Thermodynamic Relations

1 1.1 1.1.1

Thermodynamic Relations Relations for Energy Properties Internal Energy Change dU

From first law of thermodynamics dU = dQ + dW

(1.1)

For a reversible process dW = −P dV From second law of thermodynamics for a reversible process dQ = T dS Therefore Eqn.(1.1) becomes dU = T dS − P dV 1.1.2

(1.2)

Enthapy Change dH

From the definition of enthalpy H = U + PV

(1.3)

Differentiating dH = dU + P dV + V dP Substituting for dU from Eqn.(1.2) dH = V dP + T dS 1.1.3

(1.4)

Gibbs Free Energy Change dG

From the definition of Gibbs free energy G = H − TS Differentiating dG = dH − T dS − SdT Substituting for dH from Eqn.(1.4) dG = V dP − SdT 1

(1.5)

1.1.4

Helmholtz Free Energy Change dA

From the definition of Helmholtz free energy A = U − TS Differentiating dA = dU − T dS − SdT Substituting for dU from Eqn.(1.2) dA = −P dV − SdT

1.2 1.2.1

Mathamatical Concepts Exact Differential Equations

If F = F (x, y) then dF = M dx + N dy Exactness Criteria:

µ

∂M

¶

∂y

µ = x

¶

∂N ∂x

y

If F = F (x, y, z) then dF = M dx + N dy + P dz Exactness Criteria:

µ

∂M

¶

µ

∂y x,z ¶ µ ∂M ∂z x,y ¶ µ ∂N ∂z 1.2.2

=

=

¶

µ

∂x y,z ¶ ∂P

µ

∂x y,z ¶ ∂P

=

x,y

∂N

∂y

x,z

Cyclic Relation Rule

For the function in the variables x, y & z µ ¶ µ ¶ µ ¶ ∂x ∂y ∂z ∂y

z

∂z

x

2

∂x

= −1 y

(1.6)

1.2.3

Other Relations of Importance µ

∂z

¶

µ =

∂x

∂x ∂y

1.3

∂w

y

µ

∂z

¶

¶ µ y

∂w

¶

∂x

y

1 = ¡ ∂y ¢

z

∂x z

Maxwell Relations

For the fundamental property relations: dU dA dG dH

= = = =

P

T dS − P dV −P dV − SdT −SdT + V dP V dP + T dS

Applying exactness criteria of differential equation: ¶ ¶ µ µ ∂P ∂T = − ∂V S ∂S V ¶ ¶ µ µ ∂P ∂S = ∂T V ∂V T ¶ ¶ µ µ ∂S ∂V = − ∂P T ∂T P ¶ ¶ µ µ ∂V ∂T = ∂S P ∂P S

1.4 1.4.1

I @

@ ¡ @ ¡ ¡ @ ¡ @ ¡ @

U ¡

T

µ

∂U

¶

∂T V ¶ µ ∂H ∂T

P

3

= CV = CP

µ ¡

G @

H PASGVHTU

Relations for Thermodynamic Properties in terms of P V T and Specific heats Definitions

S

A

V

Volume expansivity β β=

µ

1 V

¶

∂V ∂T

P

Isothermal compressibility κ κ=− 1.4.2

µ

1 V

¶

∂V ∂P

T

Relations for dU µ

∂U

¶

µ =T

∂V

∂P

¶ −P

∂T

T

V

Considering U = U (T, V ) ¶ · µ ¸ ∂P dU = CV dT + T − P dV ∂T V For van der Waals gas, dU = CV dT + 1.4.3

