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ASSESMENT OF LUNG PARENCHYMAL DAMAGE WITH ARTERIAL BLOOD GAS ANALYSIS (ABG) PROF.N.K.AGRAWAL

ABG ANALYSIS

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Page 1: ABG ANALYSIS

ASSESMENT OF LUNG PARENCHYMAL DAMAGE WITH ARTERIAL BLOOD GAS ANALYSIS (ABG)

PROF.N.K.AGRAWAL

Page 2: ABG ANALYSIS

OBJECTIVE

The objective of the study was to assess the lung damage at alveoli level with arterial blood gas analysis (ABG)

Page 3: ABG ANALYSIS

The major task of the lung is to oxygenate the blood and remove carbon dioxide, this happens by exchange

of gas between alveoli and blood capillary. Oxygen diffuses passively from the alveoli to plasma. The

diffusion is affected by-

Surface area

available

Thickness of

membrane

Pressure difference of the gas across the

barrier

Molecular weight of the gas

Solubility of gas.

Page 4: ABG ANALYSIS

The alveoli are main respiratory surface in human being, they are small sacs of branching of bronchioles, and they are one cell thick and provide large area for gas exchange

If all other factors are kept constant the main factor are surface area and thickness of alveoli. The gas exchange is hence inversely proportionate to thickness, the more is the thickness less will be diffusion. Patients with chronic bronchitis, COPD, fibrosis pneumonia, ARDS etc have changes in thickness and surface area.2

Page 5: ABG ANALYSIS

RESPIRATORY EQUATIONS

PATM = PN 2+Po2+Pco2+PH2O

P760 = P514 + P159 + P40 + P47

PI02 = FIO2 X ( PB - -- H2 0 )

PAO2 = FiO2 x ( Pb- 47 ) - 1.25 x PCO2

Page 6: ABG ANALYSIS

The normal P ( A-a) O2 difference at patient breathing air is 5 to 20 mm of Hg while at Fio2 of 1 it may go up to 110 mm of Hg. The A-a gradient increases by5-7 mm of Hg with every 10% rise of FIO2

Expected Pao2 = PAo2 X ( 1- k ) (k = 0.05 at 20-40 %, 0.1 at 41-60%,0.15 at 61-80%,0.2 at 81-100 % FIO2 )

Page 7: ABG ANALYSIS

• The expected Pao2 will be EPao2 = PAO2 X (1-k)

(Below equation is copyright of author)

Epa02– Pa02 Lung Damage = ------------------- x100 EPao2

Page 8: ABG ANALYSIS

Case No. 1 A patient in ICU had Pao2 78 mm of hg with Fio2 =1 PAo2 = 1x (760 - 47) – 1.25x32

= 1x 713 - 40 = 673 mm of Hg EPao2= PAo2x 1-0.2 = 0.8 x 673 = 536 mm Hg

536 - 78 LD = -------------------

536 = 0.85x 100 = 85 %

Page 9: ABG ANALYSIS

CASE NO.2 A patient posted for thoracotomy the ABG showed Pao2=57

PAo2 = 0.2 x (760-47) – 1.25x32

=147-40 =107 EPao2 = 0.95x PAo2 = 0.95 x 107 =101mm of Hg

101 - 57 LD = ----------------- x 100

101 = 44 %

Page 10: ABG ANALYSIS

CONCLUSION

Here we conclude that whenever the ABG reports are read in one must take in account the Fio2 and atmospheric pressure. Using the above equation we will be able to assess the lung damage correctly to manage the patients who are appearing normal as per Pao2 available.