MECE312 Thermodynamics I
Introduction and Review
Today’s Objectives:
• Administrative matters
• The syllabus and course introduction
Administrative Matters
• 3 credit module
• Class Hours: Wed 1200-1350, Block 122, Room
001-002
• Instructor: Dr. L. Kelebopile
• Laboratory work: Tues 1400-1650(Mech) Thurs
1200-1450(Chem)
• Tutorial sessions:
(Mon 0900-0950) (Chem), Block 108/044
(Mon 10-1050) (Mech), Block 124 Room 003-004
Submission of Laboratory Work
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Within 2 weeks
Latest 15-20min before next lab starts
Late submissions will not be marked
Students will work in groups
Lab manuals will be distributed in time before
the experiments
Homework, Quizzes and Tutorials
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•
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A set of practice questions will be given every week
The questions will not be scored
Students must solve the problems independently
Unscheduled quizzes will be given during lessons
Assignments will sometimes also be given
The assignments & quizzes will be graded (they
contribute 5%)
• NO make up quizzes will be given, regardless of the
circumstances
Assessment
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•
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2 Tests 20%
Assignments & Quizzes (5%)
Laboratory work (15%)
Final Exam (60%)
Use of the Blackboard
• Course material such as Course Outlines,
Homework, Lecture Notes, Laboratory Manuals,
and others will be posted on the Blackboard
• Solutions to practice questions will not be posted
as the questions will be discussed during
tutorials
Reference textbook
• Moran J. M. & Shapiro N. H. Fundamentals of
Engineering Thermodynamics 8th Edition,
John Wiley & Sons, Inc., 2014
• Cengel A. Y. & Boles A.M. Thermodynamics, An
Engineering Approach 8th Edition,
McGrawHill., 2015
Announcements
• There will be no laboratory session / tutorial this
week
• Lecture Notes, Practice Questions, Course
Outline will be posted on the Blackboard before
the end of the week
Thermodynamics
From the Greek:
- therme, meaning “heat’’
- dynamis, meaning “power’’
ie. The power of heat or capacity of hot bodies to
produce work.
Today scope is broader and deals generally with
energy and relationships among properties of
matter.
Course Objective
• To introduce the fundamental concepts of
engineering thermodynamics, and the
techniques and methods applied to solving realworld problems
• Thermodynamics I mainly focuses on the
principles while Thermodynamics II emphasizes
application
Upon Completion, be able to:
• Explain the concepts of properties and states of a
substance, energy, entropy, processes and cycles;
and the laws of thermodynamics.
• Calculate pdV work for a closed system undergoing
quasi-equilibrium process.
• Determine the thermodynamic properties for an
ideal gas or pure substance.
• Apply the first law of thermodynamics to any
closed system or control volume.
• Apply the second law of thermodynamics to the
analysis of engineering systems or control volume.
Reading notes and structure
• The thermodynamic concepts are not very
difficult to understand.
• But the application sometimes involves abstract
concepts and calculations.
• Approach shall therefore be such that we
introduce a concept and take some examples.
• Importance of doing your homework!
Macroscopic Approach
• Also called classical thermodynamics – macroscopic
approach
• A large number of particles (molecules) make up the
substance in question
• The macroscopic approach does not require knowledge
of behaviour of individual particles
• No model of the structure of matter at the molecular,
atomic, subatomic level.
• Deals with relationships between the observable
macroscopic phenomena that we see directly
• Provides an easier and direct way to solving problems
• A more elaborate approach is the statistical
thermodynamics – microscopic approach
Today’s Objectives:
• Some definitions (a system, a process, properties,
phase, pure substance, equilibrium)
• Review units of thermodynamics related
quantities
• Review concepts of temperature, temperature
scales, absolute and gage pressure
A basic concept: System
• A System is whatever it is that we want to study.
• In thermodynamics, the first step in defining any
problem is to define exactly what is to be
monitored, examined, measured, etc.
