1. POWER PLANT TECHNOLOGY
BY. ENGR YURI G. MELLIZA
Table of Contents
Introduction..................................................................................................................................................................... 7
Introduction to Thermodynamics ..................................................................................................................................... 8
Law of Conservation of Mass ........................................................................................................................................... 8
Continuity Equation ......................................................................................................................................................... 8
Forms of Energy ............................................................................................................................................................... 9
Internal Energy: ........................................................................................................................................................... 9
Flow Energy or Flow Work:........................................................................................................................................... 9
Heat: ............................................................................................................................................................................ 9
Work: ........................................................................................................................................................................... 9
Kinetic Energy: ........................................................................................................................................................... 10
Potential Energy: ........................................................................................................................................................ 10
Zeroth Law of Thermodynamics: .................................................................................................................................... 10
Specific Heat or Heat Capacity:................................................................................................................................... 10
Sensible Heat: ............................................................................................................................................................ 11
Heat of Transformation .............................................................................................................................................. 11
A.
Latent Heat of Vaporization: ........................................................................................................................... 11
Phase Change ................................................................................................................................................................ 11
Ideal or Perfect Gas........................................................................................................................................................ 12
1

2. IDEAL GAS MIXTURE ...................................................................................................................................................... 14
1. Total moles of a mixture......................................................................................................................................... 14
2. Mole Fraction ......................................................................................................................................................... 14
3. Total mass of a mixture .......................................................................................................................................... 14
4. Mass Fraction ......................................................................................................................................................... 14
5. Equation of State.................................................................................................................................................... 14
6. Amagat's Law: ........................................................................................................................................................ 15
7. Dalton's Law: .......................................................................................................................................................... 15
8. Molecular Weight of a Mixture ............................................................................................................................... 16
9. Gas Constant of a mixture ...................................................................................................................................... 16
10. Specific Heats of a mixture ................................................................................................................................... 16
11. Gravimetric and Volumetric Analysis: ................................................................................................................... 16
Law of conservation of Energy (The First Law of Thermodynamics): ............................................................................... 17
Application of the Law of Conservation of Energy....................................................................................................... 17
B.
Open System ...................................................................................................................................................... 17
Processes of Fluids ......................................................................................................................................................... 18
1.
Isobaric Process: ................................................................................................................................................. 18
2.
Isometric Process: .............................................................................................................................................. 19
3.
Isothermal Process: ............................................................................................................................................ 20
4.
Isentropic Process: ............................................................................................................................................. 21
5.
Polytropic Process: ............................................................................................................................................. 22
6.
Throttling Process:.............................................................................................................................................. 23
2

3. Properties of Pure Substance: ........................................................................................................................................ 23
Terms and Definition .............................................................................................................................................. 25
Throttling Calorimeter ................................................................................................................................................... 26
Fuels and Combustion.................................................................................................................................................... 29
Combustion Chemistry ............................................................................................................................................... 30
Combustion of Combustible elements with Air:.......................................................................................................... 31
Theoretical Air: .......................................................................................................................................................... 32
Excess Air: .................................................................................................................................................................. 32
Hydrocarbon Fuel: ..................................................................................................................................................... 32
COMBUSTION OF HYDROCARBON FUEL(CnHm) ......................................................................................................... 32
COMBUSTION OF SOLID FUELS ................................................................................................................................... 33
DEW POINT TEMPERATURE .................................................................................................................................... 33
ULTIMATE ANALYSIS............................................................................................................................................... 33
PROXIMATE ANALYSIS ............................................................................................................................................ 33
ORSAT ANALYSIS .................................................................................................................................................... 33
MASS FLOW RATE OF FLUE GAS ............................................................................................................................. 33
a) Without considering Ash loss:............................................................................................................................. 33
b) Considering Ash loss ........................................................................................................................................... 34
MOLECULAR WEIGHT OF PRODUCTS ...................................................................................................................... 34
GAS CONSTANT OF PRODUCTS ............................................................................................................................... 34
SPECIFIC HEATS OF PRODUCTS ............................................................................................................................... 34
PARTIAL PRESSURE OF COMPONENTS .................................................................................................................... 35
3

4. HEATING VALUE ............................................................................................................................................................. 38
For Liquid Fuels ...................................................................................................................................................... 38
For Gasoline ........................................................................................................................................................... 38
For Fuel Oils ........................................................................................................................................................... 38
For Fuel Oils (From Bureau of Standard Formula) ................................................................................................... 38
Properties of Fuels and Lubricants ................................................................................................................................. 39
Cycle .............................................................................................................................................................................. 40
Steam Power Plant Cycle................................................................................................................................................ 40
Rankine Cycle ............................................................................................................................................................. 40
Reheat Cycle Steam Power Plant: ............................................................................................................................... 42
Regenerative Cycle:.................................................................................................................................................... 43
Reheat – Regenerative Cycle: ..................................................................................................................................... 45
STEAM RATE........................................................................................................................................................... 46
HEAT RATE ............................................................................................................................................................. 46
Turbine Efficiency ................................................................................................................................................... 47
Pump Efficiency ...................................................................................................................................................... 47
Boiler or Steam Generator Efficiency ...................................................................................................................... 47
GENERAL BOILER DESCRIPTION ...................................................................................................................................... 47
Boiler Auxiliaries and Accessories ............................................................................................................................... 49
BOILER PERFORMANCE .............................................................................................................................................. 50
BOILER HEAT BALANCE............................................................................................................................................... 52
CONDENSERS ................................................................................................................................................................. 57
Direct - contact or Open, condensers ......................................................................................................................... 57
Surface Condenser ..................................................................................................................................................... 58
4

5. GEOTHERMAL POWER PLANT ........................................................................................................................................ 60
The Diesel Power Plant .................................................................................................................................................. 63
ENGINE PERFORMANCE ............................................................................................................................................. 68
1. Heat supplied by fuel (Qs):.................................................................................................................................. 68
2. Indicated Power (IP): .......................................................................................................................................... 68
3. Brake or Shaft Power (BP):
68
4. Friction Power (FP): ............................................................................................................................................ 69
5. Brake Torque ...................................................................................................................................................... 69
6. Indicated Mean Effective Pressure (Pmi):............................................................................................................ 69
7. Displacement Volume (VD): ................................................................................................................................. 70
8. Specific Fuel Consumption .................................................................................................................................. 70
9. Heat Rate (HR):................................................................................................................................................... 70
10. Thermal Efficiency ............................................................................................................................................ 71
11. Mechanical Efficiency ....................................................................................................................................... 71
12. Generator Efficiency ......................................................................................................................................... 71
13. Generator Speed .............................................................................................................................................. 71
14. Volumetric Efficiency ........................................................................................................................................ 71
15. Correction Factor for Non-Standard Condition .................................................................................................. 72
16. Engine Heat Balance ......................................................................................................................................... 72
Diesel Engine Maintenance ........................................................................................................................................ 73
Hydroelectric Power Plant.............................................................................................................................................. 83
TERMS AND DEFINITION ............................................................................................................................................ 83
A. IMPULSE TYPE (Pelton type) ................................................................................................................................... 84
B. REACTION TYPE (Francis Type) ............................................................................................................................... 85
5

