AMERICAN UNIVERSITY OF MADABAAMERICAN UNIVERSITY OF MADABA

Mechanical Engineering DepartmentMechanical Engineering DepartmentThermodynamics (1) 104220Thermodynamics (1) 104220

BYBYProf. Nasri RabadiProf. Nasri Rabadi

TextbookThermodynamics: An Engineering Approach,

by Yunus A. Cengel and Michael A. Boles,

References:1) Fundamentals of Engineering Thermodynamics,

Moran, M. J., and H. N. Shapiro.

2) Fundamentals of Thermodynamics,

Sonntag, R. E., Borgnakke, C., and Van Wylen, G. J.

Chapter 1Chapter 1

Introduction and Basic Introduction and Basic ConceptsConcepts

Objectives

■ Identify the unique vocabulary associated with thermodynamics through the precise definition of basic concepts to form a sound foundation for the development of the principles of thermodynamics.

■ Review the metric SI and the English unit systems that will be used throughout the text.

■ Explain the basic concepts of thermodynamics such as system, state, state postulate, equilibrium, process, and cycle.

■ Review concepts of temperature, temperature scales, pressure, and absolute and gage pressure.

■ Introduce an intuitive systematic problem-solving technique.

Introduction

■ Thermodynamics can be defined as the science of energy and entropy

■ An alternate definition: is the science that deals with heat and work and those properties of substances that bear a relation to heat and work

■ The word thermodynamics stems from the Greek words therme (heat) and dunamies (force).

■ Thermodynamics is a science that is based on experimental findings

■ Those findings have been formalized into certain laws

Introduction

■ Energy cannot be created or destroyed it transforms (conservation of energy 1st law)

Introduction

■ Energy has a quality and a quantity (actual process is decreasing quality of energy 2nd law)

Application Areas of Thermodynamics

Macroscopic and Microscopic Views of Thermodynamics

■ Any substance consists of a large number of molecules. The properties of the substance depend on the behavior of these molecules.

■ Consider a gas in a container. The pressure of the gas is a result of the momentum transfer as the molecules hit the walls of the container.

■ However, we do need to know the force exerted by the molecule on an infinitesimal area on the wall in order to find the pressure (microscopic approach)

■ Instead, it will be sufficient to attach a pressure gauge to the wall (finite area) and read the average pressure exerted by a large number of molecules on that finite area (macroscopic approach).

■ That is, the macroscopic approach to thermodynamics is concerned with the average or overall behavior. This approach is called sometimes classical thermodynamics.

■ On the other hand, the microscopic approach to thermodynamics, known as statistical thermodynamics, is concerned directly with the structure of matter. It is objective is to find (by statistical means) the average behavior of the particles making up a system. This approach is involved and is not used any more in the remaining of this course.

■ The macroscopic approach provides a direct and easy solution to engineering problems.

Introduction

■ Thermodynamics is both a branch of physics and an engineering science.

■ The scientist is normally interested in gaining a fundamental understanding of the physical and chemical behavior of fixed quantities of matter and then, uses the principles of thermodynamics to relate the properties of matter.

■ Engineers are generally interested in studying systems and how they interact with their surroundings. To facilitate this, engineers extend the subject of thermodynamics to the study of systems through which matter flows.

Thermodynamic systems■ An important step in any engineering analysis

is to describe precisely what is being studied. ■ In mechanics, if the motion of a body is to be

determined (Figure 1), normally the first step is to define a free body and identify all the forces exerted on it by other bodies (Figure 2).

■ Newton’s second law of motion is then applied.

■ In thermodynamics, the term system is used to identify the subject of the analysis (e.g. coffee in the cup).

■ Once the system is defined and the relevant interactions with other systems are identified, one or more physical laws or relations are applied.

■ The system is whatever we want to study. It may be as simple as a free body or as complex as an entire chemical refinery. We may want to study a quantity of matter contained within a closed, rigid-walled tank or we may want to consider something such as a gas pipeline through which gas flows.

■ Everything external to the system is considered to be part of the system’s surroundings.

