2011

Pindah Panas

8/24/2011

Pindah Panas Lecture Note Principles of Food Engineering (ITP 330)

Dept of Food Science & Technology Faculty of Agricultural Technology Bogor Agricultural University BOGOR

Pindah Panas Tujuan Pembelajaran • Mengerti prinsip dasar pindah panas – untuk mengetahui bagaimana bahan pangan dipanaskan dan/atau didinginkan

• mengerti bagaimana pindah panas diukur – menentukan laju pemanasan dan pendinginan bahan pangan

• Mengerti g faktor faktor--faktor apa p saja j yang y g mempengaruhi (dan bagaimana pengaruhnya) aplikasi pindah panas dalam proses penanganan, pengolahan, distribusi dan pemanfaatan pangan

Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB

Purwiyatno Hariyadi/ITP/Fateta/IPB

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Pindah Panas • 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 (1) • May be indicated as total transfer • Identified by total heat flow (Q) with units of Btu • Identified by rate of heat flow (q) or ΔQ/ Q/Δ Δt with units of watts ot Btu/hr y be expressed p as heat transfer per p unit • Also,, may area = heat flux or q/A



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Heat Transfer (2) • Heat transfer can be classified as: 1. SteadySteady-state: o all factors are stabilized with respect to time o temperatures are constant at all locations o steadysteady-state is sometimes assumed if little error results 2. UnsteadyUnsteady-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 Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB

CONDUCTION HEAT TRANSFER • Occurs when heat moves through a material (usually solid or viscous liquid) due to molecular action only HEAT • Heat/energy is transferred 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). Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB


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CONDUCTION HEAT TRANSFER

• • • • • •

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

Purwiyatno Hariyadi/ITP/ Hariyadi/ITP/Fateta Fateta/IPB /IPB

CONDUCTION HEAT TRANSFER - one dimensional (unidirectional) • One dimensional conduction heat transfer is a function of: 1. temperature difference 2. material thickness 3. area through which heat flows 4. resistance of the material to heat flow



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CONDUCTION HEAT TRANSFER - one dimensional Fourier’s Law Of Heat Conduction:

ΔQ = qx = - kA dT dx Δt

qx

X1 X2 ΔQ = Total heat flow qx = rate of heat flow in x direction by conduction conduction, W k = thermal conductivity, W/mC A= area (normal to xx-direction) through which heat flows, m2 T = temperature, C x = distance increment, variable, m Purwiyatno Hariyadi/ITP/ Hariyadi/ITP/Fateta Fateta/IPB /IPB

SIGN CONVENTION

TEMPERATURE

direction of heat flow slope = -

ΔT Δx

dT dx Temperature profile

DISTANCE



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PH/TPG/Fateta/IPB

USING FOURIER’S LAW Integrating :

T1 T2

X1

q

X2

=- k A

x

qx

A

∫

X2

dT dx

X = X1 ...........> T = T1 X = X2 ...........> T = T2

qx

X1

dx = -kdT

qx A

d =dx

T

∫ kdT

T1

qx A ( x1 - x 2 ) = − k(T1 - T 2 )

T1 = T2 −

q1 kA

q x = − kA

(x1 - x2 ) (T1 - T2 ) (X1 - X 2 )



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HEAT CONDUCTION IN MULTILAYERED SYSTEMS Composite Rectangular Wall (In Series) kA kB kC q

Tempeerature

q

T1

xA

xB

xC Temperature profile in a multilayered system

T2 X Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB

kA

kB

kC

q

q

xA

xB

USING FOURIER’S LAW :

q = -kA Δ T = -q

dT dX Δx

kA

xC

Δ T A = -q Δ T B = -q Δ T C = -q

Δx A kAA Δx B k BA Δx C k CA



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kA q

kB

kC

T1

q

T2

xA

xB

xC

Δ T A = -q

Δ T = T1 − T 2

Δ T B = -q

Δ T = Δ TA + Δ TB + Δ TC T1 − T 2 = -

q ⎛ ΔX A ΔX B ΔX C ⎜ + + A ⎜⎝ k A kB kC

⎞ ⎟ ⎟ ⎠

Δ T C = -q

Δx A kAA Δx B k BA Δx C k CA


ro

CONDUCTION IN CYLINDRICAL OBJECTS dr

Fourier’s law in cylindrical coordinates

qr

= - kA

dT

Integrating g g: q r dr

dr

qr = -k 2 π r L

dT dr

Boundary Conditions : T = Ti

at

r = ri

T = To

at

r = ro

2πL

∫

ri

o

r

ri

= −k

∫

To dT Ti

ro To Ln r = − kT T 2πL ri i 2π Lk(T i − T o ) q = ⎛r ⎞ ln ⎜ o ⎟ ⎜ r ⎟ ⎝ i⎠ q



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COMPOSITE CYLINDRICAL TUBE r3 r1