6. PUMP STORAGE HYDRO-ELECTRIC PLANT .................................................................................................................. 86
FUNDAMENTAL EQUATIONS .................................................................................................................................. 86
GAS TURBINE POWER PLANT ......................................................................................................................................... 88
Closed Cycle Gas Turbine Cycle .................................................................................................................................. 89
WIND POWER ................................................................................................................................................................ 93
6

7. POWER PLANT TECHNOLOGY
By. Engr. Yuri G. Melliza
Introduction
This book was designed as standard learning materials intended for graduating technology students in a course of Power Plant technology. This was written with my goal
in mind to focus on the study of different types of electric generating power plant facility
used commonly in different parts of the world. With the fast growing global technological
advancement, this book gives the students a wide array of understanding the different
concepts and principles of electrical energy production as well as the analytical and
technical design of the different power plant system. On the other hand this book adopts
to used the SI system of units, which is now used worldwide as the standard system
of units.
TOPIC OUTLINE
1. Introduction to Thermodynamics
2. Fuels and Combustion
3. The Steam Power Plant Cycle
4. The Internal Combustion Engine Power Plant
5. The Hydro-Electric Power Plant
6. The Gas Turbine Power Plant
7. The Geothermal Power Plant
8. The Wind Energy
9. The Solar Energy
10. Energy From the Ocean
11. Cogeneration Power Plant
12. Environmental Aspects of Power Generation
7

8. Introduction to Thermodynamics
Law of Conservation of Mass
Mass is indestructible, in applying this law we must except nuclear processes
during which mass is converted into energy.
Verbal Form:
Mass Entering – Mass Leaving = Change of Mass stored in the system
Equation Form:
m1 – m2 = ∆m
m
1
2
For a steady-state, steady flow ∆m = 0, hence
m1= m2
Continuity Equation
For one dimensional flow
Av
υ
m1 = m 2 = m
m = ρAv =
ρ1 A 1 v 1 = ρ 2 A 2 v 2 = ρAv
A 1 v 1 A 2 v 2 Av
=
=
υ1
υ2
υ
Where:
m – mass flow rate in kg/sec
A – cross sectional area in m2
v – velocity in m/sec
ρ - density in kg/m3
υ - specific volume in m3/kg
8
m

9. Forms of Energy
• Internal Energy
• Flow Energy or Flow Work
• Heat
• Work
• Kinetic Energy
• Potential Energy
Internal Energy: It is the energy due to the overall molecular interaction.
∆U = m(u2 – u1) KJ
Where:
u – specific internal energy, KJ/kg
U – total internal energy, KJ (KW if m in kg/sec)
m – mass in kg (kg/sec, mass flow rate)
Flow Energy or Flow Work: It is the energy required in pushing a fluid into the
system or out from the system.
∆(PV) = (P2V2 – P1V1) KJ
∆(Pυ) = (P2υ2 – P1υ1) KJ/kg
Where:
P – pressure, KPa
V – volume, m3
υ - specific volume, m3/kg
PV – flow work, KJ (KW if V in m3/sec)
Heat: Heat is the energy that srosses a system’s boundary because of a temperature
difference between the system and the surrounding.
Q = m(q) KJ
Where:
Q – Total heat, KJ (KW if m in kg/sec)
q – heat in KJ/kg
Note: Q is positive if heat is added to the system and negative if heat is rejected
from the system
Work: Work is define as the force multiplied by the displacement in the direction of the
force.
W = ∫ F ⋅ dx
9

10. Kinetic Energy: It is the energy due to the motion of a body.
(
)
m v 2 − v1 )
ΔKE =
KJ
2(1000)
2
2
Where:
v – velocity, m/sec
m – mass, kg
KE – Kinetic energy, KJ (KW if m in kg/sec)
Potential Energy: It is the energy by virtue of its configuration or elevation.
ΔPE =
mg(z2 − z1 )
KJ
(1000)
Where:
z – elevation measured from a chosen datum, meters
+ z if measured above the datum
- z if measured below datum
g – gravitational acceleration, m/sec2
g = 9.81 m/sec2 (at sea level condition)
PE – potential energy, KJ (KW if m in kg/sec)
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:
Specific heat is the amount of heat required to raise the temperature of a 1 kg
mass of a substance 1°K or 1°C.
C=
dQ KJ
KJ
or
dT kg - °K kg − °C
dQ = C dT
If C is constant
Q = C(T2 – T1) KJ/kg
Considering m;
Q = mC(T2 – T1) KJ (KW if m in kg/sec)
10

11. Sensible Heat: It is the amount of heat added to heat a substance, or the amount of
heat removed to cool a substance.
Q = mC(T2 – T1) KJ (KW if m in kg/sec
Heat of Transformation
A. Latent Heat of Vaporization: It is the amount of heat added to vaporize a
liquid, or the amount of heat removed to condense a gas (Vapor)
Qv = m(Hv) KJ (KW if m in kg/sec)
Where:
m – mass in kg (kg/sec)
Hv – heat of vaporization in KJ/kg
B. Latent Heat of Fusion: It is the amount of heat added to melt a solid or
removed to solidify a liquid.
QF = m(HF) KJ (KW if m in kg/sec)
Where:
HF – latent heat of fusion in KJ/kg
Phase Change
A. Vaporization: Liquid to Vapor
B. Condensation: Vapor to liquid
C. Freezing or Solidifying: Liquid to solid
D. Melting: Solid to liquid
E. Sublimation: Change from solid directly to vapor without passing the liquid
state.
11

12. Ideal or Perfect Gas
Fundamental equations:
1. Equation of State or Perfect Gas Equation
PV = mRT
Pυ = RT
PV
=C
T
P1 V1 P2 V2
=
=C
T1
T2
2. Gas Constant
R=
8.3143 KJ
M
kg - o K
3. Boyles Law (At constant temperature, T = C)
PV = C
P1 V1 = P2 V2
4. Charle’s Law
a. At Constant Pressure
V
=C
T
V1 V2
=
T1 T2
Q = ∆h = mCP (∆T )
b. At Constant Volume
P
=C
T
P1 P2
=
T1 T2
Q = ∆U = mC V (∆T )
12

13. 5. Avogadro’s Law: All gases at the same temperature and pressure have the
same number of molecules per unit of volume. It follows that the specific
weight is directly proportional to its molecular weight.
γ 1 M1
=
γ 2 M2
6. Specific Heat
a. At Constant pressure
CP =
Rk
k −1
b. At Constant volume
R
k −1
Cp = C v + R
CV =
k=
Cp
Cv
7. Entropy Change
∆S = ∫
dQ
T
13

