Principles and Practice of Engineering PE Chemical Reference Handbook by National Council of Examiners for Engineering and Surveying (NCEES) (z-lib.org)

Chapter 2: Mass and Energy Balances

Heats of Formation and Heats of Combustion for Other Organic Compounds

Dht 0 − Dht 0
f c

Name Formula Phase HHV

Methanol CH4O g kJ Btu kJ Btu
Methanol CH4O l mol lb mol mol lb mol
Acetaldehyde C2H4O g –205 –88,100 328,500
Acetaldehyde C2H4O l –239 –103,000 764
Ethylene oxide C2H4O g –171 –73,500
Ethylene oxide C2H4O l –196 –84,300 726 312,100
Acetic Acid C2H4O2 l –53 –22,700
Ethanol C2H6O g –96 –41,200 ——
Ethanol C2H6O l –484 –208,000
Ethylene glycol C2H6O2 l –234 –100,600 ——
–276 –119,000
–460 –197,800 1306 561,500

1263 543,000

875 376,000

1366 587,300

1367 587,700

1190 511,600

Heats of Formation and Heats of Combustion for Inorganic Compounds

Dht 0 − Dht 0
f c

Name Formula Phase HHV

Ammonia NH3 g kJ Btu kJ Btu
Calcium carbide CaC2 s mol lb mol mol lb mol
Calcium carbonate CaCO3 s –45.9 –19,700
Calcium chloride CaCl2 s –62.8 –27,000 383.0 164,700
Calcium chloride CaCl2 6H2O s –1207 –518,900
Calcium hydroxide Ca(OH)2 s –795.0 –342,000 ——
Calcium oxide CaO s –2607 –1,121,000
Carbon C, graphite s –986.6 –424,200 ——
Carbon monoxide CO g –635.6 –273,200
Carbon dioxide CO2 g ——
Hydrochloric acid HCl g 0 0
Hydrogen g –110.5 –47,510 ——
Hydrogen sulfide H2 g –393.5 –169,200
Iron oxide H2S s –92.31 –39,690 ——
Iron oxide FeO s
Iron oxide Fe2O3 s 0 0 ——
Nitric acid Fe3O4 g –20.6 –8860
Nitric oxide HNO3 g –269.0 –115,700 393.5 169,200
Nitrogen dioxide NO g –822.2 –353,500
Nitrogen trioxide NO2 g –1117 –480,300 283.0 121,700
Sodium carbonate NO3 s –134.3 –57,740
Sodium carbonate NaCO3 s 90.29 38,820 ——
NaCO3 10H2O 33.10 14,200
71.13 30,580 ——
–1131 –486,300
–4082 –1,755,000 286.0 123,000

546.3 234,900

——

——

——

——

——

——

——

——

——

©2020 NCEES 90

Chapter 2: Mass and Energy Balances

Heats of Formation and Heats of Combustion for Inorganic Compounds (cont'd)

Dht 0 − Dht 0
f c

Name Formula Phase HHV

Sodium chloride NaCl s kJ Btu kJ Btu
Sodium hydroxide NaOH s mol lb mol mol lb mol
Sulfur oxide g –411.0 –176,700
Sulfur dioxide SO g –426.7 –183,500 ——
Sulfur trioxide SO2 g
Sulfur trioxide SO3 l 5.01 2150 ——
Water SO3 g –296.8 –127,600
Water H2O l –395.8 –170,200 ——
H2O –442.5 –190,300
–241.83 –103,970 ——
–285.83 –122,890
——

——

——

——

2.4.2.3 Temperature change without Phase Change
Q = mcp DT

2.4.2.4 Temperature change with Phase Change
Q = mcp DT for each phase + mDh for each phase change

©2020 NCEES 91

3 THERMODYNAMICS

3.1 Symbols and Definitions

Symbols

Symbol Description Units (U.S.) Units (SI)
lb mole
C Concentration ft3 mol
Btu m3
lbm-cF
cP, cv Specific heat capacity at constant pressure J K = m2
f or constant volume† kg : s2 : K

Ratio of vapor phase flow to feed flow dimensionless

f Fugacity of a pure component lbf =Pa m=N2 kg
fti Fugacity of component i in a mixture in 2 m : s2
G Gibbs free energy
lbf =Pa m=N2 kg
in 2 m : s2

Btu J = kg : m2
s2

g Specific Gibbs free energy† Btu J = m2
lbm kg s2

g Gravitational acceleration ft m
H Enthalpy sec2 s2
Hi Henry's Law constant of component i
Btu J
h Specific enthalpy
K Equilibrium constant lbf =Pa m=N2 kg
K Distribution coefficient in 2 m : s2

©2020 NCEES 92 Btu J = m2
lbm kg s2

varies

dimensionless

Chapter 3: Thermodynamics

Symbols (cont'd)

Symbol Description Units (U.S.) Units (SI)
k
k Isentropic coefficient dimensionless

MW Reaction rate constant _lb mole/ft3i1 − n _mol/m3i1 − n
m sec s
n
n Molecular weight (molar mass) lbm kg
lb mole mol
P
Mass lbm kg

Number of moles lb mole mol

Polytropic coefficient dimensionless

Pressure lbf or psi =Pa m=N2 kg
in 2 m : s2
Critical pressure
Pc Reduced pressure lbf = psia =Pa m=N2 kg
Pr Saturation pressure, or vapor pressure in 2 m : s2
Psat
Partial pressure dimensionless
p Heat
Q Specific heat (heat per unit mass) lbf =Pa m=N2 kg
q Ratio of liquid phase flow to feed flow in 2 m : s2
q Universal Gas Constant
R lbf =Pa m=N2 kg
Rate of reaction in 2 m : s2

Btu J

Btu J = m2
lbm kg s2

dimensionless

ft-lbf J K = kg : m2
lb mole cR mol : mol : K : s2

r lb mole mol
ft 3 −sec m3 : s
r
rc Compression ratio dimensionless
rp Cut-off ratio (Diesel) dimensionless
S Pressure ratio (Brayton) dimensionless

s Entropy Btu J
Specific entropy† oR K
SG Specific gravity Btu J
T lbm-o R kg : K
Tc dimensionless
Tr
t Temperature °R or °F K or °C
U
Critical temperature °R or °F K or °C
u Reduced temperature
Time dimensionless
Internal energy
hr s

Btu J

Specific internal energy† Btu J
lbm kg

©2020 NCEES 93

Chapter 3: Thermodynamics

Symbols (cont'd)

Symbol Description Units (U.S.) Units (SI)
u m
V Velocity ft s
v sec
m3
Volume ft3 m3
kg
Specific volume† ft 3
lbm

W Work Btu or lbf-ft J

w Specific work (work per unit mass) Btu J = m2
lbm kg s2
w Weight fraction
x Mole fraction dimensionless
y Mole fraction
z Mole fraction (typically of the feed) dimensionless
Z Compressibility factor
z Elevation, height dimensionless

a Isentropic compressibility dimensionless

aij Relative volatility for components i and j dimensionless

ft m

in2 1 = m : s2
lbf Pa kg

dimensionless

β Coefficient of expansion 1 1
g Activity coefficient cR K
g Surface tension
h Efficiency lbf dimensionless
κ Isothermal compressibility in
N = kg
µj Joule-Thompson coefficient in2 m s2
lbf
cR-in2 dimensionless
lbf
1 = m : s2
Pa kg

K = K : m : s2
Pa kg

r Density lbm kg
χ Vapor quality of a 2-phase mixture ft3 m3

dimensionless

φ Fugacity coefficient dimensionless

/ Poynting correction factor dimensionless

† Property values on molar basis are denoted by ^. For example, molar volume is vt.

©2020 NCEES 94

Chapter 3: Thermodynamics

3.2 Basic Thermodynamics

3.2.1 State Functions

Intensive properties are independent of mass.

Extensive properties are proportional to mass.

For a single-phase pure component, specifying any two intensive properties specifies the remaining intensive
properties.

Component State Functions U.S. Units SI Units
Absolute pressure psia Pa
Absolute temperature Property °R K
Specific volume P ft 3 m3
T lbm kg
Specific internal energy
v = V Btu J
Specific enthalpy m lbm kg
Btu
Specific entropy u = U lbm J
m kg
Specific Gibbs free energy Btu J
Specific Helmholtz free uh = +Pv= H lbm -°R kg : K
energy m J
Btu kg
s = S lbm
m J
Btu kg
g = h – T s = G lbm
m

ua = −T s = g − P v = A
m

Maxwell Relations c 2T m =−c 2P m
dU = TdS – PdV 2V 2S
S V

dH = TdS + VdP c 2T m = c 2V m
2P 2S
S P

dG = –SdT + VdP c 2S m = −c 2V m
2P 2T
T P

dA = –SdT – PdV c 2S m = c 2P m
2V 2T
T V

Work and Heat

Definition: Work is considered positive if it is directed outward, i.e. done by the system on the surroundings (or subtracted from
the system).

Work from various sources:

d W = P d V + F d s + m g d z + m u d u + γ d As

compression potential surface

mechanical kinetic

©2020 NCEES 95

Chapter 3: Thermodynamics

where
F = force
s = distance
g = gravitational acceleration
z = elevation, height
u = velocity
γ = surface tension
As = surface area

Reversible work: moving along equilibrium path

=Wrev #=PdV # ^PV hd ln V
Irreversible Work: moving along actual path. Example: expansion against constant external pressure.

