2011

TPG330-Prinsip Teknik Pangan

8/24/2011

PEMBEKUAN Lecture Note Principles of Food Engineering (ITP 330) Dosen : Prof. Dr Dr.. IPurwiyatno Hariyadi, Hariyadi, MSc D t off F Dept Food d Science S i & Technology T h l Faculty of Agricultural Technology Bogor Agricultural University BOGOR

PEMBEKUAN

Pengawetan pangan

Aspek engineering → design (keperluan refrigerasi, ΔT) → laju pembekuan (the (the rate at which freezing progress) progress)

Mutu produk Produktivitas Purwiyatno Hariyadi/ITP/Fateta/IPB

Purwiyatno Hariyadi/ITP/Fateta/IPB -Pembekuan

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8/24/2011

PEMBEKUAN • Penyimpanan produk pada T < suhu beku • Umumnya pada T < 28 °F ((--2 °C), atau khususnya pada < 0 °F ((--18 °C) • Sebagian besar air (~95%) beku • daya awet produk beku ` bbrp bulan --- tahun • Laju pembekuan dipengaruhi oleh bbrp faktor : perlu dikendalikan

• Pertumbuhan mikroorganisme dihambat, bbrp bahkan dirusak

Purwiyatno Hariyadi/ITP/Fateta/IPB

PENGARUH PEMBEKUAN PD PROD PANGAN PENGARUH POSITIP m.o. o • Menurunkan/menghambat pertumbuhan m • Menurunkan laju reaksi kimia/biokimia • Meningkatkan daya simpan produk • (3 (3--40 lipat untuk setiap penurunan suhu 10 °C) PENGARUH NEGATIP • Kerusakan kimia • Kerusakan fisik ((textural textural))



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Sifat Produk Pangan Beku - Penurunan titik didih = f (konsentrasi, BM) ΔTf =

2 Rg TA0 BMA .m dimana:

ΔTf = K .m

⎛ ⎞ mol solut ⎟. m = molalitas ⎜⎜ l t ⎟⎠ ⎝ 1000mg pelarut TA 0 = titik beku pelarut murni (A ) = air (°K ), 273°K

λ

R g = kons tan ta gas = 8.314 λ = panas laten pembekuan ,

Lar. X dlm air Tf = (1.86 (1 86 m)oC λ1 ⎛ 1 1 ⎜ − ⎜ R g ⎝ TA 0 TA

J mol . K

kJ kJ → air = 335 kg kg

BMA = Berat B tM Molekul l k l pelarut l t K = konstanta molal titik beku

⎞ ⎟ = ln X A ⎟ ⎠

XA = fraksi mol air λ1 = panas laten pembekuan Purwiyatno Hariyadi/ITP/Fateta/IPB

Contoh : Ice cream mix dengan komposisi sbb: 10% butterfat 12% solid solid--not not--fat (54.5%: laktosa) 15% sukrosa 0.22% stabilizer 37.22% Ditanya ΔTf = ? 2

ΔTf =

Rg TA0 BMA .m

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

L

m=? m =

Air = 62.78%

mol solut kg solven

Solut? sukrosa BM = 342 laktosa BM = 342 solut lain diabaikan !! Purwiyatno Hariyadi/ITP/Fateta/IPB


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8/24/2011

Contoh : (lanj) ∴ Fraksi gula = 0.15 + 0.12 (0.545) = 0.2154 g/g Fraksi air = 0.6278 g gula 0.2154 = 0.3431 Konsentrasi gula dlm air = 0.6278 g air g gula = 343,1 343.1 1000 g air mol gula 342 = 1.003 m m= 1000 g air ⎛ ⎞ ⎛ 1mol ⎞ ⎛ g ⎞⎛ mol ⎞⎟ ⎟⎟ (273K)2 ⎜18 ⎜⎜ 8.314 J ⎟⎟ ⎜⎜ ⎟ ⎜⎜1.003 ⎟ mol .K ⎠ ⎝ 18g ⎠ kg ⎠ ⎝ mol ⎠ ⎝ ⎝ ΔTf = J 1000. 335 kg ΔTf = 1,86 K