14. IDEAL GAS MIXTURE
Gas Mixture: A gaseous substance consisting two or more type of gases. The gases in a
gas mixture are called “components” or “constituents” of a mixture.
1. Total moles of a mixture
n = ∑ ni
2. Mole Fraction
ni
n
3. Total mass of a mixture
yi =
m = ∑ mi
4. Mass Fraction
m
xi = i
m
5. Equation of State
A. Mass Basis
a. For the mixture
PV = mRT
b. For the components
Pi Vi = miR i Ti
B. Mole Basis
a. For the mixture
PV = nRT
b. For the components
Pi Vi = ni RTi
14

15. 6. Amagat's Law: The total volume V of a mixture is equal to the sum of 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 = P 3
T = T 1 = T2 = T3
n = n1 + n2 + n3
PV PV1 PV2 PV3
=
+
+
R T R T RT RT
 PV PV1 PV2 PV3  RT 
 RT = R T + RT + RT   P 





V = V1 + V2 + V3
V = ∑ Vi
yi =
Vi
V
7. 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 the mixture volume V and temperature T.
1
n1
P1
2
n2
P2
3
n3
P3
mixture
n2
P2
n = n1 + n2 + n3
PV
RT
=
P1 V
RT
+
P2 V
RT
+
P3 V
RT
 PV P1 V P2 V P3 V  RT 

=
+
+


R T RT RT R T   V 



P = P1 + P2 + P3
V = V1 = V2 = V3
T = T 1 = T2 = T3
P = ∑ Pi
yi =
15
Pi
P

16. 8. Molecular Weight of a Mixture
M=
∑y M
M=
R 8.3143 kg
=
R
R
kgmol
i
i
9. Gas Constant of a mixture
R=
∑x R
R=
R 8.3143
KJ
=
M
M
kg - °K
i
i
10. Specific Heats of a mixture
KJ
kg - °K
KJ
C V = ∑ x iC Vi
kg - °K
KJ
CP = C V + R
kg - °K
Rk
KJ
CP =
k − 1 kg - °K
R
KJ
CV =
k − 1 kg - °K
CP = ∑ x iCPi
11. 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.
A . Volumetric or Molal analysis to Gravimetric analysis
xi =
y iM i
yM
= i i
∑ y iM i M
16

17. B. Gravimetric analysis to volumetric or Molal analysis
xi
Mi
yi =
x
∑ Mi
i
Law of conservation of Energy (The First Law of Thermodynamics):
“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
Application of the Law of Conservation of Energy
A. Closed System (Nonflow System): A system closed to matter or mass flow.
W
Q = ΔU + W
W = ∫ P ⋅ dV
dW = P ⋅ dV
Gas
dQ = dU + PdV
∆U
Q
B. Open System (Steady-State, Steady-Flow System): A system opens to
matter flow in which there’s an exchange of mass between the system and the
surrounding.
1
U1 + P1 V1 + KE1 + PE1
W
System
2
U2 + P2 V2 + KE 2 + PE 2
Q
17

18. From First Law;
E1 – E2 = ∆Es
For an Open system, ∆Es = 0, hence
E1 = E2 or
Energy entering = Energy leaving
U1 + P1 V1 + KE1 + PE1 + Q = U2 + P2 V2 + KE 2 + PE 2 + W
Q = (U2 − U1 ) + (P2 V2 − P1 V1 ) + (KE 2 − KE1 ) + (PE 2 − PE1 ) + W
Enthalpy: Sum of internal and flow energy
h= U + PV
h1 + KE1 + PE1 + Q = h2 + KE 2 + PE 2 + W
Q = (h2 − h1 ) + (KE 2 − KE1 ) + (PE 2 − PE1 ) + W
W = Q - [(h2 − h1 ) + (KE 2 − KE1 ) + (PE 2 − PE1 ) ]
W = Q - ∆h - ∆KE - ∆PE -
Processes of Fluids
1. Isobaric Process: Reversible Constant Pressure Process
A. Closed System
Q = ∆U + W
W = P(V2 – V1)
Q = m(h2 – h1)
∆U = m(u2 – u1)
For Ideal Gas
Q = mCp(T2 – T1)
∆U = mCv(T2 – T1)
W = mR(T2 – T1)
Rk
k − 1)
R
Cv =
k −1
C
k= p
Cv
Cp =
V1 V2
=
T1 T2
18

19. B. Open System (Steady-state,steady-flow)
W = -∆KE - ∆PE
If ∆KE = 0 & ∆PE = 0
W=0
Q = m(h2 – h1)
C. Entropy change
ΔS = S 2 − S1
T
ΔS = mC ln 2 → For ideal gas
P T1
2. Isometric Process: Reversible Constant Volume Process.
A. Closed System (Non-Flow)
Q = ∆U + W
W=0
∆U = m(u2 – u1)
Q = ∆U = m(u2 – u1)
For Ideal Gas
Q = mCv(T2 – T1)
Q = ∆U = mCv(T2 – T1)
W=0
Rk
Cp =
k − 1)
R
Cv =
k −1
Cp
k=
Cv
P1 P2
=
T1 T2
B. For Open System (Steady flow)
W = Q - ∆h - ∆KE - ∆PE W = -V(P2 - P1 ) - ∆KE - ∆PE
If ∆KE = 0 & ∆PE = 0
W = Q - ∆h
W = -V(P2 - P1 )
19

20. For Ideal Gas
− V(P2 − P1 ) = V(P1 − P2 )
− V(P2 − P1 ) = mR(T1 − T2 )
C. Entropy Change
ΔS = S2 − S1
ΔS = mC Vln
T2
→ For ideal gas
T1
3. Isothermal Process: Reversible Constant Temperature Process
A. Closed System (Nonflow System)
Q = ∆U + W
∆U = m(u2 – u1)
For Ideal Gas
P1V1 = P2V2 = C
∆U = mCv(T2 – T1)
T2 – T1 = 0
∆U = 0
Q=W
W = P1 V1ln
V2
P
= P1 V1ln 1
P2
V1
P1V1 = mRT1
B. For Open System (Steady Flow)
W = Q - Δh - ΔKE - ΔPE
For Ideal Gas
∆h = mC p (∆T)
ΔT = 0
Δh = 0
W = Q - ΔKE - ΔPE
Q = P1 V1 ln
P1
V
= P1 V1 ln 2
P2
V1
If ΔKE = 0 & ΔPE = 0
W=Q
W = P1 V1 ln
P1
V
= P1 V1 ln 2
P2
V1
20

