Basic Principles and Definition (Thermodynamics 1)

LESSON 1
BASIC PRINCIPLES,
CONCEPTS AND DEFINITION OF THERMODYNAMICS

OBJECTIVES
The study of this lesson will enable the students to:
1. define thermodynamics and other terms necessary in the study of thermodynamics;
2. know the approaches in the study of thermodynamics;
3. state the difference between dimension and unit;
4. explain the different thermodynamics system;
5. differentiate homogenous system to a non homogenous system;
6. know the different thermodynamic properties and state of the system;
7. explain the different thermodynamic processes and cycles.

INTRODUCTION The lessons discussed deals with the basic principles, concepts and definition of thermodynamics. Other topics such as approaches in the study of thermodynamics are given emphasis, the difference between dimension and unit, the different types of thermodynamic system, and thermodynamic properties are also introduced. Topics regarding the different thermodynamic processes and cycle are also included.
DEFINITION
Thermodynamics is the study of energy and its transformation. It is one of the most fascinating branches of science. Thermodynamics discusses the relationship between heat, work and the physical properties of working substance. It also deals with equilibrium and feasibility of a process. The science of thermodynamics is base on observations of common experience which have been formulated into laws which govern the principle of energy conversion. The word thermodynamics derives from the two Greek words “therme” which means “heat: and “dynamikos” which means “power”.
Approaches in the Study of Thermodynamics
1. Microscopic or statistical approach The knowledge of structure of matter is considered and a large number of variables are needed to describe the state of matter. The matter is composed of several molecules and behavior of each individual molecule is studied.. Each molecule is having certain position, velocity and energy at a given instant. The velocity and energy change very frequently due to collision of molecules.
2. Macroscopic approach In macroscopic approach the structure of matter is not considered, in fact it is simple, and only few variables are used to describe the state of matter. In this approach, a certain quantity of matter composed of large number of molecules is considered without the events occurring at the molecular level being taken into account. In this case, the properties of a particular mass of substance, such as its temperature, pressure, and volume are analyzed. Generally, in engineering, this analysis is used for study of heat engines and other devices. This method gives the fundamental knowledge for the analysis of a wide variety of engineering problems.
DIMENSIONS AND UNITS Dimensions implies physical quantities. Examples are length, time, mass, force, volume and velocity. In engineering analysis, it is most important to check the dimensional homogeneity of an equation relating physical quantities. This means that the dimensions of terms on one side of the equation must equal those on the other side. Primary dimensions implies units of physical quantities conceived of and used to measure other physical quantities related by definition and laws Secondary dimensions implies other physical quantities measured using primary dimensions
Whereas a dimension is a name, a unit is a definite standard or measure of a dimension For example, foot, meters and angstroms are all different units with the common dimension of length. A unit is any specified amount of a quantity by comparison with which any other quantity of the same kind is measure. In any dimensional system, the units of length, time, mass and forces are related through Newton’s second law of motion which states that the total force acting on a body is proportional to the product of the mass and the acceleration in the direction of the force, thus,
F  ma ; F = 1/k ma where k is the proportionality constant.
UNITS OF DIFFERENT DIMENSIONAL SYSTEMS
Name of system
Unit of Mass
Unit of Length
Unit of time
Unit of force
K in
F = 1/k ma
Definition of terms
SI
(mks) kgm M
Sec
N

1.0 N is the force needed to accelerate a mass of 1.0 kg at 1.0 m/s2
1.0 kgf is the force needed to accelerate a mass of 1.0 kgm at 9.8066 m/s2
English Engineering lbm Ft
Sec
lbf
32.174
1.0 lbf is the force needed to accelerate a mass of 1.0 lbm at 32.174 ft/sec2
Absolute Engineering
Slug
Ft
Sec
lbf
1.0
1.0 lbf is the force needed to accelerate a mass of 1.0 slug mass at 1.0 ft/sec2
Absolute metric (cgs) gm Cm
Sec
dyne
1.0

1.0 dyne is the force needed to accelerate a mass of 1.0 g at 1.0 cm/sec2
1.0 gf is the force needed to accelerate a mass 1.0 gm at 980.66 cm/s2
International System Institutes(SI)
(Systeme Internationale d’Unites)
1. Base Unit
Specified once a set of primary dimensions is adopted quantity unit symbol mass kilogram kg length meter m time second s