Wirr = Pext (V2 – V1)
Useful work (i.e., work that can be extracted from the system)

Wf = Wrev – Wirr
Heat and Energy: Heat is considered positive if it is inward (added to the system).

dU = dQ – dW dH = dQ – dWf
dQ = Cv dT at constant volume
dQ = CP dT at constant pressure

Heat Capacity at Constant Pressure

cP = c 2h m
2T
P

Dependence on Pressure: d 2cP n =− Tf 22v p
2P ^2T h2
T P

uHeat Capacity at Constant Volume=c2m
cv 2T
v

Dependence on Volume: d 2cv n = Tf 22p p
2v ^2T h2
T

Common Coefficients

Joule-Thompson coefficient nJ = c 2T m
Coefficient of Expansion 2P
Isothermal Compressibility h

b = 1 c 2v m
v 2T
P

κ =− 1 c 2v m
v 2P
T

©2020 NCEES 96

Chapter 3: Thermodynamics

Isentropic Compressibility a = − 1 c 2v m
Isentropic Coefficient v 2P
s

k = cP = κ
cv a

Derived Relations cP − cv = Tv b2
κ

3.2.1.1 Ideal Gases nJ =− 1 dv − Tc 2v mn =− 1 `v − Tvbj
cP 2T cP
P

c 2h m = − nJcP
2P
T

c 2h m = cPc1 − nJ b m
2P κ
v

c 2P m = b
2T κ
v

For an ideal gas,

=P vt R=T and PP12 vvtt12 T1
T2

where 1 and 2 indicate separate system states.

Alternatively,

=P V n=R T m RT and P1 V1 = n1 T1
MW P2 V2 n2 T2

Common Coefficients for Ideal Gases

Joule-Thompson coefficient nJ = 0
Coefficient of Expansion
Isothermal Compressibility b = − 1
Isentropic Compressibility T

κ = 1
P

a = 1
Pκ

cP − cv = R
For constant heat capacity:
uD = cvDT

Dh = cP DT

Ds = cP ln e T2 o− R ln e P2 o
T1 MW P1

= cv ln e T2 o + R ln d v2 n
T1 MW v1

©2020 NCEES 97

Chapter 3: Thermodynamics

u u3.2.1.2 Incompressible Fluids
= ^T h only

c=v c=P c

uFor constant heat capacity
D = cDT

Ds = c ln T2
T1

Dh = cDT + vDP

For real liquids, assume that k and b are independent of pressure and temperature:

dv = bdT − ldP
v

ln d v2 n = b_T2 − T1i − l_P2 − P1i
v1

For incompressible liquids: dv = 0

c dP m = b
dT l
v

Isothermal Compression (with constant κ)

w = κ: v ` P22 − P12j
2

Adiabatic Compression

w = v_P2 − P1i

3.2.1.3 Ideal Gas Mixtures

Dalton's Law of Partial Pressures
piV = ni RT

/n and =yi Pp=i ni
n
results in P = pi or P = pi + g + pn
i=1

where
pi = partial pressure of component i
ni = moles of component i
yi = mole fraction of component i in gas phase

Amagat's Law of Partial Volumes
PVi = ni RT

/n and y=i VV=i ni
n
results in: V = Vi or V = Vi + g + Vn
i=1

where Vi = partial volume of component i

©2020 NCEES 98

Chapter 3: Thermodynamics

u / uMolar Properties of Ideal Gas Mixtures
t mix = (yi t i)
n

/htmix = (yi hti)
n

/ctv, mix = (yictv,i)
n

/ctP, mix = (yictP,i)
n

where

ui and hi are evaluated at T

To calculate the molar volume of an ideal gas or liquid mixture:

/vtmix = _xi vtii
n

Note that this equation does not apply to density.

When mixing pure components, the entropy of the mixture is

/ /stmix = 1
n _ yi stii + R n yi ln d yi n

where sti is evaluated at T and pi

/The entropy of mixing is Dstmixing = R yi ln d 1 n
yi
n

The Gibbs free energy for the mixture is

/ /gtmix = 1
_ yi gtii − RT yi lnd yi n

n n

3.2.1.4 Equations of State for Nonideal Gases

Van der Waals Equation of State

dP + a n_v − bi = RT
v2

a = 27 R2Tc2
64 Pc

b = RTc
8Pc

where Tc and Pc are the temperature and pressure at the critical point.

Redlich-Kwong Equation of State

P = RT − a + bi
v−b T 1/2v_v

a = 0.42748 R2Tc2.5
Pc

b = 0.08664 RTc
Pc

©2020 NCEES 99

Chapter 3: Thermodynamics

Virial Equation of State

Z = P vt = 1 + B + C + D +g
RT vt vt2 vt3

where B, C, D = virial equation coefficients, accounting for two-body, three-body, and four-body interactions, respectively

Alternatively,

Z = P vt = 1 + Bl P + Cl P2 + Dl P3 g
RT

where Bl, Cl, Dl= virial equation coefficients

The two sets of virial coefficients are related by:

Bl = B
RT

Cl = C − B2
(R T) 2

Dl = D− 3BC + 2 B3
(R T) 3

Generic Cubic Equation of State

P = RT − (vt + a (T) v b)
vt − b e b) (vt +

where

a(T) = substance-dependent constant

b = substance-dependent constant

e = constant for generic cubic equation of state

s = constant for generic cubic equation of state

3.2.1.5 Compressibility

Compressibility and Expansivity

The compressibility factor is a dimensionless number defined by the equation:

Z = P vt
RT

For ideal gas, Z = 1

Theorem of Corresponding States

To first approximation, all fluids have the same compressibility factor when compared at the same reduced temperature and
reduced pressure.

Reduced temperature (Tr) and reduced pressure (Pr) are defined as

=Tr TT=c and Pr P
Pc

©2020 NCEES 100

Chapter 3: Thermodynamics

Compressibility Factor Chart

1.5 0.50
0.70 0.60
1.4
0.90 0.80
1.3 GENERALIZED COMPRESSIBILITY FACTORS 2.00 1.00
1.2 (Zce = 0.27)
3.00
5.00
10.00

15.00

1.1 Tr 15.00

COMPRESSIBILITY FACTOR, Z 1.0 0.80 0.85 0.90 0.95 1.20 2.00 5.00
0.9 SATURATED GAS 1.10 1.50 3.00
0.8 1.30
1.00 2.00
1.80
0.7
1.70
0.6 1.60
1.50
0.75
1.40
0.5 0.80 1.35
0.85
1.30
1.25

1.20

1.15

0.4 0.90 1.10 0.90
1.08
0.95 1.06
1.04

0.3 1.02

0.2 1.00

0.1 SSAATTUURRAATTEEDDLLIQIQUUIDID 0.50 1.0 2.0 3.0 4.0 5.0 10 20 30
0.0
0.2 0.3 0.4 0.5
0.1

REDUCED PRESSURE, Pr

Source: From de Nevers, Noel, Physical and Chemical Equilibrium for Chemical Engineers, 2nd ed., New York: Wiley & Sons, 2012.

3.2.1.6 Multicomponent Systems

The properties of a mixture can be estimated using the properties of its pure components, based on either a mass-fraction average
or a mole-fraction average. The one exception is entropy, which must be estimated based only on a mole-fraction average.

Mass based mixture properties

u / u / u=mix m1=mi i wi i

nn

/ /=hmix m1=mihi wihi

nn

/ /=cv,mix m1=micv,i wicv,i

nn

/ /=cP,mix m1=micP,i wicP,i

nn

©2020 NCEES 101

Chapter 3: Thermodynamics

3.2.1.7 Boiling Point Elevation and Freezing Point Depression

For dilute solutions of nonvolatile solutes in a solvent, the solution has a greater boiling point and lower freezing point than the
solvent alone.

Boiling point elevation

DTb = R Tb2px
Dhvap

where

x = solute mole fraction

Tbp = boiling point of the pure solvent at the system pressure

Δhvap = latent heat of vaporization of the pure solvent at boiling point Tbp and the system pressure

Freezing point depression

DTm = R T 2 x
mp

Dhfusion

where

x = solute mole fraction

Tmp = melting point of the pure solvent at the system pressure
Δhfusion = latent heat of fusion of the pure solvent at the melting point Tmp and the system pressure

©2020 NCEES 102

Chapter 3: Thermodynamics

For common aqueous solutions, the boiling point elevation can be determined graphically from the solution temperature and the
solute mass fraction.

Boiling-Point Rise of Aqueous Solutions

SUCROSE 12 120 340
10 110 320
60 55 50 8 100 300
CITRIC 6 90 280
45 40 80 260
ACID 70 240
KRAFT LIQUID BOILING POINT RISE, °F 60 220
35 50 200
GLYCEROL WEIGHT 40 180
25
NaNO3 30

(NH4)2SO4 PERCENT

Ca(NO3)2 K2CO3 S2O0LIDS
KCl

HNO3 15 4
LiNO3
H2SO4 SOLUTION TEMPERATURE, °F
10
60 2 BOILING POINT RISE, °F

CaCl2 55 5

NaCl 50 4W5EIGHT P4ER0CENT SOLIDS 0

NaOH KOH LiCl MgCl2

35

EXAMPLE: 30 30 160
At 270 °F, a 22% CaCl2 solution has a 25 20 140
b.p.r. of 9.7 °F 20

NOTE: 15 10 120
Points shown based mainly on atmospheric
boiling point

0 100

Source: Republished with permission of McGraw-Hill, Inc., from Perry's Chemical Engineers' Handbook, 8th ed.,
Don W. Green and Robert H. Perry, New York, 2008; permission conveyed through Copyright Clearance Center, Inc.