a V2

dV

Relations for dH µ

∂H

¶

∂P

µ =V −T

∂V ∂T

T

¶ P

Considering H = H(T, P ) ·

µ

dH = CP dT + V − T 1.4.4

∂V ∂T

Relations for dS µ µ

∂S

¶ =

∂T V ¶ ∂S ∂T

=

P

4

CV T CP T

¶ ¸ dP P

Considering S = S(T, V ) dS =

CV T

µ dT +

∂P

¶ dV

∂T

V

For van der Waals gas, dS =

CV T

dT −

R V −b

dV

Considering S = S(T, P ) dS = 1.4.5

CP T

µ dT −

∂V

¶ dP

∂T

P

Relations for Specific heats µ CP = T CV

= −T

∂P

¶ µ

∂V

¶

∂T S ∂T P ¶ µ ¶ µ ∂V ∂P ∂T

∂T

V

S

Specific heat differences: µ CP − CV = −T

∂P

¶ µ

∂V

∂V

T

∂T

For van der Waals gas, CP − CV = T

V β2 κ

Specific heat ratio: CP CV Specific heat variations: ¶ µ ∂CP

(∂P/∂V )S (∂P/∂V )T µ

∂P T ¶ µ ∂CV ∂V

=

T

∂ 2V

¶

= −T ∂T 2 µ 2 ¶ ∂ P = T ∂T 2 V 5

P

¶2 P

1.5

Two-phase Systems

Equilibrium in a closed system of constant composition: d(nG) = (nV )dP − (nS)dT During phase change T and P remains constant. Therefore, d(nG) = 0. Since dn 6= 0, dG = 0. For two phases α and β of a pure species coexisting at equilibrium: Gα = Gβ where Gα and Gβ are the molar Gibbs free energies of the individual phases. dGα = dGβ dP sat dT 1.5.1

=

dT

∆H αβ T ∆V αβ

Clausius-Clapeyron equation

ln 1.5.3

∆V αβ

Clapeyon equation dP sat

1.5.2

∆S αβ

=

P2sat P1sat

=

∆H

µ

R

1 T1

−

1

¶

T2

Vapor Pressure vs. Temperature

From Clausius-Clapeyron equation ln P sat = A −

B T

A satisfactory relation given by Antoine is of the form ln P sat = A −

B T +C

The values of the constants A, B and C are readily available for many species. 6

1.6

Gibbs Free Energy as a Generating Function µ d

¶

G

=

RT

1 RT

G

dG −

RT 2

dT

Substituting for dG from fundamental property relation, and from the definition of G: µ ¶ G V H d = dP − dT RT RT RT 2 This is a dimensionless equation. V RT H RT

· =

∂(G/RT )

= −T

¸

∂P T · ¸ ∂(G/RT ) ∂T

P

When G/RT is known as a function of T and P , V /RT and H/RT follow by simple differentiation. The remaining properties are given by defining equations. S R U RT

1.7

= =

H RT H RT

− −

G RT PV RT

Residual Properties

Any extensive property M is given by: M = M ig + M R where M ig is ideal gas value of the property, and M R is the residual value of the property. For example for the extensive property V : V = V ig + V R =

RT P

+VR

Since V = ZRT /P VR=

RT P 7

(Z − 1)

For Gibbs free enrgy µ R¶ G VR HR d = dP − dT RT RT RT 2 ·

VR

=

RT HR

∂(GR/RT )

= −T

RT

(1.7)

¸

∂P T · ¸ ∂(GR/RT ) ∂T

(1.8) (1.9)

P

From the definition of G: GR = H R − T S R SR

=

HR

−

GR

(1.10)

R RT RT At constant T Eqn.(1.7) becomes µ R¶ G VR d = dP RT RT Integration from zero pressure to the arbitrary pressure P gives Z P R V GR = dP (const. T ) RT 0 RT

(1.11)

where at the lower limit, we have set GR/RT equal to zero on the basis that the zero-pressure state is an ideal-gas state. (V R = 0) Since V R = (RT /P )(Z − 1) Eqn.(1.11) becomes Z P GR dP = (const. T ) (1.12) (Z − 1) RT P 0 Differentiating Eqn.(1.12) at with respect to T at constant P · ¸ ¶ Z Pµ ∂(GR/RT ) ∂Z dP = ∂T ∂T P P 0 P HR RT

Z = −T

0

P

µ

∂Z ∂T

¶

dP P

P

(const. T )

(1.13)

Combining Eqn.(1.12) and (1.13) and from Eqn.(1.10) ¶ Z Pµ Z P SR dP ∂Z dP = −T − (const. T ) (1.14) (Z − 1) R ∂T P P P 0 0 8

1.8

Generalized Correlations of Thermodynamic Properties for Gases

Of the two hinds of data needed for the evaluation of thermodynamic properties, heat capacities and P V T data, the latter are most frequently missing. Fortunately, the generalized methods developed for compressibility factor Z are also applicable to residual properties. Substituting for P = PcPr and T = TcTr , dP = PcdPr and dT = TcdTr

HR RTc SR R

= Z

= −Tr

0

−Tr2

Pr

µ

Z

µ

Pr

¶

∂Tr

0

∂Z

∂Z

¶

∂Tr

Pr

(const. Tr )

