Is Exponential Form Better for Balloon with Inserted Load Calculation?

[Hak]

I don't know, but it could be that the linear function approximates the power law reasonably well over these altitudes, or they simply chose a function for $\rho$ which makes the problem solvable algebraically. You have to plot the functions to see why they might have done what they have done.

I will try to trace them. However, I am noticing that in the expression of the problem a term $\alpha$ pops up, whereas here there is only one term in $\beta$. What does this mean?

 

[Hak]

The only real difference between the two functions seems to be that the function given by the text has an unlimited domain, whereas the one we have just calculated has an upper limit, so it has a limited domain. There does not seem to be any substantial difference.

 

[erobz]

I will try to trace them. However, I am noticing that in the expression of the problem a term $\alpha$ pops up, whereas here there is only one term in $\beta$. What does this mean?

$\alpha$ either characterizes the linear function they made up for $\rho ~\text{vs.}~ h$ or the linear function they observe for $\rho~ \text{vs.}~ h$, just as $\beta$ characterizes the approximately linear function they observe for $T$ vs. $h$ in the troposphere. You have to do some calculation to see what is what. I can't tell you anymore, because I'm not doing all that.

 

[Hak]

The only real difference between the two functions seems to be that the function given by the text has an unlimited domain, whereas the one we have just calculated has an upper limit, so it has a limited domain. There does not seem to be any substantial difference.

What do you think?

 

[erobz]

What do you think?

Then I think you have solved the mystery. The linear function approximates the power law ( strict ideal gas law treatment ) very well over the domain in question, and (this is the important part) ...it also makes the problem algebraically solvable. Or it too is an empirical result ( i.e. the atmosphere isn't really quite ideal ) like the temperature function, and it makes the problem solvable. It doesn't matter either way.

 

[Hak]

Then I think you have solved the mystery. The linear function approximates the power law ( strict ideal gas law treatment ) very well over the domain in question, and (this is the important part) ...it also makes the problem algebraically solvable. Or it too is an empirical result like the temperature function, and it makes the problem solvable. It doesn't matter either way.

OK, thank you, @erobz. You gave me a huge amount of help.

 

[hutchphd]

This is not rocket science.....it is balloon science. As the balloon ascends into cooler air that effect will increase the lift of the 100 deg (differential??) T this is not stated clearly at all. The air pressure is of course exponential with z so this prblem is sort of silly. Assume a uniform temperature and the simple exponential atmosphere and try for a good solution. This is a simple exercise, unless one chooses to make it otherwise. Yikes.

 

[Hak]

This is not rocket science.....it is balloon science. As the balloon ascends into cooler air that effect will increase the lift of the 100 deg (differential??) T this is not stated clearly at all.

I struggle to understand this statement. Could you explain it again? Thank you very much.

 

[Hak]

The air pressure is of course exponential with z so this prblem is sort of silly. Assume a uniform temperature and the simple exponential atmosphere and try for a good solution. This is a simple exercise, unless one chooses to make it otherwise. Yikes.

Excuse me, but I cannot understand what the crux of what you want to say is. Could you explain that too? Thank you again.

 

[hutchphd]

$$\rho(z,T)=\rho(z=0,T)e^{-\alpha z}$$and from the Ideal gas equation $$F_{lift}=\rho gV\frac{\Delta T} T$$the buoyancy is the mass of the air expelled because the ballon is heated. Also as mentioned N.B. $$\alpha\approx1/(20km)$$
REVISED

 

[Hak]

$$\rho(z,T)=\rho(z=0,T)e^{-\alpha z}$$and from the Ideal gas equation $$F_{lift}=\rho gV\frac{\Delta T} T$$the buoyancy is the mass of the air expelled because the ballon is heated. Also as mentioned N.B. $$\alpha\approx1/(20km)$$

So, what's the problem with this? I don't understand. Could you explain some more? Thanks.

 

[hutchphd]

Plug in the numbers. This will lift nearly an extra 300kg at sea level by my reckoning

 

[Hak]

Plug in the numbers. This will lift nearly an extra 300kg at sea level by my reckoning

I can't figure out which numbers I should plug in...

 

[Hak]

I can't figure out which numbers I should plug in...

In fact, I didn't even understand which method @hutchphd used. Is it an alternative method to the one we did previously? Or is it an a posteriori consideration? I don't really understand what is being demonstrated, nor what numerical values need to be considered. Since I think this theory could be very interesting, I would be very grateful if you could help me clarify it. Thank you very much.

 

[erobz]

I really wouldn't worry about it. The hornets' nest has been stirred for no good reason. We have solved the problem with reasonable theory. The linear function in the problem statement is most likely just an approximation to make the solution analytical...the problem setter wants the student to get an algebraic result...

