HEAT TRANSFER. In FOOD PROCESSING

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phariyadi.staff.ipb.ac.id -- ITP530

HEAT TRANSFER

In FOOD PROCESSING

HEAT TRANSFER

In FOOD PROCESSING

Lecture Note ITP530 Purwiyatno Hariyadi phariyadi.staff.ipb.ac.id

Dept of Food Science & Technology

Faculty of Agricultural Engineering & Technology Bogor Agricultural University

BOGOR 2018

• Heat transfer - movement of energy due to a temperature difference

• Can only occur if a temperature difference exists • Occurs through:

1. conduction, 2. convection, and 3. radiation, or

4. combination of above

• Heat transfer - movement of energy due to a temperature difference

• Can only occur if a temperature difference exists • Occurs through: 1. conduction, 2. convection, and 3. radiation, or 4. combination of above

Heat Transfer

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• May be indicated as total transfer

• Identified by total heat flow (Q) with units of Btu • Identified by rate of heat flow (q) or Q/t with units

of watts ot Btu/hr

• Also, may be expressed as heat transfer per unit area = heat flux or q/A

• May be indicated as total transfer

• Identified by total heat flow (Q) with units of Btu • Identified by rate of heat flow (q) or Q/t with units

of watts ot Btu/hr

• Also, may be expressed as heat transfer per unit area = heat flux or q/A

Heat Transfer

• Heat transfer may be classified as: 1. 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 required for the coldest spot in can to reach set temperature

• Heat transfer may be classified as: 1. 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 required for the coldest spot in can to reach set temperature

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• Occurs when heat moves through a material (usually solid or viscous liquid) due to molecular action only

CONDUCTION HEAT TRANSFER

HEAT HEAT

• 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).

• 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).

•

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

•

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

CONDUCTION HEAT TRANSFER

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• One dimensional conduction heat transfer is a function of:

1. temperature difference, 2. material thickness,

3. area through which heat flows, and 4. resistance of the material to heat flow

• One dimensional conduction heat transfer is a function of:

1. temperature difference, 2. material thickness,

3. area through which heat flows, and 4. resistance of the material to heat flow - one dimensional

- one dimensional

CONDUCTION HEAT TRANSFER

X₁

X₁ XX₂₂

q_x q_x

CONDUCTION HEAT TRANSFER

Fourier’s Law Of Heat Conduction: Fourier’s Law Of Heat Conduction:

Q = Total heat flow

q_x= rate of heat flow in x direction by conduction, W k = thermal conductivity, W/mC

A= area (normal to x-direction) through which heat flows, m2

T = temperature, C

x = distance increment, variable, m Q = Total heat flow

q_x= rate of heat flow in x direction by conduction, W k = thermal conductivity, W/mC

A= area (normal to x-direction) through which heat flows, m2

T = temperature, C

x = distance increment, variable, m

dx

dT

kA

--=

=

q

_x

q

_x

Q

t

Q

t

=

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SIGN CONVENTION

direction of heat flow

T

x

TEMPERA

TUR

E

TEMPERA

TUR

E

DISTANCE

dx dx dT dT --slope slope Temperature profile Temperature profile

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USING FOURIER’S LAW

dx

dT

- k A

q

x

=

BC : X = X₁..._{> T = T} 1 X = X₂..._{> T = T} 2

-kdT

dx

A

q

_x

=





T T X X x 1 1

kdT

-dx

A

q

Integrating : X₁ X₁ XX₂₂ q_x q_x

)

x

-x

(

kA

q

T

1 ₁ 1





)

X

-(X

)

T

-(T

kA

q

1 1 x





)

T

-k(T

)

x

-x

(

q

_x

A

1





2 q

Composite Rectangular Wall (In Series) k_A x_A k_B x_B k_C x_C q

HEAT CONDUCTION IN MULTILAYERED SYSTEMS

Temperature profile in a multilayered system T₁ T₂ X Te mperature

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phariyadi.staff.ipb.ac.id -- ITP530 q k_A x_A k_B x_B k_C x_C q

dX

dT

-kA

q 

kA

x

-q

T

_



USING FOURIER’S LAW :

k

A

x

-q

T

A A A







A

k

x

-q

T

B B B







A

k

x

-q

T

C C C







             A B C 2 1 k X k X k X A q -T T       T T_A T_B T_C   T T₁ T₂ q k_A x_A k_B x_B k_C x_C q

