Primary energy sources PRACTICE: Describe the origin of fossil fuels. Sun makes biomass through photosynthesis. Biomass gathers and grows over time. Biomass buried under great pressure and heat. Biomass becomes coal, oil and natural gas over eons. Coal, oil and natural gas are extracted. Coal, oil and natural gas are used as fuels. Primary energy sources PRACTICE: Proved reserves are resources that we are sure

we can obtain. Production means actual reserves that have been obtained and placed on the market. Use the table to estimate how long oil reserves might last. SOLUTION: Expectancy = Reserves / Production = 1200.7109 / 29.6109 = 40.6 years.

This is an estimate because both numbers are subject to yearly change. Specific energy and energy density of fuel sources Specific energy ESP is how much energy (J) you can get per unit mass (kg) from a fuel. Its units are J kg-1. Energy density ED is how much energy (J) you can get per unit volume (m 3). Its units are J m-3. EXAMPLE: Fission of each uranium-235 produces 3.510-11 J of energy. The density of U-235 is 1.8104 kg m-3. Calculate the ESP and the ED of U-235. SOLUTION: (a) m = (235 u)(1.66110-27 kg u-1) = 3.9010-25 kg.

ESP = 3.510-11 J / 3.9010-25 kg = 9.01013 J kg-1. (b) ED = (9.01013 J kg-1)(1.8104 kg m-3) = 1.61018 J m-3. Solving specific energy and energy density problems PRACTICE: If coal is transported in rail cars having a capacity of 1.5 metric tons, how many cars per day must supply the power plant of the previous example?

SOLUTION: From the previous example we calculated that we need 320000 kg of coal per day. Since a metric ton is 1000 kg, we have (320000 kg ) / (1.5 1000 kg / car) or 213 cars d-1! Solving specific energy and energy density problems PRACTICE: If a nuclear power plant powered by uranium-235 has the same output and the same

efficiency as the coal-fired plant of the previous example, how many kilograms of nuclear fuel will it burn per day? Per year? SOLUTION: From the previous example we calculated that we need 10368000 MJ of energy per day. Thus (10368000 MJ) / (1 kg / 90000000 MJ) or 0.1152 kg d-1! This is 42 kg y-1. Solving specific energy and energy density problems

PRACTICE: Explain why it is advantageous to have a submarine which is nuclear powered, as opposed to diesel. SOLUTION: There are two main reasons: (1) Nuclear reactors dont use oxygen, so the sub can stay under water for months at a time. (2) Nuclear fuel is extremely compact for the amount of energy it contains. Thus the sub can cruise far before refill.

Sankey diagrams In the balloon-water electricity engine name all of the energy conversions that occur. SOLUTION: 1 - Coal (or wood) becomes heat. 2 - Heat expands air in balloon to decrease its density. 3 - Buoyant force causes balloon to rise, increasing potential energy of the water. 4 - Falling water converts potential to kinetic energy. 5 - Moving water collides with fan blades and imparts kinetic energy to them. 6 - Moving fan blades generate electricity. In the balloon-water electricity engine name all of the energy losses that occur.

SOLUTION: 1 - Coal loses heat to air outside balloon. 2 - Hot balloon loses heat to outside air. 3 - Excess hot air escapes bottom of balloon. 4 - Falling water heats up air as it falls. 5 - Falling water continues past blades colliding with the ground. 6 - Internal friction of fan shaft impedes rotation. 7 - Internal resistance of wiring generates I 2R heat loss. Solving problems relevant to energy transformations This is a fission reaction.

Heat and some kinetic energy of the product neutrons. Looking at the reaction we see that one neutron initiates one fission but each reaction produces two neutrons. Thus the fission process can produce a self-sustaining chain reaction. Product neutrons are too fast to cause fissioning. The moderator slows them down to the right speed. The control rods absorb neutrons produced by fission. This allows a reactor to just maintain its reaction rate at the selfsustaining level, rather than becoming a dangerous chain reaction. Reactor coolant from the pile is circulated through a heat exchanger. The heat exchanger heats up water to boiling to run a steam turbine.

