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Heat Transfer Nur Istianah-THP-FTP-UB-2016

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Heat Transfer

Nur Istianah-THP-FTP-UB-2016

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Unsteady-state heat transfer

Nur Istianah-THP-FTP-UB-2016

Temperature is changing with time, it is a function of both location and time

It was in such as process: food pasteurization, sterilization, refrigeration/chilling/cooling

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9 Nur Istianah-THP-FTP-UB-2016

Introduction

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• In another words, in unsteady transient heat conduction, temperature is a function of both time and spatial coordinates

By using Fourier’s Equation governs the temperature response of a

body, and Thermal Energy equation (first law of thermodynamics

for closed system), one can fined the thermal conductivity of any

body by recording temperature change of its center with the time.

Rate of heat transfer by conduction from the center of the body out side or in opposite direction can be describe by Fourier’s Law

11

Unsteady state heat transfer

Nur Istianah-THP-FTP-UB-2016

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• When the body is a metal semi infinite slab or cylinder or sphere, for one-dimensional case the combination equation is:

Where for slab m=0

for cylinder m=1

for sphere m=2

From thermodynamics definition of the thermal diffusivity

The general equation may be

modified to:

12

Unsteady state heat transfer

Nur Istianah-THP-FTP-UB-2016

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Where

Ti Ts

T

Bi<0.1

Bi=1

Bi>>1

This Figure shows Meaning of Biot number

value (the relation between Biot number value

and the temperature gradient of any solid

object).

Convection Ts

T

Conduction Ti Ts

R

Ti Ts T

13

Biot Number

Nur Istianah-THP-FTP-UB-2016

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Three cases for unsteady-state heat transfer

• NBi < 0.1 : negligible internal resistance to heat transfer

• 0.1 < NBi < 40 : finite internal and surface resistance to heat transfer

• NBi > 40 : negligible surface resistance to heat transfer

14 Nur Istianah-THP-FTP-UB-2016

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• This indicates that the thermal resistance is negligible compared to the convection resistance and Ti Ts and the solid may have a uniform temperature.

• This phenomenon is called (Lumped Thermal Capacity).

• Where in this case the heat transfer by convection from the surface will balance the change of the internal energy of the solid.

At initial conditions

t = 0 T = Ti

the integration of of the above eq.

T -T¥

Ti -T¥

= e-h

rCpRt

1

G

15

Low Biot Number less than 0.1

Nur Istianah-THP-FTP-UB-2016

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This equation shows that the solid body

temperature approaches the surrounding

temperature T exponentially, which

means that the solid temperature change

rapidly at the beginning and slowdown

after.

For a big value of (1/) the solid reach

T in very short time (very high

conducting material).

𝑇 − 𝑇∞𝑇𝑖 − 𝑇∞

= 𝑒−𝑡Γ

G =rCpR

hTime constant

Fourier modules (Fo) • If we rearrange the exponent term in the above equation

𝑡

Γ=

ℎ𝑡

𝜌𝐶𝑝𝑅=

ℎ𝑅

𝑘

𝑘

𝜌𝐶𝑝𝑅2𝑡

Fourier number (Fo)

Fo =a

R2t

15 Nur Istianah-THP-FTP-UB-2016

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16

Contoh soal [4]

Nur Istianah-THP-FTP-UB-2016

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17 Nur Istianah-THP-FTP-UB-2016

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18 Nur Istianah-THP-FTP-UB-2016

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Nur Istianah-THP-FTP-UB-2016

Y = ϴ

X = Bi*Ʈ

19

0.1 <NBi < 40

Finite internal and surface resistances to heat transfer

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Cp

k

R

tX

TT

TTY

2

01

1

T1 = suhu medium (K) T0 = suhu awal (K) T = suhu akhir (K) = difusivitas termal t = waktu (detik) k = konduktivitas panas (W/m.K) h = koefisien transfer panas (W/m2.K) = densitas (kg/m3) cp = panas spesifik (J/kg.K) x = posisi dari pusat (m) R = dimensi relatif (1/2 ketebalan) (m)

Nur Istianah-THP-FTP-UB-2016

),,( mnXfY

20

Unsteady State (dimensionless)

