Is Exponential Form Better for Balloon with Inserted Load Calculation?

[Hak]

Which value of $T_0$ should I use? With $T_0 \approx 300 K$, I get two negative results. Therefore, this is not correct.

 

[erobz]

Which value of $T_0$ should I use? With $T_0 \approx 300 K$, I get two negative results. Therefore, this is not correct.

Maybe the relationship was with ^°C after all. So it is as you thought.

$$T = T_o [{}^{\circ} C] ( 1 - \beta h ) + 273 [K] $$

 

[Hak]

Maybe the relationship was with ^°C after all. So it is as you thought.

So? What should I do?

 

[Hak]

Maybe the relationship was with ^°C after all. So it is as you thought.

$$T = T_o [{}^{\circ} C] ( 1 - \beta h ) + 273 [K] $$

Which value of $T_0$ should I use? With $T_0 \approx 27^\circ C = 300 K$, I get two negative values, again. Perhaps $T_0$ is wrong...

 

[erobz]

Which value of $T_0$ should I use? With $T_0 \approx 27^\circ C = 300 K$, I get two negative values, again. Perhaps $T_0$ is wrong...

I would say:

$$ T_{out}[\text{K}] = 27[{}^{\circ}C] ( 1 - \beta h ) + 273[\text{K}] $$

$$T_{in} [ \text{K}] = 100[{}^{\circ}C] + 273[ \text{K}] $$

 

[Hak]

I would say:

$$ T_{out}[\text{K}] = 27[{}^{\circ}C] ( 1 - \beta h ) + 273[\text{K}] $$

$$T_{in} [ \text{K}] = 100[{}^{\circ}C] + 273[ \text{K}] $$

We have two negative results...

 

[erobz]

We have two negative results...

You mean for the height $h$?

 

[Hak]

You mean for the height $h$?

Yes.

 

[erobz]

Yes.

Does this equation rearrange to the first equation we derived?

 

[Hak]

Does this equation rearrange to the first equation we derived?

I don't think so, the results are different. Something is not right, the values are either too small or too large...

 

[erobz]

I don't think so, the results are different. Something is not right, the values are either too small or too large...

OK, thanks. So, is $\rho_{out} - \frac{m}{V} = \rho_{out} \frac{T_0}{T_{in}} (1-\beta h)$, with $\rho_{out} = \rho_0 (1 - \alpha h)$, correct?

I didn't process this $\uparrow$

The equation is.

$$ \frac{ \rho_{in}V + m }{V} = \rho_{out} $$

How did you get what you did above?

 

[Hak]

I didn't process this $\uparrow$

The equation is.

$$ \frac{ \rho_{in}V + m }{V} = \rho_{out} $$

How did you get what you did above?

I do not understand. Didn't we say we could match the average density $\rho_{avg}$ to the external density $\rho_{out}$?

 

[erobz]

I do not understand. Didn't we say we could match the average density $\rho_{avg}$ to the external density $\rho_{out}$?

Yeah, that it what I think is meant by the equation you quote? it is saying at the equilibrium position the average density of the entire balloon system will equal the density of the surrounding atmosphere at $h$.

 

[Hak]

Yeah, that it what I think is meant by the equation you quote? it is saying at the equilibrium the average density of the entire balloon system will equal the density of the surrounding atmosphere at $h$.

OK, then? I can't grasp the nettle...

 

[erobz]

OK, then? I can't grasp the nettle...

Show your work on that equation please, if it doesn't work out then so be it, but we need to see how you rearranged the equation.

 

[Hak]

All I can say about $\rho_{in}$ is: $$\frac{\rho_{in}}{\rho_{out}} = \frac{t_{out}}{t_{in}}$$. I cannot see the immediate dependence on $h$.

I did not understand. Would you like me to show how I equated this equation above with the last one you quoted?

 

[erobz]

I did not understand. Would you like me to show how I equated this equation above with the last one you quoted?

Start with the equation in post #71 and show the algebraic manipulations and substitutions you performed, because when I do those, I get the same result we got earlier.

 

[Hak]

Start with the equation in post #71 and show the algebraic manipulations and substitutions you performed, because when I do those, I get the same result we got earlier.

I get:

$$\rho_{in} + \frac{m}{V} = \rho_{out} \Rightarrow \ \rho_{in} + \frac{m}{V} = \rho_{0} (1 - \alpha h)$$.

