Energy, Society, and the Environment Unit IV: How Do Power Plants Work? Combustion and Heat Engines

Outline Combustion: Energy Generation and Pollutants 1st Law of Thermodynamics 2nd Law of Thermodynamics Efficiency

Combustion What do we do with fossil fuels: burn them Fuels Combustion: impacts Fuels Balancing combustion chemical equations Combustion products

What happens in combustion? Fuel + oxidizer -> Products + light + heat Combustion, in its simplest form, e.g: methane CH4 + 2O2  CO2 + 2H20 A clean reaction, except for the issue of carbon dioxide and the global climate This idealized reaction takes place in an ‘atmosphere’ (oxygen) free of impurities

C6H12O6 + O2  CO2 + H2O C6H12O6 + O2  6CO2 + H2O How much CO2 is produced when 1 ton of cellulose (C6H12O6) is burned? (Cellulose is the primary component of plant matter, wood etc) Write the equation: C6H12O6 + O2  CO2 + H2O Balance first the carbon: C6H12O6 + O2  6CO2 + H2O Then the hydrogen: C6H12O6 + O2  6CO2 + 6H2O Last, the oxygen (because you can change the oxygen without altering other elements): C6H12O6 + O2  6CO2 + 6H2O, So, 6 + x = 18  x = 12, or 6 O2 The balanced equation is: C6H12O6 + 6O2  6CO2 + 6H2O

How much CO2 is produced when 1 metric ton of wood is burned, continued ... 1 molecule of cellulose produces 6 molecules of CO2 How much do these weigh?

Remember your atoms C = 6 p + 6 n = 12 O = 8 p + 8 n = 16 (atomic mass unit, OR 1 mole of C is 12 grams)

How much CO2 is produced when 1 metric ton of wood is burned, continued ... 1 molecule of cellulose produces 6 molecules of CO2 How much do these weigh? C6H12O6 = (6 x 12) + (12 x 1) + (6 x16) = 180 grams (in 1 mole) 6 CO2 = 6 x [(1x12) + (2 x16)] = 264 grams (from 1 mole of cellulose) So, from C6H12O6  6CO2 + 6H2O, we see that: 180 grams cellulose  264 grams CO2 106 grams cellulose  ?? grams CO2 ??

Simple combustion equation, but put it in air Example: methane reacts with air CH4 + (O2 + 3. 76N2) -> CO2 + H2O + N2 (this is termed the ‘unbalanced’ version) CH4 + 2(O2 + 3. 76N2) -> CO2 + 2H2O + 7.52N2 (balanced: exactly the correct amount of oxidizer to convert all C to CO2, and all H to H2O) Note: Air is 78% nitrogen, 21% oxygen, + other stuff

Real Combustion If combustion occurs without complete oxidation, we get instead: CH4 + O2 + N2  mostly (CO2 + 2H20 +N2) + traces (CO + HC + NO...) This can occur when: temperature too low insufficient O combustion too rapid poor mixing of fuel and air, etc. ...

Real, Real Combustion At higher temperatures, N reacts with O: air(N2 +O2) + heat  NOx (nitrogen oxides, x can be 1 or 2) So much for pure fuels, now add impurities: enter N, S, metals and ash (non-combustibles) What we really get: Fuel (C, H, N, S, ash) + air (N2 +O2)  (CO2, H2O, CO, NOx, SOx, VOCS, particulates) + ash Volatile Organic Compounds: VOCs

Real, Real Combustion and Emissions NOx + VOCs = SMOG CO2, NOx, SOx + oxidants + water = ACID RAIN Particulates (especially ultrafine): • Create inflammatory response • Affect heart rate variability • Inducing cellular damage • May be associated with premature death

Emissions Controls Clean Air Act requires EPA to set National Ambient Air Quality Standards (NAAQS) for seven pollutants (referred to as criteria pollutants): carbon monoxide (CO), lead (Pb), ozone (ground level), nitrogen oxides (NOx), particulate matter (PM2.5 and PM10), sulfur oxides (SO2). Last amended in 1990.

