Introduction to Climatology GEOGRAPHY 300 Solar Radiation and the Seasons Tom Giambelluca University of Hawai i at Mānoa Lauren Kaiser 09/05/2012 Geography 300 Lecture Outline Energy Potential and Kinetic Conservation and Transfer Radiation Blackbodies and Grey Bodies Emissivity Radiation Laws Planck s Curve Stefan-Boltzmann Law Wien s Displacement Law Lambert s Law Energy Force = mass X acceleration Units expressed in Newtons Energy The ability to do work Force applied over a distance Units expressed in Joules Power Energy per unit time Units expressed in Watts kg m N = 2 s kg m J = N m = 2 s J W = s 2 1
Energy Transferred from the sun to the Earth in the form of electromagnetic energy Radiation provides energy for many processes: Atmospheric movement Growth of plants Evaporation of water Types of Energy Potential Energy Energy due to position in the gravitational field and space Kinetic Energy The energy of motion Kinetic Energy Kinetic energy can occur as motion associated with moving objects Kinds of Energy The greater amount of internal kinetic energy, the higher the temperature Temperature measures average kinetic energy 2
Conservation of Energy Fundamental concept of physics The total amount of energy in an isolated system remains constant over time Energy cannot be created or destroyed Transferred from one form to another Energy Transfer Conduction Heat energy moving from one molecule to another Particles gain (lose) kinetic energy by colliding with faster (slower) surrounding particles Convection Vertical movement of energy in liquids or gases Advection horizontal movement of heat energy Radiation Energy moving electromagnetic (EM) waves All objects with a temperature above 0 Kelvin emit electromagnetic radiation at some wavelength Movement of Energy Radiation Energy emitted by all matter Wave-Like characteristics EM waves are oscillating electrical and magnetic fields Oscillating electrical fields produce oscillating magnetic fields and vice-versa Frequency of oscillations has a wide range Speed of propagation (v), frequency (f), and wavelength (λ) are related as: v = f λ 3
Radiation Waves Radiation Inverse relation of wavelength and frequency Longer wavelength à lower frequency Shorter wavelength à higher frequency All forms of electromagnetic radiation travel at the speed of light at 300,000 km per second It takes 8 minutes for energy to reach the Earth Does not require a medium to move through Energy from other stars reach Earth as well Proxima Centauri is a red dwarf star 4.3 lightyears away and the closest to our solar system Wavelengths of Energy Fundamental laws of physics govern the amount of radiation emitted and its wavelengths Electromagnetic Spectrum 4
Blackbodies Perfect emitters of radiation Emits the maximum possible radiation at every wavelength and absorbs all incident radiation Hypothetical and ideal radiators The Sun is the closest to being a blackbody Everything else is considered a grey body Grey Bodies True blackbodies do not exist in nature Most liquids and solids must be treated as grey bodies which means they emit some percentage of the maximum amount of radiation possible at a given temperature The percentage of energy radiated by a substance relative to that of a blackbody is referred to as its emissivity (ε) Planck s Radiation Law Accurately describes blackbody radiation Details the amount of EM energy radiated with a certain wavelength by a blackbody Variables include I = spectral radiance υ = frequency T = temperature h = Planck constant c = speed of light 5
Stefan-Boltzmann Law Temperature determines how much energy a blackbody radiates Hotter bodies emit more energy than cooler ones The intensity of energy radiated by blackbodies increase according to the fourth power of the absolute temperature I = σt 4 I = intensity of radia6on in W m - 2 σ = Stefan- Boltzmann constant (5.67 X 10-8 W m - 2 K - 4 ) T = is the temperature of the body in Kelvin Example 1 How much radiation is emitted by a blackbody at a temperature of 25 C? 1. Convert from Celsius to Kelvin T (K) = 25 ( C) + 273.15 = 298.15 K I = σt 4 I = (5.67 X 10-8 ) X (298.15) 4 I = 448 W m -2 Emissivity The certain amount of energy radiated by a source in comparison to that of a blackbody True blackbodies have an emissivity of 1 and grey bodies have an emissivity of < 1 Emissivity is a dimensionless unit For grey bodies, this can be incorporated into the Stefan-Boltzmann Law: I = εσt4 Example 2 For a grey body at a temperature 0f 25 C, what is the radiation emitted if the emissivity is 0.7? 1. Convert from Celsius to Kelvin T (K) = 25 ( C) + 273.15 = 298.15 K I = εσt 4 I = (0.7) X (5.67 X 10-8 ) X (298.15) 4 I = 314 W m -2 6
Example 3 If given intensity, solve for temperature What is the temperature ( C) of a blackbody that is emitting 1000 W m -2? 1. Rearrange the formula so that I = σt 4 à T (K) = (I/σ) 0.25 T = (I/σ) 0.25 T = (1000/5.67 X 10-8 ) 0.25 T = 364.42 K 3. Convert from Kelvin back to Celsius K = 364.42 273.15 = 91.27 C Wien s Displacement Law Explains that hotter objects radiate energy at shorter wavelengths Shortwave radiation < 0.4 µm Longwave radiation > 0.4 µm For any radiating body, the wavelength of peak emission is given by Wien s Law Peak solar radiation intensity has a wavelength of 0.5 µm Peak Earth radiation intensity has a wavelength of 10 µm Emission Spectrum Emission Spectrum Energy radiated by a substance occurs over a wide range, or spectrum, of wavelengths Hotter objects radiate more energy at shorter wavelengths Cooler bodies radiate less energy at longer wavelengths Because the emission spectra of the Earth and the Sun are almost entirely separate, we refer to Earth s radiation as LONGWAVE RADIATION, and Solar radiation as SHORTWAVE RADIATION 7
Wien s Displacement Law λ max = wavelength of maximum radiation emission Τ = temperature of the emitting surface (K) 2897 = constant Units: µm (micrometers) Example 4 What is the wavelength of maximum radiation for a blackbody at a temperature of 25 C? 1. Convert from Celsius to Kelvin T (K) = 25 ( C) + 273.15 = 298.15 K λ max = 2897 T λ max = 2897 298.15 λ max = 9.72 µm Example 5 What is the temperature of a surface with a wavelength of maximum radiation at 1 µm? And at 15 µm? 1. Rearrange the formula so that λ max = 2897 T à T = 2897 λ max T(x) = 2897 λ max T(1) = 2897 1 = 2897 K T(15) = 2897 15 = 193 K 3. Convert from Kelvin back to Celsius K(1) = 2897 273.15 = 2623.85 C K(15) = 193 273.15 = -83.15 C Planck s Curve µ(λ) [kj/nm] Wien s Displacement Law λ max = 2897 T λ [nm] Stefan- Boltzmann Law I = εσt 4 8
More Sample Problems 1. How much radiation is emitted by a blackbody at a temperature of 15 C? 2. What is the temperature of a blackbody emitting 500 W m -2? 3. What is the radiative emission of a grey body with an emissivity of 0.95 at 24 C? 4. Use Wien s Law to estimate the temperature of the sun s surface assuming that a portion of the sun emits a maximum wavelength of 0.477 µm where peak solar irradiance is in the green portion of the EM spectrum Solutions 1. 390.8927 W m -2 15 + 273.15 = 288.15 (5.67 X 10-8 ) X (288.15) 4 = 390.89 2. 306.4409 K or 33.2909 C (500/5.67 X 10-8) 0.25 = 306.4409 273.15 = 33.29 3. 419.9616 W m -2 24 + 273.15 = 297.15 K 0.95 X (5.67 X 10-8 ) X (297.15) 4 = 419.61 4. 6.0734 X 10 3 K or 5.8002 X 10 3 C 2897/0.477 = 6.0734 X 10 3 273.15 = 5.8002 X 10 3 Radiation Laws Lambert s Law The radiant intensity of a surface is directly proportional to the cosine of the angle between the observation line of sight and the surface Lambert s Law I = I cosγ or 0 I = I sin β 0 I = radiation on a horizontal surface (incident) I 0 = radiation on a perpendicular surface (measured normal to beam) γ = angle between beam of radiation and the line normal to the surface (zenith angle) β = angle between beam of radiation and the surface (elevation angle) 9
Lambert s Law Example 6 If the radiation measured normal to the beam is 500 W m -2, what would the incident radiation be if the zenith angle were 30? 1. Plug in values to equation I = I 0 cos γ I = (500)cos(30) I = 500 X 0.866 I = 433 W m -2 Example 7 If the zenith angle is 30 and the incident radiation is 500 W m -2, what would the radiation measured normal to the beam be? 1. Rearrange the formula so that I = I 0 cos γ à I 0 = I cos γ I 0 = I cos γ I 0 = (500) cos(30) I 0 = 500 0.866 I 0 = 577 W m -2 Mahalo! Any Questions? 10