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# ثرموداينمك

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

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### ثرموداينمك

1. 1. ‫معهد اعداد المدربين التقنيين‬ ‫قسم المكائن والمعدات / السيارات‬ ‫الثرموداينمك‬ ‫الصف الول‬ ‫اعداد المدرِسة‬ ‫اوراد عبد الطيف‬‫السبوع الول : ‪Measuring units , examples force , pressure ; specific‬‬ ‫‪volume‬‬ ‫الهدف من الدرس : 1- ان يتعرف الطالب على وحدات القياس ، انواعها .‬ ‫2- ان يتعرف الطالب على معنى المفردات‬ ‫‪Force‬‬ ‫‪، Pressure ، ٍ pecific volume‬‬ ‫‪S‬‬ ‫‪Thermodynamic‬‬ ‫1‬
2. 2. 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 unitsS.I units [International System of units] Physical Qantity Symbol Unit Symbol of unit ‫الكمية الفيزيائية‬ ‫الرمز‬ ‫الوحدة‬ ‫الوحدة‬ L meter MLength ‫الطول‬ T second STime ‫الزمن‬ M kilogram KgMass ‫الكتلة‬ T kelvin K0 Temperature ‫درجة الحرارة‬ A Ampere AElectric current ‫تيار كهربائي‬The secondary SI units F Newton NForce ‫القوة‬ P Pascal NPressure ‫الضغط‬ Pa = m2 Bar Bar = 105 pa ρ kgDensity ‫كثافة‬ m3 W Joule J = N.mWork ‫شغل‬ P watt W= JPower ‫قدرة‬ SBritish unit system ‫النظام البريطاني للوحدات‬Physical Quantity Symbol Unit Symbol of unitLength L Inch InTime T Second SMass M Pound T , bmTemperature T Fahrenheit F0Force F Pound IbFPressure P - IbF/in2 psi 2
3. 3. Density ρ - Ibm/in3Work W - IbF. inPower horse power h.pLength : L 1m = 100 cm = 102 cm = 1000 mm = 103 mm 1 in = 2.54 cm 1 foot = 1 ft = 12 inTime : t : 1 hour = 60 minute = 3600 smass : m : 1 pound = 0.45 kg 1 ton = 1000 kg = 103 kgTemperature : T : F0 = 1.8 0C + 32Force : F : 1 MN = 106 N , 1 KN = 103 N N KNPressure : 1 bar = 105 pa = 105 m2 =10 3 m2Work : W 1 KJ = 1000 J = 103 J J = N.mPower : 1 h.p = 746 watt 1 KW = 1.34 h.p J KJ Watt = S , KW = S Definitions ‫تعاريف‬1- 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 Force F N N P = Area = A ‫الوحدات‬ m2 , mm 2 , KN =kpa m2 3
4. 4. 3- Density : ( ρ ) mass per unit volume ‫الكثافة‬ m ρ= V kg ‫الوحدات‬ ρ( m3 )4- Specific volume ( ) : the volume occupies by unit mass V V m3 V= m ‫الوحدات‬ kg5- power : It is work done per unit time w Power = t Units watt , kw , h.p6- Specific gravity ‫كثافة نوعية‬ ρ sub Sp . gr = ρ H 2OQuestions1- Define pressure , Specific volume .2- Convert 20 kpa to at m .3- Convert 50 ft to m . Thermodynamic terms state , process , equilibrium in : ‫السبوع الثاني‬ thermodynamic classification of system ‫الهدف من الدرس :1- ان يتعلم الطالب المصطلحات الثرموداينميكية‬ ‫2- ان يتعرف الطالب على انواع النظمة الثرموداينميكية‬ Thermodynamic terms ‫مصطلحات ثرموديناميكية‬1- 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 4
5. 5. 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 ‫النظمة الثرموديناميكية‬Thermodynamic system is defined as :-A definite area where some thermodynamic process is taking place .Any thermodynamic system has its "boundaries" and any thing outside theboundary is called "Surrounding"‫غلف‬Boundary : 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 ‫تصنيف النظمة‬ The thermodynamic system may be classified into the followingthree 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 . 5
6. 6. 3- Isolated System : It is a system of fixed mass and no heat or work cross its boundary . wtuo yradnuob w ni 1 boundary `1 ni No heat piston 1 metsyS System 2Q=0 boundary tuo N=0 Q ni Q Isolated System Isolate cylinder Closed Q Q in System Opentuosystem out Questions 1- Define : process , state . 2- write the classification of systems . : ‫السبوع الثالث‬ 1- Temperature , kind of temp. measuring and relations between them . 2- Pressure measurements and relation between them . ‫الهدف من الدرس : 1- ان يفهم الطالب معنى درجة الحرارة وانواع مقاييس درجة الحرارة‬ . ‫والعلةقة بينهما‬ . ‫2- ان يتعلم الطالب معنى الضغط ومقاييس الضغط والعلةقة بينهم‬ Temperature :- It is may be defined the degree of hot hess or the level of heat intensity of a body . Measurement of Temperature The 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 6
7. 7. 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 + 32Ex : 1) Convert 37C0 to F0 2) Convert 50 F0 to Celsius scaleAbsolute Temperature : The absolute centigrade scale is called degree "Kelvin" K0 = C0 + 273Absolute Fahrenheit scale is called degree "Rankine" R0 = F0 + 460Questions1- (20 C0) to Kelvin scale .2- Convert (400 K0) to Rankine scale .3- Convert (170 F0) to Kelvin scale . 7
8. 8. Pressure ‫ضغط‬Pressure :- 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 . Equations ‫المعادلت‬ Pabs = Patm + PgThe positive guage pressure Patm = atmospheric pressure ‫الضغط الجوي‬ h = height of the liquid ‫ارتفاع السائل‬ Pg = Pabs – Patm 8
9. 9. >0The negative guage pressure :- [ vaccume pressure ] ‫ضغط الفراغ‬Vaccume Pressure = Patm - Pabs <0 Manometer and BarometerManometer :- An instrument for measuring a pressure difference in terms of the height of a liquid . ∆= 2 P p − 1 p ∆ p = ρg ∆ h ρ = Density of liquidg = ∆h = Height of the liquidBarometer : An instrument for measuring the atmospheric pressure . 9
10. 10. Units of pressure ‫وحدات الضغط‬1 pat m = 76 cm . Hg = 760 mm Hg N1 pat m = 10 5 m2 = 10 5 pa KN1 pat m = 10 2 m2 = 10 2 Kpa1 pat m = 1.01325 barExamples1- change a pressure about 1500 mm Hg to bar .Sol.760 mm Hg = 1.0132 bar 1.01321500 mm Hg * 760 mm Hg = 1.999 bar2- A compound gauge reads (65 cm Hg ) vaccume pressure . Computeabsolute pressure 14 bar .Sol. 10
