1 Thermodynamic Relations 1.1 Relations for Energy Properties 1.1.1 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 (1.2) 1.1.2 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 (1.4) 1.1.3 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 (1.5) 1
Transcript
1 Thermodynamic Relations
1.1 Relations for Energy Properties
1.1.1 Internal Energy Change dU
From first law of thermodynamics
dU = dQ + dW (1.1)
For a reversible process
dW = −PdV
From second law of thermodynamics for a reversible process
dQ = TdS
Therefore Eqn.(1.1) becomes
dU = TdS − PdV (1.2)
1.1.2 Enthapy Change dH
From the definition of enthalpy
H = U + PV (1.3)
Differentiating
dH = dU + PdV + V dP
Substituting for dU from Eqn.(1.2)
dH = V dP + TdS (1.4)
1.1.3 Gibbs Free Energy Change dG
From the definition of Gibbs free energy
G = H − TS
Differentiating
dG = dH − TdS − SdT
Substituting for dH from Eqn.(1.4)
dG = V dP − SdT (1.5)
1
1.1.4 Helmholtz Free Energy Change dA
From the definition of Helmholtz free energy
A = U − TS
Differentiating
dA = dU − TdS − SdT
Substituting for dU from Eqn.(1.2)
dA = −PdV − SdT (1.6)
1.2 Mathamatical Concepts
1.2.1 Exact Differential Equations
If F = F (x, y) then
dF = Mdx + Ndy
Exactness Criteria: (∂M
∂y
)x
=
(∂N
∂x
)y
If F = F (x, y, z) then
dF = Mdx + Ndy + Pdz
Exactness Criteria: (∂M
∂y
)x,z
=
(∂N
∂x
)y,z(
∂M
∂z
)x,y
=
(∂P
∂x
)y,z(
∂N
∂z
)x,y
=
(∂P
∂y
)x,z
1.2.2 Cyclic Relation Rule
For the function in the variables x, y & z(∂x
∂y
)z
(∂y
∂z
)x
(∂z
∂x
)y
= −1
2
1.2.3 Other Relations of Importance
(∂z
∂x
)y
=
(∂z
∂w
)y
(∂w
∂x
)y
(∂x
∂y
)z
=1(
∂y∂x
)z
1.3 Maxwell Relations
For the fundamental property relations:
dU = TdS − PdV
dA = −PdV − SdT
dG = −SdT + V dP
dH = V dP + TdS
Applying exactness criteria of differential equation:T V
P S
U G
H
A
¡¡
¡¡
¡¡
¡µ
@@
@@
@@
@I
PASGVHTU
(∂T
∂V
)S
= −(
∂P
∂S
)V(
∂P
∂T
)V
=
(∂S
∂V
)T(
∂S
∂P
)T
= −(
∂V
∂T
)P(
∂V
∂S
)P
=
(∂T
∂P
)S
1.4 Relations for Thermodynamic Properties in termsof PV T and Specific heats
1.4.1 Definitions
(∂U
∂T
)V
= CV(∂H
∂T
)P
= CP
3
Volume expansivity β
β =1
V
(∂V
∂T
)P
Isothermal compressibility κ
κ = − 1
V
(∂V
∂P
)T
1.4.2 Relations for dU
(∂U
∂V
)T
= T
(∂P
∂T
)V
− P
Considering U = U(T, V )
dU = CV dT +
[T
(∂P
∂T
)V
− P
]dV
For van der Waals gas,
dU = CV dT +a
V 2dV
1.4.3 Relations for dH
(∂H
∂P
)T
= V − T
(∂V
∂T
)P
Considering H = H(T, P )
dH = CPdT +
[V − T
(∂V
∂T
)P
]dP
1.4.4 Relations for dS
(∂S
∂T
)V
=CV
T(∂S
∂T
)P
=CP
T
4
Considering S = S(T, V )
dS =CV
TdT +
(∂P
∂T
)V
dV
For van der Waals gas,
dS =CV
TdT − R
V − bdV
Considering S = S(T, P )
dS =CP
TdT −
(∂V
∂T
)P
dP
1.4.5 Relations for Specific heats
CP = T
(∂P
∂T
)S
(∂V
∂T
)P
CV = −T
(∂P
∂T
)V
(∂V
∂T
)S
Specific heat differences:
CP − CV = −T
(∂P
∂V
)T
(∂V
∂T
)2
P
For van der Waals gas,
CP − CV = TV β2
κ
Specific heat ratio:
CP
CV
=(∂P/∂V )S
(∂P/∂V )T
Specific heat variations:(∂CP
∂P
)T
= −T
(∂2V
∂T 2
)P(
∂CV
∂V
)T
= T
(∂2P
∂T 2
)V
5
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=
∆Sαβ
∆V αβ
1.5.1 Clapeyon equation
dP sat
dT=
∆Hαβ
T∆V αβ
1.5.2 Clausius-Clapeyron equation
lnP sat
2
P sat1
=∆H
R
(1
T1
− 1
T2
)
1.5.3 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 manyspecies.
