MASSACHUSETTS INSTITUTE OF TECHNOLOGY SPRING 2007

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY SPRING Energy Environment and Society (a Project Based First Year Subject supported by the d'arbeloff Program) Session.4. Energy Basics (continued). Discussion of team projects and project selections. Summary of thermodynamics concepts and applications Readings and Assignments D. Goodstein, "Out of Gas" (Norton, 004): Chapter, "Energy Myths and a Brief History of Energy", pp Practice Problems on First and Second Las of Thermodynamics (due Feb. ) Supplementary recommended text (on library reserve) J.B. Fenn, "Engines, Energy, and Entropy" (W.H. Freeman and Co., 98) Thermodynamics is about the flo of energy Thermodynamics as first developed to explain ho much heat could be used to do ork eventually e learned that these las apply to every material and energy transformation everyhere in the universe! Describes macroscopic properties of equilibrium systems (you don't need to kno anything about atoms and molecules but it helps!) Entirely Empirical (cannot be proven logically, but can be derived using statistical mechanics) Built on Four Las and simple mathematics (but e ill not assume you have had 8.0!) 0 th La Defines Temperature (T) "The common sense La" st La Defines Energy (U), Heat (q), Work () "You can break even, but you can't in" nd La Defines Entropy (S) "You have to go to 0 K to break even" "Everything ants to be as imperfect as possible" 3 rd La Gives Numerical Value to Entropy "You can t get to 0 K" (so you can never break even) These las are UNIVERSALLY VALID, they cannot be circumvented.

2 "From BTU's and calories Producing foot-pounds, ergs, and joules Heat Engines must, to serve our needs Obey inexorable rules." -- J.B. Fenn, "Engines, Energy, and Entropy" Definitions: System: The part of the Universe that e are interested in (everything inside the boundary) Surroundings: The rest of the Universe (everything else outside the boundary) Boundary: The surface dividing the System from the Surroundings SURROUNDINGS BOUNDARY SYSTEM

3 In order to describe a system e have to specify a small number of macroscopic properties ("state variables"), such as Pressure p Energy U (or E) Volume V Enthalpy H Temperature T Entropy S Temperature is defined by the Zeroth La of Thermodynamics: Thermal Equilibrium (hen heat stops floing) A B A B A B When a hot object is placed in thermal contact ith a cold object, heat flos from the armer to the cooler object. This continues until they are in thermal equilibrium (no more heat flo). We say, at this point, that both bodies have the same temperature. This intuitively straightforard idea is formalized in the 0 th La of thermodynamics and is made practical through the development of thermometers and temperature scales. ZERO th LAW of Thermodynamics If and are in thermal equilibrium and A B and B are in thermal equilibrium, then and are in thermal equilibrium. A C C B acts like (is) a thermometer, and A, B, and C are all at the same temperature. For thermodynamics calculations, e have to use the Absolute (Kelvin) temperature scale: T (K) = T( C)

4 Work: Work, Heat, and the First La = F l Expansion ork F = p A ext applied constant force p ext distance l p ext = -(p ext A) l = - p ext ΔV convention: > 0 means that the surroundings do ork to the system (compression, ΔV < 0). If the system does ork on the surroundings (expansion, ΔV > 0), then < 0. Work is not a property of the state of the system! = pextdv means not an exact differential hich means that the integral = p depends on the path!!! ext dv See Non Lecture # for an example calculation of the path dependence of Heat: q The quantity that flos beteen the system and the surroundings, hich results in a change of temperature of the system and/or the surroundings. Sign convention: If heat enters the system, then q is positive. Remember this: > 0 if ork done on system q > 0 if heat added to system and q are both forms of energy Units of q and : calorie = heat needed to raise g H O by C at T = 5 C Joule = 4.84 calories BTU = 5 calories =,055 Joules "quad" = 0 5 BTU =.055 exajoules Heat capacity amount of q needed to raise a substance's temperature by ΔT heat capacity of ater = cal K - g - = 4.84 J K - g - = 75.3 J K - mole - 4

5 Non Lecture #: Path dependence of ork Example: assume a reversible process so that p ext = p (p same as p ext everyhere and at all times inside the system) p ext =p Ar (g, p, V ) = Ar (g, p, V ) Compression V > V and p < p p ext =p p, V compression p, V initial final To paths: () First V V at p = p () First p p at V = V then p p at V = V then V V at p = p Ar(g, p, V ) = Ar(g, p, V ) = Ar(g, p, V ) Ar(g, p, V ) = Ar(g, p, V ) = Ar(g, p, V ) path () path () p p final path () p path () V initial V V = V V p () V extdv p V = V V ext dv pdv = p ( V V ) () = = V V V V p ext dv V V pdv = p ( V V ) p ext dv () = p ( V V ) () = p ( V V ) (Note > 0, ork is done on system to compress it)!!! (because p p ) () () Note for the closed cycle [path ()] - [path ()], d 0 closed cycle is not a state function you cannot rite = f(p,v) 5

