1. THERMODYNAMICS Part 1
By. Engr. Yuri G. Melliza
Terms & Definition
Properties of Fluids
Forms of Energy
Law of Conservation of Mass
Law of Conservation of Energy
(First Law of Thermodynamics)
Ideal Gas
Pure Substance
Processes of Fluids
Zeroth Law of Thermodynamics

2. Thermodynamics is a science that deals with
energy transformation or conversion of one
form of energy to another form
Therme – “Heat”
Dynamis – “Strength”
System: A portion in the universe, an Atom, a
Galaxy, a certain quantity of matter, or a certain
volume in space in which one wishes to study.
 It is a region enclosed by a specified boundary
that may Imaginary, Fixed or Moving.

3. System
Surrounding or
Environment

4. Open System: A system open to matter flow.
Example: Internal Combustion Engine (ICE)

5. Closed System: A system close to matter flow.
Example: Piston - in - cylinder

6. Working Substance (Working Fluid): A fluid
(Liquid or Gas) responsible for the
transformation of energy.
Example:
 air in an air compressor
 Air and fuel mixture in an internal combustion
engine

7. Pure Substance: A substance that is
homogeneous in nature and is homogeneous.
Example : Water
Phases of a Substance
A phase refers to a quantity of matter that is
homogeneous throughout in both chemical composition
and physical structure.
 Solid
 Liquid
 Gas or Vapor

8. Specific Terms To Characterized Phase
Transition
SOLIDIFYING OR FREEZING - Liquid to
Solid
MELTING - Solid to Liquid
VAPORIZATION - Liquid to Vapor
CONDENSATION - Vapor to Liquid
SUBLIMATION - a change from solid
directly to vapor phase without passing the
liquid phase.

9. Mass : It is the absolute quantity of matter in
it.
m - mass in kg
Acceleration : it is the rate of change of
velocity with respect to time t.
a = dv/dt m/sec2
Velocity: It is the distance per unit time.
v = d/t m/sec

10. Force - it is the mass multiplied by the acceleration.
F = ma/1000 KN
1 kg-m/sec2 = Newton (N)
1000 N = 1 Kilo Newton (KN)
Newton - is the force required to accelerate 1 kg mass at the rate of 1 m/sec
per second.
1 N = 1 kg-m/sec2
From Newton`s Law Of Gravitation: The force of attraction between two
masses m1 and m2 is given by the equation:
Fg = Gm1m2/r2 Newton
Where: m1 and m2 - masses in kg
r - distance apart in meters
G - Gravitational constant in N-m2/kg2
G = 6.670 x 10 -11 N-m2/kg2
WEIGHT - is the force due to gravity.
W = mg/1000 KN
Where: g - gravitational acceleration at sea level, m/sec2
g = 9.81 m/sec2

11. 3
m
kg
V
mρ
PROPERTIES OF FLUIDS
Where: - density in kg/m3
m - mass in kg
V – volume in m3
Specific Volume ( ) - it is the volume per unit mass or the
reciprocal of its density.
kg
m
m
V
3
υ
kg
m3
ρ
1υ
Density ( ) - it is the mass per unit volume.

12. Specific Weight ( ) - it is the weight per unit volume.
3
3
m
KN
m
KN
1000V
mg
γ
V
W
γ
Where: - specific weight in KN/m3
m – mass in kg
V – volume in m3
g – gravitational
At standard condition:
g = 9.81 m/sec2

13. Specific Gravity Or Relative Density (S):
FOR LIQUIDS: Its specific gravity or relative density is equal tothe ratio of its
density to that of water at standard temperature and pressure.
w
L
w
L
L
γ
γ
ρ
ρ
S
FOR GASES: Its specific gravity or relative density is equal to theratio of its
density to that of either air or hydrogen at some specified temperature and
pressure
ah
G
G
ρ
ρ
S
Where at standard condition:
w = 1000 kg/m3
w = 9.81 KN/m3

14. Temperature: It is the measure of the intensity of heat in a body.
Fahrenheit Scale:
Boiling Point = 212 F
Freezing Point = 32 F
Centigrade or Celsius Scale:
Boiling Point = 100 C
Freezing Point = 0 C
Absolute Scale:
R = F + 460 (Rankine)
K = C + 273 (Kelvin)
32F8.1F
8.1
32F
C
Conversion

15. Pressure: It is the normal component of a force per unit area.
KPaor
2m
KN
A
F
P
Where: P – pressure in KN/m2 or KPa
F – normal force in KN
A – area in m2
1 KN/m2 = 1 KPa (KiloPascal)
1000 N = 1 KN
If a force dF acts on an infinitesimal area dA, the intensity of Pressure is;
KPaor
2m
KN
dA
dF
P

16. Pascal’s Law: At any point in a homogeneous fluid at rest the pressures are
the same in all directions:
y
x
z
A
B
C
P1A1
P2A2
P3A3

17. Fx = 0
From Figure:
P1A1 - P3 A3sin = 0
P1A1 = P3A3sin Eq.1
P2A2 - P3A3cos = 0
P2A2 = P3A3 cos Eq.2
sin = A1/A3
A1 = A3sin Eq.3
cos = A2/A3
A2 = A3cos Eq.4
substituting eq. 3 to eq. 1 and eq.4 to eq.2
P1 = P2 = P3

18. Atmospheric Pressure (Pa):It is the average pressure exerted by
the atmosphere.
At sea level
Pa = 101.325 KPa
= 0.101325 MPa
= 1.01325 Bar
= 760 mm Hg
= 10.33 m of water
= 1.033 kg/cm2
= 14.7 lb/in2
Pa = 29.921 in Hg
= 33.88 ft. of water
100 KPa = 1 Bar
1000 KPa = 1 MPa

19. Absolute and Gauge Pressure
Absolute Pressure: It is the pressure measured referred to
absolute zero using absolute zero as the base.
Gauge Pressure: it is the pressure measured referred to the
existing atmospheric pressure and using atmospheric pressure as
the base.
Pgauge – if it is above atmospheric
Pvacuum – negative gauge or vacuum if it is below
atmospheric
Barometer: An instrument used to determine the absolute
pressure exerted by the atmosphere

20. Atmospheic pressure (Pa)
Absolute Zero
Pvacuum
Pgauge
Pabsolute
Pabsolute
Pabs = Pgauge + Pa
Pabs = Pvacuum - Pa

21. VARIATION OF PRESSURE
PA
(P + dP)A
W
dh
F = 0
(P + dP)A - PA - W = 0
PA + dPA - PA - W = 0
dPA - W = 0 or dPA = W Eq. 1
but : W = dV
dPA = - dV

22. where negative sign is used because distance h is measured upward and W
is acting downward.
dV = Adh then dPA = - Adh, therefore
dP = - dh
(Note: h is positive when measured upward and negative if measured
downward)

23. MANOMETERS
Manometer is an instrument used in measuring gage pressure in length of
some liquid column.
1. Open Type Manometer : It has an atmospheric surface and is capable in
measuring gage pressure.
2. Differential Type Manometer : It has no atmospheric surface and is
capable in measuring differences of pressure.
Open Type
Open end
Manometer Fluid

24. Differential Type
Fluid A
Fluid B
Fluid C

25. ENERGY FORMS
Work: It is the force multiplied by the displacement in the direction
of the force.
W =∫Fdx KJ
-W - indicates that work is done on the system
+W - indicates that work is done by the system.
Heat: It is a form of energy that crosses a system's boundary, because
of a temperature difference between the system and the surrounding.
Q - Heat KJ
+Q - indicates that heat is added to the system
-Q - indicates that heat is rejected from the system.
Internal Energy: It is the energy acquired due to the overall molecular
interaction, or the total energy that a molecule has.
U = mu KJ
U - total internal energy KJ
u - specific internal energy KJ/kg
U- change of internal energy

