ITP330 PEMBEKUAN PANGAN (Freezing of Foods)

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1 Purwiyatno Hariyadi

Email: hariyadi@seafast.org Web: phariyadi.staff.ipb.ac.id

ITP330 -- Freezing of Food - 2016

phariyadi.staff.ipb.ac.id

ITP330

PEMBEKUAN PANGAN

(Freezing of Foods)

Purwiyatno Hariyadi, PhD

• Department of Food Science & Technology, Bogor Agricultural University, BOGOR, Indonesia

Outline

 Freezing systems

 Frozen food properties (density, thermal conductivity, enthalpy, apparent specific heat, and apparent thermal diffusivity)

 Freezing time (freezing curve, Plank’s equation, prediction of freezing time, freezing rate)

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Aspek engineering



Design (keperluan refrigerasi, T)



Laju pembekuan (the rate at which freezing

progress

)

Mutu produk

Produktivitas

PEMBEKUAN PANGAN

• 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 inaktif

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Things to notice:

• Pressure and temperature

both affect the phase of

matter.

• All three phases of matter

exist at the triple point

Melting/Freezing

Boiling/Condensating

pembekuan

PEMBEKUAN PANGAN

PENGARUH POSITI

F

• Menurunkan/menghambat pertumbuhan m.o.

• Menurunkan laju reaksi kimia/biokimia

• Meningkatkan daya simpan produk



(3-40 lipat untuk setiap penurunan suhu 10°C)

PEMBEKUAN PANGAN

Pengaruhnya pada Produk Pangan :

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PENGARUH NEGATIF

• Kerusakan kimia

• Kerusakan fisik (textural)

PEMBEKUAN PANGAN

Pengaruhnya pada Produk Pangan :

+

PENGARUH NEGATIF

• Freezer burn ?

• Package properly

• Control temperature fluctuations in storage. • Oxidation? • Off-flavors • Vitamin loss • Browning • Recrystallization?

PEMBEKUAN PANGAN

Pengaruhnya pada Produk Pangan :

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- Penurunan titik beku = f (konsentrasi, BM)

m . K T  m . BM T R T f A 2 0 A g f    

 

kg kJ 335 air kg kJ , pembekuan laten panas  K . mol J 314 . 8 gas ta tan kons R K 273 , K air ) A ( murni pelarut beku titik T . pelarut mg 1000 solut mol molalitas m g 0 A                 dimana:

BM_A= Berat Molekul pelarut K = konstanta molal titik beku

Lar. X dlm air T_f= (1.86 m)o_C A A 0 A g 1 X ln T 1 T 1 R           

X_A= fraksi mol air

1_{= panas laten pembekuan}

SIFAT PRODUK PANGAN BEKU

Ice cream mix dengan komposisi sbb: 10% butterfat 12% solid-not-fat (54.5%: laktosa) 15% sukrosa 0.22% stabilizer 37.22% Air = 62.78% Ditanya T_f= ?  m . BM T R T A 2 0 A g f   m = ? solven kg solut mol m  Solut? sukrosa BM = 342 laktosa BM = 342 solut lain diabaikan !! Asumsi bahwa hanya gula (laktosa + fruktosa) yang memp. Efek

menurunkan titik beku) !!

Contoh :

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phariyadi.staff.ipb.ac.id  Fraksi gula = 0.15 + 0.12 (0.545) = 0.2154 g/g

Fraksi air = 0.6278 Konsentrasi gula dlm air =

air g 1000 gula g 1 , 343  air g gula g 3431 . 0 6278 . 0 2154 . 0  m 003 . 1 air g 1000 gula mol 342 1 . 343 m  





kg J 335 . 1000 kg mol 003 . 1 mol g 18 K 273 g 18 mol 1 K . mol J 314 . 8 T 2 f                           T_f= 1,86 K Contoh (lanjutan):

SIFAT PRODUK PANGAN BEKU

Air murni  = 335

Larutan solid x dlm air  = (335 m_w)

kg kJ

kgkJ

m_w= Fraksi massa air

Contoh: Kadar air  Selada 94.8 316.3 Strawberi 90.8 289.6 Kacang panjang 88.9 297.0 Kentang 77.8 258.0 Daging kambing 58.0 194.0

Kacang merah, biji kering 12.5 41.9

Kurma kering 24.0 79.0       kg kJ Perhitungan berdasarkan pd rumus  = 335 m_w kg kJ Air: mol J 6030 mol 1 10 18 kg J 10 335 kg kJ 335  3 3_          

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

KURVA PEMBEKUAN … untuk air murni

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

(1) Q₁= mC_p1T1 m = weight of food

C_p1=specific heat of food above freezing T = temperature difference (2) Q₂= m_w ..._{> m} w= weight of water ..._{>  = latent heat} (3) Q₃= mC_p2T₃ ..._{> m = weight of food} ..._{> C}

p2= specific heat of frozen food ..._{> T}

3= temperature difference

KURVA PEMBEKUAN … untuk air murni & energi

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Titik Beku air

Super cooling

Titik eutektik Air

Larutan Suhu

Waktu

Titik beku = f(waktu)

Driving force for nucleation/crystallization (i.e. T = T – T_f)

Removal of latent heat Removal of

sensible heat

KURVA PEMBEKUAN … untuk produk pangan

Freezing process

 AB: Food is cooled below its

freezing point (Tf), below 0oC

 B: water remains liquid

although temp below T_f supercooling

 BC: Temp rises rapidly to Tfas

ice crystals begin to form and latent heat of crystallization is released.

