BPK 306 Lecture Notes - Internal Intercostal Muscles, External Intercostal Muscles, Rib Cage

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Respiratory System #2
A. Movement of Air Molecules
1. Conduction
Movement of air into and out of lungs is due to pressure diff betw
alveolar air and atmosphere
Boyle’s Law: P1V1 = P2V2
a. Inspiration
oActive muscular process
oDiaphragm pulls connective tissue at top of bell downward to
incr volume (superior-inferior direction)
oExternal intercostal muscles incr lung volume in anterior-
posterior direction
a. Ribs originally slanted downward at rest. External
intercostals project them more anterior. Sternum pushed
20% farther away from spine. Bucket handle effect
b. Expiration
oPassive muscular process at rest
oElastic forces return lung, thoracic cavity, and diaphragm to
normal resting shape, and lower volume
oDuring exercise, abdominal muscles and internal intercostal
muscles enhance expiration
a. Abdominal muscles pull thoracic cage down (anterior-
posterior volume reduction), and push abdominal
contents up (superior-inferior vol reduction)
b. Internal intercostals reduce anterior-posterior distance
2. Diffusion
After bifurcation 17, air flow is slow due to large cross-sectional area
of respiratory tree. Here, air molecules move faster by diffusion than
by conduction
B. Resistance to Air Flow
Poiseuille’s Law describes air flow:
a. V = (dP*pi*r4)/(8nL) *21-7
b. R = (8nL)/(pi* r4)
c. Only works for laminar flow, but in lungs we also have
turbulent flow due to many bifurcations and extreme velocity
changes (b/c of enormous change in cross-sectional area)
d. Re = (2*r*v*p)/n
Largest resistance occurs in first 6 airway generations
a. Around 0.08 cm H2O/L/sec of resistance
b. Substituting a low density inspired gas for air reduces work of
c. Turbulence is in upper airways b/c of higher velocities and
large radii (higher Re)
C. Pressure Changes During the Breathing Cycle (B&L Fig 21-9)
Period betw E and I is natural resting condition
Period betw I and E is active non-resting state
Pleural pressure = -5 cm H2O at rest (E-I period)
a. Due to elastic recoil of lung and desire to expand of chest wall
b. 1 mm Hg = 1.36 cm H2O
At end of inspiration:
a. -7 to -8 cm H2O pleural pressure
b. More elastic recoil in lung = more negative pleural pressure
c. Alveolar pressure returns to 0 cm H2O (no airflow)
Expiration causes reduction in negative pleural pressure as elastic
tissues express themselves. Positive pressure in alveoli (relative to
outside) causes airflow out of lungs
D. Pressure Changes During First Breathing Cycles of an Infant
Intrapleural pressure of first breaths are about -60 to -100 cm H2O
Necessary to inflate formerly fluid-filled lungs with air
Evidence that in weeks prior to birth, babies perform breathing
1st breath volume is around 40 mL