INTRODUCTION TO FOOD ENGINEERING Lecture 5 HEAT TRANSFER IN FOOD PROCESSING

Objectives Calculate convective heat transfer coefficient Calculate overall heat transfer coefficient Calculate heat transfer area in tubular heat exchanger

Estimation of Convective Heat-Transfer Coefficient h is predicted from empirical correlation for Newtonian fluids only Forced convection

Forced Convection NNu = Nusselt number NRe = Reynold number NPr = Prandtl number

Larminar flow in pipes NRe < 2100 For (4.38) b = bulk, w = wall

For (4.39)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.41) (4.42)

Free Convection (4.43)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

Example Water flowing at 0.02 kg/s is heated from 20 to 60 C in a horizontal pipe (D = 2.5 cm). Inside T = 90 C. Estimate h if the pipe is 1 m long. Average T = (20+60)/2 = 40 C  = 992.2 kg/m3, cp = 4.175 kJ/kg C k = 0.633 W/m C,  = 658.026 x 10-6 Pa.s NPr = cp/k = 4.3, w is  at 90 C

= 1547.9 laminar flow = 166.4 > 100 NNu = 11.2

= 284 W/m2 C

Turbulent flow in pipes

Estimation of Overall Heat-Transfer Coefficient Conduction + Convection

If temperature of fluid in pipe is higher Heat flows to outside Ti > T Ui = overall heat transfer coefficient based on inside area

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING Convection from inside Conduction Convection to outside

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.48) (4.49) (4.50)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.51) (4.52) (4.53) (4.54)

Example A steel pipe (k = 43 W/mC) inside D = 2.5 cm, 0.5 cm thick, conveys liquid food at 80 C. Inside h = 10 W/m2C. Outside temp = 20 C, outside h = 100 W/m2C. Calculate overall heat transfer coefficient and heat loss from 1 m length of pipe.

ro = 0.0175 m Ri = 0.0125 m rlm = 0.01486 m 1/Ui = 0.10724 m2 C/W Ui = 9.32 W/m2 C Heat loss q = UiAi(80 – 20) = 43.9 W Uo = 6.66 W/m2 C q = 43.9 W

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING 6. Role of Insulation in Reducing Heat Loss from Process Equipment (4.55) (4.56)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.57) (4.58)

Design of a Tubular Heat Exchanger Determine desired heat-transfer area for a given application. Assuming Steady-state conditions Overall heat-transfer coefficient is constant throughout the pipe length No axial conduction of heat in metal pipe Well insulated, negligible heat loss

Design of Tubular Heat Exchanger Heat transfer from one fluid to another Energy balance for double-pipe heat exchanger (4.59) (4.60) (4.61)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING Slope of T line (4.62) (4.63)

CHAPTER 3  HEAT TRAMSFER IN FOOD PROCESSING (4.64) (4.65)

Example A liquid food (Cp = 4.0 kJ/kgC) flows in inner pipe of heat exchanger. The food enters at 20 C and exits at 60 C. Flow rate = 0.5 kg/s. Hot water at 90 C enters and flows countercurrently at 1 kg/s. Average Cp of water is 4.18 kJ/kgC. Calculate exit temp of water Calculate log-mean temperature difference If U = 2000 W/m2C and Di = 5 cm calculate L. Repeat calculations for parallel flow.

Liquid food Inlet temp = 20 C Exit temp = 60 C Cp = 4.0 kJ/kg C Flow rate = 0.5 kg/s Water Inlet temp = 90 C exit temp = ? Cp = 4.18 kJ/kgC Flow rate = 1.0 kg/s

q = mcCpc Tc = mhCph  Th Tc = 70.9 C Tlm = 39.5 C q = UA(T)lm = UDiL(T)lm = mCp T = 80 kJ/s L = 6.45 m For parallel flow L = 8 m