1. The document discusses a first week thermodynamics course covering measuring units like force and pressure, and terms like specific volume. 2. The second week covers thermodynamic terms like state and process, and classifications of systems as closed, open, or isolated. 3. The third week covers temperature measurement scales, pressure measurement units, and relations between them. It also discusses manometers and barometers. 4. The fourth week will cover work, its kinds, energy and its forms. The goal is for students to understand work and its types, and types and forms of energy.

1. ‫معهد اعداد المدربين التقنيين‬
‫قسم المكائن والمعدات / السيارات‬
‫الثرموداينمك‬
‫الصف الول‬
‫اعداد المدرِسة‬
‫اوراد عبد الطيف‬
‫السبوع 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. 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 Qantity Symbol Unit Symbol of unit
‫الكمية الفيزيائية‬ ‫الرمز‬ ‫الوحدة‬ ‫الوحدة‬
L meter M
Length ‫الطول‬
T second S
Time ‫الزمن‬
M kilogram Kg
Mass ‫الكتلة‬
T kelvin K0
Temperature ‫درجة الحرارة‬
A Ampere A
Electric current ‫تيار كهربائي‬
The secondary SI units
F Newton N
Force ‫القوة‬
P Pascal N
Pressure ‫الضغط‬ Pa = m2
Bar
Bar = 105 pa
ρ
kg
Density ‫كثافة‬ m3
W Joule J = N.m
Work ‫شغل‬
P watt W=
J
Power ‫قدرة‬ S
British unit system ‫النظام البريطاني للوحدات‬
Physical Quantity Symbol Unit Symbol of unit
Length L Inch In
Time T Second S
Mass M Pound T , bm
Temperature T Fahrenheit F0
Force F Pound IbF
Pressure P - IbF/in2 psi
2

3. Density ρ
- 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
N KN
Pressure : 1 bar = 105 pa = 105 m2
=10 3
m2
Work : W 1 KJ = 1000 J = 103 J
J = N.m
Power : 1 h.p = 746 watt
1 KW = 1.34 h.p
J KJ
Watt = S , KW = S
Definitions ‫تعاريف‬
1- Force : It is an agent which consider as a measure of mechanical
effect on the bodies .
2- Pressure : It,s the force per unit area
Force F N N
P =
Area
=
A ‫الوحدات‬ m2
,
mm 2 ,
KN
=kpa
m2
3

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 kg
5- power : It is work done per unit time
w
Power = t
Units watt , kw , h.p
6- Specific gravity ‫كثافة نوعية‬
ρ sub
Sp . gr = ρ H 2O
Questions
1- 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 ثرموديناميكية‬
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
4

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 الثرموديناميكية‬
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"
‫غلف‬
Boundary : It is a surface that separtes between the system and its
surrounding .
‫محيط‬
Surrounding : 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 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 .
5

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 2
Q=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. 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
Questions
1- (20 C0) to Kelvin scale .
2- Convert (400 K0) to Rankine scale .
3- Convert (170 F0) to Kelvin scale .
7

8. Pressure ‫ضغط‬
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 .
Equations ‫المعادلت‬
Pabs = Patm + Pg
The positive guage pressure
Patm = atmospheric pressure
‫الضغط الجوي‬
h = height of the liquid
‫ارتفاع السائل‬
Pg = Pabs – Patm
8

9. >0
The negative guage pressure :- [ vaccume pressure ]
‫ضغط الفراغ‬
Vaccume Pressure = Patm - Pabs <0
Manometer and Barometer
Manometer :- 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 liquid
g =
∆h
= Height of the liquid
Barometer : An instrument for measuring the atmospheric pressure .
9

10. Units of pressure ‫وحدات الضغط‬
1 pat m = 76 cm . Hg = 760 mm Hg
N
1 pat m = 10 5
m2
= 10 5 pa
KN
1 pat m = 10 2
m2
= 10 2 Kpa
1 pat m = 1.01325 bar
Examples
1- change a pressure about 1500 mm Hg to bar .
Sol.
760 mm Hg = 1.0132 bar
1.0132
1500 mm Hg * 760 mm Hg
= 1.999
bar
2- A compound gauge reads (65 cm Hg ) vaccume pressure . Compute
absolute pressure 14 bar .
Sol.
10