Ti

r2

To

FROM FOURIER’S LAW:

qr =

2πLk(Ti − To ) ⎛r ⎞ ln⎜ o ⎟ ⎜ ri ⎟ ⎝ ⎠


A =? Let us define logarithmic mean area Am such that

q r = kA m where A m

( Ti − To ) (ro − ri ) (ro − ri ) = 2π L ⎛r ⎞ ln ⎜⎜ o ⎟⎟ ⎝ ri ⎠

q (r − r ) Ti − To = r o i kA m



r1

Ti

r3

r2

To

T1 − T 2 = T 2 − T3 =

q r (r 2 − r1 ) (kA

m

) 12

q r (r3 − r2 ) (kA

m

) 23

adding above two equations ( T1 − T 3 ) qr = ⎛ Δr ⎞ ⎛Δ r ⎞ ⎜ ⎟ +⎜ ⎟ ⎜ kA ⎟ ⎜ kA ⎟ m ⎠ 12 m ⎠ ⎝ ⎝ 23

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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.)


CONVECTIVE HEAT TRANSFER : heat transfer to fluid q

Ta < Ts

Surface area = A

Ts

q = h A(Ts - Ta) q = rate of heat transfer h = convective heat transfer coefficient, W/m2.oC Ts= surface temperature Ta= surrounding fluid temperature Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB


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Natural Convection

Colder fluid (higher (hi h d density) it )

Fluid absorbs heat (temperature increase: density decrease)

PH/TPG/Fateta/IPB

HEAT TRANSFER TO FLUID (Forced Convection)

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



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HEAT TRANSFER TO FLUID……………> h? q = h A (Ts - Ta) y, velocity, y, diameter,, viscosity, y, specific p h = f ((density, 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? Dimensionless Numbers In Convective Heat Transfer Nusselt Number = Nnu = (hD)/k Prandtl Number = NPr = μCp/k Reynolds Number = Re = (ρ (ρvD)/ vD)/μ μ Where D = characteristic dimension k = th thermall conductivity d ti it off fluid fl id v = velocity of fluid Cp= specific heat of fluid ρ= density of fluid μ= viscosity of fluid Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB


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HEAT TRANSFER TO FLUID

….>

FORCED CONVECTION

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

(NRe R x NPr P x D/L) < 100

N Nu

D 0.14 0.085⎛⎜ NRe xNPr x ⎞⎟ ⎛ ⎞ L μ ⎠ ⎜ b⎟ ⎝ = 3.66 + 0.66 D ⎞ ⎜⎝ μ w ⎟⎠ ⎛ 1 + 0.045⎜ NRe xNPr x ⎟ L⎠ ⎝

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

N Nu

D⎞ ⎛ = 1.86⎜ N RE xN PR x ⎟ L⎠ ⎝

0.33

⎛μ ⎞ ⎜⎜ ⎟⎟ w μ ⎝ ⎠

0.14

All physical properties are evaluated at bulk fluid temperature, except μw (viscosity at the wall) Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB


….>

FORCED CONVECTION

Transition Flow in Pipes Æ NRE between 2100 and 10,000: Æ use chart to determine h : Æ diagram J Colburn factor (J) vs Re. 2

Æ

⎛ h ⎞ ⎛ Cp. Cp μ ⎞ 3 ⎛ μ w ⎟⎟ ⎜ ⎟ ⎜⎜ J = ⎜⎜ ⎝ ρCpV ⎠ ⎝ k ⎠ ⎝ μ

⎞ ⎟⎟ ⎠

0.14



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….>

FORCED CONVECTION

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

N NU = 0.023 N

0.33 Pr

⎛μ x ⎜ μw ⎜ ⎝

⎞ ⎟⎟ ⎠

0.14



….>

FREE CONVECTION

Free convection involves the dimensionless number called Grashof Number, NGr N Gr

