Comparing weights of balloons

[Orodruin]

which way the scale reading will tip is not unpredictable but depends on the specific parameters of the situation

I think ”unpredictable” or not depends on what you are given. In the OP’s case they are not given the elastic properties of condom rubber, which makes the way of tipping unpredictable given the information. Of course, the actual setup in the OP (as well as the setup weighing the full contraption) is predictable.

 

[Orodruin]

I think that a cylindrical condom would be a better fit to your model.

Also, speak for yourself!

I am a theoretical physicist - everything is a sphere!

 

[erobz]

@kuruman I was just trying to follow along with this notion proposed by @Charles Link that the buoyant force on the balloon might not be decreasing at all stages between the initial state and vacuum in the bell jar due to the complexity of the balloon expansion.

 

[kuruman]

@kuruman I was just trying to follow along with this notion proposed by @Charles Link that the buoyant force on the balloon might not be decreasing at all stages between the initial state and vacuum in the bell jar due to the complexity of the balloon expansion.

I understand. Thank you for clarifying this point.

 

[jbriggs444]

The point is that, without more information on the condom, you cannot determine $\rho_{\rm in}$. The condom volume - and therefore the density of the air inside - will depend on the elastic properties of the condom. The important thing to know if you want to know if the reading increases or decreases as you suck the air out is not knowing the density ratio - it is knowing how the density ratio changes as you change $\rho_{\rm out}$ and to know that you need to do some modelling.

The situation could be even worse. For some elastic properties, the equilibrium volume of the balloon might be multi-valued (more than one equilibrium state for a given pressure). Or there might be no stable equilibrium at all.

The volume of the balloon might not be a function of pressure alone. It might be a function of pressure history. This is not as esoteric a possibility as one might think. I claim that an ordinary rubber balloon is multi-stable in this sense. This might be usable with a pneumatic digital computer.

 

[Charles Link]

@kuruman I was just trying to follow along with this notion proposed by @Charles Link that the buoyant force on the balloon might not be decreasing at all stages between the initial state and vacuum in the bell jar due to the complexity of the balloon expansion.

@Orodruin explains this very well with his computations in post 59. You might not have seen this post, because you didn't give it a "like" yet.

 

[Charles Link]

@erobz Reading it more carefully, I think @Orodruin has a typo in post 59, and it should read $x=\rho_{out} / \rho_{in}$, (instead of $p$'s.) He's referring to @kuruman 's calculations of post 57. See also his post 61.

Edit: I need to look at it more closely though...If the gases are the same, and the temperature are the same, then the $p$ is proportional to $\rho$.... But I think in general he's trying to show it for the $\rho$ ratio.

It should also read $\sigma \, \alpha \, p_{out} / R^3$, instead of $x$. ? I need to study this further. Right now I'm just guessing...

 

[Orodruin]

@erobz Reading it more carefully, I think @Orodruin has a typo in post 59, and it should read $x=\rho_{out} / \rho_{in}$, (instead of $p$'s.) He's referring to @kuruman 's calculations of post 57. See also his post 61.

Not really, I used the pressure for finding the equilibrium of the radius. However, the ideal gas law gets you $p \propto \rho$ so the pressure ratio is equal to the density ratio. I am assuming the gas inside is the same as the outside.

 

[Orodruin]

It should also read $\sigma \, \alpha \, R^3$, instead of $x$.

It should not. That case is for $\sigma$ constant (which is kind of unphysical). In essence I base it on finding the minimum in the total potential energy from pressure and tension, which becomes $\sigma(R) = \Delta p R = R(p_{\rm in} - p_{\rm out})$. Solving for $x$ gives you the expression in #61.

If you assume constant $\sigma$ and use that $p_{\rm in} R^3$ is also constant and therefore $p_{\rm in} = \kappa/R^3$, then$$
\sigma = \frac{\kappa}{R^2} - p_{\rm out} R
$$ and therefore $$
x = \frac{p_{\rm out}R}{p_{\rm out} R + \sigma} = \frac{p_{\rm out}R}{\kappa/R^2} = \frac{p_{\rm out} R^3}{\kappa}
$$
It now strikes me that I missed the explicit $p_{\rm out}$ dependence in the numerator, which is of course also important. To determine the actual behaviour would require some additional analysis of a third degree polynomial to get $R$ as a function of $p_{\rm out}$, which I am not overly keen to do at the moment.

 

[Charles Link]

@Orodruin I also have tried to model the balloon mathematically, and it seems to be rather difficult. What is apparent though, that if you decrease $p_{out}$ a little, and the balloon suddenly expands by a factor of ten or more in volume, that you will see an increase in the buoyant force.

 

[erobz]

@Orodruin I also have tried to model the balloon mathematically, and it seems to be rather difficult. What is apparent though, that if you decrease $p_{out}$ a little, and the balloon suddenly expands by a factor of ten or more in volume, that you will see an increase in the buoyant force.

So, starting from a balloon the size of a baseball, you might need a pretty big bell jar to observe it. Also, what pressure change would this be associated with?

 

[Charles Link]

One other comment though is lowering the outside pressure is not the same thing as blowing up a balloon, where you increase the internal mass considerably. I still need to do further calculations on this.

