ERT 206/4
Thermodynamics
CHAPTER 2
The First Law and Other Basic Concepts
Miss. Rahimah Bt. Othman
Email: rahimah@unimap.edu.my
COURSE OUTCOME 1 CO1)
1. Chapter 1: Introduction to Thermodynamics
2. Chapter 2: The First Law and Other Basic Concepts
Define, discuss, apply and analyze internal energy, first law,
energy balance-closed system, thermodynamic state and state
function, equilibrium, the Phase Rule, reversible process,
constant-V and constant-P processes, enthalpy and heat
capacity.
3. Chapter 3: Volumetric properties of pure fluids
4. Chapter 4: Heat effects
5. Chapter 5: Second law of thermodynamics
6. Chapter 6: Thermodynamics properties of fluids
System
Closed system
- Known as controlled mass
system (mass constant).
- Allow energy flow (in/out)
and volume is not
necessarily constant as a
controlled.
- Often undergo processes
during the internal energy
of the system changes.
∆(Energy of the system) +
(∆Energy of surroundings) = 0
∆(Energy of the system) = Q+W
∆(Energy of the system)
=∆U+∆EK+ ∆EP
Open system
- Known as controlled
volume system.
- Allow mass and energy
cross through the
boundary of open system
(= controlled surface).
Total Energy
E = U+ KE+ PE
Microscopic Energy
Internal energy, U
- The internal energy of a system
comes from the kinetic energy of
the molecules motion and potential
energy from the atomic bonds.
- Eg: chemical energy, nuclear, sensible
energy, latent energy.
Makroscopic energy
- Kinetic energy, KE
- Potential energy, PE
Thermodynamic
state and State
functions
The phase rule
(F = 2 – π +N)
The
reversible
process
Heat Capacity
At constant Volume,
Cv and Constant
Pressure, CP
CLOSED
SYSTEM
Equilibrium
(Eg: heat, mechanical,
phase, chemical)
Entalphy
(H = U +PV)
Constant-V &
Constant-P
Processes

dmCV
 (m) fs  0
dt



(2.25)

(m) fs  m 3  m1  m 2
dmCV
 ( uA) fs  0
dt
H 
(2.26)
u 2
g

z  Q  Ws
2 gc gc
(2.32a)
Energy Balance
at Steady-state
Flow Processes
Mass
Balance
OPEN
SYSTEM
Energy
Balance
Mass Balance
at Steady-state
Flow Processes

