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CARNOT CYCLE T1 T2 Consider working substance to be an Ideal Gas Note p – v axes i) substance starts at A with temperature T2, T2 adiabatic compression to B. Since dq 0 du dw u cv (T1 T2 ) w Work done on substance increases internal energy. ii) substance expands isothermally from B C , and absorbs heat Since T T1 =constant, du 0. dq dw Q1 . v dw vvcB P dv = nR *T1 vvcB d ln v nR *T1 ln c Q1 vB iii) substance undergoes further expansion (adiabatic) from C D. Again dq 0, du dw u cv (T2 T1 ) w Work done is c v (T1 T2 ) 1 iv) substance returns to original point A by undergoing an isothermal compression from D A , and the heat exhausted is equal to Q2. Since du=0 for this isothermal process, dq = dw: dw VA VD or VD pdv nR T2 d ln V nR T2 ln Q2 VA VD VA V Q2 nRT2 ln D VA Since the working substance returns to its original state, the work done by the substance is equal to the net heat absorbed, Q1-Q2 Work done = Q1-Q2 (area inside the “rectangle”) This can be readily verified by summing the individual work terms. ? wi wii wiii wiv Q1 Q2 This can be easily verified by adding up the individual terms. Efficiency of the carnot cycle is Q1 Q2 Q1 Q: How can efficiency be increased? 2 Only by transferring heat from a warm body to a cold body can heat be converted to work in a cyclic process. For a CARNOT CYCLE Q1 T1 Q2 T2 Proof: It can be shown that pV = constant for adiabatic process Adiabatic legs For A B PAVA PBVB For C D PCVC PDVD Isothermal legs For B C PBVB PCVC For D A PAVA PDVD Combining 4 equations We already know that Vc VD VB VA V Q1 RT1 ln C VB V Q2 RT2 ln D VA 3 Therefore V T1 ln C Q1 VB T1 Q2 T ln VD T2 V 2 A maximum (Carnot) efficiency of a power cycle 1 So Q2 T T 1 2 1 cold Q1 T1 Twarm • The maximum efficiency of an energy cycle is given by the Carnot efficiency, which can be expressed as: 1 Tc Th If we consider the temperature difference between the equator and the poles: Tpole (Tc) Tequator (Th) We note that Th varies little throughout the year (estimate as ~301 K), compared with Tc Winter hemisphere: estimate Tc ~ 263 K 1 Tc 263 1 13% Th 301 Summer hemisphere: estimate Tc ~ 283 K 1 Tc 283 1 6% Th 301 --> More of the absorbed energy is converted into “work” in the winter hemisphere 4 The hurricane as an energy cycle Consider a hurricane as a Carnot cycle: http://apollo.lsc.vsc.edu/classes/met130/notes/chapter15/vertical_circ.html 3 Toutflow 4 2 1 Tsea surface Analyze each leg of the cycle: 1: absorption of energy from ocean, ~isothermal expansion 2: convection in the eyewall, ~adiabatic expansion 3: upper level outflow, ~ isothermal compression (reject heat via radiation) 4: sinking branch, ~adiabatic compression If we estimate SST~300 K, and outflow T~200 K, 1 Tc 200 1 30% Th 300 U Hawaii class notes: Latent heat released when water vapor condenses in clouds is the key. Hurricanes are giant engines that convert heat into wind energy. Consider a rain rate of 2 inches per day over an area of 300 mi radius: Over a 7 day lifecycle, the energy released is equal to 50,000 1 MT nuclear explosions, equivalent to the total explosive yield of the nuclear arsenals of the US and USSR at the height of the Cold War! www.soest.hawaii.edu/MET/Faculty/businger/.../18cHurricanes1.pdf Our next state function: ENTROPY (consequence of Second Law) There exists a function called entropy S, of the extensive variables of a system, defined for all equilibrium states, such that the values assumed by the extensive variables are those that maximize S (at equilibrium) From the viewpoint of classical thermodynamics, entropy is defined as ds dqrev T where in this case heat is added to a substance undergoing a reversible transformation. Clausius (1822-1888) stated