ITP530 PEMBEKUAN PANGAN (Freezing of Foods)

ITP530 PEMBEKUAN PANGAN (Freezing of Foods) Purwiyatno Hariyadi, PhD • Department of Food Science & Technology, Bogor Agricultural University, BOGOR, Indonesia

ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

PEMBEKUAN PANGAN

Aspek engineering  Design (keperluan refrigerasi, T)  Laju pembekuan (the rate at which freezing progress)

Mutu produk Produktivitas ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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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 ITP530 -- Freezing of Food , 2016

PurwiyatnoHariyadi/IPN/ITP/Fateta/IPB phariyadi.staff.ipb.ac.id

PEMBEKUAN PANGAN

Things to notice: • Pressure and temperature both affect the phase of matter. • All three phases of matter exist at the triple point

Melting/Freezing

pembekuan

Boiling/Condensating ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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PEMBEKUAN PANGAN Pengaruhnya pada Produk Pangan :

+

PENGARUH POSITIF

• 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)


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+

PENGARUH NEGATIF • Freezer burn ? • Package properly • Control temperature fluctuations in storage. • Oxidation? • Off-flavors • Vitamin loss • Browning • Recrystallization?


SIFAT PRODUK PANGAN BEKU - Penurunan titik beku = f (konsentrasi, BM) 2

Tf 

Rg TA0 BMA . m

Tf  K. m



dimana:

 mol solut m  molalitas   1000 mg pelarut TA0  titik beku pelarut murni R

Lar. X dlm air Tf = (1.86 m)oC

g

 kons tan ta gas  8 .314

  panas laten pembekuan ,

 .  (A)  air K , 273 K J mol . K

kJ kJ  air  335 kg kg

BMA = Berat Molekul pelarut K = konstanta molal titik beku

1  1 1     ln X  A   R g  TA0 TA 

XA = fraksi mol air 1 = panas laten pembekuan


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SIFAT PRODUK PANGAN BEKU Contoh :

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

Ditanya Tf = ? 2

Tf 

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 !!


SIFAT PRODUK PANGAN BEKU Contoh (lanjutan):  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 mol gula 342 1000 g air  1 . 003 m m 1000 g air   1mol    g  mol   273 K2 18  8.314 J     1.003 mol .K   18 g  kg   mol    Tf  J 1000 . 335 kg Tf = 1,86 K ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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PANAS LATEN PEMBEKUAN kJ

Air murni  = 335kg Larutan solid x dlm air Contoh: Selada Strawberi Kacang panjang Kentang Daging kambing Kacang merah, biji kering Kurma kering

Air:

  335

kJ

 = (335 mw) kg mw = Fraksi massa air 



Kadar air

kJ   

94.8 90.8 88.9 77.8 58.0 12.5 24.0

316.3 289.6 297.0 258.0 194.0 41.9 79.0

 kg 

Perhitungan berdasarkan pd rumus  = 335 mw

kJ kg

 335 10 3 J  18 10 3  kJ J    6030 mol  kg kg 1 mol  


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


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KURVA PEMBEKUAN … untuk air murni & energi Removal of heat (Q) from Region I (sensible heat), II (latent heat), and III (sensible heat) : (1) Q1 = mCp1T1 m = weight of food Cp1 =specific heat of food above freezing T = temperature difference

(2) Q2 = mw 

........>

(3) Q3 = mCp2T3

mw = weight of water  = latent heat ........> m = weight of food ........> C p2 = specific heat of frozen food ........> T = temperature difference 3 ........>


KURVA PEMBEKUAN … untuk produk pangan

Suhu

Removal of sensible heat

Removal of latent heat

Super cooling Titik eutetik

Titik Beku air Titik beku = f(waktu) Air Larutan Waktu

Driving force for nucleation/crystallization (i.e. T = T – Tf) ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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KURVA PEMBEKUAN … untuk produk pangan T Ti Tf

Setelah terjadi pembekuan, konsentrasi solute pada sisa larutan menjadi lebih tinggi .....> penurunan titik beku lebih besar .....> T f () t You can’t freeze all of the water (Still have unfrozen (unfreezable) water : 5-10%)


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You can’t freeze all of the water (Still have unfrozen (unfreezable) water : 5-10%) ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

KURVA PEMBEKUAN … untuk produk IKAN


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KURVA PEMBEKUAN … untuk produk IKAN


SUHU AWAL PEMBEKUAN PROD PANGAN Buah anggur (grape)

........>

1  1 1     ln X  A   R g  TA0 TA  XA=?