©2020 NCEES 103

Chapter 3: Thermodynamics

Boiling-Point Rise of Aqueous Solutions

70

7 170
60 160
SUCROSE
150
60 55 50 6 50 140
CITRIC 5
45 40 4 130
ACID 3 40 120
KRAFT LIQUID BOILING POINT RISE, °F
35 110
GLYCEROL WEIGHT 30 100
25
NaNO3 30 90
20 80
(NH4)2SO4 PERCENT

Ca(NO3)2 K2CO3 S2O0LIDS
KCl

H2SO4 HNO3 15 SOLUTION TEMPERATURE, °C
LiNO3 2

60 10 BOILING POINT RISE, °C
1
CaCl2 55 5 0

NaCl 50 4W5EIGHT P4ER0CENT SOLIDS

NaOH KOH LiCl MgCl2

35

EXAMPLE: 30 70
At 132 °C, a 22% CaCl2 solution has a 25 10 60
b.p.r. of 5.4 °F 20

NOTE: 15 50
Points shown based mainly on atmospheric
boiling point

0 40

Source: Republished with permission of McGraw-Hill, Inc., from Perry's Chemical Engineers' Handbook, 8th ed.,
Don W. Green and Robert H. Perry, New York, 2008; permission conveyed through Copyright Clearance Center, Inc.

©2020 NCEES 104

Chapter 3: Thermodynamics

3.2.2 First and Second Laws of Thermodynamics

The First Law of Thermodynamics states that energy is neither created nor destroyed but can change from one form into another.
The net energy crossing the system boundary is equal to the change in energy inside the system.

Changes in state functions are calculated by changes in Q and W, which are path-dependent. The common paths are

Isobaric DP = 0
Isochoric DV = 0

Isothermal DT = 0

Isenthalpic DH = 0
DQ = 0
Adiabatic DS = 0

Adiabatic and reversible DS > 0
(isentropic)

Adiabatic and irreversible
(polytropic)

Changes in state functions for irreversible processes are calculated by selecting an alternative, reversible process between the
same two states.

3.2.2.1 Entropy (S)

The entropy of a system represents the unavailability of the thermal energy of a system for conversion to mechanical work. It is a

measure of the microscopic disorder of a system.

uds = 1 d + P dv (Gibbs equation)
T T

ds = dQrev
T

= cvd ln T + c 2P m dv
2T
v

= cpd ln T − c 2v m dP
2T
P

Pressure dependence of entropy:

d 2s n = − bV
2p
T

Entropy requires a reference state; absolute entropy cannot be determined.

3.2.2.2 Internal Energy (U)

Energy of a system that is associated with molecular and atomic motions and forces

u = u_T,oi

uVolume dependence of internal energy2=2P −
c 2v m Tc 2T m P

T v

For ideal gases, the internal energy does not depend on volume.

uTemperature dependence of internal energyc2m=cv
2T (heat capacity at constant volume)
v

uPressure dependence of internal energy2=−
c 2P m `kP bT jv

T

©2020 NCEES 105

Chapter 3: Thermodynamics

3.2.2.3 Enthalpy (H)

uThe enthalpy of a system reflects its ability to do work.
h = + Pv
dh = Tds + vdP
Pressure dependence of enthalpy

c 2h m = v − Tc 2v m = v`1 − bT j
2P 2T
T P

Temperature dependence of enthalpy

c 2h m = cP (heat capacity at constant pressure)
2T
P

Enthalpy requires a reference state; absolute enthalpy cannot be determined.

3.2.2.4 Gibbs Free Energy (G)
Gibbs free energy indicates the potential for reversible work that a system can do at constant pressure and temperature.

g = h – Ts

dg = νdP – sdT

dd g n= v dp − h dT
RT RT RT2

At equilibrium, the Gibbs free energy attains a minimum value.

3.2.2.5 Helmholtz Free Energy (A)

The Helmholtz free energy is the potential for a system to do work at constant volume and temperature. It is the part of the
internal energy which is used for useful work.

ua = – Ts

da = – Pdv – sdT

3.2.2.6 Closed Thermodynamic Systems
In a closed thermodynamic system, no mass crosses the system boundary:

Q – W = DU + DKE + DPE

where

DU = change in internal energy DU = m(u2 – u1)
DKE = change in kinetic energy
DPE = change in potential energy DKE = 1 m`u22 − u12j, where u is velocity
2

DPE = mg(z2 – z1)

Energy can cross the system boundary only in the form of heat or work. Work can be shaft work (ws) or other work forms, such as
electrical.

©2020 NCEES 106

Chapter 3: Thermodynamics

Closed System Energy Changes for 1 Mole of an Ideal Gas

Path Isochoric Isobaric Isothermal Isentropic Polytropic
Volume (DV = 0) (DP = 0) (DT = 0) (Ds = 0)
Pressure
Temperature vt1 T1 vt1 P2 vt1 11 vt1 11
vt2 T2 vt2 P1 vt2 T2 k−1 P2 k vt2 T2 n−1 P2 n
Work (w) Dvt = 0 = = = e T1 = e P1 = e T1 = e P1
o o o o

P1 = T1 DP = 0 P1 = vt2 P1 = d vt2 k = e T2 k P1 = d vt2 n = e T2 n
P2 T2 P2 vt1 P2 vt1 T1 k−1 P2 vt1 T1 n−1
n n
o o

T1 P1 vt1 T1 T1 vt2 k−1 P2 k−1 T1 vt2 n−1 P1 n−1
T2 P2 vt2 T2 T2 vt1 P1 k T2 vt1 P2 n
= = DT = 0 = d n = e o = d n = e
o

R T ln e P1 o k − 1 −
P2 k n
0 P_vt2 − vt1i k R T1 >1 P2 o H P1vt1 P2 n 1
k −1 P1 n−1 P1
R T ln e vt2 o − e d n>1 − e o H
vt1

R T ln e P1 o
P2
Heat (q) ctv DT ctP DT 0 ctv_n − kiDT
vt2 n−1
R T ln e vt1 o

uChange in Internal ctv DT ctP DT − P (vt2 − vt1) 0 − R T1 >1 − e P2 k −1 H ctv DT
k−1 P1 k
Energy (D ) o

Change in ctv ln e T2 o ctP ln e T2 o qt 0 ctv_n − k i ln e T2 o
Entropy (Ds) T1 T1 T n−1 T1

Change in ctP DT ctP DT 0 ctP DT ctP DT
Enthalpy (Dh)

Note: Enthalpy for a closed system does not have a physical meaning, because the pv term does not represent flow work.

where k = ctP
ctv

3.2.2.7 Open Thermodynamic Systems

In an open thermodynamic system, mass does cross the system boundary. Flow work (Pv) is defined as the work for mass
entering and leaving the system.

#Reversible flow work = wrev = - v d P + DKE + DPE

Open System First Law (energy balance):

/ / umo i u 2 u 2 Qo in d_ms si
>hi + i + g ziH - mo e >he + e + g zeH + - Wo net = dt
2gc 2gc

where

subscripts i and e refer to inlet and exit states of the system

©2020 NCEES 107

Chapter 3: Thermodynamics

Wo net = rate of net or shaft work
mo = mass flow rate
h = enthalpy
g = acceleration of gravity
gc = gravitational constant
z = elevation
u = velocity
ms = mass of fluid within the system
us = specific internal energy of system
Qin = rate of heat transfer (neglecting kinetic and potential energy of the system)
The table below displays the work, heat, and internal enthalpy changes in open systems for each of the five applicable paths
for 1 mole of ideal gas. These changes assume constant heat capacities and neglect kinetic and potential energy changes.

Open System Energy Changes for 1 Mole of Ideal Gas at Steady State

Path Isochoric Isobaric Isothermal Isentropic Isenthalpic Polytropic
Work (w) − DH
Heat (q) (DV = 0) (DP = 0) (DT = 0) (DS = 0) (DH = 0)
ctv_n − kiDT
RT ln e P1 o k R T1 >1 − e P2 k − 1 n−1
P2 k −1 P1 k
o H

–vt (P2–P1) 0 RT ln e vt2 o = P2 vt2 − P1 vt1 0
ctv DT ctP DT vt1 1−k 0

=− cv DT

RT ln e P1 o
P2
0
vt2
RT ln e vt1 o

Change in ctv DT + vt_P2 − P1i ctP DT 0 − k RT1 P2 k − 1 0 ctp DT
Enthalpy (Dh) k−1 P1 k
>1 − e o H

Change in 0 =ctv_nn−−1k iG ln e T2 o
Entropy (Ds) T1

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Chapter 3: Thermodynamics

3.2.2.8 Steady-Flow Thermodynamics Systems

The system does not change state with time. This assumption is valid for the steady operation of turbines, pumps, compressors,
throttling valves, nozzles, and heat exchangers, including boilers and condensers. The letter V denotes velocity in the following
three equations:

/mo i fhi + u 2 + g zip - /mo e fhe + u 2 + g zep + Qo - Wo s = 0 /mo i = /mo e
i e
2gc 2gc

For a single fluid-flow stream at steady state, the equation reduces to:

Dh + Du 2 + g Dz + wo - qo = 0
2gc

If the fluid is incompressible with negligible friction losses, the equation reduces to:

DP + Du 2 + g Dz + wo s - qo = 0
t 2gc

3.3 Work, Heat, and Efficiency

3.3.1 Efficiency

Overall efficiency is the ratio of useful energy output of a process to the energy input

h overall = useful energy out
total energy in

Reversible process efficiency compares the work output of the actual process to that of the reversible process. This efficiency
expresses the losses in performance due to friction and other non-reversible contributions.

