Apnoeic Oxygenation Flashcards
During apnoea what is the rate of rise of PaCO2 per minute?
0.3-0.7 kPa per minute
For calculations we can take the middle value of 0.5kPa/min
What is Henry’s Law?
That the amount of gas dissolved in a liquid is directly proportional to its partial pressure.
Considering Henry’s law why does the PaCO2 rise in apnoea?
VCO2 (CO2 production) continues during apnoea.
As CO2 is not removed by ventilation (as it is apnoea) it accumulates, as quantity of dissolved gas is directly proportional to partial pressure, the PaCO2 will rise.
If we model VCO2 as constant the rate of rise of PaCO2 will also be constant.
Which forces are responsible for maintaining a persons functional residual capacity?
There are the forces of the weight of the lungs trying to collapse and the surface tension of the alveoli.
These are balanced with the forces of the rigid thoracic cage.
During anaesthesia FRC is reduced, as the muscles of the thoracic cage become reduced therefore reducing the outward force of the thoracic cage.
If we make the following assumptions:
PaCO2 =PACO2
FRC = 2L
PaCO2 rise if 0.5kPa/min during apnoea
VCO2 = 200ml
How much CO2 crosses the alveolar membrane each minute into the FRC.
PaCO2 rise = 0.5kPa/min therefore if we consider Dalton’s law of partial pressures.*
Then this is a 0.5% increase each minute. As atmospheric pressure ~100kPa.
If the total volume of the FRC is 2000ml. A 0.5% increase represents 10mls.
That means each minute 190mls of CO2 is left in the total body water.
- that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases
Create an input/output model for the FRC in an apnoeic state.
Then consider what happens if you open their airway and provide fresh gas flow.
Input:
10mls CO2*
Output
200mls O2 uptake*
*These figures are based on the following assumptions:
+FRC static at 2000mls
+O2 uptake = 200mls
+PaCO2 rise in apnoea = 0.5kPa/min
+R=1 therefore VCO2 = 200mls also
+That nitrogen input into the FRC can be ignored as it is negligble
As you can see there is a discrepancy of 190mls between the input and output.
Considering PV=nRT
Volume is constant (as stated in our assumptions), R is a constant, temp is likely to be a constant.
The amount of gas (n) is variable as is pressure. Therefore as each minute there is a 190mls less gas the pressure drops which creates a pressure gradient. Therefore if there is an open airway gas will flow into the airway.
This is known as apnoeic ventilation or apnoeic mass transfer oxygenation.
Considering the following assumptions:
+FRC static at 2000mls
+O2 uptake = 200mls
+PaCO2 rise in apnoea = 0.5kPa/min
+R=1 therefore VCO2 = 200mls also
+Open airway and fresh gas flow of 100% O2
What is the rate of fall of PAO2?
Reconsider the input/output model.
Input:
10mls CO2
190mls O2
Output:
200mls O2
Therefore there is a 10ml loss of O2 each minute.
Using Dalton’s law of partial pressures we can calculate this to be a drop of 0.5kPa/min.
In a patient that had been fully pre-oxygenated and had a starting FAO2 of 0.8.
How long would it take for the PAO2 to drop to 10kPa if they had a patent airway and gas flow with an FiO2 of 1.0.
FAO2 of 0.8 could be considered a starting PAO2 of 80kPa
If they are having apnoeic ventilation with an FiO2 of 1.0 then each minute there will be a 10ml O2 deficit* which corresponds to a 0.5kPa/min drop.
80kPa - 10kPa = 70 kPa
As the rate of change is 0.5kPa/min it would therefore take 140min
*These numbers are based on the previous figures quoted they would vary with O2 consumption and FRC.
In a patient that had been fully pre-oxygenated and had a starting FAO2 of 0.8.
How long would it take for the PAO2 to drop to 10kPa if they had a patent airway and gas flow with an FiO2 of 0.5.
FAO2 of 0.8 could be considered a starting PAO2 of 80kPa
Considering the input/output model if they have apnoeic ventilation with an FiO2 of 0.5
Input:
10ml CO2
95mls O2
95mls N2
Output:
200mls O2
The O2 discrepancy is therefore now 105mls. Which corresponds to a 5.25kPa/min*
Therefore 80kPa-10kPa = 70 kPa
70kPa/5.25kPa/min =13.3min
*2000ml (FRC) 105mls O2 deficit
Partial pressure is out of 100kPa (at atmospheric pressure)
Therefore divide both by 20 to make the ratio 100:5.25
In a patient that had been not pre-oxygenated and had a starting FAO2 of 0.15
How long would it take for the PAO2 to drop to 10kPa if they had a patent airway and gas flow with room air.
FAO2 of 0.15 = PAO2 15kPa
Output O2 = 200mls
Input CO2 = 10mls
Input N2 = 0.79x190mls
Input O2 = 0.21x190mls =40mls
Therefore each minute there is a O2 deficit of 160mls
2000:160
100:8
This translates to a drop of 8kPa/min
Therefore for the PAO2 to drop to 10kPa it would take.
5kPa/8kPa/min = 0.625min ~38secs