A

k

x

-q

T

A A A







A

k

x

-q

T

B B B







x

-q

T

_



C



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CONDUCTION IN CYLINDRICAL OBJECTS

Fourier’s law in cylindrical coordinates

dr

dT

- kA

q

_r

_

dr

dT

r L

2 

-k

q

_r

_

Boundary Conditions :

T = T

_i

at

r = r

_i

T = T

_o

at

r = r

_o   i o r       i o r ln ) T Lk(T 2 q



  o i T T dT k r dr 2L q _r o r_i i T kT Ln r 2L q   r_o T_o r_i Integrating : r_i r_o dr

COMPOSITE CYLINDRICAL TUBE

r₁

r₂

r₃

T_i

T_o

FROM FOURIER’S LAW:

















i o o i r

r

ln

)

T

Lk(T

2 

q

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A =?

Let us define logarithmic mean area A_m such that i o o i m r

)

r

(r

)

T

(

kA

q





)

r

(r

_

i o i o m

r

ln

L

2

A















where m i o r o i

kA

)

r

(

q

T

_

_



r1 r2 r3 Ti To 23 m 2 3 r 3 2 ) (kA ) r (r q T T    12 m 1 2 r 2 1 ) (kA ) r (r q T T _ _ 

adding above two equations

                 m 12 m 2 1 r kA r kA r ) T T ( q 23

Convection Heat Transfer

• Transfer of energy due to the movement of a heated fluid

• Movement of the fluid (liquid 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.)

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CONVECTIVE HEAT TRANSFER : heat transfer to fluid

q = h A(T

_s

- T

_a

)

q

q = rate of heat transfer

Surface area = A

h = convective heat transfer coefficient, W/m2_.o_C

T_s

T_s= surface temperature T_a< T_s

T_a= surrounding fluid temperature

Fluid absorbs heat (temperature increase: density decrease) Colder fluid (higher density)

Natural

Convection

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FLUID FLOW IN A 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

HEAT TRANSFER TO FLUID

q = h A (T_s- T_a)

h = f (density, velocity, diameter, viscosity, specific heat, thermal conductivity, viscosity of fluid at wall temperature



The convective heat transfer coefficient is determined by dimensional analysis.



A series of experiment are conducted to determine relationships between following dimensionless numbers.

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Dimensionless Numbers In Convective Heat

Transfer

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

HEAT TRANSFER TO FLUID

………

_{> h?}

N

_nu

= f (N

_Re

, N

_Pr

)

Laminar flow in pipes: If N_Re<2100 For (N_Rex N_Prx D/L) < 100 14 . 0 66 . 0 Pr Re Pr Re

045 .

0

1

085 .

0

66 .

3 _









































w b Nu

L

D

x

xN

N

L

D

x

xN

N



For (N_Rex N_PRx D/L) > 100 ₀_.₃₃ ₀_.₁₄ PR RE Nu

w

b

L

D

x

xN

N

86 .

1 N

_



























HEAT TRANSFER TO FLUID

….

_>

_{FORCED CONVECTION}

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Transition Flow in Pipes  NREbetween 2100 and 10,000:

use chart to determine h :

diagram J Colburn factor (J) vs Re. 

HEAT TRANSFER TO FLUID

….

_>

_{FORCED CONVECTION}

14 . 0 w 3 2 k Cp. CpV h J _                      

Turbulent Flow in Pipes: …………._{> N}

RE> 10,000: 14 . 0 0.33 Pr NU

0 .

