ITP530 PEMBEKUAN PANGAN (Freezing of Foods)

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1 ITP530 PEMBEKUAN PANGAN (Freezing of Foods) Purwiyatno Hariyadi, PhD Department of Food Science & Technology, Bogor Agricultural University, BOGOR, Indonesia PEMBEKUAN PANGAN Aspek engineering Design (keperluan refrigerasi, T) Laju pembekuan (the rate at which freezing progress) Mutu produk Produktivitas 1

2 PEMBEKUAN PANGAN Penyimpanan produk pada T < suhu beku Umumnya pada T < 8 F (- 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 pembekuan Melting/Freezing Boiling/Condensating

3 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) PEMBEKUAN PANGAN Pengaruhnya pada Produk Pangan : + PENGARUH NEGATIF Kerusakan kimia Kerusakan fisik (textural) 3

4 PEMBEKUAN PANGAN Pengaruhnya pada Produk Pangan : + 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) R T f T T K.m f A0 g Lar. X dlm air T f = (1.86 m) o C R 1 g 1 T A0 1 T BM.m A A ln X dimana: mol solut m molalitas mg pelarut T titik beku pelarut murni (A) air A0 R g panas kons tan ta laten J gas mol.k pembekuan BM A = Berat Molekul pelarut K = konstanta molal titik beku A kj, kg X A = fraksi mol air 1 = panas laten pembekuan air K kj 335 kg,73k 4

5 SIFAT PRODUK PANGAN BEKU Contoh : Ditanya T f =? T f m =? m R T g mol kg Ice cream mix dengan komposisi sbb: 10% butterfat 1% solid-not-fat (54.5%: laktosa) 15% sukrosa 0.% stabilizer 37.% Air = 6.78% A0 L solven BM.m solut A Asumsi bahwa hanya gula (laktosa + fruktosa) yang memp. Efek menurunkan titik beku)!! Solut? sukrosa BM = 34 laktosa BM = 34 solut lain diabaikan!! SIFAT PRODUK PANGAN BEKU Contoh (lanjutan): Fraksi gula = (0.545) = g/g Fraksi air = g gula Konsentrasi gula dlm air = g air g gula mol gula 343, g air m m 1000 g air T f T f = 1,86 K J mol.k 1mol 18g 73K J kg g mol mol kg 5

6 PANAS LATEN PEMBEKUAN Air murni = 335 kj kg Larutan solid x dlm air = (335 m w ) Contoh: m w = Fraksi massa air Kadar air kj kg Selada Strawberi Kacang panjang Kentang Daging kambing Kacang merah, biji kering Kurma kering kj kg Perhitungan berdasarkan pd rumus kj = 335 m w kg Air: 335 kj kg kg 3 J mol J 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). 6

7 KURVA PEMBEKUAN untuk air murni & energi Removal of heat (Q) from Region I (sensible heat), II (latent heat), and III (sensible heat) : (1) Q 1 = mc p1 T 1 m = weight of food C p1 =specific heat of food above freezing T = temperature difference () Q = m w... > m w = weight of water... > = latent heat (3) Q 3 = mc p T 3... > m = weight of food... > C p = specific heat of frozen food... > T 3 = temperature difference KURVA PEMBEKUAN untuk produk pangan Suhu Removal of sensible heat Removal of latent heat Titik Beku air Titik beku Super cooling = f(waktu) Air Titik eutetik Larutan Waktu Driving force for nucleation/crystallization (i.e. T = T T f ) 7

8 3-1 KURVA PEMBEKUAN untuk produk pangan KURVA PEMBEKUAN untuk produk pangan T T i T f 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%) 8

9 KURVA PEMBEKUAN untuk produk pangan You can t freeze all of the water (Still have unfrozen (unfreezable) water : 5-10%) KURVA PEMBEKUAN untuk produk IKAN 9

10 KURVA PEMBEKUAN untuk produk IKAN SUHU AWAL PEMBEKUAN PROD PANGAN Buah anggur (grape)... > kadar air 84.7%... > T f = -1.8 o C (71. o K) = 6003 J ln X mol A R g TA0 TA R g = J mol.k X A =? J 6003 mol J 8,314 mol.k 1 73 K K ln X A Ln X A = X A = (effective mol fraction of water ) m grape ml 10

