ER100/200, Pub Pol 184/284 Energy Toolkit III:

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1 ER100/200, Pub Pol 184/284 Energy Toolkit III: Energy Thermodynamics Lectures 6 & &

2 Outline What can the energy analyst do with Thermodynamics? 1 st Law of Thermodynamics 2 nd Law of Thermodynamics Carnot System Efficiency Thermodynamic Diagrams

3 Efficiency is important, but how does one measure it? Thermodynamic efficiency Practical measures Vehicles: EPA Mileage for vehicles (driving cycle), MPG, MPGe Solar cells: In addition to relating material to the performance of the solar cell itself, the efficiency depends on the spectrum and intensity of the incident sunlight and the temperature of the solar cell. Economic efficiencies of technology/fuel combinations

4 U.S. to Bring Gas Mileage Rules to Era of Hybrids NY Times 8/15/13: The Environmental Protection Agency said it would update its labeling rules which date to the 1970s to resolve disparities among the growing number of hybrid and electric vehicles on the market.

5 Thermo laws, simplified Zeroth: "You must play the game. First: "You can't win. Second: "You can't break even. Third: "You can't quit the game." Our focus 5

6 What can we use thermodynamics for? 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

7 What can t you do with thermodynamics? Thermodynamics provides limits (overall fuel and emissions totals), but Kinetics tells you how fast something will Kinetics tells you how fast something will occur (or if it will occur!) This matters when secondary pollutants formed, e.g. NO x, and what other reactions will take place.

8 Some Questions That Can Be Answered How much coal must be mined to meet all of the U.S. s electricity needs? How much coal could be saved by increasing the average power plant efficiency by 5%? Can we increase the plant efficiency? How much could SO 2, CO 2, and other pollutants be reduced by that efficiency increase? How many fewer power plants would need to be built in California if we increased our efficiency by 10% by 2020?

9 Coal Fired Steam Power Plant Joliet, IL

10 1st Law of Thermodynamics Water in Water out

11 1st Law of Thermodynamics Conservation of Energy Principle E in E system E out E system = E in - E out

Thermal energy can be increased within a

12 1st Law of Thermodynamics Law of energy conservation applied to a thermal system Thermal energy can change form and location, but it cannot be created or destroyed (unless we are discussing nuclear reactions, E=mc 2 ). Thermal energy can be increased within a system by adding thermal energy (heat) or by performing work in a system.

thermal energy through friction (the pump becomes hot) The total increase in internal

13 1st Law of Thermodynamics Example: Using a bicycle pump Pumping the handle results in what? Applying mechanical energy into the system Mechanical energy is converted into thermal energy through friction (the pump becomes hot) The total increase in internal energy of the system is equal to what? The applied mechanical energy istockphoto.com

14 1st Law of Thermodynamics

15 Earth Energy Balance Energy from Sun EARTH Energy Stored Energy Radiated to Space

16 Earth Energy Balance Energy from Sun EARTH Energy Stored Energy Radiated to Space E stored = E wind + E plants + E heat + etc. = E sun - E lost

17 Steady-State Condition Under steady-state conditions, there is no change in the stored energy of the system. E system = 0 = E in E out or E in = E out

18 Energy Balance for a Power Plant E fuel = 1000 MJ 1000MJ Energy from Fuel Power Plant

19 Energy Balance for a Power Plant E fuel = 1000 MJ 1000MJ Energy from Fuel Power Plant 350MJ Useful Energy (Work) E useful = 350 MJ E fuel = E useful + E waste

20 Energy Balance for a Power Plant E fuel = 1000 MJ 1000MJ Energy from Fuel Power Plant 350MJ Useful Energy (Work) E useful = 350 MJ E waste = 650 MJ waste 650MJ Wasted Energy E fuel = E useful + E waste 1000 MJ = 350 MJ MJ

21 Power Balance for a Power Plant 1 W = 1 J/s 1000MW Power Plant 350MW 650MW P in = P out P out = P out 1 + P out 2 P fuel = P useful + P waste

22 Power Balance for a Power Plant P in = 1000 MW P out = 350 MW + 650MW 1000MW Power Plant 350MW 650MW P fuel = P useful + P waste

23 Components of a power plant Energy Source Combustion: Coal, natural gas, oil, biomass, garbage Heat Generated Boiler: Coal burned, heats water, creates steam Combustion Chamber: CH 4 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 exhaust gases dumped into atmosphere Some waste heat can be recovered

24 Power Plant

25 Turbine:

26 Turbine: Blades

27 Cooling Tower

28 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

29 Heat Engine Used to approximate thermal systems High Temp. Source of Heat Q high Heat Engine W net Q low Low Temp. Sink of Waste Heat Energy Balance: Q high = W net + Q low

30 Forms of Energy 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.). Q high : 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.). W net : This is the net amount of work that exits the power plant. A turbine generates energy, but the pumps and compressors use energy. Q low : 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.

31 The Laws Zeroth Law: Two systems in thermal equilibrium with a third system are in thermal equilibrium to each other. First Law: The change in a system's internal energy is equal to the difference between heat added to the system from its surroundings and work done by the system on its surroundings. (Energy is conserved). Second Law: Entropy of an isolated system not in equilibrium always increases. (It is impossible for a process to have as its sole result the transfer of heat from a cooler body to a hotter one.) Third Law: Entropy of a system approaches a constant as the temperature approaches zero.

32 1st Law Efficiency Efficiency = æ ç è what you want what you pay for ö ø Several names: I =1st Law, Actual, or Thermal Efficiency I = W net /Q in = (Q high -Q low )/Q high = 1-(Q low /Q high )

33 Energy Balance for a Power Plant E in =1000 MJ from fuel 1000MJ Energy from Fuel Power Plant 350MJ Useful Energy (Work) E out =350 MJ useful energy +650 MJ wasted energy 650MJ Wasted Energy I = W net /Q in I =350MJ/1000MJ = 0.35 = 35%

34 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) Nothing s perfect - 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)

35 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

- a system tends to go from order to disorder Order Disorder Firewood has low entropy (molecules

36 2nd Law of Thermodynamics Entropy is the measure of how evenly distributed heat is within a system. - a system tends to go from order to disorder Order Disorder Firewood has low entropy (molecules in order) when stacked and high entropy when burning (molecules in disorder). The total amount of energy in the world does not change, but the availability of that energy constantly decreases.

37 2nd Law of Thermodynamics Thermal energy flows from hot to cold 2 types of processes: irreverisble, and reversible (in theory) istockphoto.com istockphoto.com When you touch a frozen pizza with your hand, thermal energy flows in what direction? Hand Pizza When you touch a cooked pizza with your hand, thermal energy flows in what direction? Pizza Hand

38 2 nd Law of Thermodynamics and Carnot Efficiency 2nd Law: Heat cannot be converted to work without creating some waste heat. Carnot Efficiency: The net work produced and the heat delivered into the system only depend on temperatures. No thermal system can be more efficient than the Carnot efficiency. Important: Temperatures must be in units of Kelvin or Rankin. c = W net /Q high = (T high - T low )/(T high ) c = 1 - (T low /T high ) K = ºC R = ºF And, a real system must be less efficient than this.

39 Heat Engine High Temp. Source of Heat T high Q high Heat Engine W net Q low Low Temp. Sink of Waste Heat T low c = 1 - (T low /T high ) Energy Balance: W net = Q high - Q low

40 Availability and Unavailability To transfer heat there must be a temperature difference (hot to cold - 2 nd Law) When the working fluid reaches T low no more energy can be used although the working fluid still contains energy The energy that is used is called available energy The waste energy is called unavailable energy

41 2 nd Law Efficiency The 2 nd law efficiency is a comparison of the system s thermal efficiency to the maximum possible efficiency. II = 2 nd law efficiency (effectiveness) I = 1 st law efficiency or thermal efficiency c = Carnot efficiency = II I C

42 Efficiencies of the Power Plant 1000MJ Energy from Fuel Power Plant 350MJ Useful Energy (Work) E in =1000 MJ from fuel E out =350 MJ useful energy +650 MJ wasted energy T high =1100 K 650MJ Wasted Energy T low =300 K I = W net /Q high =350MJ/1000MJ = 0.35 = 35% c = 1 T low /T high = 1 300K/1100K = 73% II = I / c =.35/.73 = 48 %

43 Cycles Initial and final states are identical E = E 2 - E 1 = 0 Q - W = 0 Net heat transfer and net work during the cycle are equal. Important Terms: Adiabatic: No heat is lost Isothermal: Constant temperature Isobaric: Constant pressure

44 Combustion Chamber Q H 2 3 Compressor Turbine W in Heat Exchanger 1 4 Q L W out 1-2: Air is compressed 2-3: Heat is added 3-4: Air turns turbine 4-1: Exhaust gases are cooled

45 Carnot Cycle PV = nrt

46 Carnot Diagram PV=nRT Isothermal: PV=const. Adiabatic: Heat const. ΔS = ΔQ reversible /T 4-1: Heat is transferd from high temp source at T high 1-2: Gas expands doing work (spinning a turbine) no heat lost 2-3: Heat is rejected to cold temp source at T low 3-4: Gas is compressed (requires work) no heat lost