m
u2 A2 u1 A1 u A


V1
V1
V


d (mU ) cv
1 2
 [(U  u  zg ) m] fs  Q  work rate
dt
2
d



(mU )CV
1
 
   H  u 2  zg  m  Q  W
dt
2
  fs

(2.28)
(2.27)
Example 2.9 (HEAT CAPACITY)
Air at 1 bar and 25oC (298.15 K) is compressed to 5 bar and 25oC (298.15 K)
by two different mechanically reversible processes:
(a) Cooling at constant pressure followed by heating at constant volume
(b) Heating at constant volume followed by cooling at constant pressure
Calculate Q and W and ∆U and ∆H of the air for each path.
Given ; Cv = 20.78 and CP = 29.10 Jmol-1K-1
Assume ;
- For air, PV/T is a constant
- At 25oC (298.15 K) and 1 bar the molar volume of air is 0.02479 m3mol-1.
Solution
In each case take the system as 1 mol of air contained in an imaginary
piston/cylinder arrangement. Because the process considered are
mechanically reversible, the piston is imagined to move in the cylinder
without friction. The final volume is;
V2  V1
P1
1
 0.02479   0.004958 m3
P2
5
Example 2.9-cont’
(a) During the first step the air is cooled at the constant pressure of 1 bar until
the final volume of 0.004958 m3 is reached. The temperature of the air at
the end of this cooling step is;
Whence;
V2
 0.004958 
T '  T1  298.15
  59.63 K
V1
 0.02479 
Q  H  C P T  (29.10)(59.63  298.15)  6941 J
U  H  ( PV )  H  PV
 6941  (1 x 105 )(0.004958  0.02479)  4958 J
During the second step the volume is held constant at V2 while the air is
heated to its final state. By Eq. (2.19),
U  Q  CV T  (20.78)(298.15  59.63)  4958 J
The complete process represents the sum of its steps. Hence,
Q  6941  4958  1983 J
And
U  4958  4958  0
Example 2.9-cont’
Because the first law applies to the entire process, H  Q  W , and
therefore,
whence;
0  1983  W
W  1983 J
Equation (2.15), H  U  PV  , also applies to the entire process. But
T1  T2 , and therefore, P1V1  P2V2 . Hence, PV   0 and
H  U  0
(b) Two different steps are used in this case to reach the same final state of
the air. In the first step the air is heated at a constant volume equal to its
initial value until the final pressure of 5 bar is reached. The air
temperature at the end of this step is;
P2
5
T '  T1  298.15   1490.75K
P1
1
For this step the volume is constant, and
Q  U  CV T  (20.78)(1490.75  298.15)  24788 J
Example 2.9-cont’
In the second step the air is cooled at P = 5bar to its final state:
Q  H  C P T  (29.10)( 298.15  1490.75)  34703 J
ΔU  H  PV   H  PV
 34703  (5 x 105 )(0.004958  0.02479)  24788 J
For the two steps combined,
Q  24788  34703  9915 J
ΔU  24788  24788  0
W
and as before
 U  Q  0   9915  9915 J
 H  U  0
Conclusion:
The property changes ∆U and ∆H calculated for the given change in
state are the same for both paths. On the other hand the answers to
parts (a) and (b) show that Q and W depend on the path.
Example 2.11 (MASS ENERGY FOR OPEN SYSTEMS)
Liquid n-hexane flows at a rate of ṁ = 0.75 kg s-1 in a pipe with inside
diameter D = 5 cm. What are q, ṅ, and u? What would these quantities be for
the same ṁ if D = 2 cm? Assume for liquid n-hexane that ρ = 659 kg m-3.
Solution
We have q  m
  1
whence
 M 1
n  m
and
0.75 kg s 1
3 1
q

0.00114
m
s
3
659 kg m
0.75 kg s 10
n 
1

g kg 1
1

8.703
mol
s
86.177 g mol 1
3
Given ṁ, these quantities are independent of D. The velocity, however,
depends on diameter through u = qA-1, where, for a circular cross-section,
A = (π/4)D2. For D = 5 cm,
2

A
whence

5 x 10
4
2

m  0.00196 m 2
0.00114 m 3s 1
1
u

0.582
m
s
0.00196 m 2
Example 2.11-cont’
Similarly, for D = 2 cm,
A  0.000314 m
2
and
0.00114
u
 3.63 m s 1
0.000314
Example 2.14 (ENERGY BALANCE FOR OPEN SYSTEM)
An insulated, electrically heated tank for hot tank for hot water contains
190 kg of liquid water at 60 oC (333.15 K) when a power outage occurs. If
water is withdrawn from the tank at a steady state of ṁ = 0.2 kg s-1, how
long will it take for the temperature of the water in the tank to drop from 60 to
35 oC (333.15 K to 308.15 K)?
Assume:
(a) Cold water enters the tank at 10 oC (283.15 K)
(b) Heat losses from the tank is negligible.
For liquid water let CV = CP = C, independent of T and P.
Solution
Here, Q  W  0 . Additional assumption;
(a) Perfect mixing of the contents of the tank.
(b) m(flow rate in) = m(flow rate out) = mcv = constant.
(c) Differences between inlet and outlet kinetic and potential energy can
be neglected.
Example 2.14-cont’
Equation (2.29) is therefore written as;
dU
dT
C
dt
dt
and
H  H1  C (T  T1 )
Where unsubscripted quantities refer to the contents of the tank and H1 is
the specific enthalpy of the water entering the tank. With CV = CP = C,
dU
m
 m ( H  H1 )  0
dt
The energy balance then becomes, on arrangement,
dt  
m dT
m T  T1
Integration from t = 0 (where T = T0) to arbitrary time t yields
m  T  T1 

