HEAT TRANSFER - phariyadi's blogphariyadi.staff.ipb.ac.id/files/2018/02/ITP530-PindahPanas-2018-handouts.pdf · Heat Transfer 2/27/2018 phariyadi.staff.ipb.ac.id --ITP530 3 phariyadi.staff.ipb.ac.id

Heat Transfer 2/27/2018

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

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

HEAT TRANSFERIn FOOD PROCESSINGHEAT TRANSFERIn FOOD PROCESSINGLecture Note ITP530

Purwiyatno Hariyadiphariyadi.staff.ipb.ac.id

Dept of Food Science & TechnologyFaculty of Agricultural Engineering & TechnologyBogor Agricultural UniversityBOGOR

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 timeo temperatures are constant at all locationso steady-state is sometimes assumed if little error

results2. Unsteady-state (transient) heat transfer occurs when:

o temperature changes with time o thermal processing of foods is an important

exampleo 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 timeo temperatures are constant at all locationso steady-state is sometimes assumed if little error

results2. Unsteady-state (transient) heat transfer occurs when:

o temperature changes with time o thermal processing of foods is an important

exampleo must know time required for the coldest spot in

can to reach set temperature

Heat Transfer



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

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

CONDUCTION HEAT TRANSFERCONDUCTION HEAT TRANSFER

HEATHEAT

• 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




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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



X1X1 X2X2

qxqx

CONDUCTION HEAT TRANSFER - one dimensionalCONDUCTION HEAT TRANSFER - one dimensional

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

Q = Total heat flowqx = rate of heat flow in x direction by conduction, Wk = thermal conductivity, W/mCA= area (normal to x-direction) through which heat flows, m2

T = temperature, Cx = distance increment, variable, m

Q = Total heat flowqx = rate of heat flow in x direction by conduction, Wk = thermal conductivity, W/mCA= area (normal to x-direction) through which heat flows, m2

T = temperature, Cx = distance increment, variable, m

dxdx

dTdTkAkA--==qxqx

QtQt

==



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SIGN CONVENTIONSIGN CONVENTION

direction of heat flowdirection of heat flow

TT

xx

TEM

PER

ATU

RE

TEM

PER

ATU

RE

DISTANCEDISTANCE

dxdxdTdT--slopeslope

Temperature profileTemperature profile




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USING FOURIER’S LAW

dx

dT- k Aq

x=

BC :X = X1

...........> T = T1X = X2

...........> T = T2

-kdTdx A

qx =

T

T

X

X

x

11

kdT-dxA

q

Integrating :

X1X1 X2X2

qxqx

)x-x(kA

qTT

11

1

)X-(X

)T-(TkAq

1

1x

)T-k(T)x-x(qx

A 21


q

Composite Rectangular Wall (In Series)kA

xA

kB

xB

kC

xC

q

HEAT CONDUCTION IN MULTILAYERED SYSTEMS

Temperature profile in a multilayered system

T1

T2

X

Tem

pera

ture



7


q

kA

xA

kB

xB

kC

xC

q

dX

dT-kAq

kA

x-qT

USING FOURIER’S LAW : Ak

x-qT

A

AA

Ak

x-qT

B

BB

Ak

x-qT

C

CC


C

C

B

B

A

A21

k

X

k

X

k

X

A

q-TT

CBATTTT

21TTT

q

kA

xA

kB

xB

kC

xC

q

Ak

x-qT

A

AA

Ak

x-qT

B

BB

Ak

x-qT

C

CC



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CONDUCTION IN CYLINDRICAL OBJECTS

Fourier’s law in cylindrical coordinates

dr

dT- kAqr

dr

dTr L2 -k qr

Boundary Conditions :

T = Ti at r = ri

T = To at r = ro

oi

r

i

or

ln

)TLk(T2q

o

i

T

T

dTkr

dr

2L

q ro

ri

iTkT Ln r

2L

q

ro To

ri

Integrating :

ri

ro

dr


COMPOSITE CYLINDRICAL TUBE

r1

r2

r3

Ti

To

FROM FOURIER’S LAW:

i

o

oir

rr

ln

)TLk(T2q



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A =?Let us define logarithmic mean area Am

such that

io

oimr

)r(r

)TT(kAq

)r(r

i

o

iom

rr

ln

L2A

where

m

ioroi

kA

)rr(qTT

r1

r2

r3

Ti

To

23m

23r

32)(kA

)r(rqTT

12m

12r

21)(kA

)r(rqTT

adding above two equations

m12m

21r

kA

r

kA

r

)TT(q

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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(Ts - Ta)

q

q = rate of heat transfer

Surface area = A

h = convective heat transfer coefficient, W/m2.oC

Ts

Ts= surface temperature

Ta < Ts

Ta= surrounding fluid temperature


Fluid absorbs heat(temperature increase:

density decrease)

Colder fluid(higher density)

NaturalConvection



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FLUID FLOW IN A PIPE

Fluid flow can occur as- laminar flow- turbulent flow- transition between laminar and turbulent flowdirection of flow …..> parallel or perpendicular to the solid

object

HEAT TRANSFER TO FLUID


q = h A (Ts - Ta)

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.