3.3.2 Compression and Expansion

3.3.2.1 Work in Reversible Compression Processes P
P2
Work for compression depends on the type of process
(equations below are for an ideal gas): P1

Isothermal compression (T = constant): ISENTROPIC (n = k)
POLYTROPIC (1 < n < k)
Pv = constant ISOTHERMAL (n = 1)

wrev, T = RT lne P2 o 1
P1 v

Isotropic compression (s = constant):

=Pvk c=onstant where k cP
cv

k k R_T2 − T1i k RT1 P2 k−1
− − MW − MW P1 k
wrev,is = _ P2v2 − P1v1i = = >1 −e o H
1 k 1 k 1 k

Polytropic compression:
=Pvn c=onstant where n polytropic coefficient (empirical)

wrev,poly = 1 n n _ P2v2 − P1v1i = 1 n n R_T2 − T1i = 1 n n RT1 >1 − e P2 n−1
− − MW − MW P1
on H

Isotropic efficiency is the efficiency of the actual process compared to an isentropic process:

his = wrev,is
wactual

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Chapter 3: Thermodynamics

Polytropic efficiency is the efficiency of the actual process compared to a polytropic process:

hpoly = wrev,poly
wactual

For the same process, the polytropic efficiency is higher than the isentropic efficiency.

For multistage compression:

Intercooling between stages reduces the work required for compression.

P WORK SAVED T P2
2 POLYTROPIC Px
P1
P2
2
Px INTERCOOLING
1 T1 1
ISOTHERMAL INTERCOOLING
P1

vs

Maximum reduction in work (in comparison to single-stage compression) occurs when the pressure ratio (PR) for each stage is
the same:

1

=PR e=PP12 om where m number of stages

3.3.2.2 Efficiency of Compressors

Compressors consume power to add energy to the working fluid. This addition of energy results in an increase in fluid pressure
(head).

Work of compression for an adiabatic compressor (assuming negligible changes in potential and kinetic energy and constant

specific heats):

Wocomp = − mo _he − hij INLET

For an ideal gas with constant specific heats:

Wocomp = − mo cp_Te − Tij COMPRESSOR •

Per unit mass: W in

wcomp = − cp_Te − Tij EXIT

If the change in kinetic energy is not negligible, the work of compression is:

Wocomp =− mo fhe − hi + ue2 − ui2 p =− mo fcp_Te − Tij + ue2 − ui2 p
2gc 2gc

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Chapter 3: Thermodynamics

The isentropic efficiency of a compressor is the ratio between the work for the isentropic process and the work of actual process.

The exit state for the actual process and the isentropic process are different, though the exit pressures are the same.

hc,is = wactual = hes − hi = Tes − Ti e PERXEISTSAUCRPTE2UAL
wisentropic he − hi Te − Ti PROCESS

Where "i" designates the inlet conditions, "e" the exit conditions and "es" the exit h
he
conditions for the isentropic process. hes

Actual work of compression for an ideal gas and negligible kinetic energy: es

Adiabatic compression: wa ws ISENTROPIC
hi PROCESS
Wocomp = − mo Pi k >e Pe 1 − 1 1H
− 1i ti hC, Pi k Pi
o − INLET
_k is STATE

where i s

Wocomp = fluid or gas power ses = si

Pi = inlet or suction pressure

Pe = exit or discharge pressure

ri = inlet gas density

hC, is = isentropic compressor efficiency

Isothermal compression: mo Pi ln Pe
=Wocomp M=W mhoCR, isToithermal ln PPei ti hC Pi

where

Wocomp, Pe, Pi, and hC = as defined for adiabatic compression, above

Ti = inlet temperature of gas

3.3.2.3 Efficiency of Turbines INLET
For an adiabatic turbine with DPE = 0 and negligible DKE:
Woturb = mo _hi − hej

For an ideal gas with constant specific heats: TURBINE •

Woturb = mo cp_Ti − Tej Wout

Per unit mass:

wturb = cp_Ti − Tej

For a turbine where DKE is included with constant specific heats: EXIT

Woturb = mo fhe − hi + ue2 − ui2 p = mo fcp_Te − Tij + ue2 − ui2 p
2gc 2gc

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Chapter 3: Thermodynamics

The isentropic efficiency of a turbine is the ratio of the actual work output of the INLET Pi
turbine to the work output that would have been achieved if the process between the h STATE
inlet state and the exit pressure were isentropic. The exit state for the actual process ACTUAL
and the isentropic process are different, though the exit pressures are the same. hi i PROCESS
e PREEXSITSUREP2
hT,is = wactual = hi − he = Ti − Te wa ws
wisentropic hi − hes Ti − Tes he ISENTROPIC
hes PROCESS
Where "i" designates the inlet conditions, "e" the exit conditions and "es" the exit es
conditions for the isentropic process. s

3.3.2.4 Efficiency of Nozzles ses = si

Nozzles are adiabatic devices to accelerate a fluid. A nozzle does not involve work
interactions and generally, changes in potential energy are minimal, so only the change in
kinetic energy is considered. Typically, inlet velocities are assumed to be negligible.

hi − he = ue2
2gc
h
The isentropic efficiency of a nozzle is defined as the ratio of the actual INLET Pi
hi STATE
kinetic energy of the fluid at nozzle exit to the kinetic energy at the exit of an V e2 /2gc ACTUAL
i PROCESS
isentropic nozzle for the same inlet state and exit pressure. The exit state for the he ISENTROPIC PROCESS
hes Ve2s /2gc e PREEXSISTURE P2
actual process and the isentropic process are different, though the exit pressures
es
are the same.

hT,is = ue2 = hi − he
ue2s hi − hes

Where "i" designates the inlet conditions, "e" the exit conditions and "es" the

exit conditions for the isentropic process.

3.4 Power cycles ses = si s

Power cycles can be characterized by

• Gas working fluid (cycle without phase change) or vapor working fluid (cycle with phase changes)

• Closed system (working fluid returned to its original state) and open system (working fluid is renewed after each cycle)

• Internal combustion (fuel burned within the system boundaries) and external combustion (energy supplied to the working fluid
from an external source)

3.4.1 Efficiency of Power Cycles

Net work:
The area enclosed by the process curve on the T-s diagram and on the P-v diagram represents the net work of the cycle.

PT

23 2 3
wnet wnet 4

1 4 1

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Chapter 3: Thermodynamics

The thermal efficiency of any power cycle is defined as:

hth = wout
Qin

Absolute-value signs are used with the heat and work quantities to make the equations independent of sign conventions.

In power production, the objective is to maximize the work obtained from the turbine.

Co-generation system:

In a co-generation system, both the work from the turbine (Wturbine) and the heat output (steam) are utilized.

Fuel Chargeable to Power (FCP) is an efficiency measure that considers both the work output from the turbine and the heat output

(Qsteamgen—i.e., the heat output used productively in the form of steam generated rather than simply the heat exhausted).

FCP = Wturbine + Qsteamgen
Qin

Units for FCP are usually Btu/kWh produced or Btu/hp produced.

3.4.2 Gas Power Cycles

For combustion engines, the following simplifications are made (air-standard cycle):

• Working fluid is air, which behaves like an ideal gas

• All processes are reversible

• The combustion process is replaced by heat addition from an external source

• The exhaust process is replaced by a heat rejection process

• Heat capacities are constant, values at room temperature are used (cold-air-standard cycle)

HEAT

AIR COMBUSTION COMBUSTION AIR HEATING AIR
FUEL CHAMBER PRODUCTS SECTION

(a) ACTUAL (b) IDEAL

Reciprocating Engines (Piston-Cylinder)
TDC = Top dead center—piston position where volume is at minimum
BDC = bottom dead center—piston position where volume is at maximum
Stroke = difference between TDC and BDC
Bore = diameter of the piston
Clearance volume = minimum volume when piston is at TDC
Displacement volume = volume between BDC and TDC

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Chapter 3: Thermodynamics

Compression ratio:

=r VV=mmainx VBDC
VTDC

Mean Effective Pressure (MEP)

MEP is a fictitious pressure that produces the same amount of net work as the actual cycle.

MEP = Wnet
Vmax − Vmin

3.4.2.1 Carnot Cycle

The Carnot cycle is used to calculate the maximum obtainable output from an ideal heat engine.

Step Process Work and Heat State Functions

w1 " 2 = − RTH ln d v2 n Ds1 " 2 = − R ln d v2 n DT1 " 2 = 0
v1 v1 P2/P1 = v1/v2
Isothermal
1→2 expansion q1 " 2 = qin = qH uD 1 " 2 = 0
(ΔT = 0) = TH _s2 − s1i = − w1 " 2

Dh1 " 2 = 0 v1 " 2 = v2 − v1

w2 "3 =− k k 1 RTH Ds2 " 3 = 0 DT2 " 3 = TC − TH
−
uD 2 " 3 = cv`TC − TH j
Isentropic P3 ^k − 1h/k P3 TH k v2 k
expansion P2 P2 TC k−1 v3
2→3 (turbine) : >1 − e o H = e = d n
o

1
v2 TH k−1
q2 " 3 = 0 Dh2 " 3 = cP`TC − TH j v3 = e TC
o

w3 " 4 = − RTc ln d v4 n Ds3 "4 =− R ln d v4 n DT3 " 4 = 0
v3 v3
Isothermal
3→4 compression q3 " 4 = qout = qc uD 3 " 4 = 0 P4 = v3
(ΔT = 0) = Tc_s4 − s3j = − w3 " 4 P3 v4

Dh3 " 4 = 0 v3 " 4 = v4 − v3

Ds4 " 1 = 0 DT4 " 1 = TH − TC

w4 "1 =− k k 1 RTc
−

Isentropic P1 ^k − 1h/k uD 4 " 1 = cv`TH − TCj P1 TC k v4 k
compression P4 P4 TH k−1 v1
4→1 (compressor) : >1 − e o H e o = e = d n
o

1
v4 Tc k−1
q4 " 1 = 0 Dh4 " 1 = cP`TH − TCj v1 = e TH
o

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Chapter 3: Thermodynamics

P qin T qin
1 TH 1 2

T = CONST.

