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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.
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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-TransferCoefficient • 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 (4.41) (4.42)
Free Convection (4.43)
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
Estimation of Overall Heat-TransferCoefficient • 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/mC) inside D = 2.5 cm, 0.5 cm thick, conveys liquid food at 80 C. Inside h = 10 W/m2C. Outside temp = 20 C, outside h = 100 W/m2C. 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 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 = mcCpcTc = mhCph Th • Tc = 70.9 C • Tlm = 39.5 C • q = UA(T)lm = UDiL(T)lm • = mCp T = 80 kJ/s • L = 6.45 m • For parallel flow L = 8 m