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

Panas Laten Pembekuan Air murni λ = 335

kJ kg

Larutan solid x dlm air

λ = (335 mw)

kJ kg

mw = Fraksi massa air C t h Contoh: Selada Strawberi Kacang panjang Kentang Daging kambing K Kacang merah, h biji kkering i Kurma kering Air:

⎛

⎞

Kadar air

kJ λ ⎜⎜ ⎟⎟

94.8 90.8 88.9 77.8 58.0 12 5 12.5 24.0

316.3 289.6 297.0 258.0 194.0 41 9 41.9 79.0

⎝ kg ⎠

(317.6) (304.5) (297.8) (260.0) (194.3) (41.9) (41 9) (80.4

Perhitungan berdasarkan pd rumus λ = 335 mw

kJ kg

− 335 10 3 J ⎛⎜ 18 10 3 ⎞⎟ kJ = ⎜ 1 mol ⎟ kg kg ⎠ ⎝ J λ = 6030 mol

λ = 335



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8/24/2011

Kurva Pembekuan Suhu Titik Beku air

Super cooling Titik eutetik t tik

Titik beku = f(waktu) Air Larutan Waktu

Driving force for nucleation/crystalyzation (i.e. ?ΔT = T – Tf) Purwiyatno Hariyadi/ITP/Fateta/IPB

Kurva Pembekuan u/ Prod Pangan T Ti Tf

Setelah terjadi pembekuan, konsentrasi solute pada sisa larutan menjadi lebih tinggi .....> penurunan titik beku lebih besar .....> T (Ë) f t


You can’t freeze all of the water (Still have unfrozen water : 55-10%)

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8/24/2011

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


ENERGY REMOVAL ASSOCIATED WITH FREEZING Removal of heat (Q) from Region I (sensible heat), II (latent heat), and III (sensible heat) : (1) Q1 = mCp1ΔT1 m = weight of food Cp1 =specific heat of food above freezing ΔT = temperature difference (2) Q2 = mw λ (3) Q3 = mCp2ΔT3

........>

mw = weight of water λ = latent heat ........> m = weight of food ........> C p2 = specific heat of frozen food ........> ΔT = temperature difference 3 ........>



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8/24/2011

Liquid

T

Solubility line glass Sugar crystal

Tf Ice

eutectic

SugarSugarIce solid


INITIAL FREEZING TEMPERATURE Buah anggur (grape)

........> ........>

λ1 ⎛ 1 1 ⎞ ⎜ ⎟ = ln X − A ⎜ ⎟ R g ⎝ TA 0 TA ⎠ XA=?

J mol J 8,314 mol . K 6003

kadar air 84.7% Tf = -1.8oC (271.2oK)

J moll J Rg = 8.314 mol . K λ1 = 6003

⎛ 1 ⎞ 1 ⎜⎜ ⎟⎟ = ln X A − ⎝ 273 K 271 .2 K ⎠

Ln XA = - 0.01755 XA = 0.9826 (effective mol fraction of water

m grape ) ml


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8/24/2011

INITIAL FREEZING TEMPERATURE XA = fraksi mol air = 0.9826 XA = 0.9826 =

84 .7 18 84 .7 15 .3 + 18 BM E

BME = 183.61 Juice anggur dapat dianggap bertingkah laku mirip/sama dgn - lar. x dlm air mol - BMx = 183.61 g - XA = 09826 - Xx = ........ dst Purwiyatno Hariyadi/ITP/Fateta/IPB

PERHITUNGAN PEMBEKUAN (DESIGN) - Pendugaan keperluan pembekuan ? ukuran sistem “mechanical compression” ? evaluasi beban refrigerasi/pembekuan -Disain peralatan + proses, untuk : ? memperoleh p p pembekuan yg diinginkan g - koef pindah panas - laju pembekuan