J mol J 8 ,314 mol . K 6003

kadar air 84.7% ........> T = -1.8oC f J 1  = 6003 mol J Rg = 8.314 mol . K

(271.2oK)

 1  1    ln X A   273 K 271 . 2 K 

Ln XA = - 0.01755 m grape XA = 0.9826 (effective mol fraction of water ) ml


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SUHU AWAL PEMBEKUAN PROD PANGAN XA = fraksi mol air = 0.9826 XA = 0.9826 = BME = 183.61

84 . 7 18 84 . 7 15 . 3  18 BM E

Juice anggur dapat dianggap bertingkah laku mirip/sama dgn - lar. x dlm air g - BMx = 183.61 mol - XA = 09826 - Xx = ........ dst ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

SUHU AWAL PEMBEKUAN PROD PANGAN


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PROD PANGAN BEKU Hubungan antara % air beku vs. suhu


LAJU PEMBEKUAN

• 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 ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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LAJU PEMBEKUAN


LAJU PEMBEKUAN

Slow Freezing


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LAJU PEMBEKUAN

Rapid Freezing


LAJU PEMBEKUAN


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LAJU PEMBEKUAN


LAJU PEMBEKUAN


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LAJU PEMBEKUAN -- Klasifikasi berdasarkan pada laju pergerakan “ice front” • 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


WAKTU/LAMA PEMBEKUAN - 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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WAKTU/LAMA PEMBEKUAN

• 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 techniques - not real-world condition

+ Time-temp. methods most common + many people use time to freeze to – 10oC 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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WAKTU/LAMA 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.


WAKTU/LAMA PEMBEKUAN Plank’s Method (for infinite slab)

Tf T1


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WAKTU/LAMA PEMBEKUAN Plank’s Method (for infinite slab) x frozen

unfrozen Tf

frozen Tf

Ts

Ts

T1

T1 q

q a


WAKTU/LAMA PEMBEKUAN Plank’s Method (for infinite slab) Convection: BTU   hr 

q 

= Qt = h (Ts – T1) ...... Pers. 1

h = convective heat transfer coeff. at the product surface. Conduction: k .A q  f Tf Ts  .......... Pers. 2 x Tf = initial freezing point x = x (t) Combine 1&2: T T q  f 1 ........... Pers. 3 x 1  kf h ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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WAKTU/LAMA PEMBEKUAN Plank’s Method (for infinite slab) Jumlah energi yang dibebaskan selama proses pembekuan qdt = mi f = f dV f qdt = f f A dx q = f f A dx/dt .............. Pers. 4

so,

Ingat Pers 3 : T T q   f 1 x 1  kf h ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

WAKTU/LAMA PEMBEKUAN Plank’s Method (for infinite slab) Kombinasi Pers. 3 dan 4

………………….>

dx T f  T1 A  x 1 dt  kf h T dt

f  f A

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

1



Pembekuan selesai lempeng jika x = a/2 a 2

Tf  x 1 f  f     dx  T f  T1  dt h 0 0 k f 2 a  f  f  a tf    T f  Ti  8 k 2 h  Ti = Suhu Pembekuan Suhu ruang pembeku ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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WAKTU/LAMA PEMBEKUAN General Plank’s Equation : tf

2 Pa   f  f  Ra    h  T f  Ti  k f

Where: Infinite slab 1/2 1/8 Thickness

P R a

Sphere 1/6 1/24 Diameter

Infinite sylinder 1/4 1/6 Diameter

Cube 1/8 1/24 Edge

 f = latent heat of fusion [=] kJ kg kJ  water = 333.22 = 144 BTU kg lb ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

WAKTU/LAMA PEMBEKUAN General Plank’s Equation : P dan R untuk bentuk bata a : dimensi terpendek c : dimensi terpanjang b

c a

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


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WAKTU/LAMA PEMBEKUAN General Plank’s Equation :

In this figure, β1 and β2 are the ratios of the two longest sides to the shortest. It does not matter in what order they are taken. Chart providing P and R constants for Plank’s equation when applied to a brick or block geometry. ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

WAKTU/LAMA PEMBEKUAN General Plank’s Equation : 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


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WAKTU/LAMA PEMBEKUAN  Pham’s Equation : Pham (1986): improved Plank’s equation : • The mean freezing temperature is defined as

T fm  1 . 8  0 . 263 Tc  0 . 105 Ta where Tc is final center temperature and Ta is freezing medium temperature. The freezing time is given by : tF 

d c  H 1 H 2  N    1  Bi     T2   2  E f h   T1


WAKTU/LAMA PEMBEKUAN  Pham’s Equation : tF 

d c  H 1 H 2  N    1  Bi     T2   2  E f h   T1

where dc = characteristic dimension ‘r’ or shortest distance Ef = a shape factor (‘1’ for slab, ‘2’ for cylinder and ‘3’ for sphere)

