ITP530 PEMBEKUAN PANGAN

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1 ITP530 PEMBEKUAN PANGAN phariyadi.sta.ipb.ac.id PEMBEKUAN Aspek engineering Design (keperluan rerigerasi, T) Laju pembekuan (the rate at which reezing progress) Mutu produk Produktivitas 1

2 PEMBEKUAN 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 aktor : perlu dikendalikan Pertumbuhan mikroorganisme dihambat, bbrp bahkan inakti PurwiyatnoHariyadi/IPN/ITP/Fateta/IPB PEMBEKUAN Things to notice: Pressure and temperature both aect the phase o matter. All three phases o matter exist at the triple point pembekuan Melting/Freezing Boiling/Condensating

3 PEMBEKUAN PENGARUH POSITIF pengaruhnya pd produk pangan 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 isik (textural) PEMBEKUAN Freezer burn? Package properly Control temperature luctuations in storage. Oxidation? O lavors Vitamin loss Browning Recrystallization? pengaruhnya pd produk pangan 3

4 Siat Produk Pangan Beku - Penurunan titik beku = (konsentrasi, BM) R T T T K.m g A0 Lar. X dlm air T = (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 = raksi mol air 1 = panas laten pembekuan air K kj 335 kg,73k Siat Produk Pangan Beku Contoh : Ditanya T =? T m =? m R g mol kg T Ice cream mix dengan komposisi sbb: 10% butterat 1% solid-not-at (54.5%: laktosa) 15% sukrosa 0.% stabilizer 37.% Air = 6.78% A0 L solven BM.m solut A Asumsi bahwa hanya gula (laktosa + ruktosa) yang memp. Eek menurunkan titik beku)!! Solut? sukrosa BM = 34 laktosa BM = 34 solut lain diabaikan!! 4

5 Siat 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 T = 1,86 K J mol.k 1mol 18g 73K J kg g mol mol kg 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 5

6 Kurva Pembekuan :.. untuk Air Murni T-t Diagram : A schematic reezing curve or water, displaying sensible heat loss (Regions I and III) and latent heat loss (Region II). ENERGY REMOVAL ASSOCIATED WITH FREEZING Removal o heat (Q) rom Region I (sensible heat), II (latent heat), and III (sensible heat) : (1) Q 1 = mc p1 T 1 m = weight o ood C p1 =speciic heat o ood above reezing T = temperature dierence () Q = m w... > m w = weight o water... > = latent heat (3) Q 3 = mc p T 3... > m = weight o ood... > C p = speciic heat o rozen ood... > T 3 = temperature dierence 6

7 Kurva Pembekuan Suhu Removal o sensible heat Removal o latent heat Titik Beku air Titik beku Super cooling = (waktu) Air Titik eutetik Larutan Waktu Driving orce or nucleation/crystallization (i.e. T = T T ) 3-1 Kurva Pembekuan 7

8 Kurva Pembekuan :.. untuk Produk Pangan T T i T Setelah terjadi pembekuan, konsentrasi solute pada sisa larutan menjadi lebih tinggi... > penurunan titik beku lebih besar... > T () t You can t reeze all o the water (Still have unrozen (unreezable) water : 5-10%) Kurva Pembekuan :.. untuk Produk Pangan You can t reeze all o the water (Still have unrozen (unreezable) water : 5-10%) 8

9 Kurva Pembekuan. or Fish Kurva Pembekuan. or Fish 9

10 INTITIAL FREEZING TEMPERATURE Buah anggur (grape)... > kadar air 84.7%... > T = -1.8 o C (71. o K) 1 1 = 6003 R g 1 T A0 1 T A ln X X A =? J 6003 mol J 8,314 mol.k A 1 73 R g = K K J mol J mol.k ln X A Ln X A = X A = (eective mol raction o water ) m grape ml INTITIAL FREEZING TEMPERATURE X A = raksi 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 10

11 INTITIAL FREEZING TEMPERATURE PRODUK BEKU? Hubungan antara % air beku vs. suhu 100 % air beku 0-40 o C 0 o C Suhu 11

12 PRODUK BEKU? Hubungan antara % air beku vs. suhu LAJU PEMBEKUAN EQUIPMENT RELATED rate o heat transer size o rerigeration unit FOOD/PRODUCT QUALITY slow reezing result in ormation o ew, large ice crystals damaging to cell structure/quality rapid reezing 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 PERHITUNGAN WAKTU PEMBEKUAN - Pendugaan keperluan pembekuan ukuran sistem mechanical compression evaluasi beban rerigerasi/pembekuan - Disain peralatan + proses, untuk : memperoleh pembekuan yg diinginkan - koe pindah panas - laju pembekuan PERHITUNGAN WAKTU PEMBEKUAN Time-temperature method Time required to reeze between two temperatures (usually T = -5 o C or 10 o C) Velocity o ice ront - rate o reezing - must be able to see ice ront Appearance o specimen - internal conditions Thermal methods - calorimetric techniques - not real-world condition +Time-temp. methods most common +many people use time to reeze to 10 o C as standard. 16