National Ambient Air Quality Standards

Criteria Pollutants & the Clean Air Act POLLUTANT Emissions Air Concentration ‘83 - ‘02 ‘93 - ‘02 ‘83 - ‘02 ‘93 - ‘02 CO - 41% - 21% - 65% - 42% Pb - 93 % - 5 % - 94% - 57% O3 [1 hr] - 40 % - 25% - 22% - 2% PM10 - 34% - 22% - - 10% PM2.5 - - 17% - - 8% SO2 - 33% - 31% -34% - 39% NOx - 15% - 9 % - 2% +.5%

Announcements 2/16 No Class THIS Friday (note the change) Homework #3 posted today on D2L, due next Monday

POWER PLANTS are Heat Engines Using Heat to Do Work (a bit of Thermodynamics)

What use is thermodynamics? For different energy sources compare: Efficiency Amount of fuel needed Pollution produced Use as a tool for improving energy systems Analyze each part of a power plant (pumps, turbine, heat exchangers, etc.) Analyze alternative energy scenarios Ethanol, biodiesel, hydrogen

Some Questions That Can Be Answered How many fewer power plants would need to be built in Arizona if we increased our efficiency by 10% by 2020? How much could SO2, CO2, and other pollutants be reduced by that efficiency increase?

Coal Fired Steam Power Plant

1st Law of Thermodynamics Water in Water out Imagine you have a bathtub – water coming in is the Ein, water going out the drain is Eout, if Ein is greater than Eout, then the bathtub will fill up, if Eout is greater than Ein, then the bathtub will empty. What is Ein ?– Fuel, Sunlight, energy stored in chemical bonds, energy as photons in sunlight, electrons in a electrical device.

1st Law of Thermodynamics Conservation of Energy Principle Esystem = Ein - Eout Ein Esystem Eout Imagine you have a bathtub – water coming in is the Ein, water going out the drain is Eout, if Ein is greater than Eout, then the bathtub will fill up, if Eout is greater than Ein, then the bathtub will empty. What is Ein ?– Fuel, Sunlight, energy stored in chemical bonds, energy as photons in sunlight, electrons in a electrical device. What is Eout? Heat, Other chemicals (like in the case of combustion), Energy balance can be used on almost anything: coffee cup, car engine, later in the class you will use the first law of thermo to analyze renewable energy systems (solar, hydro, wind), also refrigeration systems and to analyze the energy efficiency of buildings – SO it’s very important. Although the math is not very difficult – conceptually it takes practice to understand how to account for the energy. But, let’s look at an energy balance for the entire earth

Earth Energy Balance Energy from Sun EARTH Energy Stored Let’s do an energy balance about the earth. Where do we get our energy? The Sun. So our Ein is the energy we get from the sun. We store energy on the earth in a lot of ways (heat in the atmosphere, kinetic energy of wind, plants store energy, etc.) The Elost: is energy radiated into space. Average energy/time from sun = 350 W/m^2 That’s 20e16 W to the earth Assuming an average power plant ~ 1000MW – that’s 200million power plants Etotalearth = 18e16W Ereflectedspace = 53e15W Eabs in atm = 44e15W Energy Radiated to Space

Earth Energy Balance Energy from Sun EARTH Energy Stored Let’s do an energy balance about the earth. Where do we get our energy? The Sun. So our Ein is the energy we get from the sun. We store energy on the earth in a lot of ways (heat in the atmosphere, kinetic energy of wind, plants store energy, etc.) The Elost: is energy radiated into space. Average energy/time from sun = 350 W/m^2 That’s 20e16 W to the earth Assuming an average power plant ~ 1000MW – that’s 200million power plants Etotalearth = 18e16W Ereflectedspace = 53e15W Eabs in atm = 44e15W Energy Radiated to Space Estored = Ewind + Eplants + Eheat + etc. = Esun - Elost

Steady-State Condition Under steady-state conditions, there is no change in the stored energy of the system. Esystem = 0 = Ein – Eout or Ein = Eout Or, the water stays at the same level in the tub. The systems discussed in this class will be steady-state systems. Examples are: Natural gas power plant, Coal fired power plant, nuclear power plant Example of a non-steady device: Battery – if you are using a battery in calculator or Cd player, you have an Eout, but no Ein (while you are using it) – then Esystem, or the energy stored in the battery, would reduce.