11. 76 cm Hg = 1.0132 bar
1.0132 bar
Pvacc = 65 cm Hg * 76 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
Home work
1- Convert pressure 5 Kpa to bar
KN
2- Convert pressure 76 cm Hg to m2
3- Convert pressure 50 pa to mmHg
Example (3)
A manometer is used to measure the pressure in a tank . The fluid is an 0.1
with a specific gravity of (0.87) and the liquid height = 45.2 cm . If
∆h
kg
the barometric pressure = 98.4 kpa , the density of water = 1000 m3 the
m
gravity = 9.78 s2 , Determine the absolute pressure with in the tank in
kpa , at m .
Solution
ρ oil
Sp. Gr. ρ water
kg kg
ρ = 0.87 * 1000
oil = 870
m3 m3
11

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 pa
P1 = 98.4 kpa + 3.853 kpa
P1 = 102.253 kpa
1 atm
P1 = 102.253 kpa * 10 2 kpa = 1.022 atm
Note
kg .m
1 N =
s2
1 kpa = 1000 pa
1 atm = 102 kpa
Questions
KN
1- Convert 500 cm Hg to m2
2- A barometer reads (735 mm Hg) at room temp.
Determine :- the atmospheric pressure 14 bar millibar , kpa ,
kg
ρ
Hg = 13600 m3
12

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 واشكالها‬
Work :
Its may be defined as the product of a force with its corresponding
displacement
x2
W = ∫F .
x1
dx ------- (1)
J J = N.m ‫وحدات الشغل‬
Work 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* ∆x
13

14. ∆x
F body
Notes x1 x2
* 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
V2v
By taking a small element with length of p and width of dv
Area 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
A
Area = area of piston
dL = the dis placement traveled by a piston
14

15. 2
W = ∫p .
1
A dL
F =p.A
2
W = ∫F
1
dL
F = exerted force
P = exerted pressure
‫ فان‬L ‫ملحظة ليجاد الشغل المنجز خلل شوط ةقدره طوله‬
W=F.L
L = strok length (m)
Flow work ‫شغل النسياب‬ (open system)
Flow w on k = F . L
WF = F . L
WF = P . A . L
WF = P . V ‫جول‬ ∆ F
W = 2 V2
P −1 V1
P
The 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. ∆F
W = 2 . V2
P − 1 V1
P
J / kg
Types of Energies ‫انواع الطاقات‬
1- Potential energy (P.E)
The energy that system possesses by virtue of its position relative to
the surface of the earth .
P.E=mgZ
m = mass
Z = elevation , m
∆E = P.E 2 ) 2
P. ( − P.E ) 1
(
∆E =
P. mg ∆ =
Z mg ( Z 2 − 1)
Z
2- 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 2
3- Internal energy
It is the energy stored in the substance total internal energy is given
U = m Cv T
Questions
16

17. kg
1- A gas has a mass of (2 kg) and density = 21.6
m3 is transported by
pipe of height (30.25 m) from earth , the temp = 138 C0 the velocity .
KJ
Flow = 6m / sec and Cv = 0.674 kg .K 0
Compute : 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 واهميته‬
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 .
∑
dQ =∑dw
∑
: the sum for a complete cycle .
17

18. Enthalpy : ‫السبوع السادس‬
. ‫الهدف من الدرس : 1- يتعرف الطالب على معنى النثالبي واهميتها‬
Enthalpy (H) ‫المحتوى الحراري‬
Enthalpy is an extensive property is defined by the relation .
‫خاصيته تعتمد على حجم النظام‬
H=U+P.V (J , kJ)
Specific enthalpy (h)
h =+ .
u P V
V
= specific volume m3 / kg
u
= specific internal energy J/kg
p
= pressure
Heat 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/kg
Heat Sign ‫اشارة الحرارة‬
1- heat added input use ‫عكس اشارة الشغل‬
Q + (+)Q
18 (-)
Q

19. 2- heat rejected or out put us
(-Q)
Net work (Wnet) & Net heat (Qnet)
Wnet = W0 – Win
W0 = out put work
Win = input work
Qnet = Qin – Qout
Qin = input heat
Qo = output heat
19

