1 / 18

Maxwell Relations

Maxwell Relations. Thermodynamics Professor Lee Carkner Lecture 23. PAL #22 Throttling. Find enthalpies for non-ideal heat pump At point 1, P 2 = 800 kPa, T 2 = 55 C, superheated table, h 2 = 291.76 At point 3, fluid is subcooled 3 degrees below saturation temperature at P 3 = 750 K

Download Presentation

Maxwell Relations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Maxwell Relations Thermodynamics Professor Lee Carkner Lecture 23

  2. PAL #22 Throttling • Find enthalpies for non-ideal heat pump • At point 1, P2 = 800 kPa, T2 = 55 C, superheated table, h2 = 291.76 • At point 3, fluid is subcooled 3 degrees below saturation temperature at P3 = 750 K • Treat as saturated liquid at T3 = 29.06 - 3 = 26.06 C, h3 = 87.91 • At point 4, h4 = h3 = 87.91 • At point 1, fluid is superheated by 4 degrees above saturation temperature at P1 = 200 kPa • Treat as superheated fluid at T1 = (-10.09)+4 = -6.09 C, h1 = 247.87

  3. PAL #22 Throttling • COP = qH/win = (h2-h3)/(h2-h1) = (291.76-87.91)/(291.76-247.87) =4.64 • Find isentropic efficiency by finding h2s at s2 = s1 • Look up s1 = 0.9506 • For superheated fluid at P2 = 800 kPa and s2 = 0.9506, h2s = 277.26 • hC = (h2s-h1)/(h2-h1) = (277.26-247.87)/(291.76-247.87) = 0.67

  4. Mathematical Thermodynamics • We can use mathematics to change the variables into forms that are more useful • Want to find an equivalent expression that is easier to solve • We want to find expressions for the information we need

  5. Differential Relations • For a system of three dependant variables: dz = (z/x)y dx + (z/y)x dy • The total change in z is equal to the change in z due to changes in x plus the change in z due to changes in y

  6. Two Differential Theorems (x/y)z = 1/(y/x)z (x/y)z(y/z)x = -(x/ z)y • e.g., P,V and T • May allow us to rewrite equations into a form easier to solve

  7. Legendre Differential Transformation • For an equation of the form: • we can define, • and get: • We use a Legendre transform when f is not a convenient variable and we want xdu instead of udx • e.g. replace PdV with -VdP

  8. Characteristic Functions • The internal energy can be written: dU = -PdV +T dS H = U + PV dH = VdP +TdS • These expressions are called characteristic functions of the first law • We will deal specifically with the hydrostatic thermodynamic potential functions, which are all energies

  9. Helmholtz Function • From dU = T dS - PdV we can define: dA = - SdT - PdV • A is called the Helmholtz function • Used when T and V are convenient variables • Can be related to the partition function

  10. Gibbs Function • If we start with the enthalpy, dH = T dS +V dP, we can define: dG = V dP - S dT • Used when P and T are convenient variables • phase changes

  11. A PDE Theorem dz = (z/x)y dx + (z/y)x dy • or dz = M dx + N dy (M/y)x = (N/x)y

  12. Maxwell’s Relations • We can apply the previous theorem to the four characteristic equations to get: (T/V)S = - (P/S)V ( S/V)T = (P/T)V • We can also replace V and S (the extensive coordinates) with v and s

  13. Use to find characteristic functions and Maxwell relations Example: What is expression for dU? plus TdS and minus PdV (T/V)S=-(P/S)V König - Born Diagram A T V U G P S H

  14. Using Maxwell’s Relations • Example: finding entropy • Using the last two Maxwell relations we can find the change in S by taking the derivative of P or V • Example: (Ds/DP)T = -(Dv/DT)P

  15. Clapeyron Equation • For a phase-change process, P is a function of the temperature only • also for a phase change, ds = sfg and dv = vfg, so: • For a phase change, h = Tds: (dP/dT)sat = hfg/Tvfg

  16. Using Clapeyron Equation (dP/dT)sat = h12/Tv12 • v12 is the difference between the specific volume of the substance at the two phases h12 = Tv12(dP/dT)sat

  17. Clapeyron-Clausius Equation • For transitions involving the vapor phase we can approximate: • We can then write the Clapeyron equation as: (dP/dT) = Phfg/RT2 ln(P2/P1) = (hfg/R)(1/T1 – 1/T2)sat • Can use to find the variation of Psat with Tsat

  18. Next Time • Test #3 • Covers chapters 9-11 • For Friday: • Read: 12.4-12.6 • Homework: Ch 12, P: 38, 47, 57

More Related