Surrounding
Interactions
System
(all matter and
space external to
system)
Boundary(choice
governed by
convenience/individ
ual preference)
Surrounding and boundaries
• System
▫ May be a simple free body or complex such as
thermal power plant
• Boundary
▫ May be at rest or in motion
▫ May be real or imaginary
• Surrounding (external to the system)
▫ A system can interact with its surrounding by
transferring energy/matter across its boundary
System Definition
Radiation
The Earth
Radiation
System Definition
Gas
Gas cylinder
Q
Gas
Piston cylinder assembly
Gas
Q
W
System Definition
Types of Systems
• Isolated Systems – matter and energy may not
cross the boundary.
(Relevance? A combination of systems
interacting with each other can be surrounded
by another boundary to define an isolated
system).
• Closed Systems – matter may not cross the
boundary.
• Open Systems – heat, work, and matter may cross
the boundary
(more often called a Control Volume (CV)).
Closed System
• Matter (m) may not cross the boundary.
• Heat (Q) and Work (W) may cross the boundary
which change the energy (ΔE) of the system.
Po, Vo, To
Combustion
CxHy+A O2 → B CO2 + C H2O
Pf, Vf, Tf
Q
∆V →W
Control Volume / Open System
• Heat (Q), work (W), and matter (m) may cross
the boundary
Gas
Q
System Properties
• A property is a macroscopic characteristic of a system (m,
T, V, P, E, ρ, v, u, h, s)
• The state (condition) of a system is defined by its
properties
• Extensive properties – They scale with system mass (vary
directly with size or extent of the system).
• Intensive properties – Do not vary even if the mass is
increased. (are independent of size)
Give some examples of:
Extensive Properties
Intensive properties
Process vs. Cycle.
• A Process is when properties of the system
undergo a change. A process moves from one
“state” to another “state”.
• Consider a flow through an open system or
control volume, if none of its properties change
with time then system is said to be Steady State
• If a Process undergo changes that eventually
bring it back to the original state, these
transitions together are called a Cycle.
Process:
Cycle:
P
P
S1
S2
S2
S1
S4
S3
v
v
Phase and Pure Substance
• Phase – refers to a quantity of matter that is
homogeneous throughout in both chemical
composition and physical structure.
• Homogeneity in physical structure means matter
is all solid, all liquid or all gas
• A pure substance is one uniform in chemical
composition. (even if in more than one phase)
Phase - examples
• A system can contain one or more phases eg. a
system of liquid water & steam.
• Gases can be mixed in different proportions to
form a single phase.
• For liquids,
▫ mixture of alcohol-water – single liquid phase.
▫ Immiscible liquids like oil-water form two liquid
phases.
Equilibrium
•Equilibrium is a state of balance maintained by an equality
of opposing forces.
•A system is said to be in thermodynamic equilibrium if it
satisfies all the relevant types of equilibriums namely,
mechanical,
thermal, chemical and phase (mass of
phase(s) does not change).
•To test equilibrium state, isolate the system from its
surroundings and see if the properties do not change.
•At thermodynamic equilibrium, there is no energy
interaction (in form of work/heat).
•The temperature is uniform throughout the system.
•Pressure must be uniform if effect of gravity is not
significant (variation of pressure with elevation neglected)
Quasiequilibrium Processes
• Some intervening states during a process might be
non-equilibrium (eg. variation of the intensive
properties with space and time)
• A quasiequilibrium process is one in which
departure from equilibrium is at most infinitesimal.
• ‘Quasi’ means ‘almost’
• All states through which a system passes in a
quasiequilibrium process may be considered
equilibrium states.
• The quasiequilibrium process is idealized type
process
Quasiequilibrium Process Illustration
• If weights are removed one
at a time, - quasiequilibrium
(gas has sufficient time to
redistribute)
If all are removed, the piston
will move upward by allowing
sudden expansion of the gas.
Molecules far from piston
cannot cope with time lag thus
creating low pressure region
near the piston.
Incremental masses removed
during an expansion of the
gas or liquid
Gas or
liquid
Fig.2.6 Moran & Shapiro
Boundary
33
V
V
V
Non-quasi-static Process
Consider a box initially divided in half.
- Initially, one is filled with gas, the other a vacuum.
- The divider is then removed.
- The gas takes some time to fill the new volume.
During that time, there are different local values for P
in the volume. There is also likely some heat
generated, as the process is irreversible.