■ The system is distinguished from its surroundings by a specified boundary which may be at rest or in motion.

■ It is essential for the boundary to be determined carefully before proceeding with any thermodynamic analysis.

■ Two basic kinds of systems are distinguished in thermodynamics study. These are referred to as closed systems and control volumes.

Closed systems (control mass)■ A closed system refers to a fixed quantity of matter.

■ A closed system is used when a particular quantity of matter is under study. ■ A closed system always contains the same matter. ■ There can be no transfer of mass across its boundary. ■ What do we call the system if even energy is not

allowed to cross the boundary?■ The figure shows a gas in a piston-cylinder assembly. ■ Let us consider the gas to be a closed system. ■ The boundary lies just inside the piston and cylinder walls,

as shown by the dashed lines on the figure. ■ If the cylinder were placed over a flame, the gas would

expand, raising the piston. ■ The portion of the boundary between the gas and the piston

moves with the piston. ■ No mass would cross this or any other part of the boundary.

Open system (control volume)

■ An open system (control volume) is a properly selected region in space.

■ It encloses a device that involves mass flow such as nozzle, compressor, turbine.

■ Flow through such devices is best studied by selecting the region within the device as the control volume .

■ Both mass and energy can cross the boundary of the control volume.

■ There are no concrete rules for the selection of the control volume but proper choice makes the analysis much easier .

■ The boundary of the control volume is called boundary surface■ The boundary surface can be real or imaginary■ A control volume can be fixed in shape and size or it may

involve a moving boundary.

Open Systems (continued)

Open Systems (continued)

Properties of a System■ To describe a system and predict its behavior

requires knowledge of its properties and how those properties are related.

■ Properties are macroscopic characteristics of a system.

■ Any characteristic of a system is called a property. Some familiar properties are pressure P, temperature T, volume V, and mass m.

■ Properties describe the state of a system only when the system is in an equilibrium state.

■ Not all properties are independent. Density is a dependent property on pressure and temperature.

Density as a property■ Density is mass per unit volume;■ = mass/volume (kg/m3))

■ Specific gravity: the ratio of the density of a substance to the density of some standard substance at specified temperature (usually water at 4 oC for liquids and air for gases).

■ Specific volume is volume per unit mass.■ = Volume/mass, (m3/kg)■ = 1/

P

Gases

T

Gases

Volume

P

Liquids

T

Liquids

Water

Extensive and Intensive Properties■ Intensive properties are those that are

independent of the size of system, such as temperature, pressure, and density.

■ Extensive properties are dependent on the size (or extent) of the system. Mass m, volume V, and total energy E are some examples of extensive properties.

■ Criteria to differentiate extensive and intensive properties is illustrated in the Figure.

■ Extensive properties per unite mass are called specific properties (i.e. specific volume).

State■ A state is defined as a condition of a substance

that can be described by certain observable macroscopic properties. (T, P, , etc.)

■ In above figure, the system does not undergo any change. All properties can be measured throughout the system. Hence the condition of the system is completely described. This condition is called state 1.

■ Now remove some weights. If the value of even one property changes, then the state will change to different one (state 2).

■ The word State refers to the condition of a system as it is described by its properties.

State 1

m = 2kgT1= 20 °CV1= 1.5 m3

m = 2kgT1= 20 °CV1= 2.5 m3

State 2

Equilibrium

■ Thermodynamics deals with equilibrium states.

■ The word equilibrium implies a state of balance.

■ Equilibrium state means that there are no unbalanced potentials (or driving forces) within the system.

■ A system is said to be in thermodynamic equilibrium if it maintains thermal, mechanical, phase, and chemical equilibrium.

Thermal Equilibrium

■ Thermal equilibrium means that there is no temperature differential through the system.

20 °C 30 °C

30 °C

35 °C 40 °C

No thermal equilibrium

32 °C

32 °C 32 °C

32 °C

32 °C

Thermal equilibrium

Mechanical Equilibrium

■ Mechanical equilibrium means that there is no change in pressure in the system.

(a) Slow compression (quasi-equilibrium)

20 pa 20 pa

20 pa

20 pa 20 pa

(b) fast compression (non quasi-equilibrium)

20 pa

20 pa

90 pa

Phase Equilibrium

■ Phase equilibrium means that the mass of each phase reaches an equilibrium level and stays there.

Water

Vapor, P > 0

After some time

Water

Vapor, P = Pv

After long time

Water

Vacuum t= 0, P = 0

At t = 0

Chemical Equilibrium

■ Chemical equilibrium means that the chemical composition of the system does not change with time

The State Postulate ■ We mentioned earlier that a state is described uniquely by

measuring a few of its properties. The remaining properties will assume certain values. The question here is how much is this “few”?.

■ The answer depends on how simple or complex our system is.

■ If we have a system where the gravitational, electrical, magnetic, motion and surface tension effects are absent, then this system is called a simple compressible system.

■ According to what is called “state postulate”, the number of properties required to completely specify the state of such system is two independent, intensive properties.

■ The state of any system in equilibrium is determined by any two independent properties.

The State Postulate

■ If, however, the gravitational effects are important in the simple compressible system, then the elevation z needs to be specified in addition to the two properties necessary to determine the state.

■ The state postulate requires that the two properties are independent of each other.

■ Two properties are considered to be independent if one property can vary while the other one is constant.

■ Temperature and specific volume are good examples for independent properties.

■ You will see, however, in coming units that temperature and pressure are not always independent of each other. They become dependent during phase change processes.

Processes and Cycles■ Any change from one equilibrium state to

another is called a process.■ Process diagrams are very useful in

visualizing the processes.■ The series of states through which a system

passes during a process is called a path■ To describe a process completely initial and

final states as well as the path it follows, and the interactions with the surrounding should be specified

■ A process with identical end states is called a cycle

■ Process diagrams plotted by employing thermodynamic properties as coordinates are very useful in visualizing the processes.

■ Isothermal process means a Isothermal process means a process at constant T.process at constant T.

■ Isobaric process means a Isobaric process means a process at constant pressureprocess at constant pressure

■ Isochoric process means a Isochoric process means a process at constant volume process at constant volume

Quasi-Equilibrium processQuasi-Equilibrium process■ During a quasi-static or quasi-equilibrium During a quasi-static or quasi-equilibrium

process, the system remains infinitesimally process, the system remains infinitesimally close to an equilibrium state at all times. close to an equilibrium state at all times.

■ A sufficiently slow process that allow the A sufficiently slow process that allow the system to adjust itself internally so that system to adjust itself internally so that properties in one part of the system do not properties in one part of the system do not change any faster than those at other parts.change any faster than those at other parts.

■ Compression is very slow and thus equilibrium Compression is very slow and thus equilibrium is attained at any intermediate state. Therefore, is attained at any intermediate state. Therefore, the intermediate states can be determined and the intermediate states can be determined and process path can be drawn.process path can be drawn.

■ It is an idealized process but many process It is an idealized process but many process closely approximate it with negligible error.closely approximate it with negligible error.

■ Quasi-Equilibrium, Work-Producing Devices Quasi-Equilibrium, Work-Producing Devices Deliver the Most Work (it is the standard to Deliver the Most Work (it is the standard to which other processes can be compared)which other processes can be compared) (a) Slow compression (quasi-equilibrium)

20 pa 20 pa

20 pa

20 pa 20 pa

State 1

State 2 Process pathP

V

Intermediate states

20

Non-Quasi-Equilibrium process

■ Compression process is fast and thus equilibrium can not be attained.

■ Intermediate states can not be determined and the process path can not be defined. Instead we represent it as dashed line.

(b) Fast compression (non quasi-equilibrium)

20 pa

20 pa

90 pa

State 1

State 2P

V

Non-equilibruim process

20

90

?

Temperature and the Zeroth Law of Thermodynamics

■ The zeroth law of thermodynamics states that: If two bodies are in thermal equilibrium with the third body, they are also in thermal equilibrium with each other.

■ The equality of temperature is the only requirement for thermal equilibrium.