17 PERHITUNGAN WAKTU 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 siat isik bahan selama proses pembekuan ~ (T,m) PERHITUNGAN WAKTU PEMBEKUAN: Plank s Method (or ininite slab) Plank s equation is an approximate analytical solution or a simpliied phase-change model. Plank assumed that the reezing process: (a) commences with all o the ood unrozen but at its reezing temperature. (b) occurs suiciently slowly or heat transer in the rozen layer to take place under steady-state conditions. 17

18 PERHITUNGAN WAKTU PEMBEKUAN: Plank s Method (or ininite slab) T T1 PERHITUNGAN WAKTU PEMBEKUAN: Plank s Method (or ininite slab) x rozen unrozen rozen T T T s T s T 1 q q T 1 a 18

19 PERHITUNGAN WAKTU PEMBEKUAN: Plank s Method (or ininite slab) Convection: BTU q = Qt = h (Ts T1)... Pers. 1 hr h = convective heat transer coe. at the product surace. Conduction: k.a q T T s x... Pers. T = initial reezing point x = x (t) Combine 1&: T T1 q... Pers. 3 x 1 k h PERHITUNGAN WAKTU PEMBEKUAN: Plank s Method (or ininite slab) Jumlah energi yang dibebaskan selama proses pembekuan qdt = m i = dv qdt = A dx so, q = A dx/dt... Pers. 4 Ingat Pers 3 : T T1 q x 1 k h 19

20 PERHITUNGAN WAKTU PEMBEKUAN: Plank s Method (or ininite slab) Kombinasi Pers. 3 dan 4.> x 1 dx k h T dx A dt T 1 dt T T1 A x 1 k h Pembekuan selesai lempeng jika x = a/ t T i = a x 0 k T T i 1 dx h a 8 k a h Suhu Pembekuan Suhu ruang pembeku T T 1 T 0 dt PERHITUNGAN WAKTU PEMBEKUAN: General Plank s Equation t Ra Pa T Ti k h Where: Ininite slab Sphere Ininite sylinder Cube P 1/ 1/6 1/4 1/8 R 1/8 1/4 1/6 1/4 a Thickness Diameter Diameter Edge = latent heat o usion [=] kj kg kj water = 333. BTU = 144 kg lb 0

21 PERHITUNGAN WAKTU 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 PERHITUNGAN WAKTU PEMBEKUAN: General Plank s Equation In this igure, β 1 and β are the ratios o the two longest sides to the shortest. It does not matter in what order they are taken. Chart providing P and R constants or Plank s equation when applied to a brick or block geometry. 1

22 PERHITUNGAN WAKTU PEMBEKUAN: General Plank s Equation Limitation o Plank s method: no superheating or supercooling thermal properties are constant can t incorporate a variable heat transer coe. can t handle varying reezing point PERHITUNGAN WAKTU PEMBEKUAN: Pham s Equation Pham (1986): improved Plank s equation : The mean reezing temperature is deined as T T T m where T c is inal center temperature and T a is reezing medium temperature. The reezing time is given by t F d E H h T c H T c 1 N 1 Bi 1 a

23 PERHITUNGAN WAKTU PEMBEKUAN: d Pham s Equation c H 1 H N Bi tf 1 E h T1 T where d c = characteristic dimension r or shortest distance E H H = a shape actor ( 1 or slab, or cylinder and 3 or sphere) c ( T T ) 1 u u i m [ L c ( T T m c Ti T m T1 T T m T a T a )] ΔH 1 = Enthalpy change during pre-cooling, J/m 3 ΔH = Enthalpy change during phase change and post-cooling period, J/m 3 PERHITUNGAN WAKTU PEMBEKUAN: Pham s Equation In Pham s method, the value o E is adjusted (Eq. 7.16): E = G 1 + G E 1 + G 3 E where the values o G 1, G and G 3 are given in Table 7.1 and E 1 and E are calculated rom Eqs & 7.19 and Eqs & 7.0, respectively. We can now ollow Example 7. (Singh and Heldman) and compare the reezing time calculations based on Pham s approach and Plank s equation. 3

