I am planning to use an air compressor to inflate a rubber balloon. I need to size the compressor for inflating the balloon, but for sizing the compressor I need the pressure required to inflate the balloon. So, can anyone guide me to calculate the required pressure for inflating a rubber balloon?
[Physics] Balloon Inflation
everyday-lifepressure
Related Solutions
The complete stress tensor, while accurate, is largely unnecessary for solving this problem, as it is a thin walled pressure vessel
Assuming the balloon is spherical, the strain can just be calculated from the current and initial radii.
$$\epsilon=\frac{r}{r_0}-1$$
The stress can be found using the modulus of elasticity:
$$\sigma=E\,\epsilon$$
The thin wall pressure equation can get you to pressure, if you know the thickness, by balancing outward pressure inside with the inward tension along a great circle of the sphere:
$$\pi\,r^2\,P=2\,\pi\,r\,\sigma\,t$$ $$P=\frac{2\,\sigma\,t}r$$
Because balloons get thinner as they stretch, the thickness will actually vary. Rubber typically has a poisson's ratio of 0.5 meaning it keeps a constant volume while being deformed. We can then calculate the thickness in terms of the radius: $$t\,r^2=t_0\,{r_0}^2$$ $$t=t_0\,\left(\frac{r_0}{r}\right)^2$$
Putting them all together:
$$P=\frac{2\,E\,\left(r-r_0\right)\,t_0\,r_0}{r^3}$$
To see what this looks like, we can make a generic plot:
As you can see, there is a maximum pressure after which it becomes easier and easier to inflate the balloon. We can solve for this maximum pressure by equating the derivative with zero, solving for r, and plugging back in:
$$0=\frac{dP}{dr}=2\,E\,t_0\,r_0\left(\frac1{r^3}-3\frac{r-r_0}{r^4}\right)$$
$$r=\frac32\,r_0$$
$$P_{max}=\frac{8\,E\,t_0}{27\,r_0}$$
Of course this assumes a constant modulus of elasticity, which never holds true for a large enough deformation.
- Yes.
- You need to solve two problems.
First, given the natural radius of the spherical membrane $R_0$ (the radius with no tension in the membrane) and the current membrane radius $R$, membrane's modulus of elasticity $E$ and Poisson's ratio $\mu$, calculate tension stress in the membrane. If you consider an infinitesimally small square (with the side of $\delta l_0$) of the spherical membrane under no tension with thickness $d_0$, currently it will be a square with a side of $\delta l=\delta l_0\frac{R}{R_0}$. The tension stress $\sigma$ will expand the square in two directions. Calculation of tension stress from strain (using elasticity modulus and Poisson's ratio, and assuming the membrane material is isotropic) is a standard task, see e.g. https://en.wikipedia.org/wiki/Poisson%27s_ratio , although you may wish to find a better source. The exact value of the Poisson's ratio is not very important.
Second, given tension stress, calculate differential pressure in equilibrium. To do this, consider the condition of equilibrium for a half of the spherical membrane: $(p_i-p_o)\pi R^2=2\pi R d\sigma$ (assuming the membrane is thin: $d_0<<R_0$).
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Best Answer
A typical human can exert an over-pressure of ~$9.8kPa$ with their lungs. Given that balloons are designed so that a human can inflate them, I'd say go for a pump that can deliver $10kPa$ of pressure (that's around $1.5psi$ in case you were wondering).