HK
HK 3810 Package 3 Respiration.docx

# Farther away from where it wants to be so pleural

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farther away from where it wants to be so pleural pressure = -8cmH2O o Maximal inspiration Chest wall is pulled beyond it’s minimum strain of 6L so it starts to push back, causing a greater pull by the lung Pleural pressure = -12cmH2O o Expiration Passive Muscles stop contracting Elastic component of lung brings lung and chest wall back to original relationship Pleural pressure = -4cmH2O Tidal volume Resting “quiet” breathing Minute ventilation (VE) = volume/breath (VT) x breaths /min (frequency fo breathing) = 500mL x 12 = 6000/min at rest Using PPL to estimate the degree of stretch on the lung (volume in the lung, volume in the alveolus) Hysteresis – when tissue behaviour is different when challenged in one direction vs. the other air filled lungs has forces acting on it; these force is surface tension (opposes lung inflation) o Water is polar o At air-water interface molecule align and form bonds

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HK 3810 Package 3 Respiration o Surface tension will increase work of breathing Hysteresis is shown when we graph pleural pressure vs. volume; it is not a linear relationship If H2O lined the alveoli, something occurs between the alveoli Distending pressure – pressure to hold alveoli open Law of LaPlas: P = 2T/r P1 = 2T/decreased r increase P1 P2 = 2T/increase r decreased P2 So, air will move from small alveoli to large alveoli since the pressure in the small one is greater than the pressure in the large one Small alveoli (radius) require larger distending pressures P1 > P2, resulting in air Q from alveoli #1 to #2 results in the collapse of all small alveoli into larger alveoli This is a bad situation To fix this problem, we have no air-H2O interface, instead we have an air-surfactant interface Surfactant is 90% phospholipid, 10% protein, and it’s nonpolar with vert weak self attractions Surfactant functions to: o Minimize the work of breathing o Minimizes the pressure difference between alveoli with different radii; small radius alveolus will produce more surfactant P1 = 2decreaseT/decrease r; increase in surfactant = decrease T Without surfactant not enough force is generated to overcome surface tension, hyteresis is still there bc H2O is present Alveolar ventilation (volume) VA = VT-VDS = 500-150 = 350mL DS is dead space; volume in non exchanging areas (trachea, etc) Prone o Lying down, when not inspiring or expiring o PPL = -4cmH2O over whole lung o All alveoli start with a similar volume, and when we inspire -4cmH2O decreases to -8cmH2O everywhere o All alveoli increase volume equally over the lung Upright o Gravity pulls the lung mass down o At rest, the bottom of the lung has PPL of -2cmH2O, middle has -4cmH2O, top has -10cmH2O o When we inspire, these will all change by -4cmH2o resulting in -6cmH2O, -8cmH2O, -14cmH2O o So at rest, not all alveoli are starting at the same volume o There is an unequal distribution of the 350mL when standing When graphing the relationship of the movement of volume from the bottom of the lung to the top, prone has a straight horizontal line, while standing has a decreasing linear line
HK 3810 Package 3 Respiration Blood flow within the pulmonary system

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• Fall '16
• Coral Murrant
• pulmonary artery

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