=

(d

3

2

ρ gβ Δ T µ

2

hD N Nu

= a(N

= k

)

N Gr

m

) Pr

d= Dimension of the system; ρ= density; β = koeff ekspansi volumetrik (koef muai volumetrik; 1/K); µ=viscosity; µ=viscosity; g= gravity a and m = constant



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HEAT TRANSFER TO FLUID N

hD Nu

= k

= a(N G r N Pr )

….>

FREE CONVECTION

m

V l off a and Value d m =f(physical f( h i l configuration) fi ti ) Vertical surface D=vertical dim. < 1 Horizontal cylinder D = dia < 20 cm

Horizontal flat surface Facing Upward Facing downward

mNGrNPr<104

a=1.36

m=1/5

NGrNPr<10-5 10-5
a=0.49 a=0.71 a=1,09

m=0 m=1/25 m=1/10

105< NGrNPr<2x107 a=0.54 2x107< NGrNPr<3x1010 a=0.14 3x105< NGrNPr<3x1010 a=0.27

m=1/4 m=1/3 m=1/4


Equations for calculating heat transfer coefficient (h) in free convection from water to air (Toledo, p. 271, Table 7.3) Kondisi p permukaan Silinder horisontal yang dipanaskan/didinginkan Fluida di atas plate horizontal yang dipanaskan

Persamaan

Nilai untuk C Udara/Uap

Air

h = C(ΔT/D)0.25

1.3196

291.1

h = C(ΔT)0.25

2.4492

…. dst dt


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HEAT TRANSFER TO FLUID……………> U? Temperature profile : conductive and convective heat transfer through a slab T1 Ta

hi

ho

T2

Tb

Q = UA(Ta-Tb) where U = Overall heat transfer coefficient [=] W/m2C Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB

HEAT TRANSFER TO FLUID……………> U? T1 Steady State : Ta qi = qx =qo=q

q = UA(Ta-Tb) qi=q=hiA(Ta-T1) =q=kA(T1--T2)/Δ T2)/Δx qx=q=kA(T1

hi

qo=q=hoA(T2-Tb)

ho

T2

Tb

Ta-Tb = (Ta-T1)+(T1-T2)+(T2-Tb) q = q + qΔ x + q U A hiA kA h OA 1 = 1 + Δx + 1 U hi k hO

Atau, umum :

1 U iA

= i

1 + Δx + 1 hiA i kA lm h O A

O

Ai=Alm=Ao=A Purwiyatno Hariyadi Hariyadi/ITP/ /ITP/Fateta Fateta/IPB /IPB


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HEAT TRANSFER TO FLUID……………> U? T1 Ta r2 Ta r1

Surrounding fluid temp; Tb < Ta 1 = 1 + Δr + 1 U hi k hO

1 U iA

Atau, umum :

hi

ho

=

1 + Δr + 1 hiA i kA lm h O A

i

Alm =

Tb

T2

O

Ao - Ai Ao ln Ai


HEAT TRANSFER TO FLUID……………> U? ri = r1= inside radius of cylinder r1

Ti

r2

r3

ro= r3= outside radius of cylinder hi = inside heat transfer coefficient ho = outside heat transfer coefficient

To

Calculating U based on outside radius of cylinder: 1 U

=

1 + roln(r2/r1) + roln(r3/r2) r r ln(r /r ) + …+ o n n-1 + o ho k1 r Kn-1 k2 i hi



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Contoh soal • Hitung heat flux (q/A) yang melewati glass pane yang terbuat dari lapisan gelas dengan ketebalan 1.6 mm yang dipisahkan oleh 0.8 mm lapisan insulator. Koefisien pindah panas pada sisi yang satu pada 21oC adalah 2.84 W/m2K dan pada sisi yangg lain bersuhu -15oC adalah 11.4 W/m2K. y Konduktivitas panas gelas adalah 0.52 W/mK dan pada lapisan udara adalah 0.031 W/mK.

