INTRODUCTION TO FOOD ENGINEERING

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Presentation transcript:

INTRODUCTION TO FOOD ENGINEERING Lecture 6 Unsteady-State Heat Transfer

Objectives Able to calculate temperature at various time

Unsteady-state Heat Transfer Temperature changes with time and location Important in thermal processes Pasteurization Sterilization Governing Equation

External & internal resistance to heat transfer Convective resistance Conductive resistance

Biot number (4.73) (4.74) (4.75)

Characteristic dimension

Biot number NBi > 40 negligible surface resistanc (h higher than k) NBi < 0.1 negligible internal resistance (k higher than h) 0.1 < NBi < 40 finite internal & external resistance

Negligible internal resistance (NBi < 0.1) Heating & cooling of solid metals (high k) No temp gradient with location Well-stirred liquid food in a container Ta = temp of surrounding medium A = surface area of the object

Negligible internal resistance (NBi < 0.1) Separating variables, integration

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.79) (4.80)

Example Heating tomato juice from 20 C, Ta = 90 C Kettle radius = 0.5 m Cp tomato juice 3.95 kJ/kg C Density 980 kg/m3 Time of heating = 5 min A = 2r2 = 1.57 m2 V = 2/3 r3 = 0.26 m3

T = 83.3 C after 5 min of heating

Finite internal and surface resistance 0.1 < NBi < 40) Solution of equation are very complicated Use instead temperature-time charts for sphere, infinite cylinder and infinite slab D is characteristic dimension (sphere, D is radius)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

Negligible surface resistance (NBi > 40) Use temp-time chart Line k/hD = 0 (4.82)

Finite objects (4.83)

Procedures 1. Calculate NFo using characteristic dimension for infinite cylinder and infinite slab 2. Calculate Biot number 3. Use temp-time chart to find temperature ratio 4. Calculate temp ratio for finite object then calculate T (temp at geometric center)

Example Estimate temp at geometric center of food in 303 x 406 can in boiling water for 30 min Given can dimension 3 3/16 in = 0.081 m H = 4 6/16 in = 0.11 m h = 2000 W/m2C Ta = 100 C Ti = 35 C t = 30 min = 1800 s Food properties k = 0.34 W/m C Cp = 3.5 kJ/kg C  = 900 kg/m3

Infinite cylinder (from chart)

Infinite slab Close to 1, T = Ti

= 48.4 C Most of heat transfers radially

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.84)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.85) (4.86)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING Microwave Heating (4.87)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING 1. Mechanisms of Microwave Heating

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING 2. Dielectric Properties (4.88) 3. Conversion of Microwave Energy into Heat (4.89) (4.90)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING 4. Penetration Depth of Microwaves (4.91) (4.92) (4.93)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING 5. Microwave Oven 6. Microwave Heating of Foods