HEAT TRANSFER. In FOOD PROCESSING

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1 HE RNSFER In FOOD PROCESSING Lecture Note Purwiyatno Hariyadi Dept of Food Science & echnology Faculty of gricultural Engineering & echnology Bogor gricultural University BOGOR 208 Heat ransfer Heat transfer - movement of energy due to a temperature difference Can only occur if a temperature difference exists Occurs through:. conduction, 2. convection, and 3. radiation, or 4. combination of above --

2 Heat ransfer May be indicated as total transfer Identified by total heat flow (Q) with units of Btu Identified by rate of heat flow () or Q/t with units of watts ot Btu/hr lso, may be expressed as heat transfer per unit area = heat flux or / Heat ransfer Heat transfer may be classified as:. Steady-state: o all factors are stabilized with respect to time o temperatures are constant at all locations o steady-state is sometimes assumed if little error results 2. Unsteady-state (transient) heat transfer occurs when: o temperature changes with time o thermal processing of foods is an important example o must know time reuired for the coldest spot in can to reach set temperature -- 2

3 CONDUCION HE RNSFER Occurs when heat moves through a material (usually solid or viscous liuid) due to molecular action only HE Heat/energy is trasfered at molecular level No physical movement of material Heating/cooling of solid Heat flux is directly proportional to the temperature gradient, and inversely proportional to distance (thickness of material). CONDUCION HE RNSFER May occur simultaneously in one, or two, or three directions Many practical problems involve heat flow in only one or two directions Conduction along a rod heated at one end is an example of two dimensional conduction Heat flows along the length of the rod to the cooler end (one direction) If rod is not insulated, heat is also lost to surroundings Center warmer than outer surface -- 3

4 CONDUCION HE RNSFER - one dimensional One dimensional conduction heat transfer is a function of:. temperature difference, 2. material thickness, 3. area through which heat flows, and 4. resistance of the material to heat flow CONDUCION HE RNSFER - one dimensional Fourier s Law Of Heat Conduction: Q x t = = - k d dx x X X Q = otal heat flow 2 x = rate of heat flow in x direction by conduction, W k = thermal conductivity, W/mC = area (normal to x-direction) through which heat flows, m 2 = temperature, C x = distance increment, variable, m -- 4

5 SIGN CONVENION EMPERURE direction of heat flow x d slope- dx emperature profile DISNCE -- 5

6 2 X Heat ransfer 2/27/208 USING FOURIER S LW X X 2 d x = -k dx BC : X = X... > = X = X... 2 > = 2 x dx -kd = x Integrating : X x dx - kd X x ( x - x ) k( - 2 ( x - x ) k ( - ) k x (X - X ) ) HE CONDUCION IN MULILYERED SYSEMS Composite Rectangular Wall (In Series) k k B k C emperature x x B x C emperature profile in a multilayered system -- 6

7 k k B k C x x B USING FOURIER S LW : d -k dx x - k x C B C x k k x k B x C B C k k B k C 2 - x 2 X k B x B C X k B B x C X k C C B C x k k x k B x C B C -- 7

8 CONDUCION IN CYLINDRICL OBJECS Fourier s law in cylindrical coordinates r -k r -k d dr 2 r L d dr Boundary Conditions : = i at r = r i = o at r = r o dr r o Integrating : r dr o k o d 2L r r i i r o o Ln r k 2L r i i 2 Lk( i o ) r ln o ri r i COMPOSIE CYLINDRICL UBE r 3 i r r 2 o FROM FOURIER S LW: r 2Lk(i o ) ro ln ri -- 8

9 =? Let us define logarithmic mean area m such that (i o ) r k m (ro ri ) (ro ri ) where m 2L ro ln r i i o r (ro ri ) k m 2 i 2 3 o r (k r r r 2 r 3 (r 2 (r (k adding above two euations ( ) 2 r r r k m k 2 3 r ) m m ) m 2 ) r 2 23 ) 23 Convection Heat ransfer ransfer of energy due to the movement of a heated fluid Movement of the fluid (liuid or gas) causes transfer of heat from regions of warm fluid to cooler regions in the fluid Natural Convection occurs when a fluid is heated and moves due to the change in density of the heated fluid Forced Convection occurs when the fluid is moved by other methods (pumps, fans, etc.) -- 9

10 CONVECIVE HE RNSFER : heat transfer to fluid a < s Surface area = s = h ( s - a ) = rate of heat transfer h = convective heat transfer coefficient, W/m 2. o C s = surface temperature a = surrounding fluid temperature Natural Convection Colder fluid (higher density) Fluid absorbs heat (temperature increase: density decrease) -- 0

11 HE RNSFER O FLUID FLUID FLOW IN PIPE Fluid flow can occur as - laminar flow - turbulent flow - transition between laminar and turbulent flow - direction of flow.. > parallel or perpendicular to the solid object HE RNSFER O FLUID > h? = h ( s - a ) h = f (density, velocity, diameter, viscosity, specific heat, thermal conductivity, viscosity of fluid at wall temperature he convective heat transfer coefficient is determined by dimensional analysis. series of experiment are conducted to determine relationships between following dimensionless numbers. --