R

xn

hR

km

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Nur Istianah-THP-FTP-UB-2016

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Nur Istianah-THP-FTP-UB-2016

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(Ta–T) (Ta–Ti)

NFo = t / D2

1/NBi

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(Ta–T) (Ta–Ti)

NFo = t / D2

1/NBi

Nur Istianah-THP-FTP-UB-2016

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(Ta–T) (Ta–Ti)

NFo = t / D2

1/NBi

Nur Istianah-THP-FTP-UB-2016

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• For calculation of temperature at any position of the object, Gurney-Lurie Chart can be used.

21

Gurney-Lurie Chart

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2

1x

tX

t position Nur Istianah-THP-FTP-UB-2016

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Nur Istianah-THP-FTP-UB-2016

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Nur Istianah-THP-FTP-UB-2016

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Nur Istianah-THP-FTP-UB-2016

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Nur Istianah-THP-FTP-UB-2016 24

Contoh soal [5]

Daging berbentuk persegi dengan ketebalan 2.54 cm mempunyai suhu awal 10°C dimasak dalam oven dengan suhu 177°C sehingga suhu pusat mencapai 121°C. Koefisien konveksi diasumsikan konstan sebesar 25.6 W/m2.K. Hitung waktu yang diperlukan! Konduktivitas panas sebesar 0.69 W/m.K dan difusivitas panas sebesar 5.85x10-4 m2/jam.

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Cp

k

R

tX

TT

TTY

2

01

1

T1 = suhu medium (K) T0 = suhu awal (K) T = suhu akhir (K) = difusivitas termal t = waktu (detik) k = konduktivitas panas (W/m.K) h = koefisien transfer panas (W/m2.K) = densitas (kg/m3) cp = panas spesifik (J/kg.K) x = posisi dari pusat (m) R = dimensi relatif (1/2 ketebalan) (m)

Nur Istianah-THP-FTP-UB-2016

),,( mnXfY

20

Unsteady State (dimensionless)

R

xn

hR

km

Page 27: Heat Transfer - nuristianah.lecture.ub.ac.idnuristianah.lecture.ub.ac.id/files/2016/09/KP2_heat-transfer... · both location and time ... or in opposite direction can be describe

Nur Istianah-THP-FTP-UB-2016 24

Contoh soal [5]

Page 28: Heat Transfer - nuristianah.lecture.ub.ac.idnuristianah.lecture.ub.ac.id/files/2016/09/KP2_heat-transfer... · both location and time ... or in opposite direction can be describe

Produk butter dengan ketebalan 92,4 mm yang mempunyai suhu 277,6 K ditempatkan dalam ruangan bersuhu 297,1 K. Bagian samping dan bawah kontainer butter dianggap terisolasi, sedangkan bagian permukaan atas berhubungan dengan udara luar. Koefisien konveksi sebesar 8,52 W/m2. Hitung suhu pada permukaan atas, pada jarak 25,4 mm di bawah permukaan dan pada bagian tengah dari butter setelah 5 jam dibiarkan di udara luar.

k butter 0,197 W/m.K, cp = 2300 J/kg.K, = 998 kg/m3

Nur Istianah-THP-FTP-UB-2016 25

Latihan soal [2]

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Nur Istianah-THP-FTP-UB-2016

9.24 cm

x= 4.62 cm

x= 4.62-2.54=2.08cm

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Finite objects

26 Nur Istianah-THP-FTP-UB-2016

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27

Finite object

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Y x = T1 - Tx dengan Xx dan mx

T1 – T0

Y y = T1 - Ty dengan Xy dan my

T1 – T0

Y z = T1 - Tz dengan Xz dan mz

T1 – T0

Y x,y,z = (Yx)(Yy)(Yz ))= T1 – Tx,y,z

T1 – T0

Nur Istianah-THP-FTP-UB-2016

28

Finite object (Two or three dimension)

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Nur Istianah-THP-FTP-UB-2016

Kaleng silinder berisi softdrink mempunyai diameter 4,8 cm dan suhu awal 27C. Kaleng ini disterilkan dalam retort dan dikenai uap dengan suhu 177C. Hitung suhu di pusat kaleng selama pemanasan 45 menit jika bagian atas dan bawah kaleng terisolasi. Koefisien transfer panas dari uap sebesar 4540 W/m2.K. Sifat fisik dari softdrink adalah sebagai berikut; k 0,48 W/m.K dan 2,007 x 10-7 m2/s.

