1 Sun as an energy source
- Structure and composition of sun
- Energy production reactions in sun
- Estimation of energy production inside the sun
Fig 1 Anatomy of a star
Fig 1 Interior of sun (Source: https://www.nasa.gov/mission_pages/sunearth/science/Sunlayers.html)
The Corona is the Sun’s outer atmosphere. The sun’s magnetic fields rise through the convection zone and erupt through the photosphere into the chromosphere and corona. The eruptions lead to solar activity, which includes such phenomena as sunspots, flares, prominences, and coronal mass ejections.
3. Atmosphere of sun
The surface Gas Pressure (top of photosphere) is 0.868 mb and Pressure at bottom of photosphere (optical depth = 1) is 125 mb. The effective temperature is 5772 K. Temperature at top of photosphere is 4400 K and at bottom of photosphere is 6600 K. Temperature at top of chromospheres is ~30,000K.
4. Core of the sun
The core of the Sun occupies 20–25% of the solar radius from the centre. The energy production by fusion of hydrogen atoms (H) into molecules of helium (He) occur at the core. The high energy production is the result of extreme pressure and temperature that exists within the core. The pressure and temperature of the core of the sun is equivalent of 250 billion atmospheres (25.33 trillion KPa) and 15.7 million kelvin, respectively. The 99 percent of energy produced by sun happens in 24 percent of the radius i.e. core. The heat energy is transferred to other parts and outer space from the core.
5. Nuclear fusion reaction in the Sun
The temperature inside the sun is > 10,000,000 K. The stellar nuclear reactions are responsible for energy production inside the sun. The hydrogen atoms combine together to form helium atom by the nuclear fusion reaction. The Sun releases energy at a mass–energy conversion rate of 4.26 million metric tons per second, which produces the equivalent of 38,460 septillion watts (3.846×1026 W) per second.
These reactions are of two types
Proton-Proton Chain (P-P chain) & CNO cycle
The Sun emits 4 x 1026 Watts of power.
5.1 Proton-Proton Chain Reaction: It converts Hydrgen atoms (H) to Helium (He). This stellar reaction is most efficient in lower mass stars like the Sun. The reaction results in fusion of four protons into one alpha particle with the release of two positrons and two neutrinos (which further changes two of the protons into neutrons) and energy.
5.2 CNO cycle
This reaction is important in heavier stars. In this cycle, hydrogen is used to synthesize heavier elements. The heaviest elements are synthesized by fusion that occurs as a more massive star undergoes a violent supernova at the end of its life, a process known as supernova nucleosynthesis.
Fig 3 CNO cycle
This is most efficient in heavier stars having temperature greater than 16,000,000 K. This cycle was proposed by Hans Bethe in 1939.
The sun remains hot due to equilibrium maintained between Nuclear reactions; chemical or gravitational energy. But it is inadequate to maintain luminosity for billions of years. Pressure balance (gas pressure vs gravity) and energy balance (production vs outflow) are both needed to regulate heat energy inside the sun.
6. Energy production in Sun
The energy released during conversion of hydrogen to helium is mainly in the form of photons. The energy released can be estimated from the mass difference.
Mass of 4H1 = 6.69048 x 10-27 kg ……………….(1)
Mass of 1He4 = 6.64648 x10-27 kg ……………… (2)
∆m, mass difference = (1)- (2) = 0.04400 x10-27 kg
E= mc2 = 4.o x 10 -12 Joules per reaction
If 10% of hydrogen is converted into helium for the sun, it will produce 10 44 J of energy which is sufficient to keep sun burning for 10 billion years.
7. Calculations related to energy production in sun Example 1
How much mass is converted to energy every second in the sun to supply energy to the earth?
Solution:
Solar constant = 1372 W/m2
Rearth = 6.4 x 106 m
D earth-sun = 1.5 x 1011 m
Msun = 2.0 x 1030 kg
To get solar constant in units of amss (kg), we have
The total energy falling on the earth is solar constant times the projected area of the earth (from the sun), then
Which is the mass converted every second to supply light incident on earth.
Example 2
How much metric tons of H are converted to He to supply this energy?