20. Applying the first law on closed systems : ‫السبوع السابع‬
‫الهدف من الدرس : 1- ان يتعريف الطالب على تطبيق القانون الول للثرموداينميك على‬
. ‫النظمة المغلقة‬
Energy equation ‫معادلة الطاقة‬
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 .
. ‫الطاقة ل تفتى ول تستحدث وانما تتحول من شكل الى آخر‬
Q + U1 + P1V1 + K. E1 + P . E1 = W + U2 + P2V2 + K . E2 + PE2
H = U1 + P1V1
Q–W= ∆
H + K .E
∆ + P.E
∆
equation of energy
‫معادلة الطاةقة‬
We can written thia equation
1
Q – 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. 1- Non – Flow energy equation (N. F. E. E)
Closed system
For closed system PV , K.E , P.E = 0
The energy equation become :-
Q − =
W ∆U
2- Steady state flow energy equation (S. F. E. E)
(open system)
For open system and steady state
min =m out
‫الكتلة ل تتغير مع الزمن‬
By 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
2
‫معادلة 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. Examples
Closed system
Ex. 1
The change in the internal energy of closed system increase to (120 KJ)
while (150 KJ) of work , that go out of the system , Determine the amount
of heat transfer a cross system boundaries ?
Is the heat added or rejected ?
Solution
From N. F. E.E
Q1− − 1− = u
2 W 2 ∆
Q−150 =120
Q = 270 KJ
The sign (+270) therefore the heat added to the system
Note ‫ملحظة‬
If the change in the internal energy increase we use (+ ∆u
)
If the change in the internal energy decrease we use ( − u
∆
)
22

23. 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 .
Sol.
From N. F.E.E
Q −=
W ∆
u
Q = - 1500 KJ (The heat out)
W = - 5090 KJ (the work input to the system)
- 1500 – (-5090) = ∆U
∆ = 1500 +5090
U −
∆ = 3590
U KJ
Questions
1- A mass of oxygen is compressed with out friction from initial state
volume = 0.14 m3 pressure = 1.5 bar ; the temp = 15 C0 to final state the
volume = 0.06 m3 and the pressure = 1.5 bar at this process the oxygen
losses quantity of heat = 0.6 KJ to the surrounding Cv = 0.649 KJ/Kg.k
R = 0.25 KJ/Kg.k0 .
Find : 1) the change of internal energy .
2) the final temp.
23

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 المفتوحة‬
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
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 مضافة‬
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 .
1
1
Q −W =m [( h2 − 1 ) + (C 2 − 1 ) + ( Z 2 −Z 1 )
h 2
C2 g
2
Q = m (h2 – h1)
2 ter
wa
h1 > h2
Q (-1) rejected (heat out) ‫حرارة مسحوبة‬
Ex. 1
24

25. A boiler is generate steam with rate of 8 kg/s the enthalpy of liquid water
KJ KJ
is 271 kg and the enthalpy of steam is 3150 kg . For steady state
conditions and by neglecting the change in K.E and P.E . Determine the
rate of heat added to the steam in Boiler .
Solution
Q = m (h2 – h1)
KJ
h1 = 271 KJ/kg h2 = 3150 kg
m = 8 kg/s
Q = m (h2 – h1)
KJ
= 8 (3150 – 271) = 23032 S = 23032 KW
‫ ( لنها حرارة مضافة‬T ) ‫الحرارة‬
3- The Turbine
It is a mechanical device used to convert the kinetic energy of fluid into
mechanical work
input
T1 > 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 – h2
4- The Compressor
25

26. It is a mechanical device used to increase fluid pressure by using on
external mechanical work .
h1 > h2 T2 > T1
2
P2 > 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 – 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 .
Sol.
5000 KJ
h1 = 5000 J / kg = 1000 kg
KJ
h2 = 250 kg
m = 0.25 kg / s
w = m (h1 – h2) = - 0.25 (5 – 250)
w = 0.25 (-245) = - 61.25 KW
‫الشارة سالبة لنه شغل داخل الى النظام‬
Specific 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. Specific Heat ‫الحرارة النوعية‬
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 ‫الحرارة النوعية عند ثبوت الضغط‬
∆h
Cp =
∆T *
Cp
= Specific heat at constant pressure.
KJ J
Unit [ kg.k 0
or
kg . k 0 ]
= change in enthalpy ‫تغير النثالبي‬
∆
h = h2 −h1
KJ J
Unit [ kg
or
kg ]
= change in temp. ‫تغير في درجة الحرارة‬
∆ =2 −
T T T1
Unit [ k0 ]
Q1-2 = m ∆h
*
Q1-2 = m cp ∆T
*
m = mass (kg) ‫كتلة‬
27