Energy Balance for a Power Plant 1000 MJ Energy from Fuel Efuel = 1000 MJ This 1000MJ coming in could be as coal, uranium, natural gas, etc. How much fuel is 1000 MJ = 10E3 x 10E6 = 10E9 J Energy in gasoline = 115,000 BTU/Gal = 1.2 10E8J So 1000 MJ = 8 gals gas (1/2 tank full of a car).

Energy Balance for a Power Plant 1000MJ 350MJ Energy from Fuel Useful Energy (Work) Efuel = 1000 MJ Euseful = 350 MJ As I said, we will be dealing with steady-state systems: therefore, if 350MJ of the total 1000MJ comes out as useful work --- then there must be some left over (650MJ) Efuel = Euseful + Ewaste

Energy Balance for a Power Plant Useful Energy (Work) Power Plant 1000MJ 350MJ 650MJ Energy from Fuel Wasted Energy Efuel = 1000 MJ Euseful = 350 MJ Ewaste = 650 MJ Efuel = Euseful + Ewaste 1000 MJ = 350 MJ + 650 MJ

Power Balance for a Power Plant 1 W = 1 J/s Power Plant 1000MW 350MW 650MW Pin = Pout Pout = Pout 1 + P out 2 Pfuel = Puseful + Pwaste

Power Balance for a Power Plant Pin = 1000 MW Pout = 350 MW + 650MW Power Plant 1000MW 350MW 650MW Pin = Pout Pfuel = Puseful + Pwaste

Power Plant

Parts of a power plant Energy Source Heat Generated Work Produced Combustion: Coal, natural gas, oil, biomass, garbage Heat Generated Boiler: Coal burned, heats water, creates steam Combustion Chamber: CH4 burned, hot gases Work Produced High pressure steam or hot gases turn turbine blades Wasted Heat Removed/Exhaust Treatment hot water dumped into a body of water or cooling tower exhaust gases dumped into atmosphere Some waste heat can be recovered The energy balance can be used for any system (combustion, nuclear, renewables)

Turbine:

Turbine: Blades

Cooling Tower

Connection to Environment Fuel Side: Mining, drilling, transporting Waste Heat Side: Increase temperature of body of water Affect fish, algae blooms, etc. Pollutants in waste heat stream Air pollutants Pollutants in waste water stream Environmental justice Location, impact & management of power plants

Heat Engine A device that converts heat into mechanical energy Used to approximate thermal systems

Heat Engine Energy Balance: Qhot = Wnet + Qcold High Temp. Source of Heat Qhot Heat Engine Wnet Qcold Low Temp. Sink of Waste Heat Energy Balance: Qhot = Wnet + Qcold

Definitions High Temp Source of Heat: This is the source of energy that drives the power plant (heat of combustion, geothermal heat source, nuclear reactor, etc.). Qhot: This is the heat transferred from the hot source. Heat Engine: This includes the working parts of the power plant (including pumps, turbines, heat exchangers, condensers, etc.). Wnet: This is the net amount of work that exits the power plant. A turbine generates energy, but the pumps and compressors use energy. Qcold: This is the rejected or waste heat, which is dumped to a cold source (i.e. river, atmospheric air, lake, etc.). Low Temp Sink of Waste Heat: This is the reservoir (river, air, lake, etc.) that the waste heat is dumped into (often goes through a cooling tower). Low Temp Sink is where most of our problems lie (although there are definitely problems with drilling, mining, etc. for fuels).

Efficiency (Several names: I =1st Law, Actual, or Thermal Efficiency) what you want work done Efficiency = = what you pay for energy put into the system (Several names: I =1st Law, Actual, or Thermal Efficiency) Qhot  = Wnet = Qhot-Qcold

Energy Balance for a Power Plant 1000MJ 350MJ 650MJ Energy from Fuel Useful Energy (Work) Wasted Energy Ein=1000 MJ from fuel Eout=350 MJ useful energy +650 MJ wasted energy  = Wnet/Qhot  = 350 MJ / 1000 MJ = 0.35 = 35% The thermal efficiency is a great way to compare various systems – it’s not the only way (economics, pollution, land space needed, etc.). For example, thinking about other thermal systems – the diesel engine is more efficient than a regulate gasoline engine. Therefore, they put emit less CO2/mile than gasoline powered cars do – so, why don’t we use diesels exclusively then? They emit about 100times the soot particles (black stuff coming out of old buses and semi-trucks) and much more NOx (which causes photochemical smog). But, they do use less fuel – which means less drilling. Normal power plants have efficiencies of around 30%. With some extra work and money, that efficiency can be increased to 40%. So, is 50, 60 , 70 % possible. Can we get 100% efficiency out of a power plant – but we’re just to money grubbing to do it? Note: Qhot = Qin = Qfuel