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8/24/2011

Hubungan antara % air beku vs. suhu

% air beku

100

0

0oC

- 40oC

Suhu

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

LAJU PEMBEKUAN

• EQUIPMENT RELATED • rate of heat transfer • size of refrigeration g unit • FOOD/PRODUCT QUALITY • slow freezing • result in formation of few, large ice crystals • damaging to cell structure/quality • rapid freezing • results in many small ice crystals • gives best product quality • leads to IQF techniques • water ......> ice: ~ 9% increase in volume Purwiyatno Hariyadi/ITP/Fateta/IPB


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8/24/2011

FREEZING TIME

• Time Time--temperature method + • Time required to freeze between two temperatures (usually T = -5oC or –10oC) • Velocity of ice front - rate of freezing - must be able to see ice front • Appearance of specimen - internal conditions • Thermal methods - calorimetric l i t i ttechniques h i - not realreal-world condition

+ Time Time--temp. p methods most common + many people use time to freeze to – 10oC as standard. Purwiyatno Hariyadi/ITP/Fateta/IPB

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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8/24/2011

Plank’s Method (for infinite slab) x frozen

unfrozen

frozen

Tf

Tf

Ts

Ts

T1

T1 q

q

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

Plank’s Method (for infinite slab) Convection: BTU⎞ s – T1) ...... Pers. 1 ⎟ = Qt = hA (T ⎝ hr ⎠

q ⎛⎜

h = convective heat transfer coeff. coeff at the product surface surface. Conduction: k .A q = f (Tf −Ts ) ...... Pers. 2 x Tf = initial freezing point x = x ((t))

x frozen

unfrozen

Tf

Combine 1&2: T −T q = ( f 1) ....... Pers. 3 x 1 + kf h

Ts

frozen

Tf Ts

T1

T1 q

q a Hariyadi/ITP/ Purwiyatno Hariyadi/ITP/Fateta Fateta/IPB /IPB


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8/24/2011

Plank’s Method (for infinite slab) Jumlah energi yang dibebaskan selama proses pembekuan qdt = mi λf = ρf dV λf qdt = ρf λf A dx so,

q = ρf λf A dx/dt .............. Pers. 4 x frozen

Ingat g Pers 3 : T −T q = ( f 1) x 1 + kf h

unfrozen

Tf

frozen

Tf

Ts

Ts

T1

T1 q


Plank’s Method (for infinite slab) Kombinasi Pers. 3 dan 4

dx (T f − T1 )A = x 1 dt + kf h ⎛x 1⎞ ρf λ f ⎜ + ⎟ dx = (Tf − T1 )dt ⎝k f h ⎠ ρf λ f A

........>

Pembekuan selesai lempeng jika x = a/2 a 2

Tf ⎡ x 1⎤ ρf λ f ∫ ⎢ + ⎥ dx = (T f − T 1 )∫ dt h⎦ 0 0 ⎣k f x frozen

tf = Ti =

2 a ⎤ ρf λ f ⎡ a + ⎢ ⎥ T f − Ti ⎣ 8 k 2h ⎦

Suhu Pembekuan Suhu ruang pembeku

unfrozen

Tf Ts

frozen

Tf Ts

T1

T1 q



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8/24/2011

GENERAL FLANKS EQUATION

tf

2 Pa ⎤ ρ f λ f ⎡ Ra = + ⎥ n ⎦ (T f − Ti )⎢⎣ k f

Where: Infinite slab P 1/2 R 1/8 a Thickness

Sphere 1/6 1/24 Diameter

Infinite sylinder 1/4 1/6 Diameter

Cube 1/8 1/24 Edge

λ f = latent heat of fusion [=] kJ kg kJ λ water = 333.22 = 144 BTU kg lb


GENERAL FLANKS EQUATION

P dan R untuk bentuk bata a : dimensi terpendek c : dimensi terpanjang b

c a

B2 = c/a B1 = b/a Lih t chart/diagram Lihat h t/di : dengan diketahui nilai B2 dan B1 maka dapat dibaca nilai P dan R



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8/24/2011

Limitation of Plank’s method: • no superheating or supercooling • thermal properties are constant • can’t incorporate a variable heat transfer coeff. y g freezing gp point • can’t handle varying


Selesai ….………. Sekarang ke……. Psikrometrik


14

2011

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