TPG330-Prinsip Teknik Pangan 8/24/2011

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PEMBEKUAN

Lecture Note Lecture Note Principles of Food

Principles of Food EngineeringEngineering (ITP 330)(ITP 330) Dosen

Dosen : : Prof.

Prof. DrDr. . IPurwiyatnoIPurwiyatno HariyadiHariyadi, , MScMSc D t f F d

D t f F d S iS i && TT hh ll Dept of Food

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

BOGOR

PEMBEKUAN

Pengawetan pangan

Aspek engineering Aspek engineering

→

→ design (keperluan refrigerasi, design (keperluan refrigerasi, ΔΔT)T) →

→ laju pembekuan (laju pembekuan (the rate at which freezing progressthe rate at which freezing progress))

Mutu produk Mutu produk Produktivitas Produktivitas

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•• Penyimpanan produk pada T < suhu beku

Penyimpanan produk pada T < suhu beku

•• Umumnya pada T < 28

Umumnya pada T < 28 °°F (

F (--2

2 °°C), atau khususnya

C), atau khususnya

PEMBEKUAN

pada < 0

pada < 0 °°F (

F (--18

18 °°C)

C)

•• Sebagian besar air (~95%) beku

Sebagian besar air (~95%) beku

•• daya awet produk beku ` bbrp bulan daya awet produk beku ` bbrp bulan --- tahuntahun

•• Laju pembekuan dipengaruhi oleh bbrp faktor : perlu Laju pembekuan dipengaruhi oleh bbrp faktor : perlu dikendalikan

dikendalikan dikendalikan dikendalikan

•• Pertumbuhan mikroorganisme dihambat, bbrp

Pertumbuhan mikroorganisme dihambat, bbrp

bahkan dirusak

Purwiyatno Hariyadi/ITP/Fateta/IPB

PENGARUH PEMBEKUAN PD PROD PANGAN

PENGARUH POSITIP

•• Menurunkan/menghambat pertumbuhan m o

Menurunkan/menghambat pertumbuhan m o

Menurunkan/menghambat pertumbuhan m.o.

•• Menurunkan laju reaksi kimia/biokimia

Menurunkan laju reaksi kimia/biokimia

•• Meningkatkan daya simpan produk

Meningkatkan daya simpan produk

•• (3

(3--40 lipat untuk setiap penurunan suhu 10

40 lipat untuk setiap penurunan suhu 10 °°C)

C)

PENGARUH NEGATIP

•• Kerusakan kimia

Kerusakan kimia

•• Kerusakan fisik (

Kerusakan fisik (textural

textural))

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Sifat Produk Pangan Beku

-- Penurunan titik didih = f (konsentrasi, BM)Penurunan titik didih = f (konsentrasi, BM)

λλ m m .. BM BM T T R R T T AA 2 2 0 0 A A g g ff == ΔΔ .. l t l t 1000 1000 solut solut mol mol molalitas molalitas m m _⎟⎟⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜⎜⎜ ⎝⎝ ⎛⎛ == dimana: dimana: m m .. K K T T λλ ff == ΔΔ

_{(( ))}

kg kg kJ kJ 335 335 air air kg kg kJ kJ ,, pembekuan pembekuan laten laten panas panas λλ K K .. mol mol J J 314 314 .. 8 8 gas gas ta ta tan tan kons kons R R K K 273 273 ,, K K air air )) A A (( murni murni pelarut pelarut beku beku titik titik T T pelarut pelarut mg mg 1000 1000 g g 0 0 A A == → → == == == °° °° == == ⎟⎟ ⎠⎠ ⎜⎜ ⎝⎝ BM BM BB t M l k lt M l k l ll tt Lar. X dlm air Lar. X dlm air T T_ff= (1 86 m)= (1 86 m)oo_C_C BM

BMAA= Berat Molekul pelarut= Berat Molekul pelarut

K = konstanta molal titik beku K = konstanta molal titik beku

T T_ff (1.86 m) (1.86 m) CC A A A A 0 0 A A g g 1 1 X X ln ln T T 1 1 T T 1 1 R R == ⎟⎟ ⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜ ⎜⎜ ⎝⎝ ⎛⎛ −− λλ X

XAA= fraksi mol air= fraksi mol air

λλ11₌_{= panas laten pembekuan}_{panas laten pembekuan}

Purwiyatno Hariyadi/ITP/Fateta/IPB

Contoh : Contoh :

Ice cream mix dengan komposisi sbb: Ice cream mix dengan komposisi sbb:

10% butterfat 10% butterfat 12% solid

12% solid--notnot--fat (54.5%: laktosa)fat (54.5%: laktosa) 15% sukrosa 15% sukrosa 0.22% stabilizer 0.22% stabilizer 37.22% 37.22% Air = 62.78%Air = 62.78% Ditanya Ditanya ΔΔTT_ff = ?= ? m m .. BM BM T T R R T T AA 2 2 0 0 A A g g ff == ΔΔ

Asumsi bahwa hanya gula Asumsi bahwa hanya gula (laktosa+fruktosa) yang memp. (laktosa+fruktosa) yang memp. Efek menurunkan titik beku) !! Efek menurunkan titik beku) !!

L L T Tff m = ? m = ? solven solven kg kg solut solut mol mol m m == Solut?

Solut? sukrosasukrosa BM = 342BM = 342 laktosa

laktosa BM = 342BM = 342

solut lain diabaikan !! solut lain diabaikan !!

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∴

∴ Fraksi gula = 0.15 + 0.12 (0.545) = 0.2154 g/gFraksi gula = 0.15 + 0.12 (0.545) = 0.2154 g/g Fraksi air = 0.6278

Fraksi air = 0.6278 Konsentrasi gula dlm air = Konsentrasi gula dlm air =

gula gula g g air air g g gula gula g g 3431 3431 .. 0 0 6278 6278 .. 0 0 2154 2154 .. 0 0 ₌₌ Contoh : (lanj) Contoh : (lanj) air air g g 1000 1000 gula gula g g 1 1 ,, 343 343 == m m 003 003 .. 1 1 air air g g 1000 1000 gula gula mol mol 342 342 1 1 .. 343 343 m m== ==

((

))

kg kg mol mol 003 003 .. 1 1 mol mol g g 18 18 K K 273 273 g g 18 18 mol mol 1 1 K K .. mol mol J J 314 314 .. 8 8 T T 2 2 ⎟⎟⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜⎜⎜ ⎝⎝ ⎛⎛ ⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜ ⎝⎝ ⎛⎛ ⎟⎟⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜⎜⎜ ⎝⎝ ⎛⎛ ⎟⎟⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜⎜⎜ ⎝⎝ ⎛⎛ Δ Δ kg kg J J 335 335 .. 1000 1000 T T_ff == Δ Δ Δ ΔTT_ff= 1,86 K= 1,86 K Purwiyatno

Purwiyatno HariyadiHariyadi/ITP//ITP/FatetaFateta/IPB/IPB

Panas Laten Pembekuan

Air murni

Air murni λλ = 335= 335 Larutan solid x dlm air

Larutan solid x dlm air λλ = (335 m= (335 mww) )

kg kg kJ kJ kg kg kJ kJ m

mww= Fraksi massa air= Fraksi massa air

C t h C t h Contoh: Contoh: Kadar air Kadar air λλ Selada Selada 94.894.8 316.3 316.3 (317.6)(317.6) Strawberi Strawberi 90.890.8 289.6 289.6 (304.5)(304.5) Kacang panjang Kacang panjang 88.988.9 297.0 297.0 (297.8)(297.8) Kentang Kentang 77.877.8 258.0 258.0 (260.0)(260.0) Daging kambing Daging kambing 58.058.0 194.0 194.0 (194.3)(194.3) K h biji k i K h biji k i 12 512 5 41 941 9 (41 9)(41 9) ⎟⎟⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜⎜⎜ ⎝⎝ ⎛⎛ kg kg kJ kJ Perhitungan Perhitungan berdasarkan berdasarkan pd rumus pd rumus λλ = 335 m= 335 m kJkJ

Kacang merah, biji kering

Kacang merah, biji kering 12.512.5 41.9 41.9 (41.9)(41.9) Kurma kering Kurma kering 24.024.0 79.0 79.0 (80.4(80.4 λλ = 335 m= 335 mww kg kg kJ kJ Air: Air: mol mol J J 6030 6030 λλ mol mol 1 1 10 10 18 18 kg kg J J 10 10 335 335 kg kg kJ kJ 335 335 λλ 33 33 == ⎟⎟ ⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜ ⎜⎜ ⎝⎝ ⎛⎛ == == −− Purwiyatno Hariyadi/ITP/Fateta/IPB