Jawab:

1.6mm 0.8mm 1.6mm

h1 = 2.84 W/m2K

T1 = 21oC

h2 = 11.4 W/m2.K h1

h2 T2= -15oC

k1

k2

k1 = 0.52 W/mK k2 = 0.031 W/mK

k1

Heat flow

Terdapat 5 hambatan yang dialami selama proses pindah panas: 2 konveksi, 3 konduksi

1/U = 1/h1 + x1/k1 + x2/k2 + x3/k3 + 1/h2 1/U = 1/2.84 + 1.6 x 10-3/0.52 + 0.8x10-3/0.031 + 1.6x10-3/0.52 + 1/11.4 = 0.352 + 0.0031 + 0.0258 + 0.0031 + 0.0877 = 0.4718 U = 2.12 W/m2K q/A = U ΔT = 2.12 (21-(-15) = 76.32 W/m2


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Soal 2 • Hitung overall heat transfer coefficient (U) untuk heat exchanger dengan koefisien pindah panas 568/m2K di bagian dalam dan 5678 W/m2K di bagian luar. Dinding tube mempunyai konduktivitas panas (k) 55.6 W/mK. Tube mempunyai p y diameter dalam 2.21 cm dan ketebalan 1.65 mm. Jika suhu fluida di dalam tube 80oC dan di luar 120oC, hitung juga suhu pada dinding tube sebelah dalam.

ro = 1.2701 cm

(a) Menghitung U: T1 = 80oC

ri = 1.105 cm

hi = 568 W/m2K ho = 5678 W/m2.K k = 55.6 W/mK

T2 = 120oC

ri = 2.21/2 = 1.105 cm ro = 1.105 + 0.1651 = 1.2701 cm Terdapat 3 hambatan yang dialami selama proses pindah panas melalui heat exchanger: 2 konveksi, 1 konduksi 1/U = ro/rihi + + ro ln(ro/r1)/k + 1/ho 1/U =

1.2701x10-2/[(1.105x10-2)(568)]

+ 1.2701x10-2 ln(1.2701x10-2/1.105x10-2)/55.6 + 1/5678

= 20.236x10-4 + 0.318x10-4 + 1.76x10-4 = 22.315x10-4 U = 448 W/m2K


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(b) Suhu pada dinding tube sebelah dalam: q(overall) = q (yang melewati konveksi) = q (melewati konduksi) L t Tf = suhu Let: h fluida fl id di dalam d l tube; t b Tw = suhu h pada d dinding di di tube) t b ) UAoΔT = hiAi(Tw – Tf) 448(2πroL)(120-80) = 568(2πr1L)(Tw-80) (Tw – 80) = 448(ro)(40)/568(r1) = 80 + 448(1.2701)(40)/568(1.105) 448(1 2701)(40)/568(1 105) = 80 + 36.3 = 116.3oC

Uap panas 125oC, hi = 11400 W/m2K

PR…. Ke dalam sebuah pipa baja berinsulasi (ID=0.04089 m; OD=0.04826 m)) dialirkan uapp air (125oC, hi=11400 W/m2K). Ketebalan insulator adalah 5 cm (k2=0.1 W/mK). Diketahui baja memiliki k1=45 W/mK, koefisien pindah panas di lingkungan/luar pipa (ho) adalah 6 W/m2K, dan suhu lingkungan g g 20oC. Hitunglah g nilai U dan panas yang berpindah ke lingkungan (Q=UAΔT)).


r3 Ti k1 k2

r1

r2

Lingkungan (To=20C ho=6 W/m2K)

Baja Insulator

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Soal 3 • Air mengalir dengan laju 0.02 kg/s di dalam pipa penukar panas horizontal (ID= 2.5 2 5 cm, cm k=0.633 W/moC), dan dipanaskan dari 20oC menjadi 60oC (µair 658.026x10-6 Pas dan µw = 308.909x10-6 Pas). Suhu permukaan dalam pipa adalah 90oC. Perkirakan koefisien pindah panas (h) pada permukaan dalam pipa yang panjangnya 1 m. Cp air = 4175 J/kgoC

