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)
- $\dot Q\beta$ 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: $$Equation\quad 1: \qquad \dot Q \cdot \beta \cdot dP_C= (P_A-P_C) \cdot dD$$
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:
$$Equation\quad 2: \qquad {P_{Cx} – P_V \over P_A – P_V} = 1-e{^-}^{D\over\dot Q \cdot \beta}{^\cdot}{^ {x \over x_0}}$$
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 $dD=D\cdot \frac{dx}{x_0}$. Then the equation becomes $$ \dot Q \cdot \beta \cdot dP_C= (P_A-P_C) \cdot D\cdot \frac{dx}{x_0}$$ We can then move all the pressures to the left hand side $$\frac {dP_C}{P_A-P_C}=\frac D{\dot Q\beta} \frac{dx}{x_0}$$ 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 $$\ln(P_C(x)-P_A)-\ln(P_V-P_A)=\ln\frac{P_C(x)-P_A}{P_V-P_A}$$ Similarly, integrating the right hand side from $0$ to $x$ we get $$-\frac D{\dot Q\beta} \frac{x}{x_0}$$ we can now equate the two expressions, then use "If $a=b$ then $e^a=e^b$" to get $$\frac{P_C(x)-P_A}{P_V-P_A}=e^{-\frac D{\dot Q\beta} \frac{x}{x_0}}$$ Multiply both sides with $-1$ to change the denominator in the left hand side: $$\frac{P_C(x)-P_A}{P_A-P_V}=-e^{-\frac D{\dot Q\beta} \frac{x}{x_0}}$$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.