21. C. Entropy change
ΔS = S 2 − S1
Q
T
Q W
For ideal or perfect gas
ΔS = =
T T
ΔS =
4. Isentropic Process: An isentropic process is an internally reversible adiabatic
process in which the entropy remains constant (S = C or PVk = C, for ideal or
perfect gas)
P, V, & T relationships for Ideal or Perfect gas
P1 V1 = P2 V2 = C
k
T2  P2 
= 
T1  P1 
 
k
k −1
k
V
= 1
V
 2




k −1
A. Closed System (Non-Flow)
Q = ΔU + W
Q=0
W = − ΔU
For Ideal Gas
W = − ΔU = −mC v (T2 − T1 )
(P V − P1 V1 ) mRT1
W= 2 2
=
1−k
1−k
k −1


 P2  k
  − 1
 P1 

 



B. Open System (Steady state, steady flow)
W = Q − Δh − ΔKE − ΔPE
Q=0
W = -∆h - ΔKE − ΔPE
For ideal gas
k(P2 V2 - P1 V1 ) kmRT1
=
- Δh =
1- k
1−k

 P2

 P1






k −1
k

− 1



If ΔKE = 0 & ΔPE = 0
k(P2 V2 - P1 V1 ) kmRT1
=
W = -∆h =
21
1-k
1−k

 P2

 P1






k −1
k

− 1




22. C. Entropy change
∆S = 0
5. Polytropic Process: A polytropic process is an internally reversible process of
an Ideal or Perfect Gas in which PVn = C, where n stands for any constant but
not equal to zero.
P,V, & T relationship:
P1 V1 = P2 V2 = C
n
n
n−1
V 
T2  P2  n
= 
= 1
P 
V 
T1  1 
 2
A. Closed System
n−1
Q = ΔU + W
ΔU = mCv(T2 − T1 )
n−1


(P2 V2 − P1 V1 ) mRT1  P2  n
  − 1
W=
=

1−n
(1 − n)  P1 
 



Q = mCn (T2 − T1 )
k −n
Cn = C V 

1−n 
B. Open System
W = Q − Δh − ΔKE − ΔPE
Δh = mCP (T2 − T1 )
n(P2 V2 − P1 V1 ) nmRT1
Q − Δh =
=
1−n
(1 − n)
Q = mCn (T2 − T1 )
k −n
Cn = C V 

1−n 
If ∆KE = 0 & ΔPE = 0
W = Q − Δh
C. Entropy Change
ΔS = mC n ln
T2
T1
22

 P2

 P1






n−1
n

− 1




23. 6. Throttling Process: A throttling process is a steady-state, steady-flow process
in which W= 0, ∆KE = 0,∆PE = 0 where h = C.
h1 = h2
Properties of Pure Substance:
A pure substance is a substance that is
homogeneous in nature and is homogeneous.
(a)
(b)
(c)
(d)
(e)
P
P
P
P
P
t>100°C
100°C
30°C
100°C
100°C
100°C
Q
Q
Q
Q
Q
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
pressure is the saturation pressure at 100°C.
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 Mixture - a mixture of liquid and vapor at the saturation temperature.
Superheated Vapor - a vapor whose temperature is greater than the saturation
temperature.
23

24. Temperature - Specific volume Diagram (T-υ diagram)
F•
T
Critical Point
t > tsat
e
tsat
b
tsc
P=C
c
Saturation Curve
d
a
υ
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
Temperature-Entropy Diagram (T-S Diagram)
F•
T
Critical Point
t > tsat
e
I
tsat
tsc
b
III
Saturation Curve
P=C
c
II
d
a
S
Region I - sub-cooled or compressed liquid region
Region II- saturated mixture region
Region III- superheated vapor region
24

25. yurigmelliz
Enthalpy-Entropy Diagram (h-S Diagram or Mollier Chart)
F Critical Point
•
h
t = C(constant temperature curve)
III
P=C
I
Saturation Curve
II
S
The properties h,S,U,and υ at saturated liquid, saturated vapor, sub-cooled or
compressed liquid and superheated vapor condition, can be determined using the Steam
Table.
For the properties at the saturated mixture condition, its properties is equal to
r = rf + xrfg
where r stands for any property, such as h, S, U,and υ, where subscript f refers to
saturated liquid condition and fg refers to the difference in property between saturated
vapor and saturated liquid and x is called the quality.
QUALITY
x=
mv
mv + ml
=
mv
m
where: m - mass
v - refers to vapor
l - refers to liquid
Note: For sub-cooled liquid, its properties are approximately equal to the properties at
saturated liquid which corresponds to the sub-cooled temperature.
Terms and Definition
a. Saturated Liquid – a liquid existing at the saturation temperature corresponding
the pressure.
b. Saturated Vapor – a vapor existing at the saturation temperature corresponding
the pressure.
c. Superheated Vapor – a vapor whose temperature is greater than the saturation
temperature corresponding to the pressure.
d. Subcooled Liquid – a liquid whose temperature is less than the saturation
temperature corresponding to the pressure.
25

26. e. Saturated Mixture – a mixture of liquid and vapor at the saturation temperature
and pressure.
f. Saturated Temperature – it is the highest temperature reached by a liquid
heated at certain pressure in which vaporization takes place.
g. Saturated Pressure – a pressure corresponding the saturation temperature.
Example: When water is heated at standard pressure (P = 101.325 KPa) it will
boil at 100°C. This temperature is the saturation temperature corresponding
101.325 KPa and the pressure 101.325 KPa is the saturation pressure
corresponding 100°C temperature.
Throttling Calorimeter:
An apparatus that is used to determine the quality of
a desuperheated steam flowing in a steam line.
main steam
line
thermometer
main steam
line pressure
calorimeter
pressure
gauge
calorimeter
throttling
valve
to main
steam line
A throttling process is one that is a constant enthalpy process. Steam from the main
steam line expands in the calorimeter to the calorimeter pressure and temperature. A
throttling calorimeter is an instrument used to determine the quality of steam flowing in
the main steam line.
26

27. Example (Constant Pressure – Ideal Gas)
When a certain perfect gas is heated at constant pressure from 15ºC to 95ºC, the heat required
is 1136 KJ/kg. When the same gas is heated at constant volume between the same
temperatures the heat required is 808 KJ/kg. Calculate Cp, Cv, k, and M of the gas.
At P = C
Q = Cp (T2 - T1 )
T2 - T1 = t 2 - t1
1136 = C p (95 − 15)
Cp = 14.2
At V = C
Q = C v (T2 - T1 )
808 = C v (95 - 15)
C v = 10.1
14.2
= 1.406
10.1
Cp = Cv + R
R = 4.1
k=
Example 2 – (Polytropic – Ideal Gas)
A closed system consisting of 2 kg of a gas undergoes a polytropic process during which the
value of n = 1.3. The process begins with P1 = 100 KPa, υ1 = 0.5 m3/kg and ends with P2 = 25
KPa. Determine the final volume, in m3, and the work.
Given
m = 2 kg
P1 = 100 KPa ; P2 = 25 KPa
υ1 = 0.5 m3/kg
Process: PV1.3 = C
P1 V1
1.3
= P2 V2
1.3
=C
P V − P1 V1
W= 2 2
1−n
V
υ=
m
V = υm
V1 = 0.5(20) = 1 kg
 P V 1.3 
V2 =  1 1 
 P2 