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Kurva Pembekuan

Suhu Suhu

Titik Beku air Titik Beku air

Super cooling Super cooling

Titik t tik

Titik t tik AirAir

Titik beku Titik beku = f(waktu) = f(waktu) Titik eutetik Titik eutetik Larutan Larutan Waktu Waktu Purwiyatno Hariyadi/ITP/Fateta/IPB

Driving force for nucleation/crystalyzation Driving force for nucleation/crystalyzation (i.e. (i.e. ??ΔΔT = T T = T –– TT_ff)) T T_ii T T

Kurva Pembekuan u/ Prod Pangan

T

Tff _{Setelah terjadi pembekuan, konsentrasi}_{Setelah terjadi pembekuan, konsentrasi}

solute pada sisa larutan menjadi lebih solute pada sisa larutan menjadi lebih tinggi

tinggi

...

..._{> penurunan titik beku}_{> penurunan titik beku lebih besar}_{lebih besar} ... ..._{> T}_{> T} ff((ËË)) tt > T > Tf f ((ËË))

You can’t freeze all of the water You can’t freeze all of the water (Still have unfrozen water : 5 (Still have unfrozen water : 5--10%)10%)

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T T--t Diagram :t Diagram : A schematic A schematic FREEZING OF WATER FREEZING OF WATER freezing curve freezing curve for water, for water, displaying displaying sensible heat sensible heat loss (Regions I loss (Regions I and III) and and III) and latent heat loss latent heat loss latent heat loss latent heat loss (Region II). (Region II).

Removal of heat (Q) from Region I Removal of heat (Q) from Region I (sensible heat), II (latent heat), (sensible heat), II (latent heat), and III (sensible heat) :

and III (sensible heat) : ENERGY REMOVAL ASSOCIATED WITH FREEZING ENERGY REMOVAL ASSOCIATED WITH FREEZING

(1) Q

(1) Q11= mC= mCp1p1ΔΔTT11

m = weight of food m = weight of food C

C_p1_p1=specific heat of food =specific heat of food above freezing

above freezing Δ

ΔT = temperature difference T = temperature difference

(2) Q

(2) Q₂₂= m= m_w_wλλ ..._{> m}_{> m} w

w= weight of water = weight of water ...

..._>> λλ = latent heat _{= latent heat}

(3) Q

(3) Q₃₃= mC= mC_p2_p2ΔΔTT₃₃ ..._{> m = weight of food}_{> m = weight of food} ...

..._{> C}_{> C} p2

p2= specific heat of frozen food = specific heat of frozen food ...

..._>> ΔΔTT 3

3= temperature difference= temperature difference

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Liquid Liquid Solubility Solubility line line T T T Tff Ice Ice Liquid

Liquid lineline

Sugar Sugar crystal crystal glass glass eutectic eutectic Sugar Sugar--Ice solid Ice solid eutectic eutectic Purwiyatno Hariyadi/ITP/Fateta/IPB

INITIAL FREEZING TEMPERATURE INITIAL FREEZING TEMPERATURE

Buah anggur (grape)

Buah anggur (grape) ..._{> kadar air 84.7%}_{> kadar air 84.7%} ... ..._{> T}_{> T} ff= = --1.81.8ooCC (271.2(271.2ooK)K) 1 1 1 1 1 1 _⎟⎟⎞⎞ ⎜⎜ ⎛⎛ λλ λλ11_{= 6003}_{= 6003} ll J J A A A A 0 0 A A g g X X ln ln T T 1 1 T T 1 1 R R == ⎟⎟ ⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜ ⎜⎜ ⎝⎝ ⎛⎛ −− λλ λλ 6003 6003 R R_g_g= 8.314 = 8.314 K K .. mol mol J J mol mol X X_A_A=?=? A A X X ln ln K K 2 2 271 271 1 1 K K 273 273 1 1 J J mol mol J J 6003 6003 == ⎟⎟⎟⎟⎠⎠ ⎞⎞ ⎜⎜⎜⎜⎝⎝ ⎛⎛ −− K K 2 2 .. 271 271 K K 273 273 K K .. mol mol J J 314 314 ,, 8 8 ⎝⎝ ⎠⎠ Ln X Ln XAA= = -- 0.017550.01755 X

X_A_A= 0.9826 (effective mol fraction of water ) = 0.9826 (effective mol fraction of water ) ml ml grape grape m m

