ENERGY, ENERGY TRANSFER

& GENERAL ENERGY

ANALYSIS

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CHAPTER 2a
1
 CONTENTS

Forms of Energy

Energy Transfer by Heat

Energy Transfer by Work

Mechanical Forms of Work

The First Law of Thermodynamics

Energy Conversion Efficiencies
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2
 LESSON OBJECTIVES

At the end of this lesson, you should be able

to:
State the various forms of energy
Describe the nature of internal energy
Describe the energy transfer by heat and work
Define the concept of work and several forms of
mechanical work

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3
 FORMS OF ENERGY

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 FORMS OF ENERGY

Energy exists in numerous forms (thermal, mechanical, chemical, kinetic,
potential, electric, magnetic & nuclear)

The sum of the energies is the total energy, E (kJ)

Or for a unit mass, E
e= (kJ/kg)
m
Grouping of
Energy forms
energy of a system as a whole with respect to some
macroscopic outside reference frames, e.g. KE, PE

• related to molecular structure of a system and the

microscopic degree of molecular activity
• independent of outside reference frames
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• the sum is the Internal Energy, U 5
 FORMS OF ENERGY (cont’d)
Macroscopic forms of energy

Kinetic energy (KE) Potential energy (PE)

- result of motion relative to some - due to elevation in a gravitational
field (unit J)
reference frame (unit J)
1 PE = mgz
KE = mυ 2
2 ∆PE = PE2 − PE1 = mg (z 2 − z1 )
1
∆KE = KE2 − KE1 =
2
(
m υ 22 − υ12 )
where
g = gravitational acceleration, 9.81 m/s2
where h = elevation of center of gravity of a
υ = velocity of the system relative to system relative to some arbitrarily
some fixed reference frame (m/s) plane (m)
Open m = mass of an object (kg) 6
 FORMS OF ENERGY (cont’d)
Due to internal
Microscopic forms of energy
structure of the matter

Sensible Energy Chemical energy: The

Latent Energy internal energy associated
- Kinetic energy of molecules with the atomic bonds in a
- Associated with
phase of a molecule.
system Nuclear energy: The
tremendous amount of
energy associated with the
strong bonds within the
nucleus of the atom itself.

The internal energy of a system is the sum of all forms of

the microscopic energies.
Thermal = Sensible + Latent
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Internal = Sensible + Latent + Chemical + Nuclear
 FORMS OF ENERGY (cont’d)

Total Energy of a system

1
E = U + KE + PE = U + mυ 2 + mgz UNIT ?
2

Total Energy of a system per unit mass

1 2
e = u + ke + pe = u + υ + gz UNIT ?
2

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 Energy in Closed System

The macroscopic kinetic energy is an organized form of energy and is much more useful than the
disorganized microscopic kinetic energies of the molecules.

Most of the closed system remains

stationary, so for that system;
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∆E = ∆U + ∆KE + ∆PE
 FORMS OF ENERGY (cont’d)

Energy Interaction

The only two forms of energy interactions
associated with a closed system are heat
transfer and work.

Recognized as it crosses boundary,

represent energy lost or gain

The difference between heat transfer and

work: An energy interaction is heat transfer if
its driving force is a temperature difference.
Otherwise it is work.
Energy can cross the
boundaries of a closed system
in the form of heat and work.
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 ENERGY TRANSFER BY HEAT
Means of energy transfer caused by temperature
Heat Transfer difference between the system and the surroundings

Direction of heat transfer: Higher T

to Lower T

Energy is recognized as heat

transfer only as it crosses the
system boundary

Temperature difference is the driving force

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for heat transfer. The larger the temperature
difference, the higher is the rate of heat11
transfer.
 ENERGY TRANSFER BY HEAT
(cont’d)

Adiabatic process is a process during
which there is no heat transfer, Q=0

How a process can be adiabatic?

Well insulated system

No temperature difference

Adiabatic is not necessarily means
isothermal process. Temperature of the
system can still be changed by other
means

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 ENERGY TRANSFER BY HEAT
(cont’d)

Amount of heat transferred is denoted by Q (kJ)

For a unit mass: Q
q= (kJ/kg)
m

Sign Convention for Heat:
Q positive indicates heat input
Q negative indicates heat lost

Specifying the directions

of heat using in and out

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 ENERGY TRANSFER BY HEAT
(cont’d)
MECHANISMS OF HEAT TRANSFER
Conduction Convection Radiation

The transfer of energy from the The transfer of energy between

more energetic particles of a a solid surface and the adjacent The transfer of energy due to the
substance to the adjacent less fluid that is in motion, and it emission of electromagnetic waves
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energetic ones as a result of involves the combined effects of 14
(or photons).
interaction between particles. conduction and fluid motion.
 ENERGY TRANSFER BY WORK

In mechanics, work is defined as the product of the force

and the displacement in the direction of the force.

WORK Energy transfer associated with a force acting

through a distance, e.g. rotating shaft, rising piston

The work done by, or on, a system is defined as (unit kJ):

s2
W =∫ F .ds or Work done per unit mass w=
W
s1
m

The work done per unit time is Power (unit kJ/s or kW):
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 ENERGY TRANSFER BY WORK
(cont’d)

Sign convention for work:
W positive indicates work done by system (work output)
W negative indicates work done on the system (work input)

Specifying the directions

of work using in and out

The work done by, or on, a
system is defined as (unit
kJ):

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 ENERGY TRANSFER BY WORK
(cont’d)
Heat vs. Work

Both are recognized at the boundaries
of a system as they cross the
boundaries. That is, both heat and work
are boundary phenomena.

Systems possess energy, but not heat
or work.

Both are associated with a process, not
a state.

Unlike properties, heat or work has no
meaning at a state.

Both are path functions (i.e., their
magnitudes depend on the path followed Properties are point functions; but heat
during a process as well as the end and work are path functions (their
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states). magnitudes depend on the path 17
followed).
 Moving Boundary Work

If the piston is allowed to move a distance dx in a quasi-equilibrium manner, the

differential work done by the system is

δW = F dx = pA dx = p dV
2
The total work done by the system is W = ∫ p dV (kJ)
1

Work done by the system can be interpreted as the area under

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the process path on a p-V diagram. 18
 Moving Boundary Work

The total area under the curve 1-2

is: 2 2
Area = A = ∫ dA = ∫ pdV
1 1

Note:
P is the absolute pressure and is always positive.
When dV is positive, Wb is positive.
When dV is negative, Wb is negative

A gas can follow several different paths

as it expands from state 1 to state 2.
Each process path gives a different value
for the boundary work
Wb depends on the path (thus process)
and the end states
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Work is not a property 19
19
 ENERGY TRANSFER BY WORK
(cont’d)
Electrical work

NN= Coulombs of electric charge

Electrical power

When potential difference and

current change with time

Electrical power in terms of

When potential difference resistance R, current I, and
and current remain constant potential difference V.

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 EXAMPLE 2

Consider an electric
refrigerator located in a
room. Determine the
DIRECTION of work and
heat interactions (in our out)
for:
(a) Contents of the refrigerator

(b) All parts of refrigerators,

including the contents

(c) Everything contained within room

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during winter day
 MECHANICAL FORMS OF
WORK

There are two requirements for a work interaction between a system and
its surroundings to exist:
 there must be a force acting on the boundary.
 the boundary must move.

Work = Force × Distance When force is not constant

In thermo, most of the work is mechanical work, which

associated with moving boundary work

Other common forms of work

Shaft work
If there is no movement, no
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Spring work 22
work is done.
 SHAFT WORK
A force F acting through a moment arm r
generates a torque T

This force acts through a distance s for

n revolution:

Shaft work is
Shaft work: proportional to the
torque applied and the
number of revolutions
of the shaft.

The power transmitted through the shaft is the shaft work done per unit time

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 SPRING WORK
When the length of the spring changes by a
differential amount dx under the influence of a force
F, the work done is:

For linear elastic springs, the displacement x is

proportional to the force applied

k: spring constant (kN/m)

Substituting and integrating yield

The displacement of a linear

spring doubles when the force
x1 and x2: the initial and the final displacements
is doubled.
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 NET WORK DONE BY A
SYSTEM
Different forms of work transfer could occur in a system simultaneously
during a process.

The total or net work done by the system = algebraic sum of all work

Wtotal = Wb + We + Wsh + Ws + ...

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 Class Takeaway

1. How can a closed system and its surroundings interact?

2. What are the sign conventions used for energy transfer?

3. Express total energy and for each of the terms, indicate the unit.

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 ENERGY, ENERGY TRANSFER
& GENERAL ENERGY
ANALYSIS

CHAPTER 2b
1
Open
 CONTENTS

Forms of Energy

Energy Transfer by Heat

Energy Transfer by Work

Mechanical Forms of Work

The First Law of Thermodynamics

Energy Conversion Efficiencies
2
Open
 LESSON OBJECTIVES

At the end of this lesson, you should be

able to:

State the First Law of Thermodynamics

Explain the conservation of energy principle
(Energy Balance)

3
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 The First Law of Thermodynamics

The first law provides a basis for studying the relationships among the
various forms of energy and energy interactions.
Work done on a system = change in total energy of the system

4
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 The First Law of Thermodynamics
Energy Balance (cont.)

Ein − Eout = ∆Esystem

• This relation is referred to as

the energy balance
• The relationship is valid for
any system undergoing any
process

5
 Example

6
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 The First Law of Thermodynamics

Total Energy = sum of internal, kinetic, and potential energy changes

7
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 The First Law of Thermodynamics
Mechanisms of Energy Transfer, Ein and Eout
• Energy can be transferred to a system in three forms:

heat (Q)

work (W)

mass flow (Emass)
• Energy interactions are recognized at the boundary
• Taking the three forms of energy transfer into account, the energy
balance can be written as
Ein − Eout = (Qin − Qout ) + (Win − Wout ) + (Emass, in − Emass, out ) = ∆Esystem

• Or, more compactly as

Ein − Eout = ∆Esystem

Net energy transfer Change in internal, kinetic,

by heat, work, and mass potential, etc. energies 8
 CLOSED SYSTEM ENERGY BALANCE

(kJ)
In a process of a closed system the energy of the system
increases and decreases by an amount equal to the
amount of energy transferred across its boundary
Change in Energy = Net amount of -
Net amount of
of the system heat transfer in work transfer
out

E2 − E1 = Q − W

∆U + ∆KE + ∆PE = Q − W
For a cycle
9
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 RATE OF ENERGY TRANSFER
(kJ)
The energy balance can then be expressed compactly as:

Or in a rate form:

Or on a unit mass basis:

10
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 2-7 Energy Conversion Efficiencies
Performance (or Efficiency)
• An indicator of how well an energy conversion or transfer process
is accomplished
• Generally defined as
Desired Output
Performanc e =
Required Input

11
 EXAMPLE 1
A closed system containing 20 kg of air undergoes a
process in which there is a heat transfer of 1000 kJ
from the system to the surroundings. The work done on
the system is 200 kJ. If the initial specific internal
energy of the system is 300 kJ/kg, what is the final
specific internal energy? Neglect KE and PE. (260
kJ/kg)

12
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 Example 2
Water is being heated in a closed
pan on top of a range while being
stirred by a paddle wheel.
During the process, 30 kJ of heat
is transferred to the water, and
5kJ of heat is lost to the
surrounding air.
The paddle-wheel work amounts
to 500 N.m Determine the final
energy of the system if its initial
energy is 10 kJ. [Ans 35.5 kJ]
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 Example 3
Consider a fan located in a 1m x 1m
square duct. Velocities at various
points at the outlet are measured, and
the average flow velocity is
determined to be 7 m/s.

Taking the air density to 1.2 kg/m3 ,

estimate the minimum electric power
consumption of the fan motor.
[Ans 0.2058 kW]

14
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 EXAMPLE
5 kg of steam in the figure undergoes an expansion from state 1
(u1=2709.9 kJ/kg) to state 2 (u2=2659.6 kJ/kg). During the
process, there is heat transfer to the steam of 80 kJ. The paddle
wheel transfers energy to the steam by work in the amount of 18.5
kJ. Determine the energy transfer by work from the steam to the
piston during the process. ( 350 kJ)

15 1
5
 Class Takeaway
1. For any system, the first law can be written as:
a. ____________________________ (kJ) or,
b. ____________________________ (kJ/kg) or,
c. ____________________________ (kW)

2. Fill in the blank spaces in the table.

Process Q W E1 E2 ∆E
A +50 -20 +50
B +50 +20 +20
C -40 +60 +20
D -90 +50 0
E +50 +20 -100 16
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 Summary of Chapter 2

Forms of energy
 Macroscopic = kinetic + potential
 Microscopic = Internal energy (sensible + latent +
chemical + nuclear)

Energy transfer by heat

Energy transfer by work

Mechanical forms of work

The first law of thermodynamics
 Energy balance
 Energy change of a system
 Mechanisms of energy transfer (heat, work, mass flow)
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 Assignment 1
#1.37, 1-58, 2-37 , 2-43 and 2-47

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PROPERTIES OF PURE
SUBSTANCES

CHAPTER 3a
 CONTENTS

Pure Subtances

Phases of a Pure Substances

Phase Change Processes

Property Diagrams

Property Tables

The Ideal Gas Equation

Compressibility Factor
 LESSON OBJECTIVES

At the end of the lesson, you should be able

to:

Define a pure substance

Explain the phase change processes

Sketch P-v and T-v diagrams for a particular
process
 PURE SUBSTANCE

Pure substance: A substance that has a fixed chemical
composition throughout.

Air is a mixture of several gases, but it is considered to
be a pure substance.

Nitrogen and gaseous air are A mixture of liquid and gaseous

pure substances. water is a pure substance, but a
mixture of liquid and gaseous air4
is not.
PHASES OF A PURE SUBSTANCE

The molecules in a
solid are kept at
their positions by
the large springlike
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.
 PHASE-CHANGE PROCESSES OF
PURE SUBSTANCES
Let us consider the following system undergoes heating process at constant pressure.

At 1 atm and 20°C, At 1 atm pressure

water exists in the and 100°C, water
liquid phase exists as a liquid
(compressed that is ready to
liquid: not about to vaporize (saturated
vaporize). liquid: about to
vaporize).

