thermo lecture notes ch1ch-1-2
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Lecture NotesEngineering ThermodynamicsIntroductory Concepts and DefinitionsTRANSCRIPT
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-1
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Introductory Concepts and Definitions Thermodynamics Thermodynamics is the study of energy and its transformations, and its relationship to the
properties of matter. Although, it is difficult to give a precise definition for energy, it can be
viewed as the ability to cause changes.
Thermodynamics studies the behavior of how objects and systems behave as energy is
transferred between them, what happen to objects as energy is added or subtracted.
Areas of Application of Thermodynamics: All natural processes are governed by the principles of thermodynamics. However, the
following engineering devices are typically designed based on the principles of
thermodynamics.
Automotive engines, Turbines, Compressors, Pumps, Fossil and Nuclear Power Plants,
Propulsion systems for the Aircrafts, Refrigeration, Air-conditioning and Heating Devices.
The principles of thermodynamics are summarized in the form of a set of axioms. These
axioms are known as four thermodynamic laws:
The zeroth law, the first law, the second law and the third law.
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-2
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
The Zeroth Law deals with thermal equilibrium and provides a means for measuring
temperatures.
The First Law deals with the conservation of energy and introduces the concept of
internal energy.
The Second Law of thermodynamics provides with the guidelines on the conversion of
internal energy of matter into work. It also introduces the concept of entropy.
The Third Law of thermodynamics defines the absolute zero of entropy. The entropy of
a pure crystalline substance at absolute zero temperature is zero.
Thermodynamic system Thermodynamic system is defined as a quantity of matter or a region in space chosen for
study. The region outside the system is called the surroundings. The real or imaginary surface
that separates the system from its surrounding is called the boundary.
The boundary of a system can be fixed or movable, real or imaginary. The boundary is the
contact surface shared by both the system and the surroundings, and through which energy and
mass may enter or leave the system. A system together with its surroundings is said to
constitute a universe.
A control volume with
Moving and fixed boundaries
A control volume with
real and imaginary boundaries
Imaginary boundary Real boundary
System a nozzle
System
Fixed boundary
Moving boundary
SYSTEM
BOUNDARY
SURROUNDINGS
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-3
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Types of Systems Systems may be considered to be closed or open:
Closed system (also known as a control mass) is one in which the system mass cannot cross
the boundary, but energy (in the form of heat or work), can.
As a special case of closed system, even energy is not allowed to cross the boundary, that
system is called an isolated system.
Open system or a control volume, as it is often called, is one in which mass can cross the
system boundary as well as energy.
Macroscopic and Microscopic Approaches It is well-known that a substance consists of a large number of particles called molecules. The
properties of the substance naturally depend on the behavior of these particles. For example,
the pressure of a gas in a container is the result of momentum transfer between the molecules
YES
mass
energy
Closed system
m = constant
NO
YES
energy
mass
Open System
(CV)
YES
mass
energy
Isolated system
m = constant
NO
NO
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-4
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
and the walls of the container. However, one does not need to know the behavior of the gas
particles to determine the pressure in the container. It would be sufficient to attach a pressure
gage to the container. This macroscopic approach to the study of thermodynamics that based
on empirical laws to describe matter treated as a continuum, and does not require a knowledge
of the behavior of individual particles, is called classical thermodynamics. It provides a direct
and easy way to the solution of engineering problems. Another approach which concerns
directly with the structure of matter and characterize by statistical means the average behavior
of the molecules making up a system of interest, is called Microscopic approach , sometimes
called statistical thermodynamics.
Units: SI- units are used exclusively Fundamental units:
Mass kilograms kg
Length meter m
Time seconds s
Temperature Celsius / Kelvin oC / K
Derived units:
Force (F) Newton N
Pressure (P) Pascal Pa
Energy (E) Joule J
Newtons Law: Force = mass acceleration
F = m a
[N] = [kg] [m/s2] 1 N = 1 kgm/s2
P = F / A
[Pa] = [kgm/s2] [m2] 1 Pa = 1 kg/ms2
E = F x
[J] = [kgm/s2] [m] 1 J = 1 kgm2/s2
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-5
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Properties of a System Properties: are variables of a system to which a value can be assigned at given time without
knowledge of the history of the system.
Examples of properties: temperature, density, pressure, energy
Examples of non-properties: mass flow, heat, work
Not all properties are independent. Some are defined in terms of other ones. For example,
density is defined as mass per unit volume: Vm
= (kg/m3)
Sometimes the density of a substance is given relative to the density of a better known
substance. Then it is called specific gravity, or relative density (sg), and is defined as the ratio
of the density of a substance to the density of some standard substance at a specified
temperature (usually water at 4C, for which oH2 = 1000 kg/m3), That is:
OH2
sg
=
A more frequently used property in thermodynamics is the specific volume. It is the
reciprocal of density and is defined as the volume per unit mass:
==1
mV
Properties are considered to be either intensive or extensive.