Second Law of Thermodynamics Order tends to disorder, concentration tends to chaos No process can occur that only transfers energy from a cold body to a hot body (heat must flow from hot to cold) No process can occur that converts a given quantity of thermal energy into an equal quantity of mechanical work (always some degradation-always some wasted energy)

2nd Law of Thermodynamics, alternate statements: states in which direction a process can take place heat does not flow spontaneously from a cold to a hot body heat cannot be transformed completely into mechanical work it is impossible to construct an operational perpetual motion machine

2nd Law of Thermodynamics and Carnot Efficiency 2nd Law: Heat cannot be converted to work without creating some waste heat. What does this mean for a heat engine? Wnet < Qhot ALWAYS In other words, efficiency is always less than 100%. Well, how much less?

2nd Law of Thermodynamics and Carnot Efficiency Carnot Efficiency: The net work produced and the heat into the system only depend on temperatures. No thermal system can be more efficient than the Carnot efficiency. Qhot - Qcold Qcold Tcold = 1 - = 1 -  c = Qhot Qhot Thot This is the best you can achieve (under ideal conditions)

2nd Law of Thermodynamics and Carnot Efficiency Important: Temperatures must be in units of Kelvin. K = ºC + 273.15 1 ºC = 1.8 F Example: In the power plant we considered earlier, Thot = 900 K (very hot steam) and Tcold = 300 K (room temperature). What is its Carnot efficiency?  c = 1 - Tcold / Thot = 1 - (300/900) = 0.67 = 67% If this plant were an ideal heat engine, 33% of the energy would be lost (to nearby water or to atmosphere via cooling tower)

Efficiencies of the Power Plant Ein=1000 MJ from fuel Eout=350 MJ useful energy +650 MJ wasted energy Power Plant 1000MJ 350MJ 650MJ Energy from Fuel Useful Energy (Work) Wasted Energy Thigh=900 K Tlow =300 K I= Wnet/Qhigh=350 MJ / 1000 MJ = 0.35 = 35% Reality  c = 1 - Tcold / Thot = 1 - (300/900) = 0.67 = 67% Ideal

Comparison of Efficiencies Type Th (K) Tc (K) Carnot Eff. Actual Coal 800 300 62.5% 35% Nuclear 1200 75% Geo- Thermal 525 350 33% 16%

Waste Energy Note: When the working fluid reaches Tlow no more energy can be used – although the working fluid still contains energy (typically > 60% of Efuel)

Heat Pollution The circulation rate of cooling water in a typical 700 MW coal-fired power plant with a cooling tower amounts to about 71,600 cubic metres an hour (315,000 U.S. gallons per minute) and the circulating water requires a supply water make-up rate of about 5 percent (i.e., 3,600 cubic metres an hour). If that same plant had no cooling tower and used once-through cooling water, it would require about 100,000 cubic metres an hour and that amount of water would have to be continuously returned to the ocean, lake or river from which it was obtained and continuously re-supplied to the plant. Discharging such large amounts of hot water may raise the temperature of the receiving river or lake to an unacceptable level for the local ecosystem.

Cogeneration Use waste energy in another application e.g., Heaters in cars use waste energy from the engine Easy uses: Space and water heating, especially if small power plants can be built near urban areas: increase total efficiency from ~ 35% to ~ 70% Take the Qcold through another cycle with Qcolder Great potential in reducing fuel use and environmental impacts (by bringing Qcolder as close to atmospheric temperature as possible)

Gasoline and Diesel Engines Gasoline engines in our cars are also heat engines called internal combustion engines Fuel combined with gas in a closed chamber and ignited: combustion proceeds at a very high temperature and rate Typically 25% efficiency (mechanical energy/combustion energy), lots of nasty pollutants Diesel engine also an internal combustion engine Fuel and air mixed differently than gasoline engine Air compressed for heating, no electric spark Combustion temperature even higher than gasoline engine Efficiency > 30% Less CO emission, more NOx and particulate emissions