Lecture 02: STKM3212

LECTURE NOTES 02/07 STKM3212: FOOD PROCESSING TECHNOLOGY MASS BALANCE IN STEADY STATE (Imbangan Jisim dalam Keadaan Mantap) SAIFUL IRWAN ZUBAIRI PMIFT, Grad B.E.M. B. Eng. (Chemical-Bioprocess) (Hons.), UTM M. Eng. (Bioprocess), UTM ROOM NO.: 2166, CHEMISTRY BUILDING, TEL. (OFF.): 03-89215828, FOOD SCIENCE PROGRAMME, CENTRE OF CHEMICAL SCIENCES AND FOOD TECHNOLOGY, UKM BANGI, SELANGOR

1.1 OUTLINES 1.2 PROCESS CLASSIFICATION. 1.3 THE GENERAL BALANCE EQUATION. 1.4 MASS BALANCE COMPONENT. 1.5 PROBLEM EXAMPLES IN FOOD PROCESSING: A) DILUTION PROCESS (Proses Pencairan) B) EVAPORATION PROCESS (Proses Penyahidratan) = CONCENTRATION PROCESS (Proses Pemekatan) 1.6 PROBLEM EXAMPLES IN MIXTURING OF FOOD INGREDIENT.

1.2 PROCESS CLASSIFICATION Chemical processes may be classified as (1) BATCH, (2) CONTINOUS or (3) SEMIBATCH and as either STEADY STATE (mantap) OR UNSTEADY STATE (transient - tak mantap). Before writing a material balance on the process system - recognize which categories of process falls into: BATCH PROCESS : Feed is charged into the system at the beginning of the process and the product are removed all at once sometime later.  No mass (kg) crosses the system boundaries between the time the feed is charged & the time the product is removed. Extraction tank, t = 60 min F 0 (kg) H 2 O P 0 (kg) - PRODUCT (Liquid Crude Extract) System boundaries F 1 (kg) Tongkat Ali roots

CONTINUE: CONTINOUS PROCESS : The inputs and outputs flow continuously throughout the duration of the process. Holding tank, viscose MILK solution F 0 (kg/ hrs ) F 1 (kg/ hrs ) System boundaries I SPRAY DRYER system, T = 120 0 c F 2 (kg/ hrs ) - PRODUCT (Fine MILK powder) System boundaries II F 3 (kg/ hrs ) - Moist hot air

CONTINUE: If the values of all the variables in a process (i.e.: all temperature, pressure, volumes, flow rates, etc.) DO NOT CHANGE WITH TIME (t)  except possibly for minor fluctuations about constant mean values  the process is to be operating at STEADY STATE (MANTAP) . If any of the process variables CHANGE WITH TIME (t)  TRANSIENT or UNSTEADY STATE operation is said to exist. By their nature - BATCH & SEMIBATCH processes are UNSTEADY STATE OPERATION . CONTINOUS PROCESS - may be either STEADY STATE or transient. BATCH  is commonly used when relatively small quantities of a product are to be produced on any single occasion. CONTINOUS  Is better suited to large production rates. It usually run as close to STEADY STATE as possible.  The UNSTEADY STATE exist during the START-UP process.

1.3 THE GENERAL BALANCE EQUATION One of the basic laws of the physical science is  THE LAW OF CONSERVATION OF MASS . This law simply stated: “MASS CANNOT BE CREATED OR DESTROYED (excluding the nuclear or atomic reactions)” Hence, the total mass (or weight) of all materials entering any process must equal the total mass of all materials leaving + the mass of any materials accumulating or left in the process.  INPUT = OUTPUT + ACCUMULATION  In the majority of cases, there will be NO ACCUMULATION OF MATERIALS in a process.  Then, the INPUT will simply equal to the OUTPUT ----------------- “STEADY STATE PROCESS” “ What goes in must come out” ------ rules of thumb 

CONTINUE: To solve a material balance problem, it is advisable to proceed by a series of definite step as listed below: (1) SKETCH A SIMPLE DIAGRAM OF THE PROCESS  This can be simple box diagram showing each stream entering by an arrow pointing in & each stream leaving by an arrow pointing out. Include on each arrow the composition, amounts, temperature etc. of the stream. (2) WRITE THE CHEMICAL EQUATION INVOLVED (if any). (3) SELECT BASIS FOR THE CALCULATION  In most cases, the problem is concerned with a specific amount of one of the streams in the process which is selected as the BASIS . Usually, 100 kg/hrs , 100 g/min are used as a BASIS if there is NO FEED amount given.

