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Pulmonary Circulation — Starling Forces

Current · V5 (2025) → C5.i Historical · V4 (2023) → F5.i 1 exam appearance

2016B Q19

Exam question

Describe how Starling forces determine fluid flux within the pulmonary capillary bed.

CICMWrecks answer

Master answer

Introduction

Starling Forces

Starling forces across a capillary
Jv=κ([PcapilPinterstit]σ[πplasmaπinterstit])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

Reusable content

Formulae used in this answer

1

Starling Forces

CWF-0125
Jv=κ([PcapilPinterstit]σ[πplasmaπinterstit])J_v={\kappa \; ([P_{capil} - P_{interstit}] - \sigma \; [\pi_{plasma} - \pi_{interstit}])}

Past papers

Exam appearances

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Exam Exact exam wording Candidate success
2016B Q19 Describe how Starling forces determine fluid flux within the pulmonary capillary bed. 25%