26. Flow Energy Or Flow Work: It is the energy required in pushing a fluid
usually into the system or out from the system.
System or
Control Volume
P1
P2
A1
A2
L1
L2
Ef1 = F1L1
F1 = P1A1
Ef1 = P1A1L1
A1L1 = V1
Ef1 = P1V1
Ef2 = F2L2
F2 = P2A2
Ef2 = P2A2L2
A2L2 = V2
Ef2 = P2V2
Ef = Ef2 – Ef1
Ef = P2V2 – P1V1
Ef = PV
PV = P2V2 - P1V1 KJ
P = P2 2 - P1 1 m3/kg

27. Where: P – pressure in KPa
V – volume in m3
- specific volume in m3/kg
Ef = PV – Flow energy or flow work
Kinetic Energy: It is the energy acquired due to the motion of a body or a
system.
1 2
m m
F
d x
dxFdKE
dxFKE

28. KJ/kg
2(1000)
ΔKE
KJ
2(1000)
m
ΔKE
dt
dv
dx
1000
m
KE
dt
dv
1000
m
1000
ma
F
FdxKE
vv
vv
2
1
2
2
2
1
2
2
2
1
21000
m
ΔKE
dvv
1000
m
ΔKE
dt
dx
dv
1000
m
ΔKE
vv
2
1
2
2
2
1
2
1
Where:
m – mass , kg
v – velocity , m/sec
kg
KJ
10002
vv
KE
KJ
10002
vvm
KE
2
1
2
2
2
1
2
2

29. Potential Energy: It is the energy required by virtue of its configuration or
elevation.
m
m
dZ
Reference Datum
kg
KJ
1000
ZZg
PE
KJ
1000
ZZmg
PE
dZ
1000
mg
PE
dZWPE
12
12
Where:
W – Work
Q – Heat
U – Internal Energy
PV – Flow Energy or flow work
KE – Kinetic Energy
PE – Potential Energy
Note:
+Z – if measured upward
- Z –if measured downward

30. Law of Conservation of Mass
Mass is indestructible: In applying this law we must except nuclear processes
during which mass is converted into energy.
The verbal form of the law is:
Mass Entering - Mass Leaving = Change of Mass stored in the system
In equation Form:
m1 - m2 = m
1 2
m1 m2m = 0
a
b c
d
For a steady-state, steady-flow system m = 0, therefore
m1 - m2 = 0 or m1 = m2

31. For one dimensional flow, where 1 = 2 =
Let m1 = m2 = m
Continuity Equation:
υ
Av
Avρm
Where:
m - mass flw rate in kg/sec
- density in kg/m3
- specific volume inm3/kg
A - cross sectional area in m2
v - velocity in m/sec

32. Zeroth Law of Thermodynamics
If two bodies are in thermal equilibrium with a third body, they are in thermal
equilibrium with each other, and hence their temperatures are equal.
Specific Heat or Heat Capacity: It the amount of heat required to raise
the temperature of a 1 kg mass of a substance 1 C or 1 K.
tmCTmCQ
m;gConsiderin
CdtCdTdQ
C;constantFor
K-kg
KJ
or
C-kg
KJ
dt
dQ
dT
dQ
C

33. SENSIBLE HEAT: The amount of heat per unit mass that must be
transferred (added or remove) when a substance undergoes a change in
temperature
without a change in phase.
Q = mC( t) = mC( T)
where: m - mass , kg
C - heat capacity or specific heat, KJ/kg- C or
KJ/kg- K
t - temperature in C
T - temperature in K
HEAT OF TRANSFORMATION: The amount of heat per unit
mass that must be transferred when a substance completely undergoes a
phase change
without a change in temperature.
Q = mL

34. A. Heat of Vaporization: Amount of heat that must be added to
vaporize a liquid or that
must be removed to condense a gas.
Q = mL
where L - latent heat of vaporization, KJ/kg
B. Heat of Fusion : Amount of heat that must be added to melt a solid or
that must be
removed to freeze a liquid.
Q = mL
where L - latent heat of fusion, KJ/kg

35. THE FIRST LAW OF THERMODYNAMICS
(The Law of Conservation of (Energy)
“Energy can neither be created nor destroyed but can only
be converted from one form to another.”
Verbal Form:
Energy Entering – Energy Leaving = Change of Energy
stored in the system
Equation Form:
E1 – E2 = Es
1. First Corollary of the First Law: Application of first Law to a Closed System
U
Q
W
For a Closed System (Non FlowSystem),
PV, KE and PE are negligible, therefore
the changeof stored energy Es = U
Q – W = U 1
Q = U + W 2

36. By differentiation:
dQ = dU + dW 3
where:
dQ Q2 – Q1
dW W2 – W1
Work of a Closed System (NonFlow)
P
V
W = PdV
P
dV
5Eq.dVPUQ
4Eq.dVPdUdQ
3Eq.From
dVPdW
dVPW
dVAdxdxPAW
PAFdxFW

37. 2. Second Corollary of the First Law: Application of First Law to an Open System
System or
Control volume
Datum Line
Q
W
1
2
U1 + P1V1 + KE1 + PE1
U2 + P2V2 + KE2 + PE2
For an Open system (Steady state, Steady Flow system)
Es = 0, therefore
E1 – E2 = 0 or
E1 = E2 or
Energy Entering = Energy Leaving
Z1
Z2

38. U1 + P1V1 + KE1 + PE1 + Q = U2 + P2V2 + KE2 + PE2 + W 1
Q = (U2 – U1) + (P2V2 – P1V1) + (KE2 – KE1) + (PE2 – PE1) + W 2
Q = U + (PV) + KE + PE + W 3
By differentiation
dQ = dU + d(PV) + dKE + dPE + dW 4
But dQ Q2 – Q1 and dW W2 – W1
Enthalpy (h)
h = U + PV
dh = dU + d(PV) 5
dh = dU + PdV + VdP 6
But: dQ = dU + PdV
dh = dQ + VdP 7
From Eq. 3
Q = h + KE + PE + W 8
dQ = dh + dKE + dPE + dW 9
dQ = dU + PdV + VdP + dKE + dPE + dW 10
dQ = dQ + VdP + dKE + dPE + dW
0 = VdP + dKE + dPE + dW
dW = -VdP - dKE - dPE 11
By Integration
W = - VdP - KE - PE 12

39. If KE = 0 and PE = 0
Q = h + W 13
W = Q - h 14
W = - VdP 15
PEKEhVdP-W
PEKEhQW
WPEKEhQ
SYSTEMOPENanFor.B
PdVW
dWdUdQ
WUQ
SYSTEMCLOSEDaFor.A
SUMMARY

40. IDEAL OR PERFECT GAS
Prepared By: Engr Yuri G. Melliza

41. 1. Ideal Gas Equation of State
PV = mRT
P = RT
2
T
2
V
2
P
1
T
1
V
1
P
C
T
PV
RT
P
ρ
Where: P – absolute pressure in KPa
V – volume in m3
m – mass in kg
R – Gas Constant in KJ/kg- K
T – absolute temperature in K
IDEAL OR PERFECT GAS

42. 2. Gas Constant
K-
m
kg
KJ
8.3143R
K-kg
KJ
M
R
R
Where:
R- Gas Constant in KJ/kg-K
K
m
kg
KJ
constantgasuniversalR
M – Molecular weight kg/kgm
3. Boyle’s Law
If the temperature of a certain quantity of
gas is held constant the volume V is inver-
sely proportional to the absolute pressure P.

43. C
2
V
2
P
1
V
1
P
CPV
P
1
CV
PVα
4.Charle’s Law
A. At Constant Pressure (P = C)
If the pressure of a certain quantity of
gas is held constant, the volume V is directly
proportional to the temperature T during a qua-
sistatic change of state

44. 2
2
1
1
T
V
T
V
C
T
V
T;CV;TαV
B. At Constant Volume (V = C)
If the volume of a certain quantity of gas is
held constant, the pressure P varies directly
as the absolute temperature T.
2
2
1
1
T
P
T
P
C
T
P
;TCPTαP ;

45. 5. Avogadro’s Law
All gases at the same temperature and
pressure have the same number of molecules
per unit of volume, and it follows that the
specific weight is directly proportional to
its molecular weight M.
M
6.Specific Heat
Specific Heat or Heat Capacity is the amount
of heat required to raise the temperature of
a 1 kg mass 1 C or 1 K
A. SPECIFIC HEAT AT CONSTANT PRESSURE (Cp)
From: dh = dU + PdV + VdP
but dU + VdP = dQ ; therefore
dh = dQ + VdP 1

46. but at P = C ; dP = O; therefore
dh = dQ 2
and by integration
Q = h 3
considering m,
h = m(h2 - h1) 4
Q = h = m (h2 - h1) 5
From the definition of specific heat, C = dQ/T
Cp = dQ /dt 6
Cp = dh/dT, then
dQ = CpdT 7
and by considering m,
dQ = mCpdT 8
then by integration
Q = m Cp T 9
but T = (T2 - T1)
Q = m Cp (T2 - T1) 10

47. B SPECIFIC HEAT AT CONSTANT VOLUME (Cv)
At V = C, dV = O, and from dQ = dU + PdV
dV = 0, therefore
dQ = dU 11
then by integration
Q = U 12
then the specific heat at constant volume
Cv is;
Cv = dQ/dT = dU/dT 13
dQ = CvdT 14
and by considering m,
dQ = mCvdT 15
and by integration
Q = m U 16
Q = mCv T 17
Q = m(U2 - U1) 18
Q = m Cv(T2 - T1) 19

48. From:
h = U + P and P = RT
h = U + RT 20
and by differentiation,
dh = dU + Rdt 21
but dh =CpdT and dU =
CvdT, therefore
CpdT = CvdT + RdT 22
and by dividing both sides of the
equation by dT,
Cp = Cv + R 23

49. 7. Ratio Of Specific Heats
k = Cp/Cv 24
k = dh/du 25
k = h/ U 26
From eq. 32,
Cp = kCv 27
substituting eq. 27 to eq. 24
Cv = R/k-1 28
From eq. 24,
Cv = Cp/k 29
substituting eq. 29 to eq. 24
Cp = Rk/k-1 30

50. 8. Entropy Change ( S)
Entropy is that property of a substance that
determines the amount of randomness and disorder
of a substance. If during a process, an amount of
heat is taken and is by divided by the absolute
temperature at which it is taken, the result is
called the ENTROPY CHANGE.
dS = dQ/T 31
and by integration
S = ∫dQ/T 32
and from eq. 39
dQ = TdS 33

51. 2
2
1
1
2
2
1
1
2
2
1
1
2211
2
22
1
11
MM
LAWSAVOGADRO'.4
C
T
V
T
V
CPAtb.
C
T
P
T
P
CVAta.
LAWCHARLES.3
CVPVP
C)T(LAWBOYLES2.
mRTPV
C
T
VP
T
VP
StateofEquation.1
SUMMARY
T
dQ
S
CHANGEENTROPY.8
Cv
Cp
k
HEATSPECIFICOFRATIO.7
RCvCp
1-k
R
Cv;
1-k
Rk
Cp
HEATPECIFICS6.
kg
kg
R
8.3143
M
K-kg
KJ
M
8.3143
R
CONSTANTGAS.5
mol

52. GAS MIXTURE
 Total Mass of a mixture
inn
m
m
x i
i
imm
 Mass Fraction
 Total Moles of a mixture
n
n
y i
i
 Mole Fraction
Where:
m – total mass of a mixture
mi – mass of a component
n – total moles of a mixture
ni – moles of a component
xi – mass fraction of a component
yi - mole fraction of a component

53.  Equation of State
Mass Basis
A. For the mixture
iiiii TRmVP
mRTPV
TRnPV
B. For the components
iiiii TRnVP
Mole Basis
A. For the mixture
B. For the components
Where:
R – Gas constant of a mixture
in KJ/kg- K
- universal gas constant in
KJ/kgm- K
R

54.  AMAGAT’S LAW
The total volume of a mixture V is equal to the volume occupied by each
component at the mixture pressure P and temperature T.
1
n1
V1
2
n2
V2
3
n3
V3
P,T
P = P1 = P2 = P3
T = T1 = T2 = T3

55. For the components:
TR
PV
n;
TR
PV
n;
TR
PV
n 3
3
2
2
1
1
The mole fraction:
V
V
yi
TR
PV
TR
PV
y
n
n
y
i
i
i
i
i
321
321
321
321
VVVV
P
TR
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
TR
PV
nnnn
The total moles n:

56.  DALTON’S LAW
The total pressure of a mixture P is equal to the sum of the partial pressure
that each gas would exert at mixture volume V and temperature T.
1
n1
P1
2
n2
P2
3
n3
P3
MIXTURE
n
P
T1 = T2 = T3 = T
V1 = V2 = V3 = V
For the mixture
For the components
TR
VP
n
TR
VP
n
TR
VP
n
3
3
2
2
1
1
TR
PV
n

57. 321
321
321
321
PPPP
V
TR
TR
VP
TR
VP
TR
VP
TR
PV
TR
VP
TR
VP
TR
VP
TR
PV
nnnn
The total moles n: The mole fraction:
P
P
yi
TR
PV
TR
VP
y
n
n
y
i
i
i
i
i

58.  Molecular Weight of a mixture
R
R
M
MyM ii
M
R
R
RxR ii
 Gas Constant of a mixture
 Specific Heat of a mixture
RCC
CxC
CxC
vp
viiv
piip
 Ratio of Specific Heat
u
h
C
C
k
v
p

59.  Gravimetric and Volumetric Analysis
Gravimetric analysis gives the mass fractions of the components
in the mixture. Volumetric analysis gives the volumetric or molal fractions
of the components in the mixture.
i
i
i
i
i
ii
ii
i
M
x
M
x
y
My
My
x

60. PROPERTIES OF PURE SUBSTANCE
a - sub-cooled liquid
b - saturated liquid
c - saturated mixture
d - saturated vapor
e - superheated vapor
Considering that the system is heated at constant
pressure where P = 101.325 KPa, the 100 C is the
saturation temperature corresponding to 101.325 KPa,
and 101.325 KPa is the saturation pressure correspon-
ding 100 C.
P P P P P
Q
30°C
100°C
100°C 100°C
T 100°C
(a) (b) (c) (d) (e)
Q Q Q Q

61. Saturation Temperature (tsat) - is the highest temperature at a given pressure in which
vaporization takes place.
Saturation Pressure (Psat) - is the pressure corresponding to the temperature.
Sub-cooled Liquid - is one whose temperature is less than the saturation temperature
corresponding to the pressure.
Compressed Liquid - is one whose pressure is greater than the saturation pressure
corresponding to the temperature.
Saturated Liquid - a liquid at the saturation temperature
Saturated Vapor - a vapor at the saturation temperature
Saturated Mixture - a mixture of liquid and vapor at the saturation temperature.
Superheated Vapor - a vapor whose temperature is greater than the saturation temperature.
a
b c d
e
T F
Saturated Vapor
Saturated Vapor
30°C
100°C
t 100°C
Saturated Mixture
P = C
Critical Point
T- Diagram