11 SUHU AWAL PEMBEKUAN PROD PANGAN X A = fraksi mol air = X A = = BM E = BM E Juice anggur dapat dianggap bertingkah laku mirip/sama dgn - lar. x dlm air - BM x = X A = X x =... dst g mol SUHU AWAL PEMBEKUAN PROD PANGAN 11

12 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 1

13 LAJU PEMBEKUAN LAJU PEMBEKUAN Slow Freezing 13

14 LAJU PEMBEKUAN Rapid Freezing LAJU PEMBEKUAN 14

15 LAJU PEMBEKUAN LAJU PEMBEKUAN 15

16 LAJU PEMBEKUAN -- Klasifikasi berdasarkan pada laju pergerakan ice front Slow Freezers 0. cm/h - Still air and cold stores Quick Freezers cm/h - Air blast and plate freezers Rapid Freezers 5-10 cm/h - Fluidized bed freezers Ultra rapid Freezers 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 16

17 WAKTU/LAMA PEMBEKUAN Time-temperature method Time required to freeze between two temperatures (usually T = -5 o 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 10 o C as standard. WAKTU/LAMA PEMBEKUAN panas laten adalah energi utama yang hrs diperhitungkan pada proses pembekuan ~ 75% total energi pd proses pembekuan kj/kg air 144 BTU/lb air Terjadi perubahan sifat fisik bahan selama proses pembekuan ~ f (T,m) 17

18 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) T f T1 18

19 WAKTU/LAMA PEMBEKUAN Plank s Method (for infinite slab) x frozen unfrozen frozen T f T f T s T s T 1 q q T 1 a WAKTU/LAMA PEMBEKUAN Plank s Method (for infinite slab) Convection: BTU q = Qt = h (Ts T1)... Pers. 1 hr h = convective heat transfer coeff. at the product surface. Conduction: k f.a q T f T s x... Pers. Tf = initial freezing point x = x (t) Combine 1&: T f T1 q... Pers. 3 x 1 k h f 19

20 WAKTU/LAMA PEMBEKUAN Plank s Method (for infinite slab) Jumlah energi yang dibebaskan selama proses pembekuan qdt = m i f = f dv f qdt = f f A dx so, q = f f A dx/dt... Pers. 4 Ingat Pers 3 : T f T1 q x 1 k h f dx Kombinasi Pers. 3 dan 4.> f f A dt x 1 f f dx T f T1dt k f h Pembekuan selesai lempeng jika x = a/ WAKTU/LAMA PEMBEKUAN Plank s Method (for infinite slab) t f f T i = a x f 0 k f f f T T f i 1 dx h a 8 k a h Suhu Pembekuan Suhu ruang pembeku T f T 1 T f 0 dt Tf T1 A x 1 k h f 0

21 WAKTU/LAMA PEMBEKUAN General Plank s Equation : t f f f Ra Pa T f Ti k f h Where: Infinite slab Sphere Infinite sylinder Cube P 1/ 1/6 1/4 1/8 R 1/8 1/4 1/6 1/4 a Thickness Diameter Diameter Edge f = latent heat of fusion [=] kj kg kj water = 333. = 144 BTU kg lb WAKTU/LAMA PEMBEKUAN General Plank s Equation : P dan R untuk bentuk bata b a c a : dimensi terpendek c : dimensi terpanjang B = c/a B 1 = b/a Lihat chart/diagram : dengan diketahui nilai B dan B 1 maka dapat dibaca nilai P dan R 1

22 WAKTU/LAMA PEMBEKUAN General Plank s Equation : In this figure, β 1 and β 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. 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

23 WAKTU/LAMA PEMBEKUAN Pham s Equation : Pham (1986): improved Plank s equation : The mean freezing temperature is defined as T T T fm where T c is final center temperature and T a is freezing medium temperature. The freezing time is given by : t F c d c E h f H T a H T 1 N 1 Bi 1 WAKTU/LAMA PEMBEKUAN Pham s Equation : where d c = characteristic dimension r or shortest distance = a shape factor ( 1 for slab, for cylinder and 3 for sphere) E f t F d c E h f H T H T 1 N 1 Bi 1 H H T 1 T c ( T T 1 u u i fm [ L c ( T f f f fm c T i T fm T fm T a T a ) T )] ΔH 1 = Enthalpy change during precooling, J/m 3 ΔH = Enthalpy change during phase change and post-cooling period, J/m 3 3