that the cyclic integral of δQ/T is always less than or equal to zero: Q T 0 The cyclic integrals of work and of heat are not zero (engines). Entropy is a STATE VARIABLE. (We can make up an internally reversible path from A to B to compute change in entropy from the integral of δQ/T.) Since dq= Tds, the First Law can be written as, Tds du pdv Even though we imagined a reversible process to substitute for dq, this equation applies to both reversible and irreversible transitions because they are state variables Generalized Statement of the Second Law • • We need to generalize the definition of entropy since real systems are typically spontaneous and irreversible, moving from a state of non-equilibrium to a state of equilibrium. Second law can be formulated as 4 postulates: 1. There exists a STATE VARIABLE for any substance called the ENTROPY. 2. Entropy (S) may change by one of two ways; the substance may contact a thermal reservoir (heat source), or S may change due to “internal” changes in the substance. dse (externally-forced changes in S) dsi (internally-forced changes in S) Examples: dse Heat flow into a substance causing molecules to rearrange dsi Mixing a gas, forcing molecules to change position. Total entropy change ds = dse + dsi 3. Changes in S due to external influences are given by dse • dq T where dq = heat imported or exported, while substance is in contact with a thermal reservoir at temperature T dsi 0 4. For reversible transformations, For irreversible transformations, dsi 0 So reversible and irreversible processes are distinguished by entropy changes. For reversible processes, there are no entropy changes associated with internal processes, or they have ceased as the system undergoes only reversible changes, say via contact with a heat reservoir. Now consider the system + its surroundings ( = “universe”) • In a reversible process any heat flow between the system and surroundings must occur with no finite temperature difference between the system and surroundings. Otherwise the process would be irreversible dqrev Tsystem Tsurroundings UNIVERSE dqrev dsuniverse dssystem dssurroundings dqrev dqrev ( )0 Tsystem Tsurroundings Since Tsystem Tsurroundings • • Therefore Suniverse 0 for reversible process. As expected and can be shown via a similar argument, Suniverse 0 In general Irreversible process Suniverse 0 Statement of 2nd Law! Calculations of Entropy Changes • • • 2 For reversible processes, s s2 s1 dqrev T 1 dqirrev For irreversible processes, s is not necessarily equal to T However since S is a state function, S depends only upon endpoints. Hence Sirrev can be found by devising a series of reversible transformations that are equivalent to the irreversible transformations. 1. Reversible adiabatic process Hence dqrev 0 , hence s 0 2. Reversible phase change at constant T, p since T constant 2 s 1 dqrev T1 q L*m rev T T 3. Reversible, isothermal process where m=mass, L= latent heat s qrev T 4. Reversible change of state for an Ideal Gas From First Law: dqrev du dw cv dT pdv then ds dqrev cv d ln T nR * d ln V T 2 2 1 1 S cv d ln(T ) nR * d ln V T2 V2 cv ln( ) nR * ln( ) T1 V1 Heating term Volume term 5. Constant-pressure heating (irreversible) For constant pressure: dp 0 dqrev c p dT dp T2 Hence S c p d ln T T1 T S c p ln( 2 T1 ) for c p f (T ) 6. Similarly, for constant-volume heating (irreversible) dqrev cv dT pdv pdv 0 T2 S cv ln dT T1 T2 S cv ln T1 More entropy calculations at end of notes “Thermodynamic Potentials” • The internal energy, when expressed as a function of S and V, is one of the so-called “thermodynamic potentials”. The differential of a thermodynamic potential gives the first and second laws of thermodynamics combined: du dq dw du Tds Pdv • • S and V are the “native variables” for U (although we can express U as a function of any two variables) Remember enthalpy is: H U PV dh Tds vdP • • So S and P are the “native variables” for H. We define two more thermodynamic potentials that have “native variables” , respectively, of T and V and T and P. Very useful for equilibrium calcs! Helmholtz & Gibbs Functions • Consider the following system/reservoir arrangement, Q system • -Q Total work done by the system, whether it be in a reversible or irreversible process is, First Law • reservoir w (u1 u2 ) Q Q is positive if it flows into the system (1) w is positive if done by the system dq=du+dw From principle of entropy, w=Q-(u2-u1) ( s2 s1 ) sR 0 s2 s1 is entropy change of system sR is entropy change of reservoir (2) A general expression that entropy must stay constant or increase sR Q T • For the reservoir, • Therefore we have ( S 2 S1 ) • • Note T= constant; reservoir has infinite heat capacity. From (3) we have the following inequality, T ( s2 s1 ) Q heat flow out of reservoir Q 0 T > applies for irreversible processes (ssystem (3) Q ) T = applies for reversible processes • • From First Law or Eq (1) above (substitute for Q) WT (u1 u2 ) T ( s1 s2 ) Reversible process WT (u1 u2 ) T ( s1 s2 ) Irreversible process For the irreversible case the work done is less than u T s since some of the heat added to the system can go into changing the ‘internal’ entropy of the system. Combining the two equations above WT (u1 u2 ) T ( s1 s2 ) Now define the Helmholtz energy as F u Ts (4) • The difference in this energy between 2 equilibrium states at the same temperature is F ( F1 F2 ) (u1 u2 ) T ( s1 s2 ) From Eq n (4) we now have WT ( F1 F2 ) • (5) Hence change in F sets an upper limit to the work that can be done in any process between 2 equilibrium states at constant T. For a reversible process WT F1 F2 For an irreversible process WT F1 F2 • F is often referred to as the ‘free energy’ since ∆F represents the maximum amount of work that can be done in a transformation at constant temperature. • Returning to the equation WT (u1 u2 ) T ( s1 s2 ) (6) • This equation is general and can be applied to any system, change of state, phase, or even a chemical reaction. • In general, work in processes we are considering consist of pdV work and other possible forms of work, e.g. work required to form a curve interface between 2 phases, frictional dissipation, etc. • Let WT represent pdV type work and work, so we now have AT WT AT F1 F2 represent other forms of (7) • If we consider a constant volume process, WT 0, AT F1 F2 • Hence at constant T and V, Helmholtz energy sets an upper limit to the non- pdV work that can be done in such a transition. • For a reversible constant volume process AT ( F1 F2 ) • Now consider a situation where no pdV and no 0 ( F1 F2 ) AT work is done, then F2 F1 • Hence for a process at constant volume, for which AT 0 (and T=constant), F can only decrease or remain constant. A transition will not occur if F F . 2 1 Wikipedia definition: du Tds Pdv The Helmholtz free energy is a thermodynamic potential which measures the “useful” work dh Tds vdP obtainable from a closed thermodynamic system at a f u Ts (by definition) constant temperature and volume. For such a system, the negative of the difference in the df du Tds sdT Helmholtz energy is equal to the maximum amount of work extractable from a thermodynamic process sdT Pdv in which temperature and volume are held constant. Under these conditions (constant T and v), it is Constant v constraint is not minimized at equilibrium. (8) always helpful for us … • Next, consider a process that occurs at constant pressure. In this case the pdV work is WT p(V2 V1 ) • Then returning to