IPhO 1982 · Theory problem 3

Hot-air-balloon

Download PDFIPhO 1982

Consider a hot-air balloon with fixed volume \(V_{\mathrm B} = 1.1\ \mathrm{m^3}\). The mass of the balloon-envelope, whose volume is to be neglected in comparison to \(V_{\mathrm B}\), is \(m_{\mathrm H} = 0.187\ \mathrm{kg}\).

The balloon shall be started, where the external air temperature is \(\vartheta_1 = 20\ ^\circ\mathrm C\) and the normal external air pressure is \(p_{\mathrm o} = 1.013\cdot 10^5\ \mathrm{Pa}\). Under these conditions the density of air is \(\rho_1 = 1.2\ \mathrm{kg/m^3}\).

a) What temperature \(\vartheta_2\) must the warmed air inside the balloon have to make the balloon just float?

b) First the balloon is held fast to the ground and the internal air is heated to a steady-state temperature of \(\vartheta_3 = 110\ ^\circ\mathrm C\). The balloon is fastened with a rope.

Calculate the force on the rope.

c) Consider the balloon being tied up at the bottom (the density of the internal air stays constant). With a steady-state temperature \(\vartheta_3 = 110\ ^\circ\mathrm C\) of the internal air the balloon rises in an isothermal atmosphere of \(20\ ^\circ\mathrm C\) and a ground pressure of \(p_0 = 1.013\cdot 10^5\ \mathrm{Pa}\). Which height \(h\) can be gained by the balloon under these conditions?

d) At the height \(h\) the balloon (question c)) is pulled out of its equilibrium position by \(10\ \mathrm m\) and then is released again.

Find out by qualitative reasoning what kind of motion it is going to perform!

Official English text of the problem.

Answers

  • a) \(\boxed{\vartheta_2 \approx 68.4\ ^\circ\mathrm C}\), corresponding to \(T_2\approx341.5\ \mathrm K\).
  • b) Rope tension: \(\boxed{F_{\mathrm{rope}}\approx1.21\ \mathrm N}\). The balloon pulls upward on the rope.
  • c) Equilibrium height above the starting point: \(\boxed{h\approx8.4\times10^2\ \mathrm m=0.84\ \mathrm{km}}\).
  • d) Approximately simple harmonic vertical oscillations about \(h\), with amplitude \(10\ \mathrm m\) when air resistance is neglected. Air resistance damps the motion.

Solution

Treat air as an ideal gas with constant composition, \[ p=\rho R_{\mathrm s}T, \] where \(R_{\mathrm s}\) is its specific gas constant and \(T\) is absolute temperature. The relevant temperatures are \[ T_1=293.15\ \mathrm K,\qquad T_3=383.15\ \mathrm K. \]

In (a) and (b), the balloon is open at the bottom, so its internal pressure equals the surrounding pressure. We neglect pressure and density variations over the balloon’s dimensions. After sealing in (c), its enclosed air mass and prescribed volume remain constant.

Use \(g=9.81\ \mathrm{m\,s^{-2}}\), treated as constant over the heights involved.

a) Temperature required to just float

By Archimedes’ principle, the upward buoyant force equals the weight of the displaced external air: \[ F_{\mathrm B}=\rho_1V_{\mathrm B}g. \] For neutral buoyancy, this balances the combined weight of the envelope and the enclosed air: \[ \rho_1V_{\mathrm B}g =\left(m_{\mathrm H}+\rho_2V_{\mathrm B}\right)g. \] Thus the required internal density is \[ \rho_2=\rho_1-\frac{m_{\mathrm H}}{V_{\mathrm B}} =1.2-\frac{0.187}{1.1} =1.03\ \mathrm{kg\,m^{-3}}. \]

Since the internal and external pressures are equal, the ideal-gas law gives \[ \rho_2T_2=\rho_1T_1. \] Therefore \[ T_2=\frac{\rho_1T_1}{\rho_2} =\frac{1.2(293.15)}{1.03} =341.53\ \mathrm K, \] or \[ \boxed{\vartheta_2=341.53-273.15\approx68.4\ ^\circ\mathrm C}. \]

b) Force on the rope

At \(T_3=383.15\ \mathrm K\), the internal density is \[ \rho_3=\rho_1\frac{T_1}{T_3} =1.2\frac{293.15}{383.15} \approx0.91813\ \mathrm{kg\,m^{-3}}. \]