The steam turbine turns a generator to produce electricity. The reactants: U-235 and 1n: 218950 + 940 = 219890 MeV. The products: Xe-144, Sr 90 and 2n: 134080 + 83749 + 2(940) = 219709 MeV. The mass defect is 219890 219709 = 181 MeV. mass loss (This is the energy released through mass loss according to E = mc2). Solving problems relevant to energy transformations PRACTICE: Since the air is still moving

after passing through the rotor area, obviously not all of the kinetic energy can be used. The maximum theoretical efficiency of a wind turbine is about 60%. Given a turbine having a blade length of 12 m, a wind speed of 15 ms-1 and an efficiency of 45%, find the power output. The density of air is = 1.2 kg m-3. SOLUTION: A = r 2 = (122) = 452 m2. Power = (0.45)(1/2)Av3 = (0.45)(1/2)4521.2153

= 411885 W = 410 kW. Solving problems relevant to energy transformations A = r2 = (7.52) = 177. Powerin = (1/2)Av3 = (1/2)(177)(1.2)93 = 77420 W Powerthru = (1/2)Av3 = (1/2)(177)(2.2)53 = 24338 W Powerext = powerin - powerthru Powerext = 77420 24338 = 53082 W = 53 kW. Pout = (efficiency)Pextracted Pout = (0.72)(53 kW) = 38 kW

Solving problems relevant to energy transformations Power is proportional to v3. The wind speed had doubled (from 6 to 12 m s-1). Thus the power should increase by 23 = 8 times. Output will now be 8(5 kW) = 40 kW. Solving problems relevant to energy transformations Wind power doesnt produce greenhouse gas. Wind power is a renewable resource.

Wind depends on the weather. 2 GW / 0.8 MW = 2500 turbines required! Solving problems relevant to energy transformations PRACTICE: A reservoir for a hydroelectric dam is shown. (a) Calculate the potential energy yield. The total volume of water is V = 1700(2500)(50) = 2.125108 m3. The mass of the water is m = V = (1000 kg m-3)(2.125108 m3) = 2.1251011 kg. The average water height is h = 75 + 50 / 2 = 100 m. The potential energy yield will then be EP = mgh = (2.1251011)(10)(100) = 2.1251014 J. (b) If the water flow rate is 25 m3 per second, what is the power provided by the moving water?

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) = 2.5107 J. Since power is measured in W (or J s-1) the power provided is 2.5107 W = 25 MW. (c) If the water is not replenished, how long can this reservoir produce power at this rate? The total volume of water is V = 1700(2500)(50) = 2.125108 m3. The volume flow rate is 25 m3 s-1. Thus the time is given by t = (2.125108 m3) / (25 m3 s-1) = 8.5106 s = 2361 h = 100 d. Describing solar energy systems

PRACTICE: The sun radiates energy at a rate of 3.901026 W. What is the rate at which energy from the sun reaches earth if our orbital radius is 1.51011 m? The surface area of a sphere is A = 4r2. Recall that intensity is the rate at which energy is being gained per unit area. I = P/ [4r2] = 3.901026 / [4(1.51011)2] I = 1380 W m-2. This is 1380 J/s per m2. Describing solar energy systems

PRACTICE: Explain why the solar intensity is different for different parts of the earth. SOLUTION: The following diagram shows how the same intensity is spread out over more area the higher the latitude. 1380 W/m2 Solving problems relevant to energy transformations Solving problems relevant to energy transformations EXAMPLE: A photovoltaic cell of 1.00 cm2 and 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? 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. (b) If the cell is rated at 0.500 V, what is its current output? P = IV so that I = P/ V = 0.0131/.5 = 0.0262 A. (c) If ten of these cells are placed in series, what will the voltage and the current be? 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. (d) If ten of these cells are placed in parallel, what will the voltage and the current be? 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. Solving problems relevant to energy transformations 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? Pout = 0.0131 W/ cell. Thus (100 W) / (0.0131 W/ cell) = 7630 cells! Obviously, a solar-powered calculator doesnt need as much power to operate!