023 N

x

N

_















b_w

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Free convection involves the dimensionless number called Grashof Number, NGr

(

)

m G r Nu a N N k h D N Pr = =

(

)

G r N 2 2 3 _T D _ = m g 

 = koeff ekspansi volumetrik (koef muai volumetrik; 1/K) a and m = constant

All physical properties are evaluated at the film temperature …………._>T

f= (Tw + Tb)/2

HEAT TRANSFER TO FLUID

….

_>

_{FREE CONVECTION}

Value of a and m =f(physical configuration) Vertical surface

D=vertical dim. < 1 mNGrNPr<104 a=1.36 m=1/5

Horizontal cylinder

D = dia < 20 cm NGrNPr<10-5 a=0.49 m=0

10-5<_N

GrNPr<1 a=0.71 m=1/25

1<NGrNPr<104 a=1,09 m=1/10

Horizaontal flat surface

Facing Upward 105_{< N} GrNPr<2x107 a=0.54 m=1/4 2x107_{< N} GrNPr<3x1010 a=0.14 m=1/3 Facing downward 3x105_{< N} GrNPr<3x1010 a=0.27 m=1/4

HEAT TRANSFER TO FLUID

….

_>

_{FREE CONVECTION}

(

)

m G r Nu

a

N

k

h D

N

Pr

=

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Temperature profile : conductive and convective heat transfer

through a slab

T

_a

T

_b

h

_i

h

_o

Q = UA(T

_a

-T

_b

)

where

U = Overall heat transfer coefficient [=] W/m

2

_C

T

₁

T

₂

HEAT TRANSFER TO FLUID

………

_{> U?}

O O lm i i i i h A 1 kA x A h 1 A U 1 ₌ ₊  ₊ O i h 1 k x h 1 U 1 ₌ ₊  ₊ O i h A q kA x q A h q A U q ₌ ₊  ₊ Steady State : q_i= q_x=q_o=q q = UA(Ta-Tb) qi=q=hiA(Ta-T1) qx=q=kA(T1-T2)/x qo=q=hoA(T2-Tb) T_a-T_b= (T_a-T₁)+T₁-T₂)+(T₂-T_b)

T

_a

T

_b

h

_i

h

_o

T

₁

T

₂

HEAT TRANSFER TO FLUID

………

_{> U?}

Atau, umum :

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HEAT TRANSFER TO FLUID

………

_{> U?}

T_a

T

_a

T

_b

h

_i

h

_o

T

₁

T

₂ r₂ r₁

Surounding fluid temp; Tb< Ta

O O lm i i i i h A 1 kA r A h 1 A U 1 ₌ ₊  ₊ O i h 1 k r h 1 U 1 ₌ ₊  ₊ Atau, umum : A_i A_o ln A_i -A_o Alm 

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TRANSIENT

(UNSTEADY-STATE)

HEAT TRANSFER

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phariyadi.staff.ipb.ac.id -- ITP530 r Change in temperature?? T_s= f(t,r) Boiling water 100o_C Solid food material T_s,_initial=35o_C

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phariyadi.staff.ipb.ac.id -- ITP530  Importance of internal and

external resistance to heat transfer

 relative importance of conductive and conventive heat transfer  Biot number, N_Bi= hD/k k / D N_Bi ₌ h / 1 heat transfer to resistant External heat transfer to resistance Internal N or _Bi ₌ r Boiling water 100o_C

TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

q =  V Cp dT/dt = h A (Ta- T)



=

h

A

_C

d

_V

t

- T

T

dT

p a



t V) C A/ (h -o a a

_e

p

T

_

=

- Negligible internal resistance

…………._>N Bi< 0.1

V

C

A t

h

T)

ln(T

p a







T T_i t 0 TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

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 Finite Surface and Internal Resistance To Heat Transfer

…………._{> 0.1<N}

Bi< 40 ………..> m=1/NBi

 Negligible Surface Resistance To Heat Transfer…………._>

N_Bi> 40 ……….._{> m=1/N} Bi = 0

Infinite Slab, infinite cylinder and sphere Use Gurnie-Lurie Chart and/or Heisler Chart

…………_{> temperature-time (T-t) chart}

Dimensionless number : Fourier number (N_Fo)