 H 1   u c u ( T i  T fm )  H 2   f [ L f  c f (T fm  Tc )]  T  T fm  T1   i 2 

   T a 

ΔH1 = Enthalpy change during precooling, J/m3 ΔH2 = Enthalpy change during phase change and post-cooling period, J/m3

 T 2  T fm  T a ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

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WAKTU/LAMA PEMBEKUAN  Pham’s Equation : The value of Ef ? • One approach to determine the shape factor Ef in the calculation of freezing time:

2 2 ) (1  ) N Bi N Bi  E  1 21 2 2 ( 1  ) (  22  2 ) N Bi N Bi (1 

– 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) ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

WAKTU/LAMA PEMBEKUAN  Pham’s Equation : • For different shapes e.g. ellipsoid, rectangular brick, finite cylinder etc., the shape factor can be calculated:

2 2 ) (1  ) N Bi N Bi E  1  21 2 2 ( 1  ) ( 22  2 ) N Bi N Bi (1 

N Bi 

hc R k

• 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. V A • Same volume V 2  1  2 4 and  can be determined: • 1 3 2 R

 1 ( R ) 3


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PERHITUNGAN WAKTU PEMBEKUAN: CONTOH 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 oC. Estimate the time required to reduce the product temperature to -10 oC. An initial freezing temperature of -1.75 oC 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/CEK!!! (1) Determine shape factor:

0.25 0.6 0.25 0.6 A    3.056 2 0 . 25 R  (0.125) 2 )2 ( 2 V 0.25 0.6 1 2    5.999 4 3 4 3 1 ( R ) 3.056   (0.125) 3 3 (2) The Biot number is : hR 30  0.125   2.5 N Bi  c 1.5 k

1 

(3) Shape factor E: 2 2 2 2 ) (1 ) (1 ) (1 ) NBi NBi 2 .5 2 . 5   1   1.197 E  1  2 2 2 3 . 056 2 5.999   2 2 2 2 1 2 (3.056  ) (5.999  ) (1  ) (2  ) 2.5 2.5 NBi NBi (1


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PERHITUNGAN WAKTU PEMBEKUAN: CONTOH/CEK!!! (4) T3 :

T3  1 .8  0 .263 (  10 )  0 .105 (  30 )   3 .98

o

C

(5) H1: Cu(Ti-T3)

H1  Cu (Ti  T3 )  (3500J / kg.K)  (1050kg / m3 ) [5  (3.98) oC]  33001500J / m3 H 2  L  C f (T3  T f )  (333.22 kJ / kg 1000J / kJ )  (0.745) (1050kg / m3 )  1800J / kg.K   (1050kg / m3 )  (3.98  (10))  272,039,145 J / m3 (6) T1 and T2 :

(Ti  T3 ) (5  3.98)  Ta   (30)  30.51 oC 2 2 T2  T3  Ta   3.98  (30)  26.02 oC

T1 


PERHITUNGAN WAKTU PEMBEKUAN: CONTOH/CEK!!!

N Bi 0.125 33001500 272039145 R H1 H 2 2.5 (7) tslab : t slab  h [ T  T ](1  2 )  30 [ 30.51  26.02 ](1  2 ) 1 2  108,156 s

(8) t = tslab/E;

t 

t slab E



108 ,156  90355 s  25 .1 hr 1 .197

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


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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


METODE PEMBEKUAN …commercial freezing • 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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METODE PEMBEKUAN …freezing equipments

• Mechanical Freezers - Evaporate and compress the refrigerant in a continuous cycle • Cryogenic Systems - Use solid and liquid CO2, N2 directly in contact with the food




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A typical fluidized bed freezer




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Batch Freezer Blast Type

Source: Unit operations for food the food industries by: W.A. Gould ITP530 -- Freezing of Food , 2016 phariyadi.staff.ipb.ac.id

METODE PEMBEKUAN …freezing equipments Hydraulic Pump

Top Pressure plate Connecting Linkage Corner Headers Refrigerant hoses Trays

Polyurethane and polystyrene insulated doors

Contact plates

Unit operations for food the food industries by: ITP530 -- Freezing of Food ,Source: 2016 W.A. Gould

phariyadi.staff.ipb.ac.id

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Selamat belajar …


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ITP530 PEMBEKUAN PANGAN (Freezing of Foods)

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