t   ln 
m  T 0T1 
Example 2.14-cont’
Substitution of numerical values into this equation gives, for the conditions of
this problem,
t
190  308.15  283.15 
ln 
  658.5 s
0.2  333.15  283.15 
Thus, it takes about 11 minutes for the water temperature in the tank to drop
from 60 to 35 oC (333.15 K to 308.15 K).
Example 2.16
(ENERGY BALANCE FOR OPEN SYSTEM)
Air at 1 bar and 25oC (298.15 K) enters a compressor at low velocity,
discharges at 3 bar, and enters a nozzle in which it expands to a final velocity
of 600 ms-1 at the initial conditions of pressure and temperature. If the work of
compression is 240 kJ per kilogram of air, how much heat must be removed
during compression?
Solution
Because the air returns to its initial conditions of T and P, the overall
process produces no change in enthalpy of the air. Moreover, the potentialenergy change of the air is presumed negligible. Neglecting also the initial
kinetic energy of the air, we write Eq. (2.32a) as;
2
u
Q  2  Ws
2
Example 2.16-cont’
The kinetic-energy term is evaluated as follows:
2
1 2 1
m
m2
u2   600   180 000 2
2
2
s
s
m 2 kg
 180 000 2 .  180 000 N m kg 1  180 kJ kg 1
s kg
Then
Q  180  240  60 kJ kg 1
Heat in the amount of 60 kJ must be removed per kilogram of air
compressed.
Example 2.17
(ENERGY BALANCE FOR STEADY-STATE FLOW PROCESSES OPEN SYSTEM)
Water 93.5oC (366.65 K) is pumped from a storage tank at the rate of 3.15 x
10-3 m3 s-1. The motor for the pump supplies work at the rate of 1.5 kW. The
water goes through a heat exchanger , giving up heat at the rate of 700 kW,
and is delivered to a second storage tank at an elevation 15 m above the first
tank. What is the temperature of the water delivered to the second tank?
Solution
This is a steady-state flow process for Eq. (2.3b) applies. The initial and final
velocities of water in the storage tank are negligible, and the term ∆u2 / 2gc
may be omitted. The remaining terms are expressed in units of kJ kg-1
through use of appropriate conversion factors. At 366.65 K the density of
water is 958 kg m-3 thus the mass flow rate is;
(3.15)(10-3)(958) = 3.018 kJ kg-1
from which we obtain
Q = - 700/3.018 = - 231.9 kJ kg-1
Example 2.17-cont’
Since 1 kW is equivalent to 1 kJ s-1, the shaft work is;
Ws = (1.5)(1) /3.018 = 0.497 kJ kg-1
If the local acceleration of gravity is taken as the standard value of 9.807 ms-2,
the potential-energy term becomes;
9.807  159.807  0.147 kJ kg 1
g
z 
9.807  1000
gc
Equation (2.32b) now yields ∆H:
g
H  Q  Ws  z  231.9  0.497  0.147  231.55 kJ kg 1
gc
The enthalpy of saturated water at 93.5 oC (366.65 K) is given in the steam
tables as 391.6 kJ kg-1. Thus
H  H 2  H1  H 2  391.6  231.55
and
H 2  391.6  231.55  160.05 kJ kg 1
Example 2.17-cont’
The temperature of water having this enthalpy is found from the steam tables
to be;
t  38.2 o C
or
T  311.35 K
In this example Ws and (g/gc)∆z are small compared with Q, and for practical
purposes they could be neglected.
ERT 206/4
Thermodynamics
CHAPTER 3
Volumetric Properties of Pure Fluids
Miss. Rahimah Bt. Othman
Email: rahimah@unimap.edu.my
COURSE OUTCOME 1 CO1)
1. Chapter 1: Introduction to Thermodynamics
2. Chapter 2: The First Law and Other Basic Concepts
2. Chapter 3: Volumetric properties of pure fluids
DESCRIBE and EXPLAIN PVT behavior of pure substances,
Virial Equation of State, ideal gas, Virial EquationAPPLICATION, cubic equation of state, generalized correlations
for gases and liquids.
4. Chapter 4: Heat effects
5. Chapter 5: Second law of thermodynamics
6. Chapter 6: Thermodynamics properties of fluids
Objectives
• Introduce the concept of a pure substance.
• Discuss the physics of phase-change processes.
• Illustrate the P-v, T-v, and P-T property diagrams and
P-v-T surfaces of pure substances.
• Demonstrate the procedures for determining
thermodynamic properties of pure substances from
tables of property data.
• Describe the hypothetical substance “ideal gas” and the
ideal-gas equation of state.
• Apply the ideal-gas equation of state in the solution of
typical problems.
• Introduce the compressibility factor, which accounts
for the deviation of real gases from ideal-gas behavior.
• Present some of the best-known equations of state.
Pure Fluids/ Substance
 A substance that has a fixed
chemical composition throughout is
called a Pure Substance.
 Pure Substance:
- N2, O2, gaseous Air
-A mixture of liquid and gaseous
water is a pure substance, but a
mixture of liquid and gaseous air is
not.
PHASES OF A PURE SUBSTANCE
The molecules
in a SOLID are
kept at their
positions by
the large
spring like
inter-molecular
forces.
In a solid, the
attractive and
repulsive forces
between the
molecules tend to
maintain them at
relatively constant
distances from each
other.
The arrangement of atoms in different phases: (a) molecules are at relatively
fixed positions in a solid, (b) groups of molecules move about each other in the
LIQUID phase, and (c) molecules move about at random in the GAS phase.
Phase-Change Processes of Pure
Substance
 Compressed liquid or a subcooled liquid:
A liquid that is not about to vaporize.
 Saturated liquid: A liquid that is about
to vaporize.
 Saturated vapor: A vapor that is about
to condense.
 Saturated liquid-vapor mixture: the
liquid and vapor phases coexist in
equilibrium.
 Superheated vapor: A vapor that is not
about to condense
T-v diagram for the heating process of water at
constant pressure.
2-1
Phase-Change Processes of Pure
Substance
 Saturated temperature, Tsat: At a given
pressure, the temperature at which a pure
substance changes phase.
 Saturated pressure, Psat: At a given
temperature, the pressure at which a pure
substance changes phase.
 Latent heat: the amount of energy absorbed or
released during a phase-change process.
 Latent heat of fusion: the amount of energy
absorbed during melting.
 Latent heat of vaporization: the amount of
energy absorbed during vaporization.
PVT BEHAVIOR OF PURE SUBSTANCES/ FLUIDS
Critical point – highest
combination of pressure
and temperature where the
fluid exist in liq-vap
equilibrium
The 2-C line, also known as
vaporization curve is where
liquid-vapor is in equilibrium
The 1-2 line, also
known as sublimation
curve is where solidvapor is in
equilibrium
Triple point, three
phases exist in
equilibrium (F=0)
The 2-3 line, also known as
fusion curve is where solidliquid is in equilibrium
PV diagram
Boundaries in PT diagram becomes
region when illustrate with PV
diagram
Critical point becomes peak of the
curve
Triple point becomes horizontal line
Compressed liquid region
Saturated liquid line at boiling
temperature
Superheated vapor region
Saturated vapor line at condensation
temperature
T >Tc, the line do not cross the
boundary
Isotherms in sub-cooled/ compressed lliquid
region are steep, because liquid volumes
change little with large changes in pressure
SINGLE PHASE REGION
At single phase regions in PV diagram, there is a relation
connecting P,V and T.
This relation known as PVT equation of state; f(P,V,T)=0
If V is considered as a function of T and P, then V=V(T,P)
 V 
 V 
dV  
dT



 dP
 T  P
 T T
1  V 
Volume exp ansivity

 
V  T  P
1  V 
 
 
V  T T
isothermal compressib ility
The combination will yield;
dV
 dT  dP
V
3.4
Because the isotherms on the left side of PV diagram are very steep,
both β and κ are small.
Because of that, the liquid is known as incompressible fluid, where
both constants are equal to zero
However, this is just idealization, and in incompressible fluid, no
equation of state exist, since V is independent of T and P
If, we still want to calculate, for liquids, β is positive, and κ is
positive as well. Integration of 3.4 yield
ln
V2
  (T2  T1 )   ( P2  P1 )
V1
Try examples 3.1
3.5
Virial Equation of State
The coefficients a(T), b(T), c(T), and so on, that are
functions of temperature alone are called virial
coefficients.
Ideal-Gas Temperature; Universal Gas Constant
Figure 3.4: Plot of PV vs. P for 4 gaseous at
triple-point temperature of water. The limiting
value of PV as P0 is the same for all of the
gaseous.
Limiting value (asterisk)
PV* = a = f(T)
This properties of gaseous is the basis for
establishing an absolute temperature scale.
The simplest procedure to define Kelvin scale:
1. (PV)* = a ≡ RT
2. (PV)*t = R x 273.16K(3.8)
3.  PV *   T / K
 PV * 
 t 
273.16 K
*