28. T2 = Final temp. (k0) ‫درجة حرارة ثنائية‬
T1 = initial temp. (k0) ‫درجة حرارة ابتدائية‬
Q = heat ‫حرارة‬ (KJ , J)
2- Specific heat at constant volume ‫الحرارة النوعية بثبوت الحجم‬
V = constant ‫حجم ثابت‬ closed system ‫نظام مغلق‬
∆u
Cv =
∆T
= Specific heat at constant volume ‫الحرارة النوعية بثبوت الحجم‬
Cv
KJ J
Unit ( kg .k 0
) or (
kg . k
)
= change in internal energy ‫التغير في الطاةقة الداخلية‬
∆u
∆ = u 2 −1
u u
KJ J
Unit [ kg
or
kg ]
∆T
= change in temp. (k0) = T2 - T1
For 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. The relation between ( C p & Cv
)
Cp
γ =
‫بدون وحدات‬
γ
Cv
Cama (‫ -: )كاما‬Is the adiabatic expone which represent the ratio
γ
between specific heat at constant pressure and specific heat at constant
volume .
γ >1 Cp > Cv
Examples
1- compute the constant pressure specific heat of steam if the change in
enthalpy is (104.2) KJ and the change in temp. is (50 KJ) .
Solution
∆h 104.2
Cp =
∆T
=
50
= 2.084 KJ / kg . k0
2- Example (2)
Compute the change of enthalpy as (1 kg) of a gasis heated from 300 k0
KJ
to 1500 k0 γ
=
1 .4
, C v = 0.718
kg.k 0 .
Sol.
29

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
kg
3- 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
p
Sol.
Cp 1.0714 KJ
1- 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 KJ
The 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. And for ideal gas ⇒
p.v = m . R. T
For m = 1 kg
p.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
C
Home work
R =p
C −v
C
Prove 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. 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−1
Home work
R
Prove 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 وانواعه‬
The 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. 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 m2
V = volume (m3)
T = temp. (k0)
m
= mass (kg)
KJ J
R = kg .k 0
,
kg .k
R0 = R * M
R0 = universal gas constant ‫ثابت الغاز العام‬
M = Molecular weight (k mole) ‫)الوزن الجزيئي )كيلو مول‬
KJ
R0 = 8.314
k mole . k 0
Example :- A tank has a volume of (0.5 in3) and contains (10 kg) of an
ideal gas having a molecular weight = 24 , the temp = 25 C0 compute the
pressure
Solution
R0 = R . M
33

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 kpa
Ex . (5) :-
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) γ = 4
1.
Solution
P1 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. KJ
1) 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 0
Questions
What is the mass of air contained a room (6m x 10m x 4m) If the pressure
is (100 kpa) and the temp. is 25 C0 , Assume air to be an ideal gas
KJ
R = .287
0
kg . k
Ideal 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 المثالي‬
The ideal gas is defined as the state of substance that follows well – know
Bolyle,s and charle,s laws .
‫الغاز المثالي : هو حالة المادة التي تخضع لقانوني بويل وشارل‬
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 .
υ
• Temperature (T) of the gas .
The behaviour of perfect gas is governed by the following laws :-
1) Boyle,s Law
35

36. The absolute pressure of a given mass of ideal gas varies inversely of
its 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=C
Unit of pressure (Pa)
(pressure) (1)
(P) PV = C
(2)
V) volume(
Charle,s Law
The 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 = P2
Units (1) (2)
36
V

37. V (m 3 )
T (k 0 )
P ( pa )
V1 = initial volume ‫الحجم البتدائي‬
V2 = final volume ‫الحجم الثاني‬
Gay Lussac Law ‫قانون غاي لوساك‬
The absolute pressure of a given mass of ideal gas is proportional
directly 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 = constant
General Process
It is transition of a substance from state (1) [ P1 , V1 , T1] to state (2)
[P2 , V2 , T2] a cross a certain path .
. ‫وهي تغير المادة من الحالة )1( الولية الى الحالة )2( النهائية باتباع مسار معين‬
P .V= constant
=C
T
37

