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Open Systems -- Part 2

Open Systems -- Part 2. Physics 313 Professor Lee Carkner Lecture 24. First Order Phase Transitions. Consider a phase transition where T and P remain constant e.g. boiling water Entropy, volume and enthalpy will change

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Open Systems -- Part 2

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

  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 2

  3. Phase Change • Consider a substance in the middle of a phase change from initial (i) to final (f) phases • n0 is total number of moles • Can write equations for properties as the change progresses as: S = n0(1-x)si + n0xsf • Where x is fraction that has changed 3

  4. Clausius - Clapeyron Equation • Consider the first T ds equation, integrated through a phase change T ds = cV dT + T (dP/dT) dv T (sf - si) = T (dP/dT) (vf - vi) • This can be written: dP/dT = (sf -si)/(vf - vi) • But H = VdP + T ds so the isobaric change in molar entropy is T ds, yielding: dP/dT = (hf - hi)/T (vf -vi)

  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 (hf - hi) quantity is the molar latent heat • Amount of energy that needs to be added to change phase 5

  6. Changes in T and P • For small changes in T and P, the CC equation can be written: DP/DT = (hf - hi)/T (vf -vi) • or: DT = [T (vf -vi)/ (hf - hi) ] DP • The CC equation can be used to compute differences in melting T with P 6

  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? 7

  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)) Smin = Smout • For single stream m1 = m2 r1v1A1 = r2v2A2 • where v is velocity, A is area and r 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 wflow = Pv • work needed to move mass • Total energy per unit mass is: q = u + Pv + ke + pe • Since h = u +Pv q = h + ke +pe (per unit mass)

  10. Energy Balance • Rate of energy transfer in is equal to rate of energy transfer out for a steady flow system: Ein = Eout • For a steady flow situation: Sin[Q + W + mq] = Sout [Q + W + mq] • In the special case where Q = W = ke = pe = 0 Sinmh = Soutmh 10

  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: Smin = mout • Energy balance: Sinmh = mouthout • Only if Q = W = pe = ke = 0

  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 = (dU/dV)dV + (dU/dS)dS + S(dU/dnj)dnj • Where nj is the number of moles of the jth substance 12

  13. Chemical Potential • We can simplify with mj = (dU/dnj) • and rewrite the dU equation as: dU = -PdV + TdS + Smjdnj • The first term is the work • The second term is the heat • The third term is the chemical potential or: dWC = Smjdnj 13

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

  15. Mass Flow • How does mj involve mass flow? • Consider a divided chamber (sections 1 and 2) where a substance diffuses across a barrier dU = TdS +mdn dS = dU/T -(m/T)dn • The total dS is the sum of dS for each section dS = dU1/T1 -(m1/T1)dn1 + dU2/T2 -(m2/T2)dn2 15

  16. Conservation • For an isolated, insulated system • Sum of dn’s must be zero: dn1 = -dn2 • Sum of internal energies must be zero: dU1 = -dU2 • Substituting into the above dS equation: dS = [(1/T1)-(1/T2)]dU1 - [(m1/T1)-(m2/T2)]dn1 16

  17. Equilibrium • Consider the equilibrium case • No difference in S or T: (m1/T1) = (m2/T2) m1 = m2 • Chemical potentials are equal in equilibrium • T drives heat • P drives work • m drives mass flow 17

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