34
V
V
V
V
V
V
V
V
Quasi-static (quasi equilibrium)
Now we move the wall slowly, such that the gas is able to
adjust instantly. This is a reversible quasi-static process.
A quasi equilibrium process can be viewed as a sufficiently slow process
that allows the system to adjust itself internally so that properties
in one part of the system do not change any faster than those of other
parts.
Non-equilibrium process - example
The process of changing volume is NOT necessarily in
equilibrium.
- Helium balloon popping, gas does not instantly mix
with air
- Gas cylinder rupture, pressure inside is higher than
outside for some time, t
Unit Review: Weight and Mass
Weight
Mass
U.S. Customary
pound , lbf
pound of mass, lbm
Metric
Newton, N
kilogram, kg
W =m g
where
g = 9.81 m/s2
g = 32.2 ft/s2
1 N = 0.2248 lbf
1 lbf = 4.4482 N
1 kg = 2.205 lbm
1 lbm = 0.4536 kg
Relationships:
Conversions:
Unit review: Density and Specific Volume
Density
ρ
mass basis
U.S.
Customary
[lbm / ft3 ]
molar basis
mass basis
Specific Volume
v or v
[ ft3 /lbm]
[ ft3 /mol]
[kg/m3]
[m3/kg]
Metric
molar basis
Relationships
m = mass
V= Volume
M = molecular weight
n = number of moles

volume per mole
[m3/mol]
𝜌=
n=
𝑚
𝑉
m
M
=
1

 =M
Unit review: Pressure
Pressure =
U.S. Customary
Pressure head
[psi ] or [lbf /in2 ]
[psf] or [lbf /ft2 ]
[in of Hg ]
[ in of H20 ]
[ft of H20]
Metric
[N/m2]
[ Pa]
[bar]
[mm of Hg ]
[ mm of H20 ]
[cm of H20]
Other
atm
Relationships:
1 atm = 14.69 psi = 1.01325 bar
= 100 324 Pa = 760 mm of Hg
= 29.92 in of Hg = 33.96 ft of H2O
1 Pa = 1N/m2
1kPa = 103N/m2
1bar = 105N/m2
1MPa=106N/m2
Absolute, Gage and Vacuum pressure
• Absolute pressure is the actual pressure
• Pressure measuring devices often indicate
difference between absolute pressure in a system
and absolute pressure of atmosphere. This
difference is called gage pressure
pabs = patm + pgage
Or vacuum pressure if atmospheric>system
pressure
pabs = patm - pvac
Measuring temperature
Zeroth law of thermodynamics states that
when two systems are separately in thermal
equilibrium with a third one, then they are in
thermal equilibrium with each other as well.
Temperature is a physical property that tells us
when two objects are in thermal equilibrium
Temperature is measured by thermocouples,
thermometers based on liquid expansion,
electrical resistance or infrared sensing
Unit review: Temperature
Metric
U.S.
Relative
Temperature
Celcius
[ oC ]
Fahrenheit
[ o F]
Absolute
Temperature
Kelvin
[K]
Rankine
[ oR ]
Relationships
K = oC + 273.15
oR = oF + 459.67
oC =( oF - 32)/1.8
oF= 1.8 oC + 32
oR = 1.8 K
Unit Conversion
•
32.174𝑙𝑏𝑚.𝑓𝑡/𝑠2
1𝑙𝑏𝑓
1𝑘𝑔.𝑚/𝑠2
1𝑁
•
1𝑊
1𝐽/𝑠
1𝑘𝑃𝑎
1000𝑁/𝑚2
•
0.3048𝑚
1𝑓𝑡
1𝑘𝐽
1000𝑁.𝑚
1𝑚𝑖𝑛
60𝑠
1𝑙𝑏𝑚
0.45359𝑘𝑔
1𝐵𝑡𝑢
778𝑓𝑡.𝑙𝑏𝑓
Example 1: An object whose mass is 16 kg is subjected to an
applied upward force of 66.7 N. The only other force acting
on the object is the force of gravity. Determine the net
acceleration of the object in m/s2 assuming the acceleration
of gravity is constant (g = 9.81 m/s2). Is the net acceleration
up or down?