W = 91 .7 KJ
27
1
1.3
= 2 .9 m 3

28. Example 3 – (Ideal Gas)
A 5 m3 tank contained chlorine (R = 0.1172 KJ/kg-K) at 300 KPa and 300K after 3 kg of chlorine
has been used. Determine the original mass and pressure if the original temperature was 315 K.
(45.66 kg ; 337.15 KPa)
Given
V1 = V2 = 5 m3 ; R = 0.1172 KJ/kg-°K
m1 = ? ; P1 = ? ; T1 = 315°K
P2 = 300 KPa ; T2 = 300°K
m2 = (m1 – 3)
PV = mRT
P2 V2 = m2RT2
300(5) = (m1 − 3)(0.1172)(300)
m1 = 45.66 kg
P1 V1 = m1RT1
P1 = 337.15 KPa
Example 4 – (Constant Temperature/Ideal Gas)
A mass of kg of air contained in cylinder at 800 KPa, 1000°K expands in a reversible isothermal
process to 100 KPa. Calculate
a. the heat Q
b. the entropy change
Given: Process T = C or PV = C (for Air: R = 0.287 KJ/kg-°K and k = 1.4)
m = 1 kg ; P1 = 800 KPa ; T1 = 1000°K ; P2 = 100 KPa
a. At T = C for ideal Gas, Q = W
Q = W = mRT1 ln
V2
P
800
= mRT1 ln 1 = 1( 0.287 )( 1000 ) ln
V1
P2
100
Q = W = 597 KJ
Q 597
KJ
= 0.597
b. ΔS = =
T 1000
°K
Example 5 – (Polytropic Process)
One kg of oxygen are compressed polytropically from a pressure of 96.5 KPa and 21°C to 675.5
KPa. The ratio of the specific heat k = 1.395 and the compression is according to PV1.3= C.
Determine the change of entropy in KJ/K.(∆S = -0.94 KJ/K)
Given:
P1 = 96.5 KPa ; P2 =675.5 KPa
k= 1.395
PV1.3 = C
∆S = mC n ln
T2
T1
 k −n
Cn = C v 

 1−n
28

29. Fuels and Combustion
Fuel: A substance composed of chemical elements which in rapid chemical union with
oxygen produced “combustion”.
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 sufficiently fast
whereby useful quantities of heat are liberated at elevated temperature.
Types of Fuel
1. Solid Fuels
a. Coal
b. Wood
c. charcoal
2. Liquid Fuels
a. Diesel
b. Gasoline
c. Kerosene
3. Gaseous Fuels
a. LPG
b. Natural Gas
c. Methane
4. Nuclear Fuels
a. Uranium
b. Plutonium
Combustible Elements
1. Carbon (C)
2. Hydrogen (H2)
3. Sulfur (S)
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
29

30. Molecular Weight of combustion Gases
Gas
C
H
H2
O
O2
N
N2
S
Molecular
Weight
12
1
2
16
32
14
28
32
Combustion Chemistry
A. Oxidation of Carbon
C + O2 → CO2
Mole Basis
1+1→1
Mass Basis
1(12) + 1(32) → 1(44)
3 + 8 → 11
B. Oxidation of Hydrogen
H2 + ½ O2 → H2O
Mole Basis
1+½→1
Mass Basis
1(2) + ½(32) → 1(18)
2 + 16 → 18
1+8→9
C. Oxidation of Sulfur
S + O2 → SO2
Mole Basis
1+1→1
Mass Basis
1(32) + 1 (32) → 1(64)
1+1→2
30

31. Composition of Air: (in theoretical combustion)
%age by Volume (or by mole)
O2 = 21
N2 = 79
%age by mass
O2 = 23
N2 = 77
Mole Ratio
Mols N2 79
=
= 3.76
Mol O2 21
Combustion of Combustible elements 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)
3 + 8 + 3.76(7) → 11 + 3.76(7)
kg of air
8 + 3.76(7)
kg of air
=
= 11.44
kg of Carbon
3
kg of C
B. Combustion of Hydrogen with air
H2 + ½ O2 + (½)3.76N2→ H2O + (½)3.76N2
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 + (½)3.76(28) → 18 + (½)3.76(28)
1 + 8 + (½)3.76(14) → 9 + (½)3.76(14)
kg of air
8 + 12 (3.76)(14)
kg of air
=
= 34.32
kg of Hydrogen
1
kg of H2
31

32. C. Combustion of Sulfur with air
S + O2 + (3.76)N2 → 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 + 3.76(28) → 64 + 3.76(28)
kg of air
32 + (3.76)(28)
kg of air
=
= 4.29
32
kg of Sulfur
kg of S
Theoretical Air: It is the minimum amount of air required to oxidized the reactants.
With theoretical air alone, no O2 is found in the product.
Excess Air: It is an amount of air in excess of the theoretical air requirement in order to
influence complete combustion. With excess air O2 is found in the product.
Hydrocarbon Fuel: Fuels containing the element Carbon and Hydrogen.
Chemical Formula: CnHm
Family of Hydrocarbon:
1. Paraffin (CnH2n+2)
2. Olefins (CnH2n)
3. Diolefin (CnH2n-2)
4. Naphthene (CnH2n): this type of fuel has the same formula as olefins but
at different structure.
5. Aromatics ((CnH(2n-6))
COMBUSTION OF HYDROCARBON FUEL(CnHm)
A) Combustion of CnHm with 100% theoretical air
CnHm + aO2+ a(3.76)N2 → bCO2 + cH2O + a(3.76)N2
where:
a = n + 0.25m
b=n
c = 0.5m
B) With excess air
CnHm + (1+e)aO2 + (1+e)a(3.76)N2 → bCO2 + cH2O +dO2 + (1+e)a(3.76)N2
where:
d = e(n + 0.25m)
Note: The values of a,b,c, and d above in terms of n and m is applicable only for
the combustion of one type of hydrocarbon.
where: e - excess air in decimal
32

