1. INTRODUCTION TO FOOD ENGINEERING
Lecture 6
Unsteady-State Heat Transfer

2. Objectives
Able to calculate temperature at various time

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

4. External & internal resistance to heat transfer
Convective resistance
Conductive resistance

5. Biot number
(4.73)
(4.74)
(4.75)

6. Characteristic dimension

7. 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

8. 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

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

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

11. 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

12. T = 83.3 C after 5 min of heating

13. 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)

14. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

15. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

16. CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

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

18. Finite objects
(4.83)

20. 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)

21. 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 = 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

22. Infinite cylinder
(from chart)

23. Infinite slab
Close to 1, T = Ti

24. = 48.4 C
Most of heat transfers radially

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

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

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

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

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

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

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