Monday April 10, 2017 Welcome back! Last week of content! Unit 8 test on Friday Last day to retake Unit 8 test is April 28th. Tell students my last day is Friday

Renewable vs Non-Renewable Renewable resources: can be replaced in a reasonable amount of time (or are not depleted) Non-renewable resources: does not renew itself at a sufficient rate for sustainable economic extraction in meaningful human time-frames.

Energy Sources Primary: energy source used directly by the consumer Examples: wind, water, coal, natural gas Secondary: energy source that has been transformed from a primary energy source before use by the consumer Examples: electricity

Energy Sources Specific energy and energy density are useful measures of the energy that will be released from a given weight or volume of fuel when it is burned Specific energy ESP : how much energy (J) you can get per unit mass (kg) from a fuel. Units are J kg-1 Energy density ED : how much energy (J) you can get per unit volume (m3) from a fuel. Units are J m-3

Practice Fission of each uranium-235 produces 3.510-11 J of energy. The density of U-235 is 1.8104 kg m-3. Calculate the ESP and the ED of U-235. Solution: (a) m = (235 u)(1.66110-27 kg u-1) = 3.9010-25 kg. ESP = 3.510-11 J / 3.9010-25 kg = 9.01013 J kg-1. (b) ED = (9.01013 J kg-1)(1.8104 kg m-3) = 1.61018 J m-3. Ask: what do I need to find specific energy; do I have everything I need to find energy density?

Practice Coal has a specific energy of 32.5 MJ/ kg. If a city has a coal-fired power plant that needs to produce 30.0 MW of power, with an efficiency of 25%, how many kilograms of coal are needed daily? Solution: 0.25 = 30.0 MW / input  input = 120 MW But 1 day = 243600 s = 86400 s, so that input = (120 MJs-1)(86400 s) = 10368000 MJ. input = (10368000 MJ)(1 kg / 32.5 MJ) = 320000 kg. efficiency = output / input

Sankey Diagrams Continuous conversion of energy into work requires a cyclical process and the transfer of energy from the system. Second law of thermodynamics: It is possible to convert mechanical energy completely into thermal energy It is NOT possible to convert all heat energy into mechanical energy This loss of energy during conversion from one form to another is called energy degradation. Campfire: Heat released during combustion (chemical reaction) that is used; heat used directly; some heat is lost to the environment or wasted.

Sankey Diagrams Summarize all the energy transfers taking place in a process. The thicker the line or arrow, the greater the amount of energy involved Width of each arrow is proportional to the amount of energy in that section Energy degradation in systems can be shown with an energy flow diagram

Sankey Diagrams

Sankey Diagram – Electric Lamp Most of the electrical energy is transferred as heat rather than light Sankey diagrams show the efficiency of each energy conversion in Joules

Sankey Diagrams - Example Find the energy values for each of the degradations in the Sankey diagram. Solution: From conservation of energy we see that at each interface, the energy in must equal the energy out. 20 MJ 10 MJ 20 MJ Can find efficiencies using efficiency = output/input

Electricity Electricity is a secondary and versatile form of energy The most common way to generate electrical power is the coal-burning power plant. Sherco power plant in MN

Production of Electrical Power Chemical energy in coal is released by burning. Heat boils water. Steam rotates a turbine. The turbine turns a coil of wire in a magnetic field. Electrical power is produced.

Production of Electrical Power

Electrical Power & Sankey Diagrams COAL BOILER STEAM TURBINE CONDENSER GENERATOR cold water wasted heat exhaust gas friction Traditional Coal Fired Power Plants Ask: What might be a more efficient way to get chemical energy? CHEMICAL ENERGY HOT STEAM KINETIC ELECTRICITY exhaust gas wasted heat friction 40% efficient

Electrical Power & Sankey Diagrams GAS BOILER STEAM TURBINE CONDENSER GENERATOR cold water wasted heat exhaust gas friction GAS TURBINE CHEMICAL ENERGY ELECTRICITY exhaust wasted heat friction HOT STEAM KINETIC 50% efficient