12 HE RNSFER O FLUID > h? Dimensionless Numbers In Convective Heat ransfer Nusselt Number = N nu = (hd)/k Prandtl Number = N Pr = C p /k Reynolds Number = Re = (vd)/ Where D = characteristic dimension k = thermal conductivity of fluid v = velocity of fluid C p = specific heat of fluid = density of fluid = viscosity of fluid HE RNSFER O FLUID. > FORCED CONVECION N nu = f (N Re, N Pr ) Laminar flow in pipes: If N Re <200 For (N Re x N Pr x D/L) < 00 N Nu N N Re Re xn xn Pr Pr D x L D x L 0.66 For (N Re x N PR x D/L) > N Nu.86 N RE xn PR D x L 0.33 b w ll physical properties are evaluated at bulk fluid temperature, except m w b w

13 HE RNSFER O FLUID. > FORCED CONVECION ransition Flow in Pipes N RE between 200 and 0,000: use chart to determine h : diagram J Colburn factor (J) vs Re. h Cp. 2 w J CpV k HE RNSFER O FLUID. > FORCED CONVECION urbulent Flow in Pipes:. > N RE > 0,000: N NU N 0.33 Pr x b w

14 HE RNSFER O FLUID. > FREE CONVECION Free convection involves the dimensionless number called Grashof Number, N Gr N G r N Nu 3 2 ( ) D g = 2 m h D = = a G r k ( N N ) m Pr = koeff ekspansi volumetrik (koef muai volumetrik; /K) a and m = constant ll physical properties are evaluated at the film temperature. > f = ( w + b )/2 HE RNSFER O FLUID. > FREE CONVECION N Nu h D = = a k ( N N ) m G r Value of a and m =f(physical configuration) Pr Vertical surface D=vertical dim. < mn Gr N Pr <0 4 a=.36 m=/5 Horizontal cylinder D = dia < 20 cm N Gr N Pr <0-5 a=0.49 m=0 0-5< N Gr N Pr < a=0.7 m=/25 <N Gr N Pr <0 4 a=,09 m=/0 Horizaontal flat surface Facing Upward 0 5 < N Gr N Pr <2x0 7 a=0.54 m=/4 2x0 7 < N Gr N Pr <3x0 0 a=0.4 m=/3 Facing downward 3x0 5 < N Gr N Pr <3x0 0 a=0.27 m=/4 -- 4

15 HE RNSFER O FLUID > U? emperature profile : conductive and convective heat transfer through a slab a 2 b Q = U( a - b ) where U = Overall heat transfer coefficient [=] W/m 2 C h i h o HE RNSFER O FLUID > U? Steady State : i = x = o = a = U( a - b ) i ==h i ( a - ) x ==k(-2)/x o ==h o ( 2 - b ) a - b = ( a - )+ - 2 )+( 2 - b ) h i h o 2 b = + x + U h i k = + x + U h k i h O h O tau, umum : U i i = h i i + i = lm = o = x k lm + h O O -- 5

16 HE RNSFER O FLUID > U? a r 2 a r 2 b h i h o Surounding fluid temp; b < a U = h + r k + i h O tau, umum : U i lm i = h i o ln i + r k i - o i lm + h O O -- 6

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20 RNSIEN (UNSEDY-SE) HE RNSFER -- 20

21 Boiling water 00 o C Solid food material s, initial =35 o C r Change in temperature?? s = f(t,r) -- 2

22 Importance of internal and external resistance to heat transfer relative importance of conductive and conventive heat transfer Biot number, N Bi = hd/k D / k N Bi = / h or NBi = Internal External resistance resistant Boiling water 00 o C r to heat transfer to heat transfer Negligible internal resistance. >N Bi < 0. = V Cp d/dt = h ( a -) d - a = h d t C V p ln( a a a ) - - o i = e t h t C p V - (h / C V) t p

23 Finite Surface and Internal Resistance o Heat ransfer. > 0.<N Bi < 40.. > m=/n Bi Negligible Surface Resistance o Heat ransfer. > N Bi > 40.. > m=/n Bi = 0 Infinite Slab, infinite cylinder and sphere Use Gurnie-Lurie Chart and/or Heisler Chart > temperature-time (-t) chart Dimensionless number : Fourier number (N Fo ) N Fo = kt C p D 2 t = 2 D D = characteristic dimension D sphere = radius D inf cylinder = radius D inf slab = half thickness he physical meaning of Fourier Number : 2 k D t D N Fo 2 3 D CpD t Rate N Fo = of heat conduction across D in volume Rate of heat storage in volume D 3 D 3 (W/C) (W/C) Large value of N Fo indicates deeper penetration of heat into solid in a given period of time -- 23