Here are all three proposed density functions with each other:

If you are going to use a nonlinear function to get a numerical solution, just use the physical model surrounding the problem...

 

[Hak]

I really wouldn't worry about it. The hornets' nest has been stirred for no good reason. We have solved the problem with reasonable theory. The linear function in the problem statement is most likely just an approximation to make the solution analytical...the problem setter wants the student to get an algebraic result...

Here are all three proposed density functions with each other:

View attachment 332901
View attachment 332903

If you are going to use a nonlinear function to get a numerical solution, just use the physical model surrounding the problem...

Thank you very much. I understand what you mean, you have explained it many times and all very well. The crux of the matter is what you have reiterated in this message. Although it is not so important (as you rightly say) for the resolution of the problem itself, I would like to understand how @hutchphd set up his theory and arrived at these results. Based on what? I cannot understand where his equations came from and what numerical results should be included. Were you able to understand this in detail? If not, let's wait for @hutchphd's answer, nothing more can be done if neither you nor anyone else has understood it...

 

[erobz]

Thank you very much. I understand what you mean, you have explained it many times and all very well. The crux of the matter is what you have reiterated in this message. Although it is not so important (as you rightly say) for the resolution of the problem itself, I would like to understand how @hutchphd set up his theory and arrived at these results. Based on what? I cannot understand where his equations came from and what numerical results should be included. Were you able to understand this in detail? If not, let's wait for @hutchphd's answer, nothing more can be done if neither you nor anyone else has understood it...

Best of luck @Hak.

 

[Hak]

Best of luck @Hak.

? You couldn't tell me what @hutchphd did?

 

[votingmachine]

At equilibrium the weight of the gas inside the 850 m^3 will be 250 kg less than the weight of 850 m^3 of the outside gas.

As you point out there is virtually no change in temperature or density of the outside air with the the altitude change you solved for.

The mass of the internal 850 m^3 of air will be constant. You need to calculate where the altitude changes to temperature and pressure for 850 m^3 are 250 kg less. It seems that if the pressure and temperature are dependent on the height, you can plug those equations into the ideal gas law and solve for height.

I think of it in terms of the volume, and the molar mass of air. I would be trying to calculate a volume of air equal to 250 kg mass, knowing the molar mass of air. If the molar mass of air at STP is 29 g/mol and occupies 22.4 l, then calculate where the pressure and temperature (both dependent on height) move the mass of 850 m^3 to 250 kg less than the mass of the 850 m^3 at 373 K (and air also).

Maybe that is a wrong direction, but try it and see.

 

[Hak]

At equilibrium the weight of the gas inside the 850 m^3 will be 250 kg less than the weight of 850 m^3 of the outside gas.

As you point out there is virtually no change in temperature or density of the outside air with the the altitude change you solved for.

The mass of the internal 850 m^3 of air will be constant. You need to calculate where the altitude changes to temperature and pressure for 850 m^3 are 250 kg less. It seems that if the pressure and temperature are dependent on the height, you can plug those equations into the ideal gas law and solve for height.

I think of it in terms of the volume, and the molar mass of air. I would be trying to calculate a volume of air equal to 250 kg mass, knowing the molar mass of air. If the molar mass of air at STP is 29 g/mol and occupies 22.4 l, then calculate where the pressure and temperature (both dependent on height) move the mass of 850 m^3 to 250 kg less than the mass of the 850 m^3 at 373 K (and air also).

Maybe that is a wrong direction, but try it and see.

Would this be the way of @hutchphd? Thanks anyway.

 

[votingmachine]

I have not looked at every reply.

I am proposing that in the ideal gas law, you can treat V as a constant ... the 850 m^3. An you know the pressure inside equals the pressure outside.

So the ideal gas law with P and V constant reduces to
constant x n1 x T1 = constant x n2 x T2

The delta-n is from delta-250 kg. At 29 g/mol, that is delta-mol of 8621.

Now you have:
constant x n1 x 373 = constant x (n1-8621) x T2
T2 is a function of height
rearranging:
n1 / (n1-8621) = f1(h)/373

moles is density x V / 29 g/mol with V = 850 m^3
density is a function of height.
n = f2(h) x 850 m^3 / 29 g/mol

I suspect the error is omission of the load. First, you mention the load (which I now see is 200kg ... won't change my above mistake), but never include it in your calculations. You calculate the F=mg, but never include the change in mass of 200 kg (that I can see).