Jawab Soal 3 •

• •

Air mengalir pada pipa horizontal, berarti ada proses pemompaan (forced convection). Maka h akan dipengaruhi nilai NRe, NPr, D/L. NRe untuk t k Newtonian N t i fluida: fl id Re R = ρDV/µ DV/ Diketahui : µair = 658.026x10-6 Pas; µw = 308.909x10-6 Pas; Cp air = 4175 J/kgoC D = 2.5 cm = 2.5x10-2 m laju masa (ṁ) = 0.2 kg/s, L pipa = 1 m v (m/s) = ṁ (kg/s) /ρ(kg/m3)(π(D/2)2 (m2), maka = NRe = 4ρD(ṁ/ρπD2 µ) = 4ṁ/πµD = 4(0.02)/π (658.026x10-6)(2.5x10-2) = 1457.9 (Laminar) NPr = µCp/k = (658.026x10-6)(4175)/(0.633) = 4.34 D/L = 2.5x10-2/1 = 2.5x10-2

•

Maka NRe x NPr x D/L = 1457.9 x 4.34 x 2.5x10-2 = 168 (>100)

•

Maka gunakan persamaan empiris: Nu = 1.86(NRexNPrxD/L)0.33 (µ/µw)0.14= Nu = 1.86(1547.9x4.34x2.5x10-2)0.33(658.026x10-6/308.909x10-6) = 93 Nu = hD/k = 93 = 2.5x10-2/0.633 Æ h = 2355 w/m2C

• •


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PR… • Hitung berapa nilai h, h bila kecepatan aliran dinaikkan menjadi 1.5 kg/s

Soal 3 • Hitunglah tu g a kecepatan ecepata kehilangan e a ga panas pa as kee lingkungan dari pipa baja (ID=0.04089 m) berisi uap pada suhu 130oC. Koefisien konveksi (h) pada sisi uap adalah 11400 W/m2K dan di luar pipa ke udara adalah 5.7 W/m2K. Suhu lingkungan ratarata 15oC, konduktivitas panas (k) dinding pipa baja adalah 45 W/mK. • Bila Bil pipa i diberi dib i hambatan h b t (insulator) (i l t ) setebal t b l 5 cm yang memiliki k=0.07 W/mK (h uap dan udara seperti di atas), berapa energi yang dapat diselamatkan?


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TRANSIENT (UNSTEADY UNSTEADY--STATE) STATE) HEAT TRANSFER


PH/TPG/Fateta/IPB


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PH/TPG/Fateta/IPB

TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER Boiling water 100oC Solid food material Ts,initial=35oC

r

Change in temperature?? Ts = f(t,r)



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TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER

z Importance of internal and

Boiling water 100oC

external resistance to heat transfer relative importance of conductive and convective heat transfer

r

Biot number, NBi = hD/k NBi =

or NBi =

D/k 1/ h

Internal resistance to heat transfer External resistant to heat transfer



z Negligible internal resistance ………….>N Bi

< 0.1

q = ρ V Cp dT/dt = h A (Ta - T)

dT ∫ Ta - T =

hAdt ∫ ρ CpV

t T

ln(T

a

− T) = Ti

Ta - T

=e

h At

ρ C pV 0 - (h A/ ρ Cp V) t

Ta - To Purwiyatno Hariyadi/ITP/Fateta/IPB


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z Finite Surface and Internal Resistance To Heat Transfer ………….>

0.1
………..>

m=1/NBi

z Negligible Surface Resistance To Heat Transfer ………….> NBi > 40 ………..> m=1/NBi = 0 Infinite Slab, infinite cylinder and sphere Use GurnieGurnie-Lurie Chart and/or Heisler Chart …………> temperaturetemperature-time (T(T-t) chart Dimensionless number : Fourier number (NFo) N Fo =

kt 2 ρ C pD

=

α t D2

D = characteristic dimension Dsphere = radius Dinf cylinder = radius Dinf slab = half thickness



The physical meaning of Fourier Number :

1 k ⎛⎜ ⎞⎟D 2 D αt N Fo = 2 = ⎝ ⎠ 3 D ⎛ ρ Cp D ⎞ ⎜ ⎟ ⎜ ⎟ t ⎝ ⎠ NFo =

Rate of heat conduction

across D in volume D 3 (W/C)

Rate of heat storage in volume D 3

(W/C)

