Is Exponential Form Better for Balloon with Inserted Load Calculation?

[Hak]

Its seems like what you did is correct.

Then I really don't know what to come up with. Let's see if someone else disproves my thesis, but this time my unfolding seems correct....

 

[kuruman]

a value that seems exaggeratedly low to me. Where do I go wrong?

When you substituted your numbers what values did you put in for $\alpha$ and $\beta$? Show me the numbers.

 

[Hak]

When you substituted your numbers what values did you put in for $\alpha$ and $\beta$? Show me the numbers.

$\alpha = 49 \ m^{-1}$ and $\beta = 26 \ m^{-1}$.

 

[erobz]

$\alpha = 49 \ m^{-1}$ and $\beta = 26 \ m^{-1}$.

I think you went the wrong way there...

 

[Hak]

I think you went the wrong way there...

Why?

 

[erobz]

Why?

$\alpha = 0.026 \frac{1}{km} = 0.026 \frac{1}{km}\frac{1 km}{1000 m} = 0.000026 \frac{1}{m}$

etc...

 

[Hak]

$\alpha = 0.026 \frac{1}{km} = 0.026 \frac{1}{km}\frac{1 km}{1000 m} = 0.000026 \frac{1}{m}$

etc...

Damn! It's true!

 

[erobz]

Damn! It's true!

@kuruman knocks it out of the park!

 

[kuruman]

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.

 

[Hak]

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.

So, is my process incorrect?

 

[kuruman]

So, is my process incorrect?

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]

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 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?

Edit. My actual value is $h \approx 20.4 \ km$

 

[haruspex]

Its seems like what you did is correct.

I too would say:$$ \rho_o ( 1 - \alpha h ) \left( 1 - \gamma ( 1 - \beta h ) \right) - mg = 0 $$

I must be missing something. How is $\beta$ relevant? The temperature inside is constant, and the density outside is a known function of height.

 

[Hak]

I must be missing something. How is $\beta$ relevant? The temperature inside is constant, and the density outside is a known function of height.

The outside air temperature decreases as t=t_0(1-\beta h)

 

[erobz]

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.

$$ 1 = \frac{P_{out}}{P_{in}} = \frac{\rho_{out} \cancel{R} T_{out}}{ \rho_{in} \cancel{R} T_{in}}$$

 

[Hak]

Following my procedure, my value is $h \approx 20.4 \ km$. This seems excessive.

 

[kuruman]

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.

 

[Hak]

I missed it at first too. The pressure inside the balloon is changing.

$$ 1 = \frac{P_{out}}{P_{in}} = \frac{\rho_{out} \cancel{R} T_{out}}{ \rho_{in} \cancel{R} T_{in}}$$

Exactly, I looked at this very equation in my process.

 

[nasu]

I must be missing something. How is $\beta$ relevant? The temperature inside is constant, and the density outside is a known function of height.

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.

 

[Hak]

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.

Why my process is wrong, I cannot understand. I'll try to follow your way.

First, I would write down the given data and the unknown variable. Let h be the maximum height the balloon can reach, in kilometers. Then we have:
V=850m^3

m=200kg

T=100^\circ C=373K

\rho_0=1.2 kg/m^3

\alpha=0.049km^{-1}

t_0 \approx 300 K

\beta=0.026km^{-1}

Next, I would use the ideal gas law to relate the pressure, volume, temperature and number of moles of the air inside and outside the balloon. The ideal gas law is:
PV=nRT

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.

Since the balloon is open at the bottom, we can assume that the pressure inside is equal to the pressure outside at any height. Therefore, we can write:

P_{in}V_{in}=n_{in}RT_{in}

P_{out}V_{out}=n_{out}RT_{out}

where the subscripts in and out denote the inside and outside of the balloon, respectively.

Then, I would use the definition of density to express the number of moles in terms of mass and molar mass. The density is:
\rho=\frac{m}{V}

where \rho is the density, m is the mass, and V is the volume.

The molar mass is:

M=\frac{m}{n}

where M is the molar mass, m is the mass, and n is the number of moles.