29

Contoh soal [6]

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Nur Istianah-THP-FTP-UB-2016 30

Contoh soal [6]

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Kaleng silinder berisi puree mempunyai diameter 2,68 in dan tinggi 4 in serta suhu 85 F. Kaleng ini disterilkan dalam retort dan dikenai uap dengan suhu 240 F.

Hitung suhu di pusat kaleng selama pemanasan 45 menit.

Koefisien transfer panas dari uap sebesar 4542 W/m2.K. Sifat fisik dari puree adalah sebagai berikut; k 0,830 W/m.K dan 2,007 x 10-7 m2/s.

Nur Istianah-THP-FTP-UB-2016 31

Latihan soal [3]

Page 36: Heat Transfer - nuristianah.lecture.ub.ac.idnuristianah.lecture.ub.ac.id/files/2016/09/KP2_heat-transfer... · both location and time ... or in opposite direction can be describe

• Estimate the time when temperature at the geometric center of a 6 cm diameter apple held in 2C water stream reaches 3C. The initial uniform temperature of the apple is 15C. The convective heat transfer coefficient in water surrounding the apple is 50 W/m2C. The properties of the apple are thermal conductivity k = 0.355 W/mC, specific heat Cp = 3.6 kJ/kgC, and density = 820 kg/m3.

32

Latihan soal [4]

Nur Istianah-THP-FTP-UB-2016

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• Estimate the temperature at the geometric center of a food product contained in a 303X406 can exposed to boiling water at 100C for 30 min. The product is assumed to heat and cool by conduction. The initial uniform temperature of product is 35C. The properties of the food are thermal conductivity k = 0.34 W/mC, specific heat Cp = 3.5 kJ/kgC, and density = 900 kg/m3. The convective heat transfer coefficient for boiling water is estimated to be 2000 W/m2C.

34

Latihan soal [5]

Nur Istianah-THP-FTP-UB-2016

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1. Karkas daging mempunyai nilai 1073 kg/m3, cp 3,48

kJ/kg.K, dan k 0,498 W/m.K. Daging berbentuk lempeng

mempunyai ketebalan 0,203 m dan suhu awal 37.8 C

didinginkan sehingga suhu pusat mencapai 10 C.

Udara dingin bersuhu 1,7 C dan mempunyai nilai h 39,7

W/m2.K digunakan untuk mendinginkan.

Hitung waktu yang diperlukan.

39 Nur Istianah-THP-FTP-UB-2016

TUGAS

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2. Daging berbentuk persegi panjang dengan ketebalan 4 cm

dengan suhu awal 8C dimasak dengan menggunakan

oven bersuhu 185C hingga mencapai suhu pusat 121 C.

Koefisien konveksi diasumsikan konstan dengan nilai

sebesar 25,6 W/m2.K. Nilai konduktivitas panas sebesar

0,69 W/m.K dan nilai difusivitas panas () sebesar 5,85 x

10-4 m2/jam.

• Hitung waktu yang diperlukan untuk proses di atas!

40 Nur Istianah-THP-FTP-UB-2016

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3. Pipa baja berisi uap air mempunyai diameter luar 95

mm dan nilai k sebesar 45 W/m.K. Pipa ini dilapisi

dengan 75 mm isolator yang mempunyai nilai k 0,043

W/m.K. Dua termokopel ditempatkan pada antarmuka

antara dinding pipa dan isolator serta pada permukaan

luar isolator dan menunjukkan suhu berturut-turut

125C dan 35C.

• Hitung kehilangan panas yang terjadi per meter pipa.

41 Nur Istianah-THP-FTP-UB-2016

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THANKS FOR YOUR ATTENTION

The best person is one give something useful always

Nur Istianah-THP-FTP-UB-2016