Canonical Question

Pulmonary Circulation

V5 C5.i Historical V4 F5.i 1 appearance

Master answer

Introduction

Starling Forces

\[J_v={\kappa \; ([P_{capil} – P_{interstit}] – \sigma \; [\pi_{plasma} – \pi_{interstit}])}\]

where
Jv is the trans endothelial solvent filtration volume per second

( [ Pc – Pi ] – σ [ πp – πi ] ) is the net driving force
P = hydrostatic pressure
π = oncotic pressure
σ = Staverman’s reflection coefficient ie. Permeability of membrane to protein (0.5 for lung)
κ = filtration constant = LpS = Hydraulic conductivity
x Surface Area

PulmonarySystemic
Pc
Capillary hydrostatic pressure
Pressure moving fluid out of capillary13→6 mmHg
Arterial → venous

Variable due to hydrostatic effects of gravity in different parts of lung
~35→15 mmHg Arterial → venous
Pi
Interstitial hydrostatic pressure
Pressure moving fluid into capillaryVariable,
but 0 to slightly negative
5 mmHg
πp
Plasma oncotic pressure
Pressure keeping fluid within capillary25 mmHg~20 mmHg
πi
Interstitial fluid oncotic pressure
Pressure keeping fluid out of capillary17 mmHg~0 mmHg

Oncotic pressure gradient

Hydrostatic pressure gradient

Overall Effect

The balance of Starling forces in the lung is generally stated as favouring reabsorption because of the clinical fact that the lungs are generally dry and clearly need to be to facilitate gas exchange

Safety Factors Preventing Pulmonary Oedema

JC 2019

Exam appearances

ExamExact wordingRelationshipSuccess
2016B Q19 Describe how Starling forces determine fluid flux within the pulmonary capillary bed. historical_member