 CD: heat is continued to be

removed from foods freezing point is depressed  DE: solute becomes

supersaturated and crystallize  eutectic point.

 EF: Freezing continuous to

freezer temperature

Titik Beku air

Super cooling Titik eutektik Air Larutan Suhu Waktu Titik beku = f(waktu) A B C D E F T_f T_m

Freezing Process

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phariyadi.staff.ipb.ac.id T T_f Ice Sugar-Ice solid Liquid Solubility line Sugar crystal glass eutectic

PHASE DIAGRAM

Phase Diagram

3-21

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ITP330 -- Freezing of Food - 2016 phariyadi.staff.ipb.ac.id T_i Tf T t

Setelah terjadi pembekuan, konsentrasi solute pada sisa larutan menjadi lebih tinggi

..._{> penurunan titik beku lebih besar}..._{> T} f ()

You can’t freeze all of the water

(Still have unfrozen (unfreezable) water : 5-10%)

KURVA PEMBEKUAN … untuk produk pangan

You can’t freeze all of the water

(Still have unfrozen (unfreezable) water : 5-10%)

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KURVA PEMBEKUAN … untuk produk

IKAN

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Buah anggur (grape) ..._{> kadar air 84.7%} ..._{> T} f= -1.8oC (271.2oK) A A 0 A g 1 X ln T 1 T 1 R           1_{= 6003} R_g= 8.314 K . molJ molJ X_A=? A X ln K 2 . 271 1 K 273 1 K . mol J 314 , 8 mol J 6003       Ln X_A= - 0.01755

X_A= 0.9826 (effective mol fraction of water )

ml grape m

SUHU AWAL PEMBEKUAN PROD PANGAN

XA= fraksi mol air = 0.9826 X_A= 0.9826 = E BM 3 . 15 18 7 . 84 18 7 . 84  BM_E= 183.61

Juice anggur dapat dianggap bertingkah laku mirip/sama dgn - lar. x dlm air - BM_x= 183.61 - X_A= 09826 - X_x= ... dst mol g

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SUHU AWAL PEMBEKUAN PROD PANGAN

Hubungan antara % air beku vs. suhu

PROD PANGAN BEKU

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•

EQUIPMENT RELATED

•

rate of heat transfer

•

size of refrigeration 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

•

water  ice: ~ 9% increase in volume

LAJU PEMBEKUAN

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Slow Freezing

LAJU PEMBEKUAN

Rapid Freezing

LAJU PEMBEKUAN

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LAJU PEMBEKUAN

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LAJU PEMBEKUAN

• Slow Freezers 0.2 cm/h

- Still air and cold stores

• Quick Freezers 0.5-3 cm/h

- Air blast and plate freezers

• Rapid Freezers 5-10 cm/h

- Fluidized bed freezers

• Ultra rapid Freezers 10-100 cm/h

- Cryogenic freezers

-- Klasifikasi berdasarkan pada laju pergerakan “ice front”

LAJU PEMBEKUAN

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Freezing time

 Freezing time: time required to lower the temperature of a food from an initial value to a predetermined final

temperature at the thermal center.  Freezing time depends on:

 Size and shape of the product

 Thermal conductivity of the food material  Area of the food available for heat transfer  Surface heat transfer coefficient of the medium

 Temperature difference between the food and freezing medium

 Type of packaging film in the case of packaged foods

- Pendugaan keperluan pembekuan



ukuran sistem “mechanical

compression”



evaluasi beban refrigerasi/pembekuan

- Disain peralatan + proses, untuk :



memperoleh pembekuan yg diinginkan

- koef pindah panas

- laju pembekuan

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•

Time-temperature method

•

Time required to freeze between two temperatures (usually T = -5o_{C or –10}o_C)

•

Velocity of ice front

-

rate of freezing

-

must be able to see ice front

•

Appearance of specimen

-

internal conditions

•

Thermal methods

-

calorimetric techniques

-

not real-world condition

+

Time-temp. methods most common

+

many people use time to freeze to

–10o_{C as standard.}

WAKTU/LAMA 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 (

for infinite slab

)

Plank’s equation is an approximate analytical solution for a simplified phase-change model.

• Plank assumed that the freezing process: (a) commences with all of the food unfrozen

but at its freezing temperature.

(b) occurs sufficiently slowly for heat transfer in the frozen layer to take place under

steady-state conditions.