Solving problems relevant to energy transformations Efficiency = Pout / Pin so that Pin = Pout / Efficiency = 4710-3 / 0.08 = 0.5875 W. Thus I = Pin / A = 0.5875 W / 6.510-4 m2 I = 900 W m-2 = 0.90 kW m-2. Solving problems relevant to energy transformations Cloud cover variation is one reason. Season is another reason.

Do NOT use latitude as a reason for this question. The question specifically states that it is concerned with only a particular region having a variation, so its latitude does not change. Solving problems relevant to energy transformations PRACTICE: Draw a Sankey diagram for a photovoltaic cell having an efficiency of 18%. USEABLE ELECTRICITY ENERGY IN INCIDENT WASTED HEAT SUNLIGHT

Note the approximate proportions. Perhaps, like the nuclear reactor, we could also show the energy needed to process the materials that go into making a photovoltaic cell. Because of the tilt of Earths axis southern exposures get more sun in the northern hemisphere, and northern exposures get more sun in the southern hemisphere. Solving problems relevant to energy transformations

From Topic 3 the energy needed for T = 25 K is Q = mcT = 140(4200)(25) = 1.47107 J. But POUT = Q / t = 1.47107 J / (6 h)(3600 s / h) so that Pout = 681 W. Since Efficiency = POUT / PIN = 0.35 then PIN = POUT / Eff. = 681/0.35 = 1944 W. From I = PIN / A we get A = PIN / I so that A = 1944 / 840 = 2.3 m2. Topic 8: Energy production 8.2 Thermal energy transfer Conduction, convection and thermal radiation EXAMPLE: A brick has

2.50 m a thermal conductivity 0.500 m of k = 0.710 J / msC, 0.750 m 20C 100C and the dimensions shown. Two heat sources are placed in contact with the ends of the brick, as shown. Find the rate at which heat energy is conducted through the brick. SOLUTION: A = 0.7500.500 = 0.375 m2. Why?

Then Q / t = kAT / d = 0.7100.375(100 20) / 2.50 = 8.52 J s-1. Topic 8: Energy production 8.2 Thermal energy transfer Solving problems involving Wiens displacement law Higher temperature = higher intensity so Y is above X. Higher

temperature = smaller wavelength at peak Topic 8: Energy production 8.2 Thermal energy transfer Solving problems involving the Stefan-Boltzmann law EXAMPLE: Mercury has a radius of 2.50106 m. Its sunny side has a temperature of 400C (673 K) and its shady side -200C (73 K). Treating it like a black-body, find its power. SOLUTION: Asphere = 4(2.50106)2 = 7.851013 m2. For T use TAVG = (673 + 73) / 2 = 373 K

P = AT 4 = (5.6710-8)(7.851013)3734 = 8.621016 W. FYI Since no body is at absolute zero (K = 0) it follows from the Stefan-Boltzmann law that all bodies radiate. Be sure that T is in Kelvin, not Celsius. Topic 8: Energy production 8.2 Thermal energy transfer The solar constant PRACTICE: The sun radiates energy at a rate of 3.901026 W. What is the rate at which energy from the sun reaches Earth if our orbital radius is 1.51011 m? SOLUTION: ASPHERE = 4r2.

