A steam steriliser is used to sterilise liquid medium for fe

A steam steriliser is used to sterilise liquid medium for fermentation. The initial concentration of contaminating organisms is 108 per litre. For design purposes, the final acceptable level of contamination is usually taken to be 103 cells; this corresponds to a risk that one batch in a thousand will remain contaminated even after sterilisation is complete. For how long should 1 m3 of medium be treated if the sterilisation temperature is: (a) 80°C? (b) 121°C? (c) 140°C? To be safe, assume that the contaminants present are spores of Bacillus stearothermophilus, one of the most heat-resistant microorganisms known. For these spores, the activation energy for thermal death is 283 kJ gmol1 and the Arrhenius constant is 1036.2 s1[29]. Doran, Pauline M.. Bioprocess Engineering Principles (Kindle Locations 18482-18493). Elsevier Science. Kindle Edition.

Solution

A=10^36.2s^-1, E_d=283kJgmol^-1

Now, we will calculate the specific death constant K_d by using the equation K_d=Ae^-E_d^/RT

R=8.3144JK^-1gmol^-1=8.3144x10^-3 kJK^-1gmol^-1.

Now convert the temperature to Kelvin

80°C=80+273.15=353.15 K

121°C =121+273.15=394.15K

140°C =140+273.15=413.15K

Now we will start with temperature 80°C

K_d=Ae^-E_d^/RT=10^36.2s^-1e^-283kJgmol-1/(8.3144x10^-3kJK^-1gmol^-1x353.15K)=2.20x10^-6s^-1

Now we will start with temperature 121°C

K_d=Ae^-E_d^/RT=10^36.2s^-1e^-283kJgmol-1/(8.3144x10^-3kJK^-1gmol^-1x394.15K)=0.0497s^-1

Now we will start with temperature 140°C

K_d=Ae^-E_d^/RT=10^36.2s^-1e^-283kJgmol-1/(8.3144x10^-3kJK^-1gmol^-1x413.15K)=2.63s^-1

Now try to find out the relationship between the number of viable cells and time, for that convert the units of the initial concentration of contaminants x₀

x₀=10⁸ cells/l=10⁸ cells/l*/10001/1m³/=10¹¹cells/m³

Per m³, N₀=10¹¹, N=10^-3, Now substitute the above values in the below equation

10^-3=10¹¹ e^-k_d^t

10^-14= e^-k_d^t

-32.24=-k_dt

T=32.24/K_d

Using this, now estimate for the temperature 80°C

T=32.24/K_d=32.24/2.20x10^-6s^-1*/1h/3600s/=4070h

Using this, now estimate for the temperature 121°C

T=32.24/K_d=32.24/0.0497s^-1*/1min/60s/=10.8 min

Using this, now estimate for the temperature 140°C

T=32.24/K_d=32.24/2.63s^-1=12.3sec

So, at 40°C, it is 4070h, at 121°C it is 10.8 min and at 140°C it is 12.3 sec