\u03bc g \u03bc max S S D \u03bc g \u03bc max \u03bc max ln 2 \u03c4 d 0 289 hr 1 However if we dont assume

# Μ g μ max s s d μ g μ max μ max ln 2 τ d 0 289

• 17
• 80% (5) 4 out of 5 people found this document helpful

This preview shows page 11 - 15 out of 17 pages.

μ g = μ max S S D = μ g = μ max μ max = ln (2) τ d = 0 . 289 hr - 1 However if we don’t assume this: ln (2) τ d = D max S K S + S D max = ln (2)( K S + S ) τ d S As S 0, or all the substrate is consumed to produce cell mass, indicating the maximum growth rate, D max → ∞ . Thus, we need some data to deter- mine when the Monod kinetics will plateau such that μ max can be determined empirically. 6.d DX = DY X/S 0 - K S D μ max - D DX = 3 . 49 g DW/L/hr 7 Shuler 6.9 Want to maximize biomass productivity (DX) while minimizing S o ut , thus find D such that d ( DX ) dD = 0. Since, DX = DY X/S ( S 0 - S ), maximizing DX also minimizes S. 11
DX = DY X/S ( S 0 - S ) D = μ g = μ max S K S + S S = DK S μ max - D DX = DY X/S S 0 - DK S μ max - D d ( DX ) dD = Y X/S d dD DS 0 - D 2 K S μ max - D Separate and apply quotient rule to get: d ( DX ) dD = Y X/S S 0 - K S D (2 μ max ) ( μ max - D ) 2 ! Setting this equal to zero to find the extrema and solving for D: 0 = Y X/S S 0 - K S D (2 μ max ) ( μ max - D ) 2 ! D = 0 . 518 or 0 . 482 We can check whether these are maxima or minima by taking the second deriva- tive and checking the sign at each dilution rate. d 2 ( DX ) dD 2 = - Y X/S K S 2 μ 2 max ( μ max - D ) 3 ! d 2 ( DX ) dD 2 D =0 . 518 > 0 minimum d 2 ( DX ) dD 2 D =0 . 482 < 0 maximum Therefore, D optimal = 0 . 482 h - 1 . This is the same as applying equation 6.83. To determine whether we have sufficient O 2 delivery, the cell concentration at this dilution rate is needed. 12
D = μ g = μ max S K S + S S = K S D μ max D S = 0 . 803 g/L X = Y X/S ( S 0 - S ) X = 10 . 6g/L The OUR can then be compared with the OTR to determine if sufficient oxygen can be provided. OUR = q O 2 X = μ g X Y X/O 2 = D opt X Y X/O 2 Y X/O 2 = Y X/S Y O 2 /S = 0 . 25 g cell/g oxygen OUR = 20 . 4 g O 2 /L/h As OUR is greater than the OTR of the chemostat (10 g oxygen per liter per hour), not enough oxygen can be supplied to operate at the optimal dilution rate. 8 Shuler 6.12 8.a From a mass balance on cells in the CFSTR: Accum. = In - OUT + GEN - DESTROY V dX dt = QX 0 - QX + μ g XV - k d XV dX dt = DX 0 - DX + μ g X - k d X 0 = D (0) - DX + μ g X - (0) X 0 = - DX + μ g X D = μ g 13
Apply Contois equation: μ g = μ m S K SS X + S D = μ m S K SS X + S S = DK SS X μ m - D 8.b From a mass balance on substrate in the CFSTR: Accum. = In - OUT + GEN - DESTROY V dS dt = QS 0 - QS - μ g XV Y X/S dS dt = D ( S 0 - S ) - μ g X Y X/S 0 = D ( S 0 - S ) - μ g X Y X/S D ( S 0 - S ) =