Pr

Pr

dPr Pr

dPr

Z −

Pr

0

(Z − 1)

dPr Pr

(1.15)

(const. Tr ) (1.16)

1.8.1

Three Parameter Models

From three-parameter corresponding states principle developed by Pitzer Z = Z 0 + ωZ 1 Similar equations for H R and S R are: HR RTc SR R

= =

(H R)0 RTc (S R)0 R

+ω +ω

(H R)1

RTc (S R)1 R

(S R)1 (H R)0 (H R)1 (S R)0 and , , Calculated values of the quantities RTc RTc R R are shown by plots of these quantities vs. Pr for various values of Tr . (S R)0 (H R)0 used alone provide two-parameter corresponding states and RTc R correlations that quickly yield coarse estimates of the residual properties.

9

1.8.2

Correlations from Redlich/Kwong Equation of State Z=

1 1−h

−

4.934

µ

Tr1.5

h

¶

1+h

where h=

0.08664Pr ZTr

and Tr = Pr =

1.9

T Tc P Pc

Developing Tables of Thermodynamic Properties from Experimental Data

Experimental Data: (a) Vapor pressure data. (b) Pressure, specific volume, temperature (P V T ) data in the vapor region. (c) Density of saturated liquid and the critical pressure and temperature. (d) Zero pressure specific heat data for the vapor. From these data, a complete set of thermodynamic tables for the saturated liquid, saturated vapor, and super-heated vapor can be calculated as per the steps below: 1. Relation for ln P sat vs. T such as ln P sat = A −

B T +C

2. Equation of state for the vapor that accurately represents the P V T data. 3. State: 1 Fix values for H and S of saturated-liquid at a reference state. 10

4. State: 2 Enthalpy and entropy changes during vaporization are calculated from Clapeyron equation using the ln P sat vs. T data as: dP sat dT

=

∆Hlv T (Vv − Vl)

and ∆Slv =

∆Hlv

T Here Vl shall be measured, and Vv is calculated from the relation obtained in step-2. From these values of ∆Hvl and ∆Svl obtaine the values of H and S at state: 2

11

5. State: 3 Follow the constant pressure line. ¶ ¸ · µ ∂V dH = CP dT + V − T dP ∂T P µ ¶ ∂V dT − dP dS = CP T ∂T P Here for the specific heat of vapor corresponding to the pressure at state: 2 is obtained from the relation: ¶ µ µ 2 ¶ ∂CP ∂ V = −T ∂P T ∂T 2 and from the zero pressure specific heat data. With the value of CP for this state as calculated above, S and H values at state: 3 are calculated. 6. State: 4, 5 & 6 The above calculation can be done along constant temperature lineand the values at states 4, 5 and 6 can be obtained. 7. State: 7 The calculations made for state: 2 can de done for the temperature at state: 6.

12

1.10

Thermodynamic Diagrams of Importance

1.10.1

T − S diagram

13

1.10.2

P − H diagram

14

1.10.3

H − S diagram

15

2

Thermodynamics of Flow Processes

2.1

Conservation of Mass

dm dt

+ ∆(ρuA) = 0

(2.1)

where the symbol ∆ denotes the difference between exit and entrance streams. For steady flow process ∆(ρuA)fs = 0

(2.2)

Since specific volume is the reciprocal of density, m ˙ =

uA V

= constant.

This is the equation of continuity.

16

(2.3)

2.2

Conservation of Energy d(mU ) dt

˙ + ∆[(U + 12 u2 + zg)m] ˙ = Q˙ − W

W = Ws + ∆[(P V )m] ˙

d(mU ) dt

+ ∆[(H + 12 u2 + zg)m] ˙ = Q˙ − W˙ s

(2.4)

(2.5)

(2.6)

For most applications, kinetic- and potential-energy changes are negligible. Therefore d(mU )

+ ∆(H m) ˙ = Q˙ − W˙ s

dt

(2.7)

Energy balances for steady state flow processes: ∆[(H + 12 u2 + zg)m] ˙ = Q˙ − W˙ s

(2.8)

Bernoulli’s equation: P ρ

2.3

+

u2 2

+ gz = 0

(2.9)

Flow in Pipes of Constant Cross-section ∆H +

∆u2

=0

2

(2.10)

In differential form dH = −udu

(2.11)

Equation of continuity in differential form: d(uA/V ) = 0 Since A is a constant, d(u/V ) = 0. Therefore du V