11. 11. 76 cm Hg = 1.0132 bar 1.0132 barPvacc = 65 cm Hg * 76 cm Hg = 0.867 barPvacc = 0.867 barPvacc = Pat m - Pabs0.867 = 1.0132 – PabsPab = 1.0132 – 0.867 = 0.147 barHome work1- Convert pressure 5 Kpa to bar KN2- Convert pressure 76 cm Hg to m23- Convert pressure 50 pa to mmHgExample (3)A manometer is used to measure the pressure in a tank . The fluid is an 0.1with a specific gravity of (0.87) and the liquid height = 45.2 cm . If ∆h kgthe barometric pressure = 98.4 kpa , the density of water = 1000 m3 the mgravity = 9.78 s2 , Determine the absolute pressure with in the tank inkpa , at m .Solution ρ oilSp. Gr. ρ water kg kg ρ = 0.87 * 1000 oil = 870 m3 m3 11
12. 12. p1abs = p 2 +∆p p1abs = p 2 + ρ ∆ g h kg m 45.2 cm ∆ = ρ ∆ = 870 p g h * 9.8 2 * m3 s 100 cm 1m kg 1N N ∆ = 3853.752 p * = 3853.752 = 3853.75 m.s 2 kgm m2 s2 1 kpa ∆ = p 3853.752 * = .853 3 1000 paP1 = 98.4 kpa + 3.853 kpaP1 = 102.253 kpa 1 atmP1 = 102.253 kpa * 10 2 kpa = 1.022 atmNote kg .m 1 N = s21 kpa = 1000 pa1 atm = 102 kpaQuestions KN1- Convert 500 cm Hg to m22- A barometer reads (735 mm Hg) at room temp.Determine :- the atmospheric pressure 14 bar millibar , kpa , kg ρ Hg = 13600 m3 12
13. 13. 3- Convert 2 kpa to mm Hg Work and kinds of work energy and forms of energy : ‫السبوع الرابع‬ . ‫الهدف من الدرس : 1- ان يتعرف الطالب على معنى الشغل وانواعه‬ . ‫2- ان يتعرف الطالب على انواع الطاةقة واشكالها‬Work : Its may be defined as the product of a force with its correspondingdisplacement x2 W = ∫F . x1 dx ------- (1) J J = N.m ‫وحدات الشغل‬ Work is one of energies types , that we can transform it to anthertypes of energy such as (transform of mechanical work to electrical energy, kinetic energy , heat energy W = Force * displacement W=F* ∆x 13
14. 14. ∆x F bodyNotes x1 x2* If the work done by a thermodynamic system we say that is a positivework +W or W > 0 Wout = + w* If the work done on a thermodynamic system we say that is a negative wor (-w) W<0 ex [compress or , generator]Work of non of closed system V2vBy taking a small element with length of p and width of dvArea of element = dw = p . dv 2 2 W = ∫dw 1 = ∫ p. 1 dv ‫الشغل المنجز يساوي المساحة تحت المنحني‬ ‫ولحالة حركة المكبس خلل الحجم التفاضلي للشريحة فال‬ dv = . dL AArea = area of pistondL = the dis placement traveled by a piston 14
15. 15. 2 W = ∫p . 1 A dLF =p.A 2 W = ∫F 1 dLF = exerted forceP = exerted pressure ‫ فان‬L ‫ملحظة ليجاد الشغل المنجز خلل شوط ةقدره طوله‬W=F.LL = strok length (m)Flow work ‫شغل النسياب‬ (open system)Flow w on k = F . LWF = F . LWF = P . A . L WF = P . V ‫جول‬ ∆ F W = 2 V2 P −1 V1 PThe flow work (flow energy) per unit mass [specific flow work] WF = P . V ÷ m WF =P . V WF = flow work / unit mass J/kg 3 V = specific volume m / kg ∆ F W = F2 W −WF 1 15
16. 16. ∆F W = 2 . V2 P − 1 V1 P J / kgTypes of Energies ‫انواع الطاقات‬1- Potential energy (P.E) The energy that system possesses by virtue of its position relative tothe surface of the earth .P.E=mgZm = massZ = elevation , m ∆E = P.E 2 ) 2 P. ( − P.E ) 1 ( ∆E = P. mg ∆ = Z mg ( Z 2 − 1) Z2- Kinetic energy K.E The energy that a system possesses owing to its motion 1 K .E = 2 mV 2 V = velocity m/sec V12 1 ∆ .E = K 2 m (V22 − V12 ) V22 V2 = Final velocity V1 = initial velocity ∆ .E = . E 2 K K − . E1 K 1 23- Internal energy It is the energy stored in the substance total internal energy is givenU = m Cv TQuestions 16
17. 17. kg1- A gas has a mass of (2 kg) and density = 21.6 m3 is transported bypipe of height (30.25 m) from earth , the temp = 138 C0 the velocity . KJFlow = 6m / sec and Cv = 0.674 kg .K 0Compute : 1- potertial energy 2- kinetic energy 3- Internal energy 4- total energy The First Law of Thermodynamic: ‫السبوع الخامس‬ . ‫الهدف من الدرس : 1- ان يتعرف الطالب على القانون الول للثرموداينميك واهميته‬The First Law of Thermodynamics The concept of energy and hypothes is that it can be neither creatednor destroyed this is principle of the conservation of energy . The first lawof thermodynamics is merly one statement of this general principle withparticular of this general principle with particular reference to heat energyand work . When a system under goes at hermodynamics cycle then the net heatsupplied to the system from its surrounding is equal to the net work doneby the system on its surroundings . ∑ dQ =∑dw ∑ : the sum for a complete cycle . 17
18. 18. Enthalpy : ‫السبوع السادس‬ . ‫الهدف من الدرس : 1- يتعرف الطالب على معنى النثالبي واهميتها‬Enthalpy (H) ‫المحتوى الحراري‬ Enthalpy is an extensive property is defined by the relation . ‫خاصيته تعتمد على حجم النظام‬H=U+P.V (J , kJ)Specific enthalpy (h) h =+ . u P V V = specific volume m3 / kg u = specific internal energy J/kg p = pressureHeat energy (Q) ‫الطاقة الحرارية‬ 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/kgHeat Sign ‫اشارة الحرارة‬1- heat added input use ‫عكس اشارة الشغل‬ Q + (+)Q 18 (-) Q
19. 19. 2- heat rejected or out put us (-Q)Net work (Wnet) & Net heat (Qnet)Wnet = W0 – WinW0 = out put workWin = input workQnet = Qin – QoutQin = input heatQo = output heat 19
20. 20. Applying the first law on closed systems : ‫السبوع السابع‬ ‫الهدف من الدرس : 1- ان يتعريف الطالب على تطبيق القانون الول للثرموداينميك على‬ . ‫النظمة المغلقة‬Energy equation ‫معادلة الطاقة‬ By applying the first law of thermodynamic which state that theenergy can neither created nor destroid but transfer from one form toanother we obtain . . ‫الطاقة ل تفتى ول تستحدث وانما تتحول من شكل الى آخر‬Q + U1 + P1V1 + K. E1 + P . E1 = W + U2 + P2V2 + K . E2 + PE2H = U1 + P1V1Q–W= ∆ H + K .E ∆ + P.E ∆ equation of energy‫معادلة الطاةقة‬We can written thia equation 1Q – W = m [(h2 – h1) + 2 (C 2 −C1 ) + g ( Z 2 −Z 1 ) 2 2 C1 W PV 1 1 u, 1 Entry P1 u1 V1 C1 N1 Q Exit Z2 20
21. 21. 1- Non – Flow energy equation (N. F. E. E)Closed system For closed system PV , K.E , P.E = 0The energy equation become :- Q − = W ∆U2- Steady state flow energy equation (S. F. E. E)(open system) For open system and steady state min =m out ‫الكتلة ل تتغير مع الزمن‬By dividing the equation of energy by time we obtain 1 Q0 −W 0 = m 0 [( h2 −h1 ) + (C 2 −C1 ) + ( Z 2 − 1 )] 2 g Z 2‫معادلة الطاةقة للجريان المستقر‬ S. F. E E kg kg kg m0 = mass flow rate ( S , min , hr ) J Q0 = rate of heat transfer ( S = watt , kw) W0 = rate of heat transfer (w , kw) 21
