Energy in Food Processing Study Notes

Energy in Food Processing

Introduction to Energy in Food Processing

• Overview of steam generation and its uses in the food industry.

Definitions of Heat Types

• Latent Heat: The energy (heat) required for a phase transition of a body at constant temperature.

• Sensible Heat: The energy (heat) absorbed or released by a body during a change in temperature.

  • Latent Heat of Water:

  • Range of temperature: from 0°C to 100°C.

  • Latent Heat of Fusion: 335 kJ/kg

  • Latent Heat of Vaporization: 2257 kJ/kg

Detailed Calculation Example: Heating Ice to Steam

• Scenario: Heating 10 kg of ice at -20°C to melt into water at 0°C, and subsequently vaporizing the water into steam at 100°C.

• Key Processes:

  1. Heating ice from -20°C to 0°C (sensible heat)

  2. Melting ice to water at 0°C (latent heat of fusion)

  3. Heating water from 0°C to 100°C (sensible heat)

  4. Vaporizing water to steam at 100°C (latent heat of vaporization)

Required Heat Calculations

• Specific heat of materials:

  • Ice: $c_1 = 2.05 ext{ kJ/(kg°C)}$

  • Water: $c_2 = 4.19 ext{ kJ/(kg°C)}$

• Total Heat Calculation:

  • Heat to raise temperature of ice:

  • $Q<em>1 = m imes c</em>1 imes ext{ΔT} = 10 imes 2.05 imes (0 - (-20)) = 410 ext{ kJ}$

  • Heat for melting ice:

  • $Q_2 = m imes ext{Latent Heat of Fusion} = 10 imes 335 = 3350 ext{ kJ}$

  • Heat to raise temperature of water:

  • $Q<em>3 = m imes c</em>2 imes ext{ΔT} = 10 imes 4.19 imes (100 - 0) = 4190 ext{ kJ}$

  • Heat for vaporizing water:

  • $Q_4 = m imes ext{Latent Heat of Vaporization} = 10 imes 2257 = 22570 ext{ kJ}$

• Total Change in Enthalpy:

  • $Q<em>{total} = Q</em>1 + Q<em>2 + Q</em>3 + Q_4$

  • $= 410 + 3350 + 4190 + 22570 = 30520 ext{ kJ}$

General Heat Exchange Equation

• The principle used for heat exchange:

  • $ext{Heat lost} = ext{Heat gained}$

  • Formula: $m<em>1 c</em>1 ext{Δ}T<em>1 = m</em>2 c<em>2 ext{Δ}T</em>2$

Example 1: Temperature Mixture Calculation

• Scenario: Mixing 3 liters of water at 100°C with 15 liters at 40°C.

• Key Variables:

  • Mass of 1 liter of water = 1 kg

  • Specific heat capacity of water = $4.2 imes 10^3 ext{ J/(kg·K)}$

• Equations:

  • $3 imes 4.2 imes 10^3 imes (100 - T) = 15 imes 4.2 imes 10^3 imes (T - 40)$

  • Solving this gives $T = 50°C$

Example 2: Mixing Ice with Water

• Scenario: Adding 0.30 kg of ice at 0°C to 1.0 kg of water at 45°C.

• Assumptions: No heat exchange with surroundings.

• Key Variables:

  • Specific heat capacity of water = $4200 ext{ J/(kg·K)}$

  • Latent heat of fusion of ice = $3.4 imes 10^5 ext{ J/kg}$

• Heat Exchange Calculation:

  • $1 imes 4200 imes (45 - T) = (0.3 imes 3.4 imes 10^5) + (0.3 imes 4200 imes T)$

  • Rearranging and solving yields: $T = 16°C$

Energy Sources in Food Processing

• Main Energy Sources:

  • Electricity

  • Nuclear Power

  • Natural gas, coal, or oil

• Processes:

  • Water turns to steam which is then applied in the food processing industry.

Steam Generation Systems

Nuclear Steam Generator

• Component Overview:

  • Containment structure

  • Pressurizer

  • Steam generator

  • Reactor vessel

  • Turbine

  • Control rods

  • Generator

  • Condenser

Fire-tube Boiler

• How it Works: Uses fire-tube boiler design to generate steam. Main components include:

  • Steam out

  • Hot gases

  • Furnace

  • Smokestack

Cogeneration of Steam and Electricity

• Cost Analysis on Cogeneration:

  • Standby cost of purchased electricity: $35/h

  • Cost of purchased steam: $100/h

  • Cogeneration costs involve purchase and utilization of fuel:

  • Purchase fuel cost: $105/h

  • Electrical steam and process demand is calculated

  • Efficiency of cogeneration leads to reduction in cost per unit of thermal energy produced.

Steam Properties and Quality Indicators

• Steam Quality:

  • Defined as the percentage of vapor present in a mixture of steam and liquid.

  • Quality expressed as $x_S$ (%), where higher values indicate a greater proportion of vapor.

Example Calculation of Enthalpy at Given Quality

• Scenario: Determine the enthalpy of steam at 120°C and 80% quality.

• Data from steam tables:

  • Heat content of liquid at 120°C = $503.71 ext{ kJ/kg}$

  • Heat content of vapor at 120°C = $2706.3 ext{ kJ/kg}$

• Enthalpy Calculation:

  • $H = 503.71 + 0.8(2706.3 - 503.71) = 2265.78 ext{ kJ/kg}$

Additional Calculation Scenarios

Example with Milk Heating

• Given:

  • Initial temperature: $T_{initial} = 60°C$

  • Final temperature: $T_{final} = 115°C$

  • Mass flow rate: $500 ext{ kg/h}$

  • Specific Heat of Milk: $3.86 ext{ kJ/(kg°C)}$

  • Steam Quality: $0.9$

  • Assume steam temperature is at $120°C$

Another Example: Steam Injection in Food Processing

• Scenario: Heating a liquid food with 12% total solids using steam injection at a pressure of $232.1 ext{ kPa}$.

• Conditions:

  • Product enters heating system at $50°C$, heated to $120°C$.

  • The product flow rate = $100 ext{ kg/min}$

  • Product specific heat = function of composition.

Calculation Requirement

• Determine:

  • Quantity and minimum quality of steam to achieve 10% total solids in the final product.

• Specific Heat Relation:

  • $C<em>{p} = C</em>{p<em>w} imes ( ext{mass fraction H2O}) + C</em>{p_s} imes ( ext{mass fraction solid})$

  • Specific Heat for product at 12% solids = $3.936 ext{ kJ/(kg°C)}$.

Conclusion

• Understanding the energy requirements and heat transfers in food processing enhances efficiency. Steam generation and utilization are critical in achieving processing objectives while maintaining product quality.