As more heat is At 1 atm pressure,

transferred, the the temperature
temperature of remains constant As more heat is
the vapor starts at 100°C until the transferred, part of the
to rise last drop of liquid saturated liquid
(superheated is vaporized vaporizes (saturated
vapor: not (saturated vapor: liquid–vapor mixture:
about to about to liquid and vapor
condense). condense). coexist in equilibrium).
 PHASE-CHANGE PROCESSES OF
PURE SUBSTANCES (cont’d)
The constant pressure heating process (state 1 - 5) can be illustrated in the
following figure known as the T-v Diagram

Saturation state: State at which a

phase change begins or ends

T-v diagram for the

heating process of
water at constant
pressure.
Consider repeating this process for other constant pressure lines as
shown below. (Where are the saturated liquid and vapour line?)
Critical point: The point at which
the saturated liquid and saturated
vapor states are identical.
 SATURATION TEMPERATURE
AND SATURATION PRESSURE

The temperature at which water starts boiling depends
on the pressure; therefore, if the pressure is fixed, so is
the boiling temperature.

Water boils at 100°C at 1 atm pressure.

Saturation temperature Tsat: The temperature at

which a pure substance changes phase at a given
pressure.

Saturation pressure Psat: The pressure at which a

pure substance changes phase at a given
temperature.
 PROPERTY DIAGRAMS:
T-v Diagram of Pure Substance

Where are the saturation temperatures for P1 and P2?

 PROPERTY DIAGRAMS:
P-v Diagram of Pure Substance

Where are the saturation pressures for T1 and T2 ?

 EXERCISES

Sketch T-v and P-v diagram for the following processes, with
respect to saturation lines. Show initial and final states.

1. A piston cylinder device contains (by mass) 50% liquid water and
50% water vapour in equilibrium at 800 kPa (Tsat=170.4°C). Heat
is transferred at constant pressure until temperature reaches
350°C.

2. Superheated water vapour at 15 bar, T=400°C is allowed to cool

at constant volume until it becomes saturated water vapour.

3. A piston cylinder device contains compressed liquid water. Heat is

added to the system at constant pressure until the entire liquid
is vaporized.
 Extending the Diagrams to
Include The Solid Phase
P-v diagram of a substance that P-v diagram of a substance that
contracts on freezing. expands on freezing (such as water).

Triple line: All 3 phases of a pure substance coexist in equilibrium

(substance have same P & T but different v)
 PHASE DIAGRAM

At triple-point pressure and

temperature, a substance
exists in three phases in
equilibrium.

For water,
Ttp = 0.01°C
Ptp = 0.6117 kPa

At low pressures (below

the triple-point value),
solids evaporate without
melting first (sublimation).
Sublimation:: Passing
from the solid phase
P-T diagram of pure substances. directly into the vapor
phase.
 PVT SURFACE
P-v-T surface of a substance P-v-T surface of a substance that
that contracts on freezing. expands on freezing (like water).

The P-v-T surfaces present a great deal of information at once, but in a thermodynamic
analysis it is more convenient to work with two-dimensional diagrams, such as the P-v
and T-v diagrams.
 Class Takeaway

Indicate the following states of a pure substance on P-v and T-v

diagrams:
(a) Saturated liquid line
(b) Saturated vapour line
(c) Critical point
(d) Triple point line
(e) Saturation temperature lines at 1 and 10 bar
(f) Constant pressure lines at 100oC and 200oC
 PROPERTIES OF PURE
SUBSTANCES

CHAPTER 3b
Open
1
 CONTENTS
Pure Substances
Phases of a Pure Substances
Phase Change Processes
Property Diagrams
Property Tables
The Ideal Gas Equation
Compressibility Factor
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2
 LESSON OBJECTIVES

At the end of the lesson, you should be able

to:

Obtain values of v, u, h, Psat, Tsat from the
tables

Calculate x, v, u, h for saturated mixture.

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 PROPERTY TABLES

For most substances, the relationships among thermodynamic properties
are too complex to be expressed by simple equations.

Therefore, properties are frequently presented in the form of tables.

Subscript f – properties of saturated liquid

Subscript g – properties of saturated vapour
Subscript fg – difference between properties of saturated vapor and saturated liquid
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(value of g – value of f ) 4
 ENTHALPY & ENTROPY

Some thermodynamic properties can be measured easily, but others
cannot and are calculated by using the relations between them and
measurable properties.

Enthalpy—A Combination Property

The combination u + Pv is frequently
encountered in the analysis of control
volumes.

Entropy – A property associated with the 2nd Law of

Thermodynamics, and will be defined in later
chapter.
Entropy, S (kJ/K)
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Specific entropy, s (kJ/kgK) 5
 TYPES OF TABLES

PROPERTIES OF WATER

Saturated Water – Temperature Table: Table A–4 or Table F10

Saturated Water – Pressure Table: Table A–5 or Table G10

Superheated Water: Table A-6 or Table F11

Compressed Liquid Water: Table A-7

Saturated Ice-Water Vapour: Table A-8

PROPERTIES OF REFRIGERANT-134a

Saturated Refrigerant 134a – Temperature Table: Table A–11 or
Table F12

Saturated Refrigerant 134a – Pressure Table: Table A–12 or Table
F13

Superheated Refrigerant 134a: Table A-13 or Table F14
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 Saturated Liquid–Vapor
Mixture

The relative amounts of liquid and vapor
phases in a saturated mixture are specified
by the quality x.

Quality is between 0 and 1

0: sat. liquid, 1: sat. vapor.

Quality has significance for saturated mixtures only, no meaning in

Open the compressed liquid or superheated vapor regions. 7
 Saturated Liquid–Vapor
Mixture (cont’d)
v − vf vavg − v f
The relation can be expressed as x= or x=
v fg v fg

Quality, x, is related to the horizontal distances

on P-v and T-v diagrams.

The v value of a saturated liquid–vapor mixture lies

between the vf and vg values at the specified T or P,
and is given by:
v = v f + xv fg

u = u f + xu

Similarly, the other fg

properties can be found h = h f + xh fg

by the expressions:
Open s = s f + xs8 fg
 Superheated Vapor
In the region to the right of the saturated vapor line and at
temperatures above the saturation temperature, a substance
exists as superheated vapor.

Compared to saturated vapor,

superheated vapor is characterized by
A partial
listing of
Table A–6.

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 EXAMPLES

1. A rigid tank contains 50 kg of saturated liquid water at 90°C.

Determine the pressure in the tank and the volume of the tank.

2. A piston cylinder device contains 0.05 m3 of saturated refrigerant

134a vapor at 360 kPa. Determine the temperature and the mass of
EXERCISES
vapor inside the cylinder.

3. Find the enthalpy, internal energy and volume of 3 kg of steam at

10 bar and quality of 0.75.

4. Sketch the P-v diagram and identify the location of the state of a
system containing water at P = 0.3 MPa and 500°C. Then find its v,
u and h.
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Compressed liquid
• On a T-v or P-v diagram, the region to the left of the saturated
liquid line
• Tables not as common as for the superheated vapor
• For water, table A-7
• Pressure and temperature are independent variables
Compressed Liquid

Compressed liquid is characterized by

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Approximation for Compressed Liquids
• Treat the compressed liquid as a saturated liquid at the given
temperature
• Properties depend on temperature much more strongly than they do
pressure
• For specific volume, internal energy and enthalpy use,
y ≈ yf @T or y ≈ y f @T
A more accurate relation for h (for low or
moderate P & T)

12
 Interpolation
Linear Interpolation
• Used when states encountered in
solving a problem do not fall exactly on
the grid of values provided by the
property tables
• Assume that the properties vary linearly
between two known points
• Known (X1, Y1) and (X2, Y2)
• Want to find Y3 for X3

X Y Y = mX + b  Y2 − Y1 
Y3 = Y1 +  (X 3 − X1 )
X1 Y1  Y2 − Y1   X 2 − X1 
m= 
X3 Y3  X 2 − X1 
 Y3 − Y1   X 3 − X1 
X2 Y2 b = Y1 − mX1  = 
 Y2 − Y1   X 2 − X1 13
 EXAMPLE

Determine the following properties of water:

a. Specific enthalpy at 10 MPa, 40°C
b. Specific Internal Energy at 30 MPa, 100°C
c. Specific volume at 5 MPa, 20°C

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 CLASS TAKEAWAY

Determine the missing properties and the phase

descriptions in the following table for water
T (oC) P, kPa u x
1 200 0.6
2 125 1600
3 1000 2960
4 75 500
5 850 0.0
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 Example 1

Find the internal energy of water at the given states below

for 6 MPa and plot the states on T-v, and P-v diagrams.

1. Dry saturated or saturated vapor

2. Wet saturated or saturated liquid

3. Moisture = 5%

4. T = 635 °C

5. T = 100 °C

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 Solution
1. P= 6 MPa, dry saturated or saturated vapor
Using Table A-5 or G7 (saturation table), look at P=6 MPa = 60 bar

kJ
u1 = u g @ 60 bar = 2589.7
kg

2. P = 6 MPa, wet saturated or saturated liquid

Using Table A-5 or G7 (saturation table), look at P=7 MPa = 70 bar

kJ
u2 = u f @ 60 bar = 1205.4
kg
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 Solution (cont’d)
3. Moisture = 5%, P = 6 MPa
mf
moisture y defined as y = = 0.05
m
then, the quality is x = 1 − y = 1 − 0.5 = 0.95

u3 = u f + x u fg
= 1205.4 + (0.95)(2589.7 − 1205.4)
kJ
= 2520.5
kg

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 Solution (cont’d)
4. P = 6 MPa, T = 635 °C
For P = 60 bar, Table A-5 or F10 gives Tsat = 275.6 °C.
Since 635 °C > Tsat for this pressure, the state is superheated.
Use Table A-6 or F11. Since no values for P=6 MPa, use interpolation.
T (°C ) u kJ/kg
600 3267.2
635 u=?
700 3453.0

u 4 − 3267.2 635 − 600

=
3453.0 − 3267.2 700 − 600

kJ
u4 = 3332.23
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kg 19
 Solution (cont’d)
5. P = 6 MPa, T = 100 °C
For P = 60 bar, Table A-5 or G7 gives Tsat = 275.6 °C.
Since 100 °C > Tsat for this pressure, the state is compressed liquid.
Use approximation.

kJ
u5 ≅ u = 419.06
f @ T =100o C
kg

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 Example 2
Determine the enthalpy of 1.5 kg of water contained in a
volume of 1.2 m3 at 200 kPa.

Volume 12. m3 m3
v= = = 0.8
mass 15
. kg kg
Using table at P = 200 kPa = 2 bar,
vf = 0.001061 m3/kg , vg = 0.8857 m3/kg
Now,
Is v < v f ? No
Is v f < v < v g ? Yes
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Is v g < v ? No 21
 Solution to Example 2 (cont’d)
We see that the state is in the two-phase or saturation
region. So we must find the quality, x, first.
v − vf
v = v f + x (v g − v f ) x=
vg − v f
0.8 − 0.001061
=
0.8857 − 0.001061
= 0.903 (What does this mean?)
then,
h = h f + x h fg
= 504.7 + (0.903)(22019
. )
kJ
= 2493.3
Open
kg Locate this state on a T-v diagram.
22
 Example 3
Determine the internal energy of refrigerant-134a at a
temperature of 0°C and a quality of 60%.

Using Table for T = 0 °C,

uf = 49.79 kJ/kg ug =227.06 kJ/kg

then, u = u f + x (ug − u f )
= 49.79 + (0.6)(227.06 − 49.79)
kJ
= 15615
.
kg
Open
23
 Example 4
Consider a closed, rigid container of water. The pressure is 700 kPa, the
mass of the saturated liquid is 1.78 kg, and the mass of the saturated
vapor is 0.22 kg. Heat is added to the water until the pressure
increases to 8 MPa. Find the final temperature, enthalpy, and internal
energy of the water. Does the liquid level rise or fall? Plot this process
on a P-v diagram with respect to the saturation lines and the critical
point.

Solution
Let’s introduce a solution procedure that we will follow throughout the course
(discussed in Ch 1)

Known: A saturated liq-vap mixture in a closed rigid container

Find: Final Temperature, Final Enthalpy, Final internal energy, Liquid Level,
Plot on P-v
Open
24
 Schematic/Data :

T We know that state 1 is saturated

P2 = 8 MPa mixture. But we don’t know what is
state 2.
T2 P1 = 0.7 MPa
2

T1
mg1 = 0.22 kg
1
m f 1 = 1.78 kg
v
v1

Assumption : 1. It is a closed system

2. The volume of the container remains constant

Analysis :
State 1 is at saturated liq-vap mixture. Must find quality.

mg1 0.22 kg
x1 = = = 0.11
m f 1 + mg1 (1.78 + 0.22) kg
Open
25
 At state 1: P1 = 0.7 MPa = 7 bar
From Table A-3 v f 1 = 0.0011080 m 3 / kg v g1 = 0.2729 m 3 / kg
u f 1 = 696.44 kJ / kg u g1 = 2572.5 kJ / kg

For saturated mixture, v1 = v f 1 + x1 (v g1 − v f 1 )

= 0.001108 + (011
. )(0.2729 − 0.001108)
m3
= 0.031
kg
Also u1 = u f 1 + x(u g1 − u f 1 )
= 696.44 + (0.11)(2572.5 − 696.44) = 902.81 kJ / kg
At state 2: P2 = 8 MPa = 80 bar
From Table A-3 3 v g 2 = 0.02352 m 3 / kg
v f 2 = 0.0013842 m / kg
u f 2 = 1305.6 kJ / kg u g 2 = 2569.8 kJ / kg
Open
26
 Since the volume of the container remains constant during both states,
v2 = v1 = 0.031 m3 / kg
Note that v2 is greater that v g 2 = 0.02352 , which indicates that
state 2 is at superheated region.

Using Table A-4, T2 = Tsat @ 8 MPa @ v =0.031

v T u h
0.03089 360 2772.7 3019.8
0.031 ? ? ?
0.03432 400 2863.8 3138.3
By method of interpolation;

400 − 360 0.03432 − 0.03089

=
400 − T2 0.03432 − 0.031
∴ T2 = 362 °C
Open
27
 Interpolating in the superheated tables also gives,

h2 = 3024 kJ/kg u2 = 2776 kJ/kg

Since State 2 is superheated, the liquid level falls.

Plot on P-v diagram.