Nuclear Power Fuel Fuel in a nuclear power plant must be able to sustain itself: once the reaction is started, it must be able to continue on its own This is called chain reaction (Nuclear) Chain Reaction: nuclear reaction that causes one or more other nuclear reactions to take place We talked about nuclear power in the last unit

Nuclear Power Fuel Chain reaction in nuclear reactor

Nuclear Power Fuel Moderators slow down neutrons, increasing chance achieving chain reaction Control rods absorb neutrons, controlling fission reactions

Nuclear Power Nuclear power stations are the same as fossil fuel stations, from the turbine on down. Chem NRG =>produce heat => boil water = > KE => generator (rotation) => turns into electricity

Wind Generators Wind turbines can use the wind to make electricity Heated land air becomes less dense, and rises. Cooler air then fills the low pressure left behind. A convection current forms

Wind Generators Same story as before, but fuel has changed (KE => rotation => electrical energy)

Wind Generators - Power Determine the power that may be delivered by a wind generator, assuming that the wind kinetic energy is completely converted into mechanical energy. Solution: Assume a rotor blade radius of r. r d = vt

Wind Generators - Power Solution: The volume of air that moves through the blades in a time t is given by V = Ad = Avt, where v is the speed of the air and A = r2. The mass m is thus m = V = Avt. EK = (1/2)mv2 = (1/2)Avtv2 = (1/2)Av3t. Power is EK / t so that r d = vt Power equation for wind generator Power = (1/ 2)Av3 where A = r 2

Hydroelectric Systems Number of ways water can be used: Pumped Storage Plants Hydroelectric Plants Tidal Barrage Tidal Flow Systems Hoover Dam: 1.5109 W Rance Tidal Barrage: 2.4108 W Makes up for deficit of energy demands by consumers

Hydroelectric – Sun-Driven Sun-driven evaporation and rainfall place water at a high potential energy. During times of less energy demand, excess power plant electricity can be used to pump water back up into the reservoir for later use. This is called the pumped storage scheme. A typical hydroelectric dam

Pumped Storage

Pumped Storage - Example A reservoir for a hydroelectric dam is shown. (a) Calculate the potential energy yield. Solution: PE = mgh The total volume of water is V = 1700(2500)(50) = 2.125108 m3. The mass of the water is m = V = (1000 kg m-3)(2.125108 m3) = 2.1251011 kg. A typical hydroelectric dam

Pumped Storage - Example A reservoir for a hydroelectric dam is shown. (a) Calculate the potential energy yield. Solution: PE = mgh The average water height is h = 75 + 50 / 2 = 100 m. The potential energy yield will then be EP = mgh = (2.1251011)(10)(100) = 2.1251014 J. A typical hydroelectric dam

Pumped Storage - Example A reservoir for a hydroelectric dam is shown. (b) If the water flow rate is 25 m3 per second, what power is provided by the moving water? Solution: Each cubic meter has a mass of 1000 kg. Thus each second m = 25(1000) = 25000 kg falls. Thus each second the reservoir relinquishes EP = mgh = (25000)(10)(100) = 2.5107 J. Since power is measured in W (or J s-1) the power provided is 2.5107 W = 25 MW. Recall from (a) h= 100 m

Solar Power Two direct solar energy devices: Solar heating panels Solar photovoltaic cells Most energy is derived from sun, we mean power directly from the sun’s rays

Solar Heating Panels The heating panel converts sunlight directly into heat. The slower the water is circulated, the hotter it can get.

