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erobz
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## \alpha = 0.026 \frac{1}{km} = 0.026 \frac{1}{km}\frac{1 km}{1000 m} = 0.000026 \frac{1}{m} ##Hak said:Why?
etc...
## \alpha = 0.026 \frac{1}{km} = 0.026 \frac{1}{km}\frac{1 km}{1000 m} = 0.000026 \frac{1}{m} ##Hak said:Why?
Damn! It's true!erobz said:## \alpha = 0.026 \frac{1}{km} = 0.026 \frac{1}{km}\frac{1 km}{1000 m} = 0.000026 \frac{1}{m} ##
etc...
So, is my process incorrect?kuruman said:The balloon will stop rising when Its average density is equal to the density of the outside air. The average density is the total mass (air inside + envelope + load) divide by the external volume. I think we can safely ignore the buoyant force on the load. Find a relation between the density of the air inside and the density of the air outside when the pressure inside is equal to the pressure outside.
I didn't say that. I told you what I would do to make the solution transparent. Put in the numbers and see what you get.Hak said:So, is my process incorrect?
I get, with my procedure, ##h \approx 784957 km## and ##h \approx 1.77 \times 10^{-3} m##. Where am I going wrong? Have you tried entering the numbers? If so, what do you get?kuruman said:I didn't say that. I told you what I would do to make the solution transparent. Put in the numbers and see what you get.
I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.erobz said:Its seems like what you did is correct.
I too would say:$$ \rho_o ( 1 - \alpha h ) \left( 1 - \gamma ( 1 - \beta h ) \right) - mg = 0 $$
The outside air temperature decreases as [tex]t=t_0(1-\beta h)[/tex]haruspex said:I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
I missed it at first too. The pressure inside the balloon is changing.haruspex said:I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
Yes, there is an expression for ##\rho(h)##, but I think that this is the variation of density with height at constant temperature. We are given the variation of temperature with height in order to account for the effect of temperature on density.haruspex said:I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
Exactly, I looked at this very equation in my process.erobz said:I missed it at first too. The pressure inside the balloon is changing.
$$ 1 = \frac{P_{out}}{P_{in}} = \frac{\rho_{out} \cancel{R} T_{out}}{ \rho_{in} \cancel{R} T_{in}}$$
The temperature inside is constant but the pressure is not. The density inside depends on the pressure outside and this pressure depends on both density and temperature outside.haruspex said:I must be missing something. How is ##\beta## relevant? The temperature inside is constant, and the density outside is a known function of height.
Why my process is wrong, I cannot understand. I'll try to follow your way.kuruman said:The balloon will stop rising when Its average density is equal to the density of the outside air. The average density is the total mass (air inside + envelope + load) divide by the external volume. I think we can safely ignore the buoyant force on the load. Find a relation between the density of the air inside and the density of the air outside when the pressure inside is equal to the pressure outside.
A few notes, you don't need to convert by adding ##273##, just note that the temperatures used are to be absolute.Hak said:Here, I get totally stuck. Where do I go wrong?
What would be the relationship I previously found that would express ##\rho_{in}## as a function of ##h##. It currently escapes me. Could you clarify? Thank you.erobz said:A few notes, you don't need to convert by adding ##273##, just note that the temperatures used are to be absolute.
What is being suggested (as an alternative approach) is at equilibrium ##\rho_{avg} = \rho_{out} ##. You can write that immediately as:
$$\frac{\rho_{in} V + m}{V} = \rho_{out} $$
Work from there by subbing the relationships you found earlier for ##\rho_{in}## as a function of ##h## and the relationship you are given for ##\rho_{out}##.
Hak said:What would be the relationship I previously found that would express ##\rho_{in}## as a function of ##h##. It currently escapes me. Could you clarify? Thank you.
Where did I find it?erobz said:The relationship for the density of the air inside the balloon at a given altitude ##h##.
Don’t worry about where you found it. Can you find it now?Hak said:Where did I find it?
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##.erobz said:Don’t worry about where you found it. Can you find it now?
Rearrange that expression...Hak said: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##.
##\rho_{in} = \rho_{out} \frac{t_0}{t_{in}} (1-\beta h)##?erobz said:Rearrange that expression...
##\rho_{in} = ? ##
We use capital ##T## for temperature, ##t## is time.Hak said:##\rho_{in} = \rho_{out} \frac{T_{o}}{T_{in}} (1-\beta h)##?
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?erobz said:We use capital ##T## for temperature, ##t## is time.
Maybe the relationship was with °C after all. So it is as you thought.Hak said:Which value of ##T_0## should I use? With ##T_0 \approx 300 K##, I get two negative results. Therefore, this is not correct.
So? What should I do?erobz said:Maybe the relationship was with °C after all. So it is as you thought.
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 said:Maybe the relationship was with °C after all. So it is as you thought.
$$T = T_o [{}^{\circ} C] ( 1 - \beta h ) + 273 [K] $$
I would say:Hak said: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...
We have two negative results...erobz said:I would say:
$$ T_{out}[\text{K}] = 27[{}^{\circ}C] ( 1 - \beta h ) + 273[\text{K}] $$
$$T_{in} [ \text{K}] = 100[{}^{\circ}C] + 273[ \text{K}] $$
You mean for the height ##h##?Hak said:We have two negative results...
Yes.erobz said:You mean for the height ##h##?
Does this equation rearrange to the first equation we derived?Hak said:Yes.
I don't think so, the results are different. Something is not right, the values are either too small or too large...erobz said:Does this equation rearrange to the first equation we derived?