the equation: • We have: WT AT F1 F2 AT ( F1 F2 ) p(V1 V2 ) or AT (u1 u2 ) T (s1 s2 ) p(V1 V2 ) Now define the Gibbs Free Energy as, G F pV U TS pV (dG = -SdT + VdP) Then for a transition between 2 states at the same T ,p: G G1 G2 (U1 U 2 ) T (S1 S2 ) p(V1 V2 ) and substituting into (8) we have AT G1 G2 G ∆G represents the total non- pdV work for a constant T,P process. This is of fundamental importance for cloud physics. (8) quick look ahead to one application we’ll be discussing: • During nucleation, some energy is released when molecules move from the vapor to the “liquid” phase • A is the work (energy required) necessary to form the surface of the small T liquid (or ice) embryo from the vapor phase: AT Aembryo For reversible process AT , p G For irreversible process AT , p G • surface tension ( Jm 2 ) Aembryo area of embryo AT , p bounded by G If WT AT 0 then 0 G1 G2 So G can only decrease in a transition occurring at constant T, P. • If the hypothetical nucleus is too small (known as an unstable nucleus or "embryo"), the energy that would be released by forming its volume is not enough to create its surface Thermodynamic Potential • • • Given F U TS ; G U TS pV Consider a closed system which is one where no mass is allowed to cross the boundaries of the system. Taking total differentials, (1) dF dU TdS SdT dG dU TdS SdT Vdp pdV • (2) Using First Law in the form: dq dU pdV TdS dU pdV • We can substitute for (1) and (2) the dU term dF TdS pdV TdS SdT pdV SdT For dG, dG TdS pdV TdS SdT pdV Vdp dG SdT Vdp • • Consider the following series of equations, dU TdS pdV U ( S ,V ) dF SdT pdV dG SdT Vdp F (T , V ) G (T , p ) dH TdS Vdp H (S , p) It is useful to identify the coefficients of the equations on the left hand side with partial derivatives of the variables on the right hand side. • U U For: U ( S ,V ); dU dS dV S V V S • Therefore we have • U T; S V Since F F (T , V ) dF • U p V S F F dT dV T V V T Therefore F S ; T V F p V S • Since G G (T , p) G G dG dp dT T p p T • And • Since H H ( S , p ) G S T p G V p T H H dH dp ds S p p S H T S p H V p S • And • Since P, V, T and S can be expressed as partial derivatives of U, F, G and H, the latter variables are referred to as Thermodynamic Potentials. Maxwell Relations • The “Gibbs relations” are the following equations that we have already seen: du Tds Pdv dh Tds vdP da sdT Pdv dg sdT vdP • We recognize that they are exact differentials and have the form dz Mdx Ndy M N y x x y • We can deduce the “Maxwell Relations”: T P v s s v T v P s s P s P v T T v s v P T T P Can get unmeasurables in terms of measurables Stable and Unstable Equilibrium • • Stable Equilibrium: a state in which all irreversible processes have ceased, and the system is in a state of maximum entropy. Reversible transformations are possible (for which Stotal 0 ). For state equilibrium, S 0 where S S Seq MAX ENTROPY • • a) Seq S This equilibrium condition is for an isolated system (no heat transfer, no mass transfer) A typical system we will deal with is NOT in thermal isolation, but rather in contact with a heat reservoir (e.g., as approximated by the atmosphere). For a transition at constant volume, in thermal contact with a reservoir at temperature T, the stable state is defined by, F 0 F Feqm Fother b) Helmholtz free energy is a minimum For a transition at constant pressure and temperature, (in contact with a thermal reservoir), stable state is, G 0 G Geqm Gother Gibbs free energy is a minimum Applying equilibrium criteria • We can now add moisture to the air parcel we’ve computed changes for – We could have added water to the “dry air mixture” and computed a new average molecular weight, etc., BUT we know that simple approach, assuming water stays as a vapor during