33. Theoretical Air-Fuel Ratio: Ratio of Kg of Air to Kg of fuel
Actual Air-fuel Ratio: Ratio of actual kgs of Air (theoretical + excess) to kg of fuel
COMBUSTION OF SOLID FUELS
Components of Solid Fuels: C, H2, O2, N2, S, and Moisture
A) Combustion with 100% theoretical air
aC + bH2 + cO2 + dN2 + eS + fH2O + xO2 + x(3.76)N2 →
gCO2 + hH2O + iSO2 + jN2
B) Combustion with excess air (e’ - excess air in decimal)
aC + bH2 + cO2 + dN2 + eS +fH2O + (1+e’)xO2 + (1+e’)x(3.76)N2 →
gCO2 + hH2O + iSO2 + kO2 + lN2
The theoretical and actual air-fuel ratio of solid fuels can be computed based on their
balance combustion equation above.
DEW POINT TEMPERATURE
The Dew Point Temperature (tdp) is the saturation temperature corresponding the
partial pressure of the water vapor in the mixture (products of combustion).
ULTIMATE ANALYSIS
Ultimate Analysis gives the amount of C, H2, O2, N2, S and moisture in percentages by
mass, sometimes the percentage amount of Ash is given.
(A/F)t = 11.44C + 34.32(H- O/8) + 4.29S kg of air/kg of fuel
where: C, H, O and S are in decimals obtained from the Ultimate Analysis
PROXIMATE ANALYSIS
Proximate Analysis gives the percentage amount of Fixed Carbon, Volatiles, Ash and
Moisture.
ORSAT ANALYSIS
Orsat Analysis gives the volumetric or molal analysis of the products of combustion or
exhaust gases on a Dry Basis.
MASS FLOW RATE OF FLUE GAS
a) Without considering Ash loss:
A

m g = mF  + 1 
F

33

34. b) Considering Ash loss
A

m g = mF  + 1 - Ash loss 
F

where ash loss in decimal
MOLECULAR WEIGHT OF PRODUCTS
M=
M=
n CO2 M CO2 + nH2OMH2O + n O2 MO2 + n SO 2 MSO 2 + nN2 MN2 + . . . + nM
nPr oducts
kg
kgmol
R kg
R kgmol
R = 8.3143
KJ
→ Universal Gas Constant
kgmol - °K
R - Gas Constant
KJ
kg - °K
GAS CONSTANT OF PRODUCTS
R=
R=
m CO2 R CO2 + mH2OR H2O + m O2 R O2 + m SO2 R SO2 + mN2 R N2 + . . . + mR
mPr oducts
KJ
kg - °K
R
M
SPECIFIC HEATS OF PRODUCTS
mCO2 CP CO2 + mH2O CP H2O + mO2 CP O2 + mSO2 CP SO2 + mN2 CP N2 + . . . + mC P
KJ
CP =
mPr oducts
kg - °K
CP = Σx iCPi
CV =
mCO2 C V CO2 + mH2O C V H2O + mO2 C V O2 + mSO2 C V SO2 + mN2 C V N2 + . . . + mC V
mPr oducts
KJ
kg - °K
C V = Σx iC Vi
CP = C V + R
CP
CV
Where:
CP – specific heat at constant pressure in KJ/kg-°K or KJ/kg-°C
CV – specific heat at constant volume in KJ/kg-°K or KJ/kg-°C
k – ratio of specific heat
k=
34

35. PARTIAL PRESSURE OF COMPONENTS
Pi = y iP
P = ΣPi
P − total pressure of the mixture
Pi - partial pressure of the components in the mixture
EXAMPLE 1
The ultimate analysis of a coal fuel is as follows:
C = 80.7% ; H2 = 4.5% ; O2 = 2.4% ; N2 = 1.1% ; S = 1.8%; M = 3.3%
and Ash = 6.2%. Determine
a. The combustion equation
b. The air – fuel ratio
c. The HHV and LHV of the fuel
d. The M and R of the products
SOLUTION
Reduce the analysis to an ashless basis
80.7
= 86%
100 − 6.2
4.5
H2 =
= 4 .8 %
100 − 6.2
2.4
O2 =
= 2 .6
93.8
1 .1
= 1 .2 %
N2 =
93.8
1 .8
S=
= 1 .9
93.8
3.3
M=
= 3.5
93.8
C=
35

36. Converting to molal analysis
xi 86 4.8 2.6 1.2 1.9 3.5
=
+
+
+
+
+
Mi 12
2
32 28 32 18
xi
Σ
= 7.16 + 2.4 + 0.08 + 0.043 + 0.06 + 0.194 = 9.937
Mi
7.16
C=
= 72 .05 %
9.937
2 .4
H2 =
= 24 .2%
9.937
0.08
O2 =
= 0.81
9.937
0.043
N2 =
= 0.43 %
9.937
0.06
S=
= 0 .6 %
9.937
0.194
M=
= 1.91 %
9.937
Combustion with 100% theoretical air (Basis: 100 moles of fuel)
Σ
(72 .05 C + 24 .2H 2 + 0.81O 2 + 0.43N 2 + 0.6 S + 1.91H 2 O) + xO 2 + x (3.76 )N 2 →
bCO 2 + cH 2 O + dSO 2 + eN 2
CarbonBala nce
72 .05 = b
Hydrogen Balance
24.2 + 1.91 = c
c = 26 .11
O 2 Balance
1.91
26 .11
+ x = 72 .05 +
+ d → eq. 1
2
2
S Balance
0.6 = d
From eq. 1
0.81 +
x = 83 .94
N 2 Balance
0.43 + 83.94(3.76 ) = e
e = 316 .0444
(72 .05 C + 24 .2H 2 + 0.81O 2 + 0.43N 2 + 0.6 S + 1.91H 2 O) + 83 .94 O 2 + 315 .6144 N 2 →
72 .05 CO 2 + 26 .11H 2 O + 0.6SO 2 + 316 .0444 N 2
36

37. A
83.94(32) + 315.6144 (28)
11523.2832
kg of air
=
=
= 11.5
F 72.05(12) + 24.2(2) + 0.81(32) + 0.43(28) + 0.6(32) + 1.91(18)
1004.54
kg of fuel
EXAMPLE 2
An Ultimate analysis of coal yields the following composition:
C = 74% ; H2 = 5%; O2 = 6%; N2 = 1.2%; S = 1%; M = 3.8% and Ash = 9%.
If this coal is burned with 25% excess air, determine
a. The combustion equation
b. The actual air – fuel ratio in kg/kg
Fuel
Components
Ultimate
analysis
Ashless
%
M
C
H2
O2
N2
S
M
Ash
74
5
6
1.2
1
3.8
9
100
81.3
5.5
6.6
1.3
1.1
4.2
12
2
32
28
32
18
x/M
6.78
2.75
0.21
0.05
0.03
0.23
10.04
Molal
Analysis
67.5
27.4
2.1
0.5
0.3
2.3
100
Combustion w/ 100% Theo. air
O2
79.44
99.3
N2
CO2
H 2O
SO2
O2
298.7 67.47 29.66 0.34
Combustion w/ excess air e = 0.25
373.4 67.5
29.7
0.3 19.9
N2
299.17
373.8
Air-Fuel Ratio
13.7 kg/kg
EXAMPLE 3
A gas turbine generating unit produces 600 KW of power and uses a liquid fuel
represented by C8H18 and requires 300% excess air for complete combustion. For a fuel
rate of 0.234 kg/KW-hr, determine
a. The combustion equation
b. The volume of air required at P = 1500 KPa and T = 310°K
EXAMPLE 4
An unknown hydrocarbon fuel has the following Orsat Analysis:
CO2 = 12.5%; CO = 0.3%; O2 = 3.1%; N2 = 84.1%
Determine
a. The value of n and m
b. The combustion equation
c. The percent excess air (e = 15%)
d. The percent C and H in the fuel
37