Solar Heating Panels Liquid in panels is heated by sun and pumped to storage cylinder, heating up water present in cylinder

Solar Photovoltaic Panels Photovoltaic cells convert sunlight directly into electricity. Uses semiconductors to transform energy Light absorbed by semiconductor Photons transfer energy to electrons Electrons flow through the material as electrical current

Photovoltaic Panels - Example A photovoltaic cell has an area of 1.00 cm2 and an efficiency of 10.5%. (a) If the cell is placed in a position where the sun's intensity is I = 1250 W m-2, what is the power output of the cell? Solution: I = P/A A = (1 cm2)(1 m / 100 cm)2 = 0.0001 m2 PIN / A = I, so PIN = IA = 1250(0.0001) = 0.125 W. The cell is only 10.5% efficient so that POUT = 0.105PIN= 0.105(.125) = 0.0131 W. First, we need to know what the intensity of the sun means. Remember: efficiency = Pout/Pin

Photovoltaic Panels - Example A photovoltaic cell has an area of 1.00 cm2 and an efficiency of 10.5%. (b) If the cell is rated at 0.500 V, what is its current output? Solution: P = IV so I = P/ V = 0.0131/.5 = 0.0262 A. Ask: do we have a way to find current?

Photovoltaic Panels - Example A photovoltaic cell has an area of 1.00 cm2 and an efficiency of 10.5%. (c) If ten of these cells are placed in series, what will the voltage and the current be? Solution: In series the voltage increases. In series the current stays the same. Thus V = 10(0.500) = 5.00 V and I = 0.0262 A. Ask: what happens to voltage in a series circuit? Current?

Photovoltaic Panels - Example A photovoltaic cell has an area of 1.00 cm2 and an efficiency of 10.5%. (d) If ten of these cells are placed in parallel, what will the voltage and the current be? Solution: In parallel the voltage stays the same. In parallel the current increases. Thus V = 0.500 V and I = (10)(0.0262) = 0.262 A. Ask: what happens to voltage in a series circuit? Current?

Photovoltaic Panels - Example A photovoltaic cell has an area of 1.00 cm2 and an efficiency of 10.5%. (e) How many cells would you need to operate a 100 W circuit? Solution: Pout = 0.0131 W/ cell. Thus (100 W) / (0.0131 W/ cell) = 7630 cells!

Tuesday April 11, 2017 Problem sets 8.1 and 8.2 due on Thursday Unit 8 test on Friday Last day to retake Unit 8 test is April 28th. To start talking about thermal climate models, we need some background concepts…

Thermal Energy Transfer Thermal energy can be transferred from a warmer mass to a cooler mass by three means: Conduction Convection Thermal radiation

Conduction When two solids of different temperatures touch, thermal energy is transferred from the hotter object to the cooler object through a process called conduction. Essential idea: molecules vibrate and can transfer this energy to adjacent molecules Metals are good heat conductors because they have lots of free electrons (the same reason they are good electrical conductors).

Convection Convection requires a fluid as a medium of heat transfer. Convection currents form as hot air rises, cools, becomes dense, and sinks Essential idea: molecules in a liquid or gas will flow towards locations of lower density. Convection current Hot air is less dense than cold air, so it rises; as it rises it cools, and so becomes denser and sinks. When we say heat rises, we are talking about convection

Convection Convection currents drive many interesting physical systems Atmospheric convection Wind Thunderheads Oceanic convection: currents Solar convection Sunspots Flares Mantle convection: continental drift

Thermal Radiation Only thermal radiation transfers heat without any physical medium such as solid, liquid or gas. Essential idea: bodies transfer heat by emitting electromagnetic radiation. This is usually in the infrared region of the spectrum. Thermal imaging creates pictures of heat, rather than light.

Thermal Radiation Only thermal radiation transfers heat without any physical medium such as solid, liquid or gas. Example: The heat from a wood-burning stove can be felt from all the way across the room Photons carrying infrared energy can travel through space. When photons strike you, they are absorbed as heat. This process of thermal energy transfer is called thermal radiation. All energy reaching Earth from sun reaching the earth is electromagnetic radiation.

