To give the students information about basic principles of ...mti.ntu.edu.iq/wp-content/uploads/2018/12/ديناميكا-الحرارة.pdf · Thermodynamics is an axiomatic science

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وزارة التعليم العالي والبحث العلمي هيئة التعليم التقني

التخصصات التكنولوجيةالسنة االولى

اسم المادة النظام السنوي الساعات االسبوعية

( اسبوع 30 نظري عملي مجموع )

Thermodynamics

لغة التدريس/ االنكليزية 2 2

هدف المادة/

To give the students information about basic principles of thermodynamics,

application of the zeroth law, first law, and second law of thermodynamic,

deep study for Carnot power cycle& reverse Carnot

Week No. Details 1 – 6 Thermodynamic term – measuring devices – properties – state – process – cycles –

density & specific volume – the pressure ( gage , vacuum , & absolute ) –

temperature relations ( Celsius , Kelvin & ranking scale ) – energy – renewable

energy resources ( solar energy , wind energy , energy of water falling , tidal

energy ) – hydrocarbons source ( oil & gas ) – form of energy used in

thermodynamic – potential energy – kinetic – energy – heat – work . 7 Internal energy – Floe work

8 - 9 First law of thermodynamic – flow system – nun flow system – steady – un steady –

open – closed

10 – 11 - 12 Application of the first law on – nozzle , diffuser , condenser , evaporator ,

compressor , heat exchanger ( surface , open ) , turbine , boiler .

13 – 15 Thermodynamics process undergoing at constant ( pressure , volume , temperature

, enthalpy , entropy ) polytrophic process – with representation on a ( P – V ) ,

( T – S ) and ( P – H ) diagram .

16 Specific heat , kind of specific heat – gas constant

17 – 19 Carnot power cycle – Carnot revisable cycle ( refrigeration and Heat pump

applications .

20 – 21 Study of steam – Steam properties – Using Steam tables .

22 Calculations of the properties for ( Liquid – Vapor ) mixture ( Wet steam )

23 – 25 Steam process under going at constant ( pressure , volume ) – Isotropic process –

Adiabatic process & applications .

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26 - 28 Fuel – definition of accounts and properties of the fuel used in boilers and cooling

systems absorbance .

29 - 30 Boiler – types – characteristics,

المصادر :

1. Applied engineering thermodynamics by roe choudhury .

2. thermodynamics by J. P. Holman .

3. Introduction thermodynamics by Sonntag van wylem .

4. Applied thermodynamics by eastop .

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Week No. Details 1 – 6 Thermodynamic term – measuring devices – properties – state – process – cycles –

density & specific volume – the pressure ( gage , vacuum , & absolute ) –

temperature relations ( Celsius , Kelvin & ranking scale ) – energy – renewable

energy resources ( solar energy , wind energy , energy of water falling , tidal

energy ) – hydrocarbons source ( oil & gas ) – form of energy used in

thermodynamic – potential energy – kinetic – energy – heat – work .

THERMODYNAMICS - 1

INTRODUCTION

The word Thermodynamics combines the Greek word therme ( heat ) and

Dynamics ( power ) for the field involves “ heat “ and “ work “ interactions .

Thermodynamics is an axiomatic science concerned with the transformation of energy

from one form to another . However , energy and matter are closely related , in as much

as the transfer of energy results in a change of the state of matter .

Hence , thermodynamics becomes involved in describing how energy interacts

with matter , and in doing so , it becomes concerned with certain physical variables

and their interrelationships .

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The principles of thermodynamics may be summarized as four laws or axioms

known as the Zeroth , First , Second , and Third law of thermodynamics .

A sound understanding of thermodynamics will allow determining how energy

is controlled and converted in devices ranging from Ink jet printers to paper – making

machines , from lawn mower engines to thousand megawatt power plants , from

welding torches to complex chemical plants .

UNITS

Units are specified magnitudes of dimensions that are used for measurement

purposes. Units of the principal thermodynamics are given in the following table,

Quantity symbol SI Units English Units To Convert from

English to SI

Units Multiply by

Length L M ft 0.3048

Mass m Kg 1bm

(bound mass) 0.4536

Time t S sec

Area A m 2 ft 2 0.09290

Volume V M 3 ft 3 0.02832

Velocity υ m ⁄ s ft ⁄ sec 0.3048

Acceleration a m ⁄ s 2 ft ⁄ sec 2 0.3048

Angular velocity

ω rad ⁄ s 2 sec – 1 ~

Force, Weight F , W N 1bf 4,448

Density ρ Kg ⁄ m 3 1bm ⁄ ft 3 16.02

Specific weight γ N ⁄ m 3 1bf ⁄ ft 3 157.1

Pressure, Stress

P , τ Kpa 1bf ⁄ ft 2 0.04788

Work, Energy W , E , U J ft.1bf 1.356

Heat transfer Q J Btu 1055

Power W W ft.1bf ⁄ sec 1.356

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Heat flux Q W Btu ⁄ sec 1055

Mass flux m Kg ⁄ s 1bm ⁄ sec 0.4536

Flow rate fq m 3 ⁄ s ft 3 ⁄ sec 0.02832

Specific heat c kJ ⁄ kg . k Btu ⁄ 1bm .º R 4.187

Specific enthalpy

h kJ ⁄ kg Btu ⁄ 1bm 2.326

Specific entropy

s kJ ⁄ kg . k Btu ⁄ 1bm .º R 4.187

Specific volume m 3 ⁄ kg ft 3 ⁄ 1bm 0.06242

One of the mass units in the English system is the Slug:

ft

sbf

s

ft

bfslug

2

2

11

1

111

The conversion factor to pound mass is:

bmbmslug 12.321174.321

lbf is a force required to accelerate 32.2 lbm (1 slug) at a rate of one ft/sec2.

2212.3211

s

ftbm

s

ftslugbf

When expressing a quantity in SI units certain letter prefixes may be used to represent

multiplication by a power of 10, see table below:

Multiplication Factor

Prefix Symbol Multiplication

Factor Prefix Symbol

110 deci d

110 deca da

210 centi c

210 hecto h

,

5

310 milli m

310 kilo k

610 micro μ

610 mega M

910 nano n

910 giga G

1210 pico p

1210 tera T

1510 femto f

1810 atto a

The units of various quantities are interrelated via the physical laws obeyed by the quantities. It follows that, in either system, all units may be expressed as algebraic

combinations of a selected set of base units. There are seven base units in the SI: m,

kg, s, K, mol (mole), A (ampere), cd (candela). The last two are rarely encountered in

engineering thermodynamics.

BASIC CONCEPTS AND DEFINITIONS

In order to deal with the subject of Thermodynamics rigorously it is necessary to

define the concepts used.

Force: Newton’s second law may be written as:

amkF

(Where m is the mass of the body accelerated with an acceleration a , by a

force F, k is constant, and in SI k = 1) . Hence,

amF

The unit of force defined as 1 kg. m / s 2 is called a Newton, so the defining equation

is :

1 Newton (N) = 1 kilogram × 1 m / s 2 = 1 kg. m / s 2

So we can define the unit Newton ( N ) is the force required to give a mass of 1 kg

an acceleration of 1 m / s 2 .

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Or a force of one lbf accelerates 32.2 lbm (1 slug) at a rate of one ft/sec2.

Hence, the units are related as:

1 N = 1 kg. m / s2 or 1 Ibf = 32.2 Ibm. ft/sec2

The force that imparts an acceleration of 1 cm / s 2 to a mass of 1 gram is

defined as a Dyne so that:

211

s

cmgmdyne

So we obtain that:

dyneN 5101

System: is a collection of matter within prescribed and identifiable boundaries,

whose have the shape in general identifies it as the object that is.

Thermodynamics System: is any quantity of matter or region of space to which

Attention is directed for analysis , and it must be enclosed by a prescribed boundary that may be rigid or deformable.

Boundary: is the actual or hypothetical enveloped enclosing the system. The

boundary is not necessary inflexible.

Surroundings: is the region (everything) which lies outside a system boundary.

Types of the Thermodynamics’ system:

There are two main types;

1. Closed System: is the system always contains the same matter (no matter crosses

the system’s boundary, and the mass of the system is constant) .This system is surrounded by a boundary may changes position, size, or shape, but it is impervious to the flow of matter. Heat and work, which are means of energy transfer, can across the system’s boundary. As shown in Figure, the gas trapped in a cylinder by a movable piston , the boundary is deformable but no matter across it .

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An isolated system; is a special case of a closed system that in no way interacts

with its surroundings. However, if two or more systems interact with only each other, the combination of these systems can be surrounded by a boundary to define an isolated system.

2. An open System; is the system in which there is a transfer of mass across the

boundaries. For instance, the jet engine has air and fuel entering and exhausts gases leaving, as shown in figure below;

In virtually all thermodynamic analyses, the first step is to answer the question, what kind of system is involved?

A property; is any quantity which serves to describe a system.

The state; it is the condition of a system as described by giving values to its

properties at a particular instant.

The common properties are pressure, temperature, volume, velocity, and

position; but others must occasionally be considered. Shape is important when surface

Cylinder System’s

boundary

Surroundings

Jet engine

Fuel

Air

Turbine

Exhaust

Gas

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effects are significant; color is important when radiation heat transfer is being

investigated. Precisely how many properties are required to specify the state of the system

depends on the complexity of the system.

System is in an equilibrium state (or in equilibrium) if no changes can occur in

the state of the system without the aid of an external stimulus. The change in the value of a property is fixed only by the initial and final state of the system.

Thermodynamic properties are divided into two general types, intensive

and extensive properties.

An intensive property; is one which does not depend on the mass of the

system, and has the same value for any parts of a system as it does for the whole system

.Temperature, pressure, density and velocity are examples of intensive properties .If we

bring two systems together, intensive properties are not summed.

An extensive property; is one which depends on the mass of the system.

Volume, momentum, and kinetic energy, are examples of extensive properties. If two

systems are brought together the extensive property of the new system is the sum of

the extensive properties of the original two systems.

If we divide an extensive property by the mass, the result is an intensive

property, and is called a specific property. The specific volume is thus defined to be:

kg

m

m

V 31

Where is the density which defined as the mass per unit volume:

3m

kg

V

m

A capital (uppercase) letter is usually as the symbol for an extensive

property and a lowercase letter to denote the associated intensive property.

Thus V is used to represent a volume and is used for specific volume.

Weight (w): is the force of the gravity on a substance. It depends on both the mass of

the substance and the gravitational field strength.

By Newton’s second law:

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amF

As mass remains constant, the variation of W with elevation is due to changes in

the acceleration of gravity g (from about 9.77 m/s2 on the highest mountain to 9.83 m/s2

in the deepest ocean trench). We will use the standard value 9.81 m/s2 (32.2 ft/sec2 ),

unless otherwise stated. The resulting acceleration is the gravitational acceleration g , and Newton’s

second law of motion becomes: gmw

If each side divides by the volume V we obtain:

gggV

m

V

w

Where is called weight density, and is called mass density.

A specific weight is a defined as a weight of the substance divided by its volume

(or a weight per unit volume), and it is known as weight gravity:

3m

N

V

w

For water, nominal values of and are, respectively,

1000

3m

kg ( 62.4

3

1

ft

bm ) and 9810

3m

N ( 62.4

3

1

ft

bf ).

For air, the nominal values of and are, respectively,

1.21

3m

kg ( 0.0755

3

1

ft

bm ) and 11.86

3m

N ( 0.0755

3

1

ft

bf ).

Example1. Evaluate specific weight of water at 20 ºC, if its density is 998 kg / m 3.

Solution:

3232397879787)81.9()998(

m

N

sm

mkg

s

m

m

kgg

Note that is force per unit volume, as is a mass per unit volume.

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Specific gravity: is the ratio of the density of the substance to some standard density

(density of a reference substance), and can also be expressed as the ratio of a specific

weight, rather than densities, provided that the specific weight are evaluated in regions

having the same gravitational acceleration .Notice that a specific gravity is a misnomer, because it is unaffected by gravity.

For liquids and solids, the reference substance is pure water at atmospheric

pressure. For gases, the reference substance is air common.

Temperature for liquids and solids are 4 ºC ( 39.2 ºF ) , 15 ºC( 59.0 ºF ) , or 20

ºC( 68 ºF ) . At these temperatures, the density of water is respectively;

1000 kg / m 3 , 999 kg / m 3 , 998 kg / m 3.

In the case of a gas, specific gravity is the ratio of the density of the gas to the density of air at the same temperature and pressure.

EXAMPLE 2. The mass of air in a room 3 X 5 X 20 m is known to be 350 kg.

Determine the density, specific volume, and specific weight.

Solution:

33167.1

)20()5()3(

350

m

kg

m

kg

V

m

kg

mo

m

kg

3

3

857.

167.1

11

32345.11)81.9()167.1(

m

N

s

m

m

kgg

PRESSURE: it is the effect of a normal force acting on an area. If a force Δ F acts at

an angle to an area Δ A (Figure below), only the normal component ΔFn enters into the

definition of pressure:

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A

FP

n

A

0lim ……. (1)

By considering a homogeneous fluid of density ( ) in static equilibrium as shown in

figure below , a simple force balance on the fluid shows that the pressure increases toward the bottom of the tank as a result of the weight of the fluid . A pressure difference exists between two points that are separated by distance ( h ) in the vertical direction is given by the basic equation of fluid static .

Note : For gases and liquids in relative motion the pressure may vary with

direction at a point; however, this variation is extremely small and can be ignored in

most gases and in liquids with low viscosity (e.g., water).

The weight of a cylinder of fluid can be equated to the difference between forces due to pressure at two ends of the cylinder.

Summing forces in the vertical direction (up is positive) gives:

PA = W + (P + dP) A → dP = – W = – ρgdh

For a liquid, ρ is constant, and h is measured positive downward.

dP = ρgdh

Now if P is a known function of h, the above equation can be integrated

to give P ( h ) :

h

(P + dP )A Pressure force

W=ρghA

P A

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P = ρ g h ……. (2)

Where g , is a specific weight, then we can write:

hP …… (3)

This equation can be used to convert to Pa a pressure measured in meters of

water or millimeters of mercury.

Most pressure – measuring instrument measure the difference between the

pressure of a fluid and the pressure of the atmosphere. This pressure difference is

called a gauge pressure.

In most thermodynamic relations absolute pressure must be used. Absolute

pressure is gage pressure plus the local atmospheric pressure:

Pabs = Pgage + Patm

The word “gage” is generally used in statements of gage pressure; e.g.,

P = 200 kPa gage. If “gage” is not present, the pressure will, in general, be an

absolute pressure.

Atmospheric pressure is an absolute pressure, and will be taken as 100 kPa (at sea

level), unless otherwise stated.

