ArticleslgStudy

astronomy

Solar balloon

Solar balloon is a astronomy topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Solar balloon rather than just read about it. In short: A solar balloon is a balloon that gains buoyancy when the air inside is heated by solar radiation, usually with the help of black or dark balloon material. The heated air inside the solar balloon expands and has lower density than the surrounding air.

Solar balloon — main illustration
Solar balloon — illustration

Key takeaways

  • Solar balloon belongs to astronomy; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Solar balloon to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Solar balloon from memory before moving on to harder problems.

Reference excerpt

A solar balloon is a balloon that gains buoyancy when the air inside is heated by solar radiation, usually with the help of black or dark balloon material. The heated air inside the solar balloon expands and has lower density than the surrounding air. As such, a solar balloon is similar to a hot air balloon. Usage of solar balloons is predominantly in the toy market, although it has been proposed that they be used in the investigation of planet Mars, and some solar balloons are large enough for human flight. A vent at the top can be opened to release hot air for descent and deflation.

Theory of operation

Generating lift

Raising the air temperature inside the envelope makes it less dense than the surrounding (ambient) air. The balloon floats because of the buoyant force exerted on it. This force is the same force that acts on objects when they are in water and is described by Archimedes' principle. The amount of lift (or buoyancy) provided by a hot air balloon depends primarily upon the difference between the temperature of the air inside the envelope and the temperature of the air outside the envelope. The lift generated by 100,000 ft3 (2831.7 m3) of dry air heated to various temperatures may be calculated as follows:

The density of air at 20 °C, 68 °F is about 1.2 kg/m3. The total lift for a balloon of 100,000 cu ft heated to (99 °C, 210 °F) would be 1595 lbf, 723.5 kgf. In reality, the air contained in the envelope is not all the same temperature, as the accompanying thermal image shows, and so these calculations are based on averages. For typical atmospheric conditions (20 °C, 68 °F), a hot air balloon heated to (99 °C, 210 °F) requires about 3.91 m3 of envelope volume to lift 1 kilogram (62.5 cu ft/lb). The precise amount of lift provided depends not only upon the internal temperature mentioned above, but the external temperature, altitude above sea level, and humidity of the surrounding air. On a warm day, a balloon cannot lift as much as on a cool day, because the temperature required for launch will exceed the maximum sustainable for the envelope fabric. Also, in the lower atmosphere, the lift provided by a hot air balloon decreases about 3% for each 1,000 meters (1% per 1,000 ft) of altitude gained.

Solar radiation

Insolation is a measure of solar radiation energy received on a given surface area in a given time. It is commonly expressed as average irradiance in watts per square meter (W/m2). Direct insolation is the solar irradiance measured at a given location on Earth with a surface element perpendicular to the Sun's rays, excluding diffuse insolation (the solar radiation that is scattered or reflected by atmospheric components in the sky). Direct insolation is equal to the solar constant minus the atmospheric losses due to absorption and scattering. While the solar constant varies with the Earth-Sun distance and solar cycles, the losses depend on the time of day (length of light's path through the atmosphere depending on the Solar elevation angle), cloud cover, moisture content, and other impurities. Over the course of a year the average solar radiation arriving at the top of the Earth's atmosphere is roughly 1,366 watts per square meter (see solar constant). The radiant power is distributed across the entire electromagnetic spectrum, although most of the power is in the visible light portion of the spectrum. The Sun's rays are attenuated as they pass through the atmosphere, thus reducing the insolation at the Earth's surface to approximately 1,000 watts per square meter for a surface perpendicular to the Sun's rays at sea level on a clear day. A black body absorbs all the radiation that hits it. Real world objects are gray objects, with their absorption being equal to their emissivity. Black plastic might have an emissivity of around 0.95, meaning 95 percent of all radiation that hits it will be absorbed, and the remaining 5 percent reflected.

Estimating energy received

If the balloon is imagined as a sphere, the sunlight received by this sphere can be imagined as the cross-section of a cylinder with the same radius as this sphere, see diagram. The area of this circle can be calculated via: π r 2 {\displaystyle \mathrm {\pi } r^{2}\ }

For example, the energy received by a spherical, 5 metre radius, solar balloon with an envelope of black plastic on a clear day with direct insolation of 1000 W/m2, can be estimated by first calculating the area of its great circle:

A r e a = π × ( 5 m ) 2 ≈ 78 . 54 m 2 {\displaystyle \mathrm {Area} =\pi \times (5m)^{2}\approx 78{.}54m^{2}}

Then multiplying this with the emissivity of the plastic and the direct insolation of the Sun: 78.54 * 0.95 * 1000 = 74,613 Watts At sea level at 15 °C at ISA (International Standard Atmosphere), air has a density of approximately 1.22521 kg/m3. The density of air decreases with higher temperatures, at the rate of around 20 grams per m3 per 5 K. Around 1 kilojoules of energy is needed to heat 1 kilogram of dry air by one kelvin (see heat capacity). So, to increase the temperature of 1 m3 of air (at sea level and at 15 °C) 5 °C requires around 5 °C * 1 kilojoules/(kilogram*kelvin) * 1.225 kilograms = 6.125 kilojoules. By doing so, you've reduced the mass of 1 m3 of air by around 24 grams. On a clear day with a black body surface of 1 m2 perpendicular to the Sun and no heat loss, this would take a little over 6 seconds.

Estimating rate of energy lost Below is the energy balance equation of the rate of energy lost of a solar balloon when drawing the boundary line around the balloon. The Solar Balloon experiences heat transfer due to convection and heat transfer due to radiation.

Ėout= tσπr2(TS4-TF4) + hπr2(TS-TF)

Estimated Change in Entropy Tds=du+PdV Δs = ∫(cv/T)dT + Rgasln(V2/V1) Δs = cvln(T2/T1)

Equilibrium The system is in equilibrium when the energy lost from the balloon through convection, radiation and conduction, equals the energy received through radiation from the Sun.

… excerpt ends here. Continue reading the full article.

Illustrations

Solar balloon: A 10-foot solar "tetroon"
A 10-foot solar "tetroon"
Solar balloon: A 4 meters high solar balloon floats over a meadow.
A 4 meters high solar balloon floats over a meadow.
Solar balloon: Thermal image showing temperature variation in a hot air balloon
Thermal image showing temperature variation in a hot air balloon
Solar balloon: A great circle divides the sphere in two equal hemispheres
A great circle divides the sphere in two equal hemispheres
Solar balloon illustration

Worked examples

Example 1 — a first encounter with Solar balloon

Start with the simplest possible case. Write down what Solar balloon claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In astronomy, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Solar balloon before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Solar balloon ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Solar balloon

In research
Solar balloon appears in astronomy research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Solar balloon in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Solar balloon is common in secondary-school and first-year university syllabi. It links to neighbouring topics Balloons (aeronautics), so understanding it makes those chapters shorter.
In everyday life
Look for Solar balloon outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

Affiliate

Preply — study more efficiently by working with a personal tutor. 50% off.

How to study Solar balloon in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Solar balloon means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Solar balloon out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Solar balloon in simple terms?

A solar balloon is a balloon that gains buoyancy when the air inside is heated by solar radiation, usually with the help of black or dark balloon material. The heated air inside the solar balloon expands and has lower density than the surrounding air.

Why does Solar balloon matter?

Because it connects several astronomy ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Solar balloon?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Solar balloon.

Tags

  • Balloons (aeronautics)

Keep exploring