1
Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.id
ITP530
PEMBEKUAN PANGAN
Purwiyatno Hariyadiphariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.id
Aspek engineering
Design (keperluan refrigerasi, T) Laju pembekuan (the rate at which freezing
progress)
Mutu produkProduktivitas
PEMBEKUANPEMBEKUAN
2
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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
PEMBEKUANPEMBEKUAN
ITP530/2015 phariyadi.staff.ipb.a.id
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
PEMBEKUANPEMBEKUAN
pembekuan
3
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ITP530/2015 phariyadi.staff.ipb.a.id
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)
PEMBEKUAN – pengaruhnya pd produk pangan
PEMBEKUAN – pengaruhnya pd produk pangan
ITP530/2015 phariyadi.staff.ipb.a.id
– Freezer burn ?
• Package properly
• Control temperature fluctuations in storage.
–Oxidation?
• Off‐flavors
• Vitamin loss
• Browning
–Recrystallization?
PEMBEKUAN – pengaruhnya pd produk pangan
PEMBEKUAN – pengaruhnya pd produk pangan
4
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ITP530/2015 phariyadi.staff.ipb.a.id
- 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 Sifat Produk Pangan Beku
ITP530/2015 phariyadi.staff.ipb.a.id
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) !!
Sifat Produk Pangan Beku Sifat Produk Pangan Beku Contoh :
5
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ITP530/2015 phariyadi.staff.ipb.a.id
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
Sifat Produk Pangan Beku Sifat Produk Pangan Beku Contoh (lanjutan):
ITP530/2015 phariyadi.staff.ipb.a.id
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 PembekuanPanas Laten Pembekuan
6
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ITP530/2015 phariyadi.staff.ipb.a.id
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
Kurva Pembekuan :….. untuk Air Murni
ITP530/2015 phariyadi.staff.ipb.a.id
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
ENERGY REMOVAL ASSOCIATED WITH FREEZING
(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
7
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ITP530/2015 phariyadi.staff.ipb.a.id
Titik Beku air
Super cooling
Titik eutetikAir
Larutan
Suhu
Waktu
Titik beku = f(waktu)
Driving force for nucleation/crystallization(i.e. T = T – Tf)
Kurva PembekuanKurva Pembekuan
Removal of latent heat
Removal of sensible heat
ITP530/2015 phariyadi.staff.ipb.a.id
3-21
Kurva PembekuanKurva Pembekuan
8
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ITP530/2015 phariyadi.staff.ipb.a.id
Ti
Tf
T
t
Setelah terjadi pembekuan, konsentrasi solute pada sisa larutan menjadi lebih tinggi .....> penurunan titik beku lebih besar .....> Tf
()
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%)
ITP530/2015 phariyadi.staff.ipb.a.id
You can’t freeze all of the water(Still have unfrozen (unfreezable) water : 5-10%)
Kurva Pembekuan :….. untuk Produk Pangan
Kurva Pembekuan :….. untuk Produk Pangan
9
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ITP530/2015 phariyadi.staff.ipb.a.id
Kurva Pembekuan …. for FishKurva Pembekuan …. for Fish
ITP530/2015 phariyadi.staff.ipb.a.id
Kurva Pembekuan …. for FishKurva Pembekuan …. for Fish
10
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ITP530/2015 phariyadi.staff.ipb.a.id
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
INTITIAL FREEZING TEMPERATURE
ITP530/2015 phariyadi.staff.ipb.a.id
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
INTITIAL FREEZING TEMPERATURE
11
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ITP530/2015 phariyadi.staff.ipb.a.id
INTITIAL FREEZING TEMPERATURE
ITP530/2015 phariyadi.staff.ipb.a.id
100
0- 40oC 0oC
% a
ir b
eku
Suhu
Hubungan antara % air beku vs. suhu
PRODUK BEKU?
12
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ITP530/2015 phariyadi.staff.ipb.a.id
Hubungan antara % air beku vs. suhu
PRODUK BEKU?
ITP530/2015 phariyadi.staff.ipb.a.id
• 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
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ITP530/2015 phariyadi.staff.ipb.a.id
LAJU PEMBEKUAN
ITP530/2015 phariyadi.staff.ipb.a.id
LAJU PEMBEKUAN
Slow Freezing
14
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ITP530/2015 phariyadi.staff.ipb.a.id
LAJU PEMBEKUAN
Rapid Freezing
ITP530/2015 phariyadi.staff.ipb.a.id
LAJU PEMBEKUAN
15
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ITP530/2015 phariyadi.staff.ipb.a.id
LAJU PEMBEKUAN
ITP530/2015 phariyadi.staff.ipb.a.id
LAJU PEMBEKUAN
16
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ITP530/2015 phariyadi.staff.ipb.a.id
- 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
ITP530/2015 phariyadi.staff.ipb.a.id
• 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.