H

ISENTROPIC2 ISENTROPIC
ISENTROPIC ISENTROPIC

4 TC 4 3

qout TC = CONST. qout

3

v s

where

TC = temperature at which the working fluid is absorbing heat, which is the same as that of the fluid entering the turbine

TH = temperature at which heat is emitted by the working fluid, which is the same as the exit temperature from the turbine

The thermal efficiency of the Carnot Cycle is:

hth,Carnot = Wout = 1− Qout = TH − TC
Qin Qin TH

The efficiency of the Carnot cycle represents an upper limit for the efficiency (maximum possible efficiency) of any power cycle
operating between the two temperatures TH and TC.

3.4.2.2 The Stirling Cycle

The Stirling cycle is similar to the Carnot cycle, but the isentropic compression and expansion are replaced by constant volume
processes with regeneration.

Step Process Work and Heat
1→2
2→3 Isothermal expansion q1 " 2 = qin = RTH ln e P1 o =− w1 " 2
3→4 (ΔT = 0) P2

Isochoric regeneration (internal heat transfer q2 " 3 = qregen = cv`TC − TH j
from working fluid to regenerator) w2 " 3 = 0

Isothermal compression q3 " 4 = qout = RTC ln e P3 o =− w3 " 4
(ΔT = 0) P4

4→1 Isochoric regeneration (internal heat transfer q4 " 1 = qregen = cv`TH − TCj
back from the regenerator to the working fluid) w4 " 1 = 0

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Chapter 3: Thermodynamics

P T 1 qin 2
1 TH
qin REGENERATION
TC V = CONST.
T 2 4 V = CONST.

4 REGENEHRA=TCIOONNST. 3
v
qout T = CONST.

C

qout 3

s

The thermal efficiency of the Stirling cycle is:

hth,Stirling = Wout = 1− Qout = TH − TC
Qin Qin TH

The thermal efficiency of the Stirling cycle is the same as the thermal efficiency of the Carnot cycle.

3.4.2.3 The Ericsson Cycle

The Ericsson cycle is another variation of the Stirling cycle—the regeneration takes place at constant pressure rather
than constant volume.

Step Process Work and Heat
1 → 2 Isothermal expansion _DT = 0i
q1 " 2 = qin = RTH ln e P1 o =− w1 "2
P2

2→3 Isobaric regeneration (internal heat transfer q2 " 3 = qregen = cP`TC − TH j
from working fluid to regenerator) w2 " 3 = P2_v3 − v2j

3 → 4 Isothermal compression _DT = 0i q3 " 4 = qout = RTC ln e P3 o = − w3 " 4
P4

4→1 Isobaric regeneration (internal heat transfer back q4 " 1 = qregen = cP`TH − TCj
from the regenerator to the working fluid) w4 " 1 = P1_v1 − v4i

P 1 qin T
4 1 qin 2
T
REGENERATION H = CONST. TH
T L= CONST.
2 REGENERATION
v P = CONST.
P = CONST.

qout 3 TC 4 qout 3

s

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Chapter 3: Thermodynamics

The thermal efficiency of the Ericsson cycle is:

hth,Ericsson = Wout = 1− Qout = TH − TC
Qin Qin TH

The thermal efficiency of the Ericsson cycle is the same as the thermal efficiency of the Carnot cycle.

3.4.2.4 Otto Cycle
Idealized cycle to represent spark-ignition internal combustion engines

Step Process Work and Heat
1 → 2 Isentropic compression
w1 "2 =− k k 1 RT1 : >1 − e P2 ^k − 1h/k
− P1
oH

q1 " 2 = 0

2→3 Isochoric heating (represents the internal q2 " 3 = qin = cv_T3 − T2j
combustion at TDC) w2 " 3 = 0

w3 " 4 =− k k 1 RT3 : >1 − e P4 ^k − 1h/k
− P3
o H

3 → 4 Isentropic expansion q2 " 3 = 0

v4 = v1 = vmax; v3 = v2 = vmin

4→1 Isochoric cooling (represents the two q4 " 1 = qout = cv_T1 − T4i
strokes that exhaust the combustion gases w4 " 1 = 0
and draw fresh air and fuel in)

PT

3 qin 3

ISENTROPIC v = CONST. 4
qout
2 qin qout 4 2 v = CONST.
1 1
ISENTROPIC

vs

The thermal efficiency of the Otto cycle is:

hth,Otto = 1 − 1
rk−1

Typical compression ratios (r) are 7 – 10, higher ratios can lead to engine knock (premature auto-ignition of the fuel/air mixture)

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Chapter 3: Thermodynamics

3.4.2.5 Diesel cycle
Idealized cycle to represent compression-ignition internal combustion engines

Step Process Work and Heat

k P2 ^k − 1h/k
− P1
1 → 2 Isentropic compression w1 " 2 = − k 1 RT1 : >1 − e o H

q1 " 2 = 0

2→3 Isobaric heating (represents the internal w2 " 3 = P2_v3 − v2j
combustion, which occurs during initial part q2 " 3 = qin = cP_T3 − T2j
of the power stroke)

k P4 ^k − 1h/k
− P3
w3 " 4 = − k 1 RT3 : >1 − e o H

3 → 4 Isentropic expansion q3 " 4 = 0

v4 = v1 = vmax

4→1 Isochoric cooling (represents the two q4 " 1 = qout = cv_T1 − T4i
strokes that exhaust the combustion gases w4 " 1 = 0
and draw fresh air and fuel in)

P 3 T qin 3
qin 2 4
P = CONST.
2

ISENTROPIC 4 V = CONST. qout
ISENTROPIC
qout 1
1
s
v

Cut-off ratio rc

Ratio of the cylinder volumes after and before the combustion process

=rc VV=32 v3
v2

Efficiency of the Diesel Engine

hth, diesel = 1 − 1 > rck − 1H
rk−1 k_rc −
1j

For the same compression ratio,ηth,diesel < ηth,Otto, but typical compression ratios for diesel engines are higher (about 12–24), since
engine knock is not a concern for diesel engines.

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Chapter 3: Thermodynamics

3.4.2.6 Dual cycle
Combination between Otto and diesel cycle to more closely resemble the internal combustion engine

Step Process Work and Heat
1 → 2 Isentropic compression
w1 " 2 = − k k 1 RT1 : >1 − e P2 ^k − 1h/k
− P1
o H

q1 " 2 = 0

2 → X Isochoric heating (initial combustion) q2 " X = qin,v = cv_TX − T2i
w2 " X = 0

X→3 Isobaric heating (continued combustion wX " 3 = P3_v3 − v2j
during power stroke) qX " 3 = qin, P = cP_T3 − TX j

w3 " 4 = − k k 1 RT3 : >1 − e P4 ^k − 1h/k
− P3
o H

3 → 4 Isentropic expansion q3 " 4 = 0

v4 = v1 = vmax

4→1 Isochoric cooling (represents the two q4 " 1 = qout = cv_T1 − T4i
strokes that exhaust the combustion gases w4 " 1 = 0
and draw fresh air and fuel in)

P 3 T
X
qin ISENTROPIC P = CONST.
2 ISENTROPIC
v = CONST. X 3

4 2 qin
qout qin 4

1 1 v = CONST. qout
v
s

Efficiency of the Dual Cycle:

hth,dual = 1 − cv_TX cv_T4 − T1i − TX j
− T2i + cp_T3

The Otto and the diesel cycle are special cases of the dual cycle.

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Chapter 3: Thermodynamics

3.4.2.7 The Brayton Cycle
The Brayton cycle represents compression and expansion in rotating machinery, such as gas turbines.

Step Process Work and Heat
1 → 2 Isentropic compression (in a compressor)
w1 " 2 = − k k 1 RT1 : >1 − e P2 ^k − 1h/k
− P1
o H

q1 " 2 = 0

T2 P2 k−1
T1 P1 k
= e o

2→3 Isobaric heat addition (combustion chamber w2 " 3 = P2_v3 − v2j
for open systems, heat exchanger for closed q2 " 3 = qin = cP_T3 − T2j
systems) T3 = Tmax
P2 = P3

w3 " 4 = − k k 1 RT3 : >1 − e P4 ^k − 1h/k
− P3
o H

3 → 4 Isentropic expansion (in a turbine) q3 " 4 = 0

T3 P3 k−1 k−1 T2
T4 P4 k P2 k T1
= e o = e P1 o =

4→1 Isobaric heat rejection (exhaust and fresh air w4 " 1 = P4_v1 − v4i
intake for open systems, heat exchanger for q4 " 1 = qout = cP_T1 − T4i
closed systems) T1 = Tmin
P4 = P1

P S = CONST. T
qin 3

23 P = CONST.
S = CONST. qin 4

1 2 qout
qout 1 P = CONST.
4
v s

Pressure ratio for the Brayton cycle

=rp PP=12 P3
P4

Typical pressure ratios for gas turbines range from 11 to 16 and are limited by the maximum temperature at the inlet of the
turbine.

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Chapter 3: Thermodynamics

Efficiency of the Brayton cycle

hth,Brayton = 1− 1 = _T4 − T1i
_T3 − T2j
r ^k − 1h/k
p

For fixed values of Tmin and Tmax, the net work is at a maximum when:

Tmax k
Tmin
rp = e o2^k − 1h

The conversion efficiency can be expressed as "heat rate" (typically in BTU⁄kWh, i.e., the energy required to generate 1 kWh of
electricity:

heat rate = hth,Ranking : 3412 BTU
kWh

Back-work ratio is the ratio of the compressor work to the turbine work.

rpw = wcompressor = Tmax f1 − 1 p
− wturbine Tmin
r ^k − 1h/k
p

The thermal efficiency of the Brayton cycle is reduced by the efficiency of the turbine and compressor. Including the
isentropic turbine and compressor efficiencies can correct for the irreversibilities in the compressor and turbine.

