Heat Transfer 2/27/2018
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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
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• 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
Heat Transfer 2/27/2018
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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
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• 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
Heat Transfer 2/27/2018
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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).
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• 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 TRANSFERCONDUCTION HEAT TRANSFER
Heat Transfer 2/27/2018
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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 TRANSFERCONDUCTION HEAT TRANSFER
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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
==
Heat Transfer 2/27/2018
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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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Heat Transfer 2/27/2018
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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
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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
Heat Transfer 2/27/2018
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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
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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
Heat Transfer 2/27/2018
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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
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COMPOSITE CYLINDRICAL TUBE
r1
r2
r3
Ti
To
FROM FOURIER’S LAW:
i
o
oir
rr
ln
)TLk(T2q
Heat Transfer 2/27/2018
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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.)
Heat Transfer 2/27/2018
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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
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Fluid absorbs heat(temperature increase:
density decrease)
Colder fluid(higher density)
NaturalConvection
Heat Transfer 2/27/2018
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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
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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?
Heat Transfer 2/27/2018
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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?
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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
Heat Transfer 2/27/2018
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Transition Flow in Pipes NRE between 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
32
k
Cp.
CpV
hJ
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Turbulent Flow in Pipes: ………….> NRE > 10,000:
14.0
0.33PrNU xN023.0N
w
b
HEAT TRANSFER TO FLUID ….> FORCED CONVECTION
Heat Transfer 2/27/2018
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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
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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==
Heat Transfer 2/27/2018
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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?
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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
HEAT TRANSFER TO FLUID……………> U?
Atau,umum :
Ai=Alm=Ao=A
Heat Transfer 2/27/2018
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HEAT TRANSFER TO FLUID……………> U?
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
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Heat Transfer 2/27/2018
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Heat Transfer 2/27/2018
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Heat Transfer 2/27/2018
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Heat Transfer 2/27/2018
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TRANSIENT
(UNSTEADY-STATE)
HEAT TRANSFER
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Heat Transfer 2/27/2018
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r
Change in temperature??
Ts = f(t,r)
Boiling water100oC
Solidfood materialTs,initial=35oC
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Heat Transfer 2/27/2018
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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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Heat Transfer 2/27/2018
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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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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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 =
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Heat Transfer 2/27/2018
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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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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Diagram T-t : hubungan antara suhu di sumbu silinder dan NFo
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
1/Nbi = m1/Nbi = m
NFoNFo
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Heat Transfer 2/27/2018
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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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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Diagram T-t : hubungan suhu di “midplane” lempeng tak berbatas dan NFo
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
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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 :
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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Diagram Gurnie-Lurie untuk SILINDER :
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
1. Menentukan suhu setelah pemanasan/pendinginan
• cari nilai NFo=t/R2
• cari nilai Nbi dan m=1/Nbi
• 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 Nbi dan m=1/Nbi
• cari Nfo =t/R2; dan hitung t
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Diagram Gurnie-Lurie untuk BOLA :
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
1. Menentukan suhu setelah pemanasan/pendinginan
• cari nilai NFo=t/R2
• cari nilai Nbi dan m=1/Nbi
• 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 Nbi dan m=1/Nbi
• cari Nfo =t/R2; dan hitung t
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Diagram Gurnie-Lurie :(Toledo)
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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Heat Transfer 2/27/2018
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TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Heat Transfer 2/27/2018
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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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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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?
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
Heat Transfer 2/27/2018
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phariyadi.staff.ipb.ac.id -- ITP530
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
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
phariyadi.staff.ipb.ac.id -- ITP530
TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER
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
Heat Transfer 2/27/2018
phariyadi.staff.ipb.ac.id --ITP530
32
phariyadi.staff.ipb.ac.id -- ITP530