = elevation

• = heat transfer rate

• = shaft work rate

For many heat exchanger problems, kinetic and potential energy changes are negligible, and shaft work is zero.

Energy Balances in Heat Exchangers

Application to Heat Exchangers

In double-pipe or shell-and-tube heat exchangers under steady-state and no shaft work:

• Mass flow rates are constant at inlet and outlet for both hot and cold fluids.

• Energy exchange occurs via enthalpy changes in the fluids.

Enthalpy Changes in Heat Exchangers

• Hot fluid:

• Cold fluid:

If there is no phase change (no vaporization or condensation), enthalpy changes are due to temperature changes (sensible heat):

• = specific heat capacity

• = temperature change

Heat Exchanger Calculations

For a heat exchanger with no heat loss to the environment:

• = heat transferred

• = mass flow rates of hot and cold fluids

• = specific heat capacities

• = inlet and outlet temperatures of hot fluid

• = inlet and outlet temperatures of cold fluid

Log-Mean Temperature Difference (LMTD)

The LMTD is used to account for the changing temperature difference between hot and cold fluids along the length of the heat exchanger.

• = temperature difference at one end

• = temperature difference at the other end

Heat Transfer Area Calculation

The required heat transfer area (A) is calculated using:

• = overall heat transfer coefficient

• = heat transfer area

• = log-mean temperature difference

Example Problems

Example 1: Cooling a Fermentation Medium

• Given: Medium cooled from 371.9 K to 349.7 K in a double-pipe heat exchanger. Flow rate: 3630 kg/h. (medium): 2.30 kJ/kg·K. Cooling water enters at 288.6 K, flow rate: 1450 kg/h, $ c_p $ (water): 4.187 kJ/kg·K.

• Tasks:

  • Calculate water outlet temperature (countercurrent flow).

  • Calculate log-mean temperature difference.

  • Calculate heat transfer area for W/m2·K.

  • Repeat for parallel (co-current) flow.

Solution Steps:

1. Calculate heat lost by medium:

2. Set equal to heat gained by water:

3. Solve for unknown water outlet temperature.

4. Calculate and for LMTD.

5. Calculate area:

Example 2: Cooling Sterile Nutrient Medium

• Given: Medium enters at 90°C, flow rate: 2800 g/s. Cooling water enters at 15°C, volumetric flow rate: 36 m3/h, density: 1000 kg/m3. Medium has properties of water. Steady-state operation.

• Tasks:

  • Calculate required rate of heat transfer (kW) to cool medium to 25°C.

  • Find exit temperature of cooling water.

Solution Steps:

1. Convert all units to SI (kg, s, K).

2. Calculate

3. Set and solve for water exit temperature.

Summary Table: Key Energy Quantities and Units

Additional info: The above notes expand on the basic thermodynamic principles relevant to general chemistry and bioprocess engineering, including the application of energy balances to heat exchangers. The example problems illustrate practical calculations using these principles.