24 PERHITUNGAN WAKTU PEMBEKUAN: Pham s Equation Alternate approach to determine the shape actor E in the calculation o reezing time: E (1 ) (1 ) N Bi N Bi 1 1 ( 1 ) ( ) N N Bi For ininite slab, the shape actor E = 1 (since 1 =ininite, =ininite) For an ininite cylinder, the shape actor E= (since 1 =1, =ininite) For a sphere, the shape actor, E = 3 ( 1 =1, =1) Bi PERHITUNGAN WAKTU PEMBEKUAN: Pham s Equation For dierent shapes e.g. ellipsoid, rectangular brick, inite cylinder etc., the shape actor can be calculated: E (1 ) (1 ) N Bi N Bi 1 1 ( 1 ) ( ) N N Bi Bi N Bi h R c k Same characteristic dimension R: shortage distance rom thermal center to the surace o the object. Smallest cross sectional area A ; the smallest cross section that incorporates R. Same volume V A V 1 and can be determined: 1 R 4 3 1( R ) 3 4

25 PERHITUNGAN WAKTU PEMBEKUAN: Lean bee with 74.5% moisture content and 1 m length, 0.6 m width, and 0.5 m thickness is being rozen in an air blast reezer with h c = 30 W/m.K and air temperature o 30 o C. I the initial product temperature is 5 o C. Estimate the time required to reduce the product temperature to 10 o C. An initial reezing temperature o 1.75 o Chas been measured or the product. The thermal conductivity o rozen bee is 1.5 W/m.K, and the speciic heat o unrozen bee is 3.5 kj/kg.k. A product density o 1050 kg/m 3 can be assumed, and a speciic heat o 1.8 kj/kg.k or rozen bee can be estimated rom properties o ice. Product length d = 1 m Product width d 1 = 0.6 m Product thickness a = 0.5 m Convective heat transer coeicient h c = 30 W/m.k Air temperature T = 30 o C Initial product temperature T i = 5 o C Initial reezing temperature T F = 1.75 o C Product density = 1050 kg/m 3 Enthalpy change (H) = kj/kg = 48.5 kj/kg (estimate) Thermal conductivity k o rozen product = 1.5 W/m.K Speciic heat o product (C pu ) = 3.5 kj/kg.k Speciic heat o rozen product (C p ) = 1.8 kj/kg. CONTOH PERHITUNGAN WAKTU PEMBEKUAN: CONTOH/CEK!!! (1) Determine shape actor: 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 actor 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 o (3500J / kgk. ) (1050kg/ m ) [5 ( 3.98) C] 3 T ) (333.kJ / kg1000j / kj) (0.745)(1050kg/ m ) 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 or lean bee (1 m 0.6m 0.5 m) will be 5.1 hours to reeze. 6

27 METODE PEMBEKUAN 1. AIR FREEZING - Products rozen by either "still" or "blast" orced air. cheapest (investment) "still" slowest, more changes in product "blast" aster, more commonly used. INDIRECT CONTACT - Food placed in direct contact with cooled metal surace. relatively aster more expensive 3. DIRECT CONTACT - Food placed in direct contact with rerigerant (liquid nitrogen, "green" reon, carbon dioxide snow) aster expensive reeze individual ood particles METODE PEMBEKUAN commercial reezing Blast reezing a very cold air blasted on the ood cools ood very quickly. Close indirect contact ood is placed in a multi plate reezer and is rapidly rozen. Immersion ood is placed into a very cold liquid (usually salt water brine) or liquid nitrogen, this is known as cryonic reezing. 7

28 METODE PEMBEKUAN --- reezing equipment Mechanical Freezers Evaporate and compress the rerigerant in a continuous cycle Cryogenic Systems Use solid and liquid CO, N directly in contact with the ood KLASIFIKASI PEMBEKUAN -- berdasarkan pada laju pergerakan ice ront Slow Freezers 0. cm/h Still air and cold stores Quick Freezers cm/h Air blast and plate reezers Rapid Freezers 5 10 cm/h Fluidized bed reezers Ultra rapid Freezers cm/h Cryogenic reezers 8

29 ALAT PEMBEKU ALAT PEMBEKU A typical luidized bed reezer 9

30 ALAT PEMBEKU ALAT PEMBEKU Batch Freezer Blast Type Source: Unit operations or ood the ood industries by: W.A. Gould 30

31 ALAT PEMBEKU Hydraulic Pump Top Pressure plate Connecting Linkage Corner Headers Rerigerant hoses Trays Polyurethane and polystyrene insulated doors Contact plates Source: Unit operations or ood the ood industries by: W.A. Gould ALAT PEMBEKU 31

32 ALAT PEMBEKU ALAT PEMBEKU 3

33 ALAT PEMBEKU 33