# ثرموداينمك

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هد ع ماد اعد ن ي ب مدر ل ا ن ي ي ن ق ت ل ا م س ق ن ئ مكا ل ا عدات م ل وا رات ا ي س ل / ا* ك م ن ي ودا م ر1 لث ا ف ص ل ااول الاد اعد سةِ مدر ل ا اوراد د ي ع ف ت لط ا0

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الحقيبة التعليمة لمادة الثرموديناميك - قسم المكائن والمعدات - المرحلة الاولى مدرسة المادة اوراد عبد اللططيف عمر

TRANSCRIPT التقنيين المدربين اعداد معهد / السيارات والمعدات المكائن قسم

الثرموداينمكاالول الصف

المدر&سة اعدادالطيف عبد اوراد

Measuring units , examples force , pressure ; specific: االول االسبوعvolume

انواعها ، القياس وحدات على الطالب يتعرف - ان1: الدرس من الهدف.

المفردات معنى على الطالب يتعرف - ان2 Force ، Pressure ، ٍSpecific volume

Thermodynamic

0 Definition :- The field of science , which deal with the energies possessed by gases and vapours . It also includes the conversion of these energies in term of heat and mechanical work and their relationship with properties of system .

Measuring units S.I units [International System of units]

Physical Qantityالفيزيائية الكمية

Symbolالرمز

Unitالوحدة

Symbol of unitالوحدة

Length الطول L meter MTime الزمن T second SMass الكتلة M kilogram Kg

Temperature T الحرارة درجة kelvin K0

Electric كهربائي تيار current

A Ampere A

The secondary SI unitsForce القوة F Newton NPressure الضغط P Pascal

BarPa =

Work شغل W Joule J = N.mPower قدرة P watt W =

British unit system النظام البريطاني للوحدات Physical Quantity Symbol Unit Symbol of unit Length L Inch InTime T Second SMass M Pound T , bm

Temperature T Fahrenheit F0

Force F Pound IbF

Pressure P - IbF/in2 psi

Density - Ibm/in3

Work W - IbF. in

Power horse power h.p

1 Length : L 1m = 100 cm = 102 cm = 1000 mm = 103 mm

1 in = 2.54 cm 1 foot = 1 ft = 12 in

Time : t : 1 hour = 60 minute = 3600 s

mass : m : 1 pound = 0.45 kg 1 ton = 1000 kg = 103 kg

Temperature : T : F0 = 1.8 0C + 32

Force : F : 1 MN = 106 N , 1 KN = 103 N

Pressure : 1 bar = 105 pa = 105

Work : W 1 KJ = 1000 J = 103 J J = N.m

Power : 1 h.p = 746 watt 1 KW = 1.34 h.p

Watt = , KW =

Definitions تعاريف

1- Force : It is an agent which consider as a measure of mechanical effect on the bodies .

2- Pressure : It,s the force per unit area

, الوحدات

3- Density : ( ) mass per unit volume الكثافة

2 الوحدات

4- Specific volume ( ) : the volume occupies by unit mass

الوحدات

5- power : It is work done per unit time

Power =

Units watt , kw , h.p

6- Specific gravity نوعية كثافة

Sp . gr =

Questions1- Define pressure , Specific volume .2- Convert 20 kpa to at m .3- Convert 50 ft to m .

Thermodynamic terms state , process , equilibrium in: الثاني االسبوعthermodynamic classification of system

الثرموداينميكية المصطلحات الطالب يتعلم - ان1: الدرس من الهدف الثرموداينميكية االنظمة انواع على الطالب يتعرف - ان2

Thermodynamic terms ثرموديناميكية مصطلحات

1- State :- Determines the value of the thermodynamic properties for material at a certain time .

2- process :- It is that which changes the system from a certain state of thermodynamic equilibrium into anther state .

3- Thermodynamic equilibrium :- means that the thermodynamic properties of amatter is the same and constant and do not

3 change at all areas in the system .

thermodynamic systems الثرموديناميكية االنظمة

Thermodynamic system is defined as :-

A definite area where some thermodynamic process is taking place .Any thermodynamic system has its "boundaries" and any thing outside the boundary is called "Surrounding" غالفBoundary : It is a surface that separtes between the system and its surrounding . محيطSurrounding : It is a region outside the boundary of the system.

Working Substance : It is amatter that transfer the energy through the system as steam , g as …. etc .

Classification of Systems االنظمة تصنيف

The thermodynamic system may be classified into the following three groups .

1- Closed System : In this system the working substance does not cross the boundaries of the system , but heat and work can cross it .

2- Open System : In this system the working substance crosses boundary of the system . Heat and work may also cross the boundary .

3- Isolated System : It is a system of fixed mass and no heat or work cross its boundary .

4

System

Isolate

boundary

No heat

N=0

Q=0

Q in Q out

cylinder

boundary

piston 1

1`

ni

tuo

QtuoQ ni

metsyS

1

wtuo

w ni

2 Isolated System Closed System Open system

Questions

1- Define : process , state .

2- write the classification of systems .

: الثالث االسبوع1- Temperature , kind of temp. measuring and relations between them .2- Pressure measurements and relation between them .

وانواع الحرارة درجة معنى الطالب يفهم - ان1: الدرس من الهدف. بينهما والعالقة الحرارة درجة مقاييس

والعالقة الضغط ومقاييس الضغط معنى الطالب يتعلم - ان2 . بينهم

Temperature :- It is may be defined the degree of hot hess or the level of heat intensity of a body .

Measurement of TemperatureThe temperature of a body is measured by a thermometer . There are

two scales for measuring the temp. of a body .

1- Centigrade or Celsius Scale This scale is most used by engineers . The freezing point of water = 0 . The boiling point of water = 100 We use symbol (C) to describe temp.

2- Fahrenheit Scale

5 The freezing point of water = 32 The boiling point of water = 212

We use the symbol (F0) to describe temp. * The relation between centigrade scale and Fahrenheit scale is given by

F0 = 1.8 C0 + 32

Ex : 1) Convert 37C0 to F0

2) Convert 50 F0 to Celsius scale

Absolute Temperature :

The absolute centigrade scale is called degree "Kelvin"

K0 = C0 + 273

Absolute Fahrenheit scale is called degree "Rankine"

R0 = F0 + 460

Questions

1- (20 C0) to Kelvin scale . 2- Convert (400 K0) to Rankine scale .3- Convert (170 F0) to Kelvin scale .

Pressure ضغط

Pressure :- is the force exerted by the system on unit area .

