Thermo Lecture Notes CH1CH-1-2

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Course Tutor: Jasim M. Mahdi / University of Baghdad 1-1 Energy Engineering Department (Fundamentals of Thermodynamics/ 2 nd 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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Lecture NotesEngineering ThermodynamicsIntroductory Concepts and Definitions

Transcript of Thermo Lecture Notes CH1CH-1-2

  • 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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    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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    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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    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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    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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    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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    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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    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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    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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    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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    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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    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?

    --------------------------------------------------------------------------------------------------

    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.

    --------------------------------------------------------------------------------------------------

    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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    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.

    ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

    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