Blackbody Radiation A black-body absorbs all wavelengths. As it heats up it emits all wavelengths, called black-body radiation We see objects in color because they reflect (not absorb) that wavelength; black objects absorb all wavelengths

Blackbody Radiation Black body heated to incandescence shows two trends: The higher the temperature the greater the intensity at all wavelengths. The higher temperature the smaller the wavelength of the maximum intensity. Solids can be heated to incandescence (glowing); different temperatures will have different visible radiation

Blackbody Radiation Intensity Wavelength (nm) 1000 2000 3000 4000 5000 UV radiation IR radiation visible radiation Graph of different temperatures having different visible radiation

Wein’s Law Wein’s Displacement Law: tells us the wavelength of the maximum intensity max for blackbodies at temperature T in Kelvin 𝜆 𝑚𝑎𝑥 = 2.90× 10 −3 𝑇 We can find the max wavelength for blackbodies at temp T, as long as T is in Kelvins, using Wein’s Law max in nm T in K

Stefan-Boltzmann Law 𝑃=σ𝐴 𝑇 4 Stefan-Boltzmann Law: shows the relationship between the temperature of a black-body and the power emitted by the black-body’s surface area. 𝑃=σ𝐴 𝑇 4  = 5.6710-8 W m-2 K-4. Blackbody emits as much power as it absorbs; this law works for both emission and absorption problems

Practice Mercury has a radius of 2.50106 m. Its sunny side has a temperature of 400°C (673 K) and its shady side -200°C (73 K). Treating it like a black-body, find its power. Solution: Asphere = 4(2.50106)2 = 7.851013 m2 For T use TAVG = (673 + 73) / 2 = 373 K P = AT 4 = (5.6710-8)(7.851013)(373)4 = 8.621016 W No body is at absolute zero (K = 0), this law says that all bodies radiate

Solar Constant 𝑃 𝑠𝑢𝑛 =1361 W 𝑚 −2 Solar Constant: average amount of solar radiation received by the Earth's atmosphere, per unit area, when the Earth is at its mean distance from the Sun 𝑃 𝑠𝑢𝑛 =1361 W 𝑚 −2 AKA solar irradiance

Solar Constant Sun radiates energy at a rate of 3.901026 W. What is the rate at which energy from the sun reaches Earth if our orbital radius is 1.51011 m? 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦= 𝑃𝑜𝑤𝑒𝑟 𝐴𝑟𝑒𝑎 = 3.846× 10 26 𝑊 4𝜋 (1.5× 10 11 𝑚) 2 =1361 W 𝑚 −2 Using intensity = power/area, we find this value

Emissivity e= 𝑃 𝑏𝑜𝑑𝑦 𝑃 𝑏𝑙𝑎𝑐𝑘𝑏𝑜𝑑𝑦 = 𝑃 𝑏𝑜𝑑𝑦 [σ𝐴 𝑇 4 ] Emissivity, e: measure of an object's ability to emit infrared energy quantifies the emission and absorption properties of that body as compared to a blackbody of equal size 0≤𝑒≤1 e= 𝑃 𝑏𝑜𝑑𝑦 𝑃 𝑏𝑙𝑎𝑐𝑘𝑏𝑜𝑑𝑦 = 𝑃 𝑏𝑜𝑑𝑦 [σ𝐴 𝑇 4 ] Emissivity=ratio of power emitted by body, to power emitted by black-body of same size.  = 5.6710-8 W m-2 K-4.

Emissivity Can have a value from e=0 (perfectly shiny mirror) to e=1.0 (blackbody) If we know e, we can amend the Stefan-Boltzmann Law: A black-body is a perfect emitter/absorber (e = 1). 𝑃=𝑒σ𝐴 𝑇 4  = 5.6710-8 W m-2 K-4.

Albedo Albedo is a measure of how much light that hits a surface is reflected without being absorbed. PINCIDENT PSCATTERED heat If light strikes a mirror, nearly all of it will be scattered If light strikes a surface covered with lamp black, nearly all of it will be absorbed: Albedo is another word for reflectivity, but used in very specific case PINCIDENT PSCATTERED heat

Albedo 𝑎𝑙𝑏𝑒𝑑𝑜= 𝑃 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑃 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 We define albedo in terms of scattered and incident power: 𝑎𝑙𝑏𝑒𝑑𝑜= 𝑃 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑃 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 Ask: does snow have high or low albedo? Water? Answer: high (0.95); low (0.10) The mirror has an albedo of almost 1. The black-body has an albedo of almost 0.

Albedo Ocean water scatters little light (7%). Sea ice and snow scatter a lot of light (62% to 66%). Desert (36%). Landforms, vegetation, weather, and seasons affect a planet’s albedo.

Albedo Satellites are used to map out monthly albedos. White and gray areas: No data. Albedo White and gray areas: No data. Calculating albedo is complex; Earth’s mean yearly albedo is about 0.3 (or 30%); varies daily April, 2002, Terra satellite, NASA

Practice Assuming an albedo of 0.30, find, for Earth: the power of the sunlight received. Solution: Use I = 1380 W m-2 and I = P / A. The radius of Earth is r = 6.37106 m so its cross-sectional area is A = r2 = (6.37106)2 = 1.271014 m2. Ask: what can we use to find the power of the sunlight received? What area should we use?

Practice Assuming an albedo of 0.30, find, for Earth: the power of the sunlight received. Solution: Use I = 1380 W m-2 and I = P / A. An albedo of 0.30 means that 70% of the sunlight is absorbed (because 30% is scattered). P =(0.70)IA =(0.70)(1380)(1.271014) = 1.231017 W. Thus Earth intercepts energy from the sun at a rate of 1.231017 W. as: No data.

Practice Assuming an albedo of 0.30, find, for Earth: (b) the predicted temperature due to the sunlight reaching it. Solution: P = 1.231017 W. But power is distributed over the whole rotating planet, which has an area Asphere = 4r2. From Stefan-Boltzmann we have P = AT4 1.231017 = (5.6710-8)4(6.37106)2T4 T = 255 K (-18°C). as: No data. Use P from part (a) Ask: what area should we use? Do we have a way to find T?

Practice If the average temperature of Earth is 289 K, find its emissivity. Solution: a We determined that Earth absorbs energy from the sun at a rate of P = 1.231017 W. Since the temperature of Earth is relatively constant, we can assume it is radiating power at the same rate of PBODY = 1.231017 W. s: No data. Use P from previous example

Practice If the average temperature of Earth is 289 K, find its emissivity. Solution: a At 289 K the power radiated by a black-body of the same size as Earth is: PB-B = (5.6710-8)4(6.37106)22894 = 2.021017 W. Thus e = PBODY / PBLACK-BODY = 1.23 / 2.02 = 0.61. s: No data. Ask: How can we find power with the quantities that we have? What is the emissivity?

Wednesday April 12, 2017 Problem sets 8.1 and 8.2 due on Thursday Unit 8 test on Friday Last day to retake Unit 8 test is April 28th.

What is a definition for greenhouse effect? Think-Pair-Share: take 2 minutes to discuss w/partner a possible definition

How do greenhouse gases affect the climate? Greenhouse Effect How do greenhouse gases affect the climate? Search for ”phet greenhouse effect” https://phet.colorado.edu/en/simulation/greenhouse 2-3 students to a computer only!

Thursday April 13, 2017 Problem sets 8.1 and 8.2 due on today by end of day Unit 8 test tomorrow Last day to retake Unit 8 test is April 28th

Greenhouse Effect Gases in the atmosphere can absorb infrared radiation (heat). Remainder of incoming radiation then reaches the ground to either be scattered back into the atmosphere, or absorbed. Heated ground can then emit infrared radiation back into the atmosphere, which intercepts more energy on the way out. Result: atmosphere traps heat and causes the temperature of the planet to rise.

Greenhouse Effect

Greenhouse Gases Main gases that are particularly effective in absorbing infrared radiation are: methane (CH4), water vapor (H2O), carbon dioxide (CO2), and nitrous oxide (N2O). Note that there are both natural and man-made (anthropogenic) contributions and changes (flux).

Greenhouse Gases Shows correlation between increasing atmospheric concentration of CO2 & fossil fuel emissions of CO2. Trends in Atmospheric Concentrations and Anthropogenic Emissions of Carbon Dioxide

Greenhouse Gases Energy from photons can excite electrons in gases (absorption) Excited gases will de-excite and release photons Absorption and release of photons by the gases is called scattering Some of the scattered photons may be in the infrared region – able to be absorbed as internal energy and kinetic energy Scattering does not produce a net increase in heat energy in the atmosphere.

Greenhouse Gases Recall that heat energy can be stored in molecules as internal energy in the form of potential (the springs) and kinetic (the vibrations) (a) No absorption; (b) & (c) result in absorption

Greenhouse Gases Kinetic energy of the molecule that determines its temperature Three ways an extended molecule such as CO2 can store kinetic energy Vibration Translation Rotation All three modes can absorb and hold infrared radiation simultaneously Natural frequency of greenhouse gases is infrared region & they are prone to absorb such frequencies

Greenhouse Gases Different gases absorb different wavelengths

Energy Balance in Earth’s Surface/Atmosphere Average incident solar radiation: IAVG = 340 W m-2 Calculated intensity of solar radiation, not all surfaces will receive this intensity because Earth is a sphere Intensity decreases as move away from the equator

Energy Balance of Earth Atmosphere has two competing effects: Clouds raise the albedo Greenhouse gases make emissivity lower (yielding higher temperatures) The surface-atmosphere energy balance system is very complex; want equilibrium

NON-GREENHOUSE ATMOSPHERE Energy Balance in Earth’s Surface/Atmosphere Sankey diagram for a model of the earth without greenhouse gases. First draw Earth/atmosphere blocks Draw incoming solar energy (340 W m-2) Draw energy reflected by atmosphere (clouds) Draw energy reflected by ground (snow, etc.) Draw energy absorbed by ground (Q = mcT) Draw energy radiated by ground (P = AT4) 340 235 30 75 Q = heat capacity, c = specific heat GROUND NON-GREENHOUSE ATMOSPHERE 235 235

GREENHOUSE ATMOSPHERE Energy Balance in Earth’s Surface/Atmosphere Draw a Sankey diagram for a model of the earth with greenhouse gases. Put in the blocks (or sinks) first (Greenhouse gases) IIN (340 W m-2): IREFLECT(105 W m-2) IGND,abs (165 W m-2) IATMOS,abs (70 W m-2) IGND,radiate (390 W m-2) Iconvection (100 W m-2) IATM,radiate (195 W m-2) 340 195 40 30 75 Greenhouse gases (520) GROUND GREENHOUSE ATMOSPHERE 165 70 520 350 100 325 490

GREENHOUSE ATMOSPHERE Energy Balance in Earth’s Surface/Atmosphere Find the net power flow at the space / atmosphere interface and at the ground / atmosphere interface. Space/atmosphere interface: 340 IN. 195 + 75 + 30 + 40 = 340 OUT. Ground/atmosphere interface: 165 + 325 = 490 IN 350 + 40 + 100 = 490 OUT. Net = 0 W m-2. 340 75 30 165 195 70 350 40 325 100 520 GROUND GREENHOUSE ATMOSPHERE 490 This model also shows stable temperatures. Net = 0 W m-2.

GREENHOUSE ATMOSPHERE Energy Balance in Earth’s Surface/Atmosphere Find the net power flow at the ground/atmosphere interface. 340 75 30 165 195 70 360 20 335 110 540 GROUND GREENHOUSE ATMOSPHERE 500 We have: IIN = 165 + 335 = 500 W m-2. IOUT = 360 + 20 + 110 = 490 W m-2. This model shows increasing ground temperatures The net power flow is thus IIN – IOUT = 500 – 490 = + 10 W m-2