ITP530 PEMBEKUAN PANGAN

(1)

ITP530/2015 phariyadi.staff.ipb.a.id ITP530

PEMBEKUAN PANGAN

Purwiyatno Hariyadi

phariyadi.staff.ipb.ac.id

Aspek engineering



Design (keperluan refrigerasi, T)



Laju pembekuan (the rate at which freezing

progress

)

Mutu produk

Produktivitas

PEMBEKUAN

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ITP530/2015 phariyadi.staff.ipb.a.id

• 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

PurwiyatnoHariyadi/IPN/ITP/Fateta/IPB

PEMBEKUAN

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

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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)

PENGARUH NEGATIF

• Kerusakan kimia

• Kerusakan fisik (textural)

PEMBEKUAN

– pengaruhnya pd produk pangan

PEMBEKUAN

– pengaruhnya pd produk pangan

– Freezer burn

?

• Package properly

• Control temperature fluctuations in storage.

– Oxidation?

• Off‐flavors

• Vitamin loss

• Browning

– Recrystallization?

PEMBEKUAN

– pengaruhnya pd produk pangan

PEMBEKUAN

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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= ? L 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) !!

Sifat Produk Pangan Beku

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 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

Sifat Produk Pangan Beku

Contoh (lanjutan):

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_          

Panas Laten Pembekuan

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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

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

ENERGY REMOVAL ASSOCIATED WITH FREEZING

(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}

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Titik Beku air

Super cooling

Titik eutetik Air

Larutan Suhu

Waktu

Titik beku = f(waktu)

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

Kurva Pembekuan

Removal of latent heat Removal of

sensible heat

3-21

Kurva Pembekuan

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T_i Tf

T

t

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

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

()

Kurva Pembekuan :

….. untuk Produk Pangan

Kurva Pembekuan :

….. untuk Produk Pangan

You can’t freeze all of the water

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

You can’t freeze all of the water

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

Kurva Pembekuan :

….. untuk Produk Pangan

Kurva Pembekuan :

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Kurva Pembekuan …. for Fish

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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

INTITIAL FREEZING TEMPERATURE

X_A= 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 - XA= 09826 - X_x= ... dst mol g

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INTITIAL FREEZING TEMPERATURE

100 0 - 40o_C ₀o_C % air beku Suhu

Hubungan antara % air beku vs. suhu

PRODUK BEKU?

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Hubungan antara % air beku vs. suhu

PRODUK BEKU?

•

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

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LAJU PEMBEKUAN

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LAJU PEMBEKUAN

Rapid Freezing

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LAJU PEMBEKUAN

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- Pendugaan keperluan pembekuan



ukuran sistem “mechanical

compression”



evaluasi beban refrigerasi/pembekuan

- Disain peralatan + proses, untuk :



memperoleh pembekuan yg diinginkan

- koef pindah panas

- laju pembekuan

PERHITUNGAN WAKTU PEMBEKUAN

•

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.}

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• 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)

PERHITUNGAN WAKTU PEMBEKUAN

PERHITUNGAN WAKTU PEMBEKUAN:

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.

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T_f

T₁

PERHITUNGAN WAKTU PEMBEKUAN:

Plank’s Method (for infinite slab)

x frozen frozen T_f T_f Ts T₁ Ts T₁ q q unfrozen a

PERHITUNGAN WAKTU PEMBEKUAN:

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ITP530/2015 phariyadi.staff.ipb.a.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

PERHITUNGAN WAKTU PEMBEKUAN:

Plank’s Method (for infinite slab)

Jumlah energi yang dibebaskan selama proses pembekuan qdt = m_i_f= _fdV _f qdt = _f_fA dx so, q = _f_fA dx/dt ... Pers. 4 Ingat Pers 3 :





h 1 k x T T q f 1 f   

PERHITUNGAN WAKTU PEMBEKUAN:

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

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 ……….>

PERHITUNGAN WAKTU PEMBEKUAN:

Plank’s Method (for infinite slab)



























_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

PERHITUNGAN WAKTU PEMBEKUAN:

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a

b _c

P dan R untuk bentuk bata a : dimensi terpendek c : dimensi terpanjang B₂= c/a

B1= b/a

Lihat chart/diagram :

dengan diketahui nilai B₂dan B₁maka dapat dibaca nilai P dan R

PERHITUNGAN WAKTU PEMBEKUAN:

General Plank’s Equation

PERHITUNGAN WAKTU PEMBEKUAN:

General Plank’s Equation

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.