PV 
T / K  273.16
PV *t
(3.9)
•In the limit, P0, molecules separated by finite distance
•Volumes becomes negligible compare with the total volume
of the gas, and intermolecular forces approach zero
• These condition define an IDEAL GAS state & Eq. 3.9
establishes
the ideal-gas temperature scale
• The proportionality constant R = universal gas constant
Through the use of conversion factors, R may be expresses in various units.
Commonly used values are given as above table.
2 Forms of the Virial
Equation
• Auxiliary thermodynamic property =
• = Compressibility factor
•
•
•
•
•
•
Z = 1 + B’P+C’P2 +D’P3 +…
Z= 1 + B/V +C/V2 + D/V3 +…
Virial expansion = Eq 3.11 & Eq 3.12
B’, C’, D’ …, B, C, D = virial coefficient
B’ = 2nd virial coefficient
C’ = 3rd virial coefficient…etc
PV
Z
RT
(3.10)
(3.11)
(3.12)
Eq 3.11 & Eq 3.12
B
B' 
(3.13a )
RT
C  B2
C' 
( RT ) 2
(3.13b)
D  3BC  2 B 3
D' 
( RT )3
(3.13c)
THE IDEAL GAS
• Because the Eq 3.12 arise on account of molecular
interactions, the virial coefficients B,C…etc = 0
were no such interaction to exist
 Z  1 or
Z  1 or
PV  RT
PV  RT
• Phase Rule = Internal energy of a real gas is a
function of pressure as well as of temperature.
THE IDEAL GAS
• Equation of state: Any equation that
relates the pressure, temperature, and
specific volume of a substance.
• The simplest and best-known equation of
state for substances in the gas phase is the
ideal-gas equation of state. This equation
predicts the P-v-T behavior of a gas quite
accurately within some properly selected
region.
Ideal gas
equation of
state
R: gas constant
M: molar mass (kg/kmol)
Ru: universal gas constant
U = U(T) (Ideal gas)
Different substances have
(3.15) different gas constants.
Implied Property Relations for
an Ideal Gas (f(T) only)
• Heat Capacity for constant volume, Cv
 U  dU (T )
Cv  
 Cv(T )
 
dT
 T v
(3.1)
• Eq 2.11 applied to an Ideal Gas
H  U  PV  U (T )  RT  H (T )
(3.17)
Implied Property Relations
for an Ideal Gas (f(T) only)
 Heat Capacity for constant pressure, Cp
dH (T )
 H 
CP  

 CP

dT
 T  P
(3.18)
 Useful relation between Cv & Cp
CP 
dH dU

 R  CV  R
dT dT
(3.19)
NOTE: This equation does not imply that Cp and Cv are themselves
constant for an ideal gas, but only that they vary with temperature in such a
way that their differences is equal to R
For any ∆ of state of an ideal gas, Eq 3.16 and Eq.3.18 lead to:
dU  CV dT
(3.20a )
U   CV dT
(3.20b)
dH  C P dT
(3.21a )
H   C P dT
(3.21b)
• T1 & T2
• a---b = Constant volume process
• a---c & a---d ≠ constant volume

The graph show that the ∆U= CV dT
Equation for Process
Calculation for Ideal Gas
Working equation of dQ and dW depend on which pair of these variables
is selected as independent
With P=RT/V,
dV
V
dP
dQ  C P dT  RT
P
dQ  CV dT  RT
dV
V
(3.22)
dW   RT
(3.24)
dW   RdT  RT
(3.23)
dP
P
(3.25)
With V=RT/P and Cv given by Eq 3.19, dQ & dW written as Eq.3.24 & Eq.3.25
With T=PV/R, the work is simply dW=-PdV, and with Cv given by Eq.3.19,
CV
CP
dQ 
VdP 
PdV
R
R
(3.26)
Isothermal Process
• Q=-W
V2
P2
Q  W  RT ln
  R ln
V1
P1
(const T ) (3.27)
Isobaric Process
Q  H   CP dT
(Const P) (3.28)
Isochoric Process
Q  U   CV dT
(Const V ) (3.29)
Adiabatic Process;
Constant Heat Capacities
dT
R dV

