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ASSESMENT OF LUNG PARENCHYMAL DAMAGE WITH ARTERIAL BLOOD GAS ANALYSIS PROF.N.K.AGRWAL JNMC, SAWANGI,WARDHA

Assesment of lung parenchymal damage with arterial blood

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ASSESMENT OF LUNG PARENCHYMAL DAMAGE WITH ARTERIAL BLOOD GAS ANALYSISPROF.N.K.AGRWAL

JNMC, SAWANGI,WARDHA

OBJECTIVE

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

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.

THE ALVEOLI ARE SMALL SACS OF BHONCHIOLES

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

Patients with chronic bronchitis, COPD, fibrosis, pneumonia, ARDS etc have changes in thickness and surface area.2

RESPIRATORY EQUATIONS

PATM = PN 2+Po2+Pco2+PH2O NO.1

 

P760 = P514 + P159 + P40 + P47 NO. 2

 

PI02 = FIO2 X ( PB - -- H2 0 ) NO .3

PAO2 = FiO2 x ( Pb- 47 ) - 1.25 x PCO2 NO 4

 

PAO2 = FiO2 x ( 760- 47 ) - 1.25 x 32

= FiO2 x 713 -40 = FiO2 x 673

The normal P ( A-a) O2 at room air - 5 to 20 mm of Hg Fio2 - 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 % )

The expected PaO2 will be EPaO2 = PAO2 X (1-k)

(equation is copyright of author )

EpaO2– PaO2 (ABG)

Lung Damage = ------------------- x100

EPaO2

Case No. 1

A patient in ICU had PaO2 78 mm of hg with Fio2 =1

 

EPaO2= PAO2x ( 1-0.2 )

= 673 x 0.8

= 536 mm Hg ( using respiratory eq. no 4 )

536 - 78

LD = -------------------

536

= 0.85x 100

= 85 %

CASE NO.2

 A patient posted for thoracotomy the ABG showed PaO2=57 at room air FIO2= 0.2

EPaO2 = PAO2 x (1- 0.05)

= 107 x 0.95

=101mm of Hg

101 - 57

LD = ----------------- x 100

101

= 44 %

 

 

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 may be able to calculate the lung damage , in ICU as well as in pre operative phase, to assess the prognosis.

REFERENCES :

1.West JB: Respiratory physiology-- The essentials, 4th ed. Baltimore, William & Wilkins 1990

2.respiratory physiology 7th edition wyeli & Churchill page 168-169

3. Nunn JF: Alveolar air equation Anesthesiology 1996; 85:946

4.Mansoor,R. Murray and Nadel's Text book of respiratory medicine,5 the edition,philadelphia,PA 2010

5.7th edition Wylie and Churchill Davidson page 168-169

6. Gas exchange in human retrieved 19 march 2013