Wondering about how to deal with a deflating balloon?

[jeebs]

Hi,
Not sure if this is asked in the right section, but I am doing something which involves moving a small car, and I am trying to make predictions from an air resistance perspective. I'm trying to come up with a rough expression for the drag force on a balloon as it deflates. I've started with the general expression for the drag force $$F_d = 1/2\rho AC_dv^2$$.
To simplify things I'm assuming a spherical balloon, which has a radius r_max when it is fully inflated and a radius r_min when it is fully deflated, so it goes from having surface area A = pi(r_max)² to pi(r_min)². I realize the surface area of a sphere is 4pi.r² but I need the area of the object presented to the oncoming air. I also realize that the velocity v is going to increase from zero to some maximum, but I'm ignoring that for now (perhaps unjustifiably, lol).

So assuming everything else (velocity v, air density rho and drag coefficient C_d) is kept constant, the balloon radius r determines the drag force F_d.

However, this is where I start to have problems, because I am not expecting r to decrease at a constant rate - my thinking is that, from the general spring force equation F=-kx, a balloon stretched (inflated) to the maximum is going to have a greater force trying to pull its molecules back to their original positions than if it is partially inflated. This means the rate at which air is forced out (ie. pressure decreases) falls with time.

I need to find out how r varies with time but all I can think of really is that its going to depend on the "stretchiness" of the balloon, ie. something analogous to the spring constant, k, and also the size of the hole that the air is escaping from. I'm not really sure where I should go from here, so if anyone has any ideas they think might be relevant I would love to hear them.

Thankyou.

 

[Baluncore]

However, this is where I start to have problems, because I am not expecting r to decrease at a constant rate - my thinking is that, from the general spring force equation F=-kx, a balloon stretched (inflated) to the maximum is going to have a greater force trying to pull its molecules back to their original positions than if it is partially inflated. This means the rate at which air is forced out (ie. pressure decreases) falls with time.

I am not surprised you are confused. It is counter intuitive, the internal pressure rises as the balloon deflates.

The pressure difference across the balloon surface is determined by the tension in the surface and the radius of curvature of the surface, P = Ts / Rc. For that reason the pressure required to inflate a balloon is inversely proportional to the diameter of the balloon. As a balloon deflates the internal air pressure increases until tension in the skin is lost. Energy is not being stored so much in the compression of the air, as it is in the tension of the surface.

That explains why the airflow through the tube from the deflating balloon is slow at first, then increases to a maximum just before tension is lost. Since the balloon volume is the cube of the radius, the radius of a deflating balloon reduces slowly at first, then becomes very rapid towards the end.

For the purpose of propulsion, the higher velocity jet at the end of deflation is an advantage, but the area of the balloon, as far as the drag computation is concerned, will decrease approximately linearly with time.