2. Derived Unit Also termed secondary unit, these are units derived from the base units: given new names, normally after a famous scientist.
Selected SI-derived units
Newton, N 1 N = 1 kg-m/s2
Pascal, Pa 1 Pa = 1 N/m2
Joule, J 1 J = 1 N-m
Watt, W 1 W = 1 J/s
SI unit prefixes factor prefix symbol factor prefix symbol
1012 Tera T 10-2 centi c
109 Giga G 10-3 milli m
106 Mega M 10-6 micro 
103 kilo k 10-9 nano n
102 hecto h 10-12 pico p
101 deca da 10-15 femto f
10-1 deci d 10-18 atto a
English Engineering Units
1. base unit quantity unit symbol mass pound-mass lbm force foot ft time second s
2. derived unit quantity unit symbol force pound-force lbf pressure pound-force per square inch psi

Sample Problems:
Solve the following problems:
1. At standard conditions, what is the weight of 75-kgm man?
Given: m = 75 kgm g = 9.8066 m/s2

2. A box of apple weights 150 lbs what is its mass at standard condition?
Given: W = 150 lbf g = 32.174 ft/s2

Exercises:
1. How many centimeters are there in 2 miles?
2. A tank contains 1000 gallons of water, how many cubic meters of water is in the tank?
3. Express your height, mass, and weight ( g = 9.70 ,m/s2) in terms of SI units (m, kg, N).

THERMODYNAMIC SYSTEM
In thermodynamics, a system is defined as any collection of matter or any region in space bounded by a closed surface or wall. The wall may be real one, like the tank enclosing a certain amount of fluid. The wall may also be imaginary, like the boundary of a certain amount of fluid flowing along a pipe. All other systems outside the wall that interact with the system in question are known as the surroundings. The boundary distinguishes the system from its surroundings, may be at rest or in motion, usually defined by a broken line or dashed line.

Consider a piston and cylinder as arranged in above figure. The gas temperature in the cylinder can be raised by external heating and this causes the piston to move and also changes the system boundary size. It means, both heat and work crosses boundary of the system during the process, but the matter that comprises the system can always be identified.
Classification of thermodynamic system
1. Closed system- a system of fixed mass. In this system, energy may cross the boundary, and the total mass within the boundary is fixed. The system and it’s boundary may contract or expand in volume

Consider gas contained in a cylinder as shown above, the addition of heat to cylinder will raise the gas temperature and causes the piston to move. This changes the system boundary. This means, both heat and work crosses the boundary of system. But the original mass of the working substance (gas) remains unchanged.

2. Open System- one in which matter crosses the boundary of the system. There may be energy transfer also, i.e, both energy and mass crosses the boundary of the system. Most engineering devices belong to this type.

Note: If the inflow of mass is equal to the outflow of mass, then the mass in the system is constant and the system is known as steady flow.

In an air compressor, air enters at low pressure and leaves at high pressure. The working substance (gas) crosses the boundary of the system. In addition to this mass transfer, heat and work interactions take place across the system boundary.

3. Isolated System - one in which neither mass nor energy crosses the system boundary. It is of fixed mass and energy. The system is not affected by the surrounding, i.e. there is no interaction between the system and surroundings.

HOMOGENEOUS AND HETEROGENEOUS SYSTEM If the substance within the system exists in a single phase like air, steam, liquids then the system is called HOMOGENEOUS SYSTEM. In these systems, the substance exists in only one phase. If the substance within the system exists in more than one phase, then the system is HETEROGENEOUS.
THERMODYNAMIC PROPERTIES The distinguishing characteristics of a system by which its physical condition may be described are called the properties of the system. They are quantities that we must specify to give a macroscopic description of the system. Many of such quantities are familiar to us in other branches of science, such as mass, energy, pressure, volume, density, electric field, magnetic field and magnetization of matter. Two other properties -temperature and entropy- are unique to thermodynamics. Together with energy, they play a most important role in the structure of thermodynamics. A property is either a directly observable or an indirectly observable characteristic of a system. Any combination of such characteristics is also a property.
Two types of classical thermodynamic properties
1. Intensive Property
These properties are independent of mass such as pressure, temperature, voltage and density.
2. Extensive Property
These properties are dependent of mass and are total values such as total volume, total energy and entropy. Examples of thermodynamic properties, besides pressure, volume and Temperature, are: internal energy, enthalpy, and entropy. Other properties include: velocity, acceleration, moment of inertia, electric charge, conductivity (thermal and electrical), electromotive force, stress, viscosity, reflectivity, number of protons, and so on.
Definition of properties
1. Mass – amount or absolute quantity of matter in a certain body.
2. Volume – the space occupied by a certain body
3. Density
a) Mass density – the mass of substance divided by the volume the mass occupies of simply, the mass per unit volume. Water at standard conditions, ρwater = 1 gm/cm3 = 1000 kg/m3 = 62.4 lbm/ft3
Note: 9.8 N = 1 kg
b) Weight density – also known as the specific weight, it is defined as the weight per unit volume. Where : g = 9.80665 m/s2 = 32.174 ft/s2 Water at standard conditions, water = 9.81 kN/m3 = 62.4 lb/ft3
4. Specific volume - defined as the volume per unit of mass of the reciprocal of density
5. Weight, W – force exerted by gravity on a given mass, depends on both the mass of the substance and the gravitational field strength, W = mg
6. Specific gravity – the dimensionless parameter, it is defined as the ratio of the density (or specific weight) of a substance to some standard density(specific weight). Solids and liquids are referred to water as standard while gases are usually referred to air or hydrogen as standard.