A negative gage pressure is often called a vacuum pressure ( Pvac ) , and gages

capable of reading negative pressures are vacuum gages , which indicates the

magnitude of the difference between the atmospheric and absolute pressure so that :

Pabs = Patm – Pvac

A gage pressure of ( – 50 kPa ) would be referred to as a vacuum of ( 50 kPa ) , with

the sign omitted.

Figure below shows the relationships between absolute and gage pressure,

Pgage = 0

Pgage

Pabs

Pvac (negative pressure, a vacuum)

Patm

(measured by a barometer)

Pabs

Pabs = 0

Patm (Atmospheric pressure )

Absolute pressure

Vacuum pressure

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It should be noted that atmospheric pressure is highly dependent on elevation; in

Denver, Colorado, it is about 84 kPa, and in a mountain city with elevation 3000 m it is

only 70 kPa.

Because the density of any manometric liquid varies with temperature, the same pressure difference results in different manometer deflections at different temperature.

Consequently. a conventional millimeter of mercury and a conventional inch of

mercury , based on the density of mercury at 0ºC or 32ºF and a standard

acceleration of gravity , are defined by :

kPaPam

NHgmm 1334.04.1334.13398106.13

10

11

23

kPaPam

Nwaterofmm 00981.081.981.965.9806

10

11

23

(Where 1 m 3 of water weights 9810 N)

1 in Hg = 0.4912 psi ( 2

1

in

bf).

The corresponding English unit is 2

1

ft

bf, although 2

1

in

bf (psi) is commonly used.

The unit of pressure in the SI unit is the Pascal (Pa), where:

1Pa = 1N / m2 = 1kg / m.s2

A larger unit that is slightly less than the standard atmospheric pressure is the bar:

1 bar = 100 kPa

The standard atmospheric pressure is defined as the pressure produced by a

column of 760 mmHg high, the mercury density being 13.5951 gm / cm 2, and the

acceleration due to gravity being its standard value of 9.81 m/ s 2.

The standard atmospheric pressure is 14.6959 2

1

in

bf ( psi ) , 29.92 in.Hg or

101.325 kPa ( kN / m 2 ) , 1.01325 bar .

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Instruments used for measuring pressure are:

1. barometer :

is an instrument using to measure atmospheric pressure only , thus the

atmospheric pressure is often called barometric pressure .as shown below.

2. manometer :

is a simple and widely used instrument that indicates a pressure difference by balancing a measurable length of fluid column against the pressure difference.

Pressures are often expressed in units such as inches of mercury, inches of

water, or millimeters of mercury. The manometer is shown below;

3. Bourdon – gauge :

760 mmHg

Barometer

Transparent

tube

Manometer liquid

Patm

P > Patm

Transparent u - tube

Scale

A simple u – tube manometer

Pressure need

to measure

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Another instrument for measuring pressure is a Bourdon – tube pressure gauge,

which measures the pressure relative to the pressure of the surrounding atmosphere.

EXAMPLE 3. Express a pressure gage reading of 35 psi in absolute Pascal’s.

Solution:

First, we convert the pressure reading into Pascal’s. We have:

1 ft = 12 inches , 1bf / ft 2 = 0.04688 kPa

gagekPa

ftbf

kpa

ft

in

in

bf241)

104788.0()144()

135(

2

2

2

2

To find the absolute pressure we simply add the atmospheric pressure to the

above value (pressure gage). Using Patm = 100 kPa, we obtain:

P = 241 + 100 = 341 kPa

EXAMPLE 4. The manometer shown in figure below is used to measure the

pressure in the water pipe. Determine the water pressure if the

manometer reading is 0.6 m. Mercury is 13.6 times heavier than

water.

Solution:

To solve the manometer problem we use the fact that:

Pa = Pb

The pressure Pa is simply the pressure P in the water pipe plus the

pressure due to the 0.6 m of water.

The pressure Pb is the pressure due to 0.6 m of mercury. Thus,

)9810()6.13()6.0()9810()6.0(33 m

Nm

m

NmP

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This gives:

P = 74 200 Pa or P = 74.2 kPa gage.

Then, ).12

(.9.1309.32936.123 in

ftin

s

ft

ft

slughgP

.5.01

5.0144

172

17272

22222psi

in

bf

in

bf

ft

bf

fts

ftslug

TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS

The word temperature comes from the Latin word temperature meaning

“proper mixing or tempering “, implying attainment of thermal equilibrium.

Temperature is defined as the sense of hotness or coldness, when we touch

a body. Definition of temperature as Maxwell defined is “temperature of a body is its

thermal state considered with reference to its ability to communicate heat to other

bodies “.

To establish a method of defining and measuring temperature, consider two bodies A and B isolated from the surroundings but placed in contact with each other. If one is hotter than the other, the energy in form of heat will be transferred from the body at higher temperature to the body at the lower temperature. If sufficient time is allowed, bodies A and B approach a state at which no further change is observed, so that the two bodies are in a state of a thermal equilibrium. When this state is reached, we may postulate that the two bodies are at the same temperature. In other words, if two systems are in thermal equilibrium their temperatures are postulated to be equal.

A rather obvious observation is referred to as the Zeroth law of

thermodynamics; “if two systems are each in thermal equilibrium with a third

system, they are also in thermal equilibrium with each other“, its meaning is

illustrated in Figure below: Temperature can be measured only by direct methods. Generally, heat is transferred to an instrument such as body ( C ) , and the change due to temperature in same property or response of ( C ) is measured . The test body ( C ) is called

A B

C

Two systems A and B in thermal equilibrium

with system C are also in thermal equilibrium between themselves.

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thermometer , and it is important to realize that the thermometer indicates its own

temperature , which is the same as the temperature of a system in thermal equilibrium with the thermometer . A property that changes in values as a function of temperature is called a

thermometric property , example of these properties are Length , Volume , Electrical

resistance of metallic wire , and emf ( electromotive force ) of thermocouple . Using

these properties lead us to the instruments used for measuring temperature are:

1. Thermometer; is a body with a readily measurable property that is a function of

temperature. In mercury – in – glass thermometer, the volume

of the mercury depends on its temperature.

2. Thermo – couple; is one of the most frequently used temperature transducers.

Thermocouples are very rugged and inexpensive and can operate over a wide temperature range. A thermocouple is created whenever two dissimilar metals touch and the contact point produces a small open – circuit voltage as a

function of temperature .This thermoelectric voltage is known as the see beck, which

discovered it in 1821. The voltage is nonlinear with respect to temperature. However, for small changes in temperature, the voltage is approximately linear, or

∆V = S ∆T

Where ∆V is the change in voltage, S is the see beck coefficient, and ∆T is the change in temperature. There are several type of thermocouples, one of this is shown below.

Ice 0ºC

Cold junction Hot junction

Galvanometer

emf (electromotive force)

scale

capillary

bulb Hg

Transparent

tube

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3. Pyrometer; is used to measure temperature more than 500 ºC.

Relative Temperature Scale

One way to establish a temperature scale is first to assign numerical values to certain accurately reproducible temperatures. Two that have been used in past are called the ice point and the steam point.

The ice point is defined as the equilibrium temperature of a mixture of ice and air –

saturated liquid water at pressure pf one atmosphere [101.325 kPa (101 kPa) or 14.696

psi ( 2

1

in

bf )].

The steam point is defined as the equilibrium temperature of pure liquid water in

contact with its vapor at one atmosphere. These two temperatures (points) are used as a reference temperatures because they are accurately reproducible in any laboratory , between these two

points we choose the number of subdivisions , called degrees.

On the Fahrenheit scale, there are 180 degrees between these two points. On the

Celsius scale (formerly called the Centigrade), 100 degrees. On the Fahrenheit scale

the ice point is assigned the value of 32, and the steam point the value of 212, when

on the Celsius scale they are assigned the value of 0 and 100. These selections allow

us to write:

325

9 CF TT ……. (1)

)32(9

5 FC TT ……… (2)

Absolute Temperature Scale

100

º C º F

212

0 32

Absolute zero − 459.67

32.02 0.01

Boiling point of Water at 1 atm

Triple point of water

Ice point

Relationship between Celsius and Fahrenheit scales

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19

The second law of thermodynamics will allow us to define an absolute temperature scale; however, since we do not have the second law at this point and we have immediate use for absolute temperature, an empirical absolute temperature scale will be presented.

Two absolute scales commonly used are Kelvin and Rankine scales. In the

Kelvin scale, a value of 273.16 is assigned to the triple point of water ( at which

liquid ,solid and vapor can exist), and ice point is a value of 273.15, when the steam

point is found to be a value of 373.15. In the Rankine scale, the triple point of water is

at 491.69 R, whereas the ice point is 491.67 R, since the steam point is at 671.67 R.

The numerical value of a temperature on the Rankine scale is related to the value on

the Kelvin scale by the following ratio:

8.1100

180

16.273

69.491

K

R

T

T …….. (3)

The units of temperature in the SI system are º C and º K, and in the English

system are º F or º R.

RELATIONSHIP BETWEEN ABSOLUTE

AND RELATIVE TEMPERATURES

373.15

º K º R

671.67

273.15 491.67

Absolute zero 0.00

491.69 273.16

Boiling point of Water at 1 atm

Triple point of water

Ice point

Relationship between Kelvin and Rankine scales

0.00

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The relations between absolute and relative temperatures are:

67.459 FR TT …… (4)

15.273 CK TT …… (5)

Substituting equation (5) and (3) into equation (4) gives:

328.1 CF TT …… (6)

Substituting equation (1) and (2) into equation (4) and (5) gives:

RCK TTT9

515.273 ……… (7)

KFR TTT5

967.459 …….. (8)

Note: The value 273.15 and 459.67 are often replaced by the approximate

values 273 and 460 where precise accuracy is not required.

EXAMPLE 8. The temperature of a body is 50 °F. Find its temperature in º C, º K,

and O R using the conversion equations.

Solution:

CTT FC

10)3250(9

5)32(

9

5

KTT CK

2832731015.273

RTT FR

5104605067.459

Process: is the change of the system from one state to another.

Path of the process: is the series of states through which the system

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passes during the process.

Cycle or Cyclic process: is a process or series of processes that

returns the system to its original state .

Reversible process: if the initial state and all energies transferred or

transformed during the process can be completely restored in both system and its surroundings.

Irreversible process: if the either the system or its surroundings or both

can not be restored .

ENERGY

Energy is defined as “the ability to do work “. It comes in different forms,

heat (thermal), light (radiant), mechanical, electrical, chemical, and nuclear energy. All these energy we need to live our busy lives.

There are two types of energy:

1

2 •

•

Initial state

Final state

Process Path of the process

P

V

• 2

P

V

1

3

4

•

• •

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1. Potential Energy (P.E).

2. Kinetic Energy (K.E).

1. Potential Energy (P.E). It is the energy stored in a system ( mass ) as a

result of its location in a gravitational field , and its magnitude is given by:

)(.......2

2

2JouleJmN

s

mkgm

s

mkgzgmEP

1 kJ = 10 3 J

(Where z is the distance of the system’s center of gravity

from some arbitrary horizontal datum plane.

Forms of P.E are:

a. Chemical Energy : is the energy stored in food , wood , coal , petroleum ,

and other fuels.

b. Mechanical Energy: is the energy possessed by an object due to its stored

energy of position .Rubber bands and springs are good examples.

c. Nuclear Energy: is the energy locked in the nucleus of the atom. Nuclear

power plants split atoms in process called fission.

d. Gravitational Energy : is the energy stored as a result of gravitational forces

concentrated by the earth for the object. Water held back by a dam is an example.

2. Kinetic Energy (K.E). the word “ Kinetic “is derived from the Greek word

meaning “ to move “ and the word “ Energy “ is the

ability to move. Thus “Kinetic Energy “is the energy

stored in a system by virtue of the motion of the system. As the train accelerates down the hill, the

P.E is converted into K.E. Therefore , the Kinetic

energy is defined , as the energy required to move

g

m

z

•

,

23

a mass a certain velocity.

A according to Newton’s second law of motion “The force F acting on a particle is

equal to the product of its mass m and its acceleration a or:

F = m. a ….. (1)

Acceleration long x – direction can be expressed as;

xd

vdv

td

xd

xd

vd

td

vda

Substituting in equation (1) , xd

vdvmF .

The change in stored energy is equal to the work done on the body to accelerate it ;

vdvmxdxd

vdvmxdFWEK xbodyondone.

Where m = mass and it is constant, therefore by integrating we obtain;

)(.2

.2

22

JouleJmNs

mkg

vmEK

( 1 J = 10 ³ J )

Forms of K.E are:

a. Electricity: is the energy produced when something upsets the balancing

force between electrons and protons in atoms .

b. Light or Radiant energy: waves that omit energy. Examples include, radio

and television waves .Gamma rays and x – rays.

c. Heat or Thermal energy: is the energy created by heat .

d. Movement: as the power of P.E uncoils, it transforms the energy source into

K.E and its magnitude is given by :

K.E = ½ m v ² ( J , KJ )

Energy transfer Interaction between a system and its surroundings can be expressed by two methods:

1. Energy transfer by work

F m

x 0

v

,

24

Work is defined as the product of a force and the distance moved in the

direction of the force. Consider a simple closed system as a gas trapped between a piston and cylinder, assume the pressure and temperature of the gas are uniform and there is no friction between the piston and the cylinder walls as shown in Figure. Let cross – sectional area

of the piston be A , and the pressure of

the gas at any instant be P .Let the piston

move under the action of the force exerted

by the gas a distance dl to the right , then

the work done by the gas is given by force

times the distance moved ;

W done by gas = P A × dl = P dV

Where dV is a small increase in volume or considering unit mass :

W done by gas = P d υ

Where d υ is the specific volume . Hence, the work done is given by the area

under the line of the process plotted on

P – υ diagram .

2

1

2

1

dPWorVdPW

In the special case of a constant pressure , then the work is ;

2

1

121

2

1

2 )()( PWorVVPVdPVdPW

The work interaction in each case is represented by the area

2

1

VdP underneath the corresponding path. This means that the work is not

property or a state function rather, it is a path function because it depends how the

system was changed from one state to another. Therefore, the system does not posses work; instead, work is mode of transfer of energy.

dl

F P

Gas

P

P1

P2

V1 V2

V

P

1

2

V

P

P1= P2

V1 V2

1

2

P(V2 – V1)

,

25

Work done on the system by the surroundings

is considered negative ( – ev ) .

Work done by the system on the surroundings

is considered positive ( + ev ) .

If 1000 J of work is done on a system,

we can write W = – 1000 J , or W in = 1000 J ,

and Wout means work done by a system.

Units of energy transfer by work

The units of work in SI of units are the (J) and (kJ);

kg

kJor

kg

J

kg

mNmassunitperand

s

mkgmNJ

2

2

111

In the English Engineering System of Units, the unit of work is ( ft.1bf ) which is

equivalent to a force of 1bf acting through a distance of 1 foot in the direction of the

displacement .

