midterm1 2006 solutions

Midterm1 2006 - Midterm 1 Solution#1(10 i non-steady state process semi-batch and open system Fermentation broth 10 glucose ρtotal = 1 g/mL

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Midterm 1 Solution #1 (10) i. non-steady state process, semi-batch, and open system. Fermentation broth 10% glucose ρtotal = 1 g/mL Vreactor= 10L = 10 000 cc V0 = 500cc (25) ii. Overall mass balance: d ( reactorV ) in Qin out Qout dt Qout = 0 cc/h Qin = 100 cc/h reactor in 1g / cc Initial condition: at t = 0h, V0 = 500 cc dV Qin dt V V0 Qin t V 500 100t cc At 5h, V = 1000 cc Tank volume = 10 L = 10 000 cc At t = 95h, V = 500+100(95) = 10000 cc the tank is full. Equation no longer valid. (35) iii. Species balance: d ( glu cos eV ) glu cos ein Qin rate dt glu cos ein 0.1 rate 5 g h 1g cc d ( glu cos eV ) 1g 100cc g 5 dt cc h h Volume is not constant, use product rule: d glu cos e dV 1g 100cc g glu cos e V 0.1 5 dt dt cc h h d glu cos e d (500 100t ) 1g 100cc g glu cos e (500 100t ) 0.1 5 dt dt cc h h d glu cos e 1g 100cc g glu cos e 100 (500 100t ) 0.1 5 dt cc h h cc 1g 100cc g 100 0.1 5 d glu cos e g h h glu cos e cc h dt (500 100t )cc cc (500 100t )cc cc g 100 5 d glu cos e g h h glu cos e dt (500 100t )cc cc (500 100t )cc 0.1 (10) iv. First order non homogeneous differential equation, use integrating factor or method of undetermined coefficient approach to solve it (15) v. cc h a (t ) (500 100t )cc 1g 100cc g 0.1 5 h b(t ) cc h (500 100t )cc 100 cc h d ( 500 ) cc 100 100 glu cos e (t ) e cc 1g 100cc g 100 5 h d 0.1 ( 500 ) cc 100 h )d C ( cc h e 1 (500 100 )cc Evalution the integrating constant 100 cc h F(t)= ( 500100t ) cc d = 500 + 100t e 1g 100cc g 0.1 5 1 cc h h )d C glu cos e (t ) ( 500 100t )( 1 (500 100t ) (500 100t )cc 1 1g 100cc g ( 0.1 cc h 5 h )d C1 (500 100t ) 1 g glu cos e (t ) 5t cc C1 (500 100t ) glu cos e (t ) Initial condition, at t = 0, glu cos e (0) 0.1(1 g g ) 0.1 cc cc Solve for C1 C1 50 g 1 glu cos e (t ) 5t 50 cc (500 100t ) 0.12 Initial Concentration 0.1 g/cc Glucose density (g/cc) 0.1 At t = 95h tank overfilled 0.08 Equation no longer valid 0.06 Concentration approaching 0.05 g/cc 0.04 0.02 0 0 20 40 60 Tim e (h) 80 100 ...
View Full Document

This note was uploaded on 02/05/2010 for the course CHM ENG 140 taught by Professor Radke during the Fall '08 term at University of California, Berkeley.

Ask a homework question - tutors are online