Pindah Panas 2015

7/25/2019 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

MSB-FTIP-USAHID-2015 2


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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



4/40

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:



5/40

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.



6/40

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]



12/40

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



13/40

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



15/40

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



17/40

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



18/40

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



21/40

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



22/40

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



23/40

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



24/40

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



25/40

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



26/40

Heat transfer between two surfaces by

emission and later absorption of

electromagnetic radiation

requires no physical medium



27/40

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



29/40

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



30/40

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



31/40

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



32/40

In cylindrical coordinates:

r1

r2

2

1

ln

1 1 12

T

i i o o

r

rRh A Lk h A UA



33/40

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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35/40

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.



36/40

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



37/40

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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Documents

Pindah Panas 2015