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

Transcription

1 Princiles of Food and Biorocess Engineering (FS 231) Solutions to Examle Problems on Eaoration 1. The system diagram is as follows: Solids balance equation: 0.1 (m f) = 0.5 (1,000) Thus, m f = 5,000 kg/hr Oerall mass balance equation: m f = m + 1,000 Thus, m = 4,000 kg/hr m (c ) ( ) + m (H - H ) = m (H ) + 1,000 (c ) ( ) f (f) s at kpa c at 80 C at 85 C () H at kpa = 2,683.8 kj/kg H at 85 C = 2,651.9 kj/kg H c at 80 C = kj/kg c (f) = 0.1 (c (s)) (c (w)) = 4,062 J/kg K c = 0.5 (c ) (c ) = 3,590 J/kg K () (s) (w) Soling, we get: m s = 4,637.3 kg/hr. Thus, steam economy = m /m s = The system diagram is as follows:

2 a. Oerall mass balance equation: 500 = m + m Total solids balance equation: 0.2 (500) = x (m ) m f (c (f) ) (T f - 273) + m s (H at kpa - H c at 100 C) = m (H at 65 C) + m (c () ) ( ) Substituting the alues, we get: 500 (4000)( ) ( ) x 10 = m ( x 10 ) + m (3000)( ) Soling the aboe three equations for the three unknowns, we get: m = kg/hr, m = kg/hr, and x = 0.25 b. Equating the energy lost by steam to UA( T), we get: m s (H at kpa - H c at 100 C) = U A ( T) 3 i.e., (150/3600) x ( ) x 10 = 4000 A ( ) Soling, we get: A = 0.67 m 2 3. a = ( ) kj/kg = kj/kg = ( x 0.018) kj/mol = kj/mol R g = J/mol K T = 373 K, T' = 378 K Thus, ln (X ) = and hence X = 0.84 w X w = (m w/18) / [(m w/18) + (m s/200)] = 0.84 Soling, we get: m s = (m w) Thus, total solids = m /(m + m ) = = 67.9 % s s w 4. The system diagram is as follows: w Oerall mass balance equation: 10 = m + m Total solids balance equation: 0.15 (10) = x (m )

3 10 (c (f) ) ( ) + 70 (H s - H c at 105 C) = m (H at 90 C) + m (c () ) ( ) H s = 0.9 (H at 105 C) (H c at 105 C) c (f) = 0.15 (3500) (4180) c = x (3500) + (1 - x) (4180) () Once we sole the aboe set of equations, we can determine the solids fraction of the final roduct as x and the steam economy as the ratio of m to a. We start off with the Clausius-Claeyron equation: with Here, m w = 140 g, M w = 18 g, m s = 60 g, M s = 350 g Substituting, we get: X w = 0.98 Using this alue of X w in the Clausius-Claeyron equation, along with the following: T = 373 K, R g = J/mol K, a = 40, J/mol we get: T' = K (= C) The system diagram for the double-effect eaorator is as follows: b. Total solids in feed = 60/200 = 30% Oerall totals solids balance: 0.3 (1000) = 0.6 (m ) m = 500 kg/hr c. Oerall mass balance in second sub-system: (m 1) = m m 1 = 650 kg/hr Oerall mass balance in first sub-system: 1000 = m + m m = 350 kg/hr d. Energy balance in second effect: m (c )T + m (H - H ) = m (H ) + m (c ) C In the aboe equation, T 1, H, H c are the unknowns. Howeer, they are not indeendent unknowns. Sole for T by trial and error. 1

4 e. Energy balance in first effect: m (c ) 20 + m (H - H ) = m (H ) + m (c ) T f s 1 1 From steam tables, determine the temerature (T s) at which the quantity H - H c satisfies the aboe equation. 6. The system diagram is as follows: a. Oerall mass balance equation: m f = 90 + m Total solids balance equation: 0.25 (m f) = x (m ) m (c ) ( ) + m (H - H ) = m (H ) + m (c ) ( ) f (f) s s at kpa c at 90 C at 70 C () H s = 0.9 (H at kpa) (H c at kpa) = 0.9 (2660.1) (376.92) = kj/kg Substituting the alues, we get: m (4000)( ) ( ) x 10 = 90 ( x 10 ) + m (4000)( ) f Soling the aboe three equations for the three unknowns, we get: m = kg/hr, m = kg/hr, and x = (= 38.1%) f

5 7. The system diagram is as follows: Oerall mass balance equation: 5 = m + m Solids balance equation: 0.05 (5) = 0.25 (m ) 5 (3500)(15) + m (H ) = m (H ) + m (H ) + m (4000)(80) s s s c at 80 C at 80 C From the solids balance equation, we get: m = 1 kg/s Substituting this in the oerall mass balance equation, we get: m = 4 kg/s Now, H s = 0.75 (H ) at kpa (H c) at kpa = 0.75 (2683.8) (440.15) = kj/kg Also, H at 80 C = kj/kg Substituting these alues (conert kj to J) into the energy balance equation, yields: m s = 5.95 kg/s Thus, steam economy = m /m = 4/5 = The system diagram is as follows: s

6 Oerall mass balance equation: 2 = m + m Solids balance equation: 0.2 (2) = x (m ) 2 (4000)(20) (H ) = 0.5 (H ) + m (H ) + m (4000)(85) s c at 90 C at 85 C H s = 0.95 (H ) at 90 C (H c) at 90 C = 0.95 ( x 10 ) ( x 10 ) = J/kg H c at 90 C = J/kg H = J/kg at 85 C Substituting these alues for enthalies and the oerall mass balance equation in the energy balance equation, we get: m = 1.76 kg/s Thus, m = 0.24 kg/s, x = 0.23 Thus, 0.23 = m /(m + m ) This yields: m = 0.3 m s s w s w Now, Substituting M w = 18, M s = 50, and m s = 0.3 m w in the aboe equation, we get: X w = 0.90 We then use the following equation (Clausius-Claeyron eqn) to determine the boiling oint of the concentrated roduct (T'): Here, X w = 0.90, T = 373 K, Rg = J/mol K, a = 40, J/mol Substituting, we get: T' = 376 K = 103 C 9. The system diagram is as follows:

7 Oerall mass balance equation: m f = m Solids balance equation: x (m f) = 0.6 (0.5) m f (4000)(20) (H s at kpa) = 0.5 (H c at kpa) + m (H ) at 75 C (4000)(75) H s = 0.95 (H ) at kpa (H c) at kpa = 0.95 (2660.1x10 ) ( x10 ) = J/kg H c at kpa = J/kg H = J/kg at 75 C Substituting these alues for enthalies and the oerall mass balance equation in the energy balance equation, we get: m = 0.17 kg/s Thus, m f = 0.67 kg/s, x = 0.45 Thus, 0.45 = m /(m + m ) This yields: m = 0.82 m s s w s w Now, Substituting m w = 18, M s = 80, and m s = 0.82 m w in the aboe equation, we get: X w = 0.84 We then use the following equation (Clausius-Claeyron eqn) to determine the boiling oint of the concentrated roduct (T'): w g a Here, X = 0.84, T = 373 K, R = J/mol K, = 40, J/mol Substituting, we get: T' = 378 K = 105 C