Hello everybody, first of all i´m from spain and i´m sorry if my english isn´t very well, i hope you understand me.

I want to do an integrity test in HEPA filters with a particle counter in a laminar flow and biological safety cabinets. Looking the UNE 14644-3 (it is the same as ISO 14644-3 but here in Spain), in the section B.6.3, it explains how to do it, but i don´t undertand some section. In section B.6.3.6.2 it says that i have to calculate the standard leak penetration (PL) with the equation: PL = K x Ps, and I have to choose “K” from a table in function of Ps (Maximum allowable penetration). The table shows different values of Ps and the “K” you have to choose, and the minimum value to “Ps” is 5x10-4 and “K” is 10, that i think it belongs to HEPA filter U15 (in this case i think it´s called ULPA filter). My problem is that in the HEPA filters H14, Ps is 5x10-3 and i don´t know which value of “K” i have to choose.

Please help me!! I´m very desperate!!! I really hope you have understood me.

Hello,
The problem with Hepafilter-testing with a particle counter is that you don’t know how many particles you have on the pressure side.
So, you must know what level of cleanless you want in your area. That is the limit when you scan with a particle counter. To challenge your hepafilter you can ad smoke particles on the pressure side. But it is not a solid test to test your hepafilter with a particle counter. Gr Dennis

Dear all,
concerning HEPA intergrity testing - I’ve noticed this topic eralier, but was doubting to answer. Now, having more experience (proved by other professionals) I should say that problem of the ISO 14644-3:2005 is that not only question concerning K choice for PL calculations - it’s rather easy. EN 1822 contains both local and integral efficiacy and leakage, espressed in percentages. It means if you have for example H14 filters, accroding to EN 1822 it has 99,995 % efficiacy (considering MPPS) and 0,005% allowable leakage. ISO 14644-3 table B.3 express Ps values in unit fraction, it means you need to divide your percentage by 100. So for given example with H14 (0,005 % of allowable leakage) you’ll get 0,005 / 100 = 0,00005 = 5*10^(-5). You choose Factor K = 10 and that’s it.

The most inetersting details are furtheron. First of all it’s not clear in a given example (let’s talk about Fiqure B.3), why have we choose Ca for stationary measurement is equal 1? Table B.2 allows as to choose corresponding Np value only in a narrow rangefrom 1 to 11. Otherwise you are not able to calculate neither Sr nor Ts. Due to absence of Np value in B.2. Ok, let’s turn to stationary measurement - how to determine Tr? In a given example it has been chosen as 6 sec without any explanation. Then we have calculated Np and coressponding Ca equal in a given example 67 particles. It it our acceptance criteria? If so, it means that we need to compare Cc expressed in particles/cm^3 with Ca expressed in absolute units (particles). Why it is so complicated? We can easy to compare Cc with concentrations behind the filters according to EN 1822 efficiacy & allowable leakege limits using one simple proportion instead of complicated calculations.It means we need onlz to switch parcticle count in N/m^3 mode instead of absolute count (N). The leakage as a rule you can easy determine by millions of particles/m^3 passing through damaged filter.