2016-ITP530-Freezing of Foods

1

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

ITP530 PEMBEKUAN PANGAN(Freezing of Foods)

Purwiyatno Hariyadi, PhD• Department of Food Science & Technology, Bogor Agricultural

University, BOGOR, Indonesia


Aspek engineering

Design (keperluan refrigerasi, T) Laju pembekuan (the rate at which freezing

progress)

Mutu produkProduktivitas

PEMBEKUAN PANGAN

2


• 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 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

Boiling/Condensating

pembekuan

PEMBEKUAN PANGAN

3


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)

PEMBEKUAN PANGAN

Pengaruhnya pada Produk Pangan :

+


PENGARUH NEGATIF

• Kerusakan kimia• Kerusakan fisik (textural)

PEMBEKUAN PANGAN


+

4


PENGARUH NEGATIF

• Freezer burn ?• Package properly • Control temperature fluctuations in storage.

• Oxidation? • Off-flavors • Vitamin loss • Browning

• Recrystallization?

PEMBEKUAN PANGAN


+


- Penurunan titik beku = f (konsentrasi, BM)

m.KT

m.BMTRT

f

A2

0Agf

kg

kJ335air

kg

kJ,pembekuanlatenpanas

K.mol

J314.8gastatankonsR

K273,Kair)A(murnipelarutbekutitikT

.pelarutmg1000

solutmolmolalitasm

g

0A

dimana:

BMA = Berat Molekul pelarutK = konstanta molal titik beku

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

A

A0Ag

1

XlnT

1

T

1

R

XA = fraksi mol air1 = panas laten pembekuan

SIFAT PRODUK PANGAN BEKU

5


Ice cream mix dengan komposisi sbb:10% butterfat12% solid-not-fat (54.5%: laktosa)15% sukrosa0.22% stabilizer

37.22% Air = 62.78%

Ditanya Tf = ?

L

m.BMTRT

A2

0Ag

f

m = ?

solvenkg

solutmolm

Solut? sukrosa BM = 342laktosa BM = 342solut lain diabaikan !!

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

Contoh :



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

airg1000

gulag1,343

airg

gulag3431.0

6278.0

2154.0

m003.1airg1000

gulamol342

1.343

m

kg

J335.1000

kg

mol003.1

mol

g18K273

g18

mol1

K.mol

J314.8

T

2

f

Tf = 1,86 K

Contoh (lanjutan):


6


Air murni = 335

Larutan solid x dlm air = (335 mw)

kg

kJ

kgkJ

mw = Fraksi massa airContoh:

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 mwkg

kJ

Air:mol

J6030

mol1

1018

kg

J10335

kg

kJ335

33

PANAS LATEN PEMBEKUAN


T-t Diagram :

A schematic freezing curve for water, displayingsensible heat loss (Regions I and III) and latent heat loss (Region II).

KURVA PEMBEKUAN … untuk air murni

7


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

(1) Q1 = mCp1T1

m = weight of food Cp1 =specific heat of food above freezing T = temperature difference

(2) Q2 = mw ........> mw = weight of water ........> = latent heat

(3) Q3 = mCp2T3........> m = weight of food ........> Cp2 = specific heat of frozen food ........> T3 = temperature difference

KURVA PEMBEKUAN … untuk air murni & energi


Titik Beku air

Super cooling

Titik eutetikAir

Larutan

Suhu

Waktu

Titik beku = f(waktu)

Driving force for nucleation/crystallization(i.e. T = T – Tf)

Removal of latent heat

Removal of sensible heat

KURVA PEMBEKUAN … untuk produk pangan

8


3-21



Ti

Tf

T

t

Setelah terjadi pembekuan, konsentrasi solute pada sisa larutan menjadi lebih tinggi .....> penurunan titik beku lebih besar .....> Tf

()

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


9


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



KURVA PEMBEKUAN … untuk produk IKAN

10


KURVA PEMBEKUAN … untuk produk IKAN


Buah anggur (grape) ........> kadar air 84.7%........> Tf = -1.8oC (271.2oK)

A

A0Ag

1

XlnT

1

T

1

R

1 = 6003

Rg = 8.314 K.mol

JmolJ

XA=?

AXlnK2.271

1K273

1

K.molJ

314,8

molJ

6003

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

ml

grapem

SUHU AWAL PEMBEKUAN PROD PANGAN

11


XA = fraksi mol air = 0.9826

XA = 0.9826 =

EBM

3.15

18

7.8418

7.84

BME = 183.61

Juice anggur dapat dianggap bertingkah laku mirip/sama dgn- lar. x dlm air- BMx = 183.61

- XA = 09826- Xx = ........ dst

mol

g




12


Hubungan antara % air beku vs. suhu

PROD PANGAN 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

LAJU PEMBEKUAN

13


LAJU PEMBEKUAN


Slow Freezing

LAJU PEMBEKUAN

14


Rapid Freezing

LAJU PEMBEKUAN


LAJU PEMBEKUAN

15


LAJU PEMBEKUAN


LAJU PEMBEKUAN

16


• 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


- Pendugaan keperluan pembekuan ukuran sistem “mechanical

compression” evaluasi beban refrigerasi/pembekuan

- Disain peralatan + proses, untuk : memperoleh pembekuan yg diinginkan

- koef pindah panas- laju pembekuan

WAKTU/LAMA PEMBEKUAN

17


• 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.