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



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TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER Prosedur pengunaan diagram TT-t 1. Untuk silinder tak berbatas R ∞Suhu pusat (sumbu) silinder setelah pemanasan selama t? a. hitung NFo, gunakan R sebagai D b. hitung NBi, gunakan R sebagai D ………> hitung 1/NBi=m=k/hD c. gunakan diagram untuk silinder tak berbatas, dari NFo dan NBi cari ratio T Purwiyatno Hariyadi/ITP/Fateta/IPB


1/Nbi = m

NFo

Diagram T-t : hubungan antara suhu di sumbu silinder dan NFo Purwiyatno Hariyadi/ITP/Fateta/IPB


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TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER 2. Untuk lempeng tak berbatas ketebalan, X = 2D lebar = ∞; panjang =∞ =∞ Tebal=X

Suhu di tengah (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 NFo dan NBi cari ratio T Purwiyatno Hariyadi/ITP/Fateta/IPB


Diagram T-t : hubungan suhu di “midplane” lempeng tak berbatas dan NFo Purwiyatno Hariyadi/ITP/Fateta/IPB


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Diagram T-t : hubungan antara suhu di pusat bola dan NFo Purwiyatno Hariyadi/ITP/Fateta/IPB

TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER Diagram Gurnie-Lurie untuk LEMPENG : 1. Menentukan suhu setelah pemanasan/pendinginan • cari nilai NFo=αt/δ2 • cari nilai Nbi dan m=1/Nbi • tentukan posisi dimana suhu ingin diketahui, n = x/δ • cari ratio suhu 2. Menentukan waktu pemanasan/pendinginan untuk mencapai suhu ttt • cari rasio suhu suhu, pada posisi ttt yang diketahui, n = r/R • cari nilai NBi dan m=1/Nbi • cari NFo= αt/δ2; dan hitung t



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TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER Diagram Gurnie-Lurie untuk SILINDER : 1. Menentukan suhu setelah pemanasan/pendinginan • cari nilai NFo=αt/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 =αt/R2; dan hitung t Purwiyatno Hariyadi/ITP/Fateta/IPB

TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER Diagram Gurnie-Lurie untuk BOLA : 1. Menentukan suhu setelah pemanasan/pendinginan • cari nilai NFo=αt/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 =αt/R2; dan hitung t Purwiyatno Hariyadi/ITP/Fateta/IPB


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





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Finite object ….> finite slab (bentuk bata, panjang=l, lebar=w, tinggi=h) ⎛T T⎞ ⎛T T⎞ ⎛T T⎞ ⎛T T⎞ ⎜⎜ a − ⎟⎟ = ⎜⎜ a − ⎟⎟ x ⎜⎜ a − ⎟⎟ x ⎜⎜ a − ⎟⎟ ⎝ Ta − Ti ⎠Finite⎝ Ta − Ti ⎠Inf. Slab ⎝ Ta − Ti ⎠ Inf slab, ⎝ Ta − Ti ⎠ Inf slab, l

slab,, l,w,h

w

h

length depth

width Purwiyatno Hariyadi/ITP/Fateta/IPB


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

Infinite cylinder, radius R Infinite slab, thickness=h

⎛ Ta − T ⎞ ⎜⎜ ⎟⎟ T T − ⎝ a i⎠

=

⎛ Ta − T ⎞ ⎜⎜ ⎟⎟ T T − ⎝ a i⎠

Finite cylinder R, h

x

⎛ Ta − T ⎞ ⎜⎜ ⎟⎟ T T − ⎝ a i ⎠Infinite slab (h)

Infinite cylinder R


TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER Penentuan posisi pada benda berbatas ? Lokasi : tengah tutup kaleng - ditengah silinder : n=0 - dipermukaan lempeng: n=1

R

δ

r= 1/2R

X=1/2δ

X? Lokasi x - n silinder = r/R=1/2 - n lempeng = x/δ x/δ = 1/2



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TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER CONTOH SOAL Apel didinginkan dari suhu 20oC menjadi 8oC, dengan menggunakan air dingin mengalir (5oC). Aliran air dingin ini memberikan koef. Pindah panas konvensi sebesar 10 M/m2.K. Asumsikan apel sebagai bola dengan diamater 8 cm. Nilai k apel = 0.4 W/m/K, Cp apel 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? 0,140??