For solid substances

For liquid substances For gaseous substances where: ρair at STP = 1.2 kg/m3 at 1 atm, 21.1 oC
7. Pressure – defined as the normal force exerted by a system on a unit area of its boundary.
Standard reference atmospheric pressure
1 atm = 14.7 psia
= 760 mm Hg
= 29.92 in Hg at 32oF
= 760 torrs
= 101.325 kpa
= 34 ft H20
= 1.033 kg/cm2
Types of pressure
a. Gage pressure, Pg – pressure of a fluid and the atmospheric pressure, measured using manometer or bourdon gage
Note:
-Vacuum pressure is negative pressure;
-measured using fluid pressure < atmospheric pressure
b. Atmospheric pressure, Patm
- measured using a barometer, refer to standard atmospheric cited above
c. Absolute pressure, Pabs – sum of the gage pressure and atmospheric pressure.

Relationships among the types of pressure For Pabs > Patm Pabs = Patm + Pg For Pabs < Patm Pabs = Patm – Pv
Also, Pg = h = ρgh
Measuring Pressure
Manometer is commonly used to measure small or moderate pressure differences. A manometer mainly consists of a glass or plastic U-tube containing one or more fluids such as mercury, alcohol, water or all.
8. Temperature – it is an intensive property, originates with our sense perceptions, rooted in the notion of hotness or coldness of a body
Types
a. Arbitrary, t – man made calibrated
a. 1 Celsius scale, oC (used to be Centigrade scale). Named after Anders Celsius, a Swedish
- steam point - equilibrium temperature of pure liquid water in contact with its vapor at one atmosphere; 100oC
- ice point – equilibrium temperature of ice and air-saturated liquid pressure at a pressure of one atmosphere. 0oC
a. 2 Fahrenheit scale ,oF Named after Gabriel Fahrenheit, a German who devised the first mercury-in-glass thermometer; earlier thermometer fluids used were alcohol and linseed oil
- steam point : 212oF - ice point : 32oF
b. Absolute, T – measured from absolute zero, all molecular motion cease. 0Kelvin or 0Rankine b. 1 Kelvin Scale, K
Named after William Thomson, aka Lord Kelvin who related absolute scale to the Celsius scale. The ice point is assigned with a value of 273.15 K and the steam point is assigned with the value 373.15 K. The triple point of water is 273.16 K.
b. 2 Rankine Scale, R
Named after William Macquorn Rankine, a Scotish. The ice point is assigned with a value of 491.67 R and the steam point is assigned with the value 671.67 R The triple point of water is 491.69 R.
Derive the relation between degrees Fahrenheit and degrees Celsius. (EE board exam question)

THERMODYNAMIC STATE OF A SYSTEM The state of a system at any instant is it’s condition or configuration of existence at that instant. The various properties of a system define the state of the system. At any equilibrium condition, the state of the system can be described by few properties like pressure, temperature, specific volume, internal energy, etc. The state of a system can be represented by a point on the diagram whose co-ordinates are thermodynamic properties. When a system changes from one equilibrium state (state 1) to another (state 2), due to energy interaction, the system attains a new stat e which is shown by point “2” on the property diagram.