24 Prosedur pengunaan diagram -t. Untuk silinder tak berbatas R - Suhu pusat (sumbu) silinder setelah pemanasan selama t? a. hitung N Fo, gunakan R sebagai D b. hitung N Bi, gunakan R sebagai D > hitung /N Bi =m=k/hd c. gunakan diagran untuk silinder tak berbatas, dari N Fo dan N Bi cari ratio /N bi = m N Fo Diagram -t : hubungan antara suhu di sumbu silinder dan N Fo -- 24

25 2. Untuk lempeng tak berbatas ketebalan, X = 2D lebar = ; panjang = ebal=x Suhu ditengah (midplane) lempeng tak berbatas setelah pemanasan selama t?? a. hitung N Fo, gunakan (/2)X sebagai D b. hitung N Bi, gunakan (/2)X sebagai D > hitung /N Bi c. gunakan diagram untuk lempengtak berbatas, dari N Fo dan N Bi cari ratio Diagram -t : hubungan suhu di midplane lempeng tak berbatas dan N Fo -- 25

26 Diagram -t : hubungan antara suhu di pusat bola dan N Fo Diagram Gurnie-Lurie untuk LEMPENG :. Menentukan suhu setelah pemanasan/pendinginan cari nilai N Fo =t/ 2 cari nilai N bi dan m=/n bi tentukan posisi dimana suhu ingin diketahui, n = x/ cari ratio suhu 2. Menentukan waktu pemanasan/pendinginan untuk mencapai suhu ttt cari rasio suhu, pada posisi ttt yang diketahui, n = r/r cari nilai N Bi dan m=/n bi cari N Fo = t/ 2 ; dan hitung t -- 26

27 Diagram Gurnie-Lurie untuk SILINDER :. Menentukan suhu setelah pemanasan/pendinginan cari nilai N Fo =t/r 2 cari nilai N bi dan m=/n bi tentukan posisi dimana suhu ingin diketahui, n = r/r cari ratio suhu 2. Menentukan waktu pemanasan/pendinginan untuk mencapai suhu ttt cari rasio suhu, pada posisi ttt yang diketahui, n = r/r cari nilai N bi dan m=/n bi cari N fo =t/r 2 ; dan hitung t Diagram Gurnie-Lurie untuk BOL :. Menentukan suhu setelah pemanasan/pendinginan cari nilai N Fo =t/r 2 cari nilai N bi dan m=/n bi tentukan posisi dimana suhu ingin diketahui, n = r/r cari ratio suhu 2. Menentukan waktu pemanasan/pendinginan untuk mencapai suhu ttt cari rasio suhu, pada posisi ttt yang diketahui, n = r/r cari nilai N bi dan m=/n bi cari N fo =t/r 2 ; dan hitung t -- 27

28 Diagram Gurnie-Lurie :(oledo) -- 28

29 Finite object. > finite slab (bentuk bata, panjang=l, lebar=w, tinggi=h) a a a i a i Finite slab, l,w,h x Inf. Slab l a a i x Inf slab, w a a i Inf slab, h length depth width -- 29

30 Finite object. > finite slab (bentuk kaleng, jari-jari=r, tinggi=h) Infinite cylinder, radius R Infinite slab, thickness=h a a i = Finite cylinder R, h a a i Infinite cylinder R x a a i Infinite slab (h) Penentuan posisi pada benda berbatas R? Lokasi : tengah tutup kaleng - ditengah silinder : n=0 - dipermukaan lempeng: n= X=/2 r= /2R X? Lokasi x - n silinder = r/r=/2 - n lempeng = x/ = /

31 CONOH SOL pel didinginkan dari suhu 20 o C menjadi 8 o C, dengan menggunakan air dingin mengalir (5 o C). liran air dingin ini memberikan koef. Heat ransfer konvensi sebesar 0 M/m 2.K. sumsikan apel sebagai bola dengan diamater 8 cm. Nilai k apel = 0.4 W/m/K, Cp apel= 3.8 kj/kg.k dan densitasnya=960 kg/m 3. Untuk pusat geometri apel mencapai suhu 8 o C, berapa lama harus dilakukan pendinginan? Jawab :. Cek N Bi ; apakah nilainya <0.? 0,<N Bi <40? atau N Bi >40?? N Bi = (hr/k)= > 0.<N Bi <40 : gunakan diagram -t (m=/n Bi =) 2. Hitung rasio suhu yang dikehendaki : (a-)/(a-i)=(5-8)/(5-20)= Posisi? Di pusat geometri. > n=0 4. Cari nilai N Fo, dan tentukan t θ=0.2 N Fo =t/r 2 =0.78 t = 0.78R 2 / t = 0.78R 2 /[k/(.cp)] t = 0.78(0.04) 2 /[0.4/(960)(3800)] t =,38 s t = 3.6 h N Fo =t/r 2 =

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