22. 22. ExamplesClosed systemEx. 1The 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 amountof heat transfer a cross system boundaries ?Is the heat added or rejected ?SolutionFrom N. F. E.E Q1− − 1− = u 2 W 2 ∆ Q−150 =120 Q = 270 KJThe sign (+270) therefore the heat added to the systemNote ‫ملحظة‬If the change in the internal energy increase we use (+ ∆u )If the change in the internal energy decrease we use ( − u ∆ ) 22
23. 23. Ex . 2A tank containing a fluid is stirred by a paddle wheel the work input to thepaddle wheel is (5090 KJ) . The heat from the tank is (1500 KJ) .Determine the change in the internal energy .Sol.From N. F.E.E Q −= W ∆ uQ = - 1500 KJ (The heat out)W = - 5090 KJ (the work input to the system)- 1500 – (-5090) = ∆U ∆ = 1500 +5090 U − ∆ = 3590 U KJQuestions1- A mass of oxygen is compressed with out friction from initial statevolume = 0.14 m3 pressure = 1.5 bar ; the temp = 15 C0 to final state thevolume = 0.06 m3 and the pressure = 1.5 bar at this process the oxygenlosses quantity of heat = 0.6 KJ to the surrounding Cv = 0.649 KJ/Kg.kR = 0.25 KJ/Kg.k0 .Find : 1) the change of internal energy . 2) the final temp. 23
24. 24. Applying the first Law on opened systems , : ‫السبوع الثامن والتاسع‬ examples ‫الهدف من الدرس : 1- ان يتعلم الطالب كيفية تطبيق القانون الول للثرموداينميك على‬ . ‫النظمة المفتوحة‬Application of energy equation on open system :- S. F. E. E1) The Boiler ‫المرجل‬ The boiler is a heat exchange which convert the liquid water to steamat constant pressure . P = C steam 1 Q −W =m [( h2 − 1 ) + (C 2 − 1 ) + ( Z 2 −Z 1 ) h 2 C2 g 2 Q = m (h2 – h1) 1 h2 > h1 heat added water Q (+) ‫حرارة مضافة‬2) The Condenser ‫المكثف‬ It is a heat exchanger work on condenser steam of water and convertedit into a liquid by cooling the steam under constant pressure . 1 1 Q −W =m [( h2 − 1 ) + (C 2 − 1 ) + ( Z 2 −Z 1 ) h 2 C2 g 2Q = m (h2 – h1) 2 ter wah1 > h2Q (-1) rejected (heat out) ‫حرارة مسحوبة‬Ex. 1 24
25. 25. A boiler is generate steam with rate of 8 kg/s the enthalpy of liquid water KJ KJis 271 kg and the enthalpy of steam is 3150 kg . For steady stateconditions and by neglecting the change in K.E and P.E . Determine therate of heat added to the steam in Boiler .SolutionQ = m (h2 – h1) KJh1 = 271 KJ/kg h2 = 3150 kgm = 8 kg/sQ = m (h2 – h1) KJ = 8 (3150 – 271) = 23032 S = 23032 KW ‫ ( لنها حرارة مضافة‬T ) ‫الحرارة‬3- The Turbine It is a mechanical device used to convert the kinetic energy of fluid intomechanical work inputT1 > T2 h1 > h2 P1 > P 2 1 Q −W =m [( h2 − 1 ) + (C 2 − 1 ) + ( Z 2 −Z 1 ) h 2 C2 g 2 2- W = m (h2 – h1) output W = (+) work output (W = m (h1 – h24- The Compressor 25
26. 26. It is a mechanical device used to increase fluid pressure by using onexternal mechanical work .h1 > h2 T2 > T1 2P2 > P 1 1 Q −W =m [( h2 − 1 ) + (C 2 − 1 ) + ( Z 2 −Z 1 ) h 2 C2 g 2 1- W = m (h2 – h1) (W = m (h1 – h2W = (-) work inputEx.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. 5000 KJh1 = 5000 J / kg = 1000 kg KJh2 = 250 kgm = 0.25 kg / sw = m (h1 – h2) = - 0.25 (5 – 250)w = 0.25 (-245) = - 61.25 KW ‫الشارة سالبة لنه شغل داخل الى النظام‬Specific heat kinds of specific heat and relations between them :‫السبوع العاشر‬ ‫الهدف من الدرس : 1- ان يتعرف الطالب على الحرارة النوعية وانواعها والعلةقة فيما بينها‬ 26
27. 27. Specific Heat ‫الحرارة النوعية‬ Is the a mount of heat required to rise the temperature of a unit massof substance one degree .Types of Specific heat1- Specific heat at constant pressure ‫الحرارة النوعية عند ثبوت الضغط‬ ∆h Cp = ∆T * Cp = Specific heat at constant pressure. KJ JUnit [ kg.k 0 or kg . k 0 ] = change in enthalpy ‫تغير النثالبي‬ ∆ h = h2 −h1 KJ JUnit [ kg or kg ] = change in temp. ‫تغير في درجة الحرارة‬ ∆ =2 − T T T1Unit [ k0 ] Q1-2 = m ∆h * Q1-2 = m cp ∆T *m = mass (kg) ‫كتلة‬ 27
28. 28. T2 = Final temp. (k0) ‫درجة حرارة ثنائية‬T1 = initial temp. (k0) ‫درجة حرارة ابتدائية‬Q = heat ‫حرارة‬ (KJ , J)2- Specific heat at constant volume ‫الحرارة النوعية بثبوت الحجم‬V = constant ‫حجم ثابت‬ closed system ‫نظام مغلق‬ ∆u Cv = ∆T = Specific heat at constant volume ‫الحرارة النوعية بثبوت الحجم‬ Cv KJ JUnit ( kg .k 0 ) or ( kg . k ) = change in internal energy ‫التغير في الطاةقة الداخلية‬ ∆u ∆ = u 2 −1 u u KJ JUnit [ kg or kg ] ∆T = change in temp. (k0) = T2 - T1For m = 1 kg ∆= v . u C ∆ T * ‫نحصل على‬ ‫وبتعويض‬ ∆u Q1− = C v 2 m ∆ T Q1− = C v m (T2 −1 ) T * 2 28
29. 29. The relation between ( C p & Cv ) Cp γ = ‫بدون وحدات‬ γ Cv Cama (‫ -: )كاما‬Is the adiabatic expone which represent the ratio γbetween specific heat at constant pressure and specific heat at constantvolume . γ >1 Cp > CvExamples1- compute the constant pressure specific heat of steam if the change inenthalpy is (104.2) KJ and the change in temp. is (50 KJ) .Solution ∆h 104.2 Cp = ∆T = 50 = 2.084 KJ / kg . k02- Example (2) Compute the change of enthalpy as (1 kg) of a gasis heated from 300 k0 KJto 1500 k0 γ = 1 .4 , C v = 0.718 kg.k 0 .Sol. 29
30. 30. ∆h Cp = ⇒ ∆ =C p ∆ h T ∆T Cp γ = ⇒ C p = γ * Cv Cv KJ C p = .4 * 0.718 = 1.005 1 kg .k 0 KJ ∆ = 1.005 * [1500 − h 300] =1206.240 kg3- Example (3) 1- Compute the change of internal energy as (219) of a gas is heated . KJ 2- Compute the amount of heat added . γ , C =1.0714 kg = 1.39 pSol. Cp 1.0714 KJ1- Cv = γ = 1.39 =0.7708 kg .k 0 ∆ u =v C ∆ 0.7708 *[ T = 1000 − 280] KJ ∆ = u 554.976 kg Q =m C p ∆ T Q = 2 * 554.976 = 1109.952 KJThe relation between (R and Cp & Cv) h = u + pv ∆∆ p ∆ h = +v u (1)We know that ∆= p h C ∆ T (2) ∆= v ∆ u C T (3) 30