In the CGS system, the unit of work is erg, which is defined as the amount of work done

by a force of 1 dyne when it acts through a distance of 1 cm in the direction of the

displacement.

ergsJ 7101

POWER: is the time rate of energy transfer by work , or work done per unit time ,

and is denoted by ( W ).

Ws

J

t

WW

in the SI is called Watt .

________________________________________________________________

(– W )

F P

Gas

(+ W )

P

Gas

,

26

Week NO

Details

7

INTERNAL ENERGY – FLOW

WORK

Internal Energy ( Intrinsic Energy ) U , u

Internal Energy U is the form of energy stored in a system associated

with the configuration and motion of its molecules, atoms, and substance particles

relative to its center of mass. Therefore we can define Internal energy as ” it is the

energy stored in the mass of the substance within the system”.

Internal Energy is independent of gravity , motion , and is a property

consisting of the combined molecular kinetic energy and molecular potential energy , and it is determined by properties , such as pressure P and temperature T . Since internal energy is a property, then gain in it in changing from one state to another as shown in P–V diagram. This change can be written as the algebraic sum of the heat and work interaction with the surroundings: d U = d Q – d W …… (1)

Where; dQ – is the net heat transfer to the system.

dW – is the net work done on the system.

dU – is the change in internal energy.

For any closed system, there will be either heat supplied or heat rejected, but not both. Similarly, will there be rather work output or work input, but not both. Therefore, equation (1) can be written as:

V V1 V2

1

2

P

P2

P1 •

•

,

27

WQUU 12

2

1

12 ,)(.. VdPWwhenWUUQei

2

1

12 )(0 VdPUUQr …... (2)

For most substance , the Internal Energy , where no phase change is

involved , depends strongly on temperature and rather weakly on pressure or volume , and when no external work is done . Then,

systemQU …… (3)

By substituting into heat equation for system;

)( 12 TTcmQ

Equation (3) becomes:

)( 12 TTcmU …… (4)

The change in Internal Energy of an Ideal – Gas per unit mass as the gas

changes from state (1) to state (2) is:

Tdcud

T

T

2

1

2

1

Where u is the internal energy per unit mass (specific internal energy).

If c is a constant, then;

)( 1212 TTcuu …… (5)

The units of Internal Energy are usually being (J) , whereas Specific

Internal Energy is expressed in ( J / kg ) .

Example : A 2.5 kg system undergoes a process wherein its temperature

changes from 280 ºk to 340 ºk . Determine the heat interaction if

the specific heat of the system for the process is given by ;

kkg

kJ

TC

.)

320

3584.0(

, What is the average specific heat ?

Solution:

TdT

mTdcmQ

T

T

T

T

)320

3584.0(

2

1

2

1

,

28

2

1])320(ln3584.0[

T

TTTmQ

}]320)280340({ln.

35})280340(.

84.0{[5.2kkg

kJk

kkg

kJkg

kJQ 34.134

The average specific heat is :

kkg

kJ

kkg

kJ

Tm

Qc avg .

896.0)280340()5.2(

34.134

Flow Energy

Energy transfer in the form of work is associated with flow phenomena, and represents the work (energy) necessary to move a fluid at steady rate without changing its state.

Consider a fluid in a duct of mass dm and volume dv, and let an imaginary

piston be placed behind dm, as shown in Figure .

The required work done by the piston in moving ( dm ) past line a – a is : Work done = Flow Energy = F. dl

Since F = P. A

Then, Flow Energy = P. A . dl = P. d V

Pdm

dVPmassunitperEnergyFlowtheand

Thus, the flow energy per unit mass is simply equal the product of the

absolute pressure ( P ) and the specific volume of the fluid , and

P

dm

a

a

dl

Flow

Boundary

F

Imaginary piston

of cross-section

area A

,

29

sometimes is called Flow Work .

Steady Flow: is the flow where is the rate of mass transfer constant.

Unsteady Flow: is the flow where is the rate of mass transfer not constant.

EXAMPLE 1. A fluid at a pressure of 3 bar and with specific volume of

0.18 m³ /kg contained in a cylinder behind a piston expands

reversibly to a pressure of 0.6 bar according to a law

2

CP , where C is a constant . Calculate the work done

by the fluid on the system . Solution:

2

1

dPdoneWork

2

1

2

1

12

C

dC

Also 2

32 )18.0(3

kg

mbarPC

2

3

)(0972.0kg

mbarC

dl

F P

Gas

P1=3 bar

υ1 υ2 υ (m³⁄ kg)

1

2

P2 = 0.6 bar

P (bar)

2

CP

,

30

And kg

m

bar

kg

mbar

P

C 3

23

2

2 402.06.0

)(0972.0

kg

m

kg

mkg

mbardoneWork

33

23

402.0

1

18.0

1)(0972.0

kg

J

kg

mN

kg

m

m

N2982029820068.3100972.0

3

2

5

kg

kJ820.29

1. Energy transfer by Heat

Heat is a form of energy, which is transferred from one body to another at

lower temperature. Hence, it is defined as the energy that across the boundary of the system due to the difference in a temperature between the system and its surroundings.

Heat as a work can never be contained in a body or possessed , and it is a path function as a work because the amount of heat transferred depends on

how the system was changed from one state to another. Heat is denoted by the

capital letter Q .

Heat transferred (added or given) to the

System is considered positive (+ ev).

Heat transferred (rejected or taken) from

The System is considered negative (– ev).

The word adiabatic means without heat transfer, so an adiabatic process is

defined as one that involves no heat transfer.

Q add (Q in)

Q rej (Q out)

System

,

31

Units of energy transfer by heat

In the SI of units, the unit of heat is J (Joule) or ( kJ ). In the English

system of units is Btu (British thermal unit).

The quantity of heat required to raise the temperature of 1bm of water from

59.5 º F to 60.5 º F under atmospheric pressure is called the 60 º F Btu.

In the CGS system, the quantity of heat required to increase the temperature

of 1 gram of water from 14.5 º C to 15.5 º C under atmospheric pressure is called

the 15 º C Calorie.

In 1929 , the International Steam Table Conference adopted the

International Calorie ( IT ) as ;

Js

mkgCalorieIT 1868.41868.41

2

2

JCalorieITBtu 1055996.2511

Note also that ; Ckg

Jk

Cgm

Cal

Fbm

Btu .

1868.4.

1

.1

1

Heat transferred per unit mass is denoted by ( q ) ;

kg

kJor

kg

J

m

Qq

Heat Flux: is the rate time of heat transfer per unit of system surface area.

)( wattWs

J

t

QQ

22 . fth

Btuor

m

W

A

QFluxHeat

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

,

32

Week No

Details

16

Specific heat , kind of specific heat – gas constant

Specific Heat (c):

Specific Heat of a solid or liquid is defined, as the heat required raising the

temperature of a unit mass of the substance one degree. For a small quantities we have ;

TdcmQd

Where (m) is the mass, dT is the increase in temperature , and (c) is the

specific heat .

For a gas , specific heat depends on whether the gas is at constant pressure or at constant volume .

TdcmQd p , c p – specific heat at constant pressure.

)( 12 TTcmQ p

TdcmQd , c υ – specific heat at constant volume.

)( 12 TTcmQ

The average specific heat is the amount of energy transferred per unit mass divided by the accompanying increase of its temperature;

T

q

Tm

Qc avg

Where ; c avg – average specific heat ( J / kg . º k ) . Q – Heat interaction ( J ) . ∆T – temperature difference ( º k or º C ) .

,

33

m – mass ( kg ) . q – heat per unit mass ( J / kg ) . When q and ∆T are very small , then ratio q / ∆T tends to a limit that indicates the specific heat at a temperature T , thus ;

T

q

T

qc

T

0lim

Therefore, we can write the equation in form;

p

pT

qc

– Specific heat at constant pressure.

T

qc p – Specific heat at constant volume.

In the SI of units, the unit of specific heat is kkg

J.

.

In the English system of units, the unit of specific heat is Fbm

Btu.1

.

In the CGS of units, the unit of specific heat is Cgm

Cal.

.

The Ideal – Gas Equation of State When a vapor of a substance has relatively low density ( temperatures are in excess of the critical temperatures , and also at very low pressures ) , the vapor of the fluid tends to obey the equation;

,

34

RconstT

P .

….. (1)

Where; P – is absolute pressure ( 22

1,

ft

bf

m

N).

T – is absolute temperature (º k, º R).

– is specific volume (bm

ft

kg

m

1,

33

).

R – is a constant which is called the gas constant, and each gas has a

different gas constant. In practice, no gases obey this equation rigidly, but many gases tend toward it. Any

gas for which this equation is valid, is called an ideal – gas or a perfect gas, and the

equation is called the characteristic equation of state of a perfect gas.

The characteristic equation is usually written as:

TRPorTRP ….. (2)

Where, ρ – is the density ( 33

1,

ft

bm

m

kg).

For m kg, occupying V m³ equation (2) becomes:

TRmVP ….. (3)

Another form of the equation can be derived using the kilogram – mole (kmol) as

a unit. The kilogram – mole (kmol) is defined as a quantity of a gas required to M kg of

the gas, where M is the molecular weight. For example, the molecular weight of

Oxygen (O2) is 32 , then 1 kg of O2 is equivalent to 32 kg of O2 , 1 kmol of Carbon ( C

) is a mass off 12 kg .

Stated otherwise, kmol

kgM

32 for O2, kmol

kgM

12 for C.

In the English system one uses the pound – mole (1bmol) ; for O2

bmol

bmM

1

132 .

From the definition:

,

35

Mnm

Where, n – is the number of moles.

By substituting into equation (3). We obtain,

Tn

VPRMorTRMnVP ….. (4)

Avogadro’s hypothesis states that the volume of 1 mole of any gas is the same as

the volume of 1 mole of any other gas, when the gases are at the same temperature

and pressure. Therefore, n

V is the same for all gases at the same values of P and T.

That is the quantity Tn

VP is a constant for all gases. This constant is called the

universal gas constant, and is given the symbol R or R ;

TRnPVorTn

VPRMR ….. (5)

Or since M

RRRMR ….. (6)

The units of R are (Rbm

bfft

kkg

Jk

kkg

J .1

1.,

.,

.).

Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and

0ºC is approximately 22.71 m ³. Therefore,

kkmol

J

k

mm

N

Tn

VPR

.3.8314

15.2731

71.22101 3

2

5

kkmol

Jk.

3143.8Rbmol

B.1

9859.1

Rbmol

bfft.1

1.3.1545

For air bmol

bm

kmol

kgM

1

197.2897.28 , so that for air

,

36

Rbm

bfft

kkg

JkR

.1

1.3.53

.287.0 .

Care must be taken in using the simple convenient equation of state. A low – density (ρ) can be experienced by rather having a low pressure or a high temperature.

For air the ideal gas equation is surprisingly accurate for a wide range of temperatures

and pressures; less than 1 percent error is encountered for pressures as high as 3000

kPa at room temperature, or for temperatures as low as – 130 º C at atmospheric

pressure. The compressibility factor Z helps us in determining whether or not the ideal gas

equation should be used. It is defined as:

TR

PZ

….. (7)

If Z = 1 , or very nearly 1 , the ideal gas equation can be used . If Z is not

approximately 1 , then equation (7) may be used .

The compressibility factor Z can be determined for any gas by using a

generalized compressibility chart. In this chart, the reduced pressure P R and reduced

temperature TR must be used. They are calculated from:

C

R

C

RT

TT

P

PP , ….. (8)

Where PC and TC are critical – point pressure and temperature, respectively, of

table B – 3.

Example 1. An automobile tire with a volume of 0.6 m³ is inflated to a gage

pressure of 200 kPa . Calculate the mass of air in the tire if the

temperature is 20 º C .

,

37

Solution:

Air is assumed to be an ideal gas at the condition of the example . The ideal gas equation:

TR

VPmTRmVP

gageatmabs PPP

At 20 º C, Patm = 100 kPa, Pgage= 200 kPa

→ Pabs= 100 + 200 kPa = 300 kPa

Tk= TC + 273 = 20 + 273 = 293 º k

R = 287 kkg

J.

.14.2

293.

.287

6.010300 3

2

3

kg

kkkg

mN

mm

N

TR

VPm

Example 2. Air is at 25 º C and 101.325 kPa. If the gas constant R = 287 J / kg. k

, find the specific volume and the molar mass of this gas, assuming it behaves as an ideal gas. Solution: Tk = TC+ 273 = 25 + 273 = 298 º k

kg

m

kPa

kkkg

Jk

P

TRTRP

3

8445.0325.101

298.

287.0

The molar mass is : kmol

kg

kkg

Jkkmol

J

R

RM 97.28

.287

.4.8314

,

38

Week No

Details

20 – 21 Study of steam – Steam properties – Using Steam tables .

PROPERTIES OF A PURE SUBSTANCE

,

39

A Pure Substance is the substance that is chemically homogeneous and

fixed in chemical composition. Every substance in nature can theoretically exist in at

least three phases; Solid, liquid, and gaseous phase.

A phase is any homogeneous part of the system is physically distinct. Each

phase in the system is separated from other by interface called phase boundary. For

example, in the system consisting of an ice cube in liquid water, there are two phases, solid and liquid, and the phase boundary is the surface of the ice cube. Both are homogeneous and the temperature and pressure are same, but the two phases are distinct, because their densities (as well as other properties) are different.

Working Fluid is the substance contained within the boundaries of the

system, and it is stated when two independent properties are known, and then the thermodynamic state of the fluid is defined. In the thermodynamic systems the working fluid can be found in the liquid, vapor, or gaseous phase, the solid phase is not important in engineering thermodynamics. We can tend to identify all substance with the phase in which they are in equilibrium at atmospheric pressure and temperature. For instance, substance such as Oxygen (O 2 ) , Nitrogen ( N 2 ) are thought of as gases, H2O is thought of as a liquid or vapor ( i.e. water or steam ) , Mercury ( Hg ) is thought of as a liquid . All these substances can exist in different phases, O2 and N 2 can be liquefied, H2O can become a gas at very high temperature, and Hg can be vaporized and will act as a gas if the temperature raised high enough.

Week No

Details

,

40

Critical pressure

23 – 25 Steam process under going at constant ( pressure , volume ) – Isotropic process

– Adiabatic process & applications .

Liquid, Vapor, and Gas

Consider a P–V diagram for any substance, water is the typical example. When

the unit mass of water is heated at any one constant pressure, there is one fixed

temperature at which bubbles of vapor form in the liquid, this phenomenon is known as

Boiling.

The change of state is noted by the increase of the mass of the vapor and the decrease of the mass of the liquid, and the liquid is in equilibrium with its vapor. The volume occupied by 1 kg of boiling liquid at a higher pressure is slightly larger than the volume occupied by 1 kg of the same liquid when it is boiling at low pressure. A series of Boiling points (P, Q, and R) on a P – diagram will appear as a

sloping line as shown in Figure;

When a liquid at Boiling points is

heated further at constant pressure, the additional heat supplied changes the phase of the substance from liquid to vapor; during this change of phase , the pressure and the temperature remain constant. It is found that at the higher pressure is a smaller amount of heat is required, and there is a definite value of specific volume of the vapor at any one pressure at the point at which vaporization is complete.