62. a
b c d
e
T
S
F
Saturated Vapor
Saturated Vapor
30°C
100°C
t 100°C
Saturated Mixture
P = C
Critical Point
T-S Diagram
F(critical point)- at the critical point the temperature and pressure is unique.
For Steam: At Critical Point, P = 22.09 MPa; t = 374.136 C

63. a
b c d
e
T
S
F
Saturated Vapor
Saturated Vapor
ta
tsat
te
Saturated Mixture
P = C
Critical Point
T-S Diagram
tsat - saturation temperature corresponding the pressure P
ta - sub-cooled temperature which is less than tsat
te - superheated vapor temperature that is greater than tsat

64. h-S (Enthalpy-Entropy Diagram)
h
S
t = C (constant temperature curve)
P = C (constant pressure curve)
F
I
II
III
I - subcooled or compressed liquid region
II - saturated mixture region
III - superheated vapor region

65. Quality (x):
Lv
v
Lv
v
mmm
m
m
mm
m
x
Where:
mv – mass of vapor
mL – mass of liquid
m – total mass
x- quality
The properties at saturated liquid, saturated vapor, superheated
vapor and sub-cooled or compressed liquid can be determined
from tables. But for the properties at saturated mixture (liquid
and vapor) they can be determined by the equation
rc = rf + x(rfg)
rfg = rg – rf
Where: r stands for any property ( , U, h and S)
rg – property at saturated vapor (from table)
rf – property at saturated liquid
Note: The properties at siub-cooled or compressed liquid is
approximately equal to the properties at saturated liquid
corresponding the sub-cooled temperature.

66. Throttling Calorimeter
Main Steam Line
P1 – steam line
pressure
To main steam line
P2 -Calorimeter pressure
h1 = h2
h1 = hf1 + x1(hfg1)
Where:
1 – main steam line
2 - calorimeter
thermometer

67. P1
P2
1
2
T
S
h = C
T-S Diagram Throttling Process
P1 – steam line pressure
P2 – calorimeter pressure

68. 1. Isobaric Process ( P = C): An Isobaric Process is an internally
reversible constant pressure process.
A. Closed System:(Nonflow)
P
V
21P
dV
Q = U + W 1 any substance
W = PdV 2 any substance
U = m(U2 - U1) 3 any substance
W = P(V2 - V1) 4 any substance
Q = h = m(h2-h1) 5 any substance
T
S
1
2
dS
T
P = C
PROCESSES OF FLUIDS

69. For Ideal Gas:
PV = mRT
W =mR(T2-T1) 5
U = mCv(T2-T1) 6
Q = h = mCP (T2-T1) 7
Entropy Change
S = dQ/T 8 any substance
dQ = dh
For Ideal Gas
dh = mCPdT
S = dQ/T
S = mCP dT/T
S = mCP ln(T2/T1) 9
B. Open System:
Q = h + KE + PE + W 10 any substance
W = - VdP - KE - PE 11 any substance
- VdP = 0

70. Q = h 12
W = - KE - PE 13
If KE = 0 and PE = 0
W = 0 14
Q = mCP(T2-T1) 15 Ideal Gas
2. Isometric Process (V = C): An Isometric process is internally
reversible constant volume process.
A. Closed System: (Nonflow)
P
V
1
2
T
S
T
dS
1
2V = C

71. Q = U + W 1 any substance
W = PdV at V = C; dV = 0
W = 0
Q = U = m(U2 - U1) 2 any substance
h = m(h2-h1) 3 any substance
For Ideal Gas:
Q = U = mCv(T2-T1) 4
h = mCP(T2-T1) 5
Entropy Change:
S = dQ/T 6 any substance
dQ = dU
dU = mCvdT for ideal gas
S = dU/T = mCv dT/T
S = mCv ln(T2/T1) 6

72. B. Open System:
Q = h + KE + PE + W 7 any substance
W = - VdP - KE - PE 8 any substance
- VdP = -V(P2-P1) 9 any substance
Q = U = m(U2 - U1) 10 any substance
h = m(h2-h1) 11 any substance
For Ideal Gas:
- VdP = -V(P2-P1) = mR(T1-T2)
Q = U = mCv(T2-T1) 12
h = mCP(T2-T1) 13
If KE = 0 and PE = 0
Q = h + W 14 any substance
W = - VdP 15
W = - VdP = -V(P2-P1) 16 any substance
W = mR(T1-T2) 16 ideal gas
h = mCP(T2-T1) 17 ideal gas

73. 3. Isothermal Process(T = C): An Isothermal process is a reversible
constant temperature process.
A. Closed System (Nonflow)
dS
T
S
T
1 2
P
V
1
2P
dV
PV = C or
T = C
Q = U + W 1 any substance
W = PdV 2 any substance
U = m(U2 - U1) 3 any substance
For Ideal Gas:
dU = mCv dT; at T = C ; dT = 0
Q = W 4

74. W = PdV ; at PV = C ;
P1V1 = P2V2 = C; P = C/V
Substituting P = C/V to W = PdV
W = P1V1 ln(V2/V1) 5
Where (V2/V1) = P1/P2
W = P1V1 ln(P1/P2) 6
P1V1 = mRT1
Entropy Change:
dS = dQ/T 7
S = dQ/T
dQ = TdS ;at T = C
Q = T(S2-S1)
(S2-S1) = S = Q/T 8
S = Q/T = W/T 9 For Ideal Gas

75. B. Open System (Steady Flow)
Q = h + KE + PE + W 10 any substance
W = - VdP - KE - PE 11 any substance
- VdP = -V(P2-P1) 12 any substance
h = m(h2-h1) 13 any substance
For Ideal Gas:
- VdP = -P1V1ln(P2/P1) 14
- VdP = P1V1ln(P1/P2) 15
P1/P2 = V2/V1 16
dh = CPdT; at T = C; dT = 0
h = 0 16
If KE = 0 and PE = 0
Q = h + W 17 any substance
W = - VdP = P1V1ln(P1/P2) 18
For Ideal Gas
h = 0 19
Q = W = - VdP = P1V1ln(P1/P2) 20

76. 4. Isentropic Process (S = C): An Isentropic Process is an internally
“Reversible Adiabatic” process in which the entropy remains constant
where S = C and PVk = C for an ideal or perfect gas.
For Ideal Gas
1
2
1
1
1
2
1
2
2
22
1
k
k
k
k
22
k
11
11
k
V
V
P
P
T
T
VPVPand
T
VP
T
VP
CPVandC
T
PV
Using

77. A. Closed System (Nonflow)
T
S
1
2
P
V
1
2
dV
P
S = C or
PVk = C
Q = U + W 1 any substance
W = PdV 2 any substance
U = m(U2 - U1) 3 any substance
Q = 0 4
W = - U = U = -m(U2 - U1) 5

78. For Ideal Gas
U = mCV(T2-T1) 6
From PVk = C, P =C/Vk, and substituting P
=C/Vk
to W = ∫PdV, then by integration,
1
1
1
1
1
1
1
211
1
1
21
k
k
VP
k
k
12
1122
P
P
k
PdV
P
P
k
mRT
k-1
T-TmR
PdV
k
VP-VP
PdVW 7
8
9
Q = 0

79. Entropy Change
S = 0
S1 = S2
B. Open System (Steady Flow)
Q = h + KE + PE + W 10 any substance
W = - VdP - KE - PE 11 any substance
h = m(h2-h1) 12 any substance
Q = 0
W = - h - KE - PE 13
From PVk = C ,V =[C/P]1/k, substituting V to
-∫VdP, then by integration,