Regeneration T 3
qin
In the Brayton cycle with regeneration, the exhaust from the turbine is used to
pre-heat the outlet from the compressor, prior to heating it in the combustion
chamber. It reduces the required heat input and thus improves efficiency.

qregen,max = h5l – h2 = h4 – h2 qregen 5' 4
qregen,act = h5 – h2 = h4 – h6 5

REGENERATION

Effectiveness of the Regenerator 6
qout
f , T5 − T2 2 qsaved = qregen
T4 − T2 1 s

Efficiency of ideal Brayton cycle with regeneration

hth,Brayton = 1 − e Tmax or ^k − 1h/k
Tmin p

Regeneration is most effective for low pressure ratios and low temperature ratios.

Multistage compression with intercooling and multistage expansion with reheating can further increase the efficiency of the
Brayton cycle.

3.4.2.8 The Jet Propulsion Cycle T P = CONST. 4
P = CONST.
The jet propulsion cycle is similar to the Brayton cycle, but the gases are qin TURBINE
expanded in the turbine only until it produces enough work to drive the 3 5
compressor and then exit at high velocity to provide thrust to an aircraft. Thus, COMPRESSOR NOZZLE
the turbine in the Brayton cycle is replaced by a combination of turbine and 2 6
nozzle. DIFFUSOR qout

In the diffusor, the fluid changes velocity from that of the surrounding to that of 1
the moving airplane. This is a deceleration when viewed from the airplane point
of reference.

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Chapter 3: Thermodynamics

The net work for a turbojet is zero, efficiency is thus defined as propulsive efficiency based on the propulsive power _WoPi and
energy input rate _Qo ini:

hP = WoP
Qo in

WoP = mo `uexit − uinletjuaircraft

3.4.3 Vapor Power Cycles

Vapor power cycles use a working fluid that evaporates and condenses in the course of a cycle. This allows the compression to
occur in the liquid phase, where the smaller specific volume results in less work for the compression.

3.4.3.1 Rankine Cycle
The Rankine cycle is the ideal cycle for vapor power plants.

Step Process Work and Heat

1→2 Isentropic compression w1 " 2 = win = h1 − h2 = v_P1 − P2i
(of the liquid in a pump) q1 " 2 = 0

2→3 Isobaric heat addition (in a boiler, h1 = hliq@P1
3→4 vaporizing and superheating the liquid) v , v1 = vliq@P1
4→1
Isentropic expansion w2 " 3 = 0
(of the gas in the turbine) q2 " 3 = qin = h3 − h2

Isobaric heat rejection (in a condenser, w3 " 4 = wout = h3 − h4
condensing the vapor exiting the turbine) q3 " 4 = 0

w4 " 1 = 0
q4 " 1 = qout = h1 − h4

Efficiency of the Rankine Cycle T

hth,Ranking = wnet = wout − win = 1− qout
qin qin qin
3
The conversion efficiency can be expressed as "heat rate" (typically in qin wturb, out
BTU⁄kWh, i.e., the energy required to generate 1 kWh of electricity:
2 4
heat rate = hth,Ranking : 3412 BTU 1 s
kWh qout

The backwork ratio for the Rankine cycle is much smaller than the back- wpump, in

work ratio for the Brayton cycle.

=rpw ww=oiunt wpump
wturbine

Efficiency calculations for an actual (not ideal) Rankine cycle include the
isentropic efficiencies for the compressor and the pump.

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Chapter 3: Thermodynamics

3.4.3.2 Ideal Rankine Cycle with Reheat
To improve efficiency of the Rankine cycle, the steam is expanded in the turbine in 2 phase, with a reheat in between.

3 T

REHEATING
HIGH-PRESSURE 3 5
HIGH-P LOW-P TURBINE
TURBINE TURBINE LOW-PRESSURE

BOILER REHEATER TURBINE
4

4

P4 = P5 = Preheat 6

5 2
16
CONDENSER

2 PUMP

1s

With the reheat, the heat input and work output are calculated as follows:

qin = qprimary + qreheat = _h3 − h2j + _h5 − h4j
wout = wturb,HP + qturb, LP = _h4 − h3j + _h6 − h5j
3.4.3.3 Ideal Rankine Cycle with Regeneration

For regeneration, part of the steam is extracted from the turbine and used to pre-heat the feedwater. Open feedwater heaters mix
the steam directly with the water; closed feedwater heater use a heat exchanger.

Regenerating increases the inlet temperature to the boiler and thus reduces the heat required in the boiler.

5 T

BOILER TURBINE 4 5
y 1-y 23
OPEN 6
4 FWH 67 1 7

3 2 CONDENSER s
PUMP I 1
PUMP II

With the regeneration, the heat input and work output are calculated as follows (where y is the fraction of steam that is extracted

from the turbine):

y = mo 6
mo 5

qin = h5 − h4

qout = `1 − yj_h7 − h1j

wout = _h5 − h6j + `1 − yj_h6 − h7j

win = `1 − yjwPump I,in + wPump II, in = `1 − yjv1_P2 − P1i + v3_P4 − P3j

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Chapter 3: Thermodynamics

3.4.3.4 Cogeneration Cycle

A cogeneration cycle is based on the Rankine cycle, but instead of condensing the vapor leaving the turbine, it is used for process
heating. The steam is thus extracted from the turbine at higher pressures, depending on the process heating needs.

T
1,2,3

1 23

BOILER EXPANSION TURBINE 10 4
VALVE 4 5 11 5

PROCESS 6 7 6
HEATER 9 s
11
PUMP II 7 CONDENSER 8
MIXING 10
CHAMBER

98

PUMP I

The use of the higher pressure steam reduces the net work delivered by the turbine but adds the process heat to the benefit.

In actual cogeneration cycles, some losses occur due to inefficiencies of the turbine or combustion process or heat losses from the

steam piping. The utilization factor (or "fuel chargeable to power", FCP) is:

FCP = wo net + qoprocess = 1 − qoout
qoin qoin

In an ideal cogeneration cycle, the efficiency is 100%, since no heat is wasted `qoout = 0j.

Typical cogeneration cycles include a condenser in parallel to the process heater and an expansion valve in parallel to the turbine
for adjustable loads. When demand for steam is low, more power is generated by extracting less steam from the turbine. When
demand for steam is high, the turbine is bypassed.

Qo in = mo 3_h4 − h3j
Qo out = mo 7_h7 − h1j
Qo process = mo 5h5 + mo 6h6 − mo 8h8
Woturbine = _mo 4 − mo 5j_h4 − h6j + mo 7_h6 − h7j

3.4.4 Refrigeration Cycles

Refrigeration cycles are the reverse of power cycles. Power cycles transfer heat from a hot reservoir to a cold reservoir and extract
work. Refrigeration cycles use work to transfer heat from a cold reservoir to a hot reservoir. For a refrigerator (R), the objective
is to remove heat from the low temperature reservoir. For a heat pump (HP), the objective is to add heat to the high- temperature
reservoir.

3.4.4.1 Coefficient of Performance

The coefficient of performance is defined similar to the thermal efficiency, but the purpose of refrigeration cycles differs from

power cycles and thus the benefit and cost are defined differently:

=COPR C=Wooolirnkginepffuetct QC
Wnet,in

=COPHP H=Weaotirnkginefpfuetct QH
Wnet,in

Both COPR and COPHP can be > 1. When COPHP < 1, a simple resistance heater (which turns work into heat) would be more
efficient.

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Chapter 3: Thermodynamics

For fixed values of QC and QH:
COPHP = COPR + 1

3.4.4.2 Reverse Carnot Cycle and Reverse Stirling Cycle

The reverse Carnot cycle is the most efficient refrigeration cycle operating between two temperatures, but it is not a realistic
model for actual refrigeration cycles. It does provide a maximum achievable coefficient of performance against which other
cycles can be compared.

Carnot Refrigeration cycle

COPR,Carnot = 1 − 1
_TH /TC i

Carnot Heat Pump cycle

COPHP,Carnot = 1 − 1
_TC/TH i

The reverse Stirling cycle has the same coefficients of performance and is easier to implement in practice (example: Stirling

refrigerators).

3.4.4.3 The Ideal Vapor-Compression Refrigeration Cycle

The vapor-compression refrigeration cycle is the most widely used cycle for refrigerators and air conditioners. It is similar to a
reverse Rankine cycle, but uses isenthalpic throttling instead of isentropic expansion.

Step Process Work and Heat

w1 " 2 = win = − k k 1 RT1 : >1 − e P2 ^k − 1h/k
− P1
o H

1→2 Isentropic compression q1 " 2 = 0
(of the vapor in a compressor)
T2 P2 k−1
T1 P1 k
= e o

h1 = hvap_P1i

2→3 Isobaric heat rejection (in a condenser, exit w2 " 3 = 0
condition saturated liquid) q2 " 3 = qH = qout = h3 − h2

3→4 Isenthalpic throttling (of the liquid in an h=4 h=3 hliq_P3i
expansion valve or capillary tube, exit w3 " 4 = 0
conditions: 2-phase) q3 " 4 = 0

4→1 Isobaric heat adsorption (in an w4 " 1 = 0
evaporator,exit condition: saturated vapor) q4 " 1 = qC = qin = h1 − h4

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Chapter 3: Thermodynamics

T

2 P
Win
SATURATED QH QH
LIQUID 3
2
3 Win

QC 1
4

4 QC 1
SATURATED VAPOR s

h

Coefficient of Performance for the ideal Vapor-Compression Refrigeration Cycle

COPR,VCC = qc = h1 − h4
wnet,in h2 − h1

Coefficient of Performance for the ideal Vapor-Compression Heat Pump Cycle

COPHP,VCC = qH = h2 − h3
wnet,in h2 − h1

3.4.4.4 Cascade Refrigeration Systems

Cascade refrigeration system with heat exchange between the stages allows the use of different working fluids for each cycle.

