Critical Care Medicine-Pulmonary Disorders>>>>>Respiratory Diagnostic Modalities and Monitoring
Question 12#

A 65-year-old female patient presents with acute respiratory failure to the ICU. She is intubated and placed on pressure support ventilation. The driving pressure is set at 10 cm H2O. The PEEP is set at 10 cm H2O. Esophageal balloon is placed, and a pressure of negative 12 cm H2O is obtained at end inspiration.

Given these measurements please estimate transpulmonary pressure. 

A. 22 cm H2O
B. 8 cm H2O
C. 32 cm H2O
D. 20 cm H2O

Correct Answer is C

Comment:

Correct Answer: C

Transpulmonary pressure is defined as the difference between alveolar pressure and pleural pressure.

Alveolar pressure (at end inspiration) is the driving pressure plus PEEP (here, 10 cm H2O + 10 cm H2O = 20 cm H2O).

The pressure measured by the esophageal balloon is a surrogate for the intrapleural pressure and is measured at −12 cm H2O for this patient. Therefore, the estimated transpulmonary pressure is 32 cm H2O. Elevated transpulmonary pressures increase the strain on the lung and worsened lung injury. It is important for clinicians to consider the impact of the pleural pressure on the transpulmonary pressures to identify patients at risk for worsening lung injury and to institute better ventilation strategies.

References:

  1. Schmidt UH, Hess DR. Does spontaneous breathing produce harm in patients with the acute respiratory distress syndrome? Respir Care. 2010;55(6):784-786.
  2. Kassis EB, Loring SH, Talmor D. Mortality and pulmonary mechanics in relation to respiratory system and transpulmonary driving pressures in ARDS. Intensive Care Med. 2016;42(8):1206-1213.