Principles of Food and Bioprocess Engineering (FS 231) Solutions to Example Problems on Mass Balance

Transcription

1 Principles of Food and Bioprocess Engineering (FS 231) Solutions to Example Problems on Mass Balance 1. The mass balance equation for the system is: = m This yields, m = 5 kg 2. The mass balance equation for the system is: = m This yields, m = 5 kg/s 3. We have 1 mass balance equation for each sub-system, yielding a total of 2 equations. First sub-system: = m Second sub-system: = 2 + m 2 Solving, we get: m = 1 kg/s, m = 6 kg/s 4. In this system, we have 2 equations and 2 unknowns. However, the mass balance equation in the first sub-system does not give us any information except that the problem is well-posed. Thus, we are left with one mass balance equation for sub-system #2 and 2 unknowns in that equation. Thus, we can not solve for m 1 and m 2 in this problem. All that we can say is that: 4 + m = 2 + m. 5. The Overall Mass Balance (OMB) equation is: = m The solids balance equation is: 0.3 (2) + x (1) = 0.4 (m) This yields, m = 3 kg/s, x = OMB: = m Sugar: 0.2 (5) (1) = x 1 (m) Starch: 0.6 (5) (1) = x 2 (m) Solving, we get: m = 6 kg/s, x = 0.22, x = OMB for first sub-system: = m Solids balance for first sub-system: 0.6 (10) (2) = x 1 (m 1) (9) OMB for second sub-system: 9 = m Solids balance for second sub-system: 0.7 (9) = x (m ) (6) (2) 2 2 Solving, we get: m 1 = 3 kg/s, x 1 = 0.18, m 2 = 1 kg/s, x 2 = There are four components (sugar, starch, oil, and water) in each sub-system and hence there are four mass balance equations in each sub-system, yielding a total of 8 equations. OMB in first sub-system: = m 1 Starch balance in first sub-system: 0.3 (10) = x 4 (2) + x 1 (m 1) (5) Sugar balance in first sub-system: 0.06 (10) + x 2 (3) = 0.4 (m 1) Oil balance in first sub-system: x (3) = 0.3 (2) 3

2 OMB in second sub-system: m m 2 = Starch balance in second sub-system: x 1 (m 1) = 0 Sugar balance in second sub-system: 0.4 (m 1) + x 6 (m 2) (2) = x 2 (3) (9) Oil balance in second sub-system: x (m ) = x (3) Solving, we get: m 1 = 6 kg/s, m 2 = 4 kg/s, x 1 = 0, x 2 = 0.6, x 3 = 0.2, x 4 = 0.5, x 5 = 0.15, x 6 = There are four components (sugar, starch, oil, and water) in each sub-system and hence there are four mass balance equations in each sub-system, yielding a total of 8 equations. OMB in first sub-system: = 8 + m 1 Starch balance in first sub-system: 0.4 (10) = x 1 (m 1) Sugar balance in first sub-system: 0.5 (10) (2) = 0.7 (8) Oil balance in first sub-system: 0 = x (m ) 2 1 OMB in second sub-system: m 1 + m 2 = Starch balance in second sub-system: x 1 (m 1) (m 2) = x 5 (5) Sugar balance in second sub-system: 0 = x 3 (5) + x 6 (5) Oil balance in second sub-system: x (m ) (m ) = x (5) (5) 2 4 Solving, we get: m 1 = 4 kg/s, m 2 = 6 kg/s, x 1 = 1.0, x 2 = 0, x 3 = 0, x 4 = 0.4, x 5 = 0.92, x 6 = Since there are 5 components in this system, there are 5 equations. OMB: = 8 + m Sugar: 0.2 (6) = x 3 (8) Starch: x 1 (6) (4) = 0.3 (8) (m) Oil: x 6 (4) = 0.2 (m) Protein: 0 = x (8) + x (m) 2 5 In order to determine x 4, we note that the sum of the fractions of the individual components of the product exiting from the top of the system at 8 kg/s must add up to 1.0. i.e., x + x + x = Solving, we get: m = 2 kg/s, x 1 = 0.4, x 2 = 0, x 3 = 0.15, x 4 = 0.55, x 5 = 0, x 6 = Since there are 4 components in each of the 3 sub-systems, we have a total of 12 equations. Overall mass balance equation: 10 + m 1 = m2 Total Fat balance equation: 10 (x 2) + m 1 (x 3) = m 2 (0.8) Total Protein balance equation: 10 (x 1) + m 1 (0.25) = 3 (1.0) + m 2 (x 8) + 5 (0.02) Total Vitamin balance equation: 10 (0.35) = 5 (0.64) + m (x ) 2 9

3 Overall mass balance equation: = m 1 + m3 Total Fat balance equation: 2 (x 5) = m 1 (x 3) Total Protein balance equation: 5 (0.02) + 2 (x 6) + 1 (0.3) = m 1 (0.25) Total Vitamin balance equation: 5 (0.64) + 2 (x ) + 1 (x ) = m (1.0) For the third sub-system: Overall mass balance equation: m 2 = Total Fat balance equation: m 2 (0.8) = 4 (1.0) + 2 (x 5) Total Protein balance equation: m 2 (x 8) = 2 (x 6) Total Vitamin balance equation: m (x ) = 2 (x ) Solving, we get: m 1 = 4 kg/s, m 2 = 6 kg/s, m 3 = 4 kg/s, x 1 = 0.27, x 2 = 0.4, x 3 = 0.2, x 4 = 0.5, x 5 = 0.4, x = 0.3, x = 0.15, x = 0.1, x = Since there are 3 components in each of the sub-systems, we have a total of 6 equations. Overall mass balance equation: = 2 + m 1 Protein balance equation: 0.5 (5) = 0.3 (2) + x 1 (m 1) Fat balance equation: 0.3 (3) = 0.2 (2) + x (m ) Overall mass balance equation: m 1 = m Protein balance equation: x 1 (m 1) = x 3 (2) Fat balance equation: x (m ) = x (m ) Solving, we get: m 1 = 6 kg/s, m 2 = 2 kg/s, x 1 = 0.317, x 2 = 0.083, x 3 = 0.95, x 4 = Since there are 3 components in each of the 3 sub-systems, we have a total of 9 equations. Overall mass balance equation: 10 + m = Total Sugar balance equation: 0.3 (10) + x 2 (m 1) (2) = 0.1 (3) + x 1 (13) Total Starch balance equation: x (m ) = Overall mass balance equation: = m Total Sugar balance equation: x 5 (13) (2) = x 5 (m 2) (2) Total Starch balance equation: 0.05 (2) = 0.02 (5) + x (m ) 4 2 For the third sub-system: Overall mass balance equation: 5 = m 1 + m3 Total Sugar balance equation: 0 = x 7 (m 3) + x 2 (m 1) Total Starch balance equation: 0.02 (5) = x (m ) + x (m ) Solving, we get: m = 4 kg/s, m = 8 kg/s, m = 1 kg/s, x = 0.3, x = 0, x = 0, x = 0, x = 0.35, x = 0.1, x =

4 14. The system diagram for the problem is as shown below: We have 5 equations and 5 unknowns in this problem. Overall mass balance in first sub-system: 20 = m 1 + m2 Sugar balance in first sub-system: 0.15 (20) = x 1 (m 2) Overall mass balance in second sub-system: m 2 + m 3 = m Sugar balance in second sub-system: x 1 (m 2) = 0.3 (m 4) Starch balance in second sub-system: m = 0.4 (m ) 3 4 Solving, we get: m 1 = 9 kg/hr, m 2 = 11 kg/hr, m 3 = 4 kg/hr, m 4 = 10 kg/hr, x 1 = The mass flow rate (m) and moisture content on wet basis (x) are the two unknowns in the system shown below: Mass balance: 20 = 15 + m Solids balance: 0.1 (20) = x (m) m = 5 kg/hr and x = 0.4 Thus, the product consists of 2 kg/hr of solids and 3 kg/hr of water. Thus, (mc) = 3/2 = 1.5 = 150 % db

5 16. First sub-system Overall mass balance: 5 + m 1 = 3 + m 2 => 2 + m 1 = m2 Protein balance: 0.2 (5) (m 1) = 0.4 (3) + x 2 (m 2) => (m 1) = x 2 (m 2) Fat balance: 0.2 (5) (m 1) = x 1 (3) + x 3 (m 2) => (m 1) = x 1 (3) + x 3 (m 2) Carbohydrates balance: 0.3 (m ) = 0.2 (3) + x (m ) => 0.3 (m ) = x (m ) Second sub-system Overall mass balance: m = 5 + m 3 => m = m3 Protein balance: x 2 (m 2) (10) = 0.4 (5) => x 2 (m 2) = 0 Fat balance: x 3 (m 2) (10) = 0.6 (5) => x 3 (m 2) = 0 Carbohydrates balance: x (m ) + x (10) = 0.4 (m ) => x (m ) + 10 x = 0.4 (m ) Substituting x 2 (m 2) = 0 into the protein balance equation for the first sub-system, we get: (m 1) = 1.2 This yields: m = 2 kg/s 1 Substituting this value in the overall mass balance equation for the first sub-system, we get: m = 4 kg/s 2 Since m 2 0, and x 2 (m 2) = x 3 (m 2) = 0, we get: x 2 = x 3 = 0 Substituting these values of m 1 and m 2 in the fat balance equation for the first sub-system, and noting that x (m ) = 0, we get: x = Solving for the remaining variables, we get: m 3 = 9 kg/s, x 4 = 0, x 5 = 0.36