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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

PERHITUNGAN WAKTU PEMBEKUAN:

General Plank’s Equation

Pham (1986): improved Plank’s equation :

• The mean freezing temperature is defined as

a c fm

T



1 .

8 

0 .

263 

0 .

105 









 





















2

1

2 2 1 1 Bi f c F

N

T

H

T

H

h

E

d

t

where T

_c

is final center temperature and T

_a

is

freezing medium temperature. The freezing time is

given by

PERHITUNGAN WAKTU PEMBEKUAN:

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where dc= characteristic dimension ‘r’ or shortest distance E_f = a shape factor (‘1’ for slab, ‘2’ for cylinder and

‘3’ for sphere)

)

(

1 u

c

u

T

i

T

fm

H









)]

(

[

2 f

L

f

c

f

T

fm

T

c

H











a fm i T T T T _         2 1 a fm T T T    ₂

ΔH₁= Enthalpy change during pre-cooling, J/m3

ΔH₂= Enthalpy change during phase change and post-cooling period, J/m3

PERHITUNGAN WAKTU PEMBEKUAN:

Pham’s Equation











 





















2

1

2 2 1 1 Bi f c F

N

T

H

T

H

h

E

d

t

In Pham’s method, the value of E_fis adjusted (Eq. 7.16): E_f= G₁+ G₂E₁+ G₃E₂

where the values of G₁, G₂and G₃are given in Table 7.1 and E₁ and E₂are calculated from Eqs. 7.17 & 7.19 and Eqs. 7.18 & 7.20, respectively.

We can now follow Example 7.2 (Singh and Heldman) and compare the freezing time calculations based on Pham’s approach and Plank’s equation.

PERHITUNGAN WAKTU PEMBEKUAN:

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• Alternate approach to determine the shape factor Efin the calculation of freezing time:

– For infinite slab, the shape factor E = 1 (since ₁=infinite, ₂=infinite) – For an infinite cylinder, the shape factor E=2 (since ₁=1, ₂=infinite) – For a sphere, the shape factor, E = 3 (₁=1, 2=1)

)

2 (

)

2

1 (

)

2 (

)

2

1 (

1

2 2 2 1 2 1 Bi Bi Bi Bi

N

E







PERHITUNGAN WAKTU PEMBEKUAN:

Pham’s Equation

• For different shapes e.g. ellipsoid, rectangular brick, finite cylinder etc., the shape factor can be calculated: • Same characteristic dimension R: shortage distance from thermal center to the surface of the object. • Smallest cross‐sectional area A ; the smallest cross‐section that incorporates R. • Same volume V • ₁and 2can be determined: ) 2 ( ) 2 1 ( ) 2 ( ) 2 1 ( 1 2 2 2 1 2 1 Bi Bi Bi Bi N N N N E _           2 1 R A    ) 3 4 ( 3 1 2 R V     k R h N c Bi 

PERHITUNGAN WAKTU PEMBEKUAN:

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Lean beef with 74.5% moisture content and 1 m length, 0.6 m width, and 0.25 m thickness is being frozen in an air‐blast freezer with hc= 30 W/m2.K and air temperature of ‐30 oC. If the

initial product temperature is 5 o_{C. Estimate the time required to reduce the product}

temperature to ‐10 o_{C. An initial freezing temperature of ‐1.75}o_{C has been measured for the}

product. The thermal conductivity of frozen beef is 1.5 W/m.K, and the specific heat of unfrozen beef is 3.5 kJ/kg.K. A product density of 1050 kg/m3 _{can be assumed, and a specific} heat of 1.8 kJ/kg.K for frozen beef can be estimated from properties of ice. – Product length d2= 1 m – Product width d1= 0.6 m – Product thickness a = 0.25 m – Convective heat‐transfer coefficient hc= 30 W/m2.k – Air temperature T= ‐30 oC – Initial product temperature Ti= 5 oC – Initial freezing temperature TF= ‐1.75 oC – Product density  = 1050 kg/m3 – Enthalpy change (H) = 0.745333.22 kJ/kg = 248.25 kJ/kg (estimate) – Thermal conductivity k of frozen product = 1.5 W/m.K – Specific heat of product (Cpu) = 3.5 kJ/kg.K – Specific heat of frozen product (Cpf) = 1.8 kJ/kg.

PERHITUNGAN WAKTU PEMBEKUAN:

CONTOH

(1) Determine shape factor: (2) The Biot number is : (3) Shape factor E: 056 . 3 ) 125 . 0 ( 6 . 0 25 . 0 ) 2 25 . 0 ( 6 . 0 25 . 0 2 2 2 1           R A 999 . 5 ) 125 . 0 ( 3 4 056 . 3 1 6 . 0 25 . 0 ) 3 4 ( 3 3 1 2            R V 5 . 2 5 . 1 125 . 0 30 _   k R h N_Bi c 197 . 1 ) 5 . 2 999 . 5 2 999 . 5 ( ) 5 . 2 2 1 ( ) 5 . 2 056 . 3 2 056 . 3 ( ) 5 . 2 2 1 ( 1 ) 2 ( ) 2 1 ( ) 2 ( ) 2 1 ( 1 2 2 2 2 2 1 2 1                  Bi Bi N N N N E    

PERHITUNGAN WAKTU PEMBEKUAN:

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ITP530/2015 phariyadi.staff.ipb.a.id (4) T3 : (5) H1: Cu(Ti‐T3) (6) T₁and T₂: C T3  1.80.263(10)0.105(30) 3.98 o 3 3 3 1 / 33001500 ] ) 98 . 3 ( 5 [ ) / 1050 ( ) . / 3500 ( ) ( m J C m kg K kg J T T C H _u _i o         





3 3 3 3 2 / 145 , 039 , 272 )) 10 ( 98 . 3 ( ) / 1050 ( . / 1800 ) / 1050 ( ) 745 . 0 ( ) / 1000 / 22 . 333 ( ) ( m J m kg K kg J m kg kJ J kg kJ T T C L H _f _f               C T T T C T T T T o a o a i 02 . 26 ) 30 ( 98 . 3 51 . 30 ) 30 ( 2 ) 98 . 3 5 ( 2 ) ( 3 2 3 1                 

PERHITUNGAN WAKTU PEMBEKUAN:

CONTOH/CEK!!!

(7) tslab : (8) t = tslab/E; s N T H T H h R t Bi slab 156 , 108 ) 2 5 . 2 1 ]( 02 . 26 272039145 51 . 30 33001500 [ 30 125 . 0 ) 2 1 ]( [ 2 2 1 1            hr s E t t slab 90355 25.1 197 . 1 156 , 108 _ _  

Required time for lean beef (1 m  0.6m 0.25 m) will be 25.1 hours to freeze.

PERHITUNGAN WAKTU PEMBEKUAN:

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METODE PEMBEKUAN

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

• 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.

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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 equipment

• 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 PEMBEKUAN

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ALAT PEMBEKU

A typical fluidized bed freezer

ALAT PEMBEKU

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ALAT PEMBEKU

Batch Freezer

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

Blast Type

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ITP530/2015 Source: Unit operations for food the food industries by: W.A. Gouldphariyadi.staff.ipb.a.id

Hydraulic

Pump Top _Pressure plate Connecting Linkage Corner Headers Refrigerant hoses Trays Contact plates

Polyurethane and polystyrene insulated doors

ALAT PEMBEKU

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ALAT PEMBEKU

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