Open
28
 Example 5
A closed rigid container of volume 0.5 m3 is placed on a hot plate.
Initially the container holds a two-phase mixture of saturated liquid-
vapour at P1 = 1 bar with a quality of 0.5. After heating the pressure in
the container is P2 = 1.5 bar. Indicate the initial and final states on a T-v
diagram and determine the temperature and the mass of vapour present
at each state. If heating continues, determine the pressure in bar, when
the container holds only saturated vapour.

Solution
Known: A saturated liq-vap mixture in a closed rigid container is heated on a
plate.
Find: Plot T-v diagram
Temperature at state 1 and 2
Mass of vapour present at state 1 and 2
Open Pressure when only saturated vapour exists 29
 Schematic/Data:

x1 = 0.5
V = 0.5 m3

Assumption: 1. The water in the container is a closed system

2. The volume of the container remains constant

Analysis:
At State 1: P1 = 1 bar , x1 = 0.5, V = 0.5 m 3
T1 = Tsat @1bar = 99.63 °C
Open
30
 Also at State 1: v f 1 = 0.0010432 v g1 = 1.694

For saturated mixture, v1 = v f 1 + x(v g1 − v f 1 )

= 0.0010432 + (0.5)(1.694 − 0.0010432)
m3
= 0.8475
kg
V 0. 5
Then, total mass m = = = 0.59 kg
v1 0.8475

Mass of vapour at state 1 is; mg1 = xm = 0.5 * 0.59 = 0.259 kg

At state 2, quality is not 0.5 anymore.

P2 = 1.5 bar , x2 = ?, v2 = v1 = 0.8475 m 3 / kg
T2 = Tsat @1.5bar = 111.4 °C
Open
31
 Also at State 2: v f 2 = 0.0010528 v g 2 = 1.159

For saturated mixture, v2 = v f 2 + x2 ( v g 2 − v f 2 )

0.8475 = 0.0010528 + ( x2 )(1.159 − 0.0010528)
∴ x2 = 0.731
V 0.5
Then, total mass m = = = 0.59 kg
v2 0.8475

Mass of vapour at state 2 is; mg 2 = x2 m = 0.731* 0.59 = 0.431 kg

At state 3, use Table A-3 to find value of vg=0.8475.

It is found to be located between 2 bar and 2.5 bar. Use
interpolation.
2.5 − 2 0.7187 − 0.8857
=
2.5 − P3 0.7187 − 0.8475
Open
∴ P3 = 2.11 bar
32
 PROPERTIES OF PURE
SUBSTANCES

CHAPTER 3c
Open
1
 CONTENTS

Pure Subtances
Phases of a Pure Substances
Phase Change Processes
Property Diagrams
Property Tables
The Ideal Gas Equation
Compressibility Factor
Open
2
 LESSON OBJECTIVES

At the end of the lesson, you should be able

to:

Describe the ideal gas equation

Apply the ideal gas EOS in the solution of

typical problems

Define compressibility factor

Open
3
 THE IDEAL-GAS EQUATION
OF STATE

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.

Ideal gas EoS

If a gas obeys this relation it is called an ideal gas.

P: absolute pressure (kPa)

T: absolute temperature (K)
v: specific volume (m3/kg)
R: gas constant (kJ/kg.K)
Open
4
 THE IDEAL-GAS EQUATION
OF STATE (cont’d)

Recall that: Pv = RT Ideal gas EoS

Where:
Ru Universal Gas Constant
R=
Gas Constant
M
Different substances have
different gas constants.

Open
5
 THE IDEAL-GAS EQUATION
OF STATE (cont’d)
The ideal gas equation of state may be written several ways.
Pv = RT
V
V = mv P = RT PV = mRT
m
Molar mass, M is defined as the mass (kg) of one kmole of a substance
Mass of subs tan ce (kg ) m
M= =
No of mole (kmol ) N
∴ m = MN
Ru PV = NRuT
PV = mRT = ( MN ) RT = ( MN ) T
M
Open
6
 Is Water Vapor an Ideal Gas?

At pressures below 10 kPa,

water vapor can be treated as
an ideal gas, regardless of its
temperature, with negligible
error (less than 0.1 percent).

At higher pressures, however,

the ideal gas assumption yields
unacceptable errors,
particularly in the vicinity of the
critical point and the saturated
vapor line.

Fig: Percentage of error ([|vtable - videal|/vtable]

×100) involved in assuming steam to be an
ideal gas, and the region where steam can be
treated as an ideal gas with less than 1
7
percent error.
 EXAMPLE 1(a)

Determine the specific volume of R-134a at 1 MPa and 70oC,

using ideal gas EOS. Compare the result with the actual value
of 0.024261 m3/kg. (R=0.08149 kJ/kgK)

Open
8
 COMPRESSIBILITY FACTOR

Gases deviate from ideal-gas behavior significantly at states near

saturation region & critical point.

This deviation can be accounted for by a compressibility factor (Z)
Compressibility factor Z: A factor
that measures the deviation of real
gases from ideal-gas behavior.
Pv
Z=
RT
vactual
It also can be expressed as Z=
videal
RT
The compressibility factor is unity where v ideal =
for ideal gases.
P
The farther away Z is from unity, At very low pressures, all gases approach ideal-
Openthe more the gas deviates from gas behavior (regardless of their temperature).9
ideal-gas behavior.
 Generalized Compressibility
Chart (Figure A-30a, A-30b, A-30c)
Pseudo-reduced
Z is determined from the generalized compressibility chart.
specific volume

Compressibility Factor (y- axis)

Pseudo reduced v (dotted lines)

Reduced T Reduced P (x- axis)

Open
10
 Generalized Compressibility
Chart (cont’d)
The Z factor is approximately the same for all gases at the same
reduced temperature and reduced pressure, which are defined as

T P
TR = and PR =
Tcr Pcr

Values of specific volume are

included on the generalized
chart through the variable
called the pseudo reduced
volume.

Open
11
Comparison of Z factors for various gases.
 EXAMPLE 1(b)
Determine the specific volume of refrigerant-134a at 1 MPa and
50oC, using: Use Figure A-30a to read Z (R=0.08149 kJ/kgK)
a) the ideal gas equation of state
b) the generalized compressibility chart.
Compare the values obtained to the actual value of 0.02171
m3/kg and determine the error involved in each case.

(Tcr=374.2 K, Pcr=4.059 MPa). The percentage of error for the

volume ([|vtable – videal|/vtable]x100)

Ans: 0.02632, 21.2%, 0.02211, 2%

Open
 EXERCISE
Nitrogen at 150 K has a specific volume of 0.041884 m3/kg.
Determine the pressure of nitrogen, using

a) the ideal gas equation and

b) the generalized compressibility chart

Open
13
 Useful Ideal Gas Relation: The Combined
Gas Law
By writing the ideal gas equation twice for a fixed mass and simplifying, the properties
of an ideal gas at two different states are related by
m1 = m2

PV PV PV
1 1 PV
1 1
= 2 2 = 2 2
R T1 R T2 T1 T2

Example 2-7
An ideal gas having an initial temperature of 25°C undergoes the two processes
described below. Determine the final temperature of the gas.

Process 1-2: The volume is held constant while the pressure doubles.
Process 2-3: The pressure is held constant while the volume is
reduced to one-third of the original volume.
Open
14
14
 P T2
Process 1-3: 3
2
Ideal m1 = m3
Gas
T3
or T1
1
PV
1 1 PV
= 3 3
T1 T3 V

but V3 = V1/3 and P3 = P2 = 2P1

Therefore, P3 V3
T3 = T1
P1 V1
2 P1 V1 / 3 2
T3 = T1 = T1
P1 V1 3
2
Open
T3 = (25 + 273) K = 198.7 K = −74.3°C 15
15
3
 CLASS TAKEAWAY
The pressure of water vapor at T=350oC and v=0.035262 m3/kg, can be
determine by using;
a. The steam tables
b. The ideal gas equation
c. The generalized compressibility chart

Compare the values obtained and justify your answer.

Open
16
 Assignment # 2

ASSIGNMENT #2 - individual assignment

Due date: Wednesday, 11 th October 2017,

12 pm

Problems: 3-33, 3-40, 3-72 and 3-85. from

Textbook: Thermodynamics, An Engineering
Approach, 8th Ed.
Open
17
17
Reflection 1: We have been halfway through our learning journey in
this 1st half. Please reflect what we have gone through so far by
answering the following questions:
1. What are the topic/s do you like in Thermodynamics I that you
have learned so far? Please list.
2. What is the most challenging topic that you have learned so far?
What improvement do you think can be done to improve your
understanding of this topic?
3. What activity(ies) do you like in the student centre learning and
want to have it more in the remaining part?
Note: you need to write/copy and paste your reflection on
https://padlet.com/abaheta/e6daxwlgdy7z
18
 Summary of Chapter 3

Pure substance
Property tables

Phases of a pure  Enthalpy
substance
 Saturated liquid, saturated

Phase-change processes of vapor, Saturated liquid vapor
pure substances mixture, Superheated vapor,
 Compressed liquid, compressed liquid
Saturated liquid, Saturated
vapor, Superheated vapor
The ideal gas equation of
 Saturation temperature and state
Saturation pressure

Property diagrams for  Is water vapor an ideal gas?
phase change processes
Compressibility factor
 The T-v diagram, The P-v
diagram, The P-T diagram,
The P-v-T surface
Open
19
PROPERTIES OF PURE
SUBSTANCES

CHAPTER 3d
1. Define the specific heat at constant volume and the
specific heat at constant pressure.
2. Relate the specific heats to the calculation of the
changes in internal energy and enthalpy of ideal
gases.
3. Describe incompressible substances and determine
the changes in their internal energy and enthalpy.

The specific heat (C) is defined as the energy
required to raise the temperature of a unit mass of
a substance by one degree. The unit of C is
J/kg.oC or J/kg.K

Two kinds of specific heats: specific heat at
constant volume, Cv and specific heat at constant
pressure, Cp.

Cv : energy required to raise the temperature of the
unit mass of a substance by one degree as the
volume is maintained constant.

Cp : energy required to raise the temperature of the
unit mass of a substance by one degree as the
pressure is maintained constant.

Cv and Cp (values
given are for
helium gas).

Consider a fixed mass in a stationary closed
system undergoing a process @ V constant
(no expansion or compression work).

The conservation of energy principle:
ein – eout = ∆esystem (per unit mass)
δein – δeout = du (differential)

From the definition of Cv, left hand side must
be equal to Cv dT, so:
Cv dT = du (at constant V)
or

Now, consider a constant-pressure expansion
or compression process:

The conservation of energy principle:
ein – eout = ∆esystem (per unit mass)
δein – δeout = dh (differential, Ex. 4-5)

From the definition of Cp, left hand side must
be equal to Cp dT, so:
Cp dT = dh (at constant p)
or

Remind for an ideal gas: Pv = RT

For ideal gas, the internal energy is a function of
the temperature only:
u = u(T)

Using enthalpy definition and EOS of ideal gas:

Since R constant and u = u(T), then enthalpy of

ideal gas is also a function of temperature only:
h = h(T)
Since u and h depend
only on temperature
for an ideal gas, then
the specific heats Cv
and Cp also depend, at
most, on temperature
only. Therefore, at a
given temperature, u,
h, Cv, and Cp of an
ideal gas have fixed
values regardless of
the specific volume or
pressure.

The differential changes in the internal
energy and enthalpy of an ideal gas can be
expressed as:
du = Cv(T) dT and dh = Cp(T) dT

The change in internal energy and enthalpy:

Practically:

The average specific heats

Cp,avg and Cv,avg are evaluated
at the average temperature
(T1 + T2)/2

At low P, all real
gases approach
ideal-gas, and their
specific heats
depend on T only.

We need to have
relations for Cv and
Cp as functions of
temperature.

The u and h relations given previously are
not restricted to any kind of process. They
are valid for all processes.

The presence of the Cv in an equation should
not lead one to believe that this equation is
valid for a constant-volume process only.

The relation u = Cv,avg T is valid for any ideal
gas undergoing any process.

Similar argument can be given for Cp and h.
3 ways to determine the
internal energy and
enthalpy changes of ideal
gases:
1. Using the tabulated u
and h data
2. using the Cv or Cp
relations as a function
of T and integrate
them.
3. using average specific
heats
 U, H, Cv, and CP of Ideal Gases

Where to Find Specific Heat Values

•Table A-2a: specific heats for various
common gases at 300 K
•Table A-2b: specific heats for six
common gases tabulated from 250 to
1000 K in increments of 50 K
•Table A-2c: constant pressure specific
heat for various common gases
expressed empirically as a fourth-order
polynomial
U, H, Cv, and CP of Ideal Gases
U, H, Cv, and CP of Ideal Gases
U, H, Cv, and CP of Ideal Gases
 U, H, Cv, and CP of Ideal Gases
Tables
• Specific internal energy and specific enthalpy changes can also
be determined by using property tables to locate the values of
u(T1) and u(T2) or h(T1) and h(T2)

∆u = u (T2 ) − u (T1 ) and ∆h = h (T2 ) − h (T1 )

• Ideal gas properties are found in Tables A-17 through A-25

y = yM

From h = u + RT we have
dh = du + R dT

From du = Cv(T) dT and dh = Cp(T) dT
we have Cp(T) dT = Cv(T) dT + R dT

Divide by dT we have:
Cp = Cv + R

Another ideal-gas property called the specific
heat ratio k, defined as:
k varies with T. For monatomic gases,
k = 1.667. For diatomic gases (ex:
air), k = 1.4 at room T.

Air
at 300 K and 200 kPa is heated at
constant pressure to 600 K. Determine the
change in internal energy of air per unit
mass, using (a) data from the air table (Table
A–17), (b) the average specific heat value
(Table A–2b).

Aninsulated rigid tank initially contains 1.5
lbm of helium at 80°F and 50 psia. A paddle
wheel with a power rating of 0.02 hp is
operated within the tank for 30 min.
Determine (a) the final temperature and (b)
the final pressure of the helium gas.

A piston–cylinder device initially contains 0.5
m3 of nitrogen gas at 400 kPa and 27°C. An
electric heater within the device is turned on
and is allowed to pass a current of 2 A for 5
min from a 120-V source. Nitrogen expands
at constant pressure, and a heat loss of 2800
J occurs during the process. Determine the
final temperature of nitrogen.

A piston–cylinder device initially
contains air at 150 kPa and 27°C.
At this state, the piston is
resting on a pair of stops, as
shown in Fig. and the enclosed
volume is 400 L. The mass of the
piston is such that a 350-kPa
pressure is required to move it.
The air is now heated until its
volume has doubled. Determine
(a) the final temperature, (b)
the work done by the air, and (c)
the total heat transferred to the
air.

A substance whose specific volume (or density)
is constant is called an incompressible.

Cp = C v = C

Internal Energy Changes:

Enthalpy Changes:

For solid v ∆P = 0, and

For liquids:
1. Proses @ P constant
2. Proses @ T constant

Fora process between states 1 and 2, the
last relation can be expressed:

Bytaking state 2 to be the compressed liquid

state at a given T and P and state 1 to be the
saturated liquid state at the same T, the
enthalpy of the compressed liquid can be
expressed as

Determine the enthalpy of liquid water at
100°C and 15 MPa (a) by using compressed
liquid tables, (b) by approximating it as a
saturated liquid, and (c) by using the
correction given by Eq. 4–38.

A 50-kg iron block at 80°C is dropped into an
insulated tank that contains 0.5 m3 of liquid
water at 25°C. Determine the temperature
when thermal equilibrium is reached.
1. Define the specific heat at constant volume and the
specific heat at constant pressure.
2. Relate the specific heats to the calculation of the
changes in internal energy and enthalpy of ideal
gases.
3. Describe incompressible substances and determine
the changes in their internal energy and enthalpy.
 ENERGY ANALYSIS
OF CLOSED SYSTEMS
 CONTENTS

Moving Boundary Work

Energy Balance for Closed Systems

Specific Heats

Internal Energy, Enthalpy and Specific
Heats of Ideal Gases

Internal Energy, Enthalpy and Specific
Heats of Solids & Liquids

2
 LESSON OBJECTIVES
At the end of this lesson, you should be
able to:

Describe the moving boundary work or P dV work

Determine the moving boundary work for specific

processes (polytropic, constant pressure, isothermal)

Develop the general energy balance applied to

closed system

3
MOVING BOUNDARY WORK

The work associated with a moving boundary is called
boundary work.

Involves expansion and compression work

If the piston is allowed to move a distance ds in quasi-

equilibrium manner, the differential amount of work done is:

δWb = F ds = pA ds = p dV
Total Work done by the system:
2 Wb is positive → for expansion
Wb = ∫ p dV Wb is negative → for compression
1
4
 Moving Boundary Work (cont’d)

Work done by the system can be interpreted as the area under the
process path on a p-V diagram.

2 2
Area = A = ∫ dA = ∫ pdV
1 1

Note:
P is the absolute pressure and is always positive.
When dV is positive, Wb is positive.
When dV is negative, Wb is negative

5
 Moving Boundary Work (cont’d)

A gas can follow several different paths
as it expands from state 1 to state 2.

Each process path gives a different

value for the boundary work

Wb depends on the path (thus process)

and the end states

Work is NOT a property

The net work done during a cycle is the

difference between the work done by the
system and the work done on the system.

6
 CONSTANT VOLUME PROCESS

During this process, the volume is held constant

(fixed boundary).

Then the boundary work equation becomes

2
Wb = ∫ PdV How does the PV diagram for constant
1 volume looks like?
Example of constant volume system?
Wb = 0
7
 CONSTANT PRESSURE PROCESS

During this process, the pressure is held constant

(while the boundary increases or decreases).

Then the boundary work equation becomes

2 2
Wb = ∫ PdV = P ∫ dV How does the PV diagram for
1 1 constant pressure looks like?
Example of constant pressure
Wb = P(V2 − V1 ) system?

8
 ISOTHERMAL PROCESS
(of a gas)

If the temperature of an ideal gas system is held constant,
then the EOS provides the temperature-volume relation:
PV = mRT = contant
mRT
P=
V

Then the boundary work is
2 2 mRT  V2 
Wb = ∫ PdV = ∫ dV = mRT ln 
1 1 V  V1 

Note: The above equation is the result of applying the ideal gas assumption
9
for the equation of state.
 POLYTROPIC PROCESS

During an expansion or compression
process, both changes in P & V are
related by

PV n = C

Where n is a constant for that particular

process and C is a constant.
2
2 2 C 2  CV  1− n
Wb = ∫ PdV = ∫ n
dV = ∫ CV dV = 
−n

1 1 V 1
 1 − n 1
1− n 1− n 1− n
CV 2 − CV 11− n P2V 2nV2 − P1V1 nV1 P2V2 − P1V1
Wb = = =
1− n 1− n 1− n 10
 ENERGY BALANCE FOR
CLOSED SYSTEMS
First law of thermodynamics = conservation of energy
• energy can be neither created or destroyed
• energy can only change forms

The
Recall the energy balance in Chapter 2
total quantities are related to the quantities per unit time is

Qnet − Wnet + (E mass,in − E mass,out ) = ∆U + ∆KE + ∆PE

For closed system (no mass transfer),

Qnet − Wnet = ∆U + ∆KE + ∆PE 11
 ENERGY BALANCE FOR
CLOSED SYSTEMS (cont’d)

Energy balance for closed system (sign convention is used)

Qnet − Wnet = ∆U + ∆KE + ∆PE

where

The first law cannot be proven

mathematically, but no process in nature is
known to have violated the first law, and
this should be taken as sufficient proof.

Various forms of the first-law relation

for closed systems when sign
12
convention is used.
 ENERGY BALANCE FOR
CLOSED SYSTEMS (cont’d)
General analysis for a stationary closed system undergoing a quasi-
equilibrium constant-pressure process. Q is to the system and W is
from the system.

Q − W = ∆U + ∆KE + ∆PE
Q − Wother − Wb = ∆U = U 2 − U1

Q − Wother − P(V2 − V1 ) = U 2 − U1

Q − Wother = (U 2 − U1 ) + P(V2 − V1 )

Q − Wother = (U 2 + PV2 ) − (U1 + PV1 )

Q − Wother = H 2 − H1 13
 EXERCISES

1. A mass of 2.4 kg of air at 150 kPa and 12oC is contained in

a gas-tight, frictionless piston-cylinder device. The air is
now compressed to a final pressure of 600 kPa. During the
process, heat is transferred from the air such that the
temperature inside the cylinder remains constant.
Calculate the work done during this process.
(-272.1 kJ)

2. A mass of 5 kg of saturated water vapour at 300 kPa is

heated at constant pressure in a piston-cylinder assembly
until the temperature reaches 200°C. Calculate the work
done by the steam during this process.
(166 kJ)14
 EXERCISES
3. A closed rigid tank contains 2 kg of water initially at 80°C
and a quality of 0.6. Heat transfer occurs until the tank
contains only saturated vapor. Kinetic and potential energy
effects are negligible. For the water as the system,
determine the amount of heat transfer by heat, in kJ.
(ans: 1846 kJ)

4. A piston-cylinder device contains 5 kg of refrigerant-134a

at 800 kPa and 60°C. The refrigerant is now cooled at
constant pressure until it exists as a liquid at 20°C.
Determine the amount of heat loss and show the process
on a T-v diagram with respect to saturation lines.
(ans: 1087.6 kJ)
15
 EXAMPLE 1
A gas in a piston-cylinder assembly undergoes an expansion process for
which the relationship between pressure and volume is given by pVn =
constant.
The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final
volume is 0.2 m3. Determine the work for the process in kJ if (a) n = 1.5,
(b) n = 1.0, (c) n = 0

Known A gas in a piston-cylinder assembly undergoes a polytropic expansion

Find Work for the process

Schematic
Data

16
 Example 1 (cont’d)
Assumption

1. The gas is a closed system

2. The moving boundary is the only work mode
3. The expansion is a polytropic process

Analysis

For polytropic process;

P2V2 − P1V1
W=
1− n
We know the values for all parameters except P2.
1.5 1.5
V   0.1 
P2 = P1  1  = 3  = 1.06 bar
 V2   0.2 
Thus (106 * 0.2) − (300 * 0.1)
W= = 17.6 kJ
1 − 1.5 17
 Example 2
A frictionless piston-cylinder device initially contains 200L of saturated liquid
refrigerant-134a. The piston is free to move, and its mass is such that it
maintains a pressure of 800 kPa on the refrigerant. The refrigerant is now
heated until its temperature rises to 50oC. Calculate the work done during this
process.

R-134a in a piston-cylinder assembly undergoes heating process at

Known
constant pressure
Find Work for the process

Schematic
Data

18
 Example 2 (cont’d)
Assumption

1. The gas is a closed system

2. The moving boundary is the only work mode
3. Quasiequilibrium process

Analysis
2 2
Wb = ∫ PdV = P ∫ dV = Po (V2 − V1 )
1 1
We need to find V1 and V2. Refer Table A-10.
Since state 1 is saturated liquid, v = v
m3
1 f @ 8bar = 0.0008454
kg
V1 200 × 0.001
m= = = 236.57 kg
v1 0.0008454
m3
State 1 is superheated since T2 is greater than Tsat v2 = 0.02846
kg
W = Pm(v2 − v1 ) = 800 × 236.57 × (0.02846 − 0.0008454) = 5227 kJ 19
 Example 3
A piston-cylinder device initially contains 0.4 m3 of air at 100 kPa and
80°C. The air is now compressed to 0.1 m3 in such a way that the
temperature inside the cylinder remains constant. Determine the work
done during this process.
Air in a piston-cylinder device undergoes compression at constant
Known
temperature.
Find Work for the process

Schematic
Data

20
 Example 3 (cont’d)
Assumption

1. The air is a closed system

2. Process is quasiequilibrium
3. Air is an ideal gas (low pressure, high temperature)

Analysis
2 2 mRT  V2 
Wb = ∫ PdV = ∫ dV = mRT . ln 
1 1 V  V1 
We know that P1V1 = mRT = P2V2

Therefore
 V2   V2   0.1 
Wb = mRT . ln  = P1V1 ln  = 100 × 0.4 × ln 
 V1   V1   0.4 
= −55.45 kJ
21
 CLASS TAKEAWAY
2

Boundary Work: Wb = ∫ PdV

1

Derive the equations for boundary work of the following

processes:

a.Constant volume

b.Constant pressure

c.Polytropic

d.Isothermal 22
ENERGY ANALYSIS
OF CLOSED
SYSTEMS
 CONTENTS

Moving Boundary Work

Energy Balance for Closed Systems

Specific Heats

Internal Energy, Enthalpy and Specific
Heats of Ideal Gases

Internal Energy, Enthalpy and Specific
Heats of Solids & Liquids

2
 LESSON OBJECTIVES
At the end of this lesson, you should be able
to:

Develop the general energy balance applied to
closed system

3
 SPECIFIC HEATS
Definition: The energy required to raise
the temperature of a unit mass of a
substance by 1 degree.
❐ common units: kJ/kg°°C or kJ/kgK

The specific heat of a substance changes with temperature.

Specific heat at constant pressure, cp: The

 ∂h 
energy required to raise the temperature of the Cp =  
unit mass of a substance by one degree as the  ∂T  p
pressure is maintained constant.

Specific heat at constant volume, cv: The

energy required to raise the temperature of the  ∂u 
Cv =  
unit mass of a substance by one degree as the  ∂T v
volume is maintained constant. Lecture 2
 SPECIFIC HEATS (cont’d)
The differential changes in the internal energy & enthalpy can be
expressed as

du = C v ( T ) dT dh = C p ( T ) dT
The integration in these equation can be performed assuming constant specific
heats, yielding

u 2 − u1 = Cv ,av (T2 − T1 ) (kJ/kg) h2 − h1 = C p ,av (T2 − T1 ) (kJ/kg)

∆u = Cv ,av .∆T ∆h = C p ,av .∆T

∆U = mCv ,av .∆T ∆H = mC p ,av .∆T
Lecture 2
 SPECIFIC HEAT RELATIONS
OF IDEAL GASES
Three ways of calculating ∆u and ∆h for ideal gases

using tabulated u and h data –
when it is readily available

Using Cv and Cp as a function of

temperature – perform integration –
computerized

Using average Cv and Cp –

convenient when property tables not
available. Accurate if ∆T is small

For air, refer Tables

Refer Table A-2 F2 or A-16

Q − W = ∆U + ∆KE + ∆PE
Lecture 2
 Specific heat relations of ideal
gases (cont’d)
Derive relationship between specific heats.
We know that h = u + pv h = u + RT

Differentiate w.r.t T dh du
= +R
dT dT
Which means that C p (T ) = Cv (T ) + R
If CP and Cv remain
constant with C p = Cv + R (kJ/kgK)
temperature;

Specific heat ratio, k Cp The cp of an ideal gas can be

k= determined from a knowledge of
C
Lecture v
2 cv and R.
 EXAMPLE
Air at 300 K and 200 kPa is heated at constant
pressure to 600 K. Determine the change in internal
energy of air per unit mass using (a) data from table,
and (b) the average specific heat value
(ans: 220.71 kJ/kg , 219.9 kJ/kg)
Find the work done and heat transfer for the process.

8
 Internal Energy, Enthalpy, And
Specific Heats Of Solids And Liquids
Incompressible substance: A substance whose specific
volume (or density) is constant. Solids and liquids are
incompressible substances.

The cv and cp values of

incompressible substances are
The specific volumes of identical and are denoted by c.
incompressible substances
Specific heat values for common solids &
remain constant during a liquids are given in Table A-3
9
process.
 Internal Energy, Enthalpy, And
Specific Heats Of Solids And Liquids
Internal Energy Changes

Specific Enthalpy Changes

∆h = ∆u + vdP ≅ Cav ∆T + v∆P

10
 EXAMPLE
A 50-kg iron block at 80oC is dropped into an insulated
tank that contains 0.5 m3 of liquid water at 25oC.
Determine the temperature when the thermal
equilibrium is reached.
(ans: 25.6oC)

11
 Summary of Chapter 4

Moving boundary work

Wb for an isothermal process

Wb for a constant-pressure process

Wb for a polytropic process

Energy balance for closed systems

Energy balance for a constant-pressure expansion or compression
process

Specific heats

Constant-pressure specific heat, cp

Constant-volume specific heat, cv

Internal energy, enthalpy, and specific heats of ideal gases

Specific heat relations of ideal gases

Internal energy, enthalpy, and specific heats of incompressible
substances (solids and liquids)
12
 Open System
(Control Volume)-
EXAMPLES
1
#1. Steam at 0.4 MPa, 300 oC, enters an
adiabatic nozzle with a low velocity and
leaves at 0.2MPa with a quality of 90%.
Find the exit velocity, in m/s.
3
Example 2: Deceleration of Air
in a Diffuser
Air at 10oC and 80 kpa enters the diffuser of a jet
engine steadily with a velocity of 200 m/s. The inlet
area of the diffuser is 0.4 m2. The air leaves the
diffuser with a velocity that is very small compared to
the inlet velocity.