6
1.6 Gibbs Free Energy as a Generating Function
d
(G
RT
)=
1
RTdG − G
RT 2dT
Substituting for dG from fundamental property relation, and from the defi-nition of G:
d
(G
RT
)=
V
RTdP − H
RT 2dT
This is a dimensionless equation.
V
RT=
[∂(G/RT )
∂P
]T
H
RT= −T
[∂(G/RT )
∂T
]P
When G/RT is known as a function of T and P , V/RT and H/RT fol-low by simple differentiation. The remaining properties are given by definingequations.
S
R=
H
RT− G
RTU
RT=
H
RT− PV
RT
1.7 Residual Properties
Any extensive property M is given by:
M = M ig + MR
where M ig is ideal gas value of the property, and MR is the residual valueof the property.
For example for the extensive property V :
V = V ig + V R =RT
P+ V R
Since V = ZRT/P
V R =RT
P(Z − 1)
7
For Gibbs free enrgy
d
(GR
RT
)=
V R
RTdP − HR
RT 2dT (1.7)
V R
RT=
[∂(GR/RT )
∂P
]T
(1.8)
HR
RT= −T
[∂(GR/RT )
∂T
]P
(1.9)
From the definition of G:
GR = HR − TSR
SR
R=
HR
RT− GR
RT(1.10)
At constant T Eqn.(1.7) becomes
d
(GR
RT
)=
V R
RTdP
Integration from zero pressure to the arbitrary pressure P gives
GR
RT=
∫ P
0
V R
RTdP (const. T ) (1.11)
where at the lower limit, we have set GR/RT equal to zero on the basis thatthe zero-pressure state is an ideal-gas state. (V R = 0)
Since V R = (RT/P )(Z − 1) Eqn.(1.11) becomes
GR
RT=
∫ P
0
(Z − 1)dP
P(const. T ) (1.12)
Differentiating Eqn.(1.12) at with respect to T at constant P[∂(GR/RT )
∂T
]P
=
∫ P
0
(∂Z
∂T
)P
dP
P
HR
RT= −T
∫ P
0
(∂Z
∂T
)P
dP
P(const. T ) (1.13)
Combining Eqn.(1.12) and (1.13) and from Eqn.(1.10)
SR
R= −T
∫ P
0
(∂Z
∂T
)P
dP
P−
∫ P
0
(Z − 1)dP
P(const. T ) (1.14)
8
1.8 Generalized Correlations of Thermodynamic Prop-erties for Gases
Of the two hinds of data needed for the evaluation of thermodynamic proper-ties, heat capacities and PV T data, the latter are most frequently missing.Fortunately, the generalized methods developed for compressibility factor Zare also applicable to residual properties.
Substituting for P = PcPr and T = TcTr,
dP = PcdPr and dT = TcdTr
HR
RTc
= −T 2r
∫ Pr
0
(∂Z
∂Tr
)Pr
dPr
Pr
(const. Tr) (1.15)
SR
R= −Tr
∫ Pr
0
(∂Z
∂Tr
)Pr
dPr
Pr
−∫ Pr
0
(Z − 1)dPr
Pr
(const. Tr)
(1.16)
1.8.1 Three Parameter Models
From three-parameter corresponding states principle developed by Pitzer
Z = Z0 + ωZ1
Similar equations for HR and SR are:
HR
RTc
=(HR)0
RTc
+ ω(HR)1
RTc
SR
R=
(SR)0
R+ ω
(SR)1
R
Calculated values of the quantities(HR)0
RTc
,(HR)1
RTc
,(SR)0
Rand
(SR)1
Rare shown by plots of these quantities vs. Pr for various values of Tr.
(HR)0
RTc
and(SR)0
Rused alone provide two-parameter corresponding states
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
T 1.5r
(h
1 + h
)
where
h =0.08664Pr
ZTr
and
Tr =T
Tc
Pr =P
Pc
1.9 Developing Tables of Thermodynamic Propertiesfrom Experimental Data
Experimental Data:
(a) Vapor pressure data.
(b) Pressure, specific volume, temperature (PV T ) data in the vapor re-gion.
(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 saturatedliquid, saturated vapor, and super-heated vapor can be calculated as per thesteps 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 PV Tdata.
3. State: 1Fix values for H and S of saturated-liquid at a reference state.
10
4. State: 2Enthalpy and entropy changes during vaporization are calculated fromClapeyron 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 ob-tained in step-2.