WARM T DECREASE IN
ENVIRONMENT COMPRESSOR
WORK
QH
QH 6
7 CONDENSER 6

EXPANSION A 5 COMPRESSOR
VALVE COMPRESSOR
HEAT EXCHANGER
EXPANSION
VALVE 8 EVAPORATOR

HEAT 7 2
8 A5
3 CONDENSER 2
3 B
B 1
4
QC
EVAPORATOR
INCREASE IN s
41 REFRIGERATION
QC CAPACITY

COLD
REFRIGERATED

SPACE

The ratio of mass flow rates in each cycle is

mo A = h2 − h3
mo B h5 − h8

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Chapter 3: Thermodynamics

The coefficient of performance is

COPR,Cascade = Qo C = mo A_h6 mo B_h1 − h4i − h1i  
Wonet,in − h5j + mo B_h2

3.4.4.5 Multistage Compression Refrigeration Systems

Multistage compression refrigeration systems are similar to cascade system, but use the same working fluid. A mixing chamber is
used in place of the heat exchanger:

WARM T 4
ENVIRONMENT
5
QH 7
54
6
CONDENSER 8

EXPANSION HIGH-PRESSURE 2
VALVE COMPRESSOR
6
9 39
FLASH
CHAMBER 3 1
2
7 s
LOW-PRESSURE
EXPANSION COMPRESSOR
VALVE
8

EVAPORATOR 1

QC

COLD
REFRIGERATED SPACE

The coefficient of performance is:

COPR,GasRef = qoc = qoc = _1 _1 − xi_h1 − h8j
wo net,in wo comp,LP + wo comp,HP − xi_h2 − h1i + _h4 − h9j

where x is the vapor fraction after the first expansion valve (Point 6).

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Chapter 3: Thermodynamics

3.4.4.6 Gas Refrigeration Systems
The gas refrigeration cycle is the reverse Brayton cycle.

WARM T
ENVIRONMENT QH
2
QH 3

HEAT 1
EXCHANGER 4 QL

3 2 s
TURBINE Wnet, in

COMPRESSOR

4 1

HEAT
EXCHANGER

QL

COLD
REFRIGERATED

SPACE

The coefficient of performance is:

COPR,GasRef = qoc = qoc = _h2 _h1 − h4i h4j
wo net,in wo comp,in − wo turb,out − h1i + _h3 −

Coefficients of performance are lower than for the Carnot cycle and the vapor-compression refrigeration cycle. They are used in
airplanes for cooling and in liquefaction of gases.

Regeneration can be included if the turbine exit temperature is above the compressor inlet temperature. The heat transfer from the
turbine inlet stream to the compressor inlet stream reduces the amount of heat absorbed in the cold exchanger and thus lowers the
coefficient of performance. However, it also reduces the inlet temperature to the turbine and allows to achieve lower temperatures
for the cold exchanger.

COLD

REFRIGERATED

SPACE

QC REGENERATOR T QH
3
HEAT 6 Q HEAT 2
3 EXCHANGER Qregen 1
EXCHANGER

54 WARM 2 1

ENVIRONMENT 4
QH
6
TURBINE 5 QC

Wnet, in s
COMPRESSOR

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Chapter 3: Thermodynamics

3.5 Chemical Reaction Equilibria

3.5.1 Gibbs Free Energy and the Equilibrium Constant

DVg r = − RT ln K or ln K = − DVg r
RT

where K is defined in terms of concentration or pressure in the Vapor Power Cycles and Refrigeration Cycles sections in
this chapter.

3.5.2 Temperature Dependence

The change of the equilibrium constant with temperature is a function of the heat of reaction:

d^ln Kh = DVh r
dT RT2

The integrated equation is

#ln K2 = 1 T2 e DVh r od T
K1 R T1 T2

Over a range where DVh r is nearly constant, this simplifies to:

ln K2 = − DVh r d 1 − 1 n
K1 R T2 T1

Gibbs-Helmholtz equation:

− Dht = JK 2d Dgt nNO
R T2 KK RT OO
L PP
2T

At equilibrium: dP = 0

Dht0 d d Dgt0 n d^ln Kh
R T2 RT dT
= − =
dT

3.5.3 Concentration Dependence

General reaction: aA + bB * cC + dD

At equilibrium: rFWD = rREV

where

− rFWD = k1 CAa CB b
− rREV = k2 CCc CDd

The equilibrium constant is defined as

=Kc CC=ACca CCDBdb k1
k2

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Chapter 3: Thermodynamics

3.5.4 Pressure Dependence

For general reactions: aA + bB * cC + dD

At equilibrium: rFWD = rREV

where

− rFWD = k1 PAa PB b
− rREV = k2 PCc PDd

The equilibrium constant is defined as

=KP PP=ACca PPDBdb k1
k2

When a + b = c + d, KP = Kc , and both are dimensionless.
When they are not equal:

Thus: KP has units of pressure to the power (c + d – a – b).
Kc has units of concentration to the power (c + d – a – b).

KP = Kc (R T)(c+d-a-b)

3.5.5 Le Chatelier's Principle

Le Chatelier's Principle describes the qualitative effect of pressure on equilibrium: For a gaseous reaction, increasing pressure
will shift the equilibrium to the side of the reaction in the reaction equation with fewer moles.

Changes in pressure have negligible effect on liquid- or solid-phase reactions.

3.6 Phase Equilibria

3.6.1 Definitions

3.6.1.1 Phase
A phase is a homogeneous region of matter. Examples of phases include a gas, a mixture of gases, a liquid, a solution of liquids,
and a solid.

3.6.1.2 Saturation Temperature
The saturation temperature is the temperature at which both liquid and vapor exist in equilibrium at a given pressure.

3.6.1.3 Triple Point
For a pure substance, the triple point is the point at which the solid, liquid, and vapor phases exist in equilibrium.

3.6.1.4 Critical Point

For a pure substance, the critical point is the temperature and pressure at which the liquid and vapor phases exhibit identical
properties and are indistinguishable from each other.

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Chapter 3: Thermodynamics

3.6.1.5 Phase Rule
For nonreacting systems, the number of degrees of freedom F is the number of intensive variables (for example, temperature,
pressure, and composition) that must be specified in order to fix the intensive state of a system at equilibrium.

F=2–p+N

where

p = number of phases

N = number of chemical species

For reacting systems, the number of degrees of freedom F is:
F=2−r+N−r

where

r = number of independent chemical reactions at equilibrium within the system

3.6.2 Pure Substances

3.6.2.1 Phase Transitions for Pure Substances

DS = DH at constant pressure, and
T

DS = DU at constant volume
T

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Chapter 3: Thermodynamics

3.6.2.2 Phase Diagrams for Pure Substances

The pressure-temperature relationship for a pure fluid is often shown in a pressure-temperature plot. The intersection of the solid-
liquid-vapor lines is the triple point where the three phases coexist. The critical point is where vapor and liquid properties become
identical.

Four kinds of diagrams are often used for calculations involving a pure fluid. Figures below show the qualitative
behavior of fluid properties.

Thermodynamic Diagrams for a Pure Fluid

CRITICAL CRITICAL
POINT POINT

PRESSURE LIQUID CONST. T
PRESSURE SAT. VAPOR
CONST. QUALSIATT.YLIQUIDFUSIONVAPOR
CURVE PRESSURECONST. H
CURVE
CONST. V
CONST. T CONST. S

SOLID VAPOR
TRIPLE POINT

SUBLIMATION CURVE

TEMPERATURE ENTHALPY
PRESSURE-TEMPERATURE DIAGRAM FOR PURE FLUID PRESSURE-ENTHALPY DIAGRAM FOR PURE FLUID

TEMPERATURE CONST. T
SAT. LIQUID
CRITICAL CRITICAL CONST. T
CONST. PPOINT POINT CONST. QUSALAITT.YVAPOR
ENTHALPY
CONST. QUALITY SAT. LIQUID

CONST. P
SAT. VAPOR
CONST. H

ENTROPY ENTHALPY-ENTROPY
TEMPERATURE-ENTROPY DIAGRAM FOR PURE FLUID ENTHALPY-ENTROPY (MOLLIER) DIAGRAM FOR PURE FLUID

3.6.2.3 Vapor Pressure

Vapor pressure is the pressure in a closed system containing a pure fluid with both liquid and vapor in equilibrium at a given tem-
perature. The equilibrium phases are saturated.

The Antoine equation can be used to estimate the temperature dependence of vapor pressure:

log P sat = A − T B
+C

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Chapter 3: Thermodynamics

where
Psat = saturation pressure or vapor pressure
A, B, and C = constants for a given species
T = temperature

3.6.2.4 Clausius-Clapeyron Equation

The Clapeyron equation relates enthalpy change to temperature, vapor pressure, and volume in the phase change of a two-phase,

single-species system.

ddPT=sat DD=vs Dh
T Dv

where

Dh = specific latent heat for the phase change

Ds = specific entropy for the phase change

Dv = specific volume change for the phase change

For the phase transition from liquid to vapor as an ideal gas, the Clapeyron equation becomes the Clausius-Clapeyron equation:

d (ln Psat) = − Dhtvap
R
dc 1 m
T

Assuming a constant, or average, heat of vaporization between T1 and T2, the integrated form is

lnf P2sat p = − Dhtvap d 1 − 1 n
P1sat R T2 T1

3.6.2.5 Bubbles, Cavities, and Droplets

Due to surface tension, the pressure on the inside of a curved surface is higher than on the outside (Laplace Equation):

Pin = Pout + 2γr
where

r = radius of the curved surface

γ = surface tension

Droplets are spheres of the liquid phase in the vapor phase. A mist made from droplets has a higher vapor pressure than the bulk

fluid (Kelvin Equation):

Pmsaitst = P sat exp d 2cvt n
bulk rRT

Cavities are holes in a liquid filled with vapor. Cavities have a lower vapor pressure than the bulk fluid:

Pcsaavtity = P bsuatlk exp d − 2cvt n
rRT

where

v = molar volume of the liquid

Bubbles are regions of vapor or gas trapped by a thin film. Due to the double surface area, the pressure increase in bubbles is
twice that of cavities.