47 Comparison of Efficiencies Type T h (K) T l (K) Carnot Eff. 1 st Law Eff. 2 nd Law Eff. Coal % 35% 56% Nuclear % 35% 47% Geo- Thermal % 16% 48%

48 Lecture 6 stops Lecture 7 starts around here Readings: Rubin, EE, Sections (except & 5.2.3); Pages , Masters, G. (1991) Introduction to Environmental Engineering and Science (Prentice Hall: NJ), pages [ Masters_1991_Air_Pollution.pdf] Anderson (2005), Sections , pp [ Anderson_2005.pdf] Rubin, EE, Sections , Pages (skip Nuclear Energy, pp ).

Thermal Efficiency, th The thermal efficiency is the index of performance of a workproducing device or a heat engine

49 The system, or working fluid, undergoes a series of processes that constitute the heat engine cycle. The following figure illustrates a steam power plant as a heat engine operating in a thermodynamic cycle. Thermal Efficiency, th The thermal efficiency is the index of performance of a workproducing device or a heat engine and is defined by the ratio of the net work output (the desired result) to the heat input (the cost or required 49 input to obtain the desired result).

50 Entropy (Advanced) It is a measure of disorder: more disorder = more entropy It is the measure of the unavailable energy of a system Entropy is calculated as: S = Q/T (information theory: S = k log W; W = # of states ) S Steam Liquid Water Ice (Gases) (Liquids) (Solids)

51 Comparison of heat engines

52 Thermodynamics of a Car Engine The Otto Cycle, aka SIE, Spark Ignition Engine This cycle applies to two stroke and four stoke engines. It is an ideal representation of the process. 1-2: isentropic compression (compression stroke) (no change in entropy) 2-3: constant volume heat addition (power stroke) 3-4: isentropic expansion (exhaust stroke) 4-1: constant volume heat rejection (intake stroke)

53 Otto Cycle Thermodynamics

54 Thermodynamic Derivation (Extra) Model this as a closed system. No major changes in kinetic or potential energy. Energy equation becomes: (q in -q out )+(w in -w out ) = u But more importantly, let s look at the processes from 2 to 3 and 4 to 1: q in = u 3 -u 2 = C v (T 3 -T 2 ) q out = u 4 u 1 = C v (T 4 T 1 ) (no work done during these processes) th,otto = w net /q in Looking at the first law, in = out w net = q in q out Therefore, th,otto = w net /q in = (q in -q out )/q out

substitute our isentropic relationships for T and v.

55 Thermodynamics th,otto = w net /q in = (q in -q out )/q out Substituting for q in and q out th,otto = 1 ((T 4 -T 1 )/(T 3 -T 2 ) Since 1 to 2 and 3 to 4 are isentropic processes, we can substitute our isentropic relationships for T and v. Therefore, (see derivation on page 360) thotto 1 r = 1 k 1 Where the compression ratio r = V 1 /V 2 = v 1 /v 2 And k = specific heat ratio = C p /C v

56 Cycles Initial and final states are identical E = E 2 - E 1 = 0 Q - W = 0 Net heat transfer and net work during the cycle are equal. Important Terms: Adiabatic: No heat is lost Isothermal: Constant temperature Isobaric: Constant pressure

57 Property and process diagrams Thermodynamic properties are often represented graphically. The process under study can be mapped on a property diagrams, thus allowing the values of the working fluid to be determined for each state of interest. The working fluid is the medium that through its physical or chemical change provides the energy the drives the system

58 T Carnot Temperature/Entropy Diagram T high T low S 1 T = Temperature (K or R) S 2 S = Entropy (J/K, Btu/R) = Q/T S

59 T T high T low Carnot Temperature/Entropy S 1 Q high Q low T = Temperature (K or R) Diagram S = Entropy (J/K, Btu/R) = Q/T S 2 S 1 = Q high /T high S 4 S 3 = Q low /T low 2 S 2 S 1-2: Constant temp heat addition 2-3: Constant entropy (adiabatic) work generation 3-4: Constant temp heat rejection 4-1: Constant entropy (adiabatic) work

60 T Area of Box is W net T high 1 Q high 2 S = Q/T Q = S * T T low 4 3 Q low S 1 S 2 S

61 Heat Added T T high 1 Q high 2 S = Q/T Q = S * T T low 4 3 Q high = (S 2 S 1 )*T high S 1 S 2 S

62 Waste Heat T T high 1 Q high 2 S = Q/T Q = S * T T low 4 3 Q low Qlow Q high = (S 2 S 1 )*T high Q low = (S 4 S 3 )*T low S 1 S 2 S