38. 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.
For Solid Fuels
HHV = 33,820C + 144,212 (H- O/8) + 9304S KJ/kg
where: C, H2, O2, and S are in decimals from the ultimate analysis
For Coal and Oils with the absence of Ultimate Analysis
For Liquid Fuels
HHV = 31,405C + 141 647H KJ/kg
HHV = 43,385 + 93(°Be - 10) KJ/kg
Be - degrees Baume
For Gasoline
HHV = 41,160 + 93 (°API)
LHV = 38,639 + 93 (°API)
For Kerosene
HHV = 41,943 + 93 (°API)
LHV = 39,035 + 93 (°API)
KJ/kg
KJ/kg
KJ/kg
KJKkg
For Fuel Oils
HHV = 41,130 + 139.6(°API) KJ/kg
LHV = 38,105 + 139.6(°API) KJ/kg
API - American Petroleum Institute
For Fuel Oils (From Bureau of Standard Formula)
141.5
S=
131.5 + °API
140
S=
130 + Be
HHV = 51,716 – 8,793.8 (S)2 KJ/kg
LHV = HHV - QL KJ/kg
QL = 2,442.7(9H2) KJ/kg
H2 = 0.26 - 0.15(S) kg of H2/ kg of fuel
S @ t = S - 0.0007(t-15.56)
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 by the
combustion of hydrogen in the fuel
S @ t - specific gravity of fuel oil at any temperature t
38

39. Oxygen Bomb Calorimeter - instrument used in measuring heating value of solid and
liquid fuels.
Gas Calorimeter - instrument used for measuring heating value of gaseous fuels.
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.
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 isooctane in a blend with normal heptane that matches the knocking behavior of the fuel.
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.
39

40. Cycle
A cycle is a series of two or more processes in which the final state is the same as the
initial state.
Steam Power Cycle: A power generating cycle that uses steam or water vapor as the
working substance. This cycle differ with an internal combustion engine cycle because
the combustion occurs in the boiler, unlike that of an IC engine that combustion occurs
inside the working cylinders.
Steam Power Plant Cycle
Rankine Cycle
Components:
a. Steam Turbine
b. Condenser
c. Pump
d. Steam Generator or boiler
Processes:
1 to 2 – Isentropic Expansion (S = C)
2 to 3 – constant pressure Heat Rejection (P = C)
3 to 4 – Isentropic pumping (S = C)
4 to 1 – Constant pressure Heat Addition (P = C)
ms
1 kg
1
St ea m T ur b in e
Bo il e r or
St ea m
Ge ne r at or
Wt
2
QA
Co nd e ns er
Fe ed w at er
Pu mp
QR
3
WP
T
4'
1
P1
4
3
P2 = P3
2
2'
S
40

41. A. Turbine Work (Wt) (considering S = C; Q = 0; ∆KE = 0; ∆PE = 0)
Wt = ms(h1 – h2) KW
Where:
ms – steam flow rate, kg/sec
h – enthalpy, KJ/kg
Wt – turbine power, KW
B. Heat Rejected in the Condenser (QR)
QR = m(h2 – h3) KW
C. Pump Work (WP)
WP = m(h4 – h3)
D. Heat added to Boiler (QA)
QA = m(h1 – h4) KW
E. Thermal efficiency
N e t W o rk
e
= H e a t A d d e d
W
e
x 1 0 0 %
= Q
x
1 0 0 %
A
F. Net Work
W = Wt - Wp
G. Boiler Efficiency (EB)
e
B
Q
= Q
x 1 0 0 %
A
S
41

42. Reheat Cycle Steam Power Plant: In a reheat cycle, after partial expansion of steam
in the turbine the steam re-enters a section in the steam generator called the re-heater
and re-heating the steam almost the same to initial temperature and then re-expands
again to the turbine. This will result to an increase in thermal efficiency of the cycle, with
significant increase in turbine work and heat added.
From Reheater
To Reheater
1 kg
1
2
3
Wt
4
QA
QR
5
6
WP
T
1
3
2
6
5
4
S
42

43. Turbine Work
Wt = m s [(h1 − h2 ) + (h3 − h 4 ] KW
Heat Rejected
QR = ms (h4 − h5 ) KW
Pump Work
WP = ms (h6 − h5 ) KW
Heat Added
Q A = ms [h1 − h6 ) + (h3 − h2 )] KW
Where:
ms – mass flow rate of steam, kg/sec
Regenerative Cycle: In a regenerative cycle some of the steam after initial expansion
is extracted for feed-water heating by mixing the bled steam with the condensate or
drains from other heater. The remaining steam re-expands again in the turbine. The
thermal efficiency also increases due to the decrease in heat added to boiler.
1 kg
1
Wt
2
3
m1
QA
QR
4
7
6
Open
Heater
5
WP1
WP2
43

44. T
1
1 kg
7
1 kg
6
2
m
5
(1-m)
(1-m)
3
4
S
Let: m – fraction of steam extracted for feed-water heating, kg/kg
Turbine Work
Wt = ms [(h1 − h2 ) + (1 − m)(h2 − h3 )] KW
Heat Rejected
QR = ms [(1 − m)(h3 − h4 )] KW
Pump Work
a. Condensate pump (WP1)
b. Feed-water pump WP2)
WP = WP1 + WP2
WP1 = ms (1 − m)(h5 − h4 ) KW
WP2 = ms ((h7 − h6 ) KW
Heat Added
Q A = ms [(h1 − h7 )] KW
44

45. Reheat – Regenerative Cycle: In a reheat – regenerative cycle further increase in
thermal efficiency will occur because of the combine effects of reheating and
regenerative feed-water heating. Significantly heat added decreases, total pump work
decreases while turbine work increases.
Single stage reheat and single stage regenerative cycle that uses an open type feedwater heater
(1-m1)
(1-m1)
1 kg
1
2 3
Wt
4
2
m1
QA
QR
5
7
8
6
Open
Heater
WP1
WP2
T
1
1 kg
3
8
(1 kg)
m
7
6
2
(1-m)
(1-m)
5
4
S
45