Recall that intensity is the rate at which energy is being gained per unit area. intensity = power / A intensity Then I = P / [ 4r2 ] = 3.901026 / [ 4(1.51011)2 ] = 1380 W m-2. This value is called the solar constant. Psun = 1380 W m-2 the solar constant Topic 8: Energy production 8.2 Thermal energy transfer

The solar constant PRACTICE: Explain why the solar intensity is different for different latitudes. SOLUTION: The following diagram shows how the same intensity is spread out over more area the higher the latitude. 1380 W/m2 Topic 8: Energy production 8.2 Thermal energy transfer The solar constant

PRACTICE: The sun radiates energy at a rate of 3.901026 W. What is the rate at which energy from the sun reaches Jupiter if its orbital radius is 7.81011 m? (This is Jupiters solar constant.) SOLUTION: Use I = P / [ 4r2 ] I = P / [ 4r2 ] = 3.901026 / [ 4(7.81011)2 ] I = 51 W m-2. This is 51 J / s per m2. Topic 8: Energy production 8.2 Thermal energy transfer Solving problems involving albedo EXAMPLE: Assuming an albedo of 0.30,

find, for Earth: (a) the power of the sunlight received. SOLUTION: Use I = 1380 W m-2 and I = P / A. The radius of Earth is r = 6.37106 m so its cross-sectional area is A = r2 = (6.37106)2 = 1.271014 m2. An albedo of 0.30 means that 70% of the sunlight is absorbed (because 30% is scattered). Thus P =(0.70)IA =(0.70)(1380)(1.271014) = 1.231017 W. Thus Earth intercepts energy from the sun at a rate of 1.231017 W. Topic 8: Energy production

8.2 Thermal energy transfer Solving problems involving albedo EXAMPLE: Assuming an albedo of 0.30, find, for Earth: (b) the predicted temperature due to the sunlight reaching it. SOLUTION: From the previous example P = 1.231017 W. But this power is distributed over the whole rotating planet, which has an area Asphere = 4r2. From Stefan-Boltzmann we have P = AT4 1.231017 = (5.6710-8)4(6.37106)2T4

T = 255 K (-18C). Topic 8: Energy production 8.2 Thermal energy transfer Solving problems involving emissivity and Earths average temperature EXAMPLE: If the average temperature of Earth is 289 K, find its emissivity. SOLUTION: We determined that Earth absorbs energy from the sun at a rate of P = 1.231017 W. Since the temperature of Earth is relatively constant, we can assume it is radiating power at the same rate of PBODY = 1.231017 W.

At 289 K the power radiated by a black-body of the same size as Earth is: PB-B = (5.6710-8)4(6.37106)22894 = 2.021017 W. Thus e = PBODY / PBLACK-BODY = 1.23 / 2.02 = 0.61. Topic 8: Energy production 8.2 Thermal energy transfer Solving problems involving albedo albedo = PSCATTERED / PINCIDENT PINCIDENT = 340 W m-2 PSCATTERED = 100 W m-2 albedo = 100 / 340

Topic 8: Energy production 8.2 Thermal energy transfer Absorption of infrared radiation by greenhouse gases PRACTICE: Consider the absorption graphs below: (a) Which portion of the electromagnetic spectrum is represented? (b) Which greenhouse gas contributes the most to the greenhouse effect? (c) Which gas is the least significant contributor? SOLUTION:

(a) Infrared (heat). (b) Water vapor! (c) Oxygen and ozone. Topic 8: Energy production 8.2 Thermal energy transfer Energy balance in Earth surface / atmosphere system P / A = (1)(5.6710-8)2884 = 390 W m-2 (amount radiated) P / A = 140 W m-2 (amount absorbed from atmosphere) 390 140 = 250 W m-2 (amount absorbed from sun). Topic 8: Energy production

8.2 Thermal energy transfer Energy balance in Earth surface / atmosphere system P / A = (0.72)(5.6710-8)(242 + 6)4 = 154 W m-2. Topic 8: Energy production 8.2 Thermal energy transfer Energy balance in Earth surface / atmosphere system Amount absorbed from sun 250 W m-2 is still same. P / A = 154 W m-2 + 250 W m-2 = 404 W m-2. Topic 8: Energy production

8.2 Thermal energy transfer Energy balance in Earth surface / atmosphere system From Stefan-Boltzmann we have P = AT4. Thus P / A = T4 or 404 = 5.6710-8T4, and T = 291 K, or an increase of 291 288 = 3 K.