38. N
P = pressure ( m2 )
V = volume m3
T = temperature (k0)
‫ولكثر من حالة‬
2 ←
1
P V1 P V
1
= 2 2
T1 T2
Examples (1)
An air compress or is compress (2.8 m3) of air from initial pressure of
(1 bar) to final , pressure of (14 bar) calculate the final volume of air if
temp. is constant .
Solution :-
P1 V1 = P2 V2 ‫قانون بويل‬ T=C
1 bar * 2.8 m3 = 14 bar * V2
1 * 2.8
V2 = = m3
14
Examples (2)
(0.2 m3) of a gas at (50 C0) the gas is heated at a constant pressure until its
volume reached (0.4 m3) compute the final temp.
Solution :-
V1 V
= 2 =C
T1 T2
0.2 0.4
≠
50 + 273 T2
38

39. 0.2 T2 = 0.4 [ 50 + 273]
0.4 (323)
T2 = = 646 k 0
0.2
Examples (3)
(2 kg) of a gas at initial pressure of (1.4 bar) and temp. = 40 C0 , R =
J
188.34 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.
Solution
P1 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
m3
b) P = C ‫ثابت‬
V1 V
= 2
T1 T2
0.842 1.5
=
40 +273 T2
T2 = 557.6 k0
39

40. Thermodynamic Processes and applications : ‫السبوع الرابع عشر‬
. ‫الهدف من الدرس : 1- ان يتعرف الطالب على الجراءات الثرموداينمكية وتطبيقاتها‬
Some Processes for closed systems
1- 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. 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 T2
3- 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. 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 P2
Ex. 1
The pressure of a gas = 1.5 bar at 18 C0 temp. compute
1- the volume of (1 kg) of the gas .
If the gas is heated at constant pressure until the volume become [1m3]
Compute : a- the a mount of added heat .
b- the work done .
KJ
Cp = 1.005 KJ/kg.k Cv = 0.718 kg .k
Solution
R = Cp - Cv
KJ
R = 1.005 – 0.718 = 0.278 kg .k
P1 V1 = m R T1
mRT1 1 * 0.278 * (18 + 273)
V1 = =
P1 10 2 *1.5
V1 = 0.556 m3
42

43. Q12 = m Cp [T2 – T1]
V1 V
= 2
T1 T2
0.556 1
=
291 T2
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
Adiabatic 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. 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 0
compute : 1- the volume of the gas , the gas is heated at constant pressure
until the volume become (1m3).
2- the added heat .
3- the work done .
Solution
C p −v =
C R
KJ
1.005 – 0.718 = 0.278 kg .k 0
P1 V1 = m R T1
KJ
1.5 * 102 * V1 = 1 kg * 0.278 kg .k 0 * [18 +273] .
1* 0.278 * 291
V1 =
1.5 * 100
= 0.556
m3
Q = m Cp [T2 – T1]
T2 ‫ليجاد‬
V1 V
= 2
T1 T2
0.556 1
=
291 T2
291
T2 = =526.2 k 0
0.556
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 :
44

45. Given [ the pressure of air = 1.013 bar , the volume = 0.827 m3 and the
temp. = 25 C0 , the air is compressed with constant temp. until the pressure
becomes = 13.78 bar .
Compute :- the work done to compress the air .
Solution
V2
W12 =P V1 ln
1
V1 ----- (1)
‫ مجهولة‬V2 ‫ل يمكن ايجاد الشغل لن‬
P1 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. 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 =
KJ KJ
0.024 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
KJ
R = Cp – Cv = 1.005 – 0.718 = 0.287 kg .k 0
P1 V1 = m R T1
KJ
1.013 * 102 * 0.12 m3 = m * 0.287 kg .k 0 * [20 + 273]
1.013 * 10 2 * 0.12
1) 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
k0
3)
46

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
‫− 4.1‬‫1‬
‫‪W12 = - 85.282 KJ‬‬
‫السبوع الخامس عشر : ‪Reversible and irreversibility‬‬
‫الهدف من الدرس : 1- ان يتعرف الطالب على منحني النعكاسية او الرجاعية واللانعكاسية‬
‫74‬

