- #1
Hak
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- Homework Statement
- A model AX-4 hot-air balloon consists of a rigid envelope of volume [tex]V=850m^3[/tex] with an opening at the lower end. The air inside is maintained at a temperature of [tex]100^\circ[/tex]. The balloon must lift a total load (envelope plus payload) of [tex]200kg[/tex].
Knowing that air density decreases with height according to the formula [tex]\rho=\rho_0(1-\alpha h)[/tex], with [tex]\alpha=0.049km^{-1}[/tex] and that the outside air temperature decreases as [tex]t=t_0(1-\beta h)[/tex] with [tex]\beta=0.026km^{-1}[/tex], calculate the maximum height the balloon can reach.
- Relevant Equations
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My attempt is:
Archimedes' thrust on the balloon minus the weight of the air inside it is [tex]F=(\rho-\rho_{\rm in})Vg[/tex], where [tex]\rho_{in}[/tex] is the density of the hot air inside.
Now we have, in general, [tex]\rho=\mu {n \over V}=\mu {P \over RT}[/tex] from the perfect gas law ([tex]\mu[/tex] would be the molar mass of the air, but we don't care about that, it just serves as a constant of proportionality).
But because inside and outside are communicating, the pressure is the same and we have
[tex]{\rho \over \rho_{in}}={t_{in} \over t}[/tex] and thus.
[tex]F=(\rho-\rho_{in})Vg=\rho gV \left( 1-{t \over t_{in}}\right)= \rho_0gV(1-\alpha h)\left( 1-{t_0 (1-\beta h) \over t_{\rm in}}\right)[/tex]
Let us write for convenience [tex]t_0/t_{in}=\gamma[/tex], and, equalizing this expression to the weight force of the load, we obtain:
[tex]mg=F= \rho_0gV(1-\alpha h)\left( 1- \gamma (1-\beta h) \right)[/tex].
From here, I obtain a second-degree equation in h:
[tex]\rho_0 g V \alpha \beta \gamma h^2 - \rho_0 g V [\gamma (1- \beta) + \alpha] h - \rho g V (1-\beta) + mg = 0[/tex].
Now for a quick estimate of the quantities involved: the temperature on the ground is about [tex]t_0 \sim 300 K[/tex], from which [tex]\gamma \sim 300/373=0.80[/tex], while the density of air on the ground is [tex]\rho_0 \sim 1.2 kg m^{-3}[/tex]
From here, I finally get [tex]h \approx 0.029 m[/tex], a value that seems exaggeratedly low to me. Where do I go wrong?
Archimedes' thrust on the balloon minus the weight of the air inside it is [tex]F=(\rho-\rho_{\rm in})Vg[/tex], where [tex]\rho_{in}[/tex] is the density of the hot air inside.
Now we have, in general, [tex]\rho=\mu {n \over V}=\mu {P \over RT}[/tex] from the perfect gas law ([tex]\mu[/tex] would be the molar mass of the air, but we don't care about that, it just serves as a constant of proportionality).
But because inside and outside are communicating, the pressure is the same and we have
[tex]{\rho \over \rho_{in}}={t_{in} \over t}[/tex] and thus.
[tex]F=(\rho-\rho_{in})Vg=\rho gV \left( 1-{t \over t_{in}}\right)= \rho_0gV(1-\alpha h)\left( 1-{t_0 (1-\beta h) \over t_{\rm in}}\right)[/tex]
Let us write for convenience [tex]t_0/t_{in}=\gamma[/tex], and, equalizing this expression to the weight force of the load, we obtain:
[tex]mg=F= \rho_0gV(1-\alpha h)\left( 1- \gamma (1-\beta h) \right)[/tex].
From here, I obtain a second-degree equation in h:
[tex]\rho_0 g V \alpha \beta \gamma h^2 - \rho_0 g V [\gamma (1- \beta) + \alpha] h - \rho g V (1-\beta) + mg = 0[/tex].
Now for a quick estimate of the quantities involved: the temperature on the ground is about [tex]t_0 \sim 300 K[/tex], from which [tex]\gamma \sim 300/373=0.80[/tex], while the density of air on the ground is [tex]\rho_0 \sim 1.2 kg m^{-3}[/tex]
From here, I finally get [tex]h \approx 0.029 m[/tex], a value that seems exaggeratedly low to me. Where do I go wrong?
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