The rope supplies the downward force needed to balance the excess of buoyancy over weight: \[ F_{\mathrm{rope}} =\rho_1V_{\mathrm B}g -\left(m_{\mathrm H}+\rho_3V_{\mathrm B}\right)g. \] Substituting, \[ F_{\mathrm{rope}} =\left[(1.2-0.91813)(1.1)-0.187\right](9.81), \] so \[ \boxed{F_{\mathrm{rope}}\approx1.21\ \mathrm N}. \]

This is the tension: the rope pulls downward on the balloon, and the balloon pulls upward on the rope.

c) Equilibrium height in the isothermal atmosphere

Once the bottom is tied shut, the balloon’s total mass remains \[ M=m_{\mathrm H}+\rho_3V_{\mathrm B} =0.187+(0.91813)(1.1) \approx1.19694\ \mathrm{kg}. \]

To find how buoyancy changes with height, combine atmospheric hydrostatic equilibrium, \[ \frac{dp}{dz}=-\rho g, \] with the ideal-gas law at the constant external temperature \(T_1\): \[ \frac{\rho(z)}{p(z)}=\frac{\rho_1}{p_0}. \] Consequently, \[ \frac{dp}{p}=-\frac{\rho_1g}{p_0}\,dz. \] Integrating from ground level, where \(p=p_0\), gives \[ p(z)=p_0e^{-z/H}, \qquad \rho(z)=\rho_1e^{-z/H}, \] where the atmospheric scale height is \[ H=\frac{p_0}{\rho_1g} =\frac{1.013\times10^5}{(1.2)(9.81)} \approx8.605\times10^3\ \mathrm m. \]

At equilibrium, buoyancy equals the balloon’s constant weight: \[ \rho(h)V_{\mathrm B}g=Mg. \] The external density there must therefore be \[ \rho(h)=\frac{M}{V_{\mathrm B}} =\rho_3+\frac{m_{\mathrm H}}{V_{\mathrm B}} \approx1.08813\ \mathrm{kg\,m^{-3}}. \]

Using the exponential density profile, \[ \rho_1e^{-h/H} =\rho_3+\frac{m_{\mathrm H}}{V_{\mathrm B}}, \] and hence \[ h=H\ln\!\left( \frac{\rho_1}{\rho_3+m_{\mathrm H}/V_{\mathrm B}} \right). \] Numerically, \[ h=(8605.16) \ln\!\left(\frac{1.2}{1.08813}\right) \approx842\ \mathrm m. \] Thus \[ \boxed{h\approx0.84\ \mathrm{km}}. \]

This is the equilibrium height identified in part (d). A freely rising balloon can pass through it because of its inertia.

d) Motion after displacement

The balloon’s weight and volume remain constant, whereas external air density decreases with height.

  • If displaced upward, it encounters less dense air. Buoyancy becomes smaller than its weight, so the resultant force is downward.
  • If displaced downward, it encounters denser air. Buoyancy becomes greater than its weight, so the resultant force is upward.

The force therefore restores the balloon toward \(h\). On reaching \(h\), its inertia carries it through equilibrium, leading to oscillation.

The motion is approximately harmonic because the displacement is small compared with the atmospheric scale height. To see this, let \(x=z-h\). Since \(\rho(h)V_{\mathrm B}=M\), the upward resultant force is \[ F(x)=Mg\left(e^{-x/H}-1\right). \] Here \[ \frac{|x|}{H}\leq\frac{10}{8605}\approx1.2\times10^{-3}\ll1, \] so \[ F(x)\approx-\frac{Mg}{H}x. \] This is a restoring force proportional to displacement.

Thus, neglecting air resistance, the balloon performs approximately simple harmonic vertical oscillations, initially of amplitude \(10\ \mathrm m\). Air resistance dissipates energy and damps the motion, eventually bringing the balloon to rest at its equilibrium height.

The solution was checked against the official solution.