• panas laten adalah energi utama yang hrs diperhitungkan pada proses pembekuan • ~ 75% total energi pd proses pembekuan

333.3 kJ/kg air144 BTU/lb air

• Terjadi perubahan sifat fisik bahan selama proses pembekuan ~ f (T,m)


18


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.



Tf

T1



19


xfrozen frozen

Tf Tf

Ts

T1

Ts

T1

qq

unfrozen

a




Convection:

q

hr

BTU= Qt = h (Ts – T1) ...... Pers. 1

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

sff TTx

A.kq .......... Pers. 2

Tf = initial freezing pointx = x (t)

Combine 1&2:

h

1

k

xTT

q

f

1f

........... Pers. 3



20


Ingat Pers 3 :

h

1

k

xTT

q

f

1f



Jumlah energi yang dibebaskan selama proses pembekuan

qdt = mi f = f dV f

qdt = f f A dx

so, q = f f A dx/dt .............. Pers. 4


dtTTdxh

1

k

x1f

fff

fT

01f

2

a

0 fff dtTTdx

h

1

k

x

h2

a

k8

a

TTt

2

if

fff

Pembekuan selesai lempeng jika x = a/2

Ti = Suhu PembekuanSuhu ruang pembeku

h1

kx

ATT

dt

dxA

f

1fff

Kombinasi Pers. 3 dan 4 ………………….>



21


h

Pa

k

Ra

TTt

f

2

if

fff

Where:Infinite slab Sphere Infinite sylinder Cube

P 1/2 1/6 1/4 1/8R 1/8 1/24 1/6 1/24a Thickness Diameter Diameter Edge

f= latent heat of fusion [=] kJkg

kg

kJ water = 333.22 = 144

lbBTU

General Plank’s Equation :



a

b c

P dan R untuk bentuk bata

a : dimensi terpendekc : dimensi terpanjang

B2 = c/aB1 = b/a

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



22


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

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.




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



23


Pham (1986): improved Plank’s equation :

• The mean freezing temperature is defined as

acfm TTT 105.0263.08.1

where Tc is final center temperature and Ta is freezing medium temperature.

The freezing time is given by :

2

12

2

1

1 Bi

f

cF

N

T

H

T

H

hE

dt

Pham’s Equation :



where dc = characteristic dimension ‘r’ or shortest distance

Ef = a shape factor (‘1’ for slab, ‘2’ for cylinder and ‘3’ for sphere)

)(1 fmiuu TTcH

)]([2 cfmfff TTcLH

afmi T

TTT

21

afm TTT 2

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

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

2

12

2

1

1 Bi

f

cF

N

T

H

T

H

hE

dt

Pham’s Equation :


24


The value of Ef ?

Pham’s Equation :


• One approach to determine the shape factor Ef in 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(

122

212

1Bi

Bi

Bi

Bi

N

N

N

NE


• 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• 1 and 2 can be determined:

)2

(

)2

1(

)2

(

)2

1(

122

212

1Bi

Bi

Bi

Bi

N

N

N

NE

21 R

A

)3

4( 3

1

2

R

V

k

RhN c

Bi

Pham’s Equation :


25


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) = 0.745333.22 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(

6.025.0

)2

25.0(

6.025.02

221

R

A

999.5)125.0(

3

4056.3

16.025.0

)3

4( 33

1

2

R

V

5.25.1

125.030

k

RhN c

Bi

197.1)

5.2

999.52999.5(

)5.2

21(

)5.2

056.32056.3(

)5.2

21(

1)

2(

)2

1(

)2

(

)2

1(

12222

212

1

Bi

Bi

Bi

Bi

N

N

N

NE

PERHITUNGAN WAKTU PEMBEKUAN:CONTOH/CEK!!!

26


(4) T3 :

(5) H1: Cu(Ti-T3)

(6) T1 and T2 :

CT o98.3)30(105.0)10(263.08.13

3

331

/33001500

])98.3(5[)/1050()./3500()(

mJ

CmkgKkgJTTCH oiu

3

3

332

/145,039,272

))10(98.3()/1050(./1800

)/1050()745.0()/1000/22.333()(

mJ

mkgKkgJ

mkgkJJkgkJTTCLH ff

CTTT

CTTT

T

oa

oa

i

02.26)30(98.3

51.30)30(2

)98.35(

2

)(

32

31



(7) tslab :

(8) t = tslab/E;

s

N

T

H

T

H

h

Rt Bi

slab

156,108

)2

5.21](

02.26

272039145

51.30

33001500[

30

125.0)

21]([

2

2

1

1

hrsE

tt slab 1.2590355

197.1

156,108

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


27


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

28


• 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

METODE PEMBEKUAN …freezing equipments



29


A typical fluidized bed freezer




30


Batch Freezer

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

Blast Type



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

Hydraulic Pump

Top Pressure plate

Connecting Linkage

Corner Headers

Refrigerant hoses

Trays

Contact plates

Polyurethane and polystyrene insulated doors


31





32





33


Selamat belajar …

Documents

2016-ITP530-Freezing of Foods