Hence , a series of points such as

P' , Q' , and R' can be plotted and joined to

form a line as shown in the P – diagram .

When the two curves are already drawn, they will form a loop as shown in

P– diagram.

The pressure at which there is no definite transition between liquid and vapor phase , and

R'

Q'

P

Pp

PR •

•

• P'

PQ P= const.

R

Q

P

Pp

PR •

•

• P

PQ P= const.

R

Q

•

•

•

P

R'

Q'

P

Pp

PR •

•

• P'

PQ P= const.

• C

PC

,

41

Saturated Liquid State Saturated Vapor State

the liquid changes to vapor without any discontinuity is called The Critical Pressure .

The point where the liquid and vapor phase of substance are distinguishable

is called The Critical Point , which is represented in our by the point (C) , and the state

at which no discontinuities noticeable is called The Critical state .

The substance existing at a state point inside the loop consists of a mixture of liquid

and dry vapor is known as a wet vapor.

A saturation state is a state at which two or more phases of a substance can exist

together in equilibrium without change of pressure or temperature. Hence, the Boling

Points P , Q , and R are saturation states , and are called saturated liquid states . A

series of such boiling points joined up is called saturated liquid line. Similarly, the

points P' , Q' , and R' , at which the liquid is completely converted into vapor are

saturation states too , and are called saturated vapor states . A series of such points

joined up is called saturated vapor line.

Any phase of substance existing under saturation conditions is called a saturated

phase. The temperature at which vaporization takes place at a given pressure is called

The Saturation Temperature, and the pressure corresponding to

This temperature is called a Saturation Pressure. The vapor and liquid coexisting in

equilibrium are called Saturated Vapor and Saturated Liquid.

The word Saturation refers to energy saturation. For example, a slight addition of

heat to a boiling liquid changes some of it into vapor, and it is no longer a liquid but is now a wet vapor. Similarly, when a substance just on the saturated vapor line is cooled slightly, droplets of liquid will begin to form and the saturated vapor becomes a wet vapor.

A saturated vapor is usually called Dry saturated vapor to emphasizes the fact that no

liquid is represent in the vapor in this state. The heat required to change a saturated liquid state to a dry saturated vapor state is

called The Latent Heat .It can be seen that at the critical point (C.P) ,the latent heat is

zero .

Lines of constant temperature called Isothermals can be plotted on

P– diagram as shown in Figure below.

The temperature lines become

Saturated liquid

line

Saturated vapor

line

R

• R'

P

PR •

PQ

C

PC

T1

T2

T3

s

TC

TC

T4

T5

T6

•

,

42

horizontal between the saturated liquid line and the saturated vapor line (e.g. between P' , Q' , and R' ).

Thus, there is a corresponding saturation temperature for each saturation pressure . At

pressure Pp the saturation temperature is T1 , at

PQ is T2, and at PR is T3. The critical temperature line TC just touches the top of the loop at

the critical point (C).

When a dry saturated vapor is heated at constant pressure its temperature rises and it

becomes Superheated. Hence, the temperature of the vapor is higher than the saturation

temperature at the given pressure is called Superheated Temperature. The term

superheat denotes that the temperature of the vapor is in excess of the saturated value corresponding to the given pressure. The difference between the actual temperature of the superheated vapor and the

saturation temperature at the pressure of the vapor is called The Degree of Superheat.

For example, the vapor at point (S) is superheated at PQ and T3, and the degree of

superheat is (T3 – T2).

As is stated that two independent properties are sufficient to define the state of the

substance, now between points P and P' , Q and Q' , R and R' , the temperature and

pressure are not independent since they remain constant for a range of values of specific

volume ( ) . For example, the substance at PQ and T2 could be a saturated liquid, a

wet vapor , or a dry saturated vapor . The state cannot be defined until one other

property (e.g. specific volume) is given.

In the case of a liquid, if the pressure is greater than the saturation pressure corresponding to its temperature, it is called compressed liquid, it is also called

subcooled liquid, because the liquid temperature is lower than the saturation

temperature for the given pressure. In like manner, a solid at a temperature below its saturation temperature is called compressed solid.

The conditions under which three phases can exist in equilibrium are called Triple

phase state, Triple phase point, or Triple point as shown in P–T diagram, T–

diagram, and P– diagram.

T

P= const. •

Critical point

L - V

Saturated liquid line

Vapor

Saturated Vapor line

Liquid

,

43

It is conventional to call the substance in the superheated region a vapor if its

temperature is below that the critical point and a gas if it is above the critical point.

T

P

• Critical point

Liquid

• Triple point

Fusion line

Vaporization line

Vapor

Solid

P

T= const.

• C

Critical point

L - V

Saturated liquid line

Saturated Vapor line

Vapor Liquid

P

• Critical point

Solid and Vapor

Vapor

Liquid

Solid and Liquid

Solid

L - V

Superheated

vapor region

L - V

T

P1

• C.P

Liquid

Vapor

region

P2 •

Gas

PC

Compressed

Liquid

region

,

44

Week No

Details

22 Calculations of the properties for ( Liquid – Vapor ) mixture ( Wet steam )

The Liquid – Vapor Mixture (Saturation State)

A state represented by a point inside the loop such as x corresponds to an

equilibrium state of a homogeneous mixture of liquid and vapor

that is called a Wet Vapor .

The condition or quality of a wet vapor

is most frequently defined by its Dryness

Fraction, and when this is known as well as

the pressure or temperature then the state of the wet vapor is fully defined.

Dryness Fraction, x = the mass of dry vapor in 1 kg of the mixture.

R

Q

•

•

P

R'

Q'

P

Pp

PR •

• • P'

PQ

C.P PC

T1

T1

T2

T2

• T3

T3 TC

TC

•

•

• x

Liquid

Vapor

region

x

f g

,

45

Sometimes a wetness Fraction is defined as the mass of liquid in 1 kg of

the mixture , i.e.

Wetness Fraction = 1 – x

Then each kilogram of mixture contains (x) kg of vapor and (1 – x) kg of liquid.

Note that for a dry saturated vapor (x = 1), and for a saturated liquid (x = 1)

The property (x) is an intensive property defined by the equation;

m

m

mm

mx V

LV

V

Where; m L – is a mass of the saturated liquid. m V or m g – is a mass of the saturated vapor. m – is the total mass.

If x = 0 , the mixture is completely liquid and its state is represented by

points P, Q, and R.

If x =1 , the mixture is completely vapor and its state is represented by

points P', Q', and R'.

The suffix (g) is used to denote the saturated vapor or dry saturated state

the suffix (f) is used to denote the saturated liquid state, and the suffix (x) to denote a

mixture.

g

VL

Vf

VL

L

VL

gVfL

VL

VLx

mm

m

mm

m

mm

mm

mm

VV

m

V

..(1)

Using the definition of Dryness Fraction (x), we have,

VL

L

VL

V

mm

mxand

mm

mx

1 ….. (2)

By substituting (2) into equation (1), we obtain,

fgffgfgfx xxxx )()1( ….. (3)

Where; f - is the specific volume of the saturated liquid.

g - is the specific volume of the saturated vapor.

x - is a specific volume of the mixture.

The difference fgfg - is the increase in the specific volume when the

,

46

saturated liquid changes to the saturated vapor.

Equation (3) indicates that the specific volume ( x ) of two – phase mixture

increases linearly from saturated liquid ( x = 0 ) to saturated vapor ( x = 1) .

When saturated water is changed to dry saturated vapor, then,

WuuWuuQ fg )()( 12

also W is represented by the area under the horizontal line on the P– diagram .

i.e. )( fgPW

Thus , )()()()( ffggfgfg PuPuPuuQ

fg hhQPuh

The heat required to change a saturated liquid state to a dry saturated vapor

state is called The Latent Heat. Hence, Latent Heat is given in the steam tables as a

)( fgh .

In the case of Steam Tables, the internal energy of saturated liquid is taken

to be zero at the triple point (i.e. at 0.01 º C and 0.006112 bar), since,

3

5

006112.0,1. 10

0010002.010006112.00

barCooath

Where f at 0.01 º C is 0.0010002 m³ / kg .

Thus, kg

Jkh f

410112.6

This is negligibly small and the zero for enthalpy may be taken at o.o1 º C .

0h at a Triple Point 0.01 º C .

STEAM TABLES

,

47

Tabulations have been made for many substances of the thermodynamic properties P , , and T . Values are presented in the appendix in both tabular and graphical form.

Table C-1 gives the saturation properties of water as a function of saturation

temperature.

Table C-2 gives the saturation properties of water as a function of saturation

pressure. The information contained in the two tables is essentially the same, the choice

being a matter of convenience. We should note, however, that in the mixture region

pressure and temperature are dependent. Thus to establish the state of a mixture, if we

specify the pressure, we need to specify a property other than temperature. Conversely, if we specify temperature, we must specify a property other than pressure.

Table C-3 lists the properties of superheated water vapor. To establish the state of

a simple substance in the superheated region, it is necessary to specify two properties. While any two may be used, the most common procedure is to use pressure and temperature. Thus, properties such as are given in terms of the set of independent

properties P and T.

Table C-4 lists data pertaining to compressed liquid. At a given temperature, the

specific volume of a liquid is essentially independent of the pressure. For example, for a

temperature of 100°C in Table C-1, the specific volume of liquid is 0.001044 m3/kg at a

pressure of 100 kPa, whereas at a pressure of 10 MPa the specific volume is 0.001038

m3/kg, less than a 1 percent decrease in specific volume. Thus it is common in

calculations to assume that the specific volume of a compressed liquid is equal to the specific volume of the saturated liquid at the same temperature. Note, however, that the specific volume of saturated liquid increases significantly with temperature, especially at higher temperatures.

Table C-5 gives the properties of a saturated solid and a saturated vapor for an

equilibrium condition. Note that the value of the specific volume of ice is relatively

insensitive to temperature and pressure for the saturated-solid line. In addition, it has a greater value (almost 10 percent greater) than the minimum value on the saturated-liquid line.

EXAMPLE 1. Determine the volume change when 1 kg of saturated water is

completely vaporized at a pressure of ( a ) 1 kPa, ( b ) 100 kPa, and

(c ) 10000 kPa.

Solution:

,

48

Table C-2 provides the necessary values. The quantity being

sought is:

fgfg .

Note that P is given in MPa.

( a ) 1 kPa. Thus, kg

mfgfg

3

2.129001.02.129 .

( b ) 100 kPa = 0.1 MPa. kg

mfgfg

3

3.169001.0694.1

( c ) 10000 kPa=10 MPa. kg

mfg

3

01658.000145.001803.0

Notice the large change in specific volume at low pressures compared with the small change as the critical point is approached.

EXAMPLE 2. 2 Ib of water is contained in a constant-pressure container held at

540 psia. Heat is added until the temperature reaches 700 OF.

Determine the final volume of the container.

Solution:

Use Table C-3E. Since 540 psia lies between the table entry

values, the specific volume is simply:

bm

ft

bm

ft

bm

ft

bm

ft

12115.1)

13040.1

10727.1(4.0

13040.1

3333

The final volume is then: 3

3

423.21

2115.112 ftbm

ftbmmV

Properties of a Wet Vapor

,

49

For a Wet Vapor, the volume of a liquid present plus the volume of dry vapor present

give the total volume of the mixture. Therefore, the specific volume is given by;

vaporwetofmassTotal

vapordryofvolumeliquidofvolume

Now for 1 kg of wet vapor , there are ( x ) kg of dry vapor and ( 1 – x ) kg of liquid ,

where ( x ) is The Dryness Fraction , hence ,

gf xx )1(

The volume of the liquid is usually negligibly small compared to the volume of the dry saturated vapor, hence, The enthalpy of a wet vapor is:

fgffgfgf hhhhxhhxhxh )()1(

Similarly, the internal energy of wet vapor is given by:

)()1( fgfgf uuxuuuxuxu

This equation can be expressed as the equation for enthalpy, but these is more

convenient since )( fu and )( gu are tabulated and the difference

)( fgfg uuu is not tabulated.

Example 1. Find the specific volume, enthalpy, and internal energy of wet vapor at

18 bars, Dryness Fraction is 0.9.

Solution:

For the wet vapor, specific volume is:

From the pressure table C – 2 of the steam, at 18 bar = 1.8 MPa, we have,

gx

gx

kg

m

kg

mg

33

0994.01104.09.001104.0

fgf hxhh

,

50


kg

Jkh

kg

Jkh fgf 3.1912,8.884

kg

Jk

kg

Jk

kg

Jkh 87.26053.19129.08.884

gf uxuxu )1(


kg

Jk

kg

Jk

kg

Jku 83.24264.25989.07.882)9.01(

EXAMPLE 2. 4 kg of water is placed in an enclosed volume of 1m3. Heat is added

until the temperature is 150°C. Find ( a ) the pressure, ( b )the mass

of vapor, and ( c ) the volume of the vapor.

Solution:

Table C-1 is used. The volume of 4 kg of saturated vapor at 150 ºC is:

(0.3928X4) = 1.5712 m3.

Since the given volume is less than this, we assume the state to be in the

quality region ( wet vapor ).

( a ) In the quality region the pressure is given as P = 475.8 kPa.

( b ) To find the mass of the vapor we must determine the quality. It is found from:

)( fgfx x

kg

Jkh

kg

Jku gf 4.2598,7.882

,

51

using = 1/4 m3/kg, as:

)00109.03928.0(00109.025.0333

kg

m

kg

mx

kg

m

Thus, 6354.0

3917.0

2489.0

3

3

kg

m

kg

m

x

Using m

m

mm

mx

g

Lg

g

, the mass vapor is:

m g = m x = (4) kg . (0.6354) = 2.542 kg

(c) Finally, the volume of the vapor is found from:

33

9985.0542.23928.0 mkgkg

mmV ggg

Note that in mixtures where the quality is not very close to zero the vapor phase occupies most of the volume. In this example, with a quality (x) of 63.54 percent it occupies

99.85 percent of the volume.

Properties of Superheated Vapor For steam in the superheated region, temperature and pressure are independent properties. When the temperature and pressure are given for superheated vapor, then the state is defined and all other properties.

For example, vapor at 2 bar and 200 º C is superheated, since the saturation

temperature at 2 bar is 120.2 º C, which is less than the actual temperature. The vapor

in this state has a degree of superheat:

Degree of superheat = 200 – 102.2 º C = 97.8 ºC

The tables of properties of superheated vapor range in pressure from 0.006113 bar to the critical pressure of 221.2 bars, and there is an additional table of

supercritical pressure up to 1000 bar.

,

52

At each pressure, there is a range of temperatures up to high degrees of superheat, and values of specific volume, internal energy, enthalpy, and entropy are tabulated at each pressure and temperature for pressures up to and including 70 bar; above this

pressure the internal energy is not tabulated. For reference, the saturation temperatures are inserted in brackets under each pressure in the superheat table, values of

ggg hu ,, and gs are given. A specimen row of values in the table C – 3 of

superheat steam is shown below.

P, MPa

(tsat,º C) t 250 300 350 400 450 500 550

2.0

(212.4)

m³/kg 0.1114 0.1255 0.1386 0.1512 0.1635 0.1757 0.1877

u kJ/kg 2679.6 2772.6 2859.8 2945.2 3030.4 3116.2 3203.0

h kJ/kg 2902.5 3023.5 3137.0 3247.6 3357.5 3467.6 3578.3

s kJ/kg.k 6.5461 6.7672 6.9571 7.1279 7.2853 7.4325 7.5713

From superheat table above, at 20 bar and 400 º C, the specific volume is = 0.1512 m ³ / kg and the enthalpy is h = 3247.6 kJ/kg.

Example . Steam at 100 bar has a specific volume of 0.02242 m ³/kg, find the

temperature, the enthalpy, and the internal energy. Solution:

First, it is necessary to decide whether the steam is wet, dry saturated, or superheated.

Using the table C – 2 (properties of Saturated H2O – pressure table),

at 100 bar = 10 MPa , kg

mvolumeactual

kg

mg

33

02242.001803.0

The steam is superheated

,

53

Using the table C – 3 (properties of superheated steam), we have,

at 100 bar = 10 MPa , the specific volume is kg

m 3

02242.0 at a

temperature of 350 º C , and saturation temperature is 311.1 º C , then the

steam has a degree of superheat is:

Degree of superheat = 350 – 311.1 º C = 38.9 º C

From the table C – 3, kg

Jkh 4.2923 ,

kg

Jku 2.2699

Interpolation For properties that are not tabulated exactly in the tables, it is necessary to interpolate between the values tabulated. For example to find the temperature , specific volume , internal energy , and enthalpy of dry saturated steam at 9 bar , it is necessary

to interpolate between the values given in the tables . At 9 bar , the saturation temperature is equal :

)(810

8981089 baratsatbaratsatbarsatbaratsat TTTT

Using the table C – 2 (properties of Saturated H2O – pressure table),

,

54

At 8 bar = 0.8 MPa , Tsat at 8 bar = 170.4 º C

At 10 bar = 1.0 MPa , Tsat at 10 bar = 179.9 º C

)4.1709.179(5.04.1709 CCT baratsat

CT baratsat

15.1759

Similarly , kg

Jkh

kg

Jkh baratgbaratg 1.2778,1.2769 108

kg

Jk

kg

Jk

kg

Jkh baratg 6.2773)1.27691.2778(5.01.27699

An another example, consider steam at 6 bar and 320 º C. The steam is

superheat since the saturation temperature at 6 bar is 158.9 º C, but to find the specific

volume and enthalpy an interpolation is necessary:

i.e. )(300350

300320300,6350,6300,6320,6 CbaratCbaratCbaratCbarat


at 6 bar and 300 º C , is kg

m 3

4344.0 , and 350 º C is kg

m 3

4742.0

kg

m

kg

m

kg

mCbarat

333

320,645032.0)4344.04742.0(4.04344.0

Similarly; )(300350


hhhh


,

55

At 6 bar and 300 º C , is kg

Jkh 6.3061 , and 350 º C is

kg

mh

3

7.3165

kg

Jk

kg

Jk

kg

Jkh

Cbarat24.3103)6.30617.3165(4.06.3061

320,6

In same case, a double interpolation is necessary. For example, to find the enthalpy

of superheat steam at 21 bar and 432 º C, an interpolation between 20bar and 22 bars

is necessary, and an interpolation between 400 º C and 450 º C is also necessary.

First, find the enthalpy at 20 bars and 432 º C. i.e.

)(400450


hhhh


At 20 bar and 400 º C, is kg

Jkh 1.2818 , and 450 º C is

kg

mh

3

1.3060

kg

Jk

kg

Jk

kg

Jkh

Cbarat98.2972)1.28181.3060(

50

321.2818

432,20

Now find the enthalpy at 22 bars and 432 º C. I.e.

)(400450


hhhh


At 22 bar and 400 º C, is kg

Jkh 7.2733 , and 450 º C is

kg

mh

3

9.3015

,

56

kg

Jk

kg

Jk

kg

Jkh

Cbarat3.2914)7.27339.3015(

50

327.2733

432,22

Now interpolate between enthalpy at 20 bars and 432 º C and enthalpy at 22

bars and 432 º C in order to find enthalpy at 21 bar and 432 º C. i.e.

)(2022


hhhh

)98.29723.2914(5.098.2972432,21 kg

Jk

kg

Jkh

Cbarat

kg

Jk

kg

Jk

kg

Jkh

Cbarat64.294334.2998.2972

432,21

EXAMPLE . 4 kg of water is heated at a pressure of 220 kPa to produce a

mixture with quality x = 0.8. Determine the final volume occupied by

the mixture.

Solution:

Use Table C-2. To determine the appropriate numbers at 220 kPa we

linearly interpolate between 0.2 and 0.3 MPa. This provides, at 220 kPa,

kg

m

kg

m

kg

m

kg

mg

3333

8297.08857.0)8857.06058.0(200300

200220

,

57

kg

m

kg

mfg

33

0011.0,8297.0

Note that no interpolation is necessary for f , since for both pressures f is the

same to four decimal places.

Using )( fgf x , we now find:

kg

m

kg

m

kg

m

kg

m 3333

6640.0)0011.08297.0(8.00011.0

The total volume occupied by 4 kg is:

33

656.26640.04 mkg

mkgmV

.

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

13 – 15 Thermodynamics process undergoing at constant ( pressure , volume ,

temperature , enthalpy , entropy ) polytrophic process – with representation on

a ( P – V ) , ( T – S ) and ( P – H ) diagram .

The First Law Applied to Various Processes

,

58

1. The Constant – Temperature Process (Isothermal Process)

For the Isothermal process, the temperature of a system is maintained constant (T =

constant). Internal energy and enthalpy vary slightly with pressure. The energy equation

is:

UWQ

For a gas that approximates an ideal gas, the internal energy depends only on the temperature and thus ∆ U = 0 for an isothermal process; for such a process;

WQ

Using the ideal-gas equation mRTVP ,

the work for a quasiequilibrium process can be found to be:

2

1

1

2 lnln2

1

2

1P

PmRT

V

VmRT

V

VdmRTVdPW

V

V

V

V

Example 2. An ideal gas occupying a volume of 0.2 m ³ at a pressure of 1.5 MPa

expands isothermally in a quais – equilibrium process to a volume of

0.5 m ³. Find the final pressure, W, Q and Δ U.

Solution: A sketch of the system and a p – V diagram of the process is shown in

Figure.

mRTVPandmRTVP 2211

2211 VPVP

MPaMPaV

VPP 6.0

5.0

2.0)5.1(

2

112

Since the temperature remains constant, ΔU=0,

and the first law gives;

VdV

mRTVdPWQ

V

V

V

V

2

1

2

1

2121

1

211

1

2 lnlnV

VVP

V

VmRT

kJmkPa 89.2745.2ln)2.0()105.1( 33

2. The Constant – Volume Process (Isochoric Process)

1

2

•

•

T = const.

P

V

Isothermal process for an

ideal gas

1

2

•

•

PV = const.

P, MPa

V, m³ 0.5 0.2

1.5

,

59

Consider the transfer of heat to a system consisting of a fluid in a rigid vessel as shown in Figure.

Since 0. VdconstV

0 VdPW

If we assume that only internal energy comprises , for such a process the first law becomes;

UQ

Thus, the heat transfer to the system is accounted for as an increase in the internal energy of the system . Similarly , any heat transfer from the system represents an equivalent decrease in the internal energy of the system.

dTcmQ

T

T

2

1

Or , for a process for which c is essentially constant,

c ,is defined as the heat necessary to raise the temperature of one unit of

substance one degree in a constant – volume process.

Example 3. 1 kg of air (assumed ideal gas, R = 0.287 kJ / kg. ºk) is confined to a

constant – volume vessel. The volume and the initial pressure of the

air are 0.2 m³ and 350 kPa. If 120 kJ of heat are supplied to the gas,

the temperature increases to 411.5 ºk. Find

a) The work done,

b) Change of internal energy,

c) The specific heat of the gas at constant volume.

Solution: A sketch of the system and a p – V diagram of the process is shown in

Figure. a) The displacement work is zero since there is no change in volume .

0. VdPWconstV

b) First law ; kJUQ 120

c) The initial temperature may be Determined from the Ideal – Gas equation of

state;

1

2 •

•

P

V

Q

Gas Gas

W

TcmQ

1

2 •

•

P, kPa

V, m³

Q=120 kJ

Air m=1 kg V1=0.2 m³

P1=350 kPa

350

o.2

,

60

k

kkg

kJkg

mkPa

mR

VPTmRTVP

9.243

).

287.0()1(

)2.0()350( 3

111111

kkT 6.167)9.2435.411(

The average specific heat at constant volume is:

kkg

Jk

k

kg

kJ

T

uc avg

716.06.176

1

120

)(,

3.The Constant – Pressure Process (Isobaric Process)

A fluid contained at constant pressure by means of frictionless piston – cylinder assembly is shown in Figure.

UWQ

)( 12 VVPW

1212 )( UUVVPQ

HHHUPVUVPQ 121122 )()(

The first law, for a constant-pressure process is:

HQ

For a gas that behaves as an ideal gas, we have

dTcmQ

T

T

P2

1

For a process involving an ideal gas for which cp is constant there results

TcmQ p

Example 3. Air at temperature of 500 º C is compressed at a constant pressure

1 2 • •

P

V

Q

Gas

W

Gas

V1 V2

,

61

of 1.2 MPa from a volume of 2 m³ to a volume 0.4 m³. If the internal

energy decrease is 4820 kJ, find

a) The work done during the reversible compression,

b) The heat transferred,

c) The change of enthalpy,

d) The average specific heat constant pressure.

Solution: A sketch of the system and a p – V diagram of the process are shown in

Figure.

a) )( 12 VVPW

33 )24.0()102.1( mkPa

kJ1920

b) UUUWQ 122121

2121 WUQ

kJkJkJ 674019204820

c) Since the process is at constant pressure, then

kJHHQ 67401221

d) From the Ideal – gas equation of state 222111 mRTVPandmRTVP

ktTck

15.77315.27350015.273

11

km

mk

V

VTT

3.145)2(

)4.0()15.773(3

3

1

212

kg

kkkg

kJ

mkPa

RT

VPmmRTVP 816.10

)15.773().

287.0(

)2()1200( 3

1

11111

kkg

Jk

kkkg

kJ

Tm

Hc PavgP

007.1)15.7733.154()816.10(

6740)(,

1 2 • •

P,MPa

V,m³

Q

Air

0.4 2

1.2 T1=500ºC

P=1.2 MPa

=const V1=0.2m³

,

62

2. The Adiabatic Process

In an adiabatic process, no heat interaction between the system and the surroundings. The first law then becomes;

dUWQUdWQ 0,

Which , when integrated , is ;

1221 UUW

Therefore, the work done on the system is equal to the change in internal energy of the system. For an ideal gas with constant specific heats when undergoes a reversible adiabatic process, the preceding equation per unit mass can be written in differential form as:

TdcdRT

orTdcdPudW

Rearranging, we have

dR

T

Tdc

This is integrated, assuming constant c between states 1 and 2 to give;

c

R

T

T

c

R

T

T

2

1

1

2

2

1

1

2 lnln …… (1)

The temperature ratio may also be expressed in terms of the pressure ratio. From the ideal – gas equation, the volume ratio can be replaced so that;

)(

1

2

21

12

2

1

1

2

cR

R

c

R

c

R

P

P

TP

TP

T

T

….. (2)

In the case of an ideal gas , RccP . Combining equations (1) and (2),

the pressure – volume relationship in an adiabatic process is:

Pc

c

P

P

2

1

1

2

…… (3)

,

63

Referring to ,c

ck P

1,

1

k

Rcand

k

kRcP and using

the ideal – gas law, equations (2) and (3) can be written as:

kk

k

P

P

P

P

T

T

2

1

1

2

)1(

1

2

1

2 ,

…… (4)

Finally, the above three relations can be put in general forms, without reference to particular points. For the adiabatic process involving an ideal gas with constant

candcP we have:

.,.,.

)1(

1 constPconstPTconstT kk

k

k

…… (5)

The Polytropic Process

A polytropic process is a real expansion or compression process in which the

relationship between P and V is given by

.constVP n ….. (6)

Where n is the index of the process.

The work is calculated

VdVVPVdPW

V

V

V

V

nn2

1

2

1

11

n

VPVPVV

n

VP nnn

1)(

1

1221

1

1

211

….. (7)

Except 2

1

1

2 lnln2

1

2

1P

PmRT

V

VmRT

V

VdmRTVdPW

V

V

V

V

is used if n = 1 .

The heat transfer follows from the first law. Each process is associated with a

particular value for n as follows:

Isothermal : n = l

Constant – Volume : n =

Constant – pressure : n = 0

Adiabatic : n = k

,

64

The processes are displayed on a (In P )vs. (In V )plot in Figure below.

The slope of each straight line is the exponent on V in equation (6). If the slope

is none of the values , k , 1 , or zero, then the process can be referred to as a

polytropic process. For such a process any of the equations (3) , (4) ,or (5) can be

used with k simply replaced by n ; this is convenient in processes in which there is some

heat transfer but which do not maintain temperature, pressure, or volume constant.

Example 4. Determine the heat transfer necessary to increase the pressure of

70 percent quality steam from 200 to 800 kPa, maintaining the

volume constant at 2 m3. Assume a quasiequilibrium process.

Solution:

For the constant-volume quasiequilibrium process, the work is zero.

The first law reduces to:

)( 12 uumQ .

The mass is found to be;

Vm , )( fgf x

From the steam table C – 2, for pressure 200 kPa ( 0.2 MPa ) ,

kg

mand

kg

mgf

33

8857.00011.0001061.0 , then

kg

m

kg

m

kg

m 333

6203.0)0011.08857.0()7.0(0011.0

,

65

kg

kg

m

mVm 224.3

6203.0

23

3

The internal energy at state one is:

)(1 fgf uuxuu

From the steam table C – 2, for pressure 200 kPa ( 0.2 MPa ) ,

kg

Jkuand

kg

Jku gf 5.25295.504 , then

kg

Jk

kg

Jk

kg

Jku 1922)5.5045.2529()7.0(5.5041

The constant – volume process demands that kg

m3

12 6203.0 .

Now in the steam tables at 800 kPa (0.8 MPa)we look for the value of internal

energy which corresponds to the value of specific heat kg

m3

6203.0 .We find

that the approximated values of specific heat is in the table C – 3 (properties of

superheated steam) at o.8 MPa between 700 º C and 800 º C , so we can

interpolate to find the internal energy at state 2:

At 0.8 MPa , 800 ºC , kg

Jkuand

kg

m1.36616181.0

3

At 0.8 MPa, 700 ºC , kg

Jkuand

kg

m2.34765601.0

3

)(700,8.0800,8.0

700,8.0800,8.0

800,8.0

800,8.02 CMPaatCMPAat

MPaatCMPaat

CMPAat

CMPaatuu

Cuu

kg

Jk

kg

Jk

kg

Jku 3668)2.34761.3661(

5601.06181.0

6181.06203.01.36612

Note that extrapolation was necessary since the temperature at state 2 exceeds

the highest tabulated temperature of 800°C. The heat transfer is then:

Jkkg

JkkgQ 5629)19223668()224.3(

,

66

ENTHALPY

The change in the energy of the system is equal to the heat and work interaction with the surroundings .Thus,

EdWQ ….. (1)

Where; δQ – is the net heat transfer to the system .

δW – is the net work done on the system.

dE – is the difference between the final and the initial energy of

the system .

One such combination of properties can be demonstrated by considering the addition of heat to the constant-pressure situation shown in Figure below.

Heat is added slowly to the system (the gas in the cylinder), which is maintained at constant pressure by assuming a frictionless seal between the piston and the cylinder. If

the kinetic energy K.E changes and potential energy P.E changes of the system are

neglected and all other work modes are absent,then the energy comprises only internal energy U, the foregoing equation becomes:

12 UUWQ ….. (2)

The work done raising the weight for the constant-pressure process is given by:

)( 12 VVPW

Substituting in equation (1), we obtain:

Constant – Pressure Heat addition

,

67

121212 )()()( VPUVPUQUUVVPQ ….. (3)

The quantity in parentheses is a combination of properties and is thus a property

itself. It is called the enthalpy H of the system; that is,

H = U + P V ….. (4)

The specific enthalpy h is found by dividing by the mass. It is:

h = u + P v ….. (5)

Enthalpy is a property of a system and is also found in the steam tables.

If a system undergoes a constant pressure process , and if the only work involved is the flow work then the energy equation can be written pressure as:

12211221 hhqorHHQ ….. (6)

The heat transfer to the system is thus equal to the change of enthalpy . Since;

)( 122121

2

1

TTcTdcTdchTdcq P

T

T

PPP ….. (7)

PPT

hc )(

– under conditions of constant pressure and flow work .

If the enthalpy is a function of T and P, then the change in it is:

PdP

hTdcdP

P

hTd

T

hhd TPTP )()()(

…. (8)

Because only changes in enthalpy or internal energy are important, we can arbitrarily

choose the datum from which to measure h and U. We choose saturated liquid at 0°C to

be the datum point for water substance.

,

68

Win (– ev)

Wout (+ ev)

Week

NO

Details

8 - 9 First law of thermodynamic – flow system – nun flow system – steady – un

steady – open – closed

The First Law of Thermodynamics

The first law of thermodynamics is commonly called the law of conservation of energy. The study of conservation of energy emphasizes changes in kinetic and potential energy and their relationship to work. A more general form of conservation of energy includes the effects of heat transfer and internal energy

changes. This more general form is usually called the first law of thermodynamics. Other forms of energy may also be included, such as electrostatic, magnetic, strain, and surface energy.

THE FIRST LAW OF THERMODYNAMICS APPLIED TO A CYCLE

Consider a closed system that can do work on the surroundings or have work done on it, and that can have heat transfer both to and from the surroundings as shown in Figure. The necessary measurements of work and heat can be made entirely outside the system. For example, the standard system used for measure heat might be quantities of water brought into contact with the system ,and thus, if the system rejects heat, the water temperature would rise. Conversely, if heat is transferred to the system, The temperature of water would drop. Likewise, Work might be measured by counting the number of standard weights lifted a specified distance. Such measurements of heat and work for cycle of a closed system result in a plot as in Figure below.

Q re j (Q out)

Q rej (Q out)

Q add (Q in)

Standard

System

for work

(weights)

Weight

Standard

System of

heat

Gas

Work

P

V

•

•

1

2

Win (- ev)

Wout (+ev) Qrej (Qout)

Qadd (Qin)

Scale

• •

•

•

W

Q

,

69

This plot shows that the net work of the

cycle W is proportional to the net heat

transfer of the cycle Q .Where the symbol

implies an integration around a

complete cycle. We conclude that for any closed system:

QW

Thus, netnet QW

However, when a system undergoes a thermodynamic cycle, then the net

heat transfer to the system from the surroundings is proportional to the net work

done by the system on the surroundings. This is a statement of the first law of

thermodynamics.

THE FIRST LAW APPLIED TO A PROCESS

The first law of thermodynamics is often applied to a process as the system changes from one state to another. Realizing that a cycle results when a system undergoes several processes and returns to the initial state, we could consider a cycle composed of the two processes represented by A and B in Figure below. The change in the energy of the system is equal to the heat and work interaction with surroundings. Thus,

WQEd …. (1)

Where Q is the net heat transfer to the

System , W is the net work done on the system,

And Ed is the difference between the final and

the initial energy of the system .

Equation (1) can be integrated to yield;

P

V

•

•

1

2

A

B

,

70

212112 WQEE ….. (2)

Where Q1 - 2 is the heat transferred to the system during the process from

state 1 to state 2, W1 - 2 is the work done by the system on the surroundings during

the process .The symbol E represents all of the energy: kinetic energy K.E,

potential energy P.E, and internal energy U, which includes chemical energy and the energy associated with the atom. Any other form of energy is also included in

the total energy E. Its associated intensive property is designated e . The first law of thermodynamics then takes the form:

1212122121 ... UUPEEPEKEKWQ

121212 )()(2

UUZZmgm

….. (3)

If we apply the first law to an isolated system, one for which;

02121 WQ

The first law becomes the conservation of energy; that is,

12 EE ….. (4)

The internal energy U is an extensive property. Its associated intensive

property is the specific internal energy u; that is,

m

Uu .

For simple systems in equilibrium, only two properties are necessary to establish the state of a pure substance, such as air or steam. Since internal energy is a property, it depends only on, say, pressure and temperature; or, for saturated steam, it depends on quality and temperature (or pressure). Its value for a particular quality would be ;

)( fgf uuxuu ….. (5)

We can now apply the first law to systems involving working fluids with tabulated property values. Before we apply the first law to systems involving substances such as ideal gases or solids, it is convenient to introduce several additional properties that will simplify that task

Properties That Depend On the First Law

,

71

For a simple system, only two independent intensive properties define the state of the system. Thus, the internal energy and enthalpy are related to other properties, so that;

),(,),(,),(,),( PfhTPfhTPfuTfu

We can consider the specific internal energy to be a function of temperature and specific volume; that is

),( Tfu

Using the chain rule to express the partial derivatives as;

du

dTT

udu

T

….. (1)

Since Tandu ,, are all properties, the partial derivative is also a property and is

called the constant-volume specific heat c ; that is,

T

uc

…… (2)

One of the classical experiments of thermodynamics, first performed by

Joule in 1843, is illustrated in Figure below. Pressurize volume A with an ideal gas and evacuate volume B.

After equilibrium is attained, open the valve. Even though the pressure and volume of the ideal gas have changed markedly, the temperature does not change. Because there is no change in temperature, there is no net heat transfer to the water. Observing that no work is done we conclude, from the first law that the internal energy of an ideal gas does not depend on pressure or volume. For such a gas, which behaves as an ideal gas, we have ;

,

72

0

T

u

…… (3)

Combining (1) , (2) , and (3) we obtain;

dTcdu ….. (4)

This can be integrated to give

dTcuuT

T

2

112 …… (5)

For a known )(Tc this can be integrated to find the change in internal energy

over any temperature interval for an ideal gas. Likewise, considering specific enthalpy to be dependent on the two

variables T and P , we have;

dPP

hdT

T

hdh

TP

…… (6)

The constant-pressure specific heat Pc , is defined as;

P

PT

hc

…… (7)

For an ideal gas we have, returning to the definition of enthalpy;

Puh ….. (8)

The Ideal – Gas Equation of State is:

RTPorRconstT

P

.

Thus RTuPuh …… (9)

Since U is only a function of T , we see that h is also only a function of T for

an ideal gas. Hence, for an ideal gas

,

73

0

TP

h …… (10)

and we have, from (6),

dTchd P …… (11)

Over the temperature range Tl to T2 this is integrated to give

dTchh P

T

T2

112 …… (12) for an ideal gas.

It is often convenient to specify specific heats on a per-mole, rather than a

per-unit-mass, basis; these molar specific heats are notated Pcandc . Clearly,

we have the relations;

PP cMcandcMc

where M is the molar mass. Thus values of Pcandc may be simply derived

from the values of Pcandc and listed in Table B-2.

The equation for enthalpy can be used to relate, for an ideal gas, the specific

heats and the gas constant.

In differential form, equation (9) takes the form;

)( Pdudhd

Introducing the specific heat relations and the ideal-gas equation, we have

dTRTdcTdcP

This, after dividing by dT, gives;

RccP …… (13)

This relationship-or its molar equivalent RccP allows c to be

determined from tabulated values or expressions for Pc . Note that the difference

,

74

between Pcandc for an ideal gas is always a constant, even though both are

functions of temperature.

The specific heat ratio k is also a property of particular interest; it is defined as

c

ck P …… (14)

This can be substituted into (13) to give

11

k

Rcor

k

kRcP …… (15)

Obviously, since R is a constant for an ideal gas, the specific heat ratio (k) will depend only on temperature.

For gases, the specific heats slowly increase with increasing temperature.

Since they do not vary significantly over fairly large temperature differences, it is

often acceptable to treat Pcandc , as constants. For such situations there

results;

)( 1212 TTcuu …… (16)

)( 1212 TTchh P …… (17)

For air we will use )1

171.0(717.0Rbm

Btu

Ckg

Jkc

and

)1

24.0(00.1Rbm

Btu

Ckg

JkcP , unless otherwise stated. For more accurate

calculations with air, or other gases, one should consult ideal-gas tables, such as

those in Appendix F, which tabulate )()( TuandTh , or integrate using

expressions for )(TcP found in Table B-5.

For liquids and solids the specific heat Pc is tabulated in Table B-4. Since it

is quite difficult to maintain constant volume while the temperature is changing,

c values are usually not tabulated for liquids and solids; the difference ccP

is usually quite small. For most liquids, the specific heat is relatively insensitive to temperature change. For water we will use the nominal value of

)1

00.1(18.4Rbm

Btu

Ckg

JkcP . For ice the specific heat in kJ/kg.ºC is

,

75

approximately TcP 0069.01.2 , where T is measured in ºC;and in English

units of Btu / lbm.OF it is TcP 001.047.0 ,where T is measured in O F . The

variation of specific heat with pressure is usually quite slight except for special situations.

Example1. The specific heat of superheated steam at approximately 150 kPa can

be determined by the equation ;

Ckg

JkTcP .1480

40007.2

(a) What is the enthalpy change between 300 ºC and 700 ºC for 3 kg of

steam? Compare with the steam tables.

(b) What is the average value of Pc between 300 ºC and 700 ºC based on

the equation and based on the tabulated data?

Solution:

(a) The enthalpy change is found to be

JkTdT

TdcmH

T

T

P 2565)1480

40007.2(3

700

300

2

1

From the tables C – 3 we find, using P = 150 kPa ( 0.15 MPa ) ,that

for T = 300 ºC , h1 = 3073.0 kJ ,and for T = 700 º C , h2 = 3927.9 kJ,

ΔH = m ( h2 – h1 ) = (3 kg)( 3928 – 3073) kJ/ kg = 2565 kJ

(b) The average value avgPc , is found by using the relation;

TdT

corTdcmTcm avgP

T

T

PavgP )1480

4007.2()400()3(

700

300

,,

2

1

The integral was evaluated in part (a) ,hence, we have

Ckg

Jk

Ckg

Jkc avgP .

14.2)400()3(

2565,

Using the values from the steam table, we have

Ckg

Jk

C

kg

Jk

T

hc avgP .

14.2400

)30733928(

,

Because the steam tables give the same values as the linear equation of this

example, we can safely assume that the )(TcP relationship for steam over this

temperature range is closely approximated by a linear relation. This linear relation

,

76

would change, however, for each pressure chosen; hence, the steam tables are essential.

EXAMPLE. Frictionless piston is used to provide a constant pressure of 400 kPa

in

a cylinder containing steam originally at 200°C with a volume of 2m3.

Calculate the final temperature if 3500 kJ of heat is added, and then use the concept of enthalpy to solve the problem .

Solution:

using 12 UUWQ ….. (1)

The work done during the motion of the piston is:

)(400)( 1212 VVkPaVVPW

The mass before and after remains unchanged. Using the steam tables, this is expressed as:

kg

kg

m

mVm 744.3

5342.0

23

3

1

1

The volume V2 is written as: 222 .744.3 kgmV

Finding U , from the steam tables, and substituting into equation (1) ;

kgJkumkgPakJk 744.3)2647()2.744.3(4003500 2

3

2

This requires a trial-and-error process. One plan for obtaining a solution is to

guess a value for 2 and calculate u2 from the equation above. If this value

checks with the u2 from the steam tables at the same temperature, then the guess is the correct one.

For example, guess 2 = 1.0 m3/kg. Then the equation gives u2 = 3395 kJ/kg.

From the steam tables, with P = 0.4 MPa, the u2 value allows us to interpolate

T2= 654°C and the 2 gives T2 = 600°C. Therefore, the guess must be revised. Try

2 = 1.06 m3/kg. The equation gives u2 = 3372 kJ/kg.The tables are interpolated to

give T2 = 640°C; for 2 , T2= 647°C. The actual 2 is a little less than 1.06 m3/kg,

with the final temperature being approximately

,

77

T2 = 644°C . Or by using the concept of enthalpy we can solve the problem , The energy equation for a constant-pressure process is (with the subscript on the heat transfer omitted);

mkg

kJhkJorHHQ )2860(3500 21221

Using the steam tables , the mass is:

kg

kg

m

mVm 744.3

5342.0

23

3

Thus, kg

kJ

kg

kJ

kg

kJh 37952860

744.3

35002

From the steam tables this interpolates to:

CCCT 641)100(224

6.926002

Obviously, enthalpy was very useful in solving this constant-pressure problem, Trial and error was unnecessary, and the solution was rather straightforward. We illustrated that the quantity we made up, enthalpy, is not necessary, but it is quite handy. We will use it often in our calculations.

Application of the First Law to a Closed System Usually, the first law of thermodynamics applies to all energy interactions between a system and its surroundings. The term closed system where there are no effects of gravity and motion ( no move ) , there is no change in the kinetic or potential energy and the total change in the stored energy of the system is the change in the internal energy U , or ;

EWQ

The property E represents all of the energy; Kinetic KE, Potential Energy PE,

and Internal Energy U, which includes chemical energy and the energy associated

with the atom. Any other form of energy is also included in the total energy E.

UPEKEWQ

1212

2

1

2

2 )()(2

UUzzmgm

WQ

Thus, the first law applied to a closed system under these conditions becomes;

,

78

(Mass entering control volume) m1

(Mass leaving control volume) m2

(Change in mass within

control volume) Δm c.v

)( 1212 uumUUWQorWQUE ….. (1)

Example1. Radon gas initially at 65 kPa , 200 º C, is to be heated in a

Closed , rigid container until it is at 400 º C. The mass of the

Radon is 0.393 kg. A table of properties shows that at 200 º C,

the internal energy of Radon is 26.6 kJ / kg , at 400 º C it is

37.8 kJ / kg . Determine the amount of heat required.

Solution:

Since the container is rigid then the work done is 0W , the

first law of thermodynamics is then ;

)(0, 12 uumQWUWQ

Jkkg

JkkgQ 4.4)6.268.37(393.0

Application of the First Law to an Open System

The Continuity Equation

Consider a general control volume (open system) with an area A1 , where fluid

enters and an area A2 , where fluid leaves, as shown in Figure. It could have any shape

and any number of entering and exiting areas, but we will derive the continuity equation using the geometry shown. Conservation of mass requires that:

– =

V1

V2

,

79

The mass that crosses an area A over a time increment Δt can be expressed as

ρAVΔt, where νΔt is the distance the mass particles travel and AνΔt is the volume

swept out by the mass particles. Equation above can thus be put in the form:

vcmtAtA .22221111 …… (1)

Where the velocities vI and v2 are perpendicular to the areas A I and A2

respectively. We have assumed the velocity and density to be uniform over the two areas; a good assumption for the turbulent flows most often encountered entering and leaving the devices of interest.

If we divide by Δt and let Δt→0, the derivative results and we have the continuity

equation,

td

mAA

vc .

222111

…… (2)

For the steady-flow situation, in which the mass in the control volume remains constant, the continuity equation reduces to:

222111 AA …… (3)

Which will find much use in problems involving flow into and from various devices.

The quantity of mass crossing an area each second is termed the mass flux m and has

units’ kg/s (lbm/sec). It is given by the expression

Am ……. (4)

The quantity Av is the flow-rate with units of m3/ s (ft3/sec).If the velocity and density

are not uniform over the entering and exiting areas, the variation across the areas must be accounted for. This is done by recognizing that the mass flowing through a differential area element dA each second is given by ρvdA, providing v is normal to dA. In this case,

equation (4) is replaced by:

A

Adm

Observe that for incompressible flow ( ρ = constant), equation (4) holds whatever

the velocity distribution, provided only that v be interpreted as the average normal

velocity over the area A .

,

80

Net energy

transferred +

to the c.v

Q - W

Energy

entering –

the c.v

E 1

Energy

leaving = the c.v

E 2

Change of

energy in

the c.v

Δ EC . V

EXAMPLE. Water is flowing in a pipe that changes diameter from 20 to 40 mm.

If the water in the 20-mm section has a velocity of 40 m/s,

determine the velocity in the 40-mm section. Also calculate the

mass flux.

Solution:

The continuity equation 222111 AA is used. There

results, using ρ1 = ρ2 ,

2

22

22114

)04.0()40(

4

)02.0(v

m

s

mmAA

s

mv 102

The mass flux is found to be:

s

kg

s

mm

m

kgAm 57.12)40(

4

)02.0()1000(

2

311

The Energy Equation

Consider again a general control volume as sketched in Figure. The first law of thermodynamics for this control volume can be stated as:

,

81

The work W is composed of two parts: the work due to the pressure needed to

move the fluid, which is called flow work, and the work that results from a rotating shaft,

which is called shaft work Ws ,This is expressed as:

sWtAtAPW 111222 ….. (1)

where PA is the pressure force and v Δt is the distance it moves during the time

increment Δ t . The negative sign results because the work done on the system is

negative when moving the fluid into the control volume.

The energy E is composed of kinetic energy K.E, potential energy P.E , and

internal energy U. Thus,

ummgzvmE 2

2

1

The first law can now be written as:

vcs Etuzgv

vAtugzv

vAtAPtAPWQ .22

2

222211

2

1111111222 )

2()

2(

Divide through by Δt to obtain the energy equation,

td

EdPuzg

vm

Pugz

vmWQ vc

s.

1

111

2

11

2

222

2

22 )

2()

2(

….. (2)

where we have used;

vAmt

WW

t

QQ s

,,

…… (3)

For steady flow, a very common situation, the energy equation becomes:

])(2

1)([ 2

1

2

21212 vvzzghhmWQ s …… (4)

Where the enthalpy puh has been introduced. Most often used this

form when a gas or a vapor is flowing.

Quite often, the kinetic energy and potential energy changes are negligible.

The first law then takes the simplified form:

,

82

1212 )( hhwqorhhmWQ ss …… (5)

where m

Qq

and

m

Ww s

. This simplified form of the energy equation

has a surprisingly large number of applications.

For a control volume through which a liquid flows, it is most convenient to return

to .

td

EdPuzg

vm

Pugz

vmWQ vc

s.

1

111

2

11

2

222

2

22 )

2()

2(

For a steady flow with ρ2 = ρ 1 = ρ, neglecting the heat transfer and changes in

internal energy, the energy equation takes the form:

)]()2

()([ 12

2

1

2

2122 zzg

vvPPmWs

…… (6)

This is the form to use for a pump or a hydroturbine. If uandQ are not

zero, simply include them. -------------------------------------------------------------------------------------------------------------------

Week No Details

10 – 11 - 12 Application of the first law on – nozzle , diffuser , condenser , evaporator ,

compressor , heat exchanger ( surface , open ) , turbine , boiler .

APPLICATIONS OF THE ENERGY EQUATION

,

83

There are several points that must be considered in the analysis of most problems in which the energy equation is used. As a first step, it is very important to identify the control volume selected in the solution of the problems; dotted lines are used to outline the control surface. If possible, the control surface should be chosen so that the flow variables are uniform or known functions over the areas where the fluid enters or exits the control volume. For example, in Figure below the area could be chosen as in part (a) , but

the velocity and the pressure are certainly not uniform over the area.

In part (b), however, the control surface is chosen downstream from the abrupt

area change that the exiting velocity and pressure can be approximated by uniform distributions. It is also necessary to specify the process by which the flow variables change. Is it

incompressible? Isothermal? Constant-pressure? Adiabatic? . A sketch of the process

on a suitable diagram is often of use in the calculations. If the working substance behaves as an ideal gas, then the appropriate equations may be used; if not, tabulated values must be used, such as those provided for steam. For real gases that do not behave as ideal gases, specialized equations may be available for calculations . Often heat transfer from a device or an internal energy change across a device, such as flow through a pump, is not desired. For such situations, the heat transfer and

internal energy change may be lumped together as losses. In a pipeline losses occur

because of friction; in a pump, losses occur because of poor fluid motion around the rotating blades. For many devices the losses are included as an efficiency of the device. Kinetic energy or potential energy changes can often be neglected in comparison with other terms in the energy equation. Potential energy changes are usually included only in situations where liquid is involved and where the inlet and exit areas are separated by a large vertical distance. The following applications will illustrate many of the above points.

Throttling Devices

A throttling device involves a steady-flow adiabatic process that provides a

pressure drop with no significant potential energy or kinetic energy changes. The

,

84

process occurs relatively rapidly, with the result that negligible heat transfer occurs. Two such devices are sketched in Figure below.

If the energy equation is applied to such a device, obviously there is no work done; neglecting kinetic and potential energy changes, we have, for the adiabatic process:

1212 0, hhwqhhwq ss …… (1)

Where section 1 is upstream and section 2 is downstream. Most valves are throttling

devices, for which the energy equation takes the form of;

12 hh …… (2)

They are also used in many refrigeration units in which the sudden drop in pressure causes a change in phase of the working substance. The throttling process is analogous to the sudden expansion of Figure below.

In part (a) work is obviously crossing the boundary of the system by means of the

rotating shaft; yet the volume does not change. We could calculate the work input by

multiplying the weight by the distance it dropped, neglecting friction in the pulley system.

This would not, however, be equal to dVP which is zero. The paddle wheel provides

us with a nonequilibrium work mode.

Suppose the membrane in part (b) ruptures, allowing the gas to expand and fill the

evacuated volume. There is no resistance to the expansion of the gas at the moving

,

85

boundary as the gas fills the volume; hence, there is no work done. Yet there is a

change in volume. The sudden expansion is a nonequilibrium process, and again we

cannot use dVP to calculate the work.

EXAMPLE. Steam enters a throttling valve at 8000 kPa and 300°C and leaves at a

pressure of 1600 kPa .Determine the final temperature and specific

volume of the steam.

Solution:

The enthalpy of the steam as it enters is found from the superheat steam

table , at 8000 kPa ( 8 MPa ) and 300 º C to be h 1 = 2785 kJ/kg. This

must equal the exiting enthalpy as demanded by :

12 hh

The exiting steam is in the quality region, since form the steam table C – 2 at

1600 kPa , h g = 2794 kJ/kg. Thus the final temperature is T2 = 201.4ºC.

To find the specific volume we must know the quality. It is found from:

kg

Jkhhhxhh gfgf 2794, 222

From the steam table C - 2 kg

Jkh

kg

Jkh fgf 2.1935,8.858

995.02.19358.8582794 22 xkg

Jkx

kg

Jk

kg

Jk

The specific volume is then; )(22 fgf x , From the steam table C

– 3 , at 1600 kPa ( 1.6 MPa ) , kg

m

kg

mgf

33

1238.0,001159.0

kg

m

kg

m

kg

m 333

2 1232.0)0012.01238.0()995.0(001159.0

Compressors, Pumps, and Turbines

A pump is a device which transfers energy to a liquid flowing through the pump

with the result that the pressure is increased.

,

86

Compressors and blowers also fall into this category but have the primary

purpose of increasing the pressure in a gas.

A turbine, on the other hand, is a device in which work is done by the fluid on a set

of rotating blades. As a result there is a pressure drop from the inlet to the outlet of the turbine. In some situations there may be heat transferred from the device to the surroundings, but often the heat transfer can be assumed negligible. In addition the kinetic and potential energy changes are usually neglected. For such devices operating in a steady-state mode the energy equation takes the form;

0,)( 12 QhhmWQ s

1212 )( hhworhhmW s ..… (1)

where sW is negative for a compressor and positive for a gas or steam turbine. In

the event that heat transfer does occur, from perhaps a high-temperature working

fluid, it must, of course be included in the above equation.

For liquids, such as water, the energy equation;

)]()2

()([ 12

2

1

2

2122 zzg

vvPPmWs

…… (2)

neglecting kinetic and potential energy changes, becomes;

)( 12

PPWs

….. (3)

EXAMPLE. Steam enters a turbine at 4000 kPa and 500 ºC and leaves as shown

in Figure. For an inlet velocity of 200 m/s calculate the turbine power

output. (a) Neglect any heat transfer and kinetic energy change.

(b) Show that the kinetic energy change is negligible.

,

87

Solution:

(a) The energy equation in the form: )( 12 hhmWT is used.

We find m as follows: 111 vAmvAm

1

1

1

we find 1 from the steam table C – 3 ( properties of superheated steam )

, at 4000 kPa ( 4 MPa ) and 500 º C is 0.08643 m³ / kg , then;

s

kg

s

mm

m

kgm 541.4)200()

2

05.0()57.11( 2

3

Enthalpies can be found from the steam table C – 2 and C – 3 as:

From the steam table C – 3 , at 4000 KPa ( 4 MPa ) and 500 º C :

kg

Jkh 2.34451

From the steam table C – 2 , at 80 KPa ( 0.08 MPa ): kg

Jkh 7.26652

The maximum power output is :

WMrWks

Jk

kg

Jk

s

kgWT 542.30)(3542)2.34457.2665()544.4(

(b) The exiting velocity is found from the continuity equation:

331 57.11

08643.0

1

m

kg

kg

m

,

88

22

111

2222111A

vAvvAvA

2

2

1

from the steam table C – 2 , at 80 kPa ( 0.08 MPa ),

kg

mg

3

2 087.2)(

332 479.0

087.2

1

m

kg

kg

m

s

m

mm

kgs

mm

m

kg

v 23.193

)2

250.0()479.0(

)200()025.0()57.11(

2

3

2

3

2

The kinetic energy change is:

2

)200()193(

544.42

.

222

1

2

2 s

m

s

m

s

kgvvmEK

s

Jk

s

J250.66250

This is less than 0.1 percent of the enthalpy change and is indeed negligible. Kinetic

energy changes are usually omitted in the analysis of a turbine.

Nozzles and Diffusers

,

89

A nozzle is a device that is used to increase the velocity of a flowing fluid. It does

this by reducing the pressure.

A diffuser is a device that increases the pressure in a flowing fluid by reducing the

velocity. There is no work input into the devices and usually negligible heat transfer.

With the additional assumptions of negligible internal energy and potential energy changes, the energy equation takes the form:

12

2

1

2

2

220 hh

vv …… (1)

Based on our intuition we expect a nozzle to have a decreasing area in the

direction of flow and a diffuser to have an increasing area in the direction of flow

(Figure (a)).This is indeed the case for a subsonic flow in which TRkv

For a supersonic flow in which TRkv the opposite is true: a nozzle has an

increasing area and a diffuser has a decreasing (Figure (b)).

Three equations may be used for nozzle and diffuser flow; energy, continuity, and

a process equation, such as for an adiabatic quasiequilibrium flow. Thus, we may

have three unknowns at the exit, given the entering conditions. There may also be shock

waves in supersonic flows or “choked” subsonic flows.

EXAMPLE. Air flows through the supersonic nozzle shown in Figure below . The

inlet conditions are 7 kPa and 420°C. The nozzle exit diameter is

,

90

adjusted such that the exiting velocity is 700 m/s.Calculate:

(a) the exit temperature,

(b) the mass flux, and

(c) the exit diameter.

Assume an adiabatic quasiequilibrium flow. Solution: From the Figure, we have that:

We then have, using cP = 1000 J/kg.K:

CC

kkg

Js

m

s

m

Tc

vvT

P

255420

.10002

)700()400(

2

22

1

2

2

2

12

(b) To find the mass flux we must find the density at the entrance. From the inlet

conditions we have:

1

1

1TR

p

TR

pTR

pTRp

kCtkT 693273420273)()( 11

kkg

mN

kkg

JRmNJ

m

NPakPa Air

287287,,700070007

2

3

2

1

1

1 03520.0

)693().