Temperature scales

■ In thermodynamics it is desirable to have a temperature scale that is independent of the properties of any substance.

■ Note: it makes no difference to use K or C in formulas involving temperature difference. However, you should use Absolute temperature in formulas involving temperature only like the ideal gas low.

FTRT

CTKT

CTFT

KTRTFTRT

CTKT

o

o

oo

o

o

328.1

8.167.459

15.273

Dimensions and Units

The seven fundamental dimensions and their

units in SI (International System).

Dimensions and Units

SI British System Conversion

Length Meter (m) Foot (ft) 1 ft = 0.3048 m

Time Second (s) Second (s)

Mass

Kg

Slug Pound mass (lbm) 1 slug = 32.2 lbm

1 slug =14.59 kg 1 lbm = 0.4536 kg

Force

Newton (N) 1 N = (1Kg).(1 m/s2)

Pound force (lbf) 1 lbf = (1 slug)(1. ft/s2)

1 lbf = 4.448 N

Definition of

Unit force

Newton (N): is the force required to give a mass of 1 kg an acceleration of 1 m/ s2.

Pound force (lbf) is the force required to give a mass of 1 slug an acceleration of 1 ft/ s2.

°C = (5/9)*(°F –32)

°R = (9/5)*K

Tempe-rature

Degree Celsius.(°C) Absolute Temp.: Kelvin (K).

K = °C + 273.15

Degree Fahrenheit (°F) Absolute Temp.: Rankine (°R)

°R = °F + 459.67

°C = (5/9)*(°F –32)

°R = (9/5)*K

PressurePressure is defined as the normal force exerted by a fluid per unit area. Units in SI are Pa=N/m2. The pressure unit pascal is too small for pressure encountered in practice.

Therefore, kPa and MPa are commonly used.

Units in British are : psf = lbf/ft2, psi = lbf/in2

You have to convert from psi to psf ( 144 in2 = 1 ft2)

psibarskPaPaatmkPaMPaPabar

696.1401325.1325.101325,10111001.0101 5

Pressure (Continued)Pressure (Continued)

Absolute pressure, is measured relative to absolute vacuum (i.e., absolute zero pressure.)

Gauge pressure, is measured relative to atmospheric pressure

atmabsatmvac

atmatmabsgage

PPPP

PPPP

below pressurefor

above pressurefor

Pressure (continued)

Variation of Pressure with Depth

The pressure variation in a constant density fluid is given as

P + Z = constant Or P1+ Z1 = P2 + Z2

Z is the vertical coordinate ( positive upward). is the specific weight of fluids, (N/m3)

For small to moderate distances, the variation of pressure with height is negligible for gases because of their low density.

g

Pressure (continued)

Pressure at a Point

• The pressure at a point in a fluid has the same magnitude in all direction.

Pressure (continued)

Pressure Variation in horizontal planes

Pressure is constant in horizontal planes provided the fluid does not change. ( this leads to Pascal’s principle.)

. 1

2

1

2

2

2

1

121 A

AFF

AF

AFPP

Noting that P1 = P2, the area ratio A2/A1 is called the ideal mechanical advantage. Using a hydraulic car jack with A2/A1 = 10, a person can lift a 1000-kg car by applying a force just 100 kg (= 908 N).

The Manometer

A device based on P + Z = constant is called a manometer (Right), and it is commonly used to measure small and moderate pressure differences.

w

f

w

f

w

fs g

g

S

Specific gravityP2 = Patm + h

FIGURE 1–53FIGURE 1–53Schematic for Example 1–7..