80. 1
1
1
1
1
1
1
211
1
1
21
k
k
k
k
12
1122
P
P
k
VkP
VdP
P
P
k
kmRT
k-1
T-TkmR
VdP
k
VP-VPk
VdP
PdVkVdP
14
15
16
If KE = 0 and PE = 0
0 = h + W 17 any substance
W = - VdP = - h 18 any substance
h = m(h2-h1) 19 any substance
Q = 0

81. 12P
k
k
k
k
12
1122
T-TmChW
P
P
k
VkP
W
P
P
k
kmRT
k-1
T-TkmR
W
k
VP-VPk
PdVkVdPW
1
1
1
1
1
1
1
211
1
1
21
20
22
21
23

82. P
V
dP
V
Area = - VdP
S = C

83. 1
2
1
1
1
2
1
2
2
22
1
n
n
n
n
22
n
11
11
n
V
V
P
P
T
T
VPVPand
T
VP
T
VP
CPVandC
T
PV
Using
5. Polytropic Process ( PVn = C): A Polytropic Process is an
internally reversible process of an Ideal or Perfect Gas in which
PVn = C, where n stands for any constant.

84. A. Closed System: (Nonflow)
Q = U + W 1
W = PdV 2
U = m(U2 - U1) 3
Q = mCn(T2-T1) 4
U = m(U2 - U1) 5
P
V
1
2
dV
P
PVn = C
T
S
2
1
dS
T
PVn = C
K-kg
KJ
or
C-kg
KJ
heatspecificpolytropicC
n1
nk
CC
n
vn

85. From PVn = C, P =C/Vn, and substituting
P =C/Vn to W = ∫PdV, then by integration,
1
1
1
1
1
1
1
211
1
1
21
n
n
VP
n
n
12
1122
P
P
n
PdVW
P
P
n
mRT
n-1
T-TmR
PdVW
n
VP-VP
PdVW
Entropy Change
dS = dQ/T
dQ = mCndT
S = mCnln(T2/T1)
6
8
9
10

86. B. Open System (Steady Flow)
Q = h + KE + PE + W 11
W = - VdP - KE - PE 12
h = m(h2-h1) 13
Q = mCn(T2-T1) 14
dQ = mCn dT
W = Q - h - KE - PE 15
From PVn = C ,V =[C/P]1/n, substituting V to
-∫VdP, then by integration,
n
VP-VPn
VdP
PdVnVdP
1122
1
16

87. 1
1
1
1
1
1
211
1
1
21
n
n
n
n
12
P
P
n
VnP
VdP
P
P
n
nmRT
n-1
T-TnmR
VdP
If KE = 0 and PE = 0
Q = h + W 19 any substance
W = - VdP = Q - h 20 any substance
h = m(h2-h1) 21 any substance
h = mCp(T2-T1)
Q = mCn(T2-T1) 22
17
18

88. 1
1
1
1
1
1
211
1
1
21
n
n
n
n
12
P
P
n
VnP
W
P
P
n
nmRT
n-1
T-TnmR
W
24
23

89. 6. Isoenthalpic or Throttling Process: It is a steady - state, steady
flow process in which Q = 0; PE = 0; KE = 0; W = 0 and the
enthalpy remains constant.
h1 = h2 or h = C
Throttling valve
Main steam line
thermometer
Pressure Gauge
Pressure Gauge
To main steam line
Throttling Calorimeter

90. Irreversible or Paddle Work
m
W
Q
U
Wp
Q = U + W - Wp
where: Wp - irreversible or paddle work

91. THERMODYNAMICS Part 2
By. Engr. Yuri G. Melliza
2nd Law of Thermodynamics
Carnot Cycles
Steam Cycles
Fuels and Combustion
ICE Cycles

92. 2nd Law of Thermodynamics
• Second Law of Thermodynamics
• Kelvin – Planck Statement
• Carnot engine
• Carnot Refrigerator
• Sample Problems

93. Second Law of
Thermodynamics:
Whenever energy is transferred, the
level of energy cannot be conserved
and some energy must be permanently
reduced to a lower level.
When this is combined with the first law of
thermodynamics, the law of energy
conservation, the statement becomes:

94. Second Law of
Thermodynamics:
Whenever energy is transferred, energy
must be conserved, but the level of
energy cannot be conserved and some
energy must be permanently reduced
to a lower level.

95. Kelvin-Planck statement of the Second
Law:
No cyclic process is possible whose sole result
is the flow of heat from a single heat reservoir
and the performance of an equivalent amount
of work.
For a system undergoing a cycle:
The net heat is equal to the net work.
QW
dWdQ Where:
W - net work
Q - net heat

96. CARNOT CYCLE
Nicolas Leonard Sadi Carnot
1796-1832
1.Carnot Engine
Processes:
1 to 2 - Heat Addition (T = C)
2 to 3 - Expansion (S = C)
3 to 4 - Heat Rejection (T = C)
4 to 1 - Compression (S = C)

97. P
V
2
1
3
4
T = C
S = C
S = C
T = C

98. T
S
21
34
T H
T L
Q A
Q R

99. Heat Added (T = C)
QA = TH( S) 1
Heat Rejected (T = C)
QR = TL( S) 2
S = S2 - S1 = S3 – S4 3
Net Work
W = Q = QA - QR 4
W = (TH - TL)( S) 5

100. %x
Q
Q
e
%x
Q
QQ
e
%x
Q
W
e
A
R
A
RA
A
1001
100
100 6
7
8

101. Substituting eq.1 and eq. 5 to eq 6
%x
T
T
e
%x
T
TT
e
H
L
H
LH
1001
100 9
10

102. TH
TL
W
QA
QR
E
Carnot Engine

103. 2. Carnot Refrigerator:
Reversed Carnot Cycle
Processes:
1 to 2 - Compression (S =C)
2 to 3 - Heat Rejection (T = C)
3 to 4 - Expansion (S = C)
4 to 1 - Heat Addition (T = C)

104. Q
R
Q
A
1
2
4
3
S
T
TH
TL

105. Heat Added (T = C)
QA = TL( S) 1
Heat Rejected (T = C)
QR = TH( S) 2
S = S1 - S4 = S2 - S3
3
Net Work
W = Q 4
W = QR - QA 5
W = (TH - TL)( S) 6

106. Coefficient of Performance
(COP)
LH
L
A
TT
T
COP
W
Q
COP 7
8

107. 1
H
L
T
T
COP 9
Tons of Refrigeration
211 KJ/min = 1 TR
3. Carnot Heat Pump:A heat pump
uses the same components as
therefrigerator but its purpose
isto reject heat at high energy
level.

108. Performance Factor (PF)
AR
R
R
QQ
Q
PF
W
Q
PF 10
11

109. 1
1
1
COPPF
T
T
PF
Q
Q
PF
TT
T
PF
L
H
A
R
LH
H
12
13
14
15

110. TH
TL
W
QA
QR
R
Carnot Refrigerator

111. A Carnot engine operating between 775 K and
305 K produces 54 KJ of work. Determine the
change of entropy during heat addition.
TH = 775 K ; TL = 305 K
W = 54 KJ

112. TH
TL
W
QA
QR
E

113. K
KJ
0.015
775
89.04
T
Q
S-S
)S-(STQ
KJ89.04
0.606
54
e
W
Q
Q
W
e
0.606
775
305775
T
TT
e
H
A
12
12HA
A
A
H
LH

114. A Carnot heat engine rejects 230 KJ of
heat at 25 C. The net cycle work is 375 KJ.
Determine the cycle thermal efficiency and
the cycle high temperature .
Given:
QR = 230 KJ
TL = 25 + 273 = 298 K
W = 375 KJ

115. TL = 298 K
TH
WE
QR = 230 KJ
QA
K87.783
772.0
605
)S-(S
Q
T
KKJ/-0.772)S-(S
KKJ/772.0)SS(
)SS(298230
)SS(SS
)SS(TQ
)SS(TQ
62.0
605
375
QA
W
e
KJ605QA
)230375(QWQ
QQW
12
A
H
12
34
34
1234
34LR
12HA
RA
RA

116. A Carnot engine operates between temperature
reservoirs of 817 C and 25 C and rejects 25 KW to
the low temperature reservoir. The Carnot engine
drives the compressor of an ideal vapor compres-
sion refrigerator, which operates within pressure
limits of 190 KPa and 1200 Kpa. The refrigerant is
ammonia. Determine the COP and the refrigerant
flow rate.(4; 14.64 kg/min)
TH = 817 + 273 = 1090 K
TL = 25 + 273 = 298 K
QR = 25 KW

117. Internal Combustion Engine Cycles
1. Air Standard Otto Cycle (Spark Ignition Engine Cycle)
Processes
1 to 2 - Isentropic Compression (S = C)
2 to 3 - Constant Volume Heat Addition ( V = C)
3 to 4 - Isentropic Expansion (S =C)
4 to 1 - Constant Volume Heat Rejection (V = C)
P
Pm
V
VD
CVD
W
1
42
3
S = C
S = C
3QAT
S
1
2
4
V = C
V = C
QR

118. Compression Ratio
3
4
2
1
V
V
V
V
r
where:
r - compression ratio
V1 = V4 and V2 = V3
1
Heat Added (V = C)
QA = mCV(T3 - T2) 2
Heat Rejected
QR = mCV(T4 - T1) 3
Net Cycle Work
W = QA - QR
W = mCV[(T3 - T2) - (T4 - T1)] 4

119. Thermal Efficiency
100%
r
1
1e
100%
TT
TT
1e
100%x
Q
Q
1e
100%
Q
QQ
e
100%x
Q
W
e
1-k
23
14
A
R
A
RA
A
x
x
x
5
6
7
8
9

120. Mean Effective Pressure
KPa
V
W
P
D
m
where:
W - net work, KJ, KJ/kg, KW
VD - Displacement Volume, m3, m3/kg, m3/sec
VD = V1 - V2 m3
VD = 1 - 2 m3/kg
10
Percent Clearance
100%x
V
V
C
D
2
V2 = CVD
V1 = Vd + CVD
C
C1
V
V
r
2
1
11 12

121. 2. Diesel Cycle: (Compression Ignition Engine Cycle)
Processes
1 to 2 - Isentropic Compression (S = C)
2 to 3 - Constant Pressure Heat Addition (P = C)
3 to 4 - Isentropic Expansion (S = C)
4 to 1 - Constant Volume Heat Rejection (V = C)
P
V
T
S
2 3
4
1
S = C
S = C
VDCVD
1
2
3
4
V = C
P = C
QR
QA

122. Heat Added (P = C)
QA = mCP(T3 - T2) 3
QA = mkCV(T3 - T2) 4
Heat Rejected (V = C)
QR = mCV(T4 - T1) 5
Net Cycle Work
W = QA - QR
W = mCV[k(T3 - T2) - (T4 - T1)] 6
Compression Ratio
3
4
2
1
V
V
V
V
r 1
Cut - Off Ratio
2
3
c V
V
r 2

123. Thermal Efficiency
100%
1)k(r
1)(r
r
1
1e
100%
TT
TT
1e
100%x
Q
Q
1e
100%
Q
QQ
e
100%x
Q
W
e
c
k
c
1-k
23
14
A
R
A
RA
A
x
x
k
x
7
8
9
10
11

124. where:
W - net work, KJ, KJ/kg, KW
VD - Displacement Volume, m3, m3/kg, m3/sec
VD = V1 - V2 m3
VD = 1 - 2 m3/kg
KPa
V
W
P
D
m
Mean Effective Pressure
12

125. 3. Air Standard Dual Cycle
Processes:
1 to 2 - Compression (S = C)
2 to 3 -Heat Addition (V = C)
3 to 4 - Heat Addition (P = C)
4 to 5 -Expansion (S = C)
5 to 1 _ Heat Rejection (V = C)
1
1
2
2
3
3
4
4
5
5
S = C
S = C
P = C
V = C
V = C
QA1
QA2
QR
P
V
T
S
VDCVD

126. Comprssion Ratio
4
5
3
5
2
1
V
V
V
V
V
V
r
V5 = V1 ; V2 = V3
1
Pressure Ratio
2
P
4
P
2
3
p P
P
r 3
2
4
3
4
c V
V
r
V
V
2
Cut-Off Ratio

127. Heat Added (V = C)
QA1 = mCV(T3 - T2) 4
Heat Added (P =C)
QA2 = mCP(T4 - T3) 5
QA2 = mkCV(T4 - T3) 6
Heat Rejected (V = C)
QR = mCV(T5 - T1) 7
Net Cycle Work
W = QA - QR 8
QA = QA1 + QA2 9
QA = mCV[(T3 - T2) +k (T4 - T3)] 10
W = mCV[(T3 - T2) + k(T4 - T3) - (T5 - T1) ]
100%x
Q
Q
1e
100%
Q
QQ
e;100%x
Q
W
e
A
R
A
RA
A
x 12
13
Thermal Efficiency
11

128. 100%
1-(rkr1)-(r
1)-r(r
r
1
1e
100%
)]T-(Tk)T-[(T
)T-(T
1e
cpp
k
cp
1-k
3423
15
x
x
Mean Effective Pressure
Pm = W/VD KPa
where: VD = V1 - V2 m3 ; W in KJ
V1 = V5 ; V2 = V3
VD = 1 - 2 m3/kg ; W in KJ/kg
1 = 5 ; 2 = 3
For Cold Air Standard: K = 1.4
For Hot Air Standard: K = 1.3

129. Vapor Power Cycle
RANKINE CYCLE
Processes:
1 to 2 - Expansion (S = C)
2 to 3 - Heat Rejection (P = C)
3 to 4 - Compression or Pumping (S = C)
4 to 1 - Heat Addition (P = C)
Boiler or Steam
Generator
Turbine
Condenser
Pump
WP
QA
QR
Wt
1
2
3
4

130. Major Components of a Rankine Cycle
1. Steam Generator or Boiler: The working substance absorbs heat
from products of combustion or other sources of heat at constant
pressure which in turn changes the state of the working substance
(water or steam) from sub-cooled liquid and finally to superheated
vapor whence at this point it enters the turbine.
2. Steam Turbine: A steady state, steady flow device where steam
expands isentropically to a lower pressure converting some forms
of energy (h, KE, PE) to mechanical work that finally be converted
into electrical energy if the turbine is used to drive an electric gene-
rator.
3. Condenser: Steam exiting from the turbine enters this device to re-
ject heat to the cooling medium and changes its state to that of the
saturated liquid at the condenser pressure which occurred at a cons-
tant pressure process.

131. 4. Pump: It is also a steady state, steady flow machine where the
condensate leaving the condenser at lower pressure be pumped
back to the boiler in an isentropic process in order to raise the
pressure of the condensate to that of the boiler pressure.
h
S S
T
3
4
2
1
3
4
1
2
P1
P2
P1
P2
4’
2’
2’
4’

132. Turbine Work
a) Ideal Cycle
Wt = (h1 - h2) KJ/kg
Wt = ms(h1 - h2) KW
b) Actual Cycle
Wt’ = (h1 - h2’) KJ/kg
Wt’ = ms(h1 - h2’) KW
where: ms - steam flow rate in kg/sec
Turbine Efficiency
100%x
hh
hh
η
100%x
W
W
tη
21
2'1
t
t
t'

133. Pump Work
a) Ideal Cycle
WP = (h4 - h3) KJ/kg
WP = ms(h4 - h3) KW
b) Actual Cycle
WP’ = (h4’ - h3) KJ/kg
WP’ = ms(h4’ - h3) KW
Pump Efficiency
100%x
hh
hh
η
100%x
W
W
η
34'
34
p
p'
p
p

134. Heat Rejected
a) Ideal Cycle
QR = (h2 - h3) KJ/kg
QR = ms(h2 - h3) KW
QR = ms(h2 - h3) KW = mwCpw(two - twi) KW
b) Actual Cycle
QR = (h2’ - h3) KJ/kg
QR = ms(h2’ - h3) KW = mwCpw(two - twi) KW
Where: mw - cooling water flow rate in kg/sec
twi - inlet temperature of cooling water in C
two - outlet temperature of cooling water in C
Cpw - specific heat of water in KJ/kg- C or KJ/kg- K
Cpw = 4.187 KJ/kg- C or KJ/kg- K

135. Heat Added:
a) Ideal Cycle
QA = (h1 - h4) KJ/kg
QA = ms (h1 - h4) KW
b) Actual Cycle
QA = (h1 - h4’) KJ/kg
QA = ms (h1 - h4’) KW
Steam Generator or boiler Efficiency
100%x
(HV)m
)h(hm
η
100%x
Q
Q
η
f
41s
B
S
A
B
Where: QA - heat absorbed by boiler in KW
QS - heat supplied in KW
mf - fuel consumption in kg/sec
HV - heating value of fuel in KJ/kg

136. Steam Rate
KW-sec
kg
ProducedKW
rateFlowSteam
SR
Heat Rate
KW-sec
KJ
ProducedKW
SuppliedHeat
HR
Reheat Cycle
A steam power plant operating on a reheat cycle improves the thermal
efficiency of a simple Rankine cycle plant. After partial expansion of
the steam in the turbine, the steam flows back to a section in the boiler
which is the re-heater and it will be reheated almost the same to its
initial temperature and expands finally in the turbine to the con-
denser pressure.

137. Reheater
QA
WP
QR
Wt
1 kg
1 2 3
4
56
Regenerative Cycle
In a regenerative cycle, after partial expansion of the steam in the
turbine, some part of it is extracted for feed-water heating in an open or
close type feed-water heater. The bled steam heats the condensate from
the condenser or drains from the previous heater causing a decrease in
heat absorbed by steam in the boiler which result to an increase in
thermal efficiency of the cycle.

138. QA
WP1
QR
Wt
1 kg
1
2
3
456
7
WP2
m
Reheat-Regenerative Cycle
For a reheat - regenerative cycle power plant, part of the steam is re-
heated in the re-heater and some portion is bled for feed-water heating
to an open or closed type heaters after its partial expansion in the
turbine. It will result to a further increase in thermal efficiency of the
plant.

139. QA
WP1
QR
Wt
1 kg
1
2
4
5678
WP2
m
2
3
1-m
1-m
For a 1 kg basis of circulating steam, m is the fraction of steam
extracted for feed-water heating as shown on the schematic diagram
above, where the reheat and bled steam pressure are the same.

140. FUELS
and
COMBUSTION
By. Engr. Yuri G. Melliza

141. FUELS AND COMBUSTION
 Fuels and Combustion
 Types of Fuels
 Complete/Incomplete Combustion
 Oxidation of Carbon
 Oxidation of Hydrogen
 Oxidation of Sulfur
 Air composition
 Combustion with Air
 Theoretical Air
 Hydrocarbon fuels
 Combustion of Hydrocarbon Fuel

142. Fuels and Combustion
Fuel: Substance composed of
chemical elements which in rapid
chemical union with oxygen
produced combustion.

143. Combustion:
Is that rapid chemical union with
oxygen of an element, whose
exothermic heat of reaction is
sufficiently great and whose rate
of reaction is suffi-ciently fast
whereby useful quantities of heat
are liberated at elevated
temperature.

144. TYPES OF FUELS
 Solid Fuels
ex: Wood, coal, charcoal
 Liquid Fuels
ex: gasoline, diesel, kerosene
 Gaseous Fuels
ex: LPG, Natural Gas, Methane
 Nuclear Fuels
ex: Uranium
Combustible Elements
1. Carbon (C) 3. Sulfur (S)
2. Hydrogen (H2)

145. Complete Combustion: Occurs when all
the combustible elements has been fully
oxidized.
Ex:
C + O2 CO2
Incomplete Combustion: Occurs when
some of the combustible elements has not
been fully oxidized.
Ex:
C + O2 CO

146. Common Combustion Gases
GAS MOLECULAR
Weight (M)
C 12
H 1
H2 2
O 16
O2 32
N 14
N2 28
S 32

147. THE COMBUSTION CHEMISTRY
Oxidation of Carbon
1183
443612
32)1(121(16)1(12)
BasisMass
111
BasisMole
COOC 22

148. Oxidation of Hydrogen
981
18162
2)1(16(32)1(2)
BasisMass
11
BasisMole
OHOH
2
1
2
1
222 2
1

149. Oxidation of Sulfur
211
643232
32)1(32(32)1(32)
BasisMass
111
BasisMole
OSOS
1
22

150. Composition of AIR
a. Percentages by Volume (by
mole)
O2 = 21%
N2 = 79%
b. Percentages by Mass
O2 = 23%
N2 = 77%
763
21
79
.
2
2
OofMole
NofMoles

151. Combustion with Air
A. Combustion of Carbon with air
C + O2 + 3.76N2 CO2 + 3.76N2
Mole Basis:
1 + 1 + 3.76 1+ 3.76
Mass Basis:
1(12) + 1(32) + 3.76(28) 1(44) +
3.76(28)
12 + 32 + 3.76(28) 44 + 3.76(28)
3 + 8 + 3.76(7) 11+ 3.76(7)

152. kg of air per kg of Carbon:
Cofkg
airofkg
11.44=
3
3.76(7)+8
=
Cofkg
airofkg

153. B. Combustion of Hydrogen with air
H2 + ½ O2 + ½ (3.76)N2 H2O +
½(3.76)N2
Mole Basis:
1 + ½ + ½(3.76) 1 + ½(3.76)
Mass Basis:
1(2) + ½ (32) + ½(3.76)(28) 1(18) +
½ (3.76)(28)
2 + 16 + 3.76(14) 18 + 3.76(14)
1 + 8 + 3.76(7) 9 + 3.76(7)

154. kg of air per kg of Hydrogen:
22 Hofkg
airofkg
34.32=
1
3.76(7)+8
=
Hofkg
airofkg

155. C. Combustion of Sulfur with air
S + O2 + 3.76N2 SO2 + 3.76N2
Mole Basis:
1 + 1 + 3.76 1 + 3.76
Mass Basis:
1(32) + 1(32) + 3.76(28) 1(64) +
3.76(28)
32 + 32 + 105.28 64 + 105.28

156. kg of air per kg of Sulfur:
Sofkg
airofkg
4.29=
32
105.2832
=
Sofkg
airofkg

157. Theoretical Air
It is the minimum amount of air required to
oxidize the reactants or the combustible
elements found in the fuel. With theoretical
air no O2 is found in products.

158. Excess Air
It is an amount of air in excess of the
Theoretical requirements in order to
influence complete combustion. With
excess air O2 is present in the
products.