HEAT TRANSFER TO FLUID……………> h?



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Dimensionless Numbers In Convective Heat Transfer

Nusselt Number = Nnu = (hD)/kPrandtl Number = NPr = Cp/kReynolds Number = Re = (vD)/Where

D = characteristic dimensionk = thermal conductivity of fluidv = velocity of fluidCp= specific heat of fluid= density of fluid= viscosity of fluid

HEAT TRANSFER TO FLUID……………> h?


Nnu = f (NRe, NPr)Laminar flow in pipes: If NRe<2100

For (NRe x NPr x D/L) < 100

14.0

66.0

PrRe

PrRe

045.01

085.066.3

w

bNu

LD

xxNN

LD

xxNNN

For (NRe x NPR x D/L) > 100 14.033.0

PRRENuw

b

L

DxxNN86.1N

HEAT TRANSFER TO FLUID ….> FORCED CONVECTION

All physical properties are evaluated at bulk fluid temperature, except mw



13


Transition Flow in Pipes NRE between 2100 and 10,000: use chart to determine h : diagram J Colburn factor (J) vs Re.


14.0

w

32

k

Cp.

CpV

hJ


Turbulent Flow in Pipes: ………….> NRE > 10,000:

14.0

0.33PrNU xN023.0N

w

b




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Free convection involves the dimensionless number called Grashof Number, NGr

( )mG rNu

NNak

h DN

Pr==

( )G r

N2

23 TD =

m

g

= koeff ekspansi volumetrik (koef muai volumetrik; 1/K)a and m = constant All physical properties are evaluated at the film temperature ………….>Tf = (Tw + Tb)/2

HEAT TRANSFER TO FLUID ….> FREE CONVECTION


Value of a and m =f(physical configuration)

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

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

10-5<NGrNPr<1 a=0.71 m=1/251<NGrNPr<104 a=1,09 m=1/10

Horizaontal flat surfaceFacing Upward 105< NGrNPr<2x107 a=0.54 m=1/4

2x107< NGrNPr<3x1010 a=0.14 m=1/3Facing downward 3x105< NGrNPr<3x1010 a=0.27 m=1/4

HEAT TRANSFER TO FLUID ….> FREE CONVECTION

( )m

G rNuNNa

k

h DN

Pr==



15


Temperature profile : conductive and convective heat transfer through a slab

Ta

Tb

hi ho

Q = UA(Ta-Tb)whereU = Overall heat transfer coefficient [=] W/m2C

T1

T2

HEAT TRANSFER TO FLUID……………> U?


OOlmiiii Ah1

kAx

Ah1

AU1 +

+=

Oi h1

kx

h1

U1 +

+=

Oi Ahq

kAxq

Ahq

AUq +

+=

Steady State :

qi = qx =qo=q

q = UA(Ta-Tb)

qi=q=hiA(Ta-T1)

qx=q=kA(T1-T2)/x

qo=q=hoA(T2-Tb)

Ta-Tb = (Ta-T1)+T1-T2)+(T2-Tb)

Ta

Tb

hi ho

T1

T2


Atau,umum :

Ai=Alm=Ao=A



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Ta

Ta

Tb

hi ho

T1

T2

r2

r1

Surounding fluid temp; Tb < Ta

OOlmiiii Ah1

kAr

Ah1

AU1 +

+=

Oi h1

kr

h1

U1 +

+= Atau,

umum :

Ai

Aoln

Ai-AoAlm




17





18





19





20


TRANSIENT

(UNSTEADY-STATE)

HEAT TRANSFER




21



r

Change in temperature??

