Pre-Oxygenation Flashcards
When pre-oxygenating prior to an anaesthetic what is happening?
You are replacing the nitrogen in the FRC with oxygen, molecule for molecule.
If your FRC is 2L and you are pre-oxygenated with FiO2 1.0. Your alveolar ventilation is 4L/min (regardless of no. of breaths). Your starting FAN2 =0.8 and FAO2 =0.15
After 30 secs if you took 1 breath what would be the proportion of FAN2 and FAO2?
After 30 secs if you took 2 breaths what would be the proportion of FAN2 and FAO2?
If you took 1 breath you would increase the total volume to 4L from 2L.
You would therefore dilute the proportion of FAN2 by 2 therefore the FAN2 would now be 0.4. And the FAO2 would be 0.55. (as you are replacing O2 and N2 molecule for molecule the proportional change must be the same.
If you took 2 breaths, after the first breath you would increase the total volume by 1L (remember Alveolar ventilation is fixed in this model regardless of no. of breaths).
By increasing the volume by 1L you would have 3L instead of 2L. Therefore the FAN2 would be a third less after the first breath. Giving you a FAN2 of 0.53.
After the second breath you would again decrease the FAN2 by a third but the starting value is now 0.53. This would give you a FAN2 of 0.36. The FAO2 would therefore be 0.59.
If your FRC is 2L and you are pre-oxygenated with FiO2 1.0. Your alveolar ventilation is 4L/min (regardless of no. of breaths).
Create a formula for the proportional change that occurs with (FAN2 and FAO2) during pre-oxygenation over 30secs related to the number of breaths.
(n/n+1) to the power of n
If you take one breath the proportional change is 1/2 to the power of 1 which =1/2
If you take 2 breaths the proportional change is (2/3) to the power of 2. Which = 4/9 (0.44) so if the starting FAN2 was 0.8 the resulting FAN2 after 2 breaths would be 0.36.
If you take 10 breaths the proportional change is
If you take 150 breaths the proportional change is (150/151) to the power of 150 the proportional change would be 0.369
What is the maximum proportional change which can occur when modelling pre-oxygenation?
Take the formula (n/n+1) to the power of n
f you take 2 breaths the proportional change is (2/3) to the power of 2. Which = 4/9 = 0.44
If you take 10 breaths the proportional change is (10/11) to the power of 11. The proportional change would be 0.386
If you take 150 breaths the proportional change is (150/151) to the power of 150 the proportional change would be 0.369
Notice that after each breath the difference in proportional change decreases, this represents an exponential function.
As we get to infinity breaths the proportional change will = e = 0.3678
e is a mathematical constant just like Pi
1a) Draw a graph modelling FAN2 wash out over 2mins during pre-oxygenation.
1b) Write a formula for the graph.
1c) Calculate the FAN2 value at 2 mins.
The following assumptions apply:
FRC is 2L and you are pre-oxygenated with FiO2 1.0. Your alveolar ventilation is 4L/min
Note you will need a calculator with a scientific function.
y axis FAN2
x axis time
-ve exponential curve with y approaching 0 at 2 mins
Formula for the curve would be
y = A x e (to the power of -kt)
A = to the starting value or y value when x =0
k is a rate constant (number of volumes taken to fill a container over a time frame) or Q (flow) /V (volume)*
t is the time variable **
Therefore if we wanted to calculate the FAN2 at 2 mins we would have the formula.
A=0.8
k=2 as flow (Alveolar ventilation)=4 and Volume (FRC)=2
t= 2
y = 0.8 x e (to the power of -4)
= 0.014
** It can also be considered the time constant. This is the time to hit the baseline if the initial rate of fall at A had continued. Because this rate of fall is not maintained in a proportional exponential process: at 1t, 37% of A remains (e= 0.3678)
Draw a graph modelling FAO2 wash out over 2mins during pre-oxygenation.
The following assumptions apply:
FRC is 2L and you are pre-oxygenated with FiO2 1.0. Your alveolar ventilation is 4L/min. Your starting FAO2 is 0.15
Positive exponential curve (opposite to was in curve)
y axis FAO2
x axis time
y= 0.15 when x = 0
At 2 mins y = 0.85
y = A (1-e to the power of -kt)
k= Q/V 4/2 = 2