the adiabatic ascent, is only valid over small regions – That is, we’ll need to consider phase changes for the water • First, we’ll figure out where vapor, liquid and ice are in equilibrium, and the energy changes associated with the transitions – Can calculate changes such as heat released so we can treat the energy changes in our parcel – Can also figure out if the change is favored (e.g., if the transition is allowed to occur thermodynamically, given specified conditions) LATENT HEATS: During a phase change, the pressure is constant and equal to the saturation pressure (which is a function of T only). Adding heat to a substance at constant pressure and resulting in a change in physical state, i.e. change in phase, is described by, Starting with First Law in the dq dh Ldm form dq = dh when dp = 0 where L is the latent heat (J kg-1) for the particular transition and m is mass transferred between phases. The heat absorbed goes into changing the molecular configuration/structure. For water substance: L f 3.34 105 Jkg 1 Heat of fusion Heat of condensation Lc 2.25 106 Jkg 1 Heat of sublimation Ls 2.6 106 Jkg 1 SOLID Lf LIQUID Lc L is an energy change to go from one phase to another VAPOR 26 LS Thermodynamics of Moist Air • Phase transitions: what are the equilibrium conditions for a mixed-phase system, e.g. a mixture of water and water vapor? • Consider a system consisting of liquid water and H2O(v) undergoing a change in state, { V1 vapor liquid { V2 ″ denotes liquid phase ″′ denotes vapor phase vapor liquid • Define V1 total volume of state 1 V2 total volume of state 2 n1 # of moles of liquid, state 1 n1 # of moles of vapor, state 1 n2 # of moles of liquid, state 2 n2 # of moles of vapor, state 2 g specific Gibbs free energy of liquid phase g specific Gibbs free energy of vapor phase G1 Gibbs energy, state 1 • Hence G2 Gibbs energy, state 2 G1 n1 " g " n1 "' g "' G2 n2 " g " n2 "' g "' • • • Since both states are stable equilibrium states, G1=G2, and therefore g n1g n1g n g n 2 2 But since the system is closed ntotal 0 (n n1 ( n1 n )g 2) g 2 Therefore g g n1 n1 n 2 n 2 n1 n n 2 n1 2 • • • At equilibrium, the specific Gibbs free energy is the same for both phases At the triple point, g S g L gV Lets look at g between 2 phases in more detail: vapor c liquid g b,e f d a T g s T p • From earlier discussions, • Where s′″ is the specific entropy of the vapor phase. Hence the slope of g vs. T gives the specific entropy. Likewise g for the liquid phase • • s T p Note the slopes are <0, since larger values of g correspond to lower T’s. The difference between entropies is s s • Since 23 23 0, s s 0 s s vapor T liquid ( ds dq ) T Clausius-Clapeyron Equation • This equation is very important since it describes the variation in pressure with temperature, for a system consisting of two phases in equilibrium. For the case of a liquid/vapor or vapor/solid system, this “pressure” is the saturation vapor pressure. vapor liqui d • • • es ,w sat vapor pressure for water es ,i sat vapor pressure for ice For a system in equilibrium, g g Suppose the system undergoes a small change to a new pressure and temperature, p+dp, T+dT such that the new state is also a stable state. Each phase changes by dp, dT amounts. Then for each phase: dg sdT vdp dg sdT vdp • Since the new state is also an equilibrium state, dg dg • Or, sdT vdp sdT vdp ( s s)dT (v v)dp dp s s s dT v v v • Since s s CLAUSIUS-CLAPEYRON EQUATION 23 23 =latent T dp 23 dT T (v v) • • • s, v are specific quantities! heat for vapor-liquid interface For vapor-liquid system For vapor-liquid system dp de s ,w For vapor-ice system dp des ,i Geometrical Interpretation SOLID LIQUID S-L P L-V Triple point Critical point S-V VAPOR T dp dT gives the slope of the equilibrium line between the 2 phases involved. (on P-V-T diagram.) • Looking again at the Clausius-Clapeyron Equation we have three interfaces, dp 23 dT T (v v) dp 13 dT T (v v) dp 12 dT T (v v) • 1. vaporliquid solidvapor liquidsolid Since xy 0 always, the slope of the equilibrium line between the 2 phases involved is determined by the difference in specific volumes. dp is >0 for dT v v, v vaporliquid vaporsolid ] ALWAYS For all materials! 