Determine
(1) The mass flow rate of the air and
(2) The temperature of the air leaving the
diffuser.

4
5
6
Example 3: Acceleration of Steam in a Nozzle
Steam at 1.8Mpa and 350C steadily enters a nozzle whose inlet
area is 0.02 m2. The mass flow rate of the steam through the nozzle
is 5.2 kg/s. Steam leaves the nozzle at 1.2Mpa with a velocity of
300m/s. The heat losses from the nozzle per unit mass of the steam
are estimated to be 2.8 KJ/kg.
Determine:
(a) the inlet velocity and
(b) the exit temperature of the steam.

Answers: a) 40 m/s,
b) ____C
7
8
 Example (4-12): Power
Generation by a Steam Turbine
The power output of an adiabatic steam
turbine is 5 MW, and the inlet and the
exit conditions of the steam are as
indicated in the figure on the right.
a) Compare the magnitude of h, ke,
and pe.
b) Determine the work done per unit
mass of the steam flowing through the
turbine.
c) Calculate the mass flow rate of the
steam.
Answers: a) h = -885.9 kJ/kg,
ke = 14.95 kJ/kg, pe = -0.04 kJ/kg,
9
b) 871.0 kJ/kg, and c) 5.74 kg/s
10
11
12
#5. In a single-flash geothermal power plant, geothermal water
enters the flash chamber (a throttling valve) at 230°C as
a saturated liquid at a rate of 50 kg/s. The steam resulting from
the flashing process enters a turbine and leaves at 20kPa with
a moisture content of 5 percent. Determine i) the temperature of
the steam after the flashing process and ii) the power output from
the turbine if the pressure of the steam at the exit of the flash
chamber is (a) 1 MPa, (b) 500 kPa, (c) 100 kPa, (d) 50 kPa.

13
14
15
#6. In steam power plants, open feed-water heaters are
frequently utilized to heat the feed-water by mixing it with steam
bled off the turbine at some intermediate stage. Consider an
open feed-water heater that operates at a pressure of 1000kPa.
Feed-water at 50°C and 1000kPa is to be heated with
superheated steam at 200°C and 1000kPa. In an ideal feed-
water heater, the mixture leaves the heater as saturated liquid
at the feed-water pressure. Determine the ratio of the mass
flow rates of the feed-water and the superheated vapor for this
case.

16
17
 Example: Cooling of Refrigerant-134a by Water
Refrigerant-134a is to be cooled by
water in a condenser. The refrigerant
enters the condenser with a mass
flow rate of 6 kg/min at 1 MPa and
70oC and leaves at 34oC. The cooling
water enters at 300 kPa and 15oC and
leaves at 25oC. Neglecting any
pressure drop, determine
(a) the mass flow rate of the cooling
water required and (b) the heat
transfer rate from the refrigerant to
water.

<Answers: a) 0.486 kg/s, b) 20.35 kJ/s> 18

19
20
21
#8. An adiabatic air compressor is to be powered by a direct-
coupled adiabatic steam turbine that is also driving a
generator. Steam enters the turbine at 12.5MPa and 500°C
at a rate of 25 kg/s and exits at 10kPa and a quality of
0.92. Air enters the compressor at 98kPa and 295 K at a
rate of 10 kg/s and exits at 1MPa and 620 K. Determine
the net power delivered to the generator by the turbine.
23
CHAPTER 5

MASS & ENERGY ANALYSIS

OF CONTROL VOLUMES
 CONTENTS
 Conservation of Mass
 Flow Work & the Energy of a Flowing
Fluid
 Energy Analysis for Steady-Flow
Devices
 Some Steady-Flow Engineering Devices
 Energy Analysis of Unsteady Flow
 LESSON OBJECTIVES:
At the end of this lesson, you should be
able to:
 Develop the conservation of mass principle
 Apply the conservation of mass principle to
various systems

 Apply the first law of thermodynamics as the

statement of the conservation of energy
principle to control volumes
 CONSERVATION OF MASS
Conservation of mass: Mass, like energy, is a conserved property,
and it cannot be created or destroyed during a process.

 Closed systems: The mass of the system remain constant during a

process.
 Control volumes: Mass can cross the boundaries, and so we must
keep track of the amount of mass entering and leaving the control
volume.

The mass flow through a cross-sectional area per unit time is called the
 expressed as:
mass flow rate, m
Where;
m  AV 
AV  = fluid density, kg/m3
v A = flow area, m2
V = fluid velocity, m/s
v = specific volume, m3/kg
CONSERVATION OF MASS
The volume of the fluid flowing The mass and volume flow rates are
through a cross section per unit related by:
time is called the volume flow rate, V
AV V
V  AV m  
v v

Remember! V here represents Velocity (m/s), NOT Volume

The conservation of mass principle for a control volume: The net mass transfer
to or from a control volume during a time interval t is equal to the net change
(increase or decrease) in the total mass within the control volume during t.
 dmcv
  m i   m e
or dt i e

For steady flow system,

 m in   m out Multiple inlets
and exits
 Example

Steam steadily enters a turbine through a pipe

with a 150 mm diameter. The inlet steam is at
20 MPa and 600oC with a velocity of 90 m/s.
The exit pipe has a diameter of 600 mm and
the steam exits at 300 kPa and 150oC.
Determine the mass flow rate of the steam and
the steam exit velocity.
Solution
To find mass flow rate
AiVi
i 
m (1)
vi
Ai   0.15 4  0.01767 m2
2

From steam table

vi  0.01818 m3 / kg

Substituting values into (1)

0.0176790
m i   87.48 kg / s
0.01818
Solution
To find exit velocity
It is a steady state problem, so
 m in   m out
AeVe
m  m i  m e   87.48 kg / s
ve
Ae   0.6 4  0.2827 m2
2

From steam table, and interpolation

ve  0.633986 m3 / kg

me ve 87.480.633986
Ve  
Ae 0.2827
Ve  196.2 m / s
FLOW WORK AND THE ENERGY OF A
FLOWING FLUID
Flow work, or flow energy: The work (or energy) required to push the mass into
or out of the control volume. This work is necessary for maintaining a continuous
flow through a control volume.

w flow  Pv (kJ/kg)

Recall that the total energy of a system:

Therefore the total energy per unit mass for a fluid flowing in and out of a
control volume:
Total Energy of a Flowing Fluid
The flow energy is automatically taken care of by enthalpy. In fact, this is the
main reason for defining the property enthalpy.

The total energy consists of three parts for a nonflowing fluid and four parts for a
flowing fluid.
Energy Transport by Mass

The product m   is the energy

transported into or out of control
volume by mass per unit time.
ENERGY ANALYSIS OF STEADY-FLOW
SYSTEMS

Under steady-flow conditions:

Many engineering systems such
as power plants operate under  the mass and energy contents of a
steady conditions. control volume remain constant.

 the fluid properties at an inlet or exit

remain constant (do not change with
time).
ENERGY ANALYSIS OF STEADY-FLOW
SYSTEMS (cont’d)
Energy
balance

 V 2   V 2 
Qcv  Wcv   m i  hi 
  i
 gzi    m e  he  e
 gze   0
i  2  e  2 
ENERGY ANALYSIS OF STEADY-FLOW
SYSTEMS (cont’d)

Consider a steady flow device with single

inlet and outlet

 V 2   V 2 
Qcv  Wcv   m i  hi 
   i
 gzi    m e  he 
  e
 gze   0
i  2  e  2 

  
 1 2
 2
 
Qcv  Wcv  mi  hi  he  Vi  Ve  g  zi  ze   0
 2 
Example

Air enters a control volume operating at steady

state at 1.05 bar, 300 K, with a volumetric flow
rate of 12 m3/min and exit at 12 bar, 400 K.
Heat transfer occurs at a rate of 20 kW from the
control volume to the surroundings. Neglecting
kinetic and potential energy effects, determine
the power, in kW.
Solution
p1  1.05 bar
T1  300 K 1 2 p2  12 bar
 AV 1  12 m3 / min T2  400 K

Q cv  20 kW
Energy rate balance
dEcv    V 2
  V 2

 Qcv  Wcv  m i  hi   gzi   m e  he   gze   0
i e

dt  2   2 
Knowing m  m 1  m 2
  2 2
 
Wcv  Qcv  m h2  h1       g  z 2  z1 
  V2 V1

  2 2  
…cont.
The energy rate balance is reduced to
W cv  Q cv  m h2  h1 
The mass flow rate is
m 
 AV 1  p1  AV 1
v1 RT1

m 
 x

1.05 bar 12 m 3 / min 10 2 kN / m 2 1 min
x  0.2439 kg / s
0.287 kJ / kgK 300 K  1 bar 60 s
Using Table A-22 to find h1 and h2
W cv   20 kW   0.2439 kg / s 400.98  300.19  kJ / kg
W cv  44.58 kW
Summary (Steady-Flow Systems)
Mass Rate Balance

 m   m
dmcv
 i e
dt i e

 
dmcv AiVi AeVe
 
dt i vi e ve

Energy Rate Balance

 Vi 2   Ve2 
 
dEcv 
 Qcv  Wcv  m i  hi   gzi   m e  he   gze   0
dt i  2  e  2 
CHAPTER 5

MASS & ENERGY ANALYSIS

OF CONTROL VOLUMES
CONTENTS
 Conservation of Mass
 Flow Work & the Energy of a Flowing
Fluid
 Energy Analysis for Steady-Flow
Devices
 Some Steady-Flow Engineering Devices
 Energy Analysis of Unsteady Flow
LESSON OBJECTIVES:
At the end of this lesson, you should be
able to:
 Solve energy balance problems for common
steady-flow devices
Steady-Flow Engineering Devices
 For steady-flow, the mass flow rates at exits equal the mass
flow rates at inlets
 m   m
i
i
e
e

 The energy transfer by heat and work and the total

energy of the system do not vary with time – steady
 The energy rate balance can be defined as:

 Vi 2   Ve2 
 
dE cv 
 Qcv  W cv  m i  hi   gz i   m e  he   gz e   0
dt i  2  e  2 
 
V12  V22  

0  Q cv  W cv  m  h1  h2    g z1  z 2  

For single
inlet and exit
 2 
 NOZZLES & DIFFUSERS
 Nozzles and diffusers are commonly utilized
in jet engines, rockets, spacecraft, and even
garden hoses.
 A nozzle is a device that increases the
velocity of a fluid at the expense of pressure.
 A diffuser is a device that increases the
pressure of a fluid by slowing it down.
Nozzles & Diffusers have:
 Varying cross-sectional area,
 No work
 Negligible change in PE

 V12  V22  
 
0  Q cv  W cv  m  h1  h2  
 g z1  z 2 

 2 

0  Q cv  m h1  h2  
V1
2
 V 2
2 
 What happened if the nozzle/diffuser is
 2  adiabatic?
 TURBINES
Turbine drives the electric generator in steam,
gas, or hydroelectric power plants.
As the fluid passes through the turbine, work is
done against the blades, which are attached to
the shaft. As a result, the shaft rotates, and the
turbine produces work.
Energy balance for the turbine:

0  Qcv  Wcv  m  h1  h2  
  
V1
2
 V 2
2
 
 g  z1  z 2 
 2 
If PE and KE negligible, energy balance
for the turbine:
Schematic

W cv  Q cv  m  h1  h 2 
diagram
for turbine

However, most of the cases, KE is

significant (hence it cannot be ignored)
 COMPRESSORS & PUMPS
Compressors, as well as
pumps are devices used to
increase the pressure of a fluid.
Work is supplied to these devices
from an external source through
a rotating shaft.

Pumps work very much like

A compressor is capable of
compressors except that they
compressing the gas to very high
handle liquids instead of gases.
pressures.
Energy balance for the compressor in this figure:

0  Q cv  W cv  m  h1  h2  
V1
2
 V 2
2
 
 g  z1  z 2 
 2 
Most of the cases, PE and KE negligible.

Schematic diagram
W cv  Q cv  m  h1  h 2 
for compressor
HEAT EXCHANGERS
Heat exchangers are
devices where two moving
fluid streams exchange heat
without mixing. No work is
being done.

PE and KE usually negligible.

The heat transfer associated with a heat exchanger may

be zero or nonzero depending on how the control
volume is selected.

Mass and energy balances for the adiabatic heat

exchanger in the figure is:
THROTTLING DEVICES
Throttling valves are any
kind of flow-restricting
devices that cause a
significant pressure drop in
the fluid.