Example: Consider a given mass of water, which may become vapor or solid (ice) by heating or cooling. Each phase of water may exist at different pressures and temperatures or we can say water may exist in different states.
ZEROTH LAW This law states that when two bodies, isolated from other environment are in thermal equilibrium with a third body, the two are in thermal equilibrium with each other. This statement means that if a “hot” body interacts with a “cold” body and the two bodies are isolated from their surroundings, the properties of the bodies (say, the temperature, volume, conductivity, and so on) will change. However after a time, the various properties cease to change. When properties changes cease, the bodies are said to be in thermal equilibrium.
THERMODYNAMIC PROCESS If any one or more properties of a system change, the system is said to have undergone a process; there has been a change of state. If a change of state occurs during which the pressure does not change, the working substance is said to undergo constant pressure (isobaric) process. If the volume of a particular mass remains constant, but other properties change, the process is called a constant volume (isometric) process.

Thermodynamic processes that are commonly experienced in engineering practice:
1. Constant pressure/ Isobaric process
2. Constant volume/ Isochoric process
3. Constant temperature/ Isothermal process
4. Reversible adiabatic/ Isentropic process
5. Polytropic process
6. Throttling process/ Iso-enthalpic process
REVERSIBLE AND IRREVERSIBLE PROCESS
Reversible Process A reversible process for a system is an ideal process which once having taken place can be reversed in such a way that the initial state and all energies transformed during the process can be completely regained in both systems and surroundings. This process does not leave any net change in the system or in the surroundings. A reversible process is always, quasi-static.
Example: Motion without friction, expansion of compression with no pressure difference, heat transfer without temperature gradient.
Irreversible Process If the initial state and energies transformed cannot be restored without net change in the system after the process has taken place, it is called irreversible.
Consider the process 1-2 expansion of a gas cylinder. Let w12 be the amount of work done and Q12 the quantity of heat transferred between the system and surrounding. If it is possible to change the system from state 2 to state 1 by supplying back w12 and Q12, then the process 1-2 is called reversible process. If there is any change in the requirement of work and heat to bring back the system from 2 -1, then the process 1-2 is called “irreversible process”.
Quasi-Static Process If a process takes place at a faster rate, then the intermediate conditions cannot be defined. Therefore, an assumption is made such that the process is taking place at such a rate that the intermediate conditions be defined and hence must be represented on the thermodynamic property diagram. A quasi-static process is also known as quasi-equilibrium process in which the process carried out in such a manner that, at every instant the system departs only infinitesimally from an equilibrium state. It is an ideal process in which the system changes very slowly its state, under the influence of infinitesimal pressure or temperature difference. Quasi means “almost”. Infinite slowness is the characteristic feature of this process. It is also a reversible process.
THERMODYNAMIC CYCLE When a certain mass of fluid in a particular state passes through a series of processes and returns to its state, it undergoes a cycle. The processes may be ones that have names, such as constant volume and constant pressure, or they may be in series of state changes without names. Cycles are studied after a detailed study if processes.

LAW OF CONSERVATION OF MASS The law of conservation of mass states that mass is indestructible. In many systems involving thermodynamic analysis, it is quite accurate enough to assume that each point of any cross section where the fluid flows, the properties are the same and use average velocity normal to the section and assumed to be the same at each point. Thus, if the density is the same in all points of the cross section of area A, the mass rate of flow is m = ρA where m – mass ρ – density  – velocity
Accepting the statement of the law of conservation of mass, it follows with respect to any system, that the verbal form of the law is

A steady flow system is an open system in which there is no change of stored mass; having an equation called the continuity equation of steady flow; m1 = m2 = m m = ρ11A1 = ρ22A2

GENERAL METHODOLOGY FOR PROBLEM SOLVING IN ENGINEERING THERMODYNAMICS The fundamentals of engineering thermodynamics are applicable to a wide variety of problems. In order that we may maximize the results and rewards of our effort, it is appropriate in applying these fundamentals that we develop a routine which we can follow systematically. Such a general procedure for problem solving is suggested by Huang, (1988) given below:
1. Read the problem carefully
2. Since a sketch almost always aids in visualization, draw a simple diagram of all the components of the system involved. This could be a pump, a heat exchanger, gas inside a tank, or an entire power plant.
3. Select the system whose behavior we want to study by properly and clearly locating the boundary of the system. Do we have an isolated system, a closed system, or an open system.
4. Make use of the appropriate thermodynamic diagrams to locate the state points, and possibly the path of the process. These diagrams are extremely helpful as visual aids in our analysis.
5. Show all interactions (work, heat, and mass) across the boundary of the selected system.
6. Extract from the statement of the problem the unique features of the process and list them. Is the process isothermal, constant pressure, constant volume, adiabatic, isentropic, or constant enthalpy?
7. List all the assumptions that one might need to solve the problem. Are we neglecting a change of kinetic energy and change of potential energy?
8. Apply first law equation appropriate to the system we have selected.
9. Apply the principle of mass conservation appropriate to the system that we have selected.
10. Apply the second law equation appropriate to the system we have selected
11. Apply the appropriate property relations. That is, bring in data from tables, charts, or appropriate property equations.
12. Try to work with general equations as long as possible before substituting in numbers
13. Watch out for units. For example, when we use h = u +pv, h, u, and pv must all have the same units.
14. Make sure that absolute temperature, in degrees Rankine or Kelvin, is used in calculation.

