PaulTG2

Hello,

I come up with 21.7% O2. I don't know what the spreadsheet uses to come up with less than 10%.

(O2 injected - O2 used)/N2 injected =FiO2

((0.40*10.7)-2.5)/(10.5-2.5)=.217 or 21.7% FiO2.

That's injection EAN40% at 10.7lpm with an O2 consumption of 2.5lpm -- which is a consumption beyond what you'll sustain.

Can you show the formula the spreadsheet is using?

Sincerely,

-p

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