I do expect there could be a case though where you lower $p_{out}$ to start to cross a threshold, but its very difficult to model mathematically.

 

[Orodruin]

@Orodruin I also have tried to model the balloon mathematically, and it seems to be rather difficult. What is apparent though, that if you decrease $p_{out}$ a little, and the balloon suddenly expands by a factor of ten or more in volume, that you will see an increase in the buoyant force.

If you expand the volume by a factor of 10 then the pressure inside the balloon decreases by a factor of 10. This seems like a very very weird balloon.

 

[Orodruin]

but its very difficult to model mathematically.

Which part? The change in potential energy is a fairly simple mathematical model. The strain-stress relationship of rubber outside the purely elastic regime is something else.

 

[erobz]

I'm going to try something "simple" where the pressure differential between the inside of the ballon and outside is some fixed constant ( $P_b-P_j = P'_b - P'_j = C$) It obviously not going to work to vacuum, but is a really bad start? The balloon is to be undergoing adiabatic expansion, and the bell jar air is experiencing a mass change at constant temperature. Does it have a chance, or am I dead in the water?

The buoyant force will be $F_b = \rho'_j V'_b$. It seems like it already could be a bit messy.

If @Orodruin model is already producing this (more or less), I'm not clever enough to tell.

Maybe we should migrate this to a new thread?

 

[erobz]

So I get these starting equations:

For the Jar losing mass at $T$

$P_jV_j = m_jRT$

$\implies \frac{P_jV_j}{P'_jV'_j} = \frac{m_j}{m'_j}$

$\implies P'_j = \frac{P_jV_j}{V_j - (V'_b-V_b) } \frac{m'_j}{m_j} \tag{1}$

So the Pressure in the jar $P'_j$ becomes a function of the mass in the jar $m'_j$, which we can write as some fraction of the initial mass $m_j$, and function of balloon volume $V'_b$.

Then for the balloon we have the adibatic process:

$P_b {V_b}^{k }= P'_b {V'_b}^{k}$ where $k = 1.4$

Then sub in:

$P'_b - P'_j = C$

$\implies P_b {V_b}^{k} = ( C+ P'_j){V'_b}^{k} \tag{2}$

Then you have to solve those (1 and 2) simultaneously.

Thats as far as I feel like going tonight. If I wake up to find it doesn't have legs... so be it.

 

[Orodruin]

If @Orodruin model is already producing this

It won’t because I am not assuming the pressure difference to be constant. Instead, I am basing the argument on when the pressure forces and tension in the balloon are in equilibrium.

 

[Charles Link]

I think in general you might find the buoyant force to decrease on the balloon inside the bell jar as you suck the air out, contrary to what I said in post 44, and also contrary to what @Orodruin may have said in post 59. I had made the error of thinking the balloon has a threshold when you blow one up, but neglected to consider that you are also putting a lot more air into the balloon when you blow one up, as I mentioned in post 72.

From what @Orodruin has in post 69, I think it is possible to see the buoyant force increase for part of the cycle when the air is pumped out, but I think it to be somewhat unlikely.

 

[erobz]

So for $V'_b$ an analytical result looks highly improbable...

$$ \frac{P_b {V_b}^{k} - C {V'_b}^{k}}{{V'_b}^{k}} = P'_j = \frac{P_jV_j}{V_j - (V'_b-V_b) } f $$

Where $f$ is the mass fraction ($m'_j = f m_j$ )

Trying to solve for $V'_b$

$$ C {V'_b}^{k+1} - \left( C( V_j+V_b)+ P_jV_j f \right) {V'_b}^{k} - P_bV_b^{k} V'_b + P_bV_b^{k}( V_j+V_b ) = 0 \tag{3} $$

Then in principle we solve for $V'_b$ as a function of $f$.

From there we employ the Ideal Gas law to get $\rho'_j ( f ) = \frac{P'_j}{RT} = \frac{P_jV_j f }{RT\left( V_j+V_b-V'_b \right)}$

And again $F_b(f) = g\rho'_j(f)V'_b(f)$

So does anyone see some kind of analysis/method that cuts out obtaining the solution to (3) i.e. $V'_b ( f )$ or are am I stuck unable to determine how this function would behave as $f$ declines?

What happens if I say $k=2$, its like a general cubic. Can I say $k= 1$ making it a quadratic? Should/can I check terms in (3) to see if some may be negligible in comparison to others...Are there any cheats were the general function behavior wouldn't change significantly.

 

[Charles Link]

The adiabatic case will always have a lower buoyant force (higher density inside the balloon because the balloon cools with expansion) than the isothermal case, so if you are trying to show the buoyant force decreases when you let the air out, you could solve the isothermal case where $k=1$.

 

[erobz]

Ok, $F_b$ appears (not expansively tested for range of parameters) to be monotonically decreasing as air is evacuated from the bell jar ( balloon expands) under the assumptions of this model (post 79).

Inital parameters:

Functions:

Graphical representation: (I wasn't sure how to reverse the scale in the program)

I'm sure it's a gross simplification based on what has already been discussed (so I'm not sure if I've gained too much), but I couldn't leave it be.