Thermodynamics of solids 5. Unary systems Kwangheon ark Kyung Hee University Department of Nuclear Engineering
5.1. Unary heterogeneous system definition Unary system: one component system. Unary heterogeneous system: one component with several phases (water and ice, water ice steam at triple point of H O, solid and liquid iron during melting) H O Copper Carbon Allotropy: (Greek ʺotherʺ + ʺgrowʺ) is the property of some chemical elements to exist in two or more different forms, known as allotropes of these elements. 5.1. Unary heterogeneous system: definition
5.. Gibbs free energy in a unary system Unary system: one component system. Gibbs phase rule gives, f c p3 p (for one component system, c 1) where f: degree of freedom, c: no. of components, p: no. of phases Hence, maximum number of phases in a unary system is 3. system G(T,,n) dg( T,, n) SdT Vd dn where G n T, : chemical potential Let g as a molar Gibbs free energy (J/mol), then, G ( ng) n n T, T, g In a unary system, the chemical potential of the component in any state is identical with the molar Gibbs free energy for that state. 5.. Gibbs free energy in a unary system 3
Equilibrium condition between or 3 phases T SYS =T SURR ( T,, n ) ( T,, n ) G G ( T,,n ) G ( T,,n ) Since dg( T,, n) SdT Vd dn SYS = SURR At equilibrium, dg=0 (the minimum). dg dn dn ( ) dn 0 At constant T and constant, ( T T, ) dg dg dg dn dn And dn dn dn 0 Tot If two (or three) phases coexist at equilibrium, ( if third phase coexists) (also T T, ) 5.. Gibbs free energy in a unary system 4
hase stability H O g g hts Molar free energy at 1atm. Ice stable Equilibrium g ICE g LIQ s s s ICE LIQ VA T m Liquid stable T b g VA Vapor stable T 5.. Gibbs free energy in a unary system 5
5.3. The Clausius Clapeyron equation The curve that are the domains of two phase equilibria in T space in unary systems can be described mathematically by thermodynamics. Two phase curve: =(T) H O When two phases () are in equilibrium, The arbitrary change of chemical potential of one phase should be the same to that of the other to be in equilibrium, d d d d( g ) sdt vd d dg ( ) s dt v d ( s s ) dt ( v v ) d d s s s dt v v v where s s s, v v v 5.3. The Clausius Clapeyron equation 6
Since, g h s T g h s T h h h s s s T T d h dt T v ( 0 if transformation from solid to liquid) where h : latent heat for transformation from to phase. Homework 5.1. Draw G vs. T graph for three phases of H O at the triple point. Homework 5.. Calculate the slope of the curve indicating the equilibrium between water and ice at 1 atm. And find the slope of the curve between water and steam at 1 atm. Slope = d/dt. Homework 5.3. Calculate the change of melting temperature of ice when the pressure is applied from 1 atm to 100 atm. What is the boiling temperature when we apply 10 atm to the water under heating? use Tv T h 5.3. The Clausius Clapeyron equation 7
Vaporization and sublimation curves H O d dt h T v v v v v when is a vapor phase. d h h = dt RT T RT d h RT h ln RT dt constant Let s assume the vapor as an ideal gas. RT v v h RT 0 e 5.3. The Clausius Clapeyron equation 8
Vapor pressure of metals Homework 5.4. Based on the Clausius Clapeyron equation, find the vapor pressure of water at 10, 0, 50, 80 o C. And compare these results with available experimental data such as steam table. 5.3. The Clausius Clapeyron equation 9
5.4. Equation of state Thermodynamic properties, T, v of a pure substance (elemental gases, pure metal, pure compounds..) are so interrelated that one of the properties can be derived from the other two. Equation of state: ideal gas law f (, V, T, n) = 0 or f (, v, T) = 0 v T R J 8.314 mol K 3 N m where [ ] [ v] [ T] K m mol isobaric isochoric isothermal 5.4. Equation of state 10
Compressibility factor: non ideality of real gases Compressibility factor of a gas is defined as: Z v TR 1 (for ideal gas) 1 (for real gases) Compressibility factor Relative pressure 5.4. Equation of state 11
Equation of state: real gases Van der Waals equation a p )( v b) v ( Intermolecular forces RT Volume of molecules isothermal pv 3 ( pb RT ) v av ab 0 Applicable region of V. Waals equation (gas, SCF) v Unstable region: separation to phases (Liq. & Gas) 5.4. Equation of state
Critical constants of Van der Waals gases At critical point, v T 0 and v T 0 Van der Waals equation a ( )( v b ) v RT a v b v RT eq 1 or Hence, v T RT a 3 ( v b ) v v T RT ( v b ) 3 6 a 4 v At Tc, 0 v T v T 0 a C v b 7 b C 3 8a T C 7 Rb c v c T c are experimentally measurable. ( c T c are more accurately measurable than v c ) b v c 3 or b RT 8 C C 5.4. Equation of state