= = D t D C kt N ₂ ₂ p Fo   D = characteristic dimension D_sphere= radius

D_{inf cylinder}= radius D_{inf slab}= half thickness TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER



























t

D

C

D

1 k

D

t

N

₃ p 2 2 Fo





The physical meaning of Fourier Number :

Large value of N_Foindicates deeper penetration of heat into solid in a given period of time

(W/C) (W/C) D volume in storage heat of Rate D volume in D across conduction heat of Rate 3 3 N_Fo=

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Prosedur pengunaan diagram T-t 1. 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 1/N}

Bi=m=k/hD

c. gunakan diagran untuk silinder tak berbatas, dari N_Fodan N_Bicari ratio T

Diagram T-t : hubungan antara suhu di sumbu silinder dan NFo TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

1/N_bi = m 1/N_bi = m

N_Fo N_Fo

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2. Untuk lempeng tak berbatas ketebalan, X = 2D

lebar = ; panjang =

Suhu ditengah (midplane) lempeng tak berbatas setelah pemanasan selama t ??

a. hitung NFo, gunakan (1/2)X sebagai D

b. hitung NBi, gunakan (1/2)X sebagai D ………> hitung 1/NBi

c. gunakan diagram untuk lempengtak berbatas, dari NFodan NBicari ratio T

Tebal=X TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

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Diagram T-t : hubungan antara suhu di pusat bola dan NFo TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

1. Menentukan suhu setelah pemanasan/pendinginan

• cari nilai NFo=t/2

• cari nilai Nbi dan m=1/Nbi

• 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 NBi dan m=1/Nbi

• cari NFo= t/2; dan hitung t

Diagram Gurnie-Lurie untuk LEMPENG :

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Diagram Gurnie-Lurie untuk SILINDER :

• cari nilai N_Fo=t/R2

• cari nilai N_bidan m=1/N_bi

• tentukan posisi dimana suhu ingin diketahui, n = r/R

• cari nilai N_bidan m=1/N_bi

• cari N_fo=t/R2_{; dan hitung t}

Diagram Gurnie-Lurie untuk BOLA :

• cari nilai NFo=t/R2

• tentukan posisi dimana suhu ingin diketahui, n = r/R

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Diagram Gurnie-Lurie :(Toledo) TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

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Finite object

….

_>

_{finite slab (}

_{bentuk bata, panjang=l, lebar=w, tinggi=h}

₎

length depth Inf slab, h i a a Inf slab, w i a a Inf. Slab l i a a i a a T T T T x T T T T x T T T T T T T T                      Finite slab, l,w,h

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phariyadi.staff.ipb.ac.id -- ITP530 Infinite slab (h) i a a Infinite cylinder R i a a Finite cylinder R, h i a a

T











































₌

_x

Finite object

…….

_>

_{finite slab (bentuk kaleng, jari-jari=R, tinggi=h)}

Infinite cylinder, radius R

Infinite slab, thickness=h TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

Penentuan posisi pada benda berbatas

Lokasi : tengah tutup kaleng - ditengah silinder : n=0 - dipermukaan lempeng: n=1 r= 1/2R  X=1/2 Lokasi x - n silinder = r/R=1/2 - n lempeng = x/ = 1/2 R ? X?

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phariyadi.staff.ipb.ac.id -- ITP530 CONTOH SOAL

Apel didinginkan dari suhu 20o_{C menjadi 8}o_{C, dengan menggunakan air}

dingin mengalir (5o_{C). Aliran air dingin ini memberikan koef. Heat Transfer}

konvensi sebesar 10 M/m2_{.K. Asumsikan 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/m3_{. Untuk pusat geometri apel mencapai suhu 8}o_C,

berapa lama harus dilakukan pendinginan? Jawab :

1. Cek NBi; apakah nilainya <0.1?

0,1<NBi<40?

atau NBi>40??

N_Bi= (hR/k)=1 …………_{> 0.1<N}

Bi<40 : gunakan diagram T-t (m=1/NBi=1)