maxwell relations

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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 - PowerPoint PPT Presentation

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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

C = (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 + PVdH = 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

König - Born Diagram

Use to find characteristic functions and Maxwell relations

Example: What is expression for

dU?

plus TdS and minus PdV

(T/V)S=-(P/S)VH

G

A

U

S P

V T

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:

(s/P)T = -(v/T)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

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