From these values of ∆Hvl and ∆Svl obtaine the values of H and Sat state: 2
11
5. State: 3 Follow the constant pressure line.
dH = CPdT +
[V − T
(∂V
∂T
)P
]dP
dS = CP
dT
T−
(∂V
∂T
)P
dP
Here for the specific heat of vapor corresponding to the pressure atstate: 2 is obtained from the relation:(
∂CP
∂P
)T
= −T
(∂2V
∂T 2
)
and from the zero pressure specific heat data.
With the value of CP for this state as calculated above, S and Hvalues at state: 3 are calculated.
6. State: 4, 5 & 6 The above calculation can be done along con-stant temperature lineand the values at states 4, 5 and 6 can be ob-tained.
7. State: 7 The calculations made for state: 2 can de done for the tem-perature 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. (2.3)
This is the equation of continuity.
16
2.2 Conservation of Energy
d(mU)
dt+ ∆[(U + 1
2u2 + zg)m] = Q − W (2.4)
W = Ws + ∆[(PV )m] (2.5)
d(mU)
dt+ ∆[(H + 1
2u2 + zg)m] = Q − Ws (2.6)
For most applications, kinetic- and potential-energy changes are negligi-ble. Therefore
d(mU)
dt+ ∆(Hm) = Q − Ws (2.7)
Energy balances for steady state flow processes:
∆[(H + 12u2 + zg)m] = Q − Ws (2.8)
Bernoulli’s equation:
P
ρ+
u2
2+ gz = 0 (2.9)
2.3 Flow in Pipes of Constant Cross-section
∆H +∆u2
2= 0 (2.10)
In differential form
dH = −udu (2.11)
Equation of continuity in differential form:
d(uA/V ) = 0 (2.12)
Since A is a constant, d(u/V ) = 0. Therefore
du
V− udV
V 2= 0
17
or
du =udV
V(const. A) (2.13)
Substituting this in Eqn.(2.11)
dH = −u2dV
V(2.14)
From the fundamental property relations
TdS = dH − V dP
Therefore
TdS = −u2dV
V− V dP (2.15)
As gas flows along a pipe in the direction of decreasing pressure, its specificvolume increases, and also the velocity (as m = uA/V ). Thus in thedirection of increasing velocity, dP is negative, dV is positive, and the twoterms of Eqn.(2.15) contribute in opposite directions to the entropy change.According to second law dS ≥ 0.
u2maxdV
V+ V dP = 0 (const. S)
Rearranging
u2max = −V 2
(∂P
∂V
)S
(2.16)
This is the speed of sound in fluid.
2.4 General Relationship between Velocity and Cross-sectional Area
d(uA/V ) = 0
1
V(udA + Adu) − uA
dV
V 2= 0
18
or
udA + Adu
uA=
V dV
V 2
From the fundamental property relation for dH and from steady flow energyequation
−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=
(u2
u2sonic
− 1
)du
u
The ratio of actual velocity to the velocity of sound is called the Mach NumberM.
dA
A= (M2 − 1)
du
u(2.17)
Depending on whether M is greater than unity (supersonic) or less thanunity (subsonic), the cross sectional area increases or decreases with velocityincrease.includegraphicssupersonic.epsincludegraphicssubsonic.epsincludegraphicsconvergdiverg.eps
2.5 Nozzles
u22 − u2
1 = −2
∫ P2
P1
V dP =2γP1V1
γ − 1
[1 −
(P2
P1
)(γ−1)/γ]
(2.18)
19
From the definition of sound velocity
u2max = −V 2
(∂P
∂V
)S
and from the evaluation of the derivative (∂P/∂V )S for the isentropic ex-pansion of ideal gas with constant heat capacities from the relation PV γ =const,
u2throat = γP2V2 (2.19)
Substituting this value of the throat velocity for u2 in Eqn.(2.18) and solvingfor the pressure ratio with u1 = 0 gives
P2
P1
=
(2
γ + 1
)γ/(γ−1)
(2.20)
The speed of sound is attained at the throat of a conerging/diverging nozzleonly when the pressure at the throat is low enough that the critical valueof P2/P1 is reached. If insufficient pressure drop is available in the nozzlefor the velocity to become sonic, the diverging section of the nozzle acts asa diffuser.
2.6 Turbines
Ws = −m∆H (2.21)
20
and
Ws = −∆H (2.22)
Ws(isentropic) = −(∆H)S (2.23)
η =Ws
Ws(isentropic)=
∆H
(∆H)S
(2.24)
2.7 Throttling Processes
∆H = 0
Joule-Thomson Coefficient:
µJ =
(∂T
∂P
)H
=T (∂V/∂T )P − V
CP
21
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 correspondto µJ = 0. In the region inside the curve µJ is positive.