PERHITUNGAN WAKTU PEMBEKUAN
17
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ITP530/2015 phariyadi.staff.ipb.a.id
• 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)
PERHITUNGAN WAKTU PEMBEKUAN
ITP530/2015 phariyadi.staff.ipb.a.id
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.
18
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ITP530/2015 phariyadi.staff.ipb.a.id
Tf
T1
PERHITUNGAN WAKTU PEMBEKUAN:Plank’s Method (for infinite slab)
ITP530/2015 phariyadi.staff.ipb.a.id
xfrozen frozen
Tf Tf
Ts
T1
Ts
T1
unfrozen
a
PERHITUNGAN WAKTU PEMBEKUAN:Plank’s Method (for infinite slab)
19
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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:
sff TTx
A.kq .......... Pers. 2
Tf = initial freezing pointx = x (t)
Combine 1&2:
h
1
k
xTT
q
f
1f
........... Pers. 3
PERHITUNGAN WAKTU PEMBEKUAN:Plank’s Method (for infinite slab)
ITP530/2015 phariyadi.staff.ipb.a.id
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
Ingat Pers 3 :
h
1
k
xTT
q
f
1f
PERHITUNGAN WAKTU PEMBEKUAN:Plank’s Method (for infinite slab)
20
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ITP530/2015 phariyadi.staff.ipb.a.id
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 ………………….>
PERHITUNGAN WAKTU PEMBEKUAN:Plank’s Method (for infinite slab)
ITP530/2015 phariyadi.staff.ipb.a.id
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
PERHITUNGAN WAKTU PEMBEKUAN:General Plank’s Equation
21
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ITP530/2015 phariyadi.staff.ipb.a.id
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
PERHITUNGAN WAKTU PEMBEKUAN:General Plank’s Equation
ITP530/2015 phariyadi.staff.ipb.a.id
PERHITUNGAN WAKTU PEMBEKUAN:General Plank’s Equation
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.
22
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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
ITP530/2015 phariyadi.staff.ipb.a.id
Pham (1986): improved Plank’s equation :
• The mean freezing temperature is defined as
acfm TTT 105.0263.08.1
2
12
2
1
1 Bi
f
cF
N
T
H
T
H
hE
dt
where Tc is final center temperature and Ta is freezing medium temperature. The freezing time is given by
PERHITUNGAN WAKTU PEMBEKUAN:Pham’s Equation
23
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ITP530/2015 phariyadi.staff.ipb.a.id
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
PERHITUNGAN WAKTU PEMBEKUAN:Pham’s Equation
2
12
2
1
1 Bi
f
cF
N
T
H
T
H
hE
dt
ITP530/2015 phariyadi.staff.ipb.a.id
In Pham’s method, the value of Ef is adjusted (Eq. 7.16):
Ef = G1 + G2E1 + G3E2
where the values of G1, G2 and G3 are given in Table 7.1 and E1
and E2 are 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:Pham’s Equation
24
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ITP530/2015 phariyadi.staff.ipb.a.id
• Alternate 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
PERHITUNGAN WAKTU PEMBEKUAN:Pham’s Equation
ITP530/2015 phariyadi.staff.ipb.a.id
• 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
PERHITUNGAN WAKTU PEMBEKUAN:Pham’s Equation
25
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ITP530/2015 phariyadi.staff.ipb.a.id
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
ITP530/2015 phariyadi.staff.ipb.a.id
(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
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ITP530/2015 phariyadi.staff.ipb.a.id
(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
PERHITUNGAN WAKTU PEMBEKUAN:CONTOH/CEK!!!
ITP530/2015 phariyadi.staff.ipb.a.id
(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.
PERHITUNGAN WAKTU PEMBEKUAN:CONTOH/CEK!!!
27
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ITP530/2015 phariyadi.staff.ipb.a.id
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
ITP530/2015 phariyadi.staff.ipb.a.id
• 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
Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.id
• 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 equipment
ITP530/2015 phariyadi.staff.ipb.a.id
• 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-- berdasarkan pada laju pergerakan “ice front”
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Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.id
ALAT PEMBEKU
ITP530/2015 phariyadi.staff.ipb.a.id
A typical fluidized bed freezer
ALAT PEMBEKU
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Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.id
ALAT PEMBEKU
ITP530/2015 phariyadi.staff.ipb.a.id
Batch Freezer
Source: Unit operations for food the food industries by: W.A. Gould
Blast Type
ALAT PEMBEKU
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Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.idSource: 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
ALAT PEMBEKU
ITP530/2015 phariyadi.staff.ipb.a.id
ALAT PEMBEKU
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Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.id
ALAT PEMBEKU
ITP530/2015 phariyadi.staff.ipb.a.id
ALAT PEMBEKU
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Purwiyatno HariyadiEmail: [email protected]: phariyadi.staff.ipb.ac.id
ITP530/2015 phariyadi.staff.ipb.a.id
ALAT PEMBEKU