1-17

EXAMPLE 1–7 Measuring Pressure with a Multifluid Manometer The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Fig.1–53. The tank is located on a mountain at an altitude of 1400 m where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if h1 = 0.1 m, h2 = 0.2 m, and h3 = 0.35 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. Solution

The pressure in a pressurized water tank is measured by a multi- fluid manometer. The air pressure in the tank is to be determined. Assumption : The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air–water interface.Properties All densities are given

Analysis Starting with the pressure at point 1 at the air–water interface, moving along the tube by adding or subtracting the ρgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives

P1 + ρ watergh1 + ρ oil gh2 - ρ mercurygh3 = PatmSolving for P1 and substituting,

P1 = Patm - ρ watergh1 - ρ oilgh2 + ρ mercurygh3= Patm + g ( ρmercury h3 - ρwater h1 - ρoil h2 )= 85.6 kPa + 9.8m/s2(13,600 kg/m3 * 0.2m–850kg/m3*0.35m-1000kg/m3 * 0.1 m)= 130 kPa

Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis considerably. Also note that mercury is a toxic fluid, and mercury manometers and thermometers are being replaced by ones with safer fluids because of the risk of exposure to mercury vapor during an accident.

Barometer and the Atmospheric Pressure

• The atmospheric pressure is measured by a device called a barometer; thus the atmospheric pressure is often referred to as the barometric pressure.

0

( )

B B C C

C vapor

B atm C B

Hg

P Z P ZP P

P P Z Zgh

Barometer and the Atmospheric Pressure (continued)

• The standard atmospheric pressure is the pressure produced by a column of mercury 760 mm in height at 0oC. The unit of mmHG is also called the torr in honor of Evangelista Torricelli (1608−1647).

• The atmospheric pressure at a location is simply the weight of the air above that location per surface area. Patm changes with elevation and weather conditions.

The length or the cross-sectional area of the tube has no effect on the height of the fluid column of a barometer.

kPaPkPaP

kPaPkPaPPatorr

kPatorrmmHgP

mm

Denvermm

atm

5.26;05.54

4.83;88.893.1331

325.101760

000,105000

:16101000

Problem Solving Technique

Step-by-step approach:1. Problem Statement2. Schematic3. Assumptions4. Physical Laws5. Properties6. Calculations7. Reasoning, Verification, and

DiscussionThe assumptions made while solving

an engineering problem must be reasonable and justifiable.

Problem Solving Technique (continued)

A result with more significant digits than that of given data falsely implies more accuracy.

When solving problems, we will assume the given information to be accurate to at least 3 significant digits. Therefore, if the length of a pipe is given to be 40 m, we will assume it to be 40.0 m in order to justify using 3 significant digits in the final results.

Example

■ A pressure gage connected to a tank reads 500 kPa. The absolute pressure in the tank is to be determined .

Pabs

Patm = 94 kPa

500 kPa

Example

■ The vacuum gage connected to a tank reads 15 kPa at a location where the barometer reading is 750 mmHg. Determine the absolute pressure of the tank. The density of mercury is given to be = 13,590 kg/m3.

Pabs

Patm = 750 mmHg

15 kPa

Example

Example

■ The air pressure in a tank is measured by an oil manometer. For a given oil-level difference between the two columns, the absolute pressure in the tank is to be determined. The density of oil is given to be = 850 kg/m3.

■ AIR Patm = 98 kPa0.60 m

AIR

Patm = 98 kPa

0.60 m

Example730 mmHg

755 mmHg

h

Forms of Energy (continued)The portion of the internal energy of a system

associated with the1. kinetic energies of the molecules is called the

sensible energy.2. phase of a system is called the latent energy.3. atomic bonds in a molecule is called chemical

energy.4. strong bonds within the nucleus of the atom itself

is called nuclear energy.

5. Static energy (stored in a system)6. Dynamic energy: energy interactions at the

system boundary (i.e. heat and work)

kgkJeU /1073.6 10235

Forms of Energy■ In absence of magnetic, electric, and surface tension effects, the total

energy of a system consists of the kinetic, potential, and internal energies and is expressed as

■ The macroscopic form of energy are those a system possesses as a whole with respect to some outside reference (i.e. kinetics and potential).

■ The microscopic forms of energy are those related to the molecular structure of the system , independent of outside reference frames (i.e. internal).

■ The change in the total energy E of a stationary system (closed system) is identical to the change in its internal energy U.

(kJ/kg) 2

basis massunit aon or, (kJ),

2

(kJ),

2

2

gzvupekeue

mgzmvmume

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