46. Turbine Work
Wt = ms [(h1 − h2 ) + (1 − m)(h3 − h4 )] KW
Heat Rejected
QR = ms [(1 − m)(h4 − h5 )] KW
Pump Work
WP1 = ms [(1 − m)(h6 − h5 )] KW
WP2 = ms [(h8 - h7 )] KW
WP = WP1 + WP2
Heat Added
Q A = ms [(h1 − h8 ) + (1 − m)(h3 − h2 )] KW
STEAM RATE
Steam Flow Rate
kg
KW Produced KW - hr
when SR is based on the turbine power
SR =
SR =
3600m s
kg
Wt
KW - hr
where :
ms − steam flow rate in kg/sec
Wt- turbine work in KW
HEAT RATE
Heat Supplied
KJ
KW Produced KW - hr
when HR is based on the turbine power
HR =
HR =
3600Q A
KJ
Wt
KW - hr
where :
Q A Heat added in KW
Wt-turbine work in KW
46

47. Turbine Efficiency
Actual Turbine Work
x 100%
Ideal Turbine Work
Efficiency
Wt'
ηt =
x 100%
Wt
Ideal Pump Work
x 100%
ηP =
Actual Pump Work
ηt =
Pump
ηP =
WP
x 100%
WP '
Boiler or Steam Generator Efficiency
Heat Absorbed by Boiler
x 100%
Actual Heat supplied to Boiler
Q
eB = A x 100%
Qs
eB =
EXAMPLE
A coal fired steam power plant operates on the Rankine Cycle. The steam enters the turbine at 7000 KPa and 550°C with
a velocity of 30 m/sec. It discharges to the condenser at 20 KPa with a velocity of 90 m/sec. For a mass flow rate of
steam of 37.8 kg/sec, Determine
a. The ideal turbine work in KW
b. The net power produced in KW
c. The thermal efficiency of the cycle
d. The cooling water required in the condenser if cooling water enters at 20°C and leaves at 35°C
e. The coal consumption in kg/hr if the boiler efficiency is 82% and heating value of coal is 32,000 KJ/kg
From Steam Table
h1 =3529.8 ; S1 = 6.9465
h2 = 2288.3 ;x2 = 86.4%
h3 = 251.33; S3 = 0.8321
h4 = 258.43
Solution:
a. W = Q - ∆h - ∆KE - ∆PE
Q = 0 ; ∆PE = 0
Wt = (h1 – h2) - ∆KE
Wt = 46,792.6 KW
b. Wp = 268.38 KW
W = 46,524.2 KW
c. QA = 123,657.8 KW
e = 37.62%
d. QR = 76,997.5 KW
MW = 1225.99 kg/sec
e. mf = 16,965.25 KG/hr
47

48. GENERAL BOILER DESCRIPTION
1. Fire-Tube boiler: Hot gas is inside the tubes while water on the outside.
2. Water-Tube boiler: Water is inside the tube while hot gas is on the outside.
The fire-tube boiler design uses tubes to direct the hot gases from the combustion
process through the boiler to a safe point of discharge. The tubes are submerged in the
boiler water and transfer the heat from the hot gases into the water.
Inside a firetube boiler the hot gases travel down the furnace during the combustion
process, (first pass). The rear head seals the gasses in the lower portion of the head.
The gas is redirected through the second pass tubes. In the front head the hot gasses
are sealed from escaping out the stack and turned and redirected through the third pass
tubes. The hot gas travels toward the upper portion of the rear head where it’s turned
and directed through the fourth pass tubes. From there, after giving up most of the
energy from the combustion process, the gas is directed into the stack and vented to the
atmosphere.
The water-tube boiler design uses tubes to direct the boiler water through the hot
gases from the combustion process, allowing the hot gases to transfer its heat through
the tube wall into the water. The boiler water flows by convection from the lower drum
to the upper drum.
Either of the fire-tube or water-tube boiler design concepts is available in what is
popularly known as the packaged boiler, a concept introduced by Cleaver- Brooks in
1931. A packaged boiler is shipped from the manufacturer as a complete assembly, with
burner, control systems, operating and safetycontrols, all piped and/or wired into the
assembly. Equipment of this type needs only to be positioned into its intended location,
utility connections made and a means provided to direct the flue gases to a safe point of
discharge. Most packaged firetube boilers are available in capacities of 500,000 Btu/hr
up to 26,800,000 Btu/hr output. These boilers are normally rated on the basis of boiler
horsepower (BHP) output. One boiler horsepower = 33,472 Btu per hour.
Packaged water-tube boilers, designed for commercial applications, are normally
available in sizes as small as 1,200,000 Btu/hr output. Industrial watertube boilers can
be provided in packaged format in capacities of up to 134,000,000 Btu/hr.
48

49. Boiler Auxiliaries and Accessories
Superheater – a heat exchanger that is used to increase the temperature of the water
vapor greater than the saturation temperature corresponding the boiler pressure.
Evaporator – a heat exchanger that changes saturated liquid to saturated vapor.
Economizers – is the heat exchanger that raises the temperature of the water leaving
the highest pressure feedwater heater to the saturation temperature
corresponding to the boiler pressure.
Air Preheater – is a heat exchanger use to preheat air that utilizes some of the energy
left in the flue gases before exhausting them to the atmosphere.
Fans – a mechanical machine that assist to push the air in, pull the gas out or both.
Stoker – combustion equipment for firing solid fuels (used in water tube boilers)
Burners – combustion equipment for firing liquid and gaseous fuels.
Feedwater pump – a pump that delivers water into the boiler.
Pressure Gauge – indicates the pressure of steam in the boiler.
Safety Valve – A safety device which automatically releases the steam in case of over
pressure.
Temperature Gauge – indicates the temperature of steam in the boiler.
Fusible Plug – a metal plug with a definite melting point through which the steam is
released in case of excessive temperature which is usually caused by low water
level.
Water Walls – water tubes installed in the furnace to protect the furnace against high
temperature and also serve as extension of heat transfer area for the feed-water.
Gage Glass (Water column) – indicates the water level existing in the boiler.
Baffles – direct the flow of the hot gases to effect efficient heat transfer between the hot
gases and the heated water.
49

50. Furnace – encloses the combustion equipment so that the heat generated will be utilized
effectively.
Soot blower – device which uses steam or compressed air to remove the soot that has
accumulated in the boiler tubes and drums.
Blowdown Valve – valve through which the impurities that settle in the mud drum are
remove. Sometimes called blow 0ff valve.
Breeching – the duct that connects the boiler and the chimney.
Chimney or Smokestack – a structure usually built of steel or concrete that is used to
dispose the exhaust gases at suitable height to avoid pollution in the vicinity of
the plant.
BOILER PERFORMANCE
1.Heat Generated by Fuel
Qs = mf (HHV) KJ/hr
Where: mf – fuel consumption, kg/hr
HHV – higher heating value of fuel KJ/kg
2. Rated Boiