Load patterns for sterilization

Production
1

Please let me know how to decide load patterns for sterilization by considering minimum and maximum load.

2

It depends on your requirements for sterilization.

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3

yes sir i agree but which load to put first…like as i know 1st is empty load, 2nd glass wares, 3rd media. Please correct if am wrong and how do we consider for minimum and maximum load. In my previous company it was 750mL was minimum. I dont know on what basis they considered that 750mL ? Will that depend on our autoclave capacity!?

4

The minimum load would be the load that you may sterilize the minimum quantity of material for your analysis that could be a 90 ml SCDM flask only. And Maximum load would be the maximum material that you may sterilize in one cycle as 10 flasks, 5 garment packs, 6 Petri dish canes, 30 test tubes etc.
It totally depends on your requirements.

5

Thank you sir

6

There are 3 types of loads.

  1. fixed load with fixed position
    2)fixed load with variable position
    3)variable load with variable position
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7

Thank u sir.

8

I am agree with chowdhary sir kindly note that load pattern will be considered on the basis of Porous Loads And non Porous Loads.

Porous Loads required 12 log reduction 30 minutes,
However media may 8 log reduction 20 minutes,
Normally volume up to 700 ml required hold time 20mints
Up to 2700 ml required 35 minutes
Up to 5000 ml required 55 minutes
Depending upon your recipe.

9

thank you sir.

10

Sir, may i find SOP for approval of labelling material?

11

It’s a good information, but I have one doubt I.e for porous loads 12 log reduction it’s enough to get SAL.but non porous loads 6 log reduction is it enough or not.

12

6 log reduction is required in case of non porous load. That’s equivalent 6 log × BI’s d value
= 6 Log× 2.5 ( worse case)
=15 minutes.
This is SAL
Now, there are 2 conditions as per USP chapter 1229
In case non porous load
6 log reduction+ 2 additional
=6+2× ( BI’s d value)
=8×2.5= 20 minutes
This is over all kill approach

Now survival time
6-2 log reduction (Bi’s d value)
=4×2.5
=10 minutes.
Means if you sterilized your non porous load 20 minutes this will keep possiblity of spores.
And,
If you sterilized your non porous load 20 minutes this will over all kill your spore

13

Same case porous load.
6 log reduction+ 6 additional ×(BI’s d value)
12×2.5( Bi D value)

30 minutes will be over all kill approach

6 -2log reduction×(BI’s D value)
4×2.5
=10 minutes is survival time

14

When finding the kill time and survival time … why we have to add 4 for the kill time and substrate 2 for the survival time .
I came a cross white these 2 equations
Kill time = log 10 ( population +4 ) xDvalue
Survival time = log 10 ( population -2 ) x D value