I think you calculate a trivial difference from zero. My approach STARTS with the load. I think you have correctly re-arranged formulas, but never included the change necessary for the load. Your result is close to zero.
h ~ 0.029 .... ~0

 

[Hak]

I have not looked at every reply.

I am proposing that in the ideal gas law, you can treat V as a constant ... the 850 m^3. An you know the pressure inside equals the pressure outside.

So the ideal gas law with P and V constant reduces to
constant x n1 x T1 = constant x n2 x T2

The delta-n is from delta-250 kg. At 29 g/mol, that is delta-mol of 8621.

Now you have:
constant x n1 x 373 = constant x (n1-8621) x T2
T2 is a function of height
rearranging:
n1 / (n1-8621) = f1(h)/373

moles is density x V / 29 g/mol with V = 850 m^3
density is a function of height.
n = f2(h) x 850 m^3 / 29 g/mol

I suspect the error is omission of the load. First, you mention the load (which I now see is 200kg ... won't change my above mistake), but never include it in your calculations. You calculate the F=mg, but never include the change in mass of 200 kg (that I can see).

I think you calculate a trivial difference from zero. My approach STARTS with the load. I think you have correctly re-arranged formulas, but never included the change necessary for the load. Your result is close to zero.
h ~ 0.029 .... ~0

My and @erobz's solution seems correct to me. Your theory is interesting, but I can't tell if it is correct: if it were, it would upset everything. Let's wait for more responses on this, perhaps from those who have not addressed this issue, such as @kuruman, @haruspex, and all the others who are wont to participate in these threads...

 

[votingmachine]

My and @erobz's solution seems correct to me. Your theory is interesting, but I can't tell if it is correct: if it were, it would upset everything. Let's wait for more responses on this, perhaps from those who have not addressed this issue, such as @kuruman, @haruspex, and all the others who are wont to participate in these threads...

I should have read the 5 pages, and I still have not.

But look at your original post and carefully try to determine where the 200 kg load is in your final reduction. The force = 200 kg x g

You have it correct that the buoyant force reduces to the difference in densities. But the diffence in densities calculates to an absolute mass difference defined by the load.

Where you have a long equation right after:
From here, I obtain a second-degree equation in h:
you set it equal to zero. Why not the number from the load ... 200 kg x g ?

I think you are just calculating how the assorted errors in constants varies from zero.

 

[Hak]

I should have read the 5 pages, and I still have not.

But look at your original post and carefully try to determine where the 200 kg load is in your final reduction. The force = 200 kg x g

You have it correct that the buoyant force reduces to the difference in densities. But the diffence in densities calculates to an absolute mass difference defined by the load.

Where you have a long equation right after:
From here, I obtain a second-degree equation in h:
you set it equal to zero. Why not the number from the load ... 200 kg x g ?

I think you are just calculating how the assorted errors in constants varies from zero.

I don't know, I'm really confused. Let's wait for a reply.

 

[votingmachine]

What are the comment numbers that you are referring to? I could look at those.

An idle thought that just occurred to me was that you make the statement that the molar mass of air doesn't matter. But consider if the atmosphere was hydrogen. At STP, 850 m^3 of hydrogen would only weigh about 76 kg. The most sea level H2 could lift would be a pure vacuum, and then it can only displace 2 kg/22.4 m^3.

You need the 850 m^3 of hot gas to be 200 kg lighter than 850 m^3 of the surrounding gas. The composition of the gas is critical. You have fewer atoms of gas in the hot gas than the non-hot gas. The mass of each atom is what makes the gas as a whole have mass.

I may be completely wrong. I should put pen to paper and make sure I'm not just miss-reading your initial analysis.

 

[erobz]

What are the comment numbers that you are referring to? I could look at those.

An idle thought that just occurred to me was that you make the statement that the molar mass of air doesn't matter. But consider if the atmosphere was hydrogen. At STP, 850 m^3 of hydrogen would only weigh about 76 kg. The most sea level H2 could lift would be a pure vacuum, and then it can only displace 2 kg/22.4 m^3.

You need the 850 m^3 of hot gas to be 200 kg lighter than 850 m^3 of the surrounding gas. The composition of the gas is critical. You have fewer atoms of gas in the hot gas than the non-hot gas. The mass of each atom is what makes the gas as a whole have mass.

I may be completely wrong. I should put pen to paper and make sure I'm not just miss-reading your initial analysis.

You are miss reading the initial analysis. The issues were with unit transformation in $\alpha$ and $\beta$ (please read through the thread). The rest is Newtons Second at equilibrium.

$$ F_b(h) - W_{in}(h) - W_{load} = 0 $$

$$ \rho_{out}(h) V g - \rho_{in}(h) V g - m_{load}g = 0 $$

The problem has been solved with multiple approaches, and the supporting atmospheric models were examined to satisfy the OP's curiosity.