T
CV V
C P / CV
 V1 
dT
1
RdV
T2
 T  CV  V  T1   V2 
similar to Eq.3.24 & Eq.3.26 lead to :
T2
T1
 P2 

 P 

 1 
R / CP
 V1
P2

V
P1
 2




C P / CV
Also,
TV  1  const. (3.30a)
PV   const (3.30c)
TP (1 ) /   const. (3.30b)
CP
 
CV
(3.31)
IMPORTANT: Eq 3.30 are restricted in application to ideal gases with
Constant Heat capacities undergoing mechanically reversible adiabatic
expansion or compression
For ideal gas, the WORK of any adiabatic closed-system process is given
by:
dW  dU  CV dT
W  U  CV T
(3.32)
Because RT1 = P1V1 and RT2 = P2V2,
W
RT2  RT1 P2V2  P1V1

 1
 1
(3.33)
Elimination of V2 from Eq 3.33 by Eq 3.30c, valid for
mechanically reversible process, lead to:
P1V1  P2

W
  1  P1




 1 / 
 RT  P
1
 2
 1 
   1  P1



 1 / 

 1

Polytropic Process
PV   const
(3.35a )
TV  1  const
(3.35b)
TP (1 ) /   const
RT1  P2

W 
  1  P1



(3.35c )
( 1) / 
(   ) RT1  P2

Q
(  1)(  1)  P1


 1




( 1) / 
(3.36)

 1

(3.37)
(3.34)
Polytropic Process
Isobaric process
Isothermal process
Adiabatic process
Isochoric process
APPLICATION OF
VIRIAL EQ.
All isotherms originate at Z=1 for P=0
PV
BP
Z
 1
RT
RT
(3.38)
Applied to vapors at sub critical temperature up to their saturation
pressure
When the virial equation is truncated to 3 terms, the appropriate
form is:
PV
B C
Z
 1  2
RT
V V
(3.40)
CUBIC EQ. OF STATE
1. The Van der Waals Equation of state
RT
a
P
 2
V b V
(3.41)
2. A Generic Cubic Equation of State
RT
a(T )
P

V  b V  b (V  b)
(3.42)
3. Determination of Eq-of State Parameters
 P 

  0 (3.43)

V

T ;cr
 2P 
 2   0 (3.44)
 V T ;cr
4. Theorem of Corresponding State: Acentric Factor
  1.0  log( Prsat )Tr 0.7
(3.48)
NOTE: All fluid having the same value of  , when compared at the same Tr
And Pr have about the same value of Z, and all deviate from ideal gas behavior
To about he same degree
5. Vapor& Vapor-like Root of the Generic Cubic Eq of state
6. Liquid & Liquid-like Root of the Generic Cubic Eq of state
GENERALIZE
CORRELATION FOR GASES
GENERALIZE
CORRELATION FOR GASES
1. Pitzer Correlation for the
Compressibility Factor
2. Pitzer Correlation for the 2nd Virial
Coefficient
3. Correlations for the 3rd Virial
Coefficient
4. Condition of Approximate Vlidity of
the Ideal-Gas Equation
GENERALIZE CORRELATION
FOR LIQUID
Summary
• Pure substance
• Phases of a pure substance
• Phase-change processes of pure substances
– Compressed liquid, Saturated liquid, Saturated
vapor, Superheated vapor
– Saturation temperature and Saturation pressure
• Property diagrams for phase change processes
– The T-v diagram, The P-v diagram, The P-T
diagram, The P-v-T surface
• The ideal gas equation of state
– Is water vapor an ideal gas?
• Compressibility factor
• Other equations of state
Thank you
Prepared by,
MISS RAHIMAH OTHMAN