Ts = f(t,r)

Boiling water100oC

Solidfood materialTs,initial=35oC

TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER



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Importance of internal and external resistance to heat transfer

relative importance of conductive and conventive heat transfer

Biot number, NBi = hD/k

k/ DN

Bi =h/ 1

heat transfertoresistant External

heat transfertoresistanceInternalNor

Bi =

r

Boiling water100oC



q = V Cp dT/dt = h A (Ta - T)

VC

tdA h =

- TT

dT

pa

tV)CA/(h -

oa

a peTT

TT =

-

-

Negligible internal resistance ………….>NBi < 0.1

VC

A th T)ln(T

p

a

T

Ti

t

0




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Finite Surface and Internal Resistance To Heat Transfer ………….> 0.1<NBi < 40 ………..> m=1/NBi

Negligible Surface Resistance To Heat Transfer ………….> NBi > 40 ………..> m=1/NBi = 0

Infinite Slab, infinite cylinder and sphereUse Gurnie-Lurie Chart and/or Heisler Chart …………> temperature-time (T-t) chart

Dimensionless number : Fourier number (NFo)

==D

t

DC

ktN

22p

Fo

D = characteristic dimensionDsphere = radiusDinf cylinder = radiusDinf slab = half thickness



t

DC

DD1

k

Dt

N3

p

2

2Fo

The physical meaning of Fourier Number :

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

(W/C)

(W/C)

Dvolumeinstorageheat ofRate

DvolumeinDacrossconductionheat ofRate3

3

NFo =




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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 NFo, gunakan R sebagai Db. hitung NBi, gunakan R sebagai D ………> hitung 1/NBi=m=k/hDc. gunakan diagran untuk silinder tak berbatas,

dari NFo dan NBi cari ratio T



Diagram T-t : hubungan antara suhu di sumbu silinder dan NFo


1/Nbi = m1/Nbi = m

NFoNFo




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2. Untuk lempeng tak berbatasketebalan, X = 2Dlebar = ; panjang =

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

a. hitung NFo, gunakan (1/2)X sebagai Db. hitung NBi, gunakan (1/2)X sebagai D ………> hitung 1/NBi

c. gunakan diagram untuk lempengtak berbatas, dari NFo dan NBi cari ratio T

Tebal=X



Diagram T-t : hubungan suhu di “midplane” lempeng tak berbatas dan NFo





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Diagram T-t : hubungan antara suhu di pusat bola dan NFo




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 NFo=t/R2


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

• cari ratio suhu




• cari Nfo =t/R2; dan hitung t



Diagram Gurnie-Lurie untuk BOLA :



• cari nilai NFo=t/R2


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

• cari ratio suhu




• cari Nfo =t/R2; dan hitung t




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Diagram Gurnie-Lurie :(Toledo)








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Finite object ….> finite slab (bentuk bata, panjang=l, lebar=w, tinggi=h)

length

width

depth

Inf slab,h

ia

a

Inf slab,w

ia

a

Inf. Slabl

ia

a

ia

a

TTTTx

TTTTx

TTTT

TTTT

Finiteslab, l,w,h




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Infiniteslab(h)

ia

a

Infinite cylinderR

ia

a

Finite cylinderR, h

ia

a

TTTT

TTTT

TTTT

x=

Finite object …….> finite slab (bentuk kaleng, jari-jari=R, tinggi=h)

Infinite cylinder,radius R

Infinite slab,thickness=h



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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CONTOH SOAL

Apel didinginkan dari suhu 20oC menjadi 8oC, dengan menggunakan air dingin mengalir (5oC). Aliran air dingin ini memberikan koef. Heat Transfer konvensi sebesar 10 M/m2.K. Asumsikan apel sebagai bola dengandiamater 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 8oC, berapa lama harus dilakukan pendinginan?

Jawab :1. Cek NBi ; apakah nilainya <0.1?

0,1<NBi<40? atau NBi >40??

NBi= (hR/k)=1 …………> 0.1<NBi<40 : gunakan diagram T-t (m=1/NBi=1)

2. Hitung rasio suhu yang dikehendaki :(Ta-T)/(Ta-Ti)=(5-8)/(5-20)=0.2

3. Posisi? Di pusat geometri …….> n=04. Cari nilai NFo, dan tentukan t




NFo=t/R2=0.78

t = 0.78R2/

t = 0.78R2/[k/(.Cp)]

t = 0.78(0.04)2/[0.4/(960)(3800)]

t = 11,381 s

t = 3.16 h

NFo=t/R2=0.78

θ=0.2



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HEAT TRANSFER - phariyadi's blogphariyadi.staff.ipb.ac.id/files/2018/02/ITP530-PindahPanas-2018-handouts.pdf · Heat Transfer 2/27/2018 phariyadi.staff.ipb.ac.id --ITP530 3 phariyadi.staff.ipb.ac.id

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