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X

X_A_A= fraksi mol air = 0.9826= fraksi mol air = 0.9826

X X_A_A= 0.9826 = = 0.9826 = 3 3 15 15 7 7 84 84 18 18 7 7 .. 84 84

INITIAL FREEZING TEMPERATURE INITIAL FREEZING TEMPERATURE

A A E E BM BM 3 3 .. 15 15 18 18 7 7 .. 84 84 ₊₊ BM BM_E_E= 183.61 = 183.61

Juice anggur dapat dianggap Juice anggur dapat dianggap bertingkah laku mirip/sama dgn bertingkah laku mirip/sama dgn -- lar. x dlm airlar. x dlm air _mol_mol

-- BMBM_x_x= 183.61= 183.61 -- XX_A_A= 09826= 09826 -- XX_x_x= ... dst= ... dst g g mol mol Purwiyatno Hariyadi/ITP/Fateta/IPB

PERHITUNGAN PEMBEKUAN (DESIGN)

-- Pendugaan keperluan pembekuan

Pendugaan keperluan pembekuan

-- Pendugaan keperluan pembekuan

Pendugaan keperluan pembekuan

?? ukuran sistem “mechanical compression”

ukuran sistem “mechanical compression”

?? evaluasi beban refrigerasi/pembekuan

evaluasi beban refrigerasi/pembekuan

--Disain peralatan + proses, untuk :

Disain peralatan + proses, untuk :

?? memperoleh pembekuan yg diinginkan

memperoleh pembekuan yg diinginkan

p

yg

g

-- koef pindah panas

koef pindah panas

-- laju pembekuan

laju pembekuan

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100 100

Hubungan antara % air beku vs. suhu

% air beku % air beku 0 0 -- 4040oo_C_C ₀₀oo_C_C %% Suhu Suhu Purwiyatno

Purwiyatno HariyadiHariyadi/ITP//ITP/FatetaFateta/IPB/IPB

LAJU PEMBEKUAN LAJU PEMBEKUAN

••

EQUIPMENT RELATED EQUIPMENT RELATED

••

rate of heat transferrate of heat transfer

••

size of refrigeration unitsize of refrigeration unitgg

••

FOOD/PRODUCT QUALITYFOOD/PRODUCT QUALITY

••

slow freezingslow freezing

••

result in formation of few, large ice crystalsresult in formation of few, large ice crystals

••

damaging to cell structure/qualitydamaging to cell structure/quality

••

rapid freezingrapid freezing

••

results in many small ice crystalsresults in many small ice crystals

••

gives best product qualitygives best product quality

••

leads to IQF techniquesleads to IQF techniques

••

water ...> ice: ~ 9% increase in volumewater ...> ice: ~ 9% increase in volume

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FREEZING TIME FREEZING TIME

••

TimeTime--temperature methodtemperature method

••

Time required to freeze between two temperatures Time required to freeze between two temperatures (usually T =

(usually T = --55oo_{C or}_{C or –}_–10₁₀oo_C)_C)

+

••

Velocity of ice frontVelocity of ice front

--

rate of freezingrate of freezing

--

must be able to see ice frontmust be able to see ice front

••

Appearance of specimenAppearance of specimen

--

internal conditionsinternal conditions

••

Thermal methodsThermal methods

l i t i t h i

+

TimeTime--temp. methods temp. methods

--

calorimetric techniquescalorimetric techniques

--

not realnot real--world conditionworld condition

p p most common most common

+

many people use many people use time to freeze to time to freeze to –– 10

10oo_{C as standard.}_{C as standard.}

PERHITUNGAN WAKTU PEMBEKUAN

••

panas laten adalah energi utama yang hrs

diperhitungkan pada proses pembekuan

••

~ 75% total energi pd proses pembekuan

333.3 kJ/kg air

144 BTU/lb air

••

Terjadi perubahan sifat fisik bahan selama proses

pembekuan ~ f (T,m)

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Plank’s Method

Plank’s Method (for infinite slab)

(for infinite slab)

x x frozen

frozen unfrozenunfrozen frozenfrozen

T Tff TTff T T_s_s T T11 T T_s_s T T11 q q q q a a Purwiyatno

Plank’s Method

Plank’s Method (for infinite slab)

(for infinite slab)

Convection: Convection: q q ⎟⎟ ⎠⎠ ⎞⎞ ⎜⎜ ⎝⎝ ⎛⎛ hr hr BTU BTU = Qt = hA (T = Qt = hA (Tss–– TT11)) ... Pers. 1... Pers. 1

h

convective heat transfer coeff at the product surface

h

convective heat transfer coeff at the product surface

h = convective heat transfer coeff. at the product surface.