6 Absolute Pressure :- is the guage pressure plus atmospheric pressure .

Gauge pressure :- A gauge for measuring pressure records the pressure above atmospheric pressure .

Vaccume Pressure :- It is the pressure of the system below atmospheric pressure .

Equations المعادالت

Pabs = Patm + Pg

The positive guage pressure

> 0

The negative guage pressure :- [ vaccume pressure ]

7

Pg = Pabs – Patm

Patm = atmospheric pressureالجوي الضغط

h = height of the liquid السائل ارتفاع الفراغ ضغط

Vaccume Pressure = Patm - Pabs < 0

Manometer and Barometer

Manometer :- An instrument for measuring a pressure difference in terms of the height of a liquid .

= Density of liquid g =

= Height of the liquid

Barometer : An instrument for measuring the atmospheric pressure .

8 Units of pressure الضغط وحدات

1 pat m = 76 cm . Hg = 760 mm Hg

1 pat m =

1 pat m =

1 pat m = 1.01325 bar

Examples

1- change a pressure about 1500 mm Hg to bar .

Sol.

760 mm Hg = 1.0132 bar

1500 mm Hg * bar

2- A compound gauge reads (65 cm Hg ) vaccume pressure . Compute absolute pressure 14 bar .

Sol.

76 cm Hg = 1.0132 bar

Pvacc = 65 cm Hg * = 0.867 bar

Pvacc = 0.867 bar

Pvacc = Pat m - Pabs

9 0.867 = 1.0132 – Pabs

Pab = 1.0132 – 0.867 = 0.147 bar

Home work

1- Convert pressure 5 Kpa to bar

2- Convert pressure 76 cm Hg to

3- Convert pressure 50 pa to mmHg

Example (3)

A manometer is used to measure the pressure in a tank . The fluid is an 0.1 with a specific gravity of (0.87) and the liquid height = 45.2 cm . If the

barometric pressure = 98.4 kpa , the density of water = 1000 the

gravity = 9.78 , Determine the absolute pressure with in the tank in

kpa , at m .

Solution

Sp. Gr.

10 P1 = 98.4 kpa + 3.853 kpaP1 = 102.253 kpa

P1 = 102.253 kpa * = 1.022 atm

Note

1 kpa = 1000 pa1 atm = 102 kpa

Questions

1- Convert 500 cm Hg to

2- A barometer reads (735 mm Hg) at room temp.

Determine :- the atmospheric pressure 14 bar millibar , kpa ,

Hg = 13600

11 3- Convert 2 kpa to mm Hg

Work and kinds of work energy and forms of energy: الرابع االسبوع

. وانواعه الشغل معنى على الطالب يتعرف - ان1: الدرس من الهدف. واشكالها الطاقة انواع على الطالب يتعرف - ان2

Work :Its may be defined as the product of a force with its corresponding

displacement

------- (1)

J J = N.m الشغل وحداتWork is one of energies types , that we can transform it to anther

types of energy such as (transform of mechanical work to electrical energy , kinetic energy , heat energy

W = Force * displacement W = F *

Notes

* If the work done by a thermodynamic system we say that is a positive work +W or W > 0 Wout = + w

12

bodyF

x1 x2

x * If the work done on a thermodynamic system we say that is a negative w or (-w) W < 0 ex [compress or , generator]

Work of non of closed system

By taking a small element with length of p and width of dv Area of element = dw = p . dv

المنحني تحت المساحة يساوي المنجز الشغل فال للشريحة التفاضلي الحجم خالل المكبس حركة ولحالة

Area = area of pistondL = the dis placement traveled by a piston

F = p . A

F = exerted force

13

V2v P = exerted pressure

فان L طوله قدره شوط خالل المنجز الشغل اليجاد مالحظةW = F . L

L = strok length (m)

Flow work االنسياب شغل (open system)

Flow w on k = F . L

WF = F . L

WF = P . A . L

جول

The flow work (flow energy) per unit mass [specific flow work]

= flow work / unit mass J/kg = specific volume m3 / kg

J / kg

Types of Energies الطاقات انواع

1- Potential energy (P.E) The energy that system possesses by virtue of its position relative to the surface of the earth .

14

WF = P . V P . E = m g Z

m = massZ = elevation , m

2- Kinetic energy K.E The energy that a system possesses owing to its motion

= velocity m/sec

= Final velocity = initial velocity

3- Internal energy It is the energy stored in the substance total internal energy is given

U = m Cv T

Questions

1- A gas has a mass of (2 kg) and density = is transported by pipe

of height (30.25 m) from earth , the temp = 138 C0 the velocity .

Flow = 6m / sec and Cv = 0.674

Compute : 1- potertial energy 2- kinetic energy 3- Internal energy 4- total energy

15

21V 2

2V

1 2 The First Law of Thermodynamic: الخامس االسبوع

االول القانون على الطالب يتعرف - ان1: الدرس من الهدف. واهميته للثرموداينميك

The First Law of Thermodynamics

The concept of energy and hypothes is that it can be neither created

nor destroyed this is principle of the conservation of energy . The first law

of thermodynamics is merly one statement of this general principle with

particular of this general principle with particular reference to heat energy

and work .

When a system under goes at hermodynamics cycle then the net heat

supplied to the system from its surrounding is equal to the net work done

by the system on its surroundings .

: the sum for a complete cycle .

16 Enthalpy: السادس االسبوع

. واهميتها االنثالبي معنى على الطالب - يتعرف1: الدرس من الهدف

Enthalpy (H) الحراري المحتوى Enthalpy is an extensive property is defined by the relation .

النظام حجم على تعتمد خاصيتهH = U + P . V (J , kJ)Specific enthalpy (h)

= specific volume m3 / kg = specific internal energy J/kg = pressure

Heat energy (Q) الحرارية الطاقة It is the type of energy that transfer due to the different in temp .

between the system and its surrounding .

Q : heat energy (J , kJ)q : specific heat energy J/kg kg/kg

Heat Sign الحرارة اشارة 1- heat added input use الشغل اشارة عكس

Q +

2- heat rejected or out put us

(-Q)

Net work (Wnet) & Net heat (Qnet)

17

Q(+)

Q(-) Wnet = W0 – Win

W0 = out put workWin = input work

Qnet = Qin – Qout

Qin = input heatQo = output heat

Applying the first law on closed systems: السابع االسبوع

االول القانون تطبيق على الطالب يتعريف - ان1: الدرس من الهدف. المغلقة االنظمة على للثرموداينميك

18 Energy equation الطاقة معادلة By applying the first law of thermodynamic which state that the

energy can neither created nor destroid but transfer from one form to

another we obtain .