For this device, there is:

 No work being done
 Negligible change in PE, KE
 No heat transfer


0  Qcv  Wcv  m  h1  h2  
  
V12  V22  
 g  z1  z 2 
 2 
Energy
balance
Problem
Air enters an insulated diffuser operating at
steady state with a pressure of 1 bar, a
temperature of 300 K and a velocity of 250
m/s. At the exit, the pressure is 1.13 bar and
the velocity is 140 m/s. Potential energy
effects can be neglected. Using the ideal gas
model, determine:
(a) the ratio of the exit flow area to the inlet
flow area
(b) the exit temperature, in K.
Solution
p1  1 bar p2  1.13 bar
T1  300 K T2
V1  250 m / s V2  140 m / s
A1 A2

mass rate balance

A2V 2 A1V1 A2V 2 A1V1
  
v2 v1 RT 2 p 2 RT1 p1

A2  p1  T2  V1  T2 is unknown and can be obtain from

    
A1  p2  T1  V2  the energy rate balance
…cont.
energy rate balance
 V 2
 V 2

 
0  Qcv  Wcv  m
  h1  h2  1 2
 g  z1  z 2 
 2 
V12 V22
h2  h1  
2 2
From Table A-17
 250 2  140 2  1
h2   300 .19    321 .64 kJ / kg
 2  1000
Interpolating for h2 in Table A-17: T2  321.3 K
A2  1 bar  321.3 K  250 m / s 
     1.692
A1  1.13 bar  300 K  140 m / s 
Problem
A well-insulated turbine operating at steady state
develops 23 MW of power for a steam flow rate of
40.2 kg/s. The steam enters at 350oC with a velocity
of 35 m/s and exits as saturated vapor at 0.06 bar
with a velocity of 120 m/s. Neglecting potential energy
effects, determine the inlet pressure in bar.
 Solution
T1  350o C p2  0.06 bar
V1  35 m / s V2  120 m / s Mass balance
m  40.2 kg / s
sat.vapor
m  m 1  m 2
Wcv  23 MW
Q  0
cv

Energy rate balance

  V 2
 V 2
 
0  Q cv  W cv  m  h1  h 2    1 2
  g  z1  z 2 
  2  
The equation is reduced to: From Table A-6 (Superheated),
W cv V V 2 2
h1   h2  1 2
p1  16 bar
m 2
23x103 120 2  352
h1   2567.4   3146.13
40.2 21000
Problem
The cooling coil of an air-conditioning system is a heat
exchanger in which air passes over tubes through which
Refrigerant-22 flows. Air enters with a volumetric flow rate
of 40 m3/min at 27oC, 1.1 bar, and exits at 15oC, 1 bar.
Refrigerant enters the tubes at 7 bar with a quality of 16%
and exits at 7 bar, 15oC. Ignoring heat transfer from the
outside of the heat exchanger and neglecting kinetic and
potential energy effects, determine at steady state
(a) the mass flow rate of refrigerant, in kg/min
(b) the rate of energy transfer, in kJ/min, from the
air to the refrigerant.
Solution

T2  15o C
p2  1 bar 2 R  22 Mass balance
x3  16%
3 p3  7 bar m air  m 1  m 2
m R  22  m 3  m 4
4 T4  15o C
p4  7 bar
1 Energy rate balance

 
Air
 AV 1  40 m3 / min  V12  V22 
T1  27 o C 
0  Q cv  Wcv  m air  h1  h2   
 g z1  z 2 
p1  1.1 bar  2 
  V32  V42  
 
 m R  22  h3  h4  
 g z3  z 4  
 2 
 …cont.

The energy rate balance is reduced to: From Table A-8, (external resource)
 h1  h2  h3 = hf3 + x3 hfg3=58.04 + (0.16)(195.60)
m R  22  m air   h3=89.34 kJ/kg
h h 
 4 3
From Table A-9 (external resource)
The mass flow rate of air is found by
, h4 = 256.86 kJ/kg 300.19  288.15
using the ideal gas equation:  
 m R  22  51.103 
 AV 1 p1  AV 1  256.86  89.34 
m air  
v1 RT1 m R  22  3.673 kg / min
1.1x10 2 40  (b) Considering a control volume for
m air   51.103 kg / min
0.287300 
the refrigerant only:

From Table A-17, h1 = 300.19 kJ/kg

 
0  Q R  22  m R  22 h3  h4
and h2 = 288.15 kJ/kg: Q  3.673256.86  89.34 
R  22
Q R  22  615.3 kJ / min
Summary of Chapter 5
 Conservation of mass
 Mass and volume flow rates
 Mass balance for a steady-flow process
 Mass balance for incompressible flow
 Flow work and the energy of a flowing fluid
 Energy transport by mass
 Energy analysis of steady-flow systems
 Some steady-flow engineering devices
 Nozzles and Diffusers
 Turbines and Compressors
 Throttling valves
 Heat exchangers
 CHAPTER 6

THE SECOND LAW OF

THERMODYNAMICS
CONTENTS

6 - 1 Second Law of Thermodynamics

6 - 2 Heat Engine, Refrigeration & Heat
Pump
6 - 3 Irreversible and Reversible Processes
6 - 4 Carnot Cycle
6 - 5 Kelvin Temperature Scale
6 - 6 Maximum Thermal Efficiency
OBJECTIVES

 To recognise the need for the second law

of thermodynamics in real processes.

 To identify performance limits for

thermodynamic cycles.
Introduction to the Second Law of Thermodynamics

Overview of the Second Law

 From our experiences, we know that processes occur in a certain
direction and not in the reverse direction
 Examples of processes occurring in a certain direction
 A cup of hot coffee left in a cooler
room eventually cools off
 Gases expand from a high
pressure to a low pressure
 Mass falls down.

Fig. Illustrations of spontaneous processes and

the eventual attainment of equilibrium with the
surroundings. (a) Spontaneous heat transfer. (b)
Spontaneous expansion. (c) Falling mass.
Overview of the Second Law

 The first law does not place any restrictions on the direction
of a process
 Satisfying the first law alone does not guarantee that a
process will take place
 The second law of thermodynamics addresses this issue
 A process will not occur unless it satisfies both the first and
second law of thermodynamics
THERMAL RESERVOIR
• Thermal Energy Reservoir or Heat Reservoir -
a body with large thermal energy capacity
that can supply or absorb finite amounts of
heat without undergoing any change in
temperature.
• Source - a reservoir that supplies energy in
the form of heat.
• Sink - a reservoir that absorbs energy in the
form of heat Heat

• Heat transfer is the ONLY interaction between

the thermal reservoir and its surroundings.
• The reservoir temperature remains uniform
and constant during a process. Thermal energy
SINK
• Oceans, lakes, and rivers as well as the
atmospheric air can be modeled accurately
as thermal energy reservoirs.
HEAT ENGINES

Consider a rotating shaft:

 Mechanical work by shaft is

converted to heat
 BUT, heat transfer to the
water will not rotate the shaft
 UNLESS….. We use….

HEAT ENGINES
HEAT ENGINES
A heat engine – a device that operates in a cycle and
produces net positive work while heat transfer from a
high temperature source.
 Measurement of the performance for a heat engine is
known as the thermal efficiency.

Fig: Part of the heat received by a heat

engine is converted to work, while the rest
is rejected to a sink
HEAT ENGINES

1. They receive heat from a high-

temperature source (solar energy, oil
furnace, nuclear reactor, etc.).

2. They convert part of this heat to

work (usually in the form of a
rotating shaft).

3. They reject remaining waste heat to

a low-temperature sink (the
atmosphere, rivers, etc.).
Fig: Example of a Heat Engine –
4. They operate on a cycle. Steam Power Plant
HEAT ENGINES

Work Output from a Heat Engine THERMAL EFFICIENCY:

The fraction of heat input that is
W net, out  W out  W in converted to net work output
desired output
• The entire device can be analyzed performanc e 
desired input
as a closed system undergoing a
Wnet ,out
cycle (DU = 0) th 
• In this case, the net work output is Qin
also equal to the net heat transfer Wcycle QH  QL
 th  
W net, out  Q in  Q out  Q H  Q L QH QH
QL
 th  1 
QH
Example 6-1

A steam power plant produces 50 MW of net work while burning fuel

to produce 150 MW of heat energy at the high temperature.
Determine the cycle thermal efficiency and the heat rejected by the
cycle to the surroundings.
Wnet , out
 th 
QH
50 MW
  0.333 or 33.3%
150 MW

Wnet , out  QH  QL
QL  QH  Wnet , out
 150 MW  50 MW
 100 MW
STATEMENTS OF THE SECOND LAW

Kelvin-Planck Statement of the Second Law

“It is impossible for any device that operates on a cycle
to receive heat from a single reservoir and produce a net
amount of work”

“No heat engine can have a thermal

efficiency of 100 percent, or for a power
plant to operate, the working fluid must
exchange heat with environment as well
as furnace”.
Fig: A heat engine that violates the
Kelvin-Planck statement of the second law
 Refrigerators
Refrigerator
• A device that transfers heat from a low-temperature source to a
high-temperature medium
Refrigerant
• The working fluid used in a refrigeration cycle
 6-4 Refrigerators and Heat Pumps
Coefficient of Performance (Refrigerator)
• Provides a measure of the efficiency of a refrigerator

Desired Output
COPR 
Required Input

QL
COPR 
Wnet, in

W net, in  Q H  Q L

QL 1
COPR  
QH  QL QH QL  1
 Heat Pumps
The objective of the heat pump is to maintain a heated space
at a high temperature

Desired Output
COPHP 
Required Input
QH
COPHP 
Wnet, in

W net, in  Q H  Q L

QH 1
COPHP  
QH  QL 1  QL QH

• Relationship to refrigerator
COPHP  COPR  1
EXAMPLE
2. A residential heat pump is used to provide heating during the
winter season to maintain a house temperature at 21oC. On a
typical day, the heat transfer to the house is 75 MJ/h when
the outside air temperature is –4oC. The heat pump has a COP
of 3.7. Determine the power required for the heat pump and
the heat transfer rate from the outside air.
QH
COPHP  Wcycle  QH  QL
Wcycle
QH QL  QH  Wcycle
Wcycle 
COPHP
75 MJ / h QL  75  20.27  54.73 MJ / h
Wcycle   20.27 MJ / h
3.7
STATEMENTS OF THE SECOND LAW

Clausius Statement of the Second Law

“It is impossible to construct a device that operates in a cycle
and produces no effect other than the transfer of heat from a
low-temperature body to a higher temperature body”

The statement suggests that:

• When heat transfer from a cooler
body to a hotter body occurs, there
must be some other effect within the
system/surrounding accomplishing
the heat transfer.

Fig: Heat pump that violates the Clausius

statement of the second law
Example

A heat pump with a coefficient of performance of 2.5

supplies energy to a room at a rate of 63300 kJ/h.
Determine:

a. The electrical power input to the heat pump, in kW.

b. The rate of heat absorption from the outside air, in
kW.

Win = 7.03 kW, QL= 10.55 kW

Final Jan 06

A refrigeration cycle removes 18000 kJ/h of heat from

the cold space maintained at -30oC to the surrounding.
The surrounding is at a temperature of 20oC. The
coefficient of performance of the refrigerator is 25% of
that of a reversible refrigeration cycle.

a. What is the COP of the refrigerator?

b. Determine the power input to the refrigerator, in
kW.

COPrev = 4.86, COPR= 1.215, Win = 4.11 kW

Final Jan 07

A heat engine with a thermal efficiency of 35 percent

produces 750 kJ of work. Heat transfer to the engine is from a
reservoir at 550 K, and the heat transfer from the engine to
the surrounding air is at 300 K.
a. Sketch a diagram to represent the heat engine
system.
b. Determine the heat transfer to and from the heat
engine.
c. If the heat engine is replaced by a Carnot heat engine
producing the same work output, determine the heat
transfer to and from the Carnot heat engine.

QH = 2142.9 kJ, QL= 1392.9 kJ, QH = 1650 kJ, QL = 900 kJ

SUMMARY
 A process will not occur unless it satisfies both the first
and second law of thermodynamics.

 The second law of thermodynamics can be expressed by

two equivalent statements.

 The Kelvin-Planck statement of the second law places

restrictions on the operation of heat engines.
 The Clausius statement of the second law places
restrictions on the operation of heat pumps and
refrigerators.
 Devices operating in a thermodynamic cycle include heat
engines, refrigerators and heat pumps.
 CHAPTER 6

THE SECOND LAW OF

THERMODYNAMICS
CONTENTS

6 - 1 Second Law of Thermodynamics

6 - 2 Heat Engine, Refrigeration & Heat
Pump
6 - 3 Irreversible and Reversible Processes
6 - 4 Carnot Cycle
6 - 5 Kelvin Temperature Scale
6 - 6 Maximum Thermal Efficiency
Most energy extracted from the
fuel in power plants is dumped to
the environment as waste heat,
here using a large cooling tower.

Why is so much energy wasted?

2nd law: no Q  W
with 100% efficiency
OBJECTIVES

 To understand the two Carnot corollaries of

the second law.

 To describe the processes involved in a

Carnot cycle.

 To apply the Carnot corollaries to assess the

performance of heat engines, refrigerators
and heat pumps.
REVERSIBLE AND IRREVERSIBLE

Reversible Process
• A process that can be reversed without
leaving any trace on the surroundings
Eg: a change that can go
forwards or backwards, for example
melting and freezing

• Possible only when the net heat and

net work exchange between the system
and the surroundings are zero
• Reversible processes are idealizations,
they do not occur in nature
Reversible Processes

In a reversible process
the system changes
in such a way that the
system and
surroundings can be
put back in their
original states by
exactly reversing the
process.

Changes are
infinitesimally small
in a reversible
process.
 Irreversible Processes

• Irreversible processes cannot be undone

by exactly reversing the change to the
system.
• All Spontaneous processes are
irreversible.
• All Real processes are irreversible.
REVERSIBLE AND IRREVERSIBLE

Irreversible Process
•Any process that is not reversible
 Irreversibilities may be found within the system and
within its surroundings.
 Internal irreversibilities – irreversibilities within the
system.
 External irreversibilities – irreversibilities within the
surroundings.
REVERSIBLE AND IRREVERSIBLE
Irreversibilities
 Factors that cause a process to
be irreversible, Examples
 Friction
 Unrestrained expansion of a
gas
 Heat transfer through a finite 20ºC
temperature difference Heat

 Mixing of two different

substances 20ºC

5ºC

An irreversible heat transfer process

REVERSIBLE AND IRREVERSIBLE

Why are Reversible Processes of Interest?

• They serve as idealized models to which actual processes can be
compared, i.e., they provide theoretical limits for a process

• Reversible processes deliver the most work to work-producing

devices and consume the least work for work-consuming devices

• They are easy to analyze, since the system passes through a series
of equilibrium states during a reversible process
 The Carnot Cycle

Carnot Cycle
•A cycle composed of four reversible processes:
two isothermal and two adiabatic
•The processes can be executed in closed or
steady-flow devices

Process 1 → 2
Reversible isothermal heat addition at TH to the
working fluid in a piston-cylinder device that does
some boundary work.

Process 2 → 3
Reversible adiabatic expansion during which the
system does work as the working fluid temperature
decreases from TH to TL
 The Carnot Cycle
Process 3 → 4
The system is brought into contact with a heat
reservoir at TL < TH and a reversible isothermal
heat exchange takes place while work of
compression is done on the system
Process 4 → 1
Reversible adiabatic compression during which
the temperature of the working fluid increases
from TL to TH

Sadi Carnot
 The Carnot Cycle

P-V Diagram of the Carnot Cycle

• Area under 1 → 2 → 3: Work
done by the gas during the
expansion part of the cycle
• Area under 3 → 4 → 1: Work
done on the gas during the
compression part of the cycle
• Area enclosed by cycle path:
Represents the net work done
during the cycle
 The Carnot Cycle
Carnot Refrigeration Cycle
• A reversed Carnot heat engine
cycle
• All the processes involved are
reversed
• The directions of any heat and
work interactions are reversed
• Heat in the amount of QL is
absorbed from the low-
temperature reservoir
• Heat in the amount of QH is
rejected to the high-temperature
reservoir
• A work input of Wnet, in is
required to operate the cycle
 6-8 The Carnot Principles
Carnot Principles
1. The efficiency of an
irreversible heat engine is
always less than the
efficiency of a reversible one
operating between the same
two reservoirs
 th   th, rev

2. The efficiencies of all

reversible heat engines
operating between the same
two reservoirs are the same
 The Thermodynamic Temperature Scale
Thermodynamic Temperature Scale
• A temperature scale that is independent of the properties of the
substances that are used to measure temperature

Development of the Thermodynamic Temperature Scale

• The second Carnot principle states that all reversible heat engines
have the same thermal efficiency when operating between the same
two reservoirs
• It follows that the efficiency of a reversible engine is independent of
the properties of the working fluid, the manner in which the cycle is
executed, and the type of reversible engine used
• This implies that the thermal efficiency of a reversible heat engine is
a function of the reservoir temperatures only
QH
 th, rev  g T H , T L  or  f TH , TL 
QL
The Thermodynamic Temperature Scale

For the proof refer the text book

 QH  TH
  
 QL  rev TL

This temperature scale is

called the KELVIN SCALE,
and the temperatures on this
scale are called ABSOLUTE
TEMPERATURES.
 The Carnot Heat Engine
Carnot Heat Engine
• The thermal efficiency of any heat engine is given by
QL
 th  1 
QH
• For a reversible heat engine, the heat transfer ratio can be replaced
by the ratio of the absolute temperatures of the two reservoirs
TL
 th, rev  1
TH
• This relation is often referred to as the Carnot efficiency
• This quantity provides the highest efficiency a heat engine operating between
two thermal energy reservoirs at temperatures TL and TH can have
  th, rev irreversib le heat engine

 th   th, rev reversible heat engine
 
 th, rev impossible heat engine
 The Carnot Refrigerator and Heat Pump

Carnot Refrigerator and Heat Pump

• The coefficient of performance of any refrigerator or heat pump is
given by
1 1
COPR  COPHP 
QH QL  1 1  QL QH
• For a reversible refrigerator or heat pump
1 1
COPR, rev  COPHP, rev 
TH TL  1 1  TL TH
• These are the highest coefficients of performance that a
refrigerator or a heat pump operating between the temperature
limits TL and TH can have
 6-11 The Carnot Refrigerator and Heat Pump
Carnot Refrigerator and Heat Pump (cont.)

 COPx , rev irreversib le R or HP


COPx  COPx , rev reversible R or HP
 COP
 x , rev impossible R or HP
Conceptual Example: Energy Quality

You need a new water heater, & you’re trying to decide between gas & electric.
The gas heater is 85% efficient, meaning 85% of the fuel energy goes into heating water.
The electric heater is essentially 100% efficient.
Thermodynamically, which heater makes the most sense?

Ans.
Only 1/3 of fuel energy is converted to electricity at a power plant.
With this in mind, the gas heater is a better choice.
EXAMPLE
(Example 2 from Lecture 1)
A residential heat pump is used to provide heating during the
winter season to maintain a house temperature at 21oC. On a
typical day, the heat transfer to the house is 75 MJ/h when
the outside air temperature is –4oC. The heat pump has a COP
of 3.7. Determine the power required for the heat pump and
the heat transfer rate from the outside air.

If the heat pump in the previous example is replaced with a

Carnot heat pump, determine the maximum efficiency of the
cycle and the power required for the Carnot heat pump.
Discuss your answers.
SOLUTION
Coefficient of Performance

TH
COPrev 
TH  TC
294
COPrev   11.76
294  269

Power input

 Q H
Wrev  COP  3.7
COPrev
75 M J / h W cycle  20.27 MJ / h
W rev   6.378 M J / h
11.76
 EXAMPLES

• An inventor claims to have developed a power cycle capable of

delivering a net work output of 410 kJ for an energy input by
heat transfer of 1000 kJ. The source is at 500 K and the sink 300
K. Evaluate this claim.

• A Carnot heat engine operates between a source at 1000 K and a

sink at 300 K. If the heat engine is supplied with heat at a rate of
800 kJ/min, determine (a) thermal efficiency (b) power output of
this engine.

• A refrigerator is to remove heat from the cooled space at a rate

of 300 kJ/min to maintain its temperature at -8oC. If the
surrounding air is at 25oC determine the minimum power input
required for this refrigerator.
Final July 07
A Carnot heat engine receives heat from a reservoir at
900oC at a rate of 800 kJ/min and rejects the waste heat to
the ambient air at 27oC. The entire work output of the heat
engine is used to drive a refrigerator that removes heat from
the refrigerated space at -5oC and transfer it to the same
ambient air at 27oC. Determine:
a. The maximum rate of heat removal from the
refrigerated space, in kJ/min.
b. The total rate of heat rejection to the ambient air, in
kJ/min.

QL= 4986 kJ/min, Qtotal = 5786 kJ/min

Final Jan 05

A Carnot power cycle using air as a working fluid has a

thermal efficiency of 40 percent. At the beginning of the
isothermal expansion, the pressure is 620 kPa, and the
specific volume of air is 0.1 m3/kg. The heat input to the
cycle is 50 kJ/kg. Assume the air behaves as ideal gas.
a. Sketch the P-v diagram of the cycle and describe the
FOUR (4) processes of the cycle.
b. Determine the highest and lowest temperatures for
the cycle.

TH = 216.03 K, TL= 129.62 K

SUMMARY
• Corollaries of the second law stipulate that the efficiency
of a reversible cycle is greater than the efficiency of any
irreversible cycle operating between the same two thermal
reservoirs.

• The thermal efficiencies of all reversible cycles operating

between the same thermal reservoirs are the same.

• These corollaries lead to the definition of the Kelvin

temperature scale, used to obtained the maximum
performance of heat engines, refrigerators and heat
pumps.

• The Carnot cycle is an ideal cycle, composed of a series of

internally reversible processes, which have efficiencies
equal to the Carnot efficiency.
 EXERCISE 1

A heat engine is claimed to produce 200 kJ of work while

transferring 125 kJ of heat to the heat sink as shown in
FIGURE Q4. Judge whether the claim is valid based on
the evaluation of the heat engine thermal efficiency.

FIGURE Q4
EXERCISE 2

During winter, a house losses heat to the

surroundings at a rate of 20 kJ/s when the outdoor
temperature is -4oC.
• Calculate the power required for a heat pump to
maintain the temperature inside the house at 27oC.

• If the outdoor temperature increases by 5oC,

evaluate the effect on the COP and efficiency of the
heat pump.

(2.067 kW; 1.733 kW; COP increases)

 Chapter 7
ENTROPY
Second Law of
Thermodynamics-Entropy

1
CONTENTS

7-1 Entropy
7-2 The Increase of Entropy Principle
7-3 Entropy Change of Pure Substances
7-4 Isentropic Processes
7-5 Property Diagrams
7-7 The T-ds relations
7-8 Entropy Change of Liquids and Solids
7-9 The Entropy Change of Ideal Gases
7-10 Reversible Steady-Flow Work
7-11 Isentropic Efficiencies of Steady-Flow Devices
7-12 Entropy Balance
LESSON OBJECTIVES

At the end of this lesson, you should be able to:

 Define entropy, entropy change and entropy generation

 Differentiate between reversible and irreversible process

 Describe the Increase of Entropy principle

 Evaluate whether a process is irreversible, reversible or

impossible
Entropy

 Entropy measures the spontaneous dispersal of energy:

how much energy is spread out in a process, or
how widely spread out it becomes — at a specific
temperature. (Sometimes, it’s a simple equation, Entropy
change = “energy dispersed”/T, or qreversible/T , the quantitative
measure of the amount of thermal energy not available to do
work." So it's a negative kind of quantity, the opposite of
available energy.

A quantitative measure of the microscopic disorder for a system.

Definition: Theoretical measure of thermal energy that cannot be

transformed into mechanical work in a thermodynamic system.
 ENTROPY
 Clausius inequality states that for any thermodynamic cycle:.

 Q 
  T b  0
where Q = heat transfer at the system boundary
T = absolute temperature at the boundary (K)
 = integration to be performed over the entire cycle Rudolf
Clausius

 This equation is valid for ALL cycles – reversible or irreversible

 Q 
 It was found that:
  T  int  0
rev
 A quantity whose cyclic integral is zero, depends ONLY on the
states, not the process path A PROPERTY
Defining Entropy Change
 Therefore:
 Q  where S is called entropy (kJ/K)
   dS
 T int
rev

 Integrating the above equation between initial and final states:

2  Q 
S  S 2  S1     int Entropy change of a system
1
 T  rev
 Entropy change can be positive or negative, depending on the direction of
heat transfer
Q +ve S +ve

Q -ve S -ve
 Defining Entropy Change (cont’d)
 The change in entropy is given by:
N/B:
 Q  Integral of Q/T give value of dS IF the
dS    integration is carried out along an internally
 T  int,rev reversible path between 2 states

 Q 

2
S 2  S1   
 1 T  int,rev

 Entropy is a property

 Therefore entropy change S between 2

specific states is the same whether the
process is reversible or irreversible.

 Units of Entropy is kJ/K, specific entropy

kJ/kg.K
 A Special Case: Internally Reversible Isothermal
Heat Transfer Processes

 Isothermal process ---> internally reversible process

 Q  2 Q 
 S     int     int    Q  int
2 1 2
1
 T  rev 1  To  rev To 1 rev

Q
S 7.6
To

 Eq. 7.6 is useful in determining the entropy changes of thermal energy

reservoirs that can absorb or supply heat indefinitely at constant
temperature
Example
For a particular power plant, the heat added and rejected both
occur at constant temperature and no other processes
experience any heat transfer. The heat is added in the amount
of 3150 kJ at 440oC and is rejected in the amount of 1950 kJ at
20oC. Is the Clausius inequality satisfied and is the cycle
reversible or irreversible?
Calculate the net work, cycle efficiency, and Carnot efficiency based on TH and TL for
this cycle.
Wnet  Qin  Qout  (3150  1950) kJ  1200 kJ
Wnet 1200 kJ
th    0.381 or 38.1%
Qin 3150 kJ
TL (20  273) K
th, Carnot  1   1  0.589 or 58.9%
TH (440  273) K
The Clausius inequality is satisfied. Since the inequality is less than zero, the cycle has
at least one irreversible process and the cycle is irreversible.
 The Increase of Entropy Principle
 Consider a cycle made up of 2 processes
 Q 
  T b  0
From the Clausius Inequality:

 Q   Q 
or 2 1

1  T  2  T   0

b int,rev

The second integral can be expressed as

 1 Q 
 2   S1  S 2
 T  int,rev
 Q 
2
S 2  S1     Q
7.7 OR dS  7.8

1
T b T

Note: T is the temperature (in K) at boundary where Q is transferred between system

and boundary
The Increase of Entropy Principle (cont’d)
From Eq. 7.8 Q
dS 
T

If Q Process 1-2 is internally reversible

dS 
T
Q
dS  Process 1-2 is irreversible
T

 Entropy change for an irreversible process is greater than that for internally
reversible.
 Therefore, some entropy must have been generated during the irreversible
process - due to the presence of irreversibilities
 This is called entropy generation, Sgen
 It is a measure of the magnitudes of irreversibilities present during the process
 The greater the Sgen, the greater is the extent of irreversibilities
 Used to establish criteria for the performance of engineering devices
 The Increase of Entropy Principle (cont’d)
 Re-write Eq. 7.7, to include entropy generation

 Q   Q 
2 2

S 2  S1     S 2  S1      S gen 7.9
1
1
T  T b
Always positive
or zero
Entropy transfer
with heat
Sgen depends on process:
• Positive for Irreversible
• Zero for Reversible
Q
For isolated system (or simply adiabatic closed system) Q0  0
T
Sisol  S 2  S1  0
 ISOLATED SYSTEM
 May consists of any no. of subsystems
 A system and its surroundings can be viewed as 2 subsystems of
an isolated system

Sisol  Ssyst  Ssurr  0

 As there are no Q or W by an isolated system

 it is simply adiabatic closed system
Q
Q0  0
T
S gen  Sisol  S2  S1  0 Eq. 6.34

Since Sgen can never be negative, entropy of an isolated system will

increases until it reaches a maximum value (equilibrium state).
(no actual process is reversible, therefore entropy will not be constant)
 The Increase of Entropy Principle (cont’d)

S gen  Sisol  S2  S1  0

The Increase of Entropy Principle

 The entropy of an isolated system during a process always increases
or, in the limiting case of a reversible process, remains constant –
never decreases
Implication of the principle:
• Entropy of a system can decrease
• Entropy change of a system can be negative
• Entropy generation, Sgen can not be negative

0 Irreversible Process (possible)

S gen 0 Reversible Process (Ideal)

0 Impossible Process
ENTROPY

Chapter 7

1
CONTENTS

7-1 Entropy
7-2 The Increase of Entropy Principle
7-3 Entropy Change of Pure Substances
7-4 Isentropic Processes
7-5 Property Diagrams
7-7 The T-ds relations
7-8 Entropy Change of Liquids and Solids
7-9 The Entropy Change of Ideal Gases
7-10 Reversible Steady-Flow Work
7-11 Isentropic Efficiencies of Steady-Flow Devices
7-12 Entropy Balance
LESSON OBJECTIVES
At the end of this lesson, you should be able to:

 Find entropy data for water and R-134a using tables

 Describe isentropic processes

Retrieving Entropy Data
 Entropy Data for Water & Refrigerant (Table)
 Tabulated in the same way in the Tables of Thermodynamics
 Superheated, Compressed, Saturation states (quality)

 Using Graphical Entropy Data

 Locate states and plot processes on diagrams having entropy as
one of the coordinates

 Using T-ds equations (liquids, ideal gases)

Entropy Change for Pure Substances
Tabulated in the same way in the Tables of Thermodynamics

Entropy change of a specified mass,

m (of a closed system) is:

 S  m  s  m  s 2  s1 
 Example 2

 A rigid tank contains 5 kg of R-134a initially at 20oC and 140 kPa.