The only thing that hasn't been done ( a.f.a.i.k.) is numerical verification of the OP's result in the quadratic.

What the OP wants is for someone to explain the simple solution proposal that reignites the discussion with post #147 on this page.

 

[votingmachine]

My method comes up with 11 km.

I used a spreadsheet to calculate the densities at every km using your formula. Then to calculate the moles based on:
density * 850 m^3 / 29 kg/kmol

That is the outside moles in 850 m^3 for that density (at that height)

200 kg / 29 kg/kmol = 6896 mol.
So the internal mol is the outside mol (from the density calculation) - 6896

I calculated the temperature at every km using your formula.

Since pressure inside =pressure outside, and the volume is the same 850 m^3, the ideal gas law is just concerned with the number of moles, times the temperature. If I multiply I have the columns with the moles at every altitude. I have the column with the outside temperature. I know the inside is 373 K.

When the outside temperature at an altitude times the number of moles at that temperature is equal to the inside temperature times the same number of moles minus 6696.

AT 11 km, the air density is 0.5532 and the temperature is 214.2 K. That density is 16214 moles of air in a 850 m^3 volume. Inside is then 9317 moles of hot air at 373.
373 x 9317 is approximately equal to 214.2 x 16214

They are not exactly equal. You could easily take the approximation closer. I used 1 km increments in my spreadsheet.

This agrees with the answer I saw on another page. It was far easier to use a computer than to solve the very complex looking equation for height. A better math whiz might look at it and see the answer.

But ultimately, I think the agreement shows you could just look at the balloon volume net mass, and know that it has to be less by the same amount as the load.

 

[erobz]

But ultimately, I think the agreement shows you could just look at the balloon volume net mass, and know that it has to be less by the same amount as the load.

Like the equation above I wrote above ( dividing out $g$)?

$$ \overbrace{\rho_{out}(h) V - \rho_{in}(h) V }^{\text{ balloon volume net mass} } - \overbrace{m_{load}}^{ \text{less the load}} = 0 $$

or rearranging it like @kuruman suggests to find $\rho_{avg} = \rho_{out}$

These are all equivalent methods.

 

[votingmachine]

Like the equation above I wrote above ( dividing out $g$)?

$$ \overbrace{\rho_{out}(h) V - \rho_{in}(h) V }^{\text{ balloon volume net mass} } - \overbrace{m_{load}}^{ \text{less the load}} = 0 $$

or rearranging it like @kuruman suggests to find $\rho_{avg} = \rho_{out}$

These are all equivalent methods.

I skimmed after page 1, which was where I thought my comment was a useful suggestion. I see now your comment #45, which is VERY close to how I chose to approach it in a spreadsheet. When I wrote the equation all in terms of height, it looked troublesome to solve. Just tossing the formulas into a spreadsheet seemed easier. That equivalence formula could be (as you point out) done with many equivalent variations

I now see several comments that shed light on the OP question that is the same light I was trying to get to.

I still don't know that I would bother with the hairy math I see in some of the comments. A spreadsheet is the way to go. Or if I didn't have a spreadsheet program, a calculator and successive approximations would be quick. That could be what the "it's not rocket science, it's balloon science" comment was getting at. Once I saw the equations looked difficult to solve, I just used the spreadsheet.

 

[erobz]

I skimmed after page 1, which was where I thought my comment was a useful suggestion. I see now your comment #45, which is VERY close to how I chose to approach it in a spreadsheet. When I wrote the equation all in terms of height, it looked troublesome to solve. Just tossing the formulas into a spreadsheet seemed easier. That equivalence formula could be (as you point out) done with many equivalent variations

I now see several comments that shed light on the OP question that is the same light I was trying to get to.

I still don't know that I would bother with the hairy math I see in some of the comments. A spreadsheet is the way to go. Or if I didn't have a spreadsheet program, a calculator and successive approximations would be quick. That could be what the "it's not rocket science, it's balloon science" comment was getting at. Once I saw the equations looked difficult to solve, I just used the spreadsheet.

The equation is quadratic, it’s solvable by the quadratic formula. It’s really not a big deal. Just expand it out, collect like terms and turn the crank.

I think you are mixing up the solution to the problem with the math we used to look at the supporting atmospheric models in greater detail.

Anyhow, each to their own.

 

[hutchphd]

I think you are mixing up the solution to the problem with the math we used to look at the supporting atmospheric models in greater detail.

My intuition is that the exponential forms will yield a better and more transparent solution. To first order they should match. So far I am not doing well with it, but have not given it up.