6
$\begingroup$

I am developing a model for the respiratory system of a neonate (newborn child) for a BME project by using pressure/volume data of the lungs. To model these, I plan to use a rubber balloon. However. the calculations are more difficult than I thought (especially because rubber behaves non-linearly and with a hysteresis effect, but I will ignore these effects).

I will assume the balloon is perfectly spherical, that it behaves linearly and that the strain is dependent on the radius relative to the original radius ($R_0$).

Here are the equations for the basis of my model:

$$ V = \frac{4}{3} \pi R^3_1 \\ R_1 = \left( \frac{3V}{4\pi} \right)^\frac{1}{3} \\ \sigma = \frac{pR_1}{2t} \\ \epsilon = \frac{R_1 - R_0}{R_0} \\ E = \frac{\sigma}{\epsilon} = \frac{pR_0R_1}{2t(R_1-R_0)} = \frac{pR_0}{2t\left(1-\dfrac{R_0}{\left(\frac{3V}{4\pi}\right)^\frac{1}{3}}\right)} \\ $$

My conclusion is: $$E \propto \frac{p}{V^\frac{1}{3}}$$

Is this correct?

|improve this question
$\endgroup$
  • $\begingroup$ How will the sphere deform? From a uniform pressure? Is this a very thin ballon? like the outer radius and very similar to the inner radius? $\endgroup$ – willpower2727 Aug 5 '15 at 0:36
  • $\begingroup$ The sphere will expand with pressure (of course), only steady state pressure-volume relationships are known. The balloon's wall is far smaller than the radius : R >>>> t. $\endgroup$ – Alexm Aug 5 '15 at 15:10
  • $\begingroup$ It would seem that you have done a linear strain, as if the balloon is expanding along the radial direction. But it isn't, it is expanding in a normal direction and the strain would be proportional to the change in circumference. Would the strain being in two dimensions affect the modulus too? $\endgroup$ – user5367 Mar 8 '16 at 10:10
1
$\begingroup$

You need to be careful with the subscripts, e.g in the equation for $\sigma$, which $R$ do you want? However, what you seem to be saying is: $$E \propto \frac{p}{R}$$ and, if you take $p$ and $R$ as increments from initial values, that seems reasonable. Getting the constant of proportionality might be problematic.

|improve this answer
$\endgroup$
  • $\begingroup$ The R in σ should be R1, sorry for that. R0 is the starting volume of the balloon and R1 the expanded volume. $\endgroup$ – Alexm Aug 5 '15 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.