From $\frac{\rho_{in}}{\rho_{out}} = \frac{T_{out}}{T_{in}}$, we get: $\rho_{in} = \rho_{out} \frac{T_{out}}{T_{in}}$. Plugging in the previous one:

$$\rho_{out} \frac{T_{out}}{T_{in}} - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$. Finally:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

 

[erobz]

I get:

$$\rho_{in} + \frac{m}{V} = \rho_{out} \Rightarrow \ \rho_{in} + \frac{m}{V} = \rho_{0} (1 - \alpha h)$$.

From $\frac{\rho_{in}}{\rho_{out}} = \frac{T_{out}}{T_{in}}$, we get: $\rho_{in} = \rho_{out} \frac{T_{out}}{T_{in}}$. Plugging in the previous one:

$$\rho_{out} \frac{T_{out}}{T_{in}} - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$. Finally:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

Which is the same result you initially found ( applying Newtons Second )?

 

[Hak]

Which is the same result you initially found ( applying Newtons Second )?

Yes. I have discordant values, however. Now the two values come out to me as $h \approx 32.6 \ m$ and $h \approx 11 \ km$. They are not convincing at all. Could you tell me what numerical result you get?

 

[erobz]

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

As you did before:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

$\gamma = \frac{T_{0}}{T_{in}}$

$$\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

Multiply everything by ( or divide by) $-1$ :

$$-\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) +\rho_{0} (1 - \alpha h) - \frac{m}{V} = 0$$
or (rearranging)
$$ \rho_{0} (1 - \alpha h) -\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \frac{m}{V} = 0 $$

Factor:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) ) - \frac{m}{V} = 0 $$

Multiply both sides by $V$:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) )V - m = 0 $$

That is the same relationship I got! Multiply everything by $g$ and we are back at the working that used Newtons Second Law directly.

 

[Hak]

As you did before:

$$\rho_{0} (1 - \alpha h)\frac{T_{0}}{T_{in}} (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

$\gamma = \frac{T_{0}}{T_{in}}$

$$\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \rho_{0} (1 - \alpha h) + \frac{m}{V} = 0$$.

Multiply everything by ( or divide by) $-1$ :

$$-\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) +\rho_{0} (1 - \alpha h) - \frac{m}{V} = 0$$.
or
$$ \rho_{0} (1 - \alpha h) -\rho_{0} (1 - \alpha h) \gamma (1 - \beta h) - \frac{m}{V} = 0 $$

Factor:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) ) - \frac{m}{V} = 0 $$

Multiply both sides by $V$:

$$\rho_{0} (1 - \alpha h) ( 1 - \gamma (1 - \beta h) )V - m = 0 $$

That is the same relationship I got! Multiply everything by $g$ and we are back at the working that used Newtons Second Law directly.

Yes, correct. The only question is whether the method is correct, but it seems to me that it is.

 

[erobz]

Yes, correct. The only question is whether the method is correct, but it seems to me that it is.

Where did all the stuff go about you were saying both signs are flopped ect... By deleting it you are making me look like a fool that is parroting everything you have already said for no reason...

Instead, simply say "I'm sorry, you are correct"

 

[Hak]

Where did all the stuff go about you were saying both signs are flopped ect... By deleting it you are making me look like a fool that is parroting everything you have already said for no reason...

I did not understand this statement. Sorry, I'm not a native English speaker...

 

[erobz]

I did not understand this statement. Sorry, I'm not a native English speaker...

You deleted all the stuff about disagreeing with the results being the same. Instead of just saying "yeah I goofed, sorry"

 

[Hak]

You deleted all the stuff about disagreeing with the results being the same. Instead of just saying "yeah I goofed, sorry"

Yes, because I realised after not even five seconds that I had made a mistake. Of course, I didn't delete it to avoid saying 'I was wrong, sorry', but only because I realised that your statement was right. I'm not a cocky guy, next time I won't delete, sorry...

 

[Hak]

May I ask what your final numerical result is?

 

[erobz]

Let's wait for someone else to verify. I'm currently baking a cake for my youngest daughters birthday, Then I have a wedding to attend.

 

[Hak]

Let's wait for someone else to verify. I'm currently baking a cake for my youngest daughters birthday, Then I have a wedding to attend.

Sorry, I couldn't have known. Here in Italy it's almost evening, I didn't notice the time difference. Happy birthday to your daughter, and have a wonderful evening at the wedding! Felicitations.

 

[Hak]

Do you all confirm this procedure and its numerical solution? Or is there some other more subtle demonstration? Thank you very much.