Post on 12-Sep-2014
Embed Size (px)
DESCRIPTIONالحقيبة التعليمة لمادة الثرموديناميك - قسم المكائن والمعدات - المرحلة الاولى مدرسة المادة اوراد عبد اللططيف عمر
التقنيين المدربين اعداد معهد / السيارات والمعدات المكائن قسم
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
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
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
Bar = 105 paDensity كثافة
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
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 =
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
3- Density : ( ) mass per unit volume الكثافة
4- Specific volume ( ) : the volume occupies by unit mass
5- power : It is work done per unit time
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
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 .
Q in Q out
Isolated System Closed System Open system
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
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
1- (20 C0) to Kelvin scale . 2- Convert (400 K0) to Rankine scale .3- Convert (170 F0) to Kelvin scale .
Pressure :- is the force exerted by the system on unit area .
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 .
Pabs = Patm + Pg
The positive guage pressure
The negative guage pressure :- [ vaccume pressure ]
Pg = Pabs – Patm
Patm = atmospheric pressureالجوي الضغط
h = height of the liquid السائل ارتفاع
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 .
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
1- change a pressure about 1500 mm Hg to bar .
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 .
76 cm Hg = 1.0132 bar
Pvacc = 65 cm Hg * = 0.867 bar
Pvacc = 0.867 bar
Pvacc = Pat m - Pabs
0.867 = 1.0132 – Pabs
Pab = 1.0132 – 0.867 = 0.147 bar
1- Convert pressure 5 Kpa to bar
2- Convert pressure 76 cm Hg to
3- Convert pressure 50 pa to mmHg
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 .
P1 = 98.4 kpa + 3.853 kpaP1 = 102.253 kpa
P1 = 102.253 kpa * = 1.022 atm
1 kpa = 1000 pa1 atm = 102 kpa
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
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
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 *
* If the work done by a thermodynamic system we say that is a positive work +W or W > 0 Wout = + w
* 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
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
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 .
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
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
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 .
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
2- heat rejected or out put us
Net work (Wnet) & Net heat (Qnet)
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
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 :-
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)
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 ?
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 .
From N. F.E.E
Q = - 1500 KJ (The heat out)
W = - 5090 KJ (the work input to the system)
- 1500 – (-5090) =
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
1) The Boiler المرجل The boiler is a heat exchange which convert the liquid water to steam at constant pressure . P = C
h2 > h1 heat added
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
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 .
Q = m (h2 – h1)
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)
W = m (h1 – h2)
W = m (h1 – h2)
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 .
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
= 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
Q1-2 = m
Q1-2 = m cp
= 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 ( )
Cama (كاما) :- Is the adiabatic expone which represent the ratio between specific heat at constant pressure and specific heat at constant volume .
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) .
KJ / kg . k0
2- Example (2)
Compute the change of enthalpy as (1 kg) of a gasis heated from 300 k0
to 1500 k0 , .
3- Example (3)
1- Compute the change of internal energy as (219) of a gas is heated .
2- Compute the amount of heat added . ,
The relation between ( R and C p & Cv)
We know that
And for ideal gas p.v = m . R. T
For m = 1 kg
p.v = R . T
Sub . equations (2) , (3) , (4) in eq. (1) (1) معادلة في4 ،3 ،2 المعادالت عوض
Prove that : -
The relation between R and ( )
R = (1)
على ( نحصل1) معادلة ( في2) معادلة بتعويض
على ( نحصل2) معادلة ( في3) معادلة بتعويض
Cp - Cv
Prove that 1)
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
constant gas R = characteristic النحاس الغاز ثابت
المثالي للغاز العامة المعادلة (3) -----
= absolute pressure
V = volume (m3)T = temp. (k0)
= mass (kg)
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
R0 = R . M
Ex . (5) :-
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)
P1 V1 = m . R . T1
1 (kg) * R * (15 +273) k0
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
. 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)
PV = C
P1 V1 = P2 V2 = C
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
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
P1 = P2
V = constant ثابت
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
P = pressure ( )
V = volume m3 T = temperature (k0)
حالة من وألكثر
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 .
P1 V1 = P2 V2 بويل قانون T = C
1 bar * 2.8 m3 = 14 bar * V2
(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.
0.2 T2 = 0.4 [ 50 + 273]
(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.
P1 V1 = m R T1
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)
Constant volume process (Isometric)
والضغط الحرارة درجة بين العالقة اليجاد
VV1 = V2
3- Constant temp. process [Isothermal]
T = C
المعادلة على التكامل باجراء
والحجم الضغط بين العالقة اليجاد
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 .
b- the work done .
Cp = 1.005 KJ/kg.k Cv = 0.718
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
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 .
1.005 – 0.718 = 0.278
P1 V1 = m R T1
1.5 * 102 * V1 = 1 kg * 0.278 * [18 +273] .
Q = m Cp [T2 – T1] T2 اليجاد
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 .
مجهولة V2 الن الشغل ايجاد يمكن الP1 V1 = P2 V2 T = C V2 اليجاد
1.013 * 0.827 = 13.78 V2
(1) معادلة في V2 قيمة عن عوض
. 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 .
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]
1.013 (0.12)1.4 = P2 (0.024)1.4
2) P2 = 1.013 = 9.64 bar النهائي الضغط
اليجاد T2 [ النهائية [
3) المنجز الشغل اليجاد
W12 = - 85.282 KJ
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
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 .
: 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 .
(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
والمضخة الحرارية الماكنة على الطالب يتعرف ان: الدرس من الهدف
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
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
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
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
Expansion valveالتمدد صمام
االنتروبي وتغير االنتروبي معنى الطالب يتعلم - ان1: الدرس من الهدف
. المغلقة باالنظمة
. الحرارة ودرجة االنتروبي بين العالقة على الطالب - يتعرف2
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
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
Carnot Cycle: العشرون االسبوع
عملها مبدأ كارنوت دورة على الطالب يتعرف - ان1: الدرس من الهدف مع واهميتها
. والكفاءة المخطط
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 .
Fig. 1 . Steady flow Carnot engine
Fig. 2. Reciprocating Carnot engine
Fig. 3. Carnot cycle on P-v and T-s diagrams
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 ,
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 ,
Since , T1 = T2 and T4 = T3 , we have
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
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 ,
كارنون كفاءة (1
Ex . 2
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 .
1) , ,
3) the work done = Qh – Qc = 1000 – 777.7
= 222.3 J
: والعشرون والثاني والعشرون الحادي االسبوع
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
1-2 Adibatic Compression االنضغاط مرحلة
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
Constant volume , heat addition
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)
3-4 Power stroke Adibatic expansion
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
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
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 .
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.
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 .
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 .
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 .
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 .
Piston : moving from bottom dead center to top dead center .
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) :
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 .
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)
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
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
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
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
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
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
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 :
(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
(adapted from Taylor (1966))
(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
والعشرين والثامن والعشرين السابع The actual cycle : االسبوع
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 .
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
واستخداماته الغازي التوربين معنى الطالب يتعلم: الدرس من الهدفومكوناته
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
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) .
1 gas turbine components
0 1.1 ancillary equipment
2 fuel gas turbine
3 disadvantages of gas turbine
4 external links
[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
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.
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) .
1- Thermodynamic , M.M ABBott .
2- Engineering thermodynamics ,
D.B. SPALDING and E. H. Cole .
ViRGil – MORING FAIRES
KENNETH WARK , JR .
5- Gas turbine theory
H , COHEN
6- Engineering thermodynamics
G. P. Gupta and R. Praksh
7- Internal Combustion Engines ;
Harper and Row Inc.