287(

7000

m

kg

kkkg

mNm

N

TR

P

(a) To find the exit temperature the energy equation (4.72) is used. It is, using:

2

2

21

2

1

22h

vh

v

2

2

21

2

1

22Tc

vTc

vPP

,

91

The mass flux is then:

s

kg

s

mm

m

kgvAmvAm 4423.0)400()1.0(0352.0 2

3111

(c) To find the exit diameter we would use the continuity equation:

22

112

1

2

22

22

21

21

1222111 )2

()2

(v

vddv

dv

dvAvA

This requires the density at the exit. It is found by assuming adiabatic quasiequilibrium flow. Referring to:

1

1

1

2

1

21

22

1

11

1 )(1

,. kkkk

T

TTTconstT

kCtkT 528273255273)()( 22

3

)14.1(

1

3

1

1

1

2

12 01784.0693

528)0352.0()(

m

kg

k

k

m

kg

T

Tk

mmmmv

vdd 212212.0

70001784.0

4000352.02.0

22

11

12

Heat Exchangers

An important device that has many applications in engineering is the heat

exchanger. Heat exchangers are used to transfer energy from a hot body to a colder

body or to the surroundings by means of heat transfer. Energy is transferred from the

hot gases after combustion in a power plant to the water in the pipes of the boiler and from the hot water that leaves an automobile engine to the atmosphere, and electrical generators are cooled by water flowing through internal flow passages. Many heat exchangers utilize a flow passage into which a fluid enters and from which the fluid exits

at a different temperature. The velocity does not normally change, the pressure drop

through the passage is usually neglected, and the potential energy change is

assumed zero. The energy equation then results in:

,

92

)( 12 hhmQ ….. (1)

since no work occurs in the heat exchanger. Energy may be exchanged between two moving fluids. The energy equation, applied to the control volume including the combined unit, which is assumed to be insulated, as shown schematically in Figure below , would be;

)()(0 1212 BBBAAA hhmhhm ….. (2)

The energy that leaves fluid A is transferred to fluid B by means of the heat

transfer Q. For the control volumes shown in Figure below, we have;

)(,)( 1212 AAABBB hhmQhhmQ ….. (3)

EXAMPLE. Liquid sodium, flowing at 100 kg/s, enters a heat exchanger at 450°C

and exits at 350°C . The specific heat of sodium is 1.25 kJ/ kg ·ºC.

Combined unit

Separated control volumes

,

93

Water enters at 5000 kPa and 20 ºC.Determine the minimum mass

flux of the water that the water does not completely vaporize. Neglect the pressure drop through the exchanger. Also, calculate the rate of heat transfer.

Solution:

The energy equation:

)()( 1221 wwwsss hhmhhm

)()( 1221 wwwssps hhcmTTcm

Using the given values, and from the steam table C – 4 ( compessed liquid )

, at 20 º C , we find kg

Jkhw 65.881 . Since the value of 2wh at 5000 kPa

( 5 MPa ) is not tabulated in the steam table C – 2 ( properties of saturated

water – pressure table ) , we have to interpolate between 4 and 6 MPa as:

)(

46

454645,2 baratbaratbaratbaratw hhhh

from the steam table C – 2 , at 4 MPa , kg

Jkhw 4.28012 , and at 6 MPa

kg

Jkhw 3.27842 , then;

kg

Jkh baratw 85.2792)4.28013.2784(

46

454.28015,2

Now substitute in Energy equation to find mass flux as:

)65.8885.2792()350450()25.1()100(kg

JkmC

kg

Jk

s

kgw

s

kgmw 623.4

where we have assumed a saturated vapor state for the exiting steam to obtain the maximum allowable exiting enthalpy. The heat transfer is found using the energy equation applied to one of the separate control volumes.

)( 12 www hhmQ

)7.8885.2792(623.4kg

Jk

s

kgQ

,

94

MWkWQ 5.1228.12501

Week No

Details

17 – 19 Carnot power cycle – Carnot revisable cycle ( refrigeration and Heat pump

applications .

THE CARNOT ENGINE

The heat engine that operates the most efficiently between a high-temperature

reservoir and a low-temperature reservoir is the Camot engine. It is an ideal engine that

uses reversible processes to form its cycle of operation; thus it is also called a reversible

engine .

,

95

The Carnot engine is very useful, since its efficiency establishes the maximum

possible efficiency of any real engine. If the efficiency of a real engine is significantly

lower than the efficiency of a Carnot engine operating between the same limits, then

additional improvements may be possible. The cycle associated with the Carnot engine is shown in Figure below;

Using an ideal gas as the working substance. It is composed of the following four reversible processes:

1 → 2 : An isothermal expansion. Heat is transferred reversibly from the high –

temperature reservoir at the constant temperature

HT .

The piston in the cylinder is withdrawn and the volume increases.

2 → 3 : An adiabatic reversible expansion. The cylinder is completely insulated so

that no heat transfer occurs during this

,

96

reversible process. The piston continues to be withdrawn, with the volume increasing. 3 → 4 : An isothermal compression. Heat is transferred reversibly to the low –

temperature reservoir at the constant

temperature LT .

The piston compresses the working substance, with the volume decreasing. 4 → 1 : An adiabatic reversible compression. The completely insulated cylinder

allows no heat transfer during this reversible process. The piston continues to compress the working substance until the original volume, temperature, and pressure are reached, thereby completing the cycle. Applying the first law to the cycle, we note that:

netLH WQQ …… (1)

where LQ is assumed to be a positive value for the heat transfer to the low –

temperature reservoir. This allows us to write the thermal efficiency for the Carnot cycle as:

H

L

H

LH

Q

Q

Q

QQ

1

…… (2)

The following examples will be used to prove the following three postulates:

Postulate 1 : It is impossible to construct an engine, operating between two given

temperature reservoirs, that is more efficient than the Carnot engine.

Postulate 2 : The efficiency of a Carnot engine is not dependent on the working

substance used or any particular design feature of the engine.

Postulate 3 : All reversible engines, operating between two given temperature

reservoirs, have the same efficiency as a Carnot engine operating between the same two temperature reservoirs.

EXAMPLE: Show that the efficiency of a Carnot engine is the maximum possible

efficiency.

Solution:

Assume that an engine exists; operating between two reservoirs, that has an

efficiency greater than that of a Carnot engine, also, assume that a Carnot engine

operates as a refrigerator between the same two reservoirs, as sketched in Figure (a);

,

97

Let the heat transferred from the high – temperature reservoir to the engine be equal to the heat rejected by the refrigerator; then the work produced by the engine will be greater than the work required by the refrigerator that is;

LL QQ

Since the efficiency of the engine is greater than that of a Carnot engine. Now,

our system can be organized as shown in Figure (b). The engine drives the refrigerator

using the rejected heat from the refrigerator. But, there is some net work;

WW

that leaves the system. The net result is the conversion of energy from a single reservoir into work, a violation of the second law. Thus, the Carnot engine is the most efficient engine operating between two particular reservoirs.

CARNOT EFFICIENCY

Since the efficiency of a Carnot engine is dependent only on the two reservoir

temperatures, the objective of this article will be to determine that relationship. We will assume the working substance to be an ideal gas and simply perform the required calculations for the four processes of Figure below;

,

98

The heat transfer for each of the four processes is as follows:

1 → 2 : 1

221 ln

2

1V

VmRTVdPWQ

V

V

HH

2 → 3 : 032 Q

3 → 4 : 3

443 ln

4

3V

VmRTVdPWQ

V

V

LL

4 → 1 : 014 Q

Note that we want LQ to be a positive quantity, as in the thermal efficiency

relationship; hence, the negative sign. The thermal efficiency is then;

,

99

1

2

3

4

ln

ln

11

V

V

V

V

T

T

Q

Q

H

L

H

L …… (1)

During the reversible adiabatic processes 2 → 3 and 4 →1, we know that;

1

4

1

1

3

2 ,

k

H

L

k

H

L

V

V

T

T

V

V

T

T

Thus, we see that;

2

1

3

4

4

1

3

2

V

V

V

Vor

V

V

V

V

recognizing that 2

1

1

2 lnlnV

V

V

V

Substituting into (1),we obtain the result ;

H

L

T

T 1 …… (2)

We have simply replaced H

L

Q

Q with

H

L

T

T . We can do this for all reversible

engines or refrigerators. We see that the thermal efficiency of a Carnot engine is dependent only on the high and low absolute temperature of the reservoirs.

Consequently, the relationship (2) is applicable for all working substances, or for all

reversible engines, regardless of the particular design characteristics.

The Carnot engine, when operated in reverse, becomes a heat pump or a

refrigerator, depending on the desired heat transfer. The coefficient of performance for

a heat pump becomes;

H

LL

H

LH

H

net

H

T

TQ

Q

QQ

Q

W

QCOP

1

11

…… (3)

,

100

The coefficient of performance for a refrigerator takes the form;

1

11

L

HH

L

LH

L

net

L

T

TQ

Q

QQ

Q

W

QCOP

…… (4)

The above measures of performance set limits that real devices can only approach. The reversible cycles assumed are obviously unrealistic, but the fact that we have limits that we know we cannot exceed is often very helpful in evaluating proposed designs and determining the direction for further effort.

EXAMPLE 4. A Carnot engine operates between two temperature reservoirs

maintained at 200 ºC and 20 ºC, respectively. If the desired output

of the engine is 15 kW, as shown in Figure below, determine the

heat transfer from the high-temperature reservoir and the heat transfer to the low-temperature reservoir.

Solution:

The efficiency of a Carnot engine is given by:

H

L

H T

T

Q

W 1

This gives, converting the temperatures to absolute temperatures,

kW

T

T

WQ

H

LH 42.39

473

2931

15

1

Using the first law, we have:

kWWQQ HL 42.241542.39

EXAMPLE 5. A refrigeration unit is cooling a space to – 5°C by rejecting energy to

the atmosphere at 20°C. It is desired to reduce the temperature in

the refrigerated space to – 25°C. Calculate the minimum percentage

increase in work required, by assuming a Carnot refrigerator, for

,

101

the same amount of energy removed.

Solution:

For a Carnot refrigerator we know that;

1

11

rej

addadd

rej

rejadd

rejrej

T

TQ

Q

QQ

Q

W

QCOP

For the first situation we have;

rejrej

rej

addrej Q

k

kQ

T

TQW 0933.0}1

)2735(

)27320({)1(1

For the second situation there results;

rejrej

rej

addrej Q

k

kQ

T

TQW 181.0}1

)27325(

)27320({)1(2

The percentage increase in work is then;

9499.93100

0933.0

0933.0181.0100

1

12

rej

rejrej

Q

QQ

W

WW

Note the large increase in energy required to reduce the temperature in a refrigerated space. And this is a minimum percentage increase, since we have assumed an ideal refrigerator.

EXAMPLE 6. A Carnot engine operates with air, using the cycle shown in Figure

below. Determine the thermal efficiency and the work output for each cycle of operation.

,

102

Solution:

The thermal efficiency is found to be;

404.0500

30011 or

k

k

T

T

add

rej

To find the work output we can determine the heat added during the constant

temperature expansion and determine w from;

addadd q

w

Q

W

We find addq from the first law using 0u :

2

332 ln

3

2

addaddadd TRd

TRdPwq

To find 2 first we must find 1 :

kg

m

P

TR 3

1

11 076.1

80000

)300()287(

Using;

kg

m

T

T

T

T kk

314.1

1

1

1

2

112

1

1

2

2

1 300.0500

300076.1

Likewise, kg

m

T

T k 35.21

1

3

443 798.2

500

300)10(

.

Hence, kg

kJTRq addadd 0.320

300.0

798.2ln)500()287(ln

2

3

Finally, the work for each cycle is; kg

kJqw add 1283204.0

Entropy

,

103

Consider the reversible Carnot engine operating on a cycle consisting of the processes described in Figure below.

The quantity T

Q is the cyclic integral of the heat transfer divided by the

absolute temperature at which the heat transfer occurs. Since the temperature addT is

constant during the heat transfer addQ ,and rejT , is constant during heat transfer rejQ ,the

integral is given by:

rej

rej

add

add

T

Q

T

Q

T

Q

….. (1)

where the heat rejQ leaving the Carnot engine is considered to be positive.

,

104

Using add

rej

Q

Q 1 and

add

rej

T

T 1 we see that, for the Carnot cycle,

rej

rej

add

add

add

rej

add

rej

T

Q

T

Qr

T

T

Q

Q 0 …… (2)

Substituting this into (1), we find the interesting result;

0 T

Q …… (3)

Thus, the quantity T

Q is a perfect differential, since its cyclic integral is zero.

We let this perfect differential be denoted by sd , where S represents a scalar

function that depends only on the state of the system. This, in fact, was our definition of a property of a system. We shall call this extensive property entropy; its differential is given

by:

revT

QSd

…… (4)

Where the subscript “rev” emphasizes the reversibility of the process. This can be

integrated for a process to give;

revT

QS

…… (5)

From the above equation, we see that the entropy change for a reversible process can be either positive or negative depending on whether energy is added to or extracted

from the system during the heat transfer process. For a reversible adiabatic process

the entropy change is zero.

We often sketch a temperature – entropy diagram for cycles or processes of

interest. The Carnot cycle provides a simple display when plotting temperature vs.

entropy. It is shown in Figure below. The change in entropy for the first process from

state 1 to state 2 is :

add

add

T

Q

T

QSS

2

1

12

…… (6)

The entropy change for the

,

105

reversible adiabatic process

from state 2 to state 3 is zero.

For the process from state 3 to

state 4 the entropy change is

numerically equal to that of

the first process; the process

from state 4 to state 1 is also

a reversible adiabatic process

and is accompanied with a zero

entropy change.

The heat transfer during a reversible process can be expressed in differential

form as;

SdTQ ….. (7)

Hence, the area under the curve in the T – S diagram represents the heat transfer

during any reversible process. The rectangular area in Figure above thus represents the

net heat transfer during the Carnot cycle.

Since the heat transfer is equal to the work done for a cycle, the area also represents the net work accomplished by the system during the cycle. Here,

STWQ netnet

The first law of thermodynamics, for a reversible infinitesimal change, becomes,

using (7);

UdVdPSdT …… (8)

This is an important relationship in our study of simple systems. We arrived at it assuming a reversible process. However, since it involves only properties of the system, it holds for an irreversible process also. If we have an irreversible process, in general,

VdPW and SdTQ but (8) still holds as a relationship between the

properties. Dividing by the mass, we have;

uddPsdT ….. (9)

where the specific entropy is defined to be; m

Ss ….. (10)

,

106

To relate the entropy change to the enthalpy change we differentiate the equation for enthalpy;

dPdPuddhPuh

Week No

Details

26 - 28 Fuel – definition of accounts and properties of the fuel used in boilers and

cooling systems absorbance .

,

107

,

108

,

109

,

110

,

111

,

112

,

113

,

114

,

115

,

116

,

117

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