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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
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Maxwell Relations Thermodynamics Professor Lee Carkner Lecture 23
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
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
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
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
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
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
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
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
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
A PDE Theorem dz = (z/x)y dx + (z/y)x dy • or dz = M dx + N dy (M/y)x = (N/x)y
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
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
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
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
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
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
Next Time • Test #3 • Covers chapters 9-11 • For Friday: • Read: 12.4-12.6 • Homework: Ch 12, P: 38, 47, 57