The Key to Blood Gas Interpretation: Four Equations, Three Physiologic Processes Equation Physiologic Process 1) PaCO2 equation Alveolar ventilation 2) Alveolar gas equationOxygenation 3) Oxygen content equation Oxygenation 4) Henderson-Hasselbalch equation Acid-base balance These four equations, crucial to understanding and interpreting arterial blood gas data.
PaCO2 Equation: PaCO2 reflects ratio of metabolic CO2 production to alveolar ventilation PaCO 2 = VCO2 x VA = VE – VD VCO2 = CO2 production VE = minute (total) ventilation = resp. rate x tidal volume VD = dead space ventilation = resp. rate x dead space volume converts VCO2 and VA units to mm Hg PaCO 2 Condition in BloodState of Alveolar Ventilation > 45 mm HgHypercapniaHypoventilation mm HgEucapniaNormal ventilation < 35 mm HgHypocapniaHyperventilation
Dead Space Airways Alveoli Alveolus ETT Alveolus V D equip V D anat VDAVDA High PEEP
PaCO3 Equation Hypothermia Hyporthyroidism Underfeeding Neuromuscular blockade High fatty acid substrate Sepsis/inflammation Hyperthermia Hyperthyroidism High carbohydrates Seizure and agitation Low ProductionHigh Production PaCO 2 = V CO 2. VEVE * (1- V D /V T ) Respiratory Rate Tidal Volume V D equip V D anat VDAVDA Cell Metabolism HME PEEP Low BP
Hypercapnia ↑PaCO 2 ↑VCO 2 = ↔VA = VE – VD Increased CO2 production but not able to hyperventilate: Fever Sepsis Hyperthyroidism Overfeeding with carbohydrates Agitation
Hypercapnia ↑PaCO 2 ↔VCO 2 = ↓VA = ↓VE – VD Decreased Alveolar Ventilation due to Decreased Minute Ventilation (VE= ↓VT X ↓RR) Sedative drug overdose Respiratory muscle paralysis Central hypoventilation
Hypercapnia ↑PaCO 2 ↔VCO 2 = ↓VA = VE – ↑VD Decreased Alveolar Ventilation due to Increased Dead Space Ventilation (VD= Dead Space Volume X RR) Pulmonary embolism High PEEP Pulmonary hypertension Chronic obstructive pulmonary disease
Dangers of Hypercapnia An elevated PaCO2 will lower the PAO2 (Alveolar gas equation), and as a result will lower the PaO2. An elevated PaCO2 will lower the pH ( Henderson- Hasselbalch equation). The higher the baseline PaCO2, the greater it will rise for a given fall in alveolar ventilation, e.g., a 1 L/min decrease in VA will raise PaCO2 a greater amount when the baseline PaCO2 is 50 mm Hg than when it is 40 mm Hg.
Hypocapnia ↓PaCO 2 ↓VCO 2 = ↔VA = VE – VD Decreased CO2 production but same minute ventilation: Hypothermia Paralysis Hypothyroidism Underfeeding with carbohydrates Sedation
Hypocapnia ↓PaCO 2 ↔VCO 2 = ↑VA = ↑VE – VD Increased Alveolar Ventilation due to Increased Minute Ventilation (VE= ↑ VT X ↑ RR) CNS stimulants Agitation Central hyperventilation
Eucapnia ↔PaCO 2 ↑VCO 2 = Increased CO2 production and Increased Alveolar Ventilation: Fever and sepsis Hyperthyroidism Agitation ↑VA = ↑VE – VD
Eucapnia ↔PaCO 2 ↓VCO 2 = Decreased CO2 production and decreased Alveolar Ventilation Hypothermia Hypothyroidism ↓VA = ↓VE – VD
PCO2 vs. Alveolar Ventilation The relationship is shown for metabolic carbon dioxide production rates of 200 ml/min and 300 ml/min (curved lines). A fixed decrease in alveolar ventilation (x-axis) in the hypercapnic patient will result in a greater rise in PaCO2 (y-axis) than the same VA change when PaCO2 is low or normal. This graph also shows that if alveolar ventilation is fixed, an increase in carbon dioxide production will result in an increase in PaCO2.
VCO 2 X PaCO2 and Alveolar Ventilation: Test Your Understanding What is the PaCO 2 of a patient with respiratory rate 24/min, tidal volume 300 ml, dead space volume 150 ml, CO 2 production 300 ml/min? The patient shows some evidence of respiratory distress. PaCO 2 = VA = VE – VD VCO 2 =300 X.863 VA = VE (300X24) – VD (150 X 24) VCO 2 =259 VA = VE (7.2) – VD (3.6) VA = 3.6 PaCO 2 =71.9
PaCO2 and Alveolar Ventilation: Test Your Understanding What is the PaCO 2 of a patient with respiratory rate 10/min, tidal volume 600 ml, dead space volume 150 ml, CO 2 production 200 ml/min? The patient shows some evidence of respiratory distress VCO 2 X PaCO 2 = VA = VE – VD
PaCO2 and Alveolar Ventilation: Test Your Understanding A man with severe chronic obstructive pulmonary disease exercises on a treadmill at 3 miles/hr. His rate of CO 2 production increases by 50% but he is unable to augment alveolar ventilation. If his resting PaCO 2 is 40 mm Hg and resting VCO 2 is 200 ml/min, what will be his exercise PaCO 2 ? VCO 2 X PaCO 2 = VA = VE – VD PaCO 2 = X VA = 4.32 L/min ↑300 X PaCO 2 =59.9
Effective Ventilation Airways Alveoli Alveolus ETT Alveolus V D equip V D anat VDAVDA VT= 500V D equip = 50V D anat = 125V D A = 25VTe= 300RR= 10 VT= 250V D equip = 50V D anat = 125V D A = 25VTe= 50RR= 20 V E = 5 L/min
Ventilator Course in Sudan: December 15-16, 2011