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Chapter 3: Thermodynamics

3.6.3 Ideal Systems HA

3.6.3.1 Raoult's Law (for ideal solutions) PA (HENRY’S LAW)

Assuming a vapor phase that is an ideal gas and a liquid phase that is an DEVIATION DEVIATION
ideal solution: FROM HENRY FROM RAOULT

=pi y=i P xi P sat SOLUTION
i

where DILUTE IDEAL

xi = mole fraction of component i in liquid phase

P sat = vapor pressure of pure component i PAsat
i
IDEAL SOLUTION (RAOULT’S LAW)
3.6.3.2 Henry's Law (for dilute ideal solutions)
0 1
The partial pressure of a component in the gas phase is proportional to the XA
concentration of the component in the liquid phase:
"A" AS SOLVENT
pi = yi P = xi Hi "A" AS SOLUTE
where Hi = Henry's law constant for component i

3.6.3.3 Distribution of Components Between Phases in Vapor/Liquid Equilibrium

Assume Dalton's Law and Raoult's Law apply. The distribution coefficient (K-value) is defined as:

K=i xy=ii P sat
i
P

where Ki = distribution coefficient for component i

Pisat = saturation pressure of pure component i

The relative volatility is defined as:

a=ij KK=ij yi x j
y j xi

where aij = relative volatility for components i and j

For a binary system:

y1 = 1+ x1 a12 = K2 + K1 x1
x1 (a12 − 1) x1 (K1 − K2)

x1 = a12 + y1 − a12) = K1 + K2 y1 K1)
y1 (1 y1 (K2 −

3.6.4 Nonideal Systems

3.6.4.1 Fugacity
The criterion for the vapor-liquid equilibrium of non-ideal systems is
ftiV = ftiL

where

∧ = indicates properties of the component in the mixture
ftiV = fugacity of component i in the vapor phase
ftiL = fugacity of component i in the liquid phase

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Chapter 3: Thermodynamics

3.6.4.2 Fugacity of Pure Component

Vapor Fugacity of a Pure Component
f V = zP

where z= fugacity coefficient of pure component in vapor phase

The fugacity coefficient of a pure component is a function of temperature and pressure and may be determined from any of:

The residual Gibbs free energy (GR) ln z = GR
RT

An equation of state

#ln z = P (Z − 1) dP
P
0

A generalized correlation, e.g., ln z = (ln z)0 + ~ (ln z)1
where w = the acentric factor

Liquid Fugacity of a Pure Component

f L = zsat P sat / /i = exp=vt L (P − P sat) G
where RT

zsat = fugacity coefficient of pure component at saturation pressure

/ = Poynting correction factor
vt L = molar volume of pure component in the liquid phase

3.6.4.3 Fugacity of Mixtures

Vapor Fugacity of a Mixture
=ftiV zt=i Pi zt i yi P

where zt i = fugacity coefficient of component i in the vapor phase

The fugacity coefficient of a component in a mixture may be determined from an equation of state and a mixing rule.

For a pure component, using the virial equation: ln z = BP
RT
/ln zt i = d
For a mixture, using the virial equation: y j Bij − B m n P
RT
j

where

/ /Bm = yi y j Bij

ij

Bm = second virial coefficient of the mixture

Bij = virial coefficient that characterizes a bimolecular interaction between i and j

For i = j, Bij = Bji = Bii
For i Y= j , Bij must be obtained from measured values or mixing rules.

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Chapter 3: Thermodynamics

Liquid Fugacity of a Mixture

=ftiL c=i xi fiL ci xi z sat P sat /i
i i

where gi = activity coefficient of component i

Activity coefficients are normally based on experimental measurements and fitted to an activity coefficient model, for
example the Van Laar model:

ln c1 = A12 e1 + A12 x1 −2 and ln c2 = A21 e1 + A21 x2 −2
A21 x2 A12 x1
o o

where A12 and A21 = Van Laar constants, typically fitted from experimental data

3.6.4.4 Vapor-Liquid Equilibrium (Gamma/Phi Approach)

zt i yi P = ci xi z sat P sat /i
i i

The distribution coefficient for component i in a non-ideal mixture is

K=i xy=ii ci zisatPisat
zt i P

Special cases:

Ideal vapor phase, ideal liquid solution, and low pressure:

Assume=zt i 1=, ci 1, and /i = 1, (Raoult's Law)

=then yi P x=i Pisat and Ki Pisat
P

Ideal vapor phase, nonideal liquid solution, and low pressure:

Ass=ume zt i 1=and /i 1, (Henry's Law with Henry's Law Constant Hi = ci3Pisat )

=then yi P c=i xi Pisat and Ki cPisat
P

Nonideal vapor phase, nonideal liquid solution, and low to moderate pressure:

Assume /i = 1, ci zisatPisat
=then zt i yi P c=i xi zisat Pisat and Ki zt iP

3.6.5 Phase Behavior

3.6.5.1 Lever Rule

Binary Systems L

For a vapor-liquid mixture of A and B, the relative amounts of the liquid P
and vapor phases in a mixture with an overall composition of xF are given
by the following equations:

V

mL = _1 − |i = b = yA − xF
mL + mV a+b yA − xA

mV a xF − xA ab
mL + mV a+b yA − xA
= | = =

0 xA xF yA 1

where χ is the vapor quality (i.e., the mass fraction of the vapor in the CONCENTRATION OF COMPONENT A
2-phase mixture)

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Chapter 3: Thermodynamics

Ternary, Two-Phase System

In the following ternary phase diagram, two phases contain partially miscible components A, B, and C. One phase is rich in
a
component B and one is lean in component B. The fraction of the B-lean phase is a + b, where a and b represent the length of the
tie line on each side of the overall composition, denoted by the heavy black dot.

mβ = b Ternary Phase Diagram
mα + mβ a+b
100% C

mα = a
mα + mβ a+b

ab α PHASE
100% A
100% B

β PHASE

3.6.5.2 Vapor-Liquid Equilibrium in Binary, Fully Miscible System
Typical Vapor-Liquid Equilibrium Diagrams for Binary, Fully Miscible Systems

L V P = CONSTANT
V–L y
P
V–L T x
V L

x-y x-y

3.6.5.3 Fully Miscible, Binary System with Azeotropes
An azeotrope is a mixture that produces a liquid and vapor of equal composition when boiled.

For a positive azeotrope (minimum-boiling azeotrope):

• A positive deviation from Raoult's Law is exhibited on a P-xy diagram, with the P-x curve lying above that for ideal behav-
ior. This behavior results when liquid-phase intermolecular forces between like molecules are stronger than between unlike
molecules.

• The P-x curve and the P-y curve exhibit maxima at a point for which x = y.

• A T-xy diagram exhibits a minima at the point for which x = y, which represents a boiling point lower than that of any other
composition.

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Chapter 3: Thermodynamics

L Positive Azeotrope Diagrams P = CONSTANT
V–L y
V
P V–L V–L
V T
V–L L

x-y x-y x

For a negative azeotrope (maximum-boiling azeotrope):

• A negative deviation from Raoult's Law is exhibited on a P-xy diagram, with the P-x curve lying below that for ideal
behavior. This behavior results when liquid-phase intermolecular forces between unlike molecules are stronger than between
like molecules.

• The P-x curve and the P-y curve exhibit minima at a point for which x = y.

• The T-xy diagram exhibits a maxima at the point for which x = y, which represents a boiling point higher than that of any
other composition.

L Negative Azeotrope Diagrams P = CONSTANT
y
P V–L V
V T V–L
V–L
V–L
L

x-y x-y x

3.6.5.4 Partially Miscible Systems

Liquid-Liquid Equilibrium

Many mixtures of chemical species, when mixed in certain ranges of composition, form two liquid phases of different
compositions at thermodynamic equilibrium.

The criterion for the liquid-liquid equilibrium of mixtures is
ftia = ftib

where

ftia = fugacity of component i in the liquid phase designated a
ftib = fugacity of component i in the liquid phase designated b

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Chapter 3: Thermodynamics

If each species exists as a liquid at the system temperature, then:

x a c a = x ib c b
i i i

A solubility diagram is a T-x diagram at a constant pressure for a binary system. It depicts curves that indicate the
compositions of coexisting liquid phases. Such diagrams may show:

• A lower critical solution temperature, above which two liquid phases are possible and below which a single liquid phase
exists for all compositions.

• An upper critical solution temperature, below which two liquid phases are possible and above which a single liquid phase
exists for all compositions.

Upper and Lower Critical Solution Temperatures

UPPER CRITICAL 2 PHASES
SOLUTION TEMPERATURE (PARTIALLY MISCIBLE)

TT

SINGLE PHASE TWO PHASES SINGLE PHASE
(MISCIBLE) (PARTIALLY MISCIBLE) (MISCIBLE)

x LOWER CRITICAL
SOLUTION TEMPERATURE

x

Partially Miscible Ternary Systems
Most of the ternary or pseudoternary systems used in extraction are of two types:

Type I System: One binary pair has limited miscibility.

Type II System: Two binary pairs have limited miscibility.

The compositions of the two phases are equal at the plait point.

Examples of Type I and II systems are shown below.

Type I System Type II System

PLAIT ONE LIQUID ONE LIQUID
POINT PHASE PHASE

e f TWO LIQUID PHASES
c MOL FRACTION
a z d TWO LIQUID
PHASES

b

MOL FRACTION TIE LINES

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