48. . ‫او اللارجاعية‬
Reversible and Irreversible Process
: ‫الجراء الرجاعي والجراء اللارجاعي‬
: ‫الجراء النعكاسي والجراء اللانعكاسي‬
Reversible process : ‫الجراء النعكاسي‬
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 ‫الجراءات الرجاعية‬
1- Friction relative motion ‫حركة نسبية بدون احتكاك‬
48

49. 2- Extension of spring ‫تمدد النوابض‬
3- Slow frictionless adiabatic expansion ‫تمدد اديباتي بدون احتكاك‬
Irreversible process ‫الجراء اللانعكاسي‬
It is state that both the system and surrounding cannot return to original
state .
: ‫السبوع السادس عشر والسابع عشر‬
The second law of thermodynamic
Results of the second law of thermodynamic
The second law of thermodynamic
49

50. 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 .
50

51. (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
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 k
Heat engine ‫الماكنة الحرارية‬ T hot
A heat engine is a system operating in a complete cycle and
Qin
developing net work from supply heat .
W0
The second law implies that a source
51 Qout
T cold

52. 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
∑ ∑
Q= W
Net heat supplied = network done
Q1 – Q2 = W
Q1 > W the second law of thermodynamic
The 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 cycle
in this cycle heat is supplied in boiler work is developed in yurbine heat is
condenser
rejected in a condenser and small amount of work is required for the
pump .
52
pump

53. Simple steam cycle
Heat Pump ‫المضخة الحرارية‬
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
T hot
Qout
Win
Qin
T cold
Heat pump (refrigerator)
Q1 = Q2 = W
53

54. 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 .
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 الحرارة‬
Entropy
It is define as thermodynamic property that express the amount of
storage energy in the system also represent measure of reversible and
Irreversible process .
54

55. Temp. – Entropy Plane
For the reversible process the area under (T-S) plane = the heat (1)
dQ = T ds --------- (1)
2
Q = ∫T
1
ds
From equation (1)
dQ
ds =
T J / k0
1- The entropy at constant volume
T2 P2
∆ = m C v ln
S = m C v ln
T1 P1
2- The entropy at constant pressure
T2 V2
∆ = m C p ln
S = m C p ln
T1 V1
Cp > v
C
55

56. ∆
s at v=
c > sp=
∆ c
Ex : 1
Comput the entropy for reversible process at constant pressure the temp
KJ
vary from 120 C0 to 270 C0 C = 2. 1p
kg .k
Solution
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 والكفاءة‬
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
56

57. 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 .
TL T T
η =1 −
th = 1− 4 = 1 − 3
TH T1 T2
Fig. 1 . Steady flow Carnot engine
57

58. Fig. 2. Reciprocating Carnot engine
58

59. 59

60. Fig. 3. Carnot cycle on P-v and T-s diagrams
60

61. Fig. 4. Working of Carnot engine
Since the working fluid is an ideal gas with constant specific heats , we
have , 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 ratio
Carnot cycle efficiency may be written as ,
1
η =1 −
r γ−
th 1
61

62. 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 ,
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 ratio
Therefore , Carnot cycle efficiency may be written as ,
1
ηth = 1 − γ −1
rp γ
From the above equation , it can be observed that , the Carnot cycle
efficiency can be increased by increasing the pressure ratio . This means
that Carnot cycle should be operated at high peak pressure to obtain large
efficiency .
62

63. Ex. 1
The highes theortical efficiency of gasoline engine based on the Carnot
cycle is 30 y0 if this engine expels its gases into at m . which has temp of
300k . compute
1) the temp in the cylinder immediately after combustion .
2) if the engine absorbs (837 J) of heat from the hot reservoir during each
cycle how much work can it perform in each cycle ,
Solution
Tc
1) η =1 −
Th ‫كفاءة كارنون‬
Tc 300
Th = = =429 k 0
1−η 1 −0.3
W
2) η=
Qh
⇒ W =η * Qh = 0.3 * 837 = 251 J
Ex . 2
63

