Problem (Shuler and Kargi, pp 303) After a batch fermentation, the system is dismantled and approximately 75% of the cell mass is suspended in the liquid phase (2 l), while 25% is attached to the reactor walls and internals in a thick film (ca. 0.3 cm). Work with radioactive tracers shows that 50% of the target prod- uct (intracellular) is associated with each cell fraction. The productivity of this reactor is 2 g product/l at the 2 l scale. What would be the productivity at 20,000 l scale if both reactors had a height-to-diameter ratio of 2 to 1? Solution - Both tanks are geometrically similar, so we can calculate diameter, surface area and volume in both tanks 1 1 1 V = π D H = π D .2D = π D 3 2 2 4 4 2 S = π D.H = π D.2D = 2π D 2 Sandip Bankar 2 X 107 cm3 therefore. for 738 cm2 surface area • Amount of product formed at 20. D=? S=? • For 20000 L system.928g+(2g/LX1/2X20000L)=?? g Sandip Bankar .• For 2 L system . D=? S=? • At bench scale (2L) the amount of product produced by surface-attached cell is 2g/l X 2L X 1/2 (50%)= 2g.V = 2000 cm3 therefore.000 L due to surface-attached growth is 342600 cm2/738 cm2 X 2=?? g Overall yield in 2L system is 2g/l X 2L = 4 g Overall yield in 20000 L system is. 10 2 11 2. pp 325) Time (min) DO (mg/l) The air supply to a fermenter was -‐1 3.3 7.3 mg/l has been determined for the 3 0.1 16 3.4 and then restarted.3 8 1 oxygen uptake rate and kLa in this 9 1.9 14 3 15 3.1 operating conditions.3 1 2.2 17 3.3 turned off for a short period of time Air off 0 3. Use the 5 0 tabulated measurements of dissolved Air on 6 0 oxygen (DO) values to estimate the 7 0.4 12 2.6 system.3 4 0. A value for C* of 2 1. Problem (Shuler and Kargi.7 13 2.2 Sandip Bankar . t • Plotting C Vs t should yield a straight line with slope =OUR= −qO2 X • Plot a graph in “Air off” region. Slope = OUR=? Sandip Bankar . Oxygen uptake rate in “Air off” region • Plot DO Vs t dcL • Determination of OUR= = kL a (C * − CL ) − qO2 X dt • With air off: k L a (C * − CL ) = 0 Therefore dcL = −qO X dt 2 • Integrating above equation results: C − C0 = −qO2 X. Slope=kL. kLa determination in “Air-On” region dcL • = kL a (C * − CL ) − qO2 X dt dcL • Or = −kL aCL + α where α = kL a.a=? Sandip Bankar .C * − qO X dt 2 • Plotting dCL/dt Vs CL should yield straight line with slope =-kL.a • Calculate dCL/dt for values 8<t<15 • Plot a graph of dCL/dt Vs CL. mathcity.http://www.org/error .