−

udV V2 17

=0

(2.12)

or du =

udV

(const. A)

V

(2.13)

Substituting this in Eqn.(2.11) dH = −

u2 dV

(2.14)

V

From the fundamental property relations T dS = dH − V dP Therefore T dS = −

u2 dV V

− V dP

(2.15)

As gas flows along a pipe in the direction of decreasing pressure, its specific volume increases, and also the velocity (as m ˙ = uA/V ). Thus in the direction of increasing velocity, dP is negative, dV is positive, and the two terms of Eqn.(2.15) contribute in opposite directions to the entropy change. According to second law dS ≥ 0. u2max dV V

+ V dP = 0 (const. S)

Rearranging µ u2max

= −V

2

∂P ∂V

¶ (2.16) S

This is the speed of sound in fluid.

2.4

General Relationship between Velocity and Crosssectional Area d(uA/V ) = 0 1 V

(udA + Adu) − uA

18

dV V2

=0

or udA + Adu uA

V dV

=

V2

From the fundamental property relation for dH and from steady flow energy equation −V dP = udu (const. S) i.e., V = −udu/dP at constant S. Therefore dA A

+

du

=

d

udu −V 2 (∂P/∂V )S

From the relation for velocity of sound, the above equation becomes dA A

+

du u

=

udu u2sonic

Therefore dA A

=

udu u2sonic

−

du u

µ =

u2 u2sonic

¶ du −1 u

The ratio of actual velocity to the velocity of sound is called the Mach Number M. dA A

du

= (M2 − 1)

(2.17)

u

Depending on whether M is greater than unity (supersonic) or less than unity (subsonic), the cross sectional area increases or decreases with velocity increase. includegraphicssupersonic.eps includegraphicssubsonic.eps includegraphicsconvergdiverg.eps

2.5

Nozzles u22 − u21 = −2

Z

P2

V dP = P1

2γP1 V1 γ−1 19

"

µ 1−

P2 P1

¶(γ−1)/γ# (2.18)

From the definition of sound velocity u2max

= −V

2

µ

∂P ∂V

¶ S

and from the evaluation of the derivative (∂P/∂V )S for the isentropic expansion of ideal gas with constant heat capacities from the relation P V γ = const, u2throat = γP2 V2

(2.19)

Substituting this value of the throat velocity for u2 in Eqn.(2.18) and solving for the pressure ratio with u1 = 0 gives ¶γ/(γ−1) µ 2 P2 = (2.20) P1 γ+1 The speed of sound is attained at the throat of a conerging/diverging nozzle only when the pressure at the throat is low enough that the critical value of P2 /P1 is reached. If insufficient pressure drop is available in the nozzle for the velocity to become sonic, the diverging section of the nozzle acts as a diffuser.

2.6

Turbines

˙ W˙ s = −m∆H 20

(2.21)

and Ws = −∆H

(2.22)

Ws(isentropic) = −(∆H)S

(2.23)

η=

2.7

Ws Ws(isentropic)

=

∆H (∆H)S

Throttling Processes ∆H = 0

Joule-Thomson Coefficient: ¶ µ ∂T T (∂V /∂T )P − V = µJ = ∂P H CP 21

(2.24)

For ideal gases µJ = 0. For a real gas uJ can be positive, zero or negative. Any gas for which volume is linear with temperature along an isobar will have a zero Joule-Thomson coefficient. i.e., if V /T = constant = φ(P ), µJ = 0. Inversion curve: T − P diagram. The points in the curve correspond to µJ = 0. In the region inside the curve µJ is positive.

2.8

Compression

Z W =− For reversible-adiabatic compression W =

γP1 V1 γ−1

P2

V dP P1

"

µ 1−

P2

# ¶ γ−1 γ

P1

Effect of clearance on work of compression: # " µ ¶ γ−1 γP1 VI P2 γ W = 1− γ−1 P1 22

Multistage compression:

µ

Optimum compression ratio per stage =

W =

nγP1 VI γ−1

"

µ 1−

Relation between VD and VI : "

P2 P1

µ

VI = VD 1 + C − C

# ¶ γ−1 γn

P2

¶1/γ #

P1

where C = VC /VD For compression in multistages, " # 1 µ ¶ nγ P2 VI = VD 1 + C1 − C1 P1 where C1 is the clearance in the first stage.

23

P2 P1

¶1/n

2.9

Ejectors

24