1. 1 Open Systems -- Part 2 Physics 313 Professor Lee Carkner Lecture 24

2. 2 First Order Phase Transitions Consider a phase transition where T and P remain constant –e.g. boiling water Entropy, volume and enthalpy will change If the molar entropy and volume change, then the process is a first order transition

3. 3 Phase Change Consider a substance in the middle of a phase change from initial (i) to final (f) phases –n 0 is total number of moles Can write equations for properties as the change progresses as: S = n 0 (1-x)s i + n 0 xs f –Where x is fraction that has changed

4. Clausius - Clapeyron Equation Consider the first T ds equation, integrated through a phase change T ds = c V dT + T (dP/dT) dv T (s f - s i ) = T (dP/dT) (v f - v i ) This can be written: dP/dT = (s f -s i )/(v f - v i ) But H = VdP + T ds so the isobaric change in molar entropy is T ds, yielding: dP/dT = (h f - h i )/T (v f -v i )

5. 5 Phase Changes and the CC Eqn. The CC equation gives the slope of curves on the PT diagram –e.g. fusion, vaporization and sublimation The (h f - h i ) quantity is the molar latent heat –Amount of energy that needs to be added to change phase

6. 6 Changes in T and P For small changes in T and P, the CC equation can be written:  P/  T = (h f - h i )/T (v f -v i ) or:  T = [T (v f -v i )/ (h f - h i ) ]  P The CC equation can be used to compute differences in melting T with P

7. 7 Control Volumes Many engineering applications involve moving fluids Often we consider the fluid only when it is within a container called a control volume –e.g. car radiator What are the key relationships for control volumes?

8. Mass Conservation For a steady flow system, mass is conserved Rate of mass flow in equals rate of mass flow out (note italics means rate (1/s))  m in =  m out For single stream m 1 = m 2  1 v 1 A 1 =  2 v 2 A 2 –where v is velocity, A is area and  is density

9. Energy of a Moving Fluid The energy of a moving fluid (per unit mass) is the sum of the internal, kinetic, and potential energies and the flow work w flow = Pv –work needed to move mass Total energy per unit mass is:  = u + Pv + ke + pe Since h = u +Pv  = h + ke +pe (per unit mass)

10. 10 Energy Balance Rate of energy transfer in is equal to rate of energy transfer out for a steady flow system: E in = E out For a steady flow situation:  in [ Q + W + m  ] =  out [ Q + W + m  ] In the special case where Q = W = ke = pe = 0  in m h =  out m h

11. Application: Mixing Chamber A mixing chamber is where several streams of fluids come together and leave as one In general, the following holds for a mixing chamber: Mass conservation:  m in = m out Energy balance:  in m h = m out h out –Only if Q = W = pe = ke = 0

12. 12 Open Mixed Systems Consider an open system where the number of moles (n) can change The internal energy now depends on the various n’s (one for each substance) dU = (  U/  V)dV + (  U/  S)dS +  (  U/  n j )dn j –Where n j is the number of moles of the jth substance

13. 13 Chemical Potential We can simplify with  j = (dU/dn j ) and rewrite the dU equation as: dU = -PdV + TdS +   j dn j –The first term is the work –The second term is the heat –The third term is the chemical potential or: dW C =  j dn j

14. 14 The Gibbs Function Other characteristic functions can be written in a similar form Gibbs function dG = VdP - SdT +  j dn j For phase transitions with no change in P or T:  j = (dG/dn j )

15. 15 Mass Flow How does  j involve mass flow? Consider a divided chamber (sections 1 and 2 ) where a substance diffuses across a barrier dU = TdS +  dn dS = dU/T -(  /T)dn The total dS is the sum of dS for each section dS = dU 1 /T 1 -(   /T 1 )dn 1 + dU 2 /T 2 -(   /T 2 )dn 2

16. 16 Conservation For an isolated, insulated system Sum of dn’s must be zero: dn 1 = -dn 2 Sum of internal energies must be zero: dU 1 = -dU 2 Substituting into the above dS equation: dS = [(1/T 1 )-(1/T 2 )]dU 1 - [(  1 /T 1 )-(  2 /T 2 )]dn 1

17. 17 Equilibrium Consider the equilibrium case No difference in S or T: (  1 /T 1 ) = (  2 /T 2 )  1 =  2 Chemical potentials are equal in equilibrium T drives heat P drives work  drives mass flow