. آخر الى شكل من تتحول وانما تستحدث وال تفتى ال الطاقة

Q + U1 + P1V1 + K. E1 + P . E1 = W + U2 + P2V2 + K . E2 + PE2 H = U1 + P1V1

Q – W = equation of energy الطاقة معادلة

We can written thia equation

Q – W = m [(h2 – h1) +

1- Non – Flow energy equation (N. F. E. E)

Closed systemFor closed system PV , K.E , P.E = 0

The energy equation become :-

19

N1

Entry

W

P1 u1

V1 C1

ExitZ2

Q

C1

P 1V 1u , 1 2- Steady state flow energy equation (S. F. E. E)

(open system)For open system and steady state

الزمن مع تتغير ال الكتلة

By dividing the equation of energy by time we obtain

المستقر للجريان الطاقة معادلة S. F. E E

= mass flow rate ( )

= rate of heat transfer ( = watt , kw)

= rate of heat transfer (w , kw)

Examples

Closed system

Ex. 1

20 The change in the internal energy of closed system increase to (120 KJ)

while (150 KJ) of work , that go out of the system , Determine the amount

of heat transfer a cross system boundaries ?

Is the heat added or rejected ?

Solution

From N. F. E.E

The sign (+270) therefore the heat added to the system

Note مالحظة If the change in the internal energy increase we use (+ )If the change in the internal energy decrease we use ( )

Ex . 2

A tank containing a fluid is stirred by a paddle wheel the work input to the

paddle wheel is (5090 KJ) . The heat from the tank is (1500 KJ) .

Determine the change in the internal energy .

21 Sol.

From N. F.E.E

Q = - 1500 KJ (The heat out)

W = - 5090 KJ (the work input to the system)

- 1500 – (-5090) =

Questions

1- A mass of oxygen is compressed with out friction from initial state volume = 0.14 m3 pressure = 1.5 bar ; the temp = 15 C0 to final state the volume = 0.06 m3 and the pressure = 1.5 bar at this process the oxygen losses quantity of heat = 0.6 KJ to the surrounding Cv = 0.649 KJ/Kg.k R = 0.25 KJ/Kg.k0 .

Find : 1) the change of internal energy .2) the final temp.

, Applying the first Law on opened systems: والتاسع الثامن االسبوعexamples

االول القانون تطبيق كيفية الطالب يتعلم - ان1: الدرس من الهدف على للثرموداينميك

. المفتوحة االنظمة

Application of energy equation on open system :- S. F. E. E

22 1) The Boiler المرجل The boiler is a heat exchange which convert the liquid water to steam at constant pressure . P = C

Q (+) مضافة حرارة

2) The Condenser المكثف It is a heat exchanger work on condenser steam of water and converted it into a liquid by cooling the steam under constant pressure .

Q = m (h2 – h1)

h1 > h2

Q (-1) rejected (heat out) مسحوبة حرارة

Ex. 1

A boiler is generate steam with rate of 8 kg/s the enthalpy of liquid water

is 271 and the enthalpy of steam is 3150 . For steady state

conditions and by neglecting the change in K.E and P.E . Determine the rate of heat added to the steam in Boiler .

Solution

23

Q = m (h2 – h1)

steam

water1

1

2

water Q = m (h2 – h1)

h1 = 271 KJ/kg h2 = 3150

m = 8 kg/s

Q = m (h2 – h1)

= 8 (3150 – 271) = 23032 = 23032 KW

مضافة حرارة ( النهاT) الحرارة

3- The Turbine It is a mechanical device used to convert the kinetic energy of fluid into mechanical work

T1 > T2 h1 > h2 P1 > P2

- W = m (h2 – h1)

W = (+) work output

4- The Compressor It is a mechanical device used to increase fluid pressure by using on external mechanical work .

h1 > h2 T2 > T1

P2 > P1

- W = m (h2 – h1)

24

W = m (h1 – h2)

input

2output

W = m (h1 – h2)

1

2 W = (-) work input

Ex. Air enters a compressor at enthalpy (5000 J/kg) and leave with enthalpy (250 KJ / kg) the mass flow rate = 0.25 kg/s , Determine the work done .

Sol.

h1 = 5000 J / kg =

h2 = 250

m = 0.25 kg / sw = m (h1 – h2) = - 0.25 (5 – 250)w = 0.25 (-245) = - 61.25 KW

النظام الى داخل شغل النه سالبة االشارة

Specific heat kinds of specific heat and relations العاشر: االسبوعbetween them

وانواعها النوعية الحرارة على الطالب يتعرف - ان1: الدرس من الهدف بينها فيما والعالقة

Specific Heat النوعية الحرارة Is the a mount of heat required to rise the temperature of a unit mass

of substance one degree .

Types of Specific heat

1- Specific heat at constant pressure الضغط ثبوت عند النوعية الحرارة

25 *

= Specific heat at constant pressure.

Unit [ ]

= change in enthalpy االنثالبي تغير

Unit [ ]

= change in temp. الحرارة درجة في تغير

Unit [ k0 ]

*

*

m = mass (kg) كتلة T2 = Final temp. (k0) ثنائية حرارة درجة T1 = initial temp. (k0) ابتدائية حرارة درجة Q = heat حرارة (KJ , J)

2- Specific heat at constant volume الحجم بثبوت النوعية الحرارة

V = constant ثابت حجم closed system مغلق نظام

= Specific heat at constant volume الحجم بثبوت النوعية الحرارة

26

Q1-2 = m

Q1-2 = m cp Unit (

= change in internal energy الداخلية الطاقة في التغير

Unit [ ]

= change in temp. (k0) = T2 - T1

For m = 1 kg

* على نحصل وبتعويض

*

The relation between ( )

وحدات بدون

Cama (كاما) :- Is the adiabatic expone which represent the ratio between specific heat at constant pressure and specific heat at constant volume .

Examples

27 1- compute the constant pressure specific heat of steam if the change in enthalpy is (104.2) KJ and the change in temp. is (50 KJ) .

Solution

KJ / kg . k0

2- Example (2)

Compute the change of enthalpy as (1 kg) of a gasis heated from 300 k0

to 1500 k0 , .

Sol.

3- Example (3)

1- Compute the change of internal energy as (219) of a gas is heated .

2- Compute the amount of heat added . ,

Sol.

28 1-

The relation between ( R and C p & Cv)

(1)

We know that

(2)

(3)

And for ideal gas p.v = m . R. T

For m = 1 kg

p.v = R . T

(4)

Sub . equations (2) , (3) , (4) in eq. (1) (1) معادلة في4 ،3 ،2 المعادالت عوض

29 Home work

Prove that : -

The relation between R and ( )

R = (1)

(2)

على ( نحصل1) معادلة ( في2) معادلة بتعويض

(3)

على ( نحصل2) معادلة ( في3) معادلة بتعويض

30

Cp - Cv Home work

Prove that 1)

2)

Gas Constant , the universal gas constant: عشر الحادي االسبوعand specific , Examples

وانواعه الغاز ثابت معنى الطالب يفهم - ان1 : الدرس من الهدف

The general equation of Ideal Gas

المثالي للغاز العامة المعادلة

constant ----- (1)

على ( نحصلm) الكتلة ( على1) المعادلة طرفي بقسمة

Let

----- (2)

constant gas R = characteristic النحاس الغاز ثابت

المثالي للغاز العامة المعادلة (3) -----

Units

= absolute pressure

V = volume (m3)T = temp. (k0)

= mass (kg)

R =

31 R0 = universal gas constant العام الغاز ثابت M = Molecular weight (k mole) مول( )كيلو الجزيئي الوزن

Example :- A tank has a volume of (0.5 in3) and contains (10 kg) of an ideal gas having a molecular weight = 24 , the temp = 25 C0 compute the pressure

Solution

R0 = R . M

kpa

Ex . (5) :-

32

R0 = R * M one kg of a perfect gas accupies a volume of (0.85 m3) at (15C0) and at a constant pressure of (1 bar) . The gas is first heated at a constant volume and then at a constant pressure .Compute : 1) the specific heat at constant volume (Cv). 2) the specific heat at constant pressure (Cp)

Solution

P1 V1 = m . R . T1

1 (kg) * R * (15 +273) k0

1)

Questions

What is the mass of air contained a room (6m x 10m x 4m) If the pressure is (100 kpa) and the temp. is 25 C0 , Assume air to be an ideal gas

, Ideal gas, boyle,s law, chal,s law: عشر والثالث عشر الثاني االسبوعexamples المثالي الغاز مسخن على الطالب يتعرف - ان1: الدرس من الهدف وشارل بويل وقانون

33 . المثالية للغازات

Ideal gas المثالي الغاز

The ideal gas is defined as the state of substance that follows well – know Bolyle,s and charle,s laws .

وشارل بويل لقانوني تخضع التي المادة حالة : هو المثالي الغاز

Laws of Ideal gas

The physical properties of a gas are controlled by the following variables :-

pressure (P) exerted by the gas . Volume ( ) occupied by the gas . Temperature (T) of the gas .

The behaviour of perfect gas is governed by the following laws :-

1) Boyle,s Law The absolute pressure of a given mass of ideal gas varies inversely of its volume when the temp. remain constant

الغاز حجم مع عكسيا يتناسب المثالي الغاز من معينة لكتلة المطلق الضغط. الحرارة درجة ثبوت عند

T = C ثابت

T = C

Unit of pressure (Pa)

34

PV = C

P1 V1 = P2 V2 = C

)1(

)2(

(pressure) (P)

)V (volume

PV = C Charle , s Law

The volume of a given mass of ideal gas varies directly with the temp. when the absolute pressure remain constant .

P = constant ثابت

P = C

Units

V1 = initial volume االبتدائي الحجم V2 = final volume الثاني الحجم

Gay Lussac Law لوساك غاي قانون

The absolute pressure of a given mass of ideal gas is proportional directly with the temp. at constant volume .

e تناسبا مثالي غاز لكتلة المطلق الضغط يتناسب عند الحرارة درجة مع طرديا. الحجم ثبوت

35

P

P1 = P2

)1( )2(

V ثابت

V = constant ثابت

General Process

It is transition of a substance from state (1) [ P1 , V1 , T1] to state (2) [P2 , V2 , T2] a cross a certain path .

مسار باتباع ( النهائية2) الحالة الى ( االولية1) الحالة من المادة تغير وهي. معين

= constant

P = pressure ( )

V = volume m3 T = temperature (k0)

حالة من وألكثر

Examples (1)

An air compress or is compress (2.8 m3) of air from initial pressure of (1 bar) to final , pressure of (14 bar) calculate the final volume of air if temp. is constant .

Solution :-

P1 V1 = P2 V2 بويل قانون T = C

36 1 bar * 2.8 m3 = 14 bar * V2

Examples (2)

(0.2 m3) of a gas at (50 C0) the gas is heated at a constant pressure until its volume reached (0.4 m3) compute the final temp.

Solution :-

0.2 T2 = 0.4 [ 50 + 273]

Examples (3)

(2 kg) of a gas at initial pressure of (1.4 bar) and temp. = 40 C0 , R =

188.34 , Determine (a) the volume of the gas

(b) If the gas is heated at constant pressure to have a volume of (1.5 m3) . Find the final temp.

Solution

37 P1 V1 = m R T1

m3

b) P = C ثابت

T2 = 557.6 k0

Thermodynamic Processes and applications: عشر الرابع االسبوع

الثرموداينمكية االجراءات على الطالب يتعرف - ان1: الدرس من الهدف. وتطبيقاتها

Some Processes for closed systems

1- Constant pressure process (Isobaric)

38

P

V1 V2

V

1 2

WD العامة المعادلة

Constant volume process (Isometric)

والضغط الحرارة درجة بين العالقة اليجاد

39

P

P2

P1

VV1 = V2

1

2 3- Constant temp. process [Isothermal]

T = C

المعادلة على التكامل باجراء

والحجم الضغط بين العالقة اليجاد

Ex. 1

The pressure of a gas = 1.5 bar at 18 C0 temp. compute 1- the volume of (1 kg) of the gas .If the gas is heated at constant pressure until the volume become [1m3] Compute : a- the a mount of added heat .

40 b- the work done .

Cp = 1.005 KJ/kg.k Cv = 0.718

Solution

R = Cp - Cv

R = 1.005 – 0.718 = 0.278

P1 V1 = m R T1

V1 = 0.556 m3

Q12 = m Cp [T2 – T1]

T2 = 526.2 k0

Q = 1 * 1.005 * [526.2 – 291] = 243.4 KJ

W = P [V2 – V1] = 1.5 x 102 [1- 0.556] = 66.48 KJ

41 Ex. 1 : Given [ the mass of the gas = 1 kg , the pressure = 1.5 bar ,

the temp = 18C0 ] Cv = 0.718 , Cp= 1.005

compute : 1- the volume of the gas , the gas is heated at constant pressure until the volume become (1m3). 2- the added heat . 3- the work done .