v C RT C C a 3b 7 b 3 8a R 8 7 Rb 0.375 Reduced pressure, reduced volume and reduced temperature are defined as: r C v r v v C T r T T C Van der Waals equation becomes, 3 ( )( 3v 1) 8T r r v r r 5.4. Equation of state
Equation of state: T v surface for real substance isothermal process 5.4. Equation of state
Triple-point data = 0.006 atm No hase Separation hase Separation 5.4. Equation of state
-v-t surface for He with projection onto the -T plane -v-t surface showing various forms of ice 5.4. Equation of state
Homework 5.5. A cylinder provided with a movable piston contains an ideal gas at a pressure 1, specific volume v 1, and temperature T 1. The pressure and volume are simultaneously increased so that at every instant and v are related by the equation, = Av where A is a constant. (a) Express the constant A in terms of the pressure 1, the temperature T 1, and the gas constant R. (b) Construct the graph representing the process above in the v plane. (c) Find the temperature when the specific volume has doubled, if T 1 =00K. Homework 5.6. A volume V at temperature T contains n A moles of ideal gas A and n B moles of ideal gas B. The gases do not react chemically. (a) Show that the total pressure of the system is given by = A + B (it is called Dalton s law) where A and B are the pressures that each gas would exert if it were in the volume alone. A is called the partial pressure of gas, A. (b) Show that A =x A where x A is the fraction of moles of A in the system. 5.4. Equation of state 18
Stable and unstabe (metastable) equilibrium e, f: thermodynamically stable d: metastable state due to NOT forming gas nuclides in liquid. It is called superheated (liquid) state. a, b: thermodynamically stable c: metastable state due to NOT forming liquid nuclides. It is called supercooled (gas) state. 5.4. Equation of state 19
5.5. Joule Thomson experiment Gas is forced at a constant pressure and at a steady rate through a small hole, or series of holes (a form of plug) to emerge at a constant pressure. There is a finite pressure drop across the plug, so the process is irreversible. The walls of the chamber are thermally insulating and so the process is also adiabatic. Work done on the system, left side: 0 W dv V i V i i i Work done on the system, right side: Net work, W: V f i W dv V f 0 f f f W Wi Wf V i i V f f Thermodynamics 1 st law gives, U U W U V V Or, f i i i i f f U V U V i i i f f f Hence, Hi H f The throttling process is thus an isenthalpic one. 5.5. Joule Thomson experiment 0
Joule and Thomson have measured the temperature and pressure change ( f, T f ) at a steady flow condition by fixing T i and i. By changing the size of the porous plug, series of values of ( f, T f ) were obtained. Joule Thomson coefficient: T H For an ideal gas, For real gases, 0 0 Homework 5.7. Show why =o for an ideal gas using the following equation: T H 1 H T H T 5.5. Joule Thomson experiment 1
5.6. Steady flow process the turbine and the nozzle A fluid flows at a constant rate through a device so that some of the internal energy of the fluid is transformed into mechanical work (air compressor, refrigerator, turbine). 5.6. Steady flow process
The turbine Although the temperature of a gas turbine is considerably higher than that of the surroundings, the gas flow is so rapid that heat loss to surrounding is negligible. 5.6. Steady flow process 3
Flow through a nozzle When a gas flowing down a pipe encounters a change in the cross sectional area, there is a change of gas velocity. Assuming that there is no shaft work (W), the system is horizontal, and no heat (q) is entered into the system, then: 5.6. Steady flow process 4
Homework 5.8. A steam turbine takes in steam at the rate of 600kg/h and its power output is 800 kw. Neglect any heat loss from the turbine. Find the change in the specific enthalpy of the steam as it passes through the turbine if, (a) the entrance and exit are at the same elevation and the entrance and exit velocities are negligible. (b) the entrance velocity is 50 m/s and the exit velocity is 00 m/s, with the outlet pipe being m above the inlet. 5.6. Steady flow process 5
5.7. Thermodynamic cycles A thermodynamic cycle consists of a collection of thermodynamic processes transferring heat and work, while varying pressure, temperature, and other state variables, eventually returning a system to its initial state. In the process of going through this cycle, the system may perform work on its surroundings, therefore acting as a heat engine. Brayton cycle The Brayton cycle is a thermodynamic cycle that describes the workings of a constant pressure heat engine. Gas turbine engines and airbreathing jet engines use the Brayton Cycle. (Ideal) Brayton cycle efficiency, : 1 3 1 R where R 1 4 5.7. Thermodynamic cycles 6
The Carnot cycle of an ideal gas ( reversible process) Hot reservoir T a b: Q T S Q n mole ideal gas c d : Q 1 abc W cd a TS Q 1 1 Cold reservoir T a > b : isothermal expansion of the gas. T=T. The internal energy of an ideal gas is a function of temperature only. The heat flow to the system should be equal to the work done by the system. V Vb ln b V a Va Q W dv nrt b > c : isentropic expansion of the gas. T > T 1 V constant when adiabatic process TV TV 1 1 b 1 c 1 TV constant 5.7. Thermodynamic cycles 7
c > d : isothermal compression of the gas. T=T 1. The internal energy of an ideal gas is a function of temperature only. The heat flow to the surrounding should be equal to the work done by the surrounding on the system. (Q 1 >0) V Vd ln c 1 1 V 1 c Vd Q W dv nrt d > a : isentropic compression of the gas. T 1 > T TV TV 1 1 1 d a 1 1 1 T V d V c T V V a b or, V V b a V V c d Hence, Q Q T T 1 1 Total work done by the system after one cycle, W: W dv ( TdS du ) TdS du TdS T S T S Q Q 1 The efficiency of the Carnot cycle engine, : W Q Q1 Q T T T1 = 1 1 Q Q Q T T 1 1 1 1 1 No engine operating between two reservoirs can be more efficient than a Carnot engine operating between those same two reservoirs. 5.7. Thermodynamic cycles 8
Rankine cycle a b c d h g f e = ( hd ha) ( he hh) h h d ( hd he) ( ha hh) h h d a a ump q Mollier diagram Turbine q 1 5.7. Thermodynamic cycles 9
Steam table T, v, h, s 5.7. Thermodynamic cycles 30
Homework 5.9. Derive the efficiency of the Brayton cycle using the following equations; q in = h 3 h = C (T 3 T ) q out = h 4 h 1 = C (T 4 T 1 ) q q out in 1 T qin const. if the process is adiabatic. Homework 5.10. A Caront engine can be made to operate as a refrigerator. Explain in detail, with the aid of (a) a pressure volume diagram, (b) an enthalpy entropy diagram, all the processes which occur during a complete cycle or operation. This refrigerator freezes water at 0 o C and heat from the working substance is discharged into a tank containing water maintained at 0 o C. Determine the minimum amount of work required to freeze 3 kg of water. (1048) Homework 5.11. Draw a Mollier diagram and a T S diagram of the Rankine cycle for the Korean standard nuclear power plant (OR1000) using FSAR and the steam table. And calculate the efficiency of the cycle. 5.7. Thermodynamic cycles 31
5.8. hase transformation classified according to order A first order transition is one for which the free energy as a function of a given state variable (V,, T) is continuous, whereas the first derivative of the free energy with respect to the given state variable is discontinuous. Discontinuity of the first derivative of G Continuous functions: GT (, n, ) Discontinuous functions (at transition state): G S T G V T ( G/ T) (1 / T ) H Examples: fusion, vaporization and allotropic transformations 3 5.8. hase transformation according to order
A second order transition is one for which the free energy and its derivatives are continuous, whereas the second derivative of the free energy with respect to the given state variable is discontinuous. Continuous functions: GT (, n, ) G S T G V T ( G/ T) (1 / T ) H Discontinuous functions (at transition state): G S C T T T G V T T V ( G/ T) H T (1 / T) T C Examples: order disorder transformation, onset of ferromagnetism (or antiferromagnetism), onset of superconductivity when H=0. 33 5.8. hase transformation according to order
5.9. First order transition the Clausius Clapeyron equation Two phase curve: =(T) H O d dt h T v Tv Tv T h L where h L f : Latent heat, v v: specific volume change f Homework 5.1. Calculate the change in the transition temperature of the following reaction when the gray tin phase is under a hydrostatic pressure of 10 7 N/m. (1) Sn(gray) = Sn(white) T tr =86 K () Fe(ferrite) = Fe (austenite) T tr =1184 K (3) Fe(solid) = Fe(liquid) T tr =1811 K (4) ZrO (monoclinic) = ZrO (tetragonal) at T tr =1445 K 5.9. Gibbs free energy in a unary system 34
Empirically, Simon equation is used for the relation between pressure and melting point. 0 T T Melt C 1 Effect of hydrostatic pressure on transformation in iron. 5.9. Gibbs free energy in a unary system 35
The effect of external pressure on the vapor pressure of a condensed phase dg g dn g dn SYS Conds Conds Gas Gas At equilibrium state, dg 0 or, g g ( dn dn ) SYS Conds Gas Conds Gas dg dg or, Conds Gas Using dg sdt vd vd ( T=const.) dg v d Conds Conds dg v dp Gas Gas Condensed phase volume Applied hydrostatic pressure Vapor volume Vapor pressure dp vconds dp pv or, d v d RT v v Gas Conds Gas Conds if an ideal gas. -4 10 small enough to be negligible. 5.9. Gibbs free energy in a unary system 36