Lung Mechanics

Abstract

Recoil pressure in the saline-filled lung is a unique function of lung volume. This recoil is provided by forces in the tissues that form the macrostructure of the lung: the pleural membrane, the bronchial tree, and the inter-lobular membranes that connect the bronchial tree to the pleural membrane. In the air-filled lung, recoil pressure is a function of lung volume and the internal variable, surface tension, which depends on volume history. The additional recoil pressure of the air-filled lung is the result of the direct effect of surface tension and the indirect effect of inducing tension in the lines of connective tissue that form the free edges of the alveolar walls at the boundary of the lumen of the alveolar duct. Non-uniform deformations are analyzed using the methods of linear elasticity. These include the gravitational deformation of the lung and the local deformation of the parenchyma surrounding a constricted airway.

Appendix

The fundamental equation of the model is Eq. (1.2).

$$ {P}_{\gamma }=\frac{2}{3}\cdot \frac{\gamma \cdot S}{V_L}+\frac{1}{3}\cdot \frac{\tau \cdot L}{\left({V}_L+{V}_{ti}\right)} $$

(1.4)

Surface area increases with increasing V _L and decreases with increasing L. Thus, S is a function of these two variables.

$$ S=S\left({V}_L,\;L\right) $$

(1.5)

This function is subject to two conditions. The first is the condition of internal equilibrium. At equilibrium, the internal stored energy of a structure is minimum, and the change of internal stored energy for a virtual displacement of the system is zero. In this case, the internal energy is the sum of the surface energy and the stored energy in the cables, and the equilibrium condition for a virtual change in L is the following.

$$ \gamma \cdot \frac{\partial S}{\partial L}+\tau =0 $$

(1.6)

Equation (1.6) is a generalized form of the equilibrium between the outward pull of surface tension and the inward Laplace force exerted by the cable. The second condition is the equality between the pressure -volume work done by lung inflation and the increase in the stored energy in the structure.

$$ {P}_{\gamma }=\gamma \cdot \frac{\partial S}{\partial {V}_L} $$

(1.7)

Substituting from Eqs. (1.6 and 1.7) into Eq. (1.4) for τ and P _γ yields the following partial differential equation for S.

$$ \frac{\partial S}{\partial {V}_L}-\frac{2}{3}\frac{S}{V_L}+\frac{1}{3}\frac{L}{\left({V}_L+{V}_{ti}\right)}\cdot \frac{\partial S}{\partial L}=0 $$

(1.8)

The general solution to this equation is the following, where F is an arbitrary function of its argument x.

$$ S={V}_L^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cdot F\left[\frac{L}{{\left({V}_L+{V}_{ti}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right] $$

(1.9)

Here, we take $F={C}_1\left[1-{C}_2{x}^2\right]$, and S is the following.

$$ S={C}_1\cdot {V}_L^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cdot \left[1-{C}_2\cdot \frac{L^2}{{\left({V}_L+{V}_{ti}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right] $$

(1.10)

The first term in Eq. (1.10) describes the relation between surface area and volume for a volume change in which the geometry of the parenchyma remains similar. The second describes the deficit in surface area because of the volume occupied by the lumen of the duct. With this expression for S, Eq. (1.6) is the following.

$$ 2{c}_1{c}_2\frac{\gamma \cdot {V}_L^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cdot L}{{\left({V}_L+{V}_{ti}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\tau $$

(1.11)

Finally, the length-tension relation for the cables is given by the following empirical equation, where L ₀ is the resting length of the cable.

$$ \tau ={C}_3\left(\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{${L}_O$}\right.-1\right)\cdot exp\left[{C}_4{\left(\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{${L}_O$}\right.-1\right)}^2\right] $$

(1.12)

The parameter L ₀ can be absorbed into the parameters of the equations, Eqs. (1.4, 1.10, 1.11 and 1.12), by introducing the following normalized variables: $v=V/{V}_o$, $s=S/{V}_0$, $t={L}_0\cdot \tau /{V}_0$, and $l=L/{L}_0$, where the reference volume, V ₀, is taken as TLC. In terms of these variables, the governing equations, Eqs. (1.4, 1.10, 1.11 and 1.12), are the following.

$$ {P}_{\gamma }=\frac{2}{3}\frac{\gamma \cdot s}{v_L}+\frac{1}{3}\frac{t\cdot l}{\left(v+{v}_{ti}\right)} $$

(1.13)

$$ s={c}_1{v}_L^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left[1-{c}_2\frac{l^2}{{\left({v}_L+{v}_{ti}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right] $$

(1.14)

$$ 2{c}_1{c}_2\frac{\gamma \cdot {v}_L^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cdot l}{{\left({v}_L+{v}_{ti}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=t $$

(1.15)

$$ t={c}_3\left(l-1\right)\cdot \exp \left[{c}_4{\left(l-1\right)}^2\right] $$

(1.16)

For given values of γ and v _L , Eqs. (1.13–1.16) can be solved for the values of the remaining variables, l, t, s, and P _γ . To follow a volume history, starting from given values of γ and v _L , the equation governing the evolution of γ is added.

$$ s\cdot \frac{\partial \gamma }{\partial s}=200\kern0.5em \mathrm{dyn}/\mathrm{cm} $$

(1.17)

The parameter c ₁ in these equations has units of ${\mathrm{cm}}^{-1}$; its value is inversely proportional to the diameter of the alveolus. c ₂ is dimensionless; it describes the fraction of the volume of the duct that is subtended by the lumen. The parameter c ₃ has units of pressure; it sets the magnitude of the second term in Eq. (1.13), the pressure provided by the cables. c ₄ is dimensionless; it describes the nonlinearity of the length-tension curve of the cables.

The following values of the parameters were assigned.

$$ {c}_1=400\;{\mathrm{c}\mathrm{m}}^{-1}\kern1em {c}_2=0.3\kern1em {\mathrm{c}}_3=16\;\mathrm{cm}\kern0.5em {\mathrm{H}}_2\mathrm{O} $$

$$ \begin{array}{ll}{c}_4=4& {v}_{ti}=0.12\end{array} $$

The equations were solved for volume trajectories like those shown in Fig. 1.1. The outer inflation curve begins at ${v}_L=0.22$, $\gamma =2\;\mathrm{dyn}/\mathrm{cm}$, and follows the variables as v _L increases to the value at which $\gamma =28\;\mathrm{dyn}/\mathrm{cm}$. It continues, holding γ constant and ignoring Eq. (1.17), to ${v}_L=1$. The deflation limb follows the variables to the value of v _L at which $\gamma =2\;\mathrm{dyn}/\mathrm{cm}$, and continues, holding γ constant, down to ${v}_L=0.2$. Inflation limbs starting at ${v}_L=0.35$ and ${v}_L=0.5$ and $\gamma =2\;\mathrm{dyn}/\mathrm{cm}$ and going to the values of v _L at which $\gamma =28\;\mathrm{dyn}/\mathrm{cm}$ were also calculated. The predicted values of P _tp and S for these trajectories are shown in Fig. 1.4.

It is interesting to note that the values of γ and P _tp are the same for mammals of all sizes, but that alveolar diameter varies over a range of a factor of 4 with diameter roughly correlated with animal mass [48]. A larger value of c ₁ and a smaller value of c ₃ would be used to represent a species with smaller alveolar diameter, and the first term in Eq. (1.13) would be relatively bigger than the second.