Extensive properties are those whose values depend on the size or extent of the system, (i.e. the property is divided when the system is divided). Mass m, volume V, and total
energy E are some examples of extensive properties.
Intensive properties are those whose values are independent of the size or extent of the system, (i.e. the property doesnt change when the system is divided), such as
temperature, pressure, and density.
An easy way to determine whether a property is intensive or extensive is to divide the
system into two equal parts with a partition;
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-6
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Each part will have the same value of intensive properties as the original system, but half the
value of the extensive properties.
Extensive properties per unit mass are called specific properties. Some examples of specific properties are
o specific volume ( = V/m),
o specific total energy (e = E/m), and
o specific internal energy (u = U/m).
State and Equilibrium State: It is the condition of a system as described by the values
of its properties (P1,T1,.). The state of a simple compressible
system is completely specified by two independent, intensive
properties. A system is called a simple compressible system in
the absence of electrical, magnetic, gravitational, motion, and
surface tension effects. These effects are due to external force
fields and are negligible for most engineering problems.
Thermodynamics deals with equilibrium states. When the properties of the system show no
tendency to change, a state of thermodynamic equilibrium exists.
In an equilibrium state, there are no unbalanced potentials (or driving forces) within the
system.
A system that is in equilibrium experiences no changes when it is isolated from its
surroundings.
There are many types of equilibrium, and a system is not in thermodynamic equilibrium
unless the following condition satisfied: between the system and surroundings, there is no
difference in
1. temperature thermal equilibrium
2. pressure mechanical equilibrium
3. composition chemical equilibrium
m = 2 kg T2 = 20 C V2 = 2.5 m3
m = 2 kg T1 = 20 C V1 = 1.5 m3
(a) State 1 (b) State 2
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-7
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Processes and Cycles Process: It is any transformation of a system from one equilibrium state to another. The series
of states through which a system passes during a process is called the path of the process.
P- V diagram of a compression process of a gas
Cycle: It is a sequence of processes which return the system to its initial state.
Pressure Pressure (P): is defined as a normal force exerted by a fluid per unit area.
AFP = (Pa)
Units: 1 Pa = 1 N/m2
1 standard atmosphere = 101325 Pa
System
P
V
(1) (2)
1
2
V1 V2
Initial state
Final state
Process path
1
2
V
P
4
V
P
1
2
3
(a) A two-process cycle (b) A four-process cycle
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-8
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
1 bar = 105 Pa = 100 kPa = 0.1 MPa
The actual pressure at a given position is called the absolute pressure, and it is measured
relative to absolute vacuum (i.e., absolute zero pressure). Most pressure-measuring devices,
however, are calibrated to read zero in the atmosphere, and so they indicate the difference
between the absolute pressure and the local atmospheric pressure. This difference is called the
gage pressure. Pressures below atmospheric pressure are called vacuum pressures and are
measured by vacuum gages that indicate the difference between the atmospheric pressure and
the absolute pressure. Absolute, gage, and vacuum pressures are all positive quantities and are
related to each other by:
Pgage = Pabs - Patm > 0 (for pressures above Patm )
Pvac = Patm - Pabs < 0 (for pressures below Patm )
The pressure at a point in a fluid has the same magnitude in all
directions. The variation of pressure with elevation is given by
gdzdP
=
where the positive z direction is taken to be upward. When the
density of the fluid is constant, the pressure difference across a
fluid layer of thickness z is
P = P2 - P1 = gz
Patm
Patm
Pabs = 0
Patm
Pgage
Pabs Pvac
Pabs
2
z
P = Patm + gh
P = Patm 1
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-9
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Example The piston of the piston-cylinder devise containing a gas has a mass 60 kg and the
gravitational acceleration is 9.81(m.s-2). The cross-sectional area of the piston is 0.04 m2.
Determine the pressure inside the cylinder when the local atmospheric pressure is 0.98 bar.?
Solu.:
bar
AmgP
WPAPAF
patm
patmpp
117.11004.0
81.96098.0
P
0
5 =
+=
+=
+=
=
Temperature The concept of temperature is fundamental to thermodynamics. We know that a body at
high temperature will transfer energy to one at lower temperature. Consider two bodies with
different temperatures in contact with each other. Net energy transfer will be from the hotter
body to the colder body. At some point, the net energy transfer will be zero, and the bodies are
said to be in thermal equilibrium. Bodies in thermal equilibrium are defined to have the same
temperature. Temperature ( T),in units of degrees Celsius, oC, is a measure of hotness relative to
the freezing and boiling point of water. A thermometer is based on the thermal expansion of
mercury.