Problem Solving:
1) What is the pressure 8,000 ft (2000 m) below the surface of the ocean? Neglect the compressibility factor, in SI units. Specific gravity of sea water is 1.03 (ME Board Exam Oct. 1998)Ans. 20.21 Mpa
Solution:
Given: h = 2000 m S.G.sea water = 1.03 g = 9.80665 m/s2

2) What is the temperature at which water freezes using the Kelvin scale? (ME Board Exam Oct. 1998) Ans. 273
Solution
The freezing temperature of water is at 0oC
K =0oC +273
K = 273
3) The barometer reads 29.0 inches of mercury. What is the absolute pressure in SI if the vacuum gage reads 9.5 psi. (ME Board Exam Oct. 1997)Ans. 32.2 KPa
Solution:
Pabs = Patm – Pv

Pabs= 98. 21kPa- 65.48 kPa = 32.73 kPa

4) What is the atmospheric pressure in the moon if the pressure is 100 kPa and the gage pressure is 10 kPa? (ME Board Exam April 1997) And. 90 kPa
Solution:
Given: Pabs = 100 kPa Pgage = 10 Kpa
Pabs = Patm + Pgage
Patm = Pabs – Pgage
Patm = 100 kPa - 10 kPa
Patm = 90 kPa

Exercises:
1. What is the equivalent oR of 400 oK? Ans. 720.6
2. If the temperature inside the furnace is 700 oK, what is the corresponding reading in oF? Ans. 800.6
3. The specific gravity of mercury relative to water is 13.55. What is the specific weight of mercury. The specific weight of water is 62.4 lb per ft3. Ans. 132.9 kN/m3
4. An iron block weights 5 N and has a volume of 200 cm3. What is the density of the block? Ans. 2548.42 kg/m3
5. A batch of concrete consisted of 200 lbs fine aggregate, 350 lbs coarse aggregate, 94 lbs cement, and 5 gallons water. The specific gravity of sand and gravel may be taken as 2.65 and that of the cement as 3.10. What is the weight of concrete in place per cubic foot? (ME Board Exam Oct. 1997) Ans. 153.058 lb/ft3
6. Two liquids of different densities (ρ1 = 1500 kg/m3, ρ2 = 500 kg/m3), are poured together into a 100 liter tank, filling it. If the resulting density of the mixture is 800 kg/m3, find the respective amounts of liquids used. Also find the weight of the mixture, local g = 9.675 m/s2. Ans. m1 = 45 kg
7. Convert a) 122oF to oC and to K. b) -40 oC to oF and to oR, c) 942oR to oC and to K, d) 373 K to oF to oR.
8. A fluid moves in a steady-flow manner between two sections in the same flowline. At section (1): A1 = 0.10m2, vel1 = 6 m/sec, spec. vol1 = 0.33 m3/kg at section (2) A2 = 0.2m2, ρ2 =b 0.27 kg/m3. Calculate for the velocity of flow of section (2). Ans. 33.67 m/s
9. A barometer located at the ground floor registered 102 kPa. It was then transferred on top of the high-rise building where it reads 97 kPa. Assuming that the average atmospheric air density is 1.0 kgm/m3, estimate the height of the building, in meters. Ans. 509.68 m
10. If a pump discharges 75 gal/min of water, find the total time required to fill a vertical cylindrical tank 10 feet in diameter and 10 ft height. Ans. 78.31 min
11. A fluid with a vapor pressure of 0.2 Pa and a specific gravity of 12 is used in a barometer. If the fluids column is 1 m, what is the atmospheric pressure? (ME Board Exam Oct. 1997)Ans. 117.72 kPa

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