Answer:
1. 4.32 h
2. 4.12 g/L
Explanation:
1. For a batch culture, the time (tb) can be calculated by:
tb = ln (X/X0)/μmax
Where X0 is the initial mass concentration of the cells (12 g/100L = 0.12 g/L), X is the mass concentration of the cells at tb, and μmax is the maximum specific growth rate of the cells.
The biomass yield (Y) is:
Y = (X - X0)/ (S0 - S)
Where S is the mass concentration of the substrate at tb and S0 the initial mass concentration of the substrate (glucose in this case).
Reorganizing:
X = Y*(S0 -S) + X0
Let's assume that at the stationary state all substrate was consumed, so S = 0.
tb = ln[(YS0 + X0)/X0)]/μmax
tb = ln[(0.575*10 + 0.12)/0.12]/0.9
tb = 4.32 h
2. If 70% of the substrate is consumed, S = 10 - 0.7*10 = 3 g/L
tb = ln[(0.575*(10-3) + 0.12)/0.12]/0.9
tb = 3.93 h
The initial concentration is X0 = 0.12 g/L, the X:
tb = ln (X/X0)/μmax
3.93 = ln(X/0.12)/0.9
ln(X/0.12) = 3.537
X/0.12 = [tex]e^{3.537}[/tex]
X/0.12 = 34.36
X = 4.12 g/L
Answer:
a) to = 4.3 h
b) xf = 4.1 g/L
Explanation:
We have the following data:
So = 10 g*L^-1 = initial concentration
Ys = biomass yield = 0.575 g*g^-1
umax = maximum specific growth rate = 0.9 h^-1
The initial cell concentration is equal to:
xo = 12 g/100 L = 0.12 g/L
a) The batch time it takes to reach the stationary phase equals:
to = (1/umax)*ln(1 + (Ys*(so-sf)/xo))) = (1/0.9)*ln(1+(0.575*(10-0))/0.12)))) = 4.3 h
b) Since the fermentation stops after consuming only 70% of the glucose, we have the following:
sf = (1-07)*so = 0.3*10 = 3 g/L
tb = (1/0.9)*ln(1+(0.575*(10-3))/(0.12)))) = 3.94 h
Finally, the final cell concentration can be found by the following equation:
xf = xo*e^(umax * tb) = 0.12 * e^(0.9 * 3.94) = 4.1 g/L
