Krdytkyu

Diffusion model

Could anyone help with my understanding of the maths in this excerpt from a physiology textbook (please see link).

The authors describe a model of gas diffusion across the length of an alveolar capillary.

• 
  $dot Q$ is blood flow through the capillary in ml/min


• 
  $beta$ is the capacitance coefficient for blood. It is the increment of total content of oxygen in blood per
  increment in partial pressure - mmol/(kPa*ml)


• 
  $dot Qbeta$ is known as the perfusion conductance - mmol/(min* kPa)


• D is the diffusing capacity of the lung - mmol/min*kPa


• P is partial pressure of gas


• 
  $P_A$ is the partial pressure of gas in the alveoli


• 
  $P_V$ is the partial pressure of gas in the veins


• 
  $P_C$ is the 'partial pressure' of gas in the capillary


• 
  $x$ is a distance along the capillary


• 
  $x_0$ is distal end the capillary


• 
  $P_{Cx}$ is the partial pressure of gas in the capillary at position $x$
  

The equation in figure A of the book is this:
$$Equationquad 1: qquad dot Q cdot beta cdot dP_C= (P_A-P_C) cdot dD$$

This makes sense to me as a mass balance equation:

$dot Q cdot beta cdot dP_C$ as above is the perfusion conductance - this is the amount of gas that is removed from the alveoli per kPa of partial pressure per minute

$(P_A-P_C) cdot dD$ is the amount of gas that diffuses across the alveolar membrane per kPa of partial pressure per minute

The authors then describe how this equation is 'integrated' for $P_C$ at a distance $x$ from the distal end of the capillary i.e. $P_{Cx}$. This integration yields the equation seen in figure B:

$$Equationquad 2: qquad {P_{Cx} – P_V over P_A – P_V} = 1-e{^-}^{Doverdot Q cdot beta}{^cdot}{^ {x over x_0}}$$

This is where my rudimentary maths lets me down. I don't understand how the authors get from equation 1 to equation 2. Perhaps, it's quite simple but could you explain how the integration is performed?

Thank you very much

M

There seems to be an assumption that $dD=Dcdot frac{dx}{x_0}$. Then the equation becomes $$ dot Q cdot beta cdot dP_C= (P_A-P_C) cdot Dcdot frac{dx}{x_0}$$ We can then move all the pressures to the left hand side
$$frac {dP_C}{P_A-P_C}=frac D{dot Qbeta} frac{dx}{x_0}$$
You need to integrate both sides. We can multiply both sides with $-1$. The limits of integration can be found from the figure. At $x=0$ you have $P_C(0)=P_V$. If you would have an infinite amount of $x$ you would get to $P_C(infty)=P_A$
Integrating left hand side between $P_V$ and $P_C(x)$ we get $$ln(P_C(x)-P_A)-ln(P_V-P_A)=lnfrac{P_C(x)-P_A}{P_V-P_A}$$
Similarly, integrating the right hand side from $0$ to $x$ we get $$-frac D{dot Qbeta} frac{x}{x_0}$$
we can now equate the two expressions, then use "If $a=b$ then $e^a=e^b$" to get $$frac{P_C(x)-P_A}{P_V-P_A}=e^{-frac D{dot Qbeta} frac{x}{x_0}}$$
Multiply both sides with $-1$ to change the denominator in the left hand side:
$$frac{P_C(x)-P_A}{P_A-P_V}=-e^{-frac D{dot Qbeta} frac{x}{x_0}}$$Then add and subtract $P_V$ in the numerator on the left
$$frac{P_C(x)-P_A}{P_A-P_V}=frac{P_C(x)-P_V+P_V-P_A}{P_A-P_V}=frac{P_C(x)-P_V}{P_A-P_V}-frac{P_A-P_V}{P_A-P_V}=frac{P_C(x)-P_V}{P_A-P_V}-1$$
Now moving the $-1$ from the left hand side to the right hand side yields your equation.