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Pathophysiology of hypoxic respiratory failure
1. PATHOPHYSIOLOGY OF ACUTE RESPIRATORY FAILURE Dr. Andrew Ferguson Consultant in Anaesthetics & Intensive Care Medicine Craigavon Area Hospital, U.K. http://www.slideshare.net/fergua
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6. CASE SCENARIO John is a 43 year old with mild asthma. He attends the Emergency Department with a 72 hour history of myalgias and fever, with increasing dyspnoea and a productive cough. His room air SpO 2 is 84% and his respiratory rate is 37/minute and shallow. He is using his accessory muscles and there is evidence of muscle use on exhalation. ABG shows pH 7.34, pCO 2 6.1 kPa, pO 2 7.8 kPa on room air He is sweaty and tiring rapidly. You detect crepitations at his right lung base and widespread wheeze. CXR confirms a right lower zone infiltrate.
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10. RESPONSE TO INCREASED FIO 2 IN SHUNT Shunt fraction (%) Alveolar pO 2 As shunt fraction increases, higher inspired (and alveolar) pO 2 has less and less effect on arterial pO 2
21. BOHR EQUATION You might want to get coffee….we are going to derive the Bohr equation and there are some mathematics and mental gymnastics involved! You have been warned! And yes…it does have some clinical relevance…read on!
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27. THE ALVEOLAR GAS EQUATION (COMPLEX) If R < 1 (and especially if FiO 2 is high) then more O 2 is taken up from the alveolus than CO 2 is released into it, and alveolar volume would fall if more gas did not move in passively from the airway to replace it. This gas moving in can increase the P A O 2 very slightly and if we want to correct for this we need to use this larger equation. OUCH! Luckily the effect is so small that in the real world we can pretty much ignore it. P A O 2 = (P atm – P H2O ) x FiO 2 – PaCO 2 /R + (FiO 2 x P a CO 2 x (1-R)/R))
28. Back to the A-aDO 2 A-aDO 2 = P A O 2 -P a O 2 So John, assuming R is 0.8 and breathing 80% O 2 should have an alveolar O 2 of: 94.75 x 0.8 – 1.25 x 6.6 = 75.8 – 8.25 = 67.6 kPa John’s A-aDO 2 is 67.6 – 8.6 = 59 kPa Which is WAY above the normal of 2 kPa, so his hypoxia is plainly not due to his hypercapnia!!!
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33. THE SHUNT EQUATION OK, so we have Qt = Qs + (Qt – Qs) , but we are interested in the total O 2 running in this system, which equals flow x content Let’s add in the content and go from there: Qt.CaO 2 = Qs.CvO 2 + (Qt – Qs).CcO 2 (since Qt – Qs = Qc) Qt.CaO 2 = Qs.CvO 2 + Qt.CcO 2 – Qs.CcO 2 Arrange the equation so Qs and Qt are on opposite sides: Qs.CcO 2 – Qs.CvO 2 = Qt.CcO 2 – Qt.CaO 2 now factorise… Qs(CcO 2 – CvO 2 ) = Qt(CcO 2 – CaO 2 ) and rearrange to get Qs/Qt Shunt fraction Qs/Qt = (CcO 2 – CaO 2 ) / (CcO 2 – CvO 2 )
41. SO BACK TO JOHN… John has been ventilated for 2 hours. It looks like you have stabilised him!!! Initial + 1 hour Current PEEP (cmH 2 O) 8 8 10 PC (cmH 2 O) 20 16 18 Peak (cmH 2 O) 28 28 28 Mean AP (cmH 2 O) 12.5 12 14 RR / min 15 16 16 I:E ratio 1:2 1:1 1:1 FiO 2 80% 60% 60% pO 2 (kPa) 9.9 9.1 9.7 pCO 2 (kPa) 4.9 5.3 5.2
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43. THANK YOU FOR TAKING THIS TUTORIAL! Please let me know if you found it helpful, or if you have other areas you would like to see covered. You can email me at: fergua at gmail.com
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