31. 31. And for ideal gas ⇒ p.v = m . R. TFor m = 1 kgp.v = R . T p ∆ v = ∆ R T (4)Sub . equations (2) , (3) , (4) in eq. (1) (1) ‫عوض المعادلت 2 ، 3 ، 4 في معادلة‬ Cp ∆ = v ∆= ∆ T C T R T Cp = v C +R R =p C −v CHome work R =p C −v CProve that : -The relation between R and ( γ, C p , Cv )R= Cp - Cv (1) Cp Cp = v . C γ γ= Cv ⇒ (2) ‫بتعويض معادلة )2( في معادلة )1( نحصل على‬ R= v . C γ− C v R = v C (γ 1) − 31
32. 32. Cv = R γ −1 (3) ‫بتعويض معادلة )3( في معادلة )2( نحصل على‬ Cp = v . C γ R γ. R Cp = .γ = γ−1 γ −1 γR Cp = γ−1Home work RProve that 1) Cv = γ −1 γR 2) Cp = γ −1 Gas Constant , the universal gas constant and : ‫السبوع الحادي عشر‬ specific , Examples ‫الهدف من الدرس : 1- ان يفهم الطالب معنى ثابت الغاز وانواعه‬The general equation of Ideal Gas ‫المعادلة العامة للغاز المثالي‬ P .V T =C = constant ----- (1) ‫( نحصل على‬m) ‫بقسمة طرفي المعادلة )1( على الكتلة‬ P .V C C m .T = m Let R= m 32
33. 33. P .V m .T ≠ R ----- (2) constant gas R = characteristic ‫ثابت الغاز النحاس‬ ‫المعادلة العامة للغاز المثالي‬ P. V =m . R . T ----- (3)Units N P = absolute pressure m2V = volume (m3)T = temp. (k0) m = mass (kg) KJ JR = kg .k 0 , kg .k R0 = R * MR0 = universal gas constant ‫ثابت الغاز العام‬M = Molecular weight (k mole) ‫)الوزن الجزيئي )كيلو مول‬ KJ R0 = 8.314 k mole . k 0Example :- A tank has a volume of (0.5 in3) and contains (10 kg) of anideal gas having a molecular weight = 24 , the temp = 25 C0 compute thepressureSolutionR0 = R . M 33
34. 34. KJ P 8.3144 k mole . k 0 KJ R = = = = 0.346 M 24 kg kg.k 0 k mole P . V =m .R . T KJ P * 0.5 m 3 = 10 kg * 0.346 * ( 25 +273) k 0 kg . k 0 10 * 0.346 * 298 P = 0.5 =2066 kpaEx . (5) :-one kg of a perfect gas accupies a volume of (0.85 m3) at (15C0) and at aconstant pressure of (1 bar) . The gas is first heated at a constant volumeand then at a constant pressure .Compute : 1) the specific heat at constant volume (Cv). 2) the specific heat at constant pressure (Cp) γ = 4 1.SolutionP1 V1 = m . R . T1 KN 1*10 9 m2 * 0.85 m 3 = 1 (kg) * R * (15 +273) k0 1 * 10 5 * 0.85 R = 1 * 288 KJ R = 0.295 kg . k 0 R 0.295 Cv = = γ−1 1.4 −1 34
35. 35. KJ1) C v = 0.788 kg . k 0 Cp γ ≠ Cv Cp = γ. Cv C p = 4 (0.788) 1. KJ C p = 1.033 kg . k 0QuestionsWhat is the mass of air contained a room (6m x 10m x 4m) If the pressureis (100 kpa) and the temp. is 25 C0 , Assume air to be an ideal gas KJ R = .287 0 kg . kIdeal gas, boyle,s law, chal,s law , examples: ‫السبوع الثاني عشر والثالث عشر‬ ‫الهدف من الدرس : 1- ان يتعرف الطالب على مسخن الغاز المثالي وةقانون بويل وشارل‬ . ‫للغازات المثالية‬ Ideal gas ‫الغاز المثالي‬The ideal gas is defined as the state of substance that follows well – knowBolyle,s and charle,s laws . ‫الغاز المثالي : هو حالة المادة التي تخضع لقانوني بويل وشارل‬Laws of Ideal gasThe physical properties of a gas are controlled by the following variables :- • pressure (P) exerted by the gas . • Volume ( ) occupied by the gas . υ • Temperature (T) of the gas .The behaviour of perfect gas is governed by the following laws :-1) Boyle,s Law 35
36. 36. The absolute pressure of a given mass of ideal gas varies inversely ofits volume when the temp. remain constant ‫الضغط المطلق لكتلة معينة من الغاز المثالي يتناسب عكسيا مع حجم الغاز عند ثبوت درجة‬ . ‫الحرارة‬ 1 Pα V T=C ‫ثابت‬ PV = C P1 V1 = P2 V2 = C T=CUnit of pressure (Pa) (pressure) (1) (P) PV = C (2) V) volume(Charle,s LawThe volume of a given mass of ideal gas varies directly with the temp.when the absolute pressure remain constant . V αT V T = C P = constant ‫ثابت‬ P V1 V T1 = 2 =C T2 P=C P1 = P2Units (1) (2) 36 V
37. 37. V (m 3 ) T (k 0 ) P ( pa )V1 = initial volume ‫الحجم البتدائي‬V2 = final volume ‫الحجم الثاني‬Gay Lussac Law ‫قانون غاي لوساك‬ The absolute pressure of a given mass of ideal gas is proportionaldirectly with the temp. at constant volume . . ‫يتناسب الضغط المطلق لكتلة غاز مثالي تناسبا طرديا مع درجة الحرارة عند ثبوت الحجم‬ ً P αT P T =C ‫ثابت‬ P P = 2 =C ‫ثابت‬ 1 T1 T2 V = constantGeneral Process It is transition of a substance from state (1) [ P1 , V1 , T1] to state (2)[P2 , V2 , T2] a cross a certain path . . ‫وهي تغير المادة من الحالة )1( الولية الى الحالة )2( النهائية باتباع مسار معين‬ P .V= constant =C T 37
38. 38. NP = pressure ( m2 )V = volume m3T = temperature (k0) ‫ولكثر من حالة‬ 2 ← 1 P V1 P V 1 = 2 2 T1 T2Examples (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 iftemp. is constant .Solution :-P1 V1 = P2 V2 ‫قانون بويل‬ T=C1 bar * 2.8 m3 = 14 bar * V2 1 * 2.8 V2 = = m3 14Examples (2)(0.2 m3) of a gas at (50 C0) the gas is heated at a constant pressure until itsvolume reached (0.4 m3) compute the final temp.Solution :- V1 V = 2 =C T1 T2 0.2 0.4 ≠ 50 + 273 T2 38
39. 39. 0.2 T2 = 0.4 [ 50 + 273] 0.4 (323) T2 = = 646 k 0 0.2Examples (3)(2 kg) of a gas at initial pressure of (1.4 bar) and temp. = 40 C0 , R = J188.34 kg .k 0 , 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.SolutionP1 V1 = m R T1 N J 1.4 * 10 5 * V1 = 2 kg * 188.34 * ( 40 +273) k 0 m2 kg .k 0 2 * 188.34 V1 = 1.4 * 10 5 V1 = .842 0 m3b) P = C ‫ثابت‬ V1 V = 2 T1 T2 0.842 1.5 = 40 +273 T2 T2 = 557.6 k0 39
40. 40. Thermodynamic Processes and applications : ‫السبوع الرابع عشر‬ . ‫الهدف من الدرس : 1- ان يتعرف الطالب على الجراءات الثرموداينمكية وتطبيقاتها‬Some Processes for closed systems1- Constant pressure process (Isobaric) P Q = W +∆ u 2 1 2 ∫ W = p dv = p (V2 − 1 ) =m R (T2 − 1 ) P2 = 1 1 V T P WD Q1− =1− = u1− 2 W 2 ∆ 2 V Q1− = C 2 m p [T2 −1 ] T V1 V2 ∆ u = C v [T2 − ] m T1 Q1− = [V 2 −] + C v [T2 −] 2 p V1 m T1 P V1 PV = 2 2 ‫المعادلة العامة‬ 1 P = 2 1 P T1 T2 V1 V2 = T1 T2 40
41. 41. Constant volume process (Isometric) P 2 Q1− = 1− = u1− 2 W 2 ∆ 2 P2 W1− =0 1 2 P1 Q1− = 2 u2 − = C v [T2 − ] u1 m T1 V V1 = V2 ‫ليجاد العلةقة بين درجة الحرارة والضغط‬ P V1 PV 1 = 2 2 V1 = 2 V T1 T2 P P 1 = 2 T1 T23- Constant temp. process [Isothermal]T=C V2 W1−2 = A = ∫ p dv V1 Q1− − 12 2 W = u ∆ ∆ m u = CV [T2 − ] = T1 0 Q12 = 12 W 41
42. 42. V2 W12 = P V1 ln ‫باجراء التكامل على المعادلة‬ 1 V1 ‫ليجاد العلةقة بين الضغط والحجم‬ P V1 P V 1 = 2 2 T1 T2 T1 = 2 T P V1 = V 2 1 P2Ex. 1The pressure of a gas = 1.5 bar at 18 C0 temp. compute1- 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 . KJCp = 1.005 KJ/kg.k Cv = 0.718 kg .kSolutionR = Cp - Cv KJR = 1.005 – 0.718 = 0.278 kg .kP1 V1 = m R T1 mRT1 1 * 0.278 * (18 + 273) V1 = = P1 10 2 *1.5V1 = 0.556 m3 42
43. 43. Q12 = m Cp [T2 – T1] V1 V = 2 T1 T2 0.556 1 = 291 T2T2 = 526.2 k0Q = 1 * 1.005 * [526.2 – 291] = 243.4 KJW = P [V2 – V1] = 1.5 x 102 [1- 0.556] = 66.48 KJAdiabatic Process P V γ=C Cp γ = CV Q1− = 2 0 Q1− − 2 W = u1− ∆ 2 − 1− = u1− W 2 ∆ 2 − 1− = W 2 m C v (T2 −1) T V2 W1−2 = ∫ p dv V1 P V1γ = P 2 V2 1 γ P V1 −P2 V2 m R (T1 − 2 ) T W = 1 = γ− 1 γ− 1 43
44. 44. Ex. 1 : Given [ the mass of the gas = 1 kg , the pressure = 1.5 bar , KJ KJ the temp = 18C0 ] Cv = 0.718 kg .k 0 , Cp= 1.005 kg .k 0compute : 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 C p −v = C R KJ1.005 – 0.718 = 0.278 kg .k 0P1 V1 = m R T1 KJ1.5 * 102 * V1 = 1 kg * 0.278 kg .k 0 * [18 +273] . 1* 0.278 * 291 V1 = 1.5 * 100 = 0.556 m3Q = m Cp [T2 – T1] T2 ‫ليجاد‬ V1 V = 2 T1 T2 0.556 1 = 291 T2 291 T2 = =526.2 k 0 0.556Q = 1 * 1.005 (526.2 – 291)Q = 243.4 KJW = P [ V2 – V1]W = 1.5 * 102 [1 - 0.566]W = 66.48 KJEx . 2 : 44
45. 45. Given [ the pressure of air = 1.013 bar , the volume = 0.827 m3 and thetemp. = 25 C0 , the air is compressed with constant temp. until the pressurebecomes = 13.78 bar .Compute :- the work done to compress the air .Solution V2 W12 =P V1 ln 1 V1 ----- (1) ‫ مجهولة‬V2 ‫ل يمكن ايجاد الشغل لن‬P1 V1 = P2 V2 T=C V2 ‫ليجاد‬ ⇐1.013 * 0.827 = 13.78 V2 1.013 * 0.827 V2 = 13.78 = 0.060 m3 (1) ‫ في معادلة‬V2 ‫عوض عن ةقيمة‬ 0.060 W12 = .013 * 10 2 * 0.827 1 ln 0.827 W12 =83.775 ln 0.073 W12 = .775 83 ( −61 2. ) W12 =−218.653 KJ . ‫الشارة سالبة تدل على ان الشغل داخل الى النظام‬ 45
46. 46. Ex . 3 : Given [ the volume of gas = 0.12 m3 , the temp. = 20 C0 and thepressure = 1.013 bar is compressed adibatically until the volume become = KJ KJ0.024 m3 Cv = 0.718 kg .k 0 , Cp = 1.005 kg.k 0 .Compute : 1) the mass of the gas . 2) the final pressure and temp. 3) the work done .Solution KJR = Cp – Cv = 1.005 – 0.718 = 0.287 kg .k 0P1 V1 = m R T1 KJ1.013 * 102 * 0.12 m3 = m * 0.287 kg .k 0 * [20 + 273] 1.013 * 10 2 * 0.121) m = 0.287 * 293 =0.145 kg Cp 1.005 γ = = = 1.4 CV 0.718 P V1γ =P2 V2γ 1 1.013 (0.12)1.4 = P2 (0.024)1.4  0.12 2) P2 = 1.013    0.024  = 9.64 bar ‫الضغط النهائي‬ P V1 P V = 2 2 ] ‫ ] النهائية‬T2 ‫ليجاد‬ 1 T1 T2 1.013 * 0.12 9.64 * 0.024 = 293 T2 293 * 9.64 * 0.024 T2 = 1.013 * 0.12 =557.7 k03) 46
47. 47. ‫ليجاد الشغل المنجز‬ ‫2‪P V1 − P2 V‬‬ ‫= 21‪W‬‬ ‫1‬ ‫−‪γ‬‬ ‫1‬ ‫420.0 * 2 01* 46. − 21.0 * 2 01 * 310.1‬ ‫9‬ ‫= 21‪W‬‬ ‫− 4.1‬‫1‬‫‪W12 = - 85.282 KJ‬‬ ‫السبوع الخامس عشر : ‪Reversible and irreversibility‬‬‫الهدف من الدرس : 1- ان يتعرف الطالب على منحني النعكاسية او الرجاعية واللانعكاسية‬ ‫74‬
48. 48. . ‫او اللارجاعية‬ Reversible and Irreversible Process : ‫الجراء الرجاعي والجراء اللارجاعي‬ : ‫الجراء النعكاسي والجراء اللانعكاسي‬Reversible process : ‫الجراء النعكاسي‬ The process in which the system and surrounding can be restored tothe initial state without producing any change in the thermodynamicproperties :-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 smallReversible processes ‫الجراءات الرجاعية‬ 1- Friction relative motion ‫حركة نسبية بدون احتكاك‬ 48
49. 49. 2- Extension of spring ‫تمدد النوابض‬ 3- Slow frictionless adiabatic expansion ‫تمدد اديباتي بدون احتكاك‬Irreversible process ‫الجراء اللانعكاسي‬It is state that both the system and surrounding cannot return to originalstate . : ‫السبوع السادس عشر والسابع عشر‬ The second law of thermodynamic Results of the second law of thermodynamicThe second law of thermodynamic 49
50. 50. The first law of thermodynamic indicates that the net heat supplied in acycle is equal to the net work done the gross heat supplied must be greaterthan 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 cycleif 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 andto convert it into work without simultaneously transferring heat from a hotto a cold reservoir . This statement of the second law is related toequilibrium , i.e. work can be obtained from a system only when thesystem is not already at equilibrium . If a system is at equilibrium , nospontaneous process occurs and no work is produced .Evidently , Kelvins principle indicates that the spontaneous process is theheat flow from a higher to a lower temperature , and that only from such aspontaneous process can work be obtained .(2) The principle of Clausius States : It is impossible to devise an engine which , working in a cycle , shallproduce no effect other than the transfer of heat from a colder to a hotterbody . A good example of this principle is the operation of a refrigerator . 50
51. 51. (3) The principle of Planck states : It is impossible to construct an engine which , working in a completecycle , will produce no effect other than raising of a weight and the coolingof a heat reservoir .(4) The Kelvin – Planck Principle : May be obtained by combining the principles of Kelvin and of Planckinto one equivalent statement as the Kelvin – Planck statement of thesecond law . It states : No process is possible whose sole result is theabsorption of heat from a reservoir and the conversion this heat into work . Heat engine Source Q1 Heat engine W Heat engine , heat pump : ‫السبوع الثامن عشر‬ Q2 . ‫الهدف من الدرس : ان يتعرف الطالب على الماكنة الحرارية والمضخة الحرارية‬ Sin kHeat engine ‫الماكنة الحرارية‬ T hot A heat engine is a system operating in a complete cycle and Qindeveloping net work from supply heat . W0 The second law implies that a source 51 Qout T cold