63 T Area of Box is W net T high 1 Q high 2 S = Q/T Q = S * T T low 4 3 Q low Q high = (S 2 S 1 )*T high Q low = (S 4 S 3 )*T low W net = Q high - Q low S 1 S 2 S

64 T Area of Box is W net T high 1 Q high 2 S = Q/T Q = S * T T low 4 3 Q low Q high = (S 2 S 1 )*T high Q low = (S 4 S 3 )*T low W net = Q high - Q low S 1 S 2 S

65 The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. Natural Gas Power Plant T T high T low 1 Q high Q low Carnot: W net = area of rectangle Natural Gas Plant: W net = area of blue shape S 1 S 2 S Realistic systems (coal, natural gas, nuclear power plants) do not fill the entire box therefore, they have not reached the Carnot efficiency Recall Ideal Gas Law: PV = nrt

66 Problem solving Draw a sketch and identify the system List the given information Check for special processes and state any assumptions Apply the conservation equations Draw a process diagram Determine the required properties and unknowns

67 Step-by-step-by-step by Energy Transfer Q = m Cp ΔT Q = energy transfer (Joules) m = mass of the material(kilograms) C = specific heat capacity of the material (J / kg C ) p T = temperature Δ = difference

68 Thermal Energy Transfer Equations Q = m Cp ΔT Q = energy transfer (Joules) m = mass of the material(kilograms) C = specific heat capacity of the material (J / kg C ) p T = temperature Δ = difference

69 Thermal Energy Transfer Equations Q P = Δt ΔT P = ka L PL k = AΔT P = rate of energy transfer (Watts) Q = energy transfer (Joules) t = time(seconds) k = thermal conductivity A = area of thermal conductivity L = thickness of material T = temperature Δ = difference

Calculating Energy Transfer Calculate the energy transferred when a block of aluminum at 80.0 C is placed in 1.00 liter (1kg) of water at 25.0 C if the final temperature becomes 30.0 C. Step 1.

70 Calculating Energy Transfer Calculate the energy transferred when a block of aluminum at 80.0 C is placed in 1.00 liter (1kg) of water at 25.0 C if the final temperature becomes 30.0 C. Step 1. List all known values Q Mass of water = C p of water = 1 kg J 4184 o kg C Difference in temperature = J C p of Al = 900. o kg C Difference in temperature = ΔT = 30.0 C 25.0 C = 5.0 C ΔT = 80.0 C 30.0 C = 50.0 C

71 Calculating Energy Transfer Step 2. List all unknown values Q = energy transferred m Al = mass of the Al block Step 3. Select equations to solve unknown values Q = m C ΔT Q Al = Q water p Step 4. Solve for Q water J kg C o Q water = (1.00kg) C = 21,000 J gained o

72 Calculating Energy Transfer Step 5. Solve for m Al Q (lost) = Q Al water (gained) = 20,920 J Q = m Cp ΔT Al Q Al 20, 920J Al m Al = = C p ΔT J 900. o o 50.0 C kg C m Al = 0.46 kg = 460g

73 Energy Transfer, Ex #2 Calculate the energy transfer in a wall section measuring 2m by 1m by 0.04m thick with a thermal conductivity of Q Opposing sides of the wall section have a temperature of 10 C and 5 C after one hour. Step 1. List all known values Area of thermal conductivity = Thermal conductivity = Thickness of material = k =0.10 L = 0.04m A = 2m * 1m = 2m 2 o s m C Difference in temperature = ΔT = 10 C - 5 C = 5 C Difference in Time = Δt = 1 hour = 3600s J

74 Calculating Energy Transfer Step 2. List all unknown values P = Rate of energy transfer Q = Energy transfer Step 3. Select equations to solve unknown values Q P = Δt Step 4. Solve in terms of Q Q = P Δt Step 5. Combine equations ΔT Q = (ka ) Δt L ΔT P = ka L

75 Calculating Energy Transfer Step 6. Apply known values Q = (ka ΔT L ) Δt Q = ææ J 2 ö æ 5 C ö ö ç ç m ç 3600s èè s m C ø è 0.04m ø ø Q = 90,000J

Existing Resources: Supplemental/reference for students with thermodynamics background and interests:

Existing Resources: Supplemental/reference for students with thermodynamics background and interests: Existing Resources: Masters, G. (1991) Introduction to Environmental Engineering and Science (Prentice Hall: NJ), pages 15 29. [ Masters_1991_Energy.pdf] Supplemental/reference for students with thermodynamics