2. dp can be of either sign for liquid solid interface. dT a) for v v 0 v v For substance that expands upon freezing (like water). Hence dp 0 dT P dp 0 dT L S Triple V T b) for v v 0 v v substance contracts upon freezing. MOST SUBSTANCES! dp 0 dT • Integrate the Clausius-Clapeyron Eqn/ to get es as a function of T. For a vapor/liquid system, dp 23 dT T (v v) 1) need 23 (T ) 2)v v as a function of T . • • constant RvT v v v p Assume 23 Therefore dp dT GAS LAW L23 RvT 2 P L23 ln p constant RvT p es es 6.11mb @ T 0C after integration More entropy calculations 35 Rapid expansion 7. Irreversible change of state of an Ideal Gas Let n moles of an ideal gas of P1,V1, T1 irreversibly change state to P2 ,V2 , T2 This irreversible process can be broken into 2 reversible processes: 1.) A slow expansion to V2 at constant T1 -put gas in a frictionless piston-cylinder, place in constant temperature bath; slowly move piston out 2.) Hold gas at V2 , heat gas to temperature T2 and pressure P2 S S1 S2 Hence 2 2 1 1 Sirrev cv (T )d ln T nR * d ln V Constant-volume heating Isothermal expansion 8. Irreversible phase change Let 10g of supercooled water at -10°C change to ice at -10°C. Consider constant pressure. This process is IRREVERSIBLE e.g. if the ice is slowly heated, it will not become water at -10°C. Water at -10°C IRREVERSIBLE Ice at -10°C 1 3 2 Water at 0°C Ice at 0°C Find equivalent reversible path and compute entropy change 2 1.) Isobaric heating of water 2.) Phase change S2 T dqrev 2 S1 mc p , w d ln T T 1 T1 Lf m T H m T 2 3.) Isobaric cooling of ice S3 m c p ,i d ln T 1 STotal S1 S2 S3 STotal Lf m T2 T mc p , w ln( ) mc p ,i ln( 1 ) T2 T1 T 273 J ) 1.57 263 deg 2 J 10 kg 5 J S2 3.34 10 12.23 kg 273 K deg 263 J S3 102 kg (2106 Jkg 1 deg 1 ) ln( ) 0.8 273 deg J STotal S 11.46 deg S1 102 kg (4218 Jkg 1 deg 1 ) ln( T2 0 C T1 10 C m 10 g c p , w 4218 Jkg 1 K 1 c p ,i 2106 Jkg 1 K 1 More entropy calculations • Calculate the change in entropy when 5g of water at 0°C are raised to 100°C and converted into vapor (steam) at that temperature. Lv 2.25 106 Jkg 1 1. Raise temperature of H2O(l) from 0°C to 100°C S S373 S273 T2 dqrev T1 Let dq mcwdT T Specific heat of water cw 4.18 103 J deg 1 kg 1 T2 T S1 mcw d ln T mcw ln( 2 ) 6.58 J deg 1 T1 T1 2. Convert to vapor (steam) at 100°C Lv m (2.25 106 Jkg 1 )(5 103 kg ) S2 30.2 J deg 1 T 373 K Stotal S1 S 2 36.8 J deg 1 • Find S for the melting of 5g of ice at 0°C at 1 atm. ( L f 79.7cal g 1 ) This process is reversible since both ice and liquid states are possible at 0°C. (Equilibrium states!) S • Lf m T 79.7cal ( g )(5 g ) 1.46 cal 6.1 J K K 273K For freezing of liquid water, qrev 0 and Why the specific signs for melting vs. freezing? S 6.1 J K • Let n moles of a perfect gas undergo an adiabatic free expansion into a vacuum (this is Joule’s experiment). – Express ∆S in terms of the initial and final temperature and volume. – Calculate ∆S if V2 2V1 • The initial state is T1 ,V1 and the final state is T2 ,V2 Although the process is adiabatic, (dq=0), entropy change is non-zero because this process is irreversible. Entropy of course changes (and increases) since the molecules changed their configuration expanding into the vacuum. Hence V S nR* ln( 2 V1 ) nR ln 2 * This is the so-called Entropy of Mixing. With V2 2V1