64. A Carnot engine is operated between two heat reservoirs at temp of 450 k0
and 350 k0 , if the engine receive (1000 J) of heat in each cycle .
Compute : 1) the amount of heat reject .
2) the efficiency of the engine .
3) the work done by the engine in each cycle .
Solution
1) 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
Tc
2) η = 1 −
Th
350
η= 1 −
450
= 0.22 = 22 %
‫الكفاءة‬
3) 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. ‫.‬ ‫وكفاءة الدورة‬
‫] ‪Otto Cycle : [ Constant Volume Cycle‬‬
‫‪The cycle consist of four reversible process‬‬
‫تتالف الدورة من اربعة اجراءات انعكاسية‬
‫مرحلة النضغاط ‪1-2 Adibatic Compression‬‬
‫يتم 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 ‫1‬
‫شوط 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 – T‬‬
‫‪3-4 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. ‫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 ‪. Qc‬‬
‫‪Rejected heat‬‬ ‫] 4‪Q c = m C v [ T1 – T‬‬
‫66‬

67. ‫الحرارة المنعكسة‬
‫‪The Heat efficiency (eff.) of otto cycle‬‬
‫‪Qc‬‬ ‫1‪T4 −T‬‬
‫− 1= ‪η‬‬ ‫− 1=‬
‫‪Qh‬‬ ‫2‪T3 − T‬‬
‫معادلة 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 ‫1‬
‫1-5‬
‫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 ignition‬‬
‫‪engine . Fuel is sprayed into the cylinder at P2 (high pressure) when the‬‬
‫76‬

68. 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 :
QL C (T − 4 )
T
η Dicsel =1 + =1 + v 1
QH C p (T3 − 2 )
T
T1 (T4 / T1 − )
1
= T2 (T3 / T2 − )
1
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 .
Muddy Points
68

69. 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.
1 rE λ − rC γ
− −
η= 1 −
γ rE 1 − rc−1
−
Design of a Diesel Cycle
69

70. The General Idea
The Diesel cycle is very similar to the Otto cycle in that both are
closed cycles commonly used to model internal combustion engines . The
difference between them is that the Diesel cycle is a compression –
ignition cycle use fuels that begin combustion when they reach a
temperature and pressure that occurs naturally at some point during the
cycle and , therefore , do not require a separate energy source (e.g. from a
spark plug) to burn . Diesel fuels are mixed so as to combust reliably at the
proper thermal state so that Diesel cycle engines run well .
(We might note that most fuels will start combustion on their own at
some temperature and pressure . But this is often not intended to occur and
can result in the fuel combustion occurring too early in the cycle . For
instance , when a gasoline engine – ordinarily an Otto cycle device – is run
at overly high compression ratios , it can start "dieseling" where the fuel
ignites before the spark is generated . It is often difficult to get such an
engine to turn off since the usual method of simply depriving it of a spark
may not work .
Stages of Diesel Cycles
Diesel Cycles have four stages : compression , combustion ,
expansion , and cooling .
Compression
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
70

71. compress it by moving the piston down the cylinder . It is in this part of
the cycle that we contribute work to the air . In the ideal Diesel cycle , this
compression is considered to be isentropic .
It is at this stage that we set the volumetric compression ratio , r
which is the ratio of the volume of the working fluid before the
compression process to its volume after .
Piston : moving from top dead center to bottom dead center .
Combustion
Next , heat is added to the air by fuel combustion . This process
begins just as the piston leaves its bottom dead center position . Because
the piston is moving during this part of the cycle , we say that the heat
addition is isochoric , like the cooling process .
Piston : starts at bottom dead center , begins moving up .
Expansion
In the Diesel cycle , fuel is burned to heat compressed air and the hot
gas expands forcing the piston to travel up in the cylinder . It is in this
phase that the cycle contributes its useful work , rotating the automobile 's
crankshaft . We make the ideal assumption that this stage in an ideal
Diesel cycle is isentropic .
Piston : moving from bottom dead center to top dead center .
71

72. Cooling
Next , the expanded air is cooled down to ambient conditions . In an
actual automobile engine , this corresponds to exhausting the air from the
engine to the environment and replacing it with fresh air . Since this
happens when the piston is at the top dead center position in the cycle and
is not moving , we say this process is isochoric (no change in volume) .
Piston : at top dead center .
Dual Cycle : ‫السبوع الخامس والعشرون‬
. ‫الهدف من الدرس : 1- يتعرف الطالب على منحنى دورة ديول والمخطط والكفاءة‬
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
72