Gas
m = 60 kg
Ice Bath Boiling Water
0oC
100oC
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-10
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Microscopic point of view:
Temperature is a measure of the internal molecular motion, e.g., average molecule
kinetic energy, At a temperature of 273.15oC molecular motion ceases.
Temperature in units of degrees kelvin, K, is measured relative to this absolute zero
temperature, so
0 K = -273oC in general, T in K = T in oC + 273.13
Zeroth Law of Thermodynamics
If a system A is in thermal equilibrium with another system B and also with a third
system C, then all of the systems are in thermal equilibrium with each other. This is called the
zeroth law of thermodynamics. This is how a thermometer works. If a thermometer is placed in
a substance for temperature measurement, the thermometer's glass comes into thermal
equilibrium with the substance. The glass then comes into thermal equilibrium with the liquid
(mercury, alcohol, etc . . .) inside the thermometer. Because the substance is in thermal
equilibrium with the glass and the glass is in thermal equilibrium with the inner liquid, the
substance and liquid must be in thermal equilibrium by the zeroth law. And because they are
thermally equivalent, they must have the same temperature.
TA = TB
TA = TC
A B C
Zeroth law of thermodynamics: TA = TB = TC
Ice bath
0oC
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-11
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Energy Energy (E) is the capacity either latent or apparent to exert a force through a distance.
There are four primary forms of energy that are of interest in macroscopic thermodynamics:
a) Potential energy (P.E): is the energy due to elevation.
P.E = mgz (J)
b) Kinetic energy (K.E): is the energy of a moving mass.
K.E = 2mc2
1 (J)
c) Internal energy (U): is a measure of kinetic energy of the molecules and atoms that
make up the matter. It is different from the bulk kinetic energy associated with the movement of the entire solid. The internal energy of matter is exhibited by molecular
motion .The internal energy of matter is measured by its temperature. Hot water has
more internal energy than the same amount of cold water.
d) Enthalpy (H): is the amount of energy in a system capable of doing mechanical work.
it is the sum of the internal energy and the flow work (PV)
H = U + PV (J)
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-12
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
Tutorial # (1)
Problem 1
The level of the water in an enclosed water tank is (40 m) above ground level.
The pressure in the air space above the water is (120 kPa), and the density of the water
is (1000 kg/m3). What is the water pressure at a ground level?
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Problem 2
A gas is contained in a vertical frictionless piston-cylinder device. The piston
has a mass of (4 kg) and cross sectional area of (35 cm2). A compressed spring above
the piston exerts a force of (60 N). If the atmospheric pressure is (95 kPa), Determine
the pressure inside the cylinder.
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Problem 3
Two piston/cylinder arrangements, A and B, have their gas chambers connected
by pipe, see following figure. Cross-sectional areas are AA = 75 cm2 and AB = 25 cm2,
with the piston mass in A being mA = 25 kg. Assume outside pressure is 100 kPa and
standard gravitation. Find the mass mB so that none of the pistons have to rest on the
bottom.
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Course Tutor: Jasim M. Mahdi / University of Baghdad
1-13
Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)
H.W # (1)
Q1: Four cubic meters of water at 25 C and 1 bar have a mass of 3990 kg. (a) List the values of two extensive and three intensive properties of the system. (b) If the local gravity g for the system is 9.7 m/s2, evaluate the specific weight.
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Q2: The acceleration of gravity is given as a function of elevation above sea level by the relation g = 9.81 - 3.3210-6 h m/s2, with h measured in meters. What is the weight of an airplane at 10 km elevation when its weight at sea level is 40 kN.
Ans.: 39.9 kN ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Q3: The lower half of a 10-m-high cylindrical container is filled with water ( = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and bottom of the cylinder.
Ans.: 90.7 kPa ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Q4: A vertical, frictionless pistoncylinder device contains a gas at 250 kPa absolute pressure. The atmospheric pressure outside is 100 kPa, and the piston area is 30 cm2. Determine the mass of the piston. Ans.: 45.9 kg ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Q5: A valve in the cylinder pictured to the right has a cross-sectional area of 11 cm2 with a pressure of 735 kPa inside the cylinder and 99 kPa outside. How large a force is needed to open the valve?
Ans.: 699.6 N ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Q6: A vertical pistoncylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 5 kg and a diameter of 12 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will double the pressure of the gas inside the cylinder. Ans.: 95.7 kPa, 115.3 kg ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Q7: Two cylinders A and B are connected by a piston of two different diameters as shown in figure. Cylinder B contains a gas, while cylinder A contains oil that is pumped up to a pressure of 300 kPa by a hydraulic pump. The mass of the piston is 10 kg. Calculate the gas pressure in cylinder B. Ans.: 3.5 MPa
DB = 25 mm
Piston
Patm =100 kPa
DA = 100 mm
Pump
g
B
A