The refrigerant is now cooled while being stirred until its pressure
drops to 100 kPa. Determine the entropy change of the
refrigerant during this process.
(x2 = 0.8587, s2 = 0.8275 kJ/kgK, ∆S = -1.1744 kJ/K)
Property Diagrams
 Two diagrams that are commonly used
 Temperature-entropy (T-s)
 enthalpy-entropy (h – s)

 The diagrams are important in visualizing the second law aspect of

thermodynamics.

 Recall Eq. 7.4  Q   Q in t rev  TdS

dS   
 T  int,rev

 Total area under the T-s curve is the total heat transfer
during an internally reversible process.

 The area has no meaning for an irreversible process.

Temperature – entropy diagram (T-s)

 Constant volume lines are steeper

than constant Pressure lines

 Constant P lines almost coincides

with saturated liquid line in the
compressed liquid region
Enthalpy – entropy (h – s) diagram (Mollier Diagram)

 Valuable in the analysis of steady-

flow devices
 h : indicate work
 s : indicate irreversibilities
ISENTROPIC PROCESSES
 Entropy of a fixed mass can be changed by
 Heat transfer
 Irreversibilities
 Thus, for internally reversible and adiabatic process:
entropy is constant ISENTROPIC process

s  0
or

s 2  s1
ISENTROPIC PROCESSES (cont’d)
 Many engineering systems are adiabatic and performs best when
irreversibilities are minimized.
 Isentropic process can serve as an appropriate model for actual
processes – define efficiencies to compare actual with idealized
conditions

NOTE
Internally reversible and adiabatic process means isentropic
process.
 T-ds Equations
 Previously, we find entropy change for any substance using:
 Q 

2
S 2  S1    Or from tables for water & refrigerant
 1 T  int,rev

 Need to find relationship between T and Q.

 Consider a closed system containing a simple compressible
substance:
Energy balance:  Q in t,rev  dU   W in t,rev
For simple compressible fluid:  Q in t,rev  TdS and
 W in t,rev  pdV
Substituting into the energy balance equation:

T dS  dU  pdV ( k J ) 1st T-ds equation

T-ds Equations (cont’d)
Second T-dS equation is derived from dH  dU  d  pV 
dH  dU  pdV  V dp
From 1st T-ds equation:
dH  V dp  dU  pdV T dS  dU  pdV ( k J )

T dS  dH  V dp 2nd T-ds equation

Writing both T-dS equations on a unit mass basis:

T ds  du  pdv the equations are also valid for irreversible

processes because they involve only properties and
so are path-independent.
T ds  dh  v dp
 Example – Application of Tds relations
 Consider steam undergoing a phase transition from liquid to
vapor at a constant temperature of 20°C. Determine the entropy
change sfg=sg-sf using the Tds equations and compare the value
to that read directly from the thermodynamic table.
du P
ds   dv, change from liquid to vapor
T T
1 P
s fg  s g  s f  (u g  u f )  (v g  v f )
T T
From table A-4:
T=20°C, P = 2.338 kPa, vf = 0.001002(m3/kg), vg=57.79(m3/kg),
uf=83.9(kJ/kg), ug=2402.9(kJ/kg)
Substituting into the Tds relation:
sfg= (1/293)(2402.9-83.9) + (2.338/293)(57.79-0.001002) = 8.375(kJ/kg K)
ug - uf vg - vf
1/T P/T
Compares favorably with the tabulated value sfg=8.3715(kJ/kg K)
ENTROPY

Chapter 7

1
CONTENTS

7-1 Entropy
7-2 The Increase of Entropy Principle
7-3 Entropy Change of Pure Substances
7-4 Isentropic Processes
7-5 Property Diagrams
7-7 The T-ds relations
7-8 Entropy Change of Liquids and Solids
7-9 The Entropy Change of Ideal Gases
7-10 Reversible Steady-Flow Work
7-11 Isentropic Efficiencies of Steady-Flow Devices
7-12 Entropy Balance
LESSON OBJECTIVES
At the end of this lesson, you should be able to:

 To derive T-dS equation for an ideal gas.

 To apply the equations for analysis of entropy change of

an ideal gas.
T-ds Equations for Liquids & Solids
 Recall that the equations are:
Tds  du  pdv and Tds  dh  vdp
 Liquids & solids can be approximated as incompressible
substances.
 Therefore, the entropy change for liquids & solids is:
du c dT
ds  
T T
 From integration:
2
c (T ) dT T2
s2  s1    cavg ln (kJ/kg.K)
1
T T1
Note: According to the above equation, an isothermal process for a pure
incompressible substance is isentropic.
Example – Entropy Change of Incomp. Substances
A 1-kg metal bar, initially at 1000 K, is removed from an oven and
quenched by immersing in a closed tank containing 20 kg of water,
initially at 300 K. Assume both substances are incompressible and
Cwater= 4(kJ/kg K), Cmetal = 0.4(kJ/kg K). Neglect heat transfer between
the tank and its surroundings.
(a) Determine the final temperature of the metal bar, (b) entropy
generation during the process.
Tm=1000 K, mm=1kg, cm=0.4 kJ/kg K

Tw=300 K, mw=20 kg, cw=4 kJ/kg K

 Solution
(a) Energy balance from the first law:
U  Q-W  0, no heat transfer and no work done
U water  U metal bar  0, both bar and water reach final temperature Tf
mwcw (T f  Tw )  mm cm (Tm  T f )  0
mw (cw / cm )Tw  mmTm (20)(10)(300)  (1)(1000)
Tf    303.5( K )
mw (cw / cm )  mm (20)(10)  1
(b) No heat transfer with the outside Q  0, the entropy
balance of the system s  s(generation)  s g
Tf Tf
s g  s(water)  s(bar)  mw cw ln  mm cm ln
Tw Tm
303.5 303.5
s g  (20)(4) ln  (1)(0.4) ln  0.928  0.477  0.451(kJ / K )
300 1000
The total entropy of the system increases, thus satisfy the second law
Ideal Gas T-ds equations
 Recall that
Tds  du  pdv and Tds  dh  vdp

du p dh v
ds   dv ds   dp
T T T T
 For an ideal gas:
du  cv dT , dh  c p dT , pv  RT
 Substituting into the above equations, yields:

dT dv dT dp
ds  cv R ds  c p R
T v T p
ENTROPY CHANGE OF IDEAL GAS
Previously, we obtain the T-ds equations for ideal gases as:

dT dv dT dp
ds  cv R ds  c p R
T v T p
The entropy change for a process is obtained by integrating:

2
dT v2
1

s2  s1  cv T 
T
 R ln
v1
Eq. 7.31

2
dT p2

s2  s1  c p T 
1
T
 R ln
p1
Eq. 7.32
ENTROPY CHANGE OF AN IDEAL GAS –
Approximate Analysis

 Approximation by assuming constant specific heats for

ideal gases

 So, Eq. 7.31 and Eq. 7.32 can be written as:

T2 v2
s2  s1  cv ln  R ln Eq. 7.33
T1 v1

T2 p2
s2  s1  c p ln  R ln Eq. 7.34
T1 p1
ENTROPY CHANGE OF AN IDEAL GAS –
Exact Analysis
The value of specific entropy is evaluated with respect to the reference
conditions at 0 K and 1 atm. T Table A-17 to A-25
dT

s  c p T 
o

0
T

Therefore in terms of so:

2
s2  s1   c p T 
dT P2
 R ln
1
T P1
T2 T
dT 1 dT p2
s2  s1   c p   cp  R ln
0
T 0
T p1
p2 p2
s2  s1  s  s  R ln
o
2
o
1 s2  s1  s  s  Ru ln
o
2 1
o
p1 p1
(kJ/kg.K) (kJ/kmol.K)
Example
 Air is compressed from an initial state of 100 kPa and
300 K to 500 kPa and 360 K. Determine the entropy
change using constant cp=1.003 (kJ/kg K)

T2 P2
s2  s1  cP ln( )  R ln( ) if cPis constant
T1 P1
360 500
s2  s1  1.003ln  (0.287) ln  0.279(kJ / kg K )
300 100

• Negative entropy due to heat loss to the surroundings

 Example 3
Air undergoes a change of state from an initial temperature and
o o
pressure of 22 C and 100 kPa to a final state of 107 C and 100
kPa. Determine the entropy change of the gas by using;
(a) Constant specific heats
(b) Variable specific heats

Exercise
Carbon dioxide gas undergoes a change of state from an initial
o
temperature and pressure of 45 C and 190 kPa to a final state of
o
80 C and 375 kPa. Determine the entropy change of the gas by
using constant specific heats and variable specific heats
Isentropic Processes of Ideal Gases
Approximate Analysis
For two states with same entropy, equations 7.33 and 7.34
are reduced to:
T2 v2 T2 p2
0  cv ln  R ln 0  c p ln  R ln
T1 v1 T1 p1
Re-arrange the equations:
R / cv R /cp
 T2   v1   T2   p2 
         
 T1   v2   T1   p1 
Since R  c p  c , k  c p / c
k 1  k 1 k k
T2  v1  T2  p2  p2  v1 
        
T1  v2  T1  p1  p1  v2 
Eq. 7.42 Eq. 7.43 Eq. 7.44
Isentropic Processes of Ideal Gases
Exact Analysis

For two states with same entropy

p2
s2  s1  0  s2o  s1o  R ln
p1
To find entropy of state 2 when other properties are known,

p2  s o (T2 )  s o (T1 )  exp[ s o (T2 ) / R] Pr 2

 exp  o
 Eq. 7.48
p1  R  exp[ s (T1 ) / R] Pr1
If volume ratio is given instead of pressure ratio, use relative
volume, vr and relative pressure, Pr equations.

 2    P2  P Relative values for air

   r2    r2 are given in table A-17
  1  s const  r1  P1  s const Pr1 or G3 (formula book)
Eq. 7.50 Eq. 7.49
ENTROPY

Chapter 7

1
CONTENTS

7-1 Entropy
7-2 The Increase of Entropy Principle
7-3 Entropy Change of Pure Substances
7-4 Isentropic Processes
7-5 Property Diagrams
7-7 The T-ds relations
7-8 Entropy Change of Liquids and Solids
7-9 The Entropy Change of Ideal Gases
7-10 Reversible Steady-Flow Work
7-11 Isentropic Efficiencies of Steady-Flow Devices
7-12 Entropy Balance
LESSON OBJECTIVES
At the end of this lesson, you should be able to:

 Determine isentropic efficiencies of engineering devices

Reversible Steady-Flow Work
 Recall relation between heat transfer and entropy for an
int. rev. process:
2
Qint 
rev

1
TdS , or
2
qrev   Tds
1

  dh  vdP
2

1
2
 h  h   vdP
2 1
1
Reversible Steady-Flow Work
 From CV energy balance,
q  w  h2  h1  12 V22  V12   g ( z2  z1 )
or
w  q  h1  h2  12 V12  V22   g ( z1  z2 )
 for a reversible process,
2
wrev  h2  h1   vdP  h1  h2 
1
1
2 V
1
2
 V22   g ( z1  z2 )

   vdP  12 V12  V22   g ( z1  z2 )

2

1
Reversible Steady-Flow Work

1
2
wrev    vdP  1
2 V
1
2
 V22   g ( z1  z2 )
 For turbines, compressors, and pumps with negligible
KE, PE effects:
2
wrev    vdP
1

 The larger the specific volume, the greater the work produced
(or consumed) by the steady-flow device

 Desirable for work-producing devices

 Undesirable for work-consuming devices (so, should try to reduce
the specific volume)
Reversible Steady-Flow Work
1) Pumps - fluid is incompressible (i.e., liquid) so v
= constant:
wrev  v ( P2  P1 )

2) Ideal gas (Pv = RT) compressor:

2  RT 
wrev     dP
1
 P 
ISENTROPIC EFFICIENCIES

 Isentropic process involves no irreversibilities and serves as

the ideal process for adiabatic devices

 Irreversibilities will downgrade the performance of the device

(actual situation)

 The more closely the device approximates the idealized

isentropic process, the better the device performs.

 Parameter that quantify how efficiently a device approximates

an idealized one is called ISENTROPIC or ADIABATIC
efficiency
ISENTROPIC EFFICIENCIES –
Adiabatic Turbine
Isentropic efficiency of a turbine is:
 the ratio of the actual work output of the turbine to the
work output that would be achieved if the process
were isentropic

actual turbine work

T 
isentropic turbine work
wa h1  h2 a
 
ws h1  h2 s

Eq. 7.61
ISENTROPIC EFFICIENCIES –
Pumps / Compressors
Isentropic efficiency of a compressor is:
 the ratio of the work input required in an isentropic
manner to the actual work input

isentropic compressor work

C 
actual compressor work
ws h2 s  h1
 
wa h2 a  h1
Eq. 7.63

ws  P2  P1 
P  
wa h2 a  h1 Eq. 7.64
Example 7-15

Air enters a compressor and is compressed adiabatically from 0.1 MPa, 27oC, to a
final state of 0.5 MPa. Find the work done on the air for a compressor isentropic
efficiency of 80 percent. Assume ideal gas and constant specific heat Cp=1.005
kJ/kg.K
System: The compressor control volume

Property Relation: Ideal gas equations, assume constant properties.

2a
T 2s P2

P1
1

s
Using the ideal gas assumption with constant specific heats, the isentropic work per
unit mass flow is
wcs  h2 s  h1  C p (T2 s  T1 )

The isentropic temperature at state 2 is found from the isentropic relation

The compressor isentropic

efficiency is defined as

The conservation of energy becomes ws

C 
wa
wCs  C p (T2 s  T1 ) wcs
wCa 
 1005
.
kJ
(475.4  300) K C
kg  K kJ
kJ 176
 176.0 kg kJ
kg
  220
0.8 kg
Example 7-14
Steam enters an adiabatic turbine steadily at 3 MPa and 400oC
and leaves at 50 kPa and 100oC. If the power output of the
turbine is 2 MW, determine

a. Isentropic efficiency of the turbine

b. The mass flow rate of the steam flowing through the turbine
Example 7-15
Air is compressed by an adiabatic compressor from 100 kPa and
12oC to a pressure of 800 kPa at a steady rate of 0.2 kg/s. If
the isentropic efficiency of the compressor is 80 percent,
determine the exit temperature of air.