Solution

1.005 – 0.718 = 0.278

P1 V1 = m R T1

1.5 * 102 * V1 = 1 kg * 0.278 * [18 +273] .

m3

Q = m Cp [T2 – T1] T2 اليجاد

42 Q = 1 * 1.005 (526.2 – 291)Q = 243.4 KJ

W = P [ V2 – V1]W = 1.5 * 102 [1 - 0.566]W = 66.48 KJ

Ex . 2 : Given [ the pressure of air = 1.013 bar , the volume = 0.827 m3 and the temp. = 25 C0 , the air is compressed with constant temp. until the pressure becomes = 13.78 bar . Compute :- the work done to compress the air .

Solution

----- (1)

مجهولة V2 الن الشغل ايجاد يمكن الP1 V1 = P2 V2 T = C V2 اليجاد

1.013 * 0.827 = 13.78 V2

m3

(1) معادلة في V2 قيمة عن عوض

43 KJ

. النظام الى داخل الشغل ان على تدل سالبة االشارة

Ex . 3 : Given [ the volume of gas = 0.12 m3 , the temp. = 20 C0 and the pressure = 1.013 bar is compressed adibatically until the volume become =

0.024 m3 Cv = 0.718 , Cp = 1.005 .

Compute : 1) the mass of the gas .2) the final pressure and temp. 3) the work done .

Solution

R = Cp – Cv = 1.005 – 0.718 = 0.287

P1 V1 = m R T1

1.013 * 102 * 0.12 m3 = m * 0.287 * [20 + 273]

1) kg

1.013 (0.12)1.4 = P2 (0.024)1.4

2) P2 = 1.013 = 9.64 bar النهائي الضغط

اليجاد T2 [ النهائية [

44 k0

3) المنجز الشغل اليجاد

W12 = - 85.282 KJ

45 Reversible and irreversibility: عشر الخامس االسبوع

او االنعكاسية منحني على الطالب يتعرف - ان1: الدرس من الهدف والالانعكاسية االرجاعية

. الالارجاعية او

Reversible and Irreversible Process

: الالارجاعي واالجراء االرجاعي االجراء: الالانعكاسي واالجراء االنعكاسي االجراء

Reversible process : االنعكاسي االجراء

The process in which the system and surrounding can be restored to the initial state without producing any change in the thermodynamic properties :-

Conditions of reversible process

1- All the initial and final state of system should be in equilibrium with the each other .

2- The process should occur in very small

46 Reversible processes االرجاعية االجراءات

1- Friction relative motion احتكاك بدون نسبية حركة

2- Extension of spring النوابض تمدد

3- Slow frictionless adiabatic expansion احتكاك بدون اديباتي تمدد

Irreversible process الالانعكاسي االجراء

It is state that both the system and surrounding cannot return to original state .

47 : عشر والسابع عشر السادس االسبوع The second law of thermodynamic

Results of the second law of thermodynamic

The second law of thermodynamic

The first law of thermodynamic indicates that the net heat supplied in a

cycle is equal to the net work done the gross heat supplied must be greater

than the net work done some of heat must be rejected by the system .

Kelven blank definition :-

It is impossible for heat engine to produce net work in a complete cycle

if it exchange heat only with bodies at a single fixed temperature .

The second law has been stated in several ways .

(1) The principle of Thomson (Lord Kelvin) states :

It is impossible by a cyclic process to take heat from a reservoir and

to convert it into work without simultaneously transferring heat from a hot

to a cold reservoir . This statement of the second law is related to

equilibrium , i.e. work can be obtained from a system only when the

system is not already at equilibrium . If a system is at equilibrium , no

spontaneous process occurs and no work is produced .

Evidently , Kelvin's principle indicates that the spontaneous process is the

heat flow from a higher to a lower temperature , and that only from such a

spontaneous process can work be obtained .

48 (2) The principle of Clausius States :

It is impossible to devise an engine which , working in a cycle , shall

produce no effect other than the transfer of heat from a colder to a hotter

body . A good example of this principle is the operation of a refrigerator .

(3) The principle of Planck states :

It is impossible to construct an engine which , working in a complete

cycle , will produce no effect other than raising of a weight and the cooling

of a heat reservoir .

(4) The Kelvin – Planck Principle :

May be obtained by combining the principles of Kelvin and of Planck

into one equivalent statement as the Kelvin – Planck statement of the

second law . It states : No process is possible whose sole result is the

absorption of heat from a reservoir and the conversion this heat into work .

Heat engine , heat pump: عشر الثامن االسبوع

49 والمضخة الحرارية الماكنة على الطالب يتعرف ان: الدرس من الهدف

. الحرارية

Heat engine الحرارية الماكنة

A heat engine is a system operating in a complete cycle and

developing net work from supply heat .

The second law implies that a source

of heat supply and sink for the rejection

of heat are both necessary since some

heat must be always be rejected by

the system .

By first law

Net heat supplied = network done

Q1 – Q2 = W

Q1 > W the second law of thermodynamic

The thermal efficiency of heat engine للماكنة الحرارية الكفاءة

الحرارية

50

Source

Heat engine

Sin k

T hot

T cold

Qout

Qin

W0

Q2

Q1

W

Heat engine One good example in practice of heat engine is a simple steam cycle

in this cycle heat is supplied in boiler work is developed in yurbine heat is

rejected in a condenser and small amount of work is required for the

pump .

Simple steam cycle

Heat Pump الحرارية المضخة

The heat pump is reversed heat engine in the heat pump (or

refrigerator)cycle an amount of heat Q2 is supplied from the cold reservoir

and an amount of heat Q1 is rejected to the hot reservoir and there must be

a work done on the cycle W

51

Boiler

condenser

pump

T hot

Qout

Win

T cold

Qin Heat pump (refrigerator)

Q1 = Q2 = W

There for W > 0 , the heat pump requires an input energy in order to

transfer heat from the cold chamber and reject it at higher temp .

Heat pump system

- Entropy , changes on closed systems and temp: عشر التاسع االسبوع

Entropy plan

52

Expansion valveالتمدد صمام االنتروبي وتغير االنتروبي معنى الطالب يتعلم - ان1: الدرس من الهدف

. المغلقة باالنظمة

. الحرارة ودرجة االنتروبي بين العالقة على الطالب - يتعرف2

Entropy

It is define as thermodynamic property that express the amount of

storage energy in the system also represent measure of reversible and

Irreversible process .

Temp. – Entropy Plane

For the reversible process the area under (T-S) plane = the heat (1)

dQ = T ds --------- (1)

From equation (1)

J / k0

53 1- The entropy at constant volume

2- The entropy at constant pressure

Ex : 1 Comput the entropy for reversible process at constant pressure the temp

vary from 120 C0 to 270 C0

Solution

Carnot Cycle: العشرون االسبوع

عملها مبدأ كارنوت دورة على الطالب يتعرف - ان1: الدرس من الهدف مع واهميتها

. والكفاءة المخطط

54 Carnot Cycle A Carnot gas cycle operating in a given temperature range is shown in

the T-s diagram in Fig. 1(a) . One way to carry out the processes of this

cycle is through the use of steady – state , steady – flow devices as shown

in Fig. 1(b) . The isentropic expansion process 2-3 and the isentropic

compression process 4-1 can be simulated quite well by a well – designed

turbine and compressor respectively , but the isothermal expansion process

1-2 and the isothermal compression process 3-4 are most difficult to

achieve . Because of these difficulties , a steady – flow Carnot gas cycle is

not practical .

The Carnot gas cycle could also be achieved in a cylinder – piston

apparatus (a reciprocating engine) as shown in Fig. 4.2 (b) . The Carnot

cycle on the p-v diagram is as shown in Fig 4.2 (a) , in which processes

1-2 and 3-4 are isothermal while processes 2-3 and 4-1 are isentropic . We

know that the Carnot cycle efficiency is given by the expression .

55 Fig. 1 . Steady flow Carnot engine

56 Fig. 2. Reciprocating Carnot engine

57 Fig. 3. Carnot cycle on P-v and T-s diagrams

58 Fig. 4. Working of Carnot engine

Since the working fluid is an ideal gas with constant specific heats , we have , for the isentropic process ,

Now , T1 = T2 and T4 = T3 , therefore

compression or expansion ratio

Carnot cycle efficiency may be written as ,

59 From the above equation , it can be observed that the Carnot cycle

efficiency increases as "r" increases . This implies that the high thermal

efficiency of a Carnot cycle is obtained at the expense of large piston

displacement . Also, for isentropic processes we have ,

and

Since , T1 = T2 and T4 = T3 , we have

pressure ratio

Therefore , Carnot cycle efficiency may be written as ,

From the above equation , it can be observed that , the Carnot cycle

efficiency can be increased by increasing the pressure ratio . This means

that Carnot cycle should be operated at high peak pressure to obtain large

efficiency .

60 Ex. 1

The highes theortical efficiency of gasoline engine based on the Carnot

cycle is 30 y0 if this engine expels its gases into at m . which has temp of

300k . compute

1) the temp in the cylinder immediately after combustion .

2) if the engine absorbs (837 J) of heat from the hot reservoir during each

cycle how much work can it perform in each cycle ,

Solution

كارنون كفاءة (1

2)

Ex . 2

61 A Carnot engine is operated between two heat reservoirs at temp of 450 k0

and 350 k0 , if the engine receive (1000 J) of heat in each cycle .

Compute : 1) the amount of heat reject .

2) the efficiency of the engine .

3) the work done by the engine in each cycle .

Solution

1) , ,

2)

الكفاءة

3) the work done = Qh – Qc = 1000 – 777.7

= 222.3 J

: والعشرون والثاني والعشرون الحادي االسبوع

62 والحرارة اوتو دورة معنى الطالب يتعلم - ان1: الدرس من الهدف

الشغل وصافي المرجعة

. الدورة وكفاءة

Otto Cycle : [ Constant Volume Cycle ]

The cycle consist of four reversible process

اجراءات اربعة من الدورة تتالف

انعكاسية

درجة وترتفعV2 الى V1 من الحجم يقل اديباتية عملية في الغاز ضغط يتم

.T2 الىT1 من الحرارة

Win الغاز على المبذول الشغل مقدار ويكون

2–3 Combustion stroke االحتراق شوط

T3 الىT2 من الحرارة ودرجة الضغط من كل يزداد االحتراق مرحلة تسمى

المضافة الحرارةQh (added heat) االشتعال من حرارة كمية النظام ويمتص

.

Added heat Qh = m Cv (T3 – T2)

63 من الحرارة وتقلV1 الىV2 من الحجم يوداد اديباتية عملية في الغاز يتمدد

T3الى T4( القوة مرحلة )تسمى

A standard dtt cycle

Exhaust stroke العادم شوط

4-1 Constant volume , heat rejection

الحرارة( من )التخلص الحجم ثبوت اجراء

64

P

V2V1

P0 T1 الىT4 من الحرارة درجة العادم( تنخفض صمام )مرحلة ويسمى

الجوي الضغط الى الضغط ويعود العادم صمام لفتح نتيجة الضغط وينخفض .Qc حرارة كمية النظام ويفقد

Rejected heat Qc = m Cv [ T1 – T4 ]المنعكسة الحرارة

The Heat efficiency (eff.) of otto cycle

الحرارية اوتو لدورة الكفاءة معادلة

5-1العادم( )مرحلة وهو افقي خط يالحظ المخطط : في مالحظة

. متعاكستان النهما مؤثرتان غير األخذ( تكونان )مرحلة1-5 و

: والعشرين والرابع والعشرين الثالث االسبوعDiesel Cycle net work out put and its eff.

ديزل دورة منحنى على الطالب يتعرف - ان1: الدرس من الهدف الخارج الشغل وصافي

. والكفاءة

Diesel Cycle

65 The Diesel cycle is a compression ignition (rather than spark ignition)

engine . Fuel is sprayed into the cylinder at P2 (high pressure) when the

compression is complete , and there is ignition without spark . An

idealized Diesel engine cycle is shown in figure 1.

Figure 1. The ideal Diesel cycle

The thermal efficiency is given by :

=

This cycle can operate with a higher compression ratio than the Otto cycle

because only air is compressed and there is no risk of auto – ignition of the

fuel . Although for a given compression ratio the Otto cycle has higher

efficiency , because the Diesel engine can be operated to higher

compression ratio , the engine can actually have higher efficiency than an

Otto cycle when both are operated at compression ratios that might be

achieved in practice .

66 Muddy Points

When and where do we use Cv and Cp ? Some definitions use dU=Cv dT is

it ever dU = Cp dT ? (MP 3.8)

Explanation of the above comparison between Diesel and Otto. (MP 3.9)

Air standard diesel engine cycle

The term "compression ignition" is typically used in technical literature to

describe the modern engines commonly called "Diesel engines" . This is in

contrast to "spark ignition" for the typical automobile gasoline engines that

operate on a cycle derived from the Otto cycle . Rudolph Diesel patented

the compression – ignition cycle which bears his name in the 1890s.

