pindah panas 2015
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PRINSIP TEKNIK PANGAN
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Pindah panas adalah perpindahan energi dari
suatu titik ke titik yang lain berdasarkan atas
perbedaan suhu
Contoh: pemanasan, pendinginan, pembekuan Perbedaan suhu (antar bahan) merupakan
faktor yang sangat penting dalam
menentukan laju pindah panas
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1. Konduksi: pindah panas darimolekul ke molekul
2. Konveksi:pindah panas dengan
adanya gerakan bahan secara
curah dari bahan bersuhu tinggi
ke bahan bersuhu rendah
3. Radiasi: pindah panas dalambentuk gelombang
elektromagnetik
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Pindah panas pada tingkat molekul
Tidak ada perpindahan fisik pada bahan
Terjadi pada bahan padat (solid)
Berlaku hukum Fourier:
q = laju aliran panas A = luas an tempat panas dipindahkan (tegak lurus arah
aliran) q/A = fluks panas K = konduktivitas panas [W/(m.K)] dT/dx = gradien suhu Tanda negatif berarti bahwa aliran panas positif akan terjadi
dalam arah suhu yang menurun
q kAdTdx
x Hukum Fourier I: Fluks panas (laju pindahpanas per satuan luas) pada konduksiberbanding lurus dengan gradien suhu:
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Thermal Conductivity, k unit: W/mC Metals: k = 50-400 W/mC
Water: k = 0.597 W/mC
Air : k = 0.0251 W/mC
Insulating materials: k = 0.035 - 0.173 W/mC
For foods k = 0.25 mc+ 0.155 mp+ 0.16 mf+ 0.135 ma+
0.58 mm
Where m is mass fraction and subscripts c:carbohydrate, p: protein, f: fat, a: ash, m:moisture.
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Pendugaan nilai k pada produk pangan (Choi
dan Okos, 1987)
k = (ki Xvi)
Xvi = (Xi )/i
= 1/(xi/ i)
k = konduktivitas
ki = konduktivitas tiap komponenXvi = fraksi volume tiap komponen
Xi = fraksi massa tiap komponen
= densitas
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Hubungan k komponen (ki) dengan suhu T (C)
Hubungan komponen (i) dengan suhu T (C)
[satuan = kg/m3]
kair = kw= 0,57109 + 0,0017625 T 6,7306 x 10-6T2
kes = kic= 2,2196 - 0,0062489 T + 1,0154 x 10-4T2
krotein = kp= 0,1788 + 0,0011958 T 2,7178 x 10-6T2klemak= kf= 0,1807 - 0,0027604 T 1,7749 x 10
-7T2
kkarbohidrat= kc= 0,2014 + 0,0013874 T 4,3312 x 10-6T2
kserat= kfi= 0,18331 + 0,0012497 T 3,1683 x 10-6T2
kabu = ka= 0,3296 + 0,001401 T 2,9069 x 10-6T2
air = w= 997,18 + 0,0031439 T 0,0037574 T2
es = ic= 916,89 - 0,13071 Trotein = p= 1329,9 - 0,51814 T
lemak= f= 925,59 - 0,41757 T
karbohidrat= c= 1599,1 - 0,31046 T
serat= fi= 1311,5 - 0,36589 T
abu = a= 2423,8 - 0,28063 T
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Steady state berarti suhu bahan dapat
berubah pada berbagai tempat tetapi tidak
berubah karena waktu
Konduksi melalui lempeng:
qx
x1
x2
x
dTq kA
dx
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xq dx kdT A
1 1
x Tx
x T
qdx kdT
A
1 1( )xq x x k T T
A
Boundary Conditionsx = x
1 T = T
1
x = x2 T = T
2
Separating the variables,
Integrating from x1to x (some interior location within the
slab)
1 1( )xqT T x x
kA
1
1
( )
( )x
T Tq kA
x x
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if integrated from x1to x2
2 1 1 2
2 1
( ) ( )
( )x
T T T T q kA kA
x x L
where L = thickness of the slab.
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COMPOSITE RECTANGULAR WALL (IN SERIES)
T3
Temperature
q
q
kB
kC
kD
LB
LC
LD
T0
x0
x1
x2
x3
x
T0
T3
B
C
D
q/A = kB[(T1-T0)/LB] = kC[(T2-T1)/LC] = kD[(T3-T2)/LD]
q/A = T/[X1/k1 + X2/k2 + ...+ Xn/kn]
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In the flow of electricity, the electrical
resistance is determined from the ratio of
electric potential divided by the electric
current. Similarly, we may consider heatresistance as a ratio of the driving potential
(temperature difference) divided by the rate
of heat transfer.
Thus, 1 2conductionx
T TR
q
2 1conduction
x x LR
kA kA
or,
units of resistance = C/W
T1
T2
L/kA
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HEAT TRANSFER IN A TUBULAR PIPE
r
i
ro
Fourier's Law in cylindrical coordinates
r
dTq kA
dr
2rdT
q k rLdr
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Boundary Conditions
T =Ti at r = ri
T=To at r = r0
Separating the variables
2
rq dr kdT rL
2
o o
i i
r T
r T
q drk dT
L r
ln2
o o
i i
r T
r T
qr k T
L
Integrating
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Recall for a single-layer pipe:
and resistance due to conduction
2 ( )
ln
i or
o
i
Lk T Tq
r
r
ln
2
o
i
c
r
rR
Lk
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Thermal resistance diagram:
r1
r2
r3
T1
T3
Material A
Material B
T1 T2 T3RcA RcB
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For the composite pipe shown in the diagram
The individual resistance values are:
and, referring to the resistance diagram,
Substituting the resistance terms
2
1ln
2cA
A
r
rR
Lk
3
2ln
2cB
B
r
rR
Lk
1 3r
CA CB
T TqR R
1 3
2 3
1 2
ln ln
2 2
r
A B
T Tq r r
r r
Lk Lk
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Fluid flow over a solid body -- heat transfer
between a solid and a fluid.
