How To Calculate The Thermal Efficiency Of A Heat Engine

Transcription

1 Lecture 7 Entropy and the Second Law of Thermodynamics 15/08/07 The Adiabatic Expansion of Gases In an adiabatic process no heat is transferred, Q=0 = C P / C V is assumed to be constant during this process The pressure and volume of an ideal gas at any time during an adiabatic process are related by PV = constant All three variables in the ideal gas law (P, V, T ) can change during an adiabatic process. We get P i V i = Pf V f

2 (Q=0)

3 Assume an ideal gas is in an equilibrium state and so PV = nrt is valid. We get in differential form PdV + VdP = nrdt Since R= C P C V we have PdV + VdP C P C V = ndt We also have de int =nc V dt=qpdv= PdV (Q=0) or ndt = -PdV/C V Using both equations gives dp P + Integrating gives or PV = const. ln P + C P C V C P C V lnv = const. dv V = 0

4 The adiabatic curve in the PV diagram depends on For a monatomic gas C V =3/2R, C P =5/2R and = 5/3 For a diatomic C V =5/2R, C P =7/2R and = 7/5 For a polyatomic C V =3R, C P =4R and = 4/3 By finding one can determine the nature of the gas P i V i = Pf V f P i V i P f V f = 1 ln P i V i P f V f = ln1

5 Example: Cylinder of 50 cm 3 of air at 27 ºC and 1 atm is compressed very rapidly (adiabatically) to 10 cm 3. What is the final temperature? P i V i = Pf V f nrt i V i V i = nrt f V f V f which gives T i V i 1 = T f V f 1 T f = T i V i V f 1 = 300K = 571K No heat in or out The air is heating up to 298ºC, it becomes very hot

6 The Heat Engine A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work A heat engine carries some working substance through a cyclical process The working substance absorbs energy by heat from a high temperature energy reservoir (Q h ) Work is done by the engine (W eng ) Energy is expelled as heat to a lower (colder) temperature reservoir (Q c ) Since it is a cyclical process, E int = 0 V

7 E int = 0 Q net = W eng The work done by the engine equals the net energy absorbed by the engine Q c can be thought of the heat loss to the environment, that is Q h is not transformed 100% to work Q h Q c

8 Thermal Efficiency of a Heat Engine Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature = W eng Q h = Q h Q c Q h = 1 Q c Q h = 1 T c T h We can think of the efficiency as the ratio of what you gain to what you give If Q c =0 we get =1, that is a 100% efficiency In a similar way one defines the coefficient of performance (COP) COP = Q h W eng

9 Second Law: Kelvin Form It is impossible to construct a heat engine that, operating in a cycle, produces no other effect than the absorption of energy from a reservoir and the performance of an equal amount of work Means that Q c cannot equal 0 Some Q c must be expelled to the environment Means that cannot equal 100%

10 Second Law: Clausius Form It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work Energy does not transfer spontaneously by heat from a cold object to a hot object

11 Impossible Engines Perfect Heat Engine Perfect Heat Pump

12 The Carnot Engine A theoretical engine developed by Sadi Carnot A heat engine operating in an ideal, reversible cycle (now called a Carnot cycle) between two reservoirs is the most efficient engine possible This sets an upper limit on the efficiencies of all other engines No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle Sadi Carnot

13 The Carnot Cycle E int = 0 for the entire cycle W eng = Q h Q c

14 Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs == 1 T c T h Temperatures must be in Kelvins All Carnot engines operating between the same two temperatures will have the same efficiency Efficiency is 0 if T h = T c Efficiency is 100% only if T c = 0 K Such reservoirs are not available, as the absolute zero temperature cannot be reached Efficiency is always less than 100%

15 The efficiency increases as T c is lowered and as T h is raised In most practical cases, T c is near room temperature, 300 K So generally T h is raised to increase efficiency Theoretically, a Carnot-cycle heat engine can run in reverse This would constitute the most effective heat pump available This would determine the maximum possible COPs for a given combination of hot and cold reservoirs

16 In heating mode: COP h = Q h W eng = T h T h T c In cooling mode: COP c = Q c W eng = T c T h T c A good refrigerator should have a high COP Typical values are 5 or 6

17 The Combustion (Gasoline) Engine In a gasoline engine, six processes occur during each cycle: for a given cycle, the piston moves up and down twice This represents a four-stroke cycle

18 The processes in the cycle can be approximated by the Otto cycle OA B C D A O NB: We are not on the isotherms, this process deviates substantially from a Carnot cycle

19 O A in the Otto cycle During the intake stroke, the piston moves downward A gaseous mixture of air and fuel is drawn into the cylinder Energy enters the system as potential energy in the fuel Intake

20 A B in the Otto cycle In the compression stroke The piston moves upward The air-fuel mixture is compressed adiabatically The temperature increases The work done on the gas is positive and equal to the negative area under the curve Compression

21 B C in the Otto cycle Combustion occurs when the spark plug fires This is not one of the strokes of the engine It occurs very quickly while the piston is at its highest position Conversion from chemical energy of the fuel + O 2 to internal energy Spark

22 C D in the Otto cycle In the power stroke, the gas expands adiabatically This causes a temperature drop Work is done by the gas The work is equal to the area under the curve Power

23 D A in the Otto cycle Valve Opens: An exhaust valve opens as the piston reaches its bottom position The pressure drops suddenly The volume is approximately constant So no work is done Energy begins to be expelled from the interior of the cylinder (through the exhaust of the engine)

24 A O in the Otto cycle In the exhaust stroke, the piston moves upward while the exhaust valve remains open Residual gases are expelled to the atmosphere The volume decreases Exhaust Exhaust

25 If the air-fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle is connected approximately to an adiabatic process: T 1 V 1 1 = T 2 V 2 1 = 1 is the ratio of the molar specific heats V 1 / V 2 is called the compression ratio Typical values: Compression ratio = 8, = 1.4, = 56% Efficiencies of real engines are 15% to 20% Mainly due to friction, energy transfer by conduction, incomplete combustion of the air-fuel mixture etc. 1 ( ) 1 V 1 / V 2

26 Diesel Engines Operate on a cycle similar to the Otto cycle without a spark plug The compression ratio is much greater and so the cylinder temperature at the end of the compression stroke is much higher Fuel is injected and the temperature is high enough for the mixture to ignite without the spark plug Diesel engines are more efficient than gasoline engines

27 The Future: H 2 or CH 3 OH Fuel Cells Proton exchange membrane fuel cell The Toyota FCHV is a series of prototype hydrogen fuel cell vehicle, presented in A fuel cell is an electrochemical energy conversion device. It produces electricity from external supplies of fuel (on the anode side) and oxidant (on the cathode side). The tank-to-wheel efficiency of a fuel cell vehicle is about 45% at low loads and shows average values of about 36% when a driving cycle like the NEDC (New European Driving Cycle) is used as test procedure. The comparable NEDC value for a Diesel vehicle is 22%. Methanol Fuel Cell