FAF= 66.7 N
mass= 16 kg
Fg= ?
Example 1: An object whose mass is 16 kg is subjected to an
applied upward force of 66.7N. The only other force acting
on the object is the force of gravity. Determine the net
acceleration of the object in m/s2 assuming the acceleration
of gravity is constant (g = 9.81 m/s2). Is the net acceleration
up or down?
F = ma
FAF= 66.7 N
m= 16 kg
Fg= ?
FAF + ( − Fg ) = ma
FAF − Fg
FAF − mg FAF
a=
=
=
−g
m
m
m
2
 66.7  N kg.m / s
2
a=
−
9.81
m
s

16
N

 kg
a = −5.64 m s
2
Example 2: A system consists of N2 in a piston-cylinder
assembly, initially at P1 = 140 kPa, and occupying a volume
of 0.068 m3. The N2 is compressed to P2 = 690 kPa and a
final volume of 0.041 m3. During the process, the
relationship between P and V is linear. Determine the P in
kPa at an intermediate state where the volume is 0.057 m3
and sketch the process on a graph of P vs V.
P1= 140 kPa
V1 =0.068 m3
P2= 690 kPa
V2 =0.041 m3
Example 2: A system consists of N2 in a piston-cylinder
assembly, initially at P1 = 140 kPa, and occupying a volume
of 0.068 m3. The N2 is compressed to P2 = 690 kPa and a
final volume of 0.041 m3. During the process, the
relationship between P and V is linear. Determine the P in
kPa at an intermediate state where the volume is 0.057 m3
and sketch the process on a graph of P vs V.
Since the relationship is linear (y=mx+b)
 P − P   690 − 140 
3
m= 2 1=
=
−
20370.3
kPa
/
m

V
−
V
0.041
−
0.068


 2 1
P − P1 = m (V − V1 ) → P = m (V − V1 ) + P1
P = −20370.3
kPa
m3
( 0.057 − 0.068) m3 + 140kPa = 364kPa
Example 3: A manometer is attached to a tank of gas in which the pressure is
104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3.
If g = 9.81 m/s2 and the atmospheric pressure is 101.33 kPa, calculate
(a) The difference in mercury levels in the manometer, in cm.
(b) The gauge pressure of the gas in kPa and bar
(c) The absolute pressure of the gas kPa, atm, and psi
Example 3: A manometer is attached to a tank of gas in which the pressure is
104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3.
If g = 9.81 m/s2 and the atmospheric pressure is 101.33 kPa, calculate
(a) The difference in mercury levels in the manometer, in cm.
(b) The gauge pressure of the gas in kPa and bar
(c) The absolute pressure of the gas atm, and psi
Solution: a)
P = Patm + gL → L =
P − Patm
g
(104.0 − 101.33) kPa
103 N / m 2 103 g 1 kg  m / s 2
(1m)3
L=
(13.59b)
g / cm3 )(9.81m / s 2 ) 1 kPa
1 kg
1N
(100cm)3
L = 0.02m
b)
c)
𝑃𝑔𝑎𝑔𝑒 = 𝑃 − 𝑃𝑎𝑡𝑚 = 104.0 − 101.33 = 2.67𝑘𝑃𝑎
=0.0267bar
1 atm
Pabs = 104.0 kPa
= 1.026 atm
101.33 kPa
14.7 psi
= 1.026 atm
= 15.09 psi
1 atm
Quizz 1
State whether each of the following properties is
an intensive property or an extensive property.
Mass, temperature, density, concentration, volume
pressure,
molecular weight, number of moles
Extensive Properties
Intensive properties
Homework
Question 1
The pressure in a compressed air storage tank is
1500 kPa.
What is the tank’s pressure in:
(a) kN and m units.
(b) kg, m, and s units, and
(c) kg, km, and s units.
Question 2 The absolute pressure in water at a
depth of 5m is read to be 145 kPa .
Determine:
(a) The local atmospheric pressure, and
(b) The absolute pressure at a depth of 5m in a
liquid whose specific gravity is 0.85 at the same
location.