CONTINUE: (4) MAKE A MATERIAL BALANCE  The arrows into the process will be INPUT items and the arrows going out will be the OUTPUT items. The balance can be a total material balance (in = out [steady state]) or a balance on each component present (if no chemical reaction occurs). Typical processes that DO NOT undergo chemical reactions are: a) Drying b) Evaporation c) Dilution of solution d) Distillation e) Extraction

CONTINUE: Example 1: Concentration of Orange fruit In the concentration of orange juice, a fresh extracted and strained juice containing 7.08% (w/w) solids is fed to a vacuum evaporator. In the evaporator, water is removed and the solids content increased to 58% (w/w) solids. For 1000 kg/hrs entering, calculate the amounts of the outlet streams of concentrated juice and water. Assume that the process is in steady state ANS: Evaporator W (kg/hrs) H 2 O C (kg/hrs) - PRODUCT (Concentrated Juice) 58% (w/w) solids 1000 (kg/hrs) Juice 7.08% (w/w) solids

CONTINUE:  Note that the letter (W) represents the unknown amount of H 2 O.  (C) is the amount of concentrated juice.  No chemical reactions are given.  BASIS (the amount have been given): 1000 kg/hrs entering juice Write down the TOTAL material balance : In = Out (steady state): 1000 = W + C --------------------- (1) SOLID BALANCE: This gives one equation and two unknowns. Hence, a component balance on SOLIDS will be made: 1000 kg/hrs (7.08/100) = W kg/hrs (0) + C kg/hrs (58/100) 70.8 kg/hrs = 0.58 kg solid/kg C kg/hrs ---------------------- (2)  C = 70.8/0.58 = 122.1 kg/hrs concentrated juice

CONTINUE: Direct substitute into equation (1): 1000 = W + 122.1 And we obtain  W = 877.9 kg/hrs H 2 O AS A CHECK ON OUR CALCULATIONS, WE CAN WRITE A BALANCE ON THE WATER COMPONENT: WATER BALANCE: 1000(100 - 7.08/100) = 877.9 + 122.1(100 - 58/100) 929.2 = 877.9 + 51.3 = 929.2 kg/hrs H 2 O 1000 (7.08/100) = 0 + 122.1(58/100) 70.8 = 70.8 kg/hrs solids

1 st QUIZ [10 mins] {5 marks} In the evaporation unit, a diluted material enters and condensed material (concentrated material) out from the system. During the process, H 2 O is evaporated. If (I) is the mass (kg/hrs) of the diluted material, (W) is the evaporated H 2 O mass (kg/hrs) and (C) is the mass of the concentrated material (kg/hrs), write down the equation of the overall system. Assume that the system in steady state.

2 nd Assignment Built one flowchart and put in the overall mass balance for the evaporation process system. Air inlet is at (A) Ib m /min and the wet material is at (I) Ib m /min. The dry material out from the system at (D) Ib m /min and the evaporated H 2 O, at (W) Ib m /min. Assume the system is a steady state.

1.4 MASS BALANCE COMPONENT In mass balance CALCULATION , make sure to use ONLY AMOUNT IN MASS UNIT : (A) Mass flow rate: kg/hrs, g/min (B) Mass fraction component  e.g.: A %(w/w) = kg A/kg, g A/g If the given AMOUNT (flow rate especially) is in VOLUME UNIT , CONVERT IT INTO MASS FLOW RATE BY USING IT DENSITY : example: Feed volume flow rate of liquid syrup B is 20.0 cm 3 /min. Convert it into mass flow rate of g/min. The density of the liquid syrup B is 2.00 g/cm 3 ANS:  Mass flow rate B = = 40 g/min 20.0 cm 3 min 2.00 g cm 3

CONTINUE: Example 1: BALANCES ON A BATCH MIXING PROCESS Two food grade methanol-water mixtures are contained in separate flasks. The first mixture contains 40% (w/w) methanol and the second contains 70% (w/w) methanol. If 200 g of the first mixture are combined with 150 g of the second, what are the mass and composition of the product? Assume the system is in steady state. ANS: BATCH MIXER F 1 = 200 g 0.4 g CH 3 OH/g 0.6 g H 2 O/g F 2 = 150 g 0.7 g CH 3 OH/g 0.3 g H 2 O/g Q (g) x (g CH 3 OH/g)

CONTINUE:  No chemical reactions are given.  BASIS (the amount have been given).  So, “Input = Out” --------------- STEADY STATE TOTAL MASS BALANCE: 200 g + 150 g = Q (g)  Q (g) = 350 g METHANOL MASS BALANCE: 0.4 g CH 3 OH 200 g g + 0.7 g CH 3 OH 150 g g = x (g CH 3 OH) Q (g) g 185 g CH 3 OH/ 350 g = x (g CH 3 OH/g)  x (g CH 3 OH/g) = 0.529 = 50.29% (w/w) To verify: use WATER BALANCE (200)(0.6) + (150)(0.3) = (350)(1-0.529)

