Ppl = Pressure of the Pleural Space

Pleural Pressure
(the pressure between the two pleural membranes)

Deffinition:

Pleural pressure, or Ppl, is the pressure surrounding the lung, within the pleural space. This pressure is contained between the two membranes surrounding the lung and the chest cavity, namely the visceral pleura and the parietal pleura. The two membranes contain only a few mm's of fluid, serous interpleural fluid. During normal breathing the plural pressure is negative, or < 760 mmHg.

The Ppl directly affects the alveolar pressure which in turn affects the transpulmonary pressure. Mathematically speaking the overall functionionality of the lungs does depend on the Ppl at some point, as the gas intake varies with the specific pressures mentioned above. Since the totall lung size is determined by the transpulmonary pressure we can certainly see how important the Ppl really is. Since the serous interpleural fluid does obey all laws of gravity, when the individual is standing up the Ppl is greater at the base of the lung, leaving the top section of the lung with a lower transpulmonary pressure and therefore more expanded and less compliant than the base. During passive expiration Ppl helps eliminate gass from the lungs; at the same time it can cause temporary collapse of the bronchi during active expiration if the Ppl is allowed to become positive, or > 760 mmHg. In humans, the Ppl can be estimated with the help of a device called an esophageal baloon.

Equation of the day:

with Pm = respiratory muscle pressure forcing
       R   = respiratory system resistance
       C = respiratory system compliance
       V = lung volume

Interesting questions:
(all taken from pg. 176)

11. Why is airway resistance decreased at high lung volumes?

14. What breathing pattern might a patient with a normal airway resistance but very stiff lungs (low compliance) adopt which would reduce his work of breathing?

Sample problem:
(compliment to Dr. Y's exam problem)

Based on a linear model of respiratory mechanics consisting of resistance R = 2cm water/LPS and compliance C = 0.1 L /cm water, if the respiratory muscles provide 10 cm water driving pressure, how long will it take to inflate the lungs from FRC to 0.5L above FRC?

Solution:

Using the equation of the day, we can easily see that most of the unknowns are provided:

Pm = 10 cm water
R   =  2 cm water/LPS
C   = 0.1 L/cm water
V   = 0.5L (DON'T plug this in until the equation is solved first)

therefore:                          10 = 2 * dV/dt + V/0.1    or      5 = dV/dt + 5V
separation of variables:      integral[dt] = integral[dV/(5-5V)]     then      t + C1 = -1/5 * ln(5-5V) + C2
solving for t:                      t = -1/5 * ln(5-5V) + C     (collect all constants into one)
from initial conditions we can take FRC as 0 and therefore 0.5 + FRC as 0.5,
yielding:                            0 = -1/5 * ln(5) + C,     C = 0.321
completing the equation:    t = -1/5 * ln(5-5V) + 0.321    and NOW plug V = 0.5 in to get
final result:                         t = -1/5 * ln(5-5 * 0.5) + 0.321,    t = 0.1377 s

Useful links:

Johns Hopkins Pleural Pressure Page,     not that Doug hasn't already mentioned it...
Related Diseases,                                   something to watch out for
Pretty pictures of what we talked about,  just something for the visually inclined
My page ,                                                 in case you get really bored

A sample problem:

Answers:

For more information on this topic, please refer to West , page 99.

Also, check out the following links that may be helpful:

• And, as always, if you have questions, or if you can suggest a helpful link, please e-mail me.

No!
Please don't make me go to any of those pages!
I want to go somewhere completely different!

This page was written by Emil Istoc , a student in this course.

BME 403 Pages maintained by the T.A., Douglas Miles.