67 Design of a Diesel Cycle

The General Idea

The Diesel cycle is very similar to the Otto cycle in that both are

closed cycles commonly used to model internal combustion engines . The

difference between them is that the Diesel cycle is a compression –

ignition cycle use fuels that begin combustion when they reach a

temperature and pressure that occurs naturally at some point during the

cycle and , therefore , do not require a separate energy source (e.g. from a

spark plug) to burn . Diesel fuels are mixed so as to combust reliably at the

proper thermal state so that Diesel cycle engines run well .

(We might note that most fuels will start combustion on their own at

some temperature and pressure . But this is often not intended to occur and

can result in the fuel combustion occurring too early in the cycle . For

instance , when a gasoline engine – ordinarily an Otto cycle device – is run

at overly high compression ratios , it can start "dieseling" where the fuel

ignites before the spark is generated . It is often difficult to get such an

engine to turn off since the usual method of simply depriving it of a spark

may not work .

Stages of Diesel Cycles

Diesel Cycles have four stages : compression , combustion ,

expansion , and cooling .

Compression

68 We start out with air at ambient conditions – often just outside air

drawn into the engine . In preparation for adding heat to the air , we

compress it by moving the piston down the cylinder . It is in this part of

the cycle that we contribute work to the air . In the ideal Diesel cycle , this

compression is considered to be isentropic .

It is at this stage that we set the volumetric compression ratio , r

which is the ratio of the volume of the working fluid before the

compression process to its volume after .

Piston : moving from top dead center to bottom dead center .

Combustion

Next , heat is added to the air by fuel combustion . This process

begins just as the piston leaves its bottom dead center position . Because

the piston is moving during this part of the cycle , we say that the heat

addition is isochoric , like the cooling process .

Piston : starts at bottom dead center , begins moving up .

Expansion

In the Diesel cycle , fuel is burned to heat compressed air and the hot

gas expands forcing the piston to travel up in the cylinder . It is in this

phase that the cycle contributes its useful work , rotating the automobile's

crankshaft . We make the ideal assumption that this stage in an ideal

Diesel cycle is isentropic .

69 Piston : moving from bottom dead center to top dead center .

Cooling

Next , the expanded air is cooled down to ambient conditions . In an

actual automobile engine , this corresponds to exhausting the air from the

engine to the environment and replacing it with fresh air . Since this

happens when the piston is at the top dead center position in the cycle and

is not moving , we say this process is isochoric (no change in volume) .

Piston : at top dead center .

Dual Cycle: والعشرون الخامس االسبوع

ديول دورة منحنى على الطالب - يتعرف1: الدرس من الهدف

. والكفاءة والمخطط

Limited Pressure Cycle (or Dual Cycle) :

70 This cycle is also called as the dual cycle , which is shown in Fig. 1

Here the heat addition occurs partly at constant volume and partly at

constant pressure . This cycle is a closer approximation to the behavior of

the actual Otto and Diesel engines because in the actual engines , the

combustion process does not occur exactly at constant volume or at

constant pressure but rather as in the dual cycle .

Process 1-2 : Reversible adiabatic compression .

Process 2-3 : Constant volume heat addition .

Process 3-4 : Constant pressure heat addition .

Process 4-5 : Reversible adiabatic expansion .

Process 5-1 : Constant volume heat rejection .

71 Fig. 1 Dual cycle on p-v and T-s diagrams

Air Standard Efficiency

Heat supplied = m Cv (T3 – T2) + m Cp (T4 – T3)

Heat rejected = m Cv (T5 – T1)

Net work done = m Cv (T3 – T2) + m Cp (T4 – T3) – m Cv (T5 – T1)

72 Let ,

=

=

From the above equation , it is observed that , a value of rp > 1 results in

an increased efficiency for a given value of rc and . Thus the efficiency

of the dual cycle lies between that of the Otto cycle and the Diesel cycle

having the same compression ratio .

Mean Effective Pressure

73 =

=

=

والعشرون : السادس االسبوع

Comparing between Fuel – air and the air standard cycles

الدرس : من وتود- 1 الهدف دورة بين المقارنة على الطالب يتعرف –ان

الهواء ودورة هواء

القياسية .

Fuel – air cycle

The simple ideal air standard cycles overestimate the engine efficiency

by a factor of about 2. A significant simplification in the air standard

cycles is the assumption of constant specific heat capacities . Heat

74 capacities of gases are strongly temperature dependent , as shown by

figure (1) .

The molar constant – volume heat capacity will also vary , as will

the ratio of heat capacities :

If this is allowed for , air standard Otto cycle efficiency falls from 57

per cent to 49.4 per cent for a compression ratio of 8 .

When allowance is made for the presence of fuel and combustion

products , there is an even greater reduction in cycle efficiency . This leads

to the concept of a fuel – air cycle which is the same as the ideal air

standard Otto cycle , except that allowance is made for the real

thermodynamic behaviour of the gases . The cycle assumes instantaneous

complete combustion , no heat transfer , and reversible compression and

expansion . Taylor (1966) discusses these matters in detail and provides

results in graphical form . Figure (2) and (3) .

INTRODUCTION TO INTERNAL COMBUSTION ENGINES

75 Figure (1) : Molar heat capacity at constant pressure of gases above 150C

quoted as averages between 150C and abscissa temperature

Show the variation in fuel –air cycle efficiency as a function of

equivalence ratio for a range of compression ratios . Equivalence ratio is

defined as the chemically correct (stoichiometric)air / fuel ratio divided by

the actual air / fuel ratio . The datum conditions at the start of the

compression stroke are pressure (p1) 1.013 bar , temperature (T1) 1150C ,

mass fraction of combustion residuals (f) 0.05 , and specific humidity ( )

0.02 – the mass fraction of water vapour .

The fuel 1- octane has the formula C8H16 , and structure

76 Figure (2) shows the pronounced reduction in efficiency of the fuel – air

cycle for rich mixtures . The improvement in cycle efficiency with

increasing compression ratio is shown in figure 3 , where the ideal air

standard Otto cycle efficiency has been included for comparison .

In order to make allowances for the losses due to phenomena such as

heat transfer and finite combustion time , it is necessary to develop

computer models .

Prior to the development of computer models , estimates were made for

the various losses that occur in real operating cycles . Again considering

the Otto

77 Figure (2) : Variation of efficiency with equivalence ratio for a constant –

volume fuel – air cycle with 1 – octane fuel for different compression

ratios ( adapted from Taylor (1966))

Cycle , these are as follows :

78 (a) "Finite piston speed losses" occur since combustion takes a finite time

and cannot occur at constant volume . This leads to the rounding of the

indicator diagram and Taylor (1960) estimates these losses as being about

6 per cent .