WAKTU/LAMA PEMBEKUAN

Tf

T1

Plank’s Method (

for infinite slab

)

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phariyadi.staff.ipb.ac.id x frozen frozen T_f T_f T_s T1 T_s T1 q q unfrozen a

Plank’s Method (

for infinite slab

)

WAKTU/LAMA PEMBEKUAN

phariyadi.staff.ipb.ac.id Convection: q       hr BTU = Qt = h (Ts– T1) ... Pers. 1

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



f s



f T T x A . k q  _{... Pers. 2} Tf= initial freezing point

x = x (t) Combine 1&2:





h 1 k x T T q f 1 f    ... Pers. 3

Plank’s Method (

for infinite slab

)

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ITP330 -- Freezing of Food - 2016 phariyadi.staff.ipb.ac.id Ingat Pers 3 :





h 1 k x T T q f 1 f   

Plank’s Method (

for infinite slab

)

WAKTU/LAMA PEMBEKUAN

Jumlah energi yang dibebaskan selama proses pembekuan qdt = mif= fdV f qdt = _f_fA dx so, q = _f_fA dx/dt ... Pers. 4



T T



dt dx h 1 k x 1 f f f f           







   _         Tf 0 1 f 2 a 0 f f f _h dx T T dt 1 k x            h 2 a k 8 a T T t 2 i f f f f

Pembekuan selesai lempeng jika x = a/2

T_i= Suhu Pembekuan Suhu ruang pembeku





h 1 k x A T T dt dx A f 1 f f f      Kombinasi Pers. 3 dan 4 ……….>

Plank’s Method (

for infinite slab

)

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

























_T

Ra

_k

Pa

_h

t

f 2 i f f f f Where:

Infinite slab Sphere Infinite sylinder Cube

P 1/2 1/6 1/4 1/8

R 1/8 1/24 1/6 1/24

a Thickness Diameter Diameter Edge

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

General Plank’s Equation :

WAKTU/LAMA PEMBEKUAN

a

b _c

P dan R untuk bentuk bata a : dimensi terpendek c : dimensi terpanjang ₂= c/a

₁= b/a

Lihat chart/diagram :

dengan diketahui nilai ₂dan ₁maka dapat dibaca nilai P dan R

General Plank’s Equation :

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phariyadi.staff.ipb.ac.id Chart providing P and R constants for Plank’s equation when applied to a brick or block geometry.

In this figure,

β₁and β₂are the ratios of the two longest sides to the shortest.

It does not matter in what order they are taken.

General Plank’s Equation :

WAKTU/LAMA PEMBEKUAN

Limitation of Plank’s method:

• no superheating or supercooling

• thermal properties are constant

• can’t incorporate a variable heat transfer

coeff.

• can’t handle varying freezing point

General Plank’s Equation :

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phariyadi.staff.ipb.ac.id 1. AIR FREEZING - Products frozen by either "still" or "blast" forced

air.

• cheapest (investment)

• "still" slowest, more changes in product • "blast" faster, more commonly used

2. INDIRECT CONTACT - Food placed in direct contact with cooled

metal surface. • relatively faster • more expensive

3. DIRECT CONTACT - Food placed in direct contact with

refrigerant (liquid nitrogen, "green" freon, carbon dioxide snow) • faster

• expensive

• freeze individual food particles

METODE PEMBEKUAN

• Blast freezing

– a very cold air blasted on

the food cools food very quickly.

• Close indirect contact

– food is placed in a

multi-plate freezer and is rapidly frozen.

• Immersion

– food is placed into a very cold

liquid (usually salt water – brine) or liquid

nitrogen, this is known as cryonic freezing.

METODE PEMBEKUAN …commercial freezing

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• Mechanical Freezers

- Evaporate and compress the refrigerant in

a continuous cycle

• Cryogenic Systems

- Use solid and liquid CO

₂

, N

₂

directly in

contact with the food

METODE PEMBEKUAN …freezing equipments

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A typical fluidized bed freezer

METODE PEMBEKUAN …freezing equipments

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Batch Freezer

Source: Unit operations for food the food industries by: W.A. Gould

Blast Type

METODE PEMBEKUAN …freezing equipments

Hydraulic

Pump Top _Pressure

plate Connecting Linkage Corner Headers Refrigerant hoses Trays Contact plates

Polyurethane and polystyrene insulated doors

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METODE PEMBEKUAN …freezing equipments

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METODE PEMBEKUAN …freezing equipments

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Selamat belajar …

1. Water evaporation from the surface of the product can result in considerable weight loss in blast air freezers. 2. Individually quick frozen (IQF) products can be

produced in belt freezers and fluidized bed freezers. 3. Cryogenic freezing results in frozen products with large

ice crystals.

4. The rate of freezing in plate freezers is high.

5. Water activity of a frozen product is only a function of temperature.

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6. Bound water does not freeze.

7. Bound water affects water activity of a frozen product. 8. As water in a solution changes to ice, solute

concentration increases and freezing point decreases. 9. Heat capacity and thermal conductivity below the

freezing point change significantly with temperature. 10.Thawing is slower than freezing because the thermal

conductivity of ice is lower than that of water.