159. HYDROCARBON FUELS
Fuels containing the element s Carbon
and Hydrogen.
Chemical Formula: CnHm

160. Family Formula Structure Saturated
Paraffin CnH2n+2 Chain Yes
Olefin CnH2n Chain No
Diolefin CnH2n-2 Chain No
Naphthene CnH2n Ring Yes
Aromatic
Benzene CnH2n-6 Ring No
Naphthalene CnH2n-12 Ring No
Alcohols Note: Alcohols are not pure
hydrocarbon, because one of its
hydrogen atom is replace by an
OH radical. Sometimes it is used
as fuel in an ICE.
Methanol CH3OH
Ethanol C2H5OH

161. Saturated Hydrocarbon: All the carbon atoms are
joined by a single bond.
Unsaturated Hydrocarbon: It has two or more
adjacent Carbon atoms joined by a double or
triple bond.
Isomers: Two hydrocarbons with the same
number of carbon and hydrogen atoms but at
different structures.

162. H H H H
H C C C C H
H H H H
Chain structure Saturated
H H
H C C=C C H
H H H H
Chain Structure Unsaturated
Ring structure Saturated
H H
H C H
C C
H C H
H H

163. Theoretical Air: It is the minimum or theoretical amount
of air required to oxidized the reactants. With theoretical
air no O2 is found in the products.
Excess Air: It is an amount of air in excess of the theo-
retical air required to influence complete combustion.
With excess air O2 is found in the products.
Combustion of Hydrocarbon
Fuel(CnHm)
A. Combustion with 100% theoretical air
CnHm + aO2 + a(3.76)N2 bCO2 + cH2O +
a(3.76)N2
fuel
air
t kg
kg
m12n
)a(3.76)(28a(32)
F
A

164. Combustion of Hydrocarbon Fuel
Formula: (CnHm)
A. Combustion with 100% theoretical air
CnHm + aO2 + a(3.76)N2 bCO2 + cH2O +
a(3.76)N2
fuel
air
t kg
kg
m12n
)a(3.76)(28a(32)
F
A

165. fuel
air
a kg
kg
m12n
)a(3.76)(28a(32)
e)(1
F
A
B. Combustion with excess air e
CnHm +(1+e) aO2 + (1+e)a(3.76)N2 bCO2
+ cH2O + dO2 + (1+e)a(3.76)N2
Actual Air – Fuel Ratio
fuel
air
ta kg
kg
F
A
e)(1
F
A
Where: e – excess air in decimal
Note: Sometimes excess air is expressible in terms of
theoretical air.
Example: 25% excess air = 125% theoretical air

166. Orsat Analysis: Orsat analysis gives the
volumetric or molal analysis of the
PRODUCTS on a DRY BASIS, (no amount of
H2O given).
Proximate Analysis: Proximate analysis
gives the amount of Fixed
Carbon, Volatiles, Ash and Moisture, in
percent by mass. Volatiles are those
compounds
that evaporates at low temperature when
the solid fuel is heated.

167. ULTIMATE ANALYSIS: Ultimate analysis
gives the amount of C, H, O, N, S in
percentages by mass, and sometimes the
amount of moisture and ash are given.

168. SOLID FUELS
Components of Solid Fuels:
1. Carbon (C)
2. Hydrogen (H2)
3. Oxygen (O2)
4. Nitrogen (N2)
5. Sulfur (S)
6. Moisture (M)
7. Ash (A)

169. A.Combustion with 100% theoretical air
aC + bH2 + cO2 + dN2 + eS + fH2O + gO2 +
g(3.76)N2 hCO2 + iH2O + jSO2 + kN2
B.Combustion with excess air x:
aC + bH2 + cO2 + dN2 + eS + fH2O +
(1+x)gO2 +(1+x)g(3.76)N2 hCO2 +
iH2O + jSO2 + lO2 + mN2
WHERE: a, b, c, d, e, f, g, h, I, j, k, x are the
number of moles of the elements.
x – excess air in decimal

170. fuelkg
airkg
18f32e28d32c2b12a
3.76(28)g32g
F
A
t
Theoretical air-fuel ratio:
Actual air-fuel ratio:
fuelkg
airkg
18f32e28d32c2b12a
3.76(28)g32gx)(1
aF
A

171. MASS FLOW RATE OF FLUE GAS (Products)
Air+Fuel Products
A. Without considering Ash loss
1
F
A
mm Fg
B. Considering Ash loss
lossAsh1
F
A
mm Fg

172. Heating Value
Heating Value - is the energy released by
fuel when it is completely burned and the
products of combustion are cooled to the
original fuel temperature.
Higher Heating Value (HHV) - is the
heating value obtained when the water in
the products is liquid.
Lower Heating Value (LHV) - is the
heating value obtained when the water
in the products is vapor.

173. For Solid Fuels with the presence of
Fuel’s ULTIMATE ANALYSIS
kg
KJ
S9304
8
O
H212,144C820,33HHV 2
2
where: C, H2, O2, and S are in decimals from the
ultimate analysis

174. HHV = 31 405C + 141 647H KJ/kg
HHV = 43 385 + 93(Be - 10) KJ/kg
For Liquid Fuels
where: Be - degrees Baume
 For Coal and Oils with the absence of
Ultimate Analysis
fuelofkg
airofKg
3041
HHV
F
A
t

175. For Gasoline
kg
KJ)API(93639,38LHV
kg
KJ)API(93160,41HHV
kg
KJ)API(93035,39LHV
kg
KJ)API(93943,41HHV
For Kerosene

176. For Fuel Oils
InstitutePetroleumAmericanAPI
kg
KJ)API(6.139105,38LHV
kg
KJ)API(6.139130,41HHV

177. For Fuel Oils (From Bureau of Standard
Formula)
).t(.St@S 561500070
API131.5
141.5
S
HHV = 51,716 – 8,793.8 (S)2 KJ/kg
LHV = HHV - QL KJ/kg
QL = 2442.7(9H2) KJ/kg
H2 = 0.26 - 0.15(S) kg of H2/ kg of
fuel

178. Where
S - specific gravity of fuel oil at 15.56 C
H2 - hydrogen content of fuel oil
QL - heat required to evaporate and superheat the water
vapor formed bythe combustion of hydrogen in the fuel
S @ t - specific gravity of fuel oil at any temperature t
Oxygen Bomb Calorimeter - instrument used in mea-
suring heating value of solid and liquid fuels.
Gas Calorimeter - instrument used for measuring
heating value of gaseous fuels.

179. Properties of Fuels and Lubricants
a) Viscosity - a measure of the resistance to flow that a
lubricant offers when it is subjected to shear stress.
b) Absolute Viscosity - viscosity which is determined by
direct measurement of shear resistance.
c) Kinematics Viscosity - the ratio of the absolute viscosity
to the density
d) Viscosity Index - the rate at which viscosity changes with
temperature.
e) Flash Point - the temperature at which the vapor above
a volatile liquid forms a combustible mixture with air.
f) Fire Point - The temperature at which oil gives off vapor
that burns continuously when ignited.

180. g) Pour Point - the temperature at which oil will no longer pour freely.
h) Dropping Point - the temperature at which grease
melts.
i) Condradson Number(carbon residue) - the percentage amount by mass of the
carbonaceous residue remaining after destructive distillation.
j) Octane Number - a number that provides a measure of the ability of a fuel to
resist knocking when it is burnt in a gasoline engine. It is the percentage by volume
of iso-octane in a blend with normal heptane that matches the knocking behavior of
the fuel.

181. k) Cetane Number - a number that provides a measure of the ignition
characteristics of a diesel fuel when it is burnt in a standard diesel engine. It is the
percentage of cetane in the standard fuel.
Prepared By: ENGR YURI G. MELLIZA, RME