52. 52. of heat supply and sink for the rejectionof heat are both necessary since someheat must be always be rejected bythe system .By first law ∑ ∑ Q= WNet heat supplied = network doneQ1 – Q2 = WQ1 > W the second law of thermodynamicThe thermal efficiency of heat engine ‫الكفاءة الحرارية للماكنة الحرارية‬ W η= Q1 Q1 − Q2 Q2 η= =1− Q1 Q1 Boiler One good example in practice of heat engine is a simple steam cyclein this cycle heat is supplied in boiler work is developed in yurbine heat is condenserrejected in a condenser and small amount of work is required for thepump . 52 pump
53. 53. Simple steam cycleHeat Pump ‫المضخة الحرارية‬ The heat pump is reversed heat engine in the heat pump (orrefrigerator)cycle an amount of heat Q2 is supplied from the cold reservoirand an amount of heat Q1 is rejected to the hot reservoir and there must bea work done on the cycle W T hot Qout Win Qin T cold Heat pump (refrigerator)Q1 = Q2 = W 53
54. 54. There for W > 0 , the heat pump requires an input energy in order totransfer heat from the cold chamber and reject it at higher temp . Expansion valve ‫صمام التمدد‬ Heat pump system Entropy , changes on closed systems and temp - : ‫السبوع التاسع عشر‬ Entropy plan . ‫الهدف من الدرس : 1- ان يتعلم الطالب معنى النتروبي وتغير النتروبي بالنظمة المغلقة‬ . ‫2- يتعرف الطالب على العلةقة بين النتروبي ودرجة الحرارة‬Entropy It is define as thermodynamic property that express the amount ofstorage energy in the system also represent measure of reversible andIrreversible process . 54
55. 55. Temp. – Entropy PlaneFor the reversible process the area under (T-S) plane = the heat (1)dQ = T ds --------- (1) 2 Q = ∫T 1 dsFrom equation (1) dQ ds = T J / k01- The entropy at constant volume T2 P2 ∆ = m C v ln S = m C v ln T1 P12- The entropy at constant pressure T2 V2 ∆ = m C p ln S = m C p ln T1 V1 Cp > v C 55
56. 56. ∆ s at v= c > sp= ∆ cEx : 1 Comput the entropy for reversible process at constant pressure the temp KJvary from 120 C0 to 270 C0 C = 2. 1p kg .kSolution T2 ∆ = m C p ln s T1 KJ 270 +273 ∆ = * 2.1 s 1 ln kg .k 120 + 273 ∆ s =1 2. ln 1.38 KJ ∆ = 2.1* 0.322 = 0.676 s k0 Carnot Cycle : ‫السبوع العشرون‬ ‫الهدف من الدرس : 1- ان يتعرف الطالب على دورة كارنوت مبدأ عملها واهميتها مع‬ . ‫المخطط والكفاءة‬Carnot Cycle A Carnot gas cycle operating in a given temperature range is shown inthe T-s diagram in Fig. 1(a) . One way to carry out the processes of thiscycle is through the use of steady – state , steady – flow devices as shownin Fig. 1(b) . The isentropic expansion process 2-3 and the isentropiccompression process 4-1 can be simulated quite well by a well – designedturbine and compressor respectively , but the isothermal expansion process1-2 and the isothermal compression process 3-4 are most difficult to 56
57. 57. achieve . Because of these difficulties , a steady – flow Carnot gas cycle isnot practical . The Carnot gas cycle could also be achieved in a cylinder – pistonapparatus (a reciprocating engine) as shown in Fig. 4.2 (b) . The Carnotcycle on the p-v diagram is as shown in Fig 4.2 (a) , in which processes1-2 and 3-4 are isothermal while processes 2-3 and 4-1 are isentropic . Weknow that the Carnot cycle efficiency is given by the expression . TL T T η =1 − th = 1− 4 = 1 − 3 TH T1 T2 Fig. 1 . Steady flow Carnot engine 57
58. 58. Fig. 2. Reciprocating Carnot engine 58
59. 59. 59
60. 60. Fig. 3. Carnot cycle on P-v and T-s diagrams 60
61. 61. Fig. 4. Working of Carnot engineSince the working fluid is an ideal gas with constant specific heats , wehave , for the isentropic process , γ−1 γ−1 T1 V  T2 V  = 4 V   ; = 3 V   T4  1  T3  3 Now , T1 = T2 and T4 = T3 , therefore V4 V V1 = 3 = r V2 = compression or expansion ratioCarnot cycle efficiency may be written as , 1 η =1 − r γ− th 1 61
62. 62. From the above equation , it can be observed that the Carnot cycleefficiency increases as "r" increases . This implies that the high thermalefficiency of a Carnot cycle is obtained at the expense of large pistondisplacement . Also, for isentropic processes we have , γ−1 γ−1 P γ P γ T1 T4 = 1 P  4    and T2 T3 = 2 P  3   Since , T1 = T2 and T4 = T3 , we have P P 1 P4 = 2 = rp = P3 pressure ratioTherefore , Carnot cycle efficiency may be written as , 1 ηth = 1 − γ −1 rp γFrom the above equation , it can be observed that , the Carnot cycleefficiency can be increased by increasing the pressure ratio . This meansthat Carnot cycle should be operated at high peak pressure to obtain largeefficiency . 62
63. 63. Ex. 1The highes theortical efficiency of gasoline engine based on the Carnotcycle is 30 y0 if this engine expels its gases into at m . which has temp of300k . compute1) the temp in the cylinder immediately after combustion .2) if the engine absorbs (837 J) of heat from the hot reservoir during eachcycle how much work can it perform in each cycle ,Solution Tc1) η =1 − Th ‫كفاءة كارنون‬ Tc 300 Th = = =429 k 0 1−η 1 −0.3 W2) η= Qh ⇒ W =η * Qh = 0.3 * 837 = 251 JEx . 2 63
64. 64. A Carnot engine is operated between two heat reservoirs at temp of 450 k0and 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 .Solution1) Th =450 k0 , Tc = 350 k0 , Qh =1000 J Qc T = c Qh Th Tc 350 Qc = Qh = 1000 * = 777.7 k 0 Th 450 Tc2) η = 1 − Th 350 η= 1 − 450 = 0.22 = 22 % ‫الكفاءة‬3) the work done = Qh – Qc = 1000 – 777.7 = 222.3 J : ‫السبوع الحادي والعشرون والثاني والعشرون‬ ‫الهدف من الدرس : 1- ان يتعلم الطالب معنى دورة اوتو والحرارة المرجعة وصافي الشغل‬ 64