Therefore, we can write:

n=\frac{m}{M}=\frac{\rho V}{M}

Substituting this into the ideal gas law equations, we get:

P_{in}V_{in}=\frac{\rho_{in} V_{in}}{M_{in}}RT_{in}

P_{out}V_{out}=\frac{\rho_{out} V_{out}}{M_{out}}RT_{out}

Next, I would simplify these equations by canceling out some terms. Since we are assuming that both the inside and outside air are composed of dry air with a constant molar mass of about 29 grams per mole, we can write:
M_{in}=M_{out}=M=0.029kg/mol

Also, since we are given that the volume of the envelope is constant, we can write:

V_{in}=V_{out}=V=850m^3

Therefore, we can simplify the equations as:

P_{in}=\frac{\rho_{in}}{M}RT_{in}

P_{out}=\frac{\rho_{out}}{M}RT_{out}

Then, I would equate these equations and solve for \rho_{in}/\rho_{out}. Since we are assuming that the pressure inside and outside are equal at any height, we can write:
\frac{\rho_{in}}{M}RT_{in}=\frac{\rho_{out}}{M}RT_{out}

Canceling out some terms and rearranging, we get:

\frac{\rho_{in}}{\rho_{out}}=\frac{T_{out}}{T_{in}}

Next, I would use the given formulas for \rho_{out} and T_{out} as functions of height to substitute them into this equation. We have:
\rho=\rho_0(1-\alpha h)

t=t_0(1-\beta h)

Therefore,

\rho_{out}=\rho_0(1-\alpha h)

T_{out}=t_0(1-\beta h)+273

where we have added 273 to convert the temperature from Celsius to Kelvin.

Substituting these into the equation for \rho_{in}/\rho_{out}, we get:

\frac{\rho_{in}}{\rho_0(1-\alpha h)}=\frac{t_0(1-\beta h)+273}{T_{in}}

Finally, I would use the definition of the average density of the balloon to express \rho_{in} in terms of the total mass and volume. We have:
\rho_{avg}=\frac{m_{total}}{V}

where \rho_{avg} is the average density, m_{total} is the total mass, and V is the volume.

The total mass is the sum of the mass of the air inside, the mass of the envelope, and the mass of the load. We can write:

m_{total}=m_{air}+m_{env}+m_{load}

The mass of the air inside is equal to the density of the air inside times the volume of the envelope. We can write:

m_{air}=\rho_{in}V

The mass of the envelope and the load are given as 200 kg. Therefore, we can write:

m_{env}+m_{load}=200kg

Substituting these into the equation for \rho_{avg}, we get:

\rho_{avg}=\frac{\rho_{in}V+200}{V}

Solving for \rho_{in}, we get:

\rho_{in}=\rho_{avg}-\frac{200}{V}

Substituting this into the equation for \rho_{in}/\rho_{out}, we get:

\frac{\rho_{avg}-\frac{200}{V}}{\rho_0(1-\alpha h)}=\frac{t_0(1-\beta h)+273}{T_{in}}

Here, I get totally stuck. Where do I go wrong?

 

[erobz]

Here, I get totally stuck. Where do I go wrong?

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]

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 $\rho_{out}$.

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]

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.

The relationship for the density of the air inside the balloon at a given altitude $h$.

 

[Hak]

The relationship for the density of the air inside the balloon at a given altitude $h$.

Where did I find it?

 

[erobz]

Where did I find it?

Don’t worry about where you found it. Can you find it now?

 

[Hak]

Don’t worry about where you found it. Can you find it now?

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]

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 $h$.

Rearrange that expression...

$\rho_{in} = ?$

 

[Hak]

Rearrange that expression...

$\rho_{in} = ?$

$\rho_{in} = \rho_{out} \frac{t_0}{t_{in}} (1-\beta h)$?

 

[erobz]

$\rho_{in} = \rho_{out} \frac{T_{o}}{T_{in}} (1-\beta h)$?

We use capital $T$ for temperature, $t$ is time.

Keep going... There is another substitution to make.

 

[Hak]

We use capital $T$ for temperature, $t$ is time.

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?