6 The equivalence of ork and heat as demonstrated by Joule's experiments on raising the temperature of a knon amount of ater (see Goodstein, Chapter, for detail) QuickTime Graphics are needed d (a) ith only heat T T (b) ith only ork (eight falls, propeller rotates, viscous friction heats ater): are QuickTime Graphics needed dec to T T The First La of Thermodynamics: Conservation of Energy Mathematical statement: du = d q + d or Δ U = q+ or system dq= d Δ U = q+ Δ U = q surroundings Δ U =Δ U +Δ U = 0 universe system surroundings U is a "state function", value depends only on properties of system (p, V, T, etc.) Enthalpy H = U + pv, useful to measure energy changes in processes taking place at constant pressure For example, the ΔH of vaporization of ater at p = atmosphere = 400 J/gram = 43. kj/mole. 6

7 Clausius statement of st La: The energy of the universe is conserved. The First La tells us that a properly designed engine can run in a cycle by converting heat into useful ork. The ork obtained cannot exceed the heat. But can all of the heat be converted into ork? Heat reservoir A very large system of uniform T. This T of the reservoir does not change regardless of the amount of heat added or ithdran. Also called heat bath. Real systems can come close to this idealized definition. Different statements of the Second La Kelvin: It is impossible for any system to operate in a cycle that takes heat from a hot reservoir and converts it to ork in the surroundings ithout simultaneously transferring some heat to a colder reservoir. T (hot) q q > 0 (heat in) < 0 (ork out) = q T (hot) > 0 < 0 q < 0 (heat out) = q - q T (cold) Arros sho direction of actual energy flo, q and are positive hen heat is added to system or ork is done on system. Clausius: It is impossible for any system to operate in a cycle that takes heat from a cold reservoir and transfers it to a hot reservoir ithout at the same time converting some ork into heat. Clausius statement is similar to Kelvin s, but for an engine operating in the opposite direction (as a refrigerator or heat pump). The to statements of the second la may be proven to be equivalent by connecting a Kelvin heat engine to a Clausius heat pump, then shoing that violation of either statement ill lead to violation of the other. 7

8 T (hot) q > 0 (heat in) T (hot) < 0 (heat out) = q q > 0 > 0 (ork in) < 0 = + q q T (cold) q T (cold) Alternative Clausius statement: All spontaneous processes are irreversible (e.g. heat flos from hot to cold spontaneously and irreversibly). dq T dqirrev = 0 and < 0 T rev d q rev q is a state function = ds ds = d rev T T S ENTROPY Kelvin and Clausius statements are specialized to heat engines. Mathematical statement seems very abstract. Link them through analytical treatment of a heat engine. It turns out that the Carnot Cycle is both a fundamental standard against hich all other cycles may be evaluated, and all properties of a Carnot Cycle may be calculated. (See Non-Lecture for explicit calculation of ork and heat changes for a Carnot cycle in hich the orking fluid is an ideal gas. The overall result does not depend on the choice of the orking fluid see Non Lecture 3. ) Nicholas Leonhard Sadi Carnot. b Published "Reflections on the Motive Poer of Heat and on Machines" (his only publication) in 84. d. of cholera in 83. 8

9 The Carnot Cycle hot isotherm (T ) > - an idealized heat engine in hich all paths are reversible T (hot) adiabat compression 4 cold isotherm (T 3) adiabat expansion q T (cold) isothermal expansion at T (hot) U = + ( > 0, < 0) 3 adiabatic expansion (q = 0) U = ( < 0) 3 4 isothermal compression at T (cold) U = q + (q < 0, > 0) 4 adiabatic compression (q = 0) U = ( > 0) The signs of, q, and cannot all be positive. When run as a heat engine (converts heat input to ork output), e ill find that > 0, q and < 0. When run as a heat pump (converts ork input to transfer of heat from a cold to a hot reservoir), q and are > 0, < 0. st La du = 0 q + q = ( ) ( ) total ork out Efficiency as heat engine = ε = = = + q =+ q heat in from hot reserv. at T Efficiency ε =+ q = T T = T T T but ε 00% only as T 0 K For a heat engine (Kelvin) (heat from hot reservoir converted to ork by system) > 0, < 0, T < T Total ork out =-=ε = T T ( )< T Note: In the limit T 0 K, (-) and ε 00% conversion of heat into ork. Can e cool the cold reservoir to T = 0 K? 9

10 For a heat pump (i.e. refrigerator) (Clausius) (pumps heat from cold reservoir q > 0 to q > 0, > 0, T < T hot reservoir < 0) Total ork in = = T T T < 0, > 0, T T < 0 But = q = T T T q = T T q T T T T T But in the limit T 0 K,. This means it takes an infinite amount of ork to extract heat from a reservoir at 0 K. This implies that 0 K cannot be reached (a form of the Third La ). Finally, the entropy of an isolated system never decreases. (A) irreversible (isolated) (A) The system is isolated and irreversibly (spontaneously) changes from [] to [] (B) The system is brought into contact ith a heat reservoir and reversibly brought back from [] to [] (B) reversible (not isolated) Path (A): q irrev = 0 (isolated) q Clausius: d 0 dq dq irrev rev + 0 T T T Definition of ΔS for the change of state of an isolated system: dqrev =0! = S S = ΔS 0 T Δ S = S S 0 For isolated systems This gives the direction of spontaneous change! If S > S, isolated system goes spontaneously and irreversibly from. ² S > 0 Spontaneous, irreversible process ² S = 0 Reversible process ² S < 0 Impossible! No exceptions! 0

11 Non Lecture : Carnot Cycle for an Ideal Gas q isotherm T ) adiabatic (expansion) isotherm (compression at T < T ) adiabatic (compression) = T ln V 4 V 3 T ln V V V V 4 γ ² U = 0; = = pdv = RT ln V > 0 V q = 0; = C V ( T T )< 0 T Reversible adiabatic = V γ < T V 3 4 ² U = 0; q = = pdv = RT 3 ln V 4 < 0 Reversible adiabatic ( ) ( ) = T = V T V 3 q = 0; = C V ( T T )> 0 T T here comes the crucial trick! γ V 4 = V q = T V 3 V = V 4 T V γ > V 3 q q dq or + = 0 rev = 0 T T T e have a state function! This provides a link beteen (ideal gas) heat engines and the mathematical statement of the Second La! But it is not a proof of the Second La. It tells us that entropy, efficiency, and reversibility are inter-related via the Second La.

12 Non-Lecture 3: Entropy and the Clausius Inequality We ant to sho that the efficiency of any reversible engine is the same as that for a reversible Carnot engine here the gas in the Carnot engine is ideal. We do this by connecting the general engine to the Carnot engine, one acting as a heat engine (using heat from a hot reservoir to produce ork) and the other acting as a heat pump or refrigerator (using ork to pump heat from a cold to a hot reservoir). The to engines connected together produce no net ork. Work output from the heat engine is used as ork input to the heat pump. First e set up the Carnot engine as a heat engine and run the general engine in reverse as a heat pump. We assume that the general engine is less efficient than the Carnot engine and sho that this assumption is in violation of the Second La. Then e set up the general engine as a heat engine and run the Carnot engine in reverse as a heat pump. We assume that the general engine is more efficient than the Carnot engine and sho that this assumption violates the Second La. Since the general engine cannot be less or more efficient than the Carnot engine (because each ould violate the Second La), the efficiencies are identical. general (reverse) T (hot) = q q T (cold) Carnot q cycle Right Engine: ε = Left Engine ε= ork out heat from T in = = T ork in heat to T out = Since = ', the total ork from the to engines = ' + = 0 We assume that ε< ε > T (both quantities are positive) > ( replaced by ) > < (both sides divided by > 0)

13 We add to both sides of inequality + q < = 0 Net heat has been input into hot reservoir ithout net ork! Assumption ε < ε' must be invalid. No reverse the direction of operation of both engines. general T (hot) q = q T (cold) q Carnot (reverse) Right Engine: ε = Left Engine ε= ork in heat to T out = = T ork out heat from T in = Since = ', the total ork from the to engines = ' + = 0 We assume that ε> ε > > > q q < T (both quantities are positive) We add to both sides of the inequality and multiply both sides by, reversing the sense of the inequality. + q < 0 Net heat is input into hot reservoir ithout net ork! Assumption ε > ε must be invalid. Thus e have shon that both ε < ε and ε > ε lead to violation of the second la. Thus ε = ε. 3

Lecture 2 Entropy and Second Law

Lecture 2 Entropy and Second Law Etymology: Entropy, entropie in German. En from energy and trope turning toward Turning to energy Motivation for a Second Law!! First law allows us to calculate the energy