Question

An experiment is conducted to measure bacterial growth. The bacteria are continually monitored, but the initial data is lost. It is known that at t=1.4d the bacteria concentration is 0.9 mg VSS/L, and at t = 1.9 d the bacteria concentration is 2.1 mg VSS/L.

i) Calculate the doubling time

ii) assuming exponential growth applies, estimate the exponential growth constant using the exponential growth equation                    

iii) using the exponential growth model estimate the initial bacteria concentration at t=0 d,

iv) estimate the biomass concentration at t=3d.

                                                                                                             

Answer 1

calculating exponential growth constant (k)

initial concentration Ci = 0.9 mg VSS/L

final concentration Cf = 2.1 mg VSS/L

initial time ti = 1.4 d

final time tf= 1.9 d

Cf= Ci* e^k*(tf-ti_i)

Cf / Ci = e ^k(tf-ti)

ln (Cf/Ci) = k ( tf-ti )

ln ( 2.1 / 0.9 ) = k ( 1.9 - 1.4 )

ln 2.3 = k ( 0.5 )

0.83 = k ( 0.5 )

k = 1.6

calculating doubling time (T)

T = ln 2 / k

= 0.693 / 1.6

= 0.43 d

calculating initial bacteria concentration at t=0 d

Ci = ?

Cf = 0.9 mg VSS/L

ti = 0 d

tf = 1.4 d

k = 1.6

substituiting these values in the following equation

Cf= Ci* e^k*(tf-ti_i)

0.9 = Ci * e ^{1.6 (1.4-0)}

0.9 = Ci * e^2.24

0.9 = Ci * 9.39

Ci = 0.09  mg VSS/L

calculating biomass concentration at t=3d

ti = 0 d

tf= 3 d

Ci = 0.09 mg VSS/L ( as calculated above )

Cf = ?

substituting these values in the equation :

Cf= Ci* e^k*(tf-ti_i)

Cf = 0.09 * e ^1.6(3-0)

Cf = 0.09 * e ^4.8

Cf = 0.09 * 121.5

Cf = 10.93 mg VSS/L