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