Conduction: Conduction:

((

ff ss

))

ff T T T T x x A A .. k k q q== −− _{... Pers. 2}_{... Pers. 2} T

Tff= initial freezing point= initial freezing point

x = x (t) x = x (t)

x x frozen

frozen unfrozenunfrozen frozenfrozen

( ) ( ) Combine 1&2: Combine 1&2:

((

))

h h 1 1 k k x x T T T T q q ff 1 1 ff ++ −− == ... Pers. 3... Pers. 3 Purwiyatno

T Tff TTff T Tss T T11 T Tss T T11 q q q q a a

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Jumlah energi yang dibebaskan selama proses Jumlah energi yang dibebaskan selama proses pembekuan

pembekuan

Plank’s Method

Plank’s Method (for infinite slab)

(for infinite slab)

qdt = m

qdt = m_iiλλ_ff= = ρρ_ffdV dV λλ_ff qdt =

qdt = ρρ_ff λλ_ffA dxA dx

so,

so, q q = ρρ= _ff λλ_ffA dx/dt ... Pers. 4A dx/dt ... Pers. 4

Ingat Pers 3 Ingat Pers 3 ::

x x frozen

g g

((

))

h h 1 1 k k x x T T T T q q ff 1 1 ff ++ −− == Purwiyatno

((

TT TT

))

dtdt dx dx 1 1 x x 1 1 ff ff ff ⎟⎟ == ⎞⎞ ⎜⎜ ⎛⎛ ++ λλ ρρ Pembekuan selesai Pembekuan selesai

Plank’s Method

Plank’s Method (for infinite slab)

(for infinite slab)

₍₍

₎₎

h h 1 1 k k x x A A T T T T dt dt dx dx A A ff 1 1 ff ff ff ++ −− == λλ ρρ Kombinasi Pers. 3 dan 4

Kombinasi Pers. 3 dan 4 ..._>_>

((

TT TT

))

dtdt dx dx h h k k_ff ff 11 ff ff ⎟⎟ == −− ⎠⎠ ⎜⎜ ⎝⎝ ++ λλ ρρ

((

−−

))

∫∫ == ∫∫ _⎥⎥ ⎦⎦ ⎤⎤ ⎢⎢ ⎣⎣ ⎡⎡ ++ λλ ρρ TTff 0 0 1 1 ff 2 2 a a 0 0 ff ff ff dxdx TT TT dtdt h h 1 1 k k x x ⎤⎤ ⎡⎡ λλ aa22 aa Pembekuan selesai Pembekuan selesai lempeng jika x = a/2 lempeng jika x = a/2

x x frozen

⎥⎥ ⎦⎦ ⎤⎤ ⎢⎢ ⎣⎣ ⎡⎡ ++ −− λλ ρρ == h h 2 2 a a k k 8 8 a a T T T T tt ii ff ff ff ff T

Ti i = = Suhu Suhu PembekuanPembekuan

Suhu ruang pembeku Suhu ruang pembeku

Purwiyatno

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GENERAL FLANKS EQUATION GENERAL FLANKS EQUATION

((

))

⎥⎥

⎦⎦

⎤⎤

⎢⎢

⎣⎣

⎡⎡

++

−−

λλ

ρρ

==

n

Pa

k

Ra

T

tt

ff 2 2 ii ff ff ff ff Where: Where: Where: Where: Infinite slab

Infinite slab SphereSphere Infinite sylinderInfinite sylinder CubeCube

P

P 1/21/2 1/61/6 1/41/4 1/81/8

R

R 1/81/8 1/241/24 1/61/6 1/241/24

a

a ThicknessThickness DiameterDiameter DiameterDiameter EdgeEdge

λλ = latent heat of fusion [=]= latent heat of fusion [=] kJkJ λλ_{ff= latent heat of fusion [=]}_{= latent heat of fusion [=]}

kg kg kg kg kJ kJ λλ water water = 333.22 = 333.22 = 144 = 144 lb lb BTU BTU Purwiyatno

P dan R untuk bentuk bata P dan R untuk bentuk bata GENERAL FLANKS EQUATION

GENERAL FLANKS EQUATION

a a b b _c_c a : dimensi terpendek a : dimensi terpendek c : dimensi terpanjang c : dimensi terpanjang B B₂₂= c/a= c/a B B11= b/a= b/a Lih t h t/di Lih t h t/di Lihat chart/diagram : Lihat chart/diagram : dengan diketahui nilai B

dengan diketahui nilai B₂₂dan Bdan B₁₁maka maka dapat dibaca nilai P dan R