65. 65. ‫.‬ ‫وكفاءة الدورة‬‫] ‪Otto Cycle : [ Constant Volume Cycle‬‬‫‪The cycle consist of four reversible process‬‬ ‫تتالف الدورة من اربعة اجراءات انعكاسية‬‫مرحلة النضغاط ‪1-2 Adibatic Compression‬‬‫يتم ضغط الغاز في عملية اديباتية يقل الحجم من 1‪ V‬الى 2‪ V‬وترتفع درجة الحرارة من 1‪ T‬الى‬ ‫2‪. T‬‬ ‫ويكون مقدار الشغل المبذول على الغاز ‪Win‬‬ ‫2‪P V1 −P2 V‬‬ ‫= ‪Win‬‬ ‫1‬ ‫−‪γ‬‬ ‫1‬‫شوط التحتراق ‪2–3 Combustion stroke‬‬ ‫‪Constant volume , heat addition‬‬ ‫تسمى مرحلة الحتراق يزداد كل من الضغط ودرجة الحرارة من 2‪ T‬الى 3‪ T‬ويمتص النظام‬ ‫كمية حرارة من الشتعال ‪ (Qh (added heat‬الحرارة المضافة .‬‫)2‪Added heat Qh = m Cv (T3 – T‬‬‫‪3-4 Power stroke Adibatic expansion‬‬ ‫يتمدد الغاز في عملية اديباتية يوداد الحجم من 2‪ V‬الى 1‪ V‬وتقل الحرارة من 3‪ T‬الى 4‪) T‬تسمى‬ ‫مرحلة القوة (‬ ‫56‬
66. 66. ‫4‪P V3 − P4 V‬‬ ‫= ‪Wout‬‬ ‫3‬ ‫−‪γ‬‬ ‫1‬ ‫‪P‬‬ ‫0‪P‬‬ ‫2‪V‬‬ ‫1‪V‬‬ ‫‪A standard dtt cycle‬‬ ‫شوط العادم ‪Exhaust stroke‬‬ ‫‪4-1 Constant volume , heat rejection‬‬ ‫اجراء ثبوت الحجم )التخلص من الحرارة(‬ ‫ويسمى )مرحلة صمام العادم( تنخفض درجة الحرارة من 4‪ T‬الى 1‪ T‬وينخفض الضغط نتيجة‬ ‫لفتح صمام العادم ويعود الضغط الى الضغط الجوي ويفقد النظام كمية حرارة ‪. Qc‬‬‫‪Rejected heat‬‬ ‫] 4‪Q c = m C v [ T1 – T‬‬ ‫66‬
67. 67. ‫الحرارة المنعكسة‬‫‪The Heat efficiency (eff.) of otto cycle‬‬ ‫‪Qc‬‬ ‫1‪T4 −T‬‬ ‫− 1= ‪η‬‬ ‫− 1=‬ ‫‪Qh‬‬ ‫2‪T3 − T‬‬‫معادلة الكفاءة لدورة اوتو الحرارية‬ ‫الجمالي ‪Wnet‬‬ ‫الشغل‬ ‫صافي‬ ‫‪Wnet = Wout − Win‬‬ ‫) 2 ‪( P V3 − 4V 4 ) − P V1 − 2V‬‬ ‫‪P‬‬ ‫1 (‬ ‫‪P‬‬ ‫= ‪Wnet‬‬ ‫3‬ ‫−‪γ‬‬ ‫1‬‫1-5‬ ‫ملحظة : في المخطط يلحظ خط افقي وهو )مرحلة العادم(‬ ‫و 5-1 )مرحلة الخذ( تكونان غير مؤثرتان لنهما متعاكستان .‬ ‫السبوع الثالث والعشرين والرابع والعشرين :‬‫.‪Diesel Cycle net work out put and its eff‬‬ ‫الهدف من الدرس : 1- ان يتعرف الطالب على منحنى دورة ديزل وصافي الشغل الخارج‬ ‫والكفاءة .‬‫‪Diesel Cycle‬‬‫)‪The Diesel cycle is a compression ignition (rather than spark ignition‬‬‫‪engine . Fuel is sprayed into the cylinder at P2 (high pressure) when the‬‬ ‫76‬
68. 68. compression is complete , and there is ignition without spark . Anidealized Diesel engine cycle is shown in figure 1. Figure 1. The ideal Diesel cycleThe thermal efficiency is given by : QL C (T − 4 ) T η Dicsel =1 + =1 + v 1 QH C p (T3 − 2 ) T T1 (T4 / T1 − ) 1 = T2 (T3 / T2 − ) 1This cycle can operate with a higher compression ratio than the Otto cyclebecause only air is compressed and there is no risk of auto – ignition of thefuel . Although for a given compression ratio the Otto cycle has higherefficiency , because the Diesel engine can be operated to highercompression ratio , the engine can actually have higher efficiency than anOtto cycle when both are operated at compression ratios that might beachieved in practice .Muddy Points 68
69. 69. When and where do we use Cv and Cp ? Some definitions use dU=Cv dT isit ever dU = Cp dT ? (MP 3.8)Explanation of the above comparison between Diesel and Otto. (MP 3.9) Air standard diesel engine cycleThe term "compression ignition" is typically used in technical literature todescribe the modern engines commonly called "Diesel engines" . This is incontrast to "spark ignition" for the typical automobile gasoline engines thatoperate on a cycle derived from the Otto cycle . Rudolph Diesel patentedthe compression – ignition cycle which bears his name in the 1890s. 1 rE λ − rC γ − − η= 1 − γ rE 1 − rc−1 − Design of a Diesel Cycle 69
70. 70. The General Idea The Diesel cycle is very similar to the Otto cycle in that both areclosed cycles commonly used to model internal combustion engines . Thedifference between them is that the Diesel cycle is a compression –ignition cycle use fuels that begin combustion when they reach atemperature and pressure that occurs naturally at some point during thecycle and , therefore , do not require a separate energy source (e.g. from aspark plug) to burn . Diesel fuels are mixed so as to combust reliably at theproper thermal state so that Diesel cycle engines run well . (We might note that most fuels will start combustion on their own atsome temperature and pressure . But this is often not intended to occur andcan result in the fuel combustion occurring too early in the cycle . Forinstance , when a gasoline engine – ordinarily an Otto cycle device – is runat overly high compression ratios , it can start "dieseling" where the fuelignites before the spark is generated . It is often difficult to get such anengine to turn off since the usual method of simply depriving it of a sparkmay not work .Stages of Diesel Cycles Diesel Cycles have four stages : compression , combustion ,expansion , and cooling .Compression We start out with air at ambient conditions – often just outside airdrawn into the engine . In preparation for adding heat to the air , we 70
71. 71. compress it by moving the piston down the cylinder . It is in this part ofthe cycle that we contribute work to the air . In the ideal Diesel cycle , thiscompression is considered to be isentropic . It is at this stage that we set the volumetric compression ratio , rwhich is the ratio of the volume of the working fluid before thecompression 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 processbegins just as the piston leaves its bottom dead center position . Becausethe piston is moving during this part of the cycle , we say that the heataddition 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 hotgas expands forcing the piston to travel up in the cylinder . It is in thisphase that the cycle contributes its useful work , rotating the automobile scrankshaft . We make the ideal assumption that this stage in an idealDiesel cycle is isentropic .Piston : moving from bottom dead center to top dead center . 71