Newtons Law of Cooling:
where: h is convective heat transfer
coefficient (W/m2C), A is area (m2), Tpis
plate surface temperature (C), Tais
surrounding fluid temperature (
C).
q = h A (Tp-Ta)
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where h = convective heat transfer coefficient,
W/m2C
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 experiments are conducted to
determine relationships between following
dimensionless numbers.
q=hA(TW-Ta)
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Forced Convection - artificially induced fluid flow
Free (Natural) Convection -- caused due to density
differences
Fluid condition h (W/m2C) Air, free convection 5-25
Air, forced convection 10-200
Water, free convection 20-100
Water, forced convection 50-10,000 Boiling water 3,000-100,000
Condensing water vapor 5,000-100,000
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Nusselt Number = NNu= hD/k
Prandtl Number = NPr= mcp/k
Reynolds Number = NRe = rvD/m
where
D = characteristic dimension
k = thermal conductivity of fluid
v = velocity of fluid
cp= specific heat of fluid
r = density of fluid
m = viscosity of f
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FORCED CONVECTION: NNu = f(NRe, NPr)
Laminar Flow in Pipes: If NRe < 2100:
For (NRex NPr x D/L ) < 100
For (NRex NPr x D/L) >100
0.14Re Pr
0.66
Re Pr
0.085
3.66
1 0.045
bNu
w
DN NL
ND
N NL
0.140.33
Re Pr 1.86 b
Nu
w
DN N N
L
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All physical properties are evaluated at bulk
fluid temperature, except mw.
Transition Flow in Pipes: NRebetween 2100and 10,000: use figure 4.26 to determine h.
Turbulent Flow in Pipes:NRe> 10000:
0.14
0.8 0.33
Re Pr 0.023 b
Nu
w
N N N
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FREE CONVECTION
Free convection involves the dimensionless
number called Grashof Number, NGr
All physical properties are evaluated at the
filmtemperature Tf= (Tw+ Tb)/2
3 22Gr
D g TN
Prm
Nu Gr
hDN a N N
k
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Forced Convection - artificially induced fluid flow
Free (Natural) Convection -- caused due to density
differences
Fluid condition h (W/m2C) Air, free convection 5-25
Air, forced convection 10-200
Water, free convection 20-100
Water, forced convection 50-10,000 Boiling water 3,000-100,000
Condensing water vapor 5,000-100,000
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Heat transfer between two surfaces by
emission and later absorption of
electromagnetic radiation
requires no physical medium
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Stefen-Boltzmann Equation:
where s = Stefen-Boltzmann's constant, 5.669x10-8W/m2K4
e = emissivity, (varies from 0 to 1)
dimensionless
A = area, m2
T1= temperature of surface 1, Absolute
T2= temperature of surface 2, Absolute
q = A s e (T24T1
4)
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q = UA(Ta- Tb) (1)
Heat transfer through individual layers
q = hiA(Ta- T1)
q = hoA(T2- Tb)
Ta
hi
T2
Tb
T1
ho
1 2( )kA T T
q x
1a
i
qT T
h A
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Adding above three equations
From (1)
1 2
qT T
kA
x
2 b
o
qT T
h A
a b
qT T
UA
a b
i o
q q x qT T
h A kA h A
a b
qT T
UA
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i o
q q q x q
UA h A kA h A
1 1 1
i o
x
U h k h
1 1 1
i i i i m o o
x
U A h A kA h A
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Similar expression obtained when an overall thermal
resistance is calculated
RT= Rconv1+ Rcond+ Rconv2
Then
Rconv1
Rcond
Rconv2
Ta
Tb
1
1conv
i
Rh A
cond
xR
kA
2
1conv
oR h A
1 1T
i o
xR
h A kA h A
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In cylindrical coordinates:
r1
r2
2
1
ln
1 1 12
T
i i o o
r
rRh A Lk h A UA
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1
2
Fluid H
Fluid H
Fluid C
Fluid C
Insulation
dTH
dTC
dq
Temperature
Area
TH, inlet
TC, inlet
TH, exit
TC, exit
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The primary objective in using a heat
exchanger is to transfer thermal energy from
one fluid to another. We will use the
following assumptions:
1. Heat transfer is under steady-state
conditions.
2. The overall heat-transfer coefficient is
constant throughout the length of pipe.
3. There is no axial conduction of heat in the
metal pipe.
4. The heat exchanger is well insulated. The
heat exchange is between the two liquid streamsflowing in the heat exchanger. There is negligible
heat loss to the surroundings.
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Recall that change in heat energy in a fluid
stream, if its temperature changes from T1to
T2, is:
where = mass flow rate of a fluid (kg/s), cp=
specific heat of a fluid (kJ/kgC), and the
temperature change of a fluid is from some inlet
temperature T1to an exit temperature T2.
q mc T T p ( )1 2
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Tlmis called the log mean temperature difference.Equation is used to design a heat exchanger and determine
its area and the overall resistance to heat transfer, as
illustrated in the following examples.
, , 1
, , 2
H inlet C inlet
H exit C exit
T T T
T T T
q UA T T
T
T
2 1
2
1
ln
q UA T lm ( )
T T TT
T
lm 2 1
2
1
ln
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