NBi= (hR/k)=1 …………> 0.1

TRANSIENT (UNSTEADY-STATE) HEAT TRANSFER 2. Hitung rasio suhu yang dikehendaki : (Ta-T)/(Ta-Ti) = (58)/(5-20) = 0.2

θ=0.2

3. Posisi? Di pusat geometri Æ n=0 4. Cari nilai NFo, dan tentukan t n=0

m=1

NFo=αt/R2=0.78

NFo=αt/R2=0.78 t = 0.78R 0 78R2/α / 2 t = 0.78R /[k/(ρ.Cp)] t = 0.78(0.04)2/[0.4/(960)(3800)] t = 11,381 s t = 3.16 h



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PH/TPG/Fateta/IPB

PH/TPG/Fateta/IPB


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PH/TPG/Fateta/IPB

PH/TPG/Fateta/IPB


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PH/TPG/Fateta/IPB

PH/TPG/Fateta/IPB


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PH/TPG/Fateta/IPB

Selesai ….………. NEXT Heat Exchangers


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Soal 3 • Hitung laju kehilangan panas (q) dari sebuah retort horizontal dengan diameter dalam 1.524 m dan panjang 9.144 m. Uap di dalam retort bersuhu 121oC. Udara luar bersuhu 25oC. Retort dibuat dari baja (k = 42 W/mK) dan mempunyai ketebalan 0.635 m.

Jawaban Soal 3 • Diketahui Diketahui, udara melewati silinder horisontal, maka h = 1.3196(ΔT/Do)0.25


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Soal 3 • Hitung overall heat transfer coefficient (U) untuk saus tomat (densitas 995 kg/m3, viskositas 0.676 Pas) yang dipanaskan dari suhu 20oC ke 80oC dalam stainless steel tube dengan panjang 5 m dengan inside diameter 1.034 cm dan ketebalan 2.77 mm. Uap p panas p di luar tube bersuhu 120oC. Koefisien pindah panas steam di dalam tube 6000 W/m2K. Laju aliran (v) adalah 0.1 m/s.

Soal 3 • Hitung tu g nilai a koefisien oe s e ppindah da panas pa as (h) ( ) dan da overall ove a heat eat transfer coefficient (U) untuk saus tomat (densitas 995 kg/m3, n=0.34 dan K= 10.42 Pas; equivalent Newtonian viscosity (µ)= 0.5 Pa.s, µw= 0.45 Pa.s, panas jenis=3817 J/kg.K) yang dipompa dan dipanaskan dalam stainless steel tube. Saus masuk pada suhu 20oC dan keluar pada suhu 80oC. Stainless steel tube berdimensi panjang 5 m, inside diameter 1.034 cm, dan ketebalan 2.77 mm, konduktivitas panas (k) tube wall 17.3 W/mK. Uap panas di luar tube bersuhu 120oC. Koefisien pindah panas steam di dalam tube 6000 W/m2K. Laju aliran rata-rata (v) adalah 0.1 m/s.


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Jawaban Soal 3 n 8 ( v ) 2 -n ( R ) ρ

• •

Dipompa, berarti forced convection Reynolds Number:

• • •

NRe = 8(0.1)2-0.34+(1.034x10-2)0.34*995 = 11.02 (Laminar) 10.42 [(3*0.34+1)/0.34]0.34 Prandtl Number = NPr = μCp/k = 0.5*3817/17.3 = 110.3 D/L = 1.034E-2/5 = 0.02

•

Maka: (NRe x NPr x D/L) = 11 11.02 02 * 110.3 110 3 * 0.02 0 02 = 24.31 24 31 < 100

•

0 . 085 Maka, N = 3. 66 + Nu

⎛⎜ ⎝ N

NRe =

D ⎞ xN x ⎠ Re Pr L

D ⎛ 1 + 0 . 045 ⎜N xN x ⎝ Re Pr L

0.66 ⎞ ⎟ ⎠

⎡ 3 n + 1⎤ K⎢ ⎣ n ⎥⎦

⎛ μ ⎞ b ⎟ ⎜ ⎜ ⎟ ⎝ μ w ⎠

n

0 .14 = (hD)/k

h = ….

Overall heat transfer coeffiecient: 1/U = ro/rihi + + ro ln(ro/r1)/k + 1/ho Maka : U???

PR…..


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