24 WAKTU/LAMA PEMBEKUAN Pham s Equation : The value of E f? One approach to determine the shape factor E f in the calculation of freezing time: E (1 ) (1 ) N Bi N Bi 1 1 ( 1 ) ( ) N N Bi For infinite slab, the shape factor E = 1 (since 1 =infinite, =infinite) For an infinite cylinder, the shape factor E= (since 1 =1, =infinite) For a sphere, the shape factor, E = 3 ( 1 =1, =1) Bi WAKTU/LAMA 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 A V 1 1 and can be determined: R 4 3 1( R ) 3 (1 ) (1 ) N Bi N Bi E 1 1 ( 1 ) ( ) N N Bi Bi N Bi h R c k 4

25 PERHITUNGAN WAKTU PEMBEKUAN: Lean beef with 74.5% moisture content and 1 m length, 0.6 m width, and 0.5 m thickness is being frozen in an air-blast freezer with h c = 30 W/m.K and air temperature of -30 o C. 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 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/m 3 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 d = 1 m Product width d 1 = 0.6 m Product thickness a = 0.5 m Convective heat-transfer coefficient h c = 30 W/m.k Air temperature T = -30 o C Initial product temperature T i = 5 o C Initial freezing temperature T F = o C Product density = 1050 kg/m 3 Enthalpy change (H) = kj/kg = 48.5 kj/kg (estimate) Thermal conductivity k of frozen product = 1.5 W/m.K Specific heat of product (C pu ) = 3.5 kj/kg.k Specific heat of frozen product (C pf ) = 1.8 kj/kg. CONTOH PERHITUNGAN WAKTU PEMBEKUAN: CONTOH/CEK!!! (1) Determine shape factor: A R 0.5 (0.15) ( ) V ( R ) (0.15) 3 3 () The Biot number is : hc R N Bi.5 k 1.5 (3) Shape factor E: (1 ) (1 ) (1 ) (1 ) NBi NBi E ( 1 ) ( ) (3.056 ) (5.999 ) N N.5.5 Bi Bi 5

26 PERHITUNGAN WAKTU PEMBEKUAN: CONTOH/CEK!!! (4) T 3 : T ( 10) ( 30) 3.98 o C H (5) H 1 : C u (T i -T 3 ) H C ( T T ) 1 u i J / m L C ( T f o (3500J / kgk. ) (1050kg/ m ) [5 ( 3.98) C] 3 T ) (333.kJ / kg1000j / kj) (0.745)(1050kg/ m ) f 1800J / kg. K 3 7,039,145 J / m (6) T 1 and T : ( Ti T3 ) (5 3.98) T1 Ta ( 30) o T T T 3.98( 30) 6.0 C 3 a 3 (1050kg/ m ) ( 3.98( 10)) o C PERHITUNGAN WAKTU PEMBEKUAN: CONTOH/CEK!!! (7) t slab : t slab R H1 H NBi [ ](1 ) [ ](1 ) h T T ,156 s (8) t = t slab /E; t t slab E 108, s 5.1 hr Required time for lean beef (1 m 0.6m 0.5 m) will be 5.1 hours to freeze. 6

27 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. 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. 7

28 METODE PEMBEKUAN freezing equipments Mechanical Freezers - Evaporate and compress the refrigerant in a continuous cycle Cryogenic Systems - Use solid and liquid CO, N directly in contact with the food METODE PEMBEKUAN freezing equipments 8

29 METODE PEMBEKUAN freezing equipments A typical fluidized bed freezer METODE PEMBEKUAN freezing equipments 9

30 METODE PEMBEKUAN freezing equipments Blast Type Batch Freezer Source: Unit operations for food the food industries by: W.A. Gould METODE PEMBEKUAN freezing equipments Hydraulic Pump Top Pressure plate Connecting Linkage Corner Headers Refrigerant hoses Trays Polyurethane and polystyrene insulated doors Contact plates ITP Freezing of Food, Source: 016 Unit operations for food the food industries by: W.A. Gould 30

31 METODE PEMBEKUAN freezing equipments METODE PEMBEKUAN freezing equipments 31

32 METODE PEMBEKUAN freezing equipments METODE PEMBEKUAN freezing equipments 3

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