73. 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 .
73

74. 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)
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. P3 V4 V1
Let , 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 in
an increased efficiency for a given value of rc and γ
. 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
Workdone
mep =
Displacement volume
75

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 والعشرون‬
Comparing between Fuel – air and the air standard cycles
‫الهدف من الدرس : 1- ان يتعرف الطالب على المقارنة بين دورة وتود – هواء ودورة الهواء‬
. ‫القياسية‬
Fuel – air cycle
The simple ideal air standard cycles overestimate the engine efficiency
by a factor of about 2. A significant simplification in the air standard
cycles is the assumption of constant specific heat capacities . Heat
76

77. capacities of gases are strongly temperature dependent , as shown by
figure (1) .
The molar constant – volume heat capacity will also vary , as will γ
the 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 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
77

78. 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 φ
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 ( ω
)
0.02 – the mass fraction of water vapour .
The fuel 1- octane has the formula C8H16 , and structure
78

79. Figure (2) shows the pronounced reduction in efficiency of the fuel – air
cycle for rich mixtures . The improvement in cycle efficiency with
increasing compression ratio is shown in figure 3 , where the ideal air
standard Otto cycle efficiency has been included for comparison .
In order to make allowances for the losses due to phenomena such as
heat transfer and finite combustion time , it is necessary to develop
computer models .
Prior to the development of computer models , estimates were made for
the various losses that occur in real operating cycles . Again considering
the Otto
79

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. (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))
81

82. (b) "Heat losses" , in particular between the end of the compression stroke
and the beginning of the expansion stroke . Estimates of up to 12 per cent
have been made by both Taylor (1966) and Ricardo and Hempson (1968) .
However, with no heat transfer the cycle temperatures would be raised and
the fuel – air cycle efficiencies would be reduced slightly because of
increasing gas specific heats and dissociation .
(c ) Exhaust losses due to the exhaust valve opening before the end of the
expansion stroke . This promotes gas exchange but reduces the expansion
work . Taylor (1966) estimates these losses as 2 per cent .
Since the fuel is injected towards the end of the compression stroke in
compression ignition engines (unlike the spark ignition engine where it is
pre-mixed with the air) the compression process will be closer to ideal in
the compression ignition engine than in the spark ignition engine . This is
another reason for the better fuel economy of the compression ignition
engine .
The actual cycle : ‫السبوع السابع والعشرين والثامن والعشرين‬
82

83. Comparing between actual cycles and air standard cycles
. ‫الهدف من الدرس : 1- ان يتعرف الطالب مفهوم الدراسة الحقيقية‬
. ‫2- يقارن بين الدراسة الحقيقية ودورات الهواء القياسية‬
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 .
83

84. 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 :‫السبوع التاسع والعشرين والثلثون‬
84

85. ‫الهدف من الدرس : يتعلم الطالب معنى التوربين الغازي واستخداماته ومكوناته‬
Gas Turbine
Gas turbine . types which burn fuels such as oil and natural gas .
Instead of using heat to produce steam , as in steam turbines , the gas
turbine hot gas is used directly . Used for the operation of gas turbine
generators , ships , race cars , as used in jet engines .
Most of the gas turbine systems of the three main parts
1- the air compressor , 2- the combustion chamber , 3- turbine
The so–called air compressor with the combustion chamber , usually ,
the gas generator . In most systems , gas turbine , the air compressor and
turbine boats on both sides of a horizontal axis , located between the
combustion chamber . And is part of the power turbine air compressor .
Absorbs quantity of air compressor and air ldguetha , so pressure is
increasing . In the combustion chamber , compressed air mixes with fuel
and burning the mixture . The more air pressure , improved combustion of
fuel mixed with air . The burning gases expand rapidly and to flow into the
turbine , leading to the rotation of the wheels of the turbine . And moving
hot gases through the various stages in the same manner as the gas turbine
flows through the steam turbine vapor . The Downs fixed mobile gas resat
member to change the rotor speed .
And benefit most from the gas turbine systems , the hot gases
emerging from the turbine . In some systems are some of these gases , and
go to a so – called renewed . And these gases are used to heat the
85