CONTINUE: Example 2: Draw a diagram and make an equation to represent the total mass balance and mass balance component for the system that involved a mixture of chicken meat [15% (w/w) protein, 20% (w/w) fats & 65% (w/w) H 2 O] and synthetic fats [5% (w/w) protein, 15% (w/w) H 2 O & 80% (w/w) fats] for making 100 kg mixture of new food product that contain 25% (w/w) fats. Assume the system in steady state. ANS: BATCH MIXER P = x kg chicken meat 0.15 kg protein/kg 0.20 kg fats/kg 0.65 kg H 2 O/kg M = 100 kg NEW FOOD PRODUCT 0.25 kg fats/kg a kg protein/kg b kg H 2 O/kg B = y kg synthetic fats 0.05 kg protein/kg 0.80 kg fats/kg 0.15 kg H 2 O/kg

CONTINUE:  No chemical reactions are given.  BASIS (the amount have been given at the OUTPUT STREAM).  So, “Input = Out” ---------- STEADY STATE TOTAL MASS BALANCE: P + B = 100 kg ------------- (1) FATS MASS BALANCE: (0.2)(P) + (0.8)(B) = (0.25)(100) 0.2P + 0.8B = 25 ------------- (2) Substitute (1) into (2) (0.2)(100 - B) + 0.8B = 25 20 - 0.2B + 0.8B = 25 B(0.8 - 0.2) = 25 - 20  B = 8.33 kg  P = 100 - 8.33 = 91.67 kg

CONTINUE: PROTEIN MASS BALANCE: (0.15)(91.67) + (0.05)(8.33) = (a)(100) 13.75 + 0.42 = (a)(100) 14.17/100 = (a)  a = 0.142 kg protein/kg = 14 % (w/w) H 2 O MASS BALANCE: (0.65)(91.67) + (0.15)(8.33) = (b)(100) 59.59 + 1.25 = (b)(100) 60.84/100 = (b)  a = 0.61 kg H 2 O/kg = 61 % (w/w)

1.5 PROBLEM EXAMPLES IN FOOD PROCESSING Example 3: DILUTION PROCESS How much kg solution which contain 10% NaCI can be obtained by diluting 15 kg solution which contain 20% NaCI with H 2 O? Assume the system is in steady state. ANS: **TAKE NOTE: Dilution process will not altered the mass of the solution. Only the mass fraction component will be changed. BATCH MIXER P = 15 kg solution 0.20 kg NaCI/kg 0.80 kg H 2 O/kg M = y kg NEW DILUTED SOLUTION 0.10 kg NaCI/kg 0.90 kg H 2 O/kg B = x kg H 2 O

CONTINUE:  No chemical reactions are given.  BASIS (the amount have been given at the INPUT STREAM).  So, “Input = Out” ---------- STEADY STATE TOTAL MASS BALANCE: 15 + B = M ------------- (1) NaCI MASS BALANCE: (0.2)(15) = (0.10)(M) 3 = 0.1M ------------- (2)  M = 30 kg  B = 30 kg - 15 kg = 15 kg

CONTINUE: Example 4: EVAPORATION PROCESS How much reduction of mass will be obtained when the solution of grape juices are evaporated from 80% to 50% of its moisture content? ANS: Evaporator W (kg/hrs) H 2 O C (kg/hrs) - PRODUCT (Concentrated grape Juice) 0.50 kg solids/kg 0.50 kg H 2 O/kg 100 (kg/hrs) grape juice 0.2 kg solids/kg 0.8 kg H 2 O/kg

CONTINUE:  No chemical reactions are given.  NO BASIS are given -------- Put a BASIS of 100 kg/hrs of grape juice  So, “Input = Out” ---------- STEADY STATE TOTAL MASS BALANCE: 100 = W + C ------------- (1) H 2 O MASS BALANCE: (0.2)(100) = (0.50)(C) 20 = 0.5C  40 kg/hrs = C ------------- (2) Substitute (2) into (1) 100 = W + 40  W = 60 kg/hrs REDUCTION OF MASS: 100 kg - 40 kg  = 60 kg % REDUCTION OF MASS: 60 kg/100 kg × 100%  = 60% (w/w)

CONTINUE: Example 5: EVAPORATION PROCESS One evaporator have the capability of evaporating H 2 O at 500 kg/hrs. Calculate the production rate of concentrated mango juice which contain 45% solids from the original juice which contain 12% solids. Assume the system in steady state. ANS: Evaporator 500 (kg/hrs) H 2 O C (kg/hrs) - PRODUCT (Concentrated mango Juice) 0.45 kg solids/kg 0.55 kg H 2 O/kg F (kg/hrs) mango juice 0.12 kg solids/kg 0.88 kg H 2 O/kg

CONTINUE:  No chemical reactions are given.  BASIS (the amount have been given at the H 2 O STREAM).  So, “Input = Out” ---------- STEADY STATE TOTAL MASS BALANCE: F = 500 + C ------------- (1) H 2 O MASS BALANCE: (0.88)(F) = 500(1) + (0.55)(C) --------- (2) Substitute (1) into (2): (0.88)(500 + C) = 500 + 0.55C 440 + 0.88C = 500 + 0.55C 0.88C - 0.55C = 500 - 440 0.33C = 60  C = 181.8 kg/hrs  F = 500 + 181.8 = 681.8 kg/hrs

1.6 PROBLEM EXAMPLES IN MIXTURING OF FOOD INGREDIENT Example 6: BATCH MIXING PROCESS Identify the amount of concentrated fruit juice in each of the concentrated juice A which contain 65% solids and the concentrated juice B which contain 15% solids to be mixed in the mixer to produce 100 kg of NEW CONCENTRATED JUICE PRODUCT which contain only 45% solids. Assume the system is in steady state. ANS: BATCH MIXER x kg juice A 0.65 kg solids/kg 0.35 kg H 2 O/kg M = 100 kg NEW CONCENTRATED JUICE 0.45 kg solids/kg 0.55 kg H 2 O/kg y kg juice B 0.15 kg solids/kg 0.85 kg H 2 O/kg

CONTINUE:  No chemical reactions are given.  BASIS (the amount have been given at the OUTPUT STREAM).  So, “Input = Out” ---------- STEADY STATE TOTAL MASS BALANCE: x + y = 100 ------------- (1) SOLIDS MASS BALANCE: (0.65)( x ) + (0.15)( y ) = (0.45)(100) 0.65 x + 0.15 y = 45 ------------- (2) Substitute (1) into (2): 0.65(100 - y ) + 0.15 y = 45 65 - 0.65 y + 0.15 y = 45 y(0.15 - 0.65) = 45 - 65 y = -20/-0.5  y = 40 kg  x = 100 kg - 40 kg = 60 kg

CONTINUE: Example 7: BATCH MIXING PROCESS Identify the amount of beef meat, chicken fats and H 2 O that should be used to produce 100 kg formulation of frankfurters. About z; kg of chilled water is pumped and mixed into the mixer to facilitate fat solidification process. The ingredient or raw materials and frankfurters are: Beef meat - 14% fats, 67% H 2 O & 19% protein Chicken fats - 89% fats, 8% H 2 O & 3% protein Frankfurters - 20% fats, 65% H 2 O & 15% protein ANS: BATCH MIXER x kg beef meat 0.14 kg fats /kg 0.67 kg H 2 O/kg 0.19 kg protein/kg M = 100 kg NEW FORMULATION FRANKFURTERS 0.20 kg fats/kg 0.65 kg H 2 O/kg 0.15 kg protein/kg y kg chicken fats 0.89 kg fats/kg 0.08 kg H 2 O/kg 0.03 kg protein/kg z kg H 2 O (Chilled water)

CONTINUE:  No chemical reactions are given.  BASIS (the amount have been given at the OUTPUT STREAM).  So, “Input = Out” ---------- STEADY STATE TOTAL MASS BALANCE: x + y + z = 100 ------------- (1) FATS MASS BALANCE: (0.14)( x ) + (0.89)( y ) = (0.20)(100) 0.14 x + 0.89 y = 20 ------------- (2) H 2 O MASS BALANCE: (0.67) ( x ) + (0.08)( y ) + (1)( z ) = (100)(0.65) 0.67 x + 0.08 + z = 65 --------- (3) Substitute (1) into (3): 0.67 x + 0.08 + 100 - x - y = 65 Substitute (2) into (3): 0.67 x + 0.08 + 100 - x - (20 - 0.14 x /0.89) = 65

CONTINUE: 0.67 x + 0.08 + 100 - x - (20 - 0.14 x /0.89) = 65 0.67 x + 0.08 + 100 - x - ({20/0.89} - 0.14 x /0.89) = 65 0.67 x + 0.08 + 100 - x - 22.47 + {0.14 x /0.89} = 65 -0.33 x + 77.61 + {0.14 x /0.89} = 65 -0.33 x + {0.14 x /0.89} = 65 - 77.61 {-0.294 x + 0.14 x /0.89} = -12.61 -0.154 x /0.89 = -12.61 -0.154 x = -11.22  x = -11.22/-0.154 = 72.88 kg 0.14 (72.88) + 0.89 y = 20 ------------- (2) 10.20 + 0.89y = 20  y = 9.8/0.89 = 11.0 kg x + y + z = 100 ------------- (1) 72.88 + 11.0 + z = 100  Z = 16.12 kg

Lecture 02: STKM3212

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