72. 72. Cooling Next , the expanded air is cooled down to ambient conditions . In anactual automobile engine , this corresponds to exhausting the air from theengine to the environment and replacing it with fresh air . Since thishappens when the piston is at the top dead center position in the cycle andis not moving , we say this process is isochoric (no change in volume) .Piston : at top dead center . Dual Cycle : ‫السبوع الخامس والعشرون‬ . ‫الهدف من الدرس : 1- يتعرف الطالب على منحنى دورة ديول والمخطط والكفاءة‬Limited Pressure Cycle (or Dual Cycle) : This cycle is also called as the dual cycle , which is shown in Fig. 1Here the heat addition occurs partly at constant volume and partly atconstant pressure . This cycle is a closer approximation to the behavior ofthe actual Otto and Diesel engines because in the actual engines , the 72
73. 73. combustion process does not occur exactly at constant volume or atconstant 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 . 73
74. 74. Fig. 1 Dual cycle on p-v and T-s diagramsAir Standard EfficiencyHeat 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) m C v (T3 − 2 ) +m C T (T4 −T3 ) − m C v (T5 − 1 ) T η = th p m C v (T3 − T2 ) +m C p (T4 − 3 ) T T5 −T1 η =−1 (T3 −T2 ) +γ (T4 −T3 ) th 74
75. 75. P3 V4 V1Let , P2 = rp ; V3 =rc ; V2 =r T2 =T1 r γ− 1 T3 = 2 r p =1 r γ 1 r p T T − T4 = rc = r γ 1 r p rc T3 T1 − γ−1 γ−1 γ−1 T5 V  V V  r  = 4 V   = 4 . 2 V   = c  T4  5   2 V5  r  γ−1 r  T5 = T4  c  = 1 r p rcγ T r  T1 rp rcγ − 1 T η =1 − th {(T 1 r γ − r p − 1 r γ − ) + γ (T1 r γ − rp rc − T1 r γ − rp )} 1 T 1 1 1 ( rp rcγ − ) 1 = 1− {(r p r γ−1 −r γ−1 ) + γ (r p rc r γ −1 − rp r γ −1 )} 1   r p rcγ −1   = η 1 − th r γ−1  ( 1 1    r p − ) + γ r p ( rc − ) From the above equation , it is observed that , a value of rp > 1 results inan increased efficiency for a given value of rc and γ . Thus the efficiencyof the dual cycle lies between that of the Otto cycle and the Diesel cyclehaving the same compression ratio .Mean Effective Pressure Workdone mep = Displacement volume 75
76. 76. m C v (T3 − 2 ) +m C p (T4 − 3 ) −m C v (T5 − 1 ) T T T = V1 − 2 V m C v (γ − ) T1 r −  1 1 V1 − 2 = V   P1  r  p1 r  3− 2 T T γ(T4 − 3 ) T T − 1T map =  + − 5  ( r − )(γ − )  T1 1 1 T1 T1  = p1 r ( r − )(γ − ) 1 1 {r γ −1 ( r p − ) +γ r γ− r p ( rc − ) − r p rcγ − )} 1 1 1 ( 1 = p1 r ( r − )(γ − ) 1 1 {r γ {(r −1 p − ) +γ rp ( rc − )} − r p rcγ − )} 1 1 ( 1 : ‫السبوع السادس والعشرون‬Comparing between Fuel – air and the air standard cycles ‫الهدف من الدرس : 1- ان يتعرف الطالب على المقارنة بين دورة وتود – هواء ودورة الهواء‬ . ‫القياسية‬Fuel – air cycle The simple ideal air standard cycles overestimate the engine efficiencyby a factor of about 2. A significant simplification in the air standardcycles is the assumption of constant specific heat capacities . Heat 76
77. 77. capacities of gases are strongly temperature dependent , as shown byfigure (1) . The molar constant – volume heat capacity will also vary , as will γthe ratio of heat capacities : C p − v C =0 R , γ=C p / Cv If this is allowed for , air standard Otto cycle efficiency falls from 57per cent to 49.4 per cent for a compression ratio of 8 . When allowance is made for the presence of fuel and combustionproducts , there is an even greater reduction in cycle efficiency . This leadsto the concept of a fuel – air cycle which is the same as the ideal airstandard Otto cycle , except that allowance is made for the realthermodynamic behaviour of the gases . The cycle assumes instantaneouscomplete combustion , no heat transfer , and reversible compression andexpansion . Taylor (1966) discusses these matters in detail and providesresults in graphical form . Figure (2) and (3) . INTRODUCTION TO INTERNAL COMBUSTION ENGINES 77
78. 78. Figure (1) : Molar heat capacity at constant pressure of gases above 150C quoted as averages between 150C and abscissa temperatureShow the variation in fuel –air cycle efficiency as a function ofequivalence ratio for a range of compression ratios . Equivalence ratio φis defined as the chemically correct (stoichiometric)air / fuel ratio dividedby the actual air / fuel ratio . The datum conditions at the start of thecompression stroke are pressure (p1) 1.013 bar , temperature (T1) 1150C ,mass fraction of combustion residuals (f) 0.05 , and specific humidity ( ω )0.02 – the mass fraction of water vapour . The fuel 1- octane has the formula C8H16 , and structure 78
79. 79. Figure (2) shows the pronounced reduction in efficiency of the fuel – aircycle for rich mixtures . The improvement in cycle efficiency withincreasing compression ratio is shown in figure 3 , where the ideal airstandard Otto cycle efficiency has been included for comparison . In order to make allowances for the losses due to phenomena such asheat transfer and finite combustion time , it is necessary to developcomputer models . Prior to the development of computer models , estimates were made forthe various losses that occur in real operating cycles . Again consideringthe Otto 79
80. 80. 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 : 80
81. 81. (a) "Finite piston speed losses" occur since combustion takes a finite timeand cannot occur at constant volume . This leads to the rounding of theindicator diagram and Taylor (1960) estimates these losses as being about6 per cent . Figure (3) : Variation of efficiency with compression ratio for a constantvolume fuel – air cycle with 1 – octane fuel for different equivalence ratios (adapted from Taylor (1966)) 81