Solution:
Take mp =1.67262 x 10-27 kg
D earth-sun = 1.5 x 1011 m
M He = 6.64648 x 10-27 kg
The mass difference between four protons (6.69048 x 10-27 kg) and one helium nucleus (2 proton +2 neutrons)
is 0.04400 x 10-27 kg
The ratio of H-burned to converted mass is 6.69048/0.04400 = 150 0r 0.7 % of rest energy (mass) of original H converted to He.
Hence total mass burn rate of the sun is 150 x 4.2 x 10 9 kg/s = 6.3 x 1011 kg/s = 630 million metric tons each second
Example 3
How long will the sun continue to produce energy? (Ignoring other processes and emissions from the sun)
Dividing the burn rate into fuel supply (total mass) gives an estimate of how long sun will last
The sun convert only 10 % of its H mass to He, so the above estimate o=is high by a factor of 10. And main life time of sun is ~ 10 10 yr
8. Energy Transport in Sun
The energy produced at core of the sun by fusion reaction is transported by Conduction, Convection and Radiation.
Table 2 Energy transport in sun
The stars generally use the convective and radiative method.
9. Solar energy to earth
The sunlight that reaches the earth depends on the revolution of the earth around the sun.
Orbital plane : The orbit of the earth around the sun lies in a geometrical plane called the orbital plane. The ecliptic plane is the plane of the orbit that intersects the sun. The line of intersection between the orbital plane and the ecliptic plane is the line of nodes. Planetary orbits, including the earth’s orbit, lie in the ecliptic plane.
The luminosity of a star is the total energy radiated per second by the star.
The amount of radiation from the sun that reaches the earth’s atmosphere is called the solar constant. The solar constant varies with time because the earth follows an elliptical orbit around the sun and the axis of rotation of the earth is inclined relative to the plane of the earth’s orbit. Distances between points on the surface of the earth and the sun vary throughout the year.
The flux of solar radiation incident on a surface placed at the edge of the earth’s atmosphere depends on the time of day and year, and the geographical location of the surface. Some incident solar radiation is reflected by the earth’s atmosphere.
The fraction of solar radiation that is reflected back into space by the earth-atmosphere system is called the albedo. The solar radiation is absorbed or reflected by clouds (20%), atmospheric particles (10%), and reflection by the earth’s surface (5%). Thereby prevents from reaching the earth surface.
The solar flux that enters the atmosphere is reduced by the albedo. So the solar radiation undergoes absorption, reflection or scattering by air, water vapor, dust particles, and aerosols while travelling to earth. The solar radiation that reaches directly to earth is direct solar radiation and that reaches after scattering is known as diffuse radiation.
Light is electromagnetic radiation. It can be used to transfer energy by the propagation of electromagnetic waves from one location to another. Two other energy transfer processes are conduction and convection. Conduction is the transfer of energy as the result of a temperature difference between substances that are in contact. Convection is the transfer of energy by the movement of a heated substance.
10. Solar radiation spectrum
Fig 4 Solar radiation spectrum
11.Sun-Earth relationship
The distance between sun and earth is measured in Astronomical Unit (AU). One AU is the distance travelled in 8.31 minutes at the speed of light. The mean distance of sun from Earth is 149.6 x 106 km with minimum of 147.1 x106 km and maximum of 152.1 x 106 km.
12. Solar energy received on earth
Example 4: The solar power incident on a surface averages 400 W/m2 for 12 hours. How much energy is received?
Solution:
400 W/m2 x 12 hours = 4800 Wh/m2 = 4.8 kWh/m2
Example 5: The amount of solar energy collected on a surface over 8 hours is 4 kWh/m2. What is the average solar power received over this period?
Solution
4 kWh/m2/8 h = 0.5 kW/ m2 = 500 W/m2
Conclusion
The sun produces the energy by nuclear fusion reaction of hydrogen to helium atoms in the core. The energy is transported by means of radiation and convection. Only a small portion of sun is reaching the earth and all the energy sources are dependent on sun.
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Suggested readings
- Singer, Neal (2011). Wonders of Nuclear Fusion: Creating an Ultimate Energy Source. University of New Mexico Press, 130 p.