Figure (3) : Variation of efficiency with compression ratio for a constant

volume fuel – air cycle with 1 – octane fuel for different equivalence ratios

79 (b) "Heat losses" , in particular between the end of the compression stroke

and the beginning of the expansion stroke . Estimates of up to 12 per cent

have been made by both Taylor (1966) and Ricardo and Hempson (1968) .

However, with no heat transfer the cycle temperatures would be raised and

the fuel – air cycle efficiencies would be reduced slightly because of

increasing gas specific heats and dissociation .

(c ) Exhaust losses due to the exhaust valve opening before the end of the

expansion stroke . This promotes gas exchange but reduces the expansion

work . Taylor (1966) estimates these losses as 2 per cent .

Since the fuel is injected towards the end of the compression stroke in

compression ignition engines (unlike the spark ignition engine where it is

pre-mixed with the air) the compression process will be closer to ideal in

the compression ignition engine than in the spark ignition engine . This is

another reason for the better fuel economy of the compression ignition

engine .

والعشرين والثامن والعشرين السابع The actual cycle : االسبوع

80 Comparing between actual cycles and air standard cycles

الدرس : من الحقيقية- .1الهدف الدراسة مفهوم الطالب يتعرف ان

القياسية- .2 الهواء ودورات الحقيقية الدراسة بين يقارن

The Actual Cycle

The eff of the actual cycle is low than the air standard cycle .

1- The fluid used is a mixture of air and fuel and the exhaust gases .

2- Energy loss is caused by time when the valves are opened or closed .

3- The specific heat varies with temp.

4- Due to the chemical dissociation there will energy loss .

5- The combustion isn't complete

6- The engine loses heat directly .

7- There are heat losses with exhaust gases .

8- The engine loses heat directly .

81 Air Standard Cycle

1- The efficiency is high .

2- The fluid used in the cycle is air follows the ideal gas PV= m R T

3- The used gas has a constant mass of air in the closed system or moves

with constant flourate in closed cycle .

4- The value of specific heat is constant .

5- Chemical reactions don't occur .

6- The compression and expansion occurs with constant entropy .

7- The received heat and rejected heat occurs reversibly .

8- There is no change I the kinetic & the potential energies that is why

they are neglected .

9- The engine works without friction .

Gas Turbine: والثالثون والعشرين التاسع االسبوع

82 واستخداماته الغازي التوربين معنى الطالب يتعلم: الدرس من الهدفومكوناته

Gas Turbine

Gas turbine . types which burn fuels such as oil and natural gas .

Instead of using heat to produce steam , as in steam turbines , the gas

turbine hot gas is used directly . Used for the operation of gas turbine

generators , ships , race cars , as used in jet engines .

Most of the gas turbine systems of the three main parts

1- the air compressor , 2- the combustion chamber , 3- turbine The so–called air compressor with the combustion chamber , usually ,

the gas generator . In most systems , gas turbine , the air compressor and

turbine boats on both sides of a horizontal axis , located between the

combustion chamber . And is part of the power turbine air compressor .

Absorbs quantity of air compressor and air ldguetha , so pressure is

increasing . In the combustion chamber , compressed air mixes with fuel

and burning the mixture . The more air pressure , improved combustion of

fuel mixed with air . The burning gases expand rapidly and to flow into the

turbine , leading to the rotation of the wheels of the turbine . And moving

hot gases through the various stages in the same manner as the gas turbine

flows through the steam turbine vapor . The Downs fixed mobile gas resat

member to change the rotor speed .

And benefit most from the gas turbine systems , the hot gases

emerging from the turbine . In some systems are some of these gases , and

83 go to a so – called renewed . And these gases are used to heat the

compressed air after leaving the air compressor . Before entering the

combustion chamber reduces heat the compressed air in this way the

amount of fuel used for combustion process . In jet engines , most of the

gas used to produce momentum . See : Jet propulsion .

Gas turbines operate at a temperature higher than the steam turbines .

And increase the efficiency of the turbine the more the temperature of

operation ; a standard operating temperature of most gas turbines is 8750C

or more .

Gas turbine (Gas Turbine) for this type of turbine has many uses , it

is used in aircraft and jet – propelled means of transport and maritime and

land in the area of oil as well for use in power plants , especially in peak

hours exceeded . Advantages of speed of operation (unlike the vapor

turbine , which needs to prepare a preliminary arrangements) .

Contents

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1 gas turbine components

84 0 1.1 ancillary equipment

2 fuel gas turbine

[Changed] the components of gas turbine

Gas turbine consists of the following main parts :

air compressor (The Air Compressor) : take the air from the

atmosphere and the pressure to raise tens of air pressure .

the combustion chamber (The Combustion Chamber) : the mixed

compressed air from the pressurized air with fuel and burn together

by means of special Balachtal , and the combustion products of

various gases and high temperatures up to 1000 degrees Fahrenheit ,

the pressure is high .

Turbine (The Turbine) : It would be centered with the horizontal axis

attached directly to the air pressure on the one hand and from other

mechanical contraceptives to be recycled (such as when a generator

85 for example) and through the gearbox (Gear Box) to reduce the speed

as the speed of rotation of the turbine are very high .

Interference resulting from the combustion gases in the turbine

Vtstdm quill of the many and then to the chimney.

Ancillary Equipment

Gas turbines need to operate safely and the safety of certain machinery

and equipment assistance (Auxiliaries) , as follows :

* Filters the air before entering the air pressure (air in take filters).

* initial operation of any assistant (the first operation of Starter) , which

is either electric motor or diesel engine or a steam turbine (starting

steam turbine) .

* the means or system setting .

* cooling system .

* system control and measurement equipment , temperature and pressure

at each stage of the work and an integrated operating system , such as

(mark 4, mark 5, mark 6) contains a processor , or more .

Fuel Gas turbine

Gas turbine operating on many types of fuel , it operates on natural

gas (Natural Gas) and on diesel , gasoline and even on the crude oil (with

some additions , chemical and arrangements) .

References

1- Thermodynamic , M.M ABBott .

86 2- Engineering thermodynamics ,

D.B. SPALDING and E. H. Cole .

3- Thermodynamics

ViRGil – MORING FAIRES

4- Thermodynamics

KENNETH WARK , JR .

5- Gas turbine theory

H , COHEN

G.F.C. ROGERS

6- Engineering thermodynamics

G. P. Gupta and R. Praksh

7- Internal Combustion Engines ;

Harper and Row Inc.

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