3rd Group Meeting VIVA M.Phil Transfer 2010 2nd Draft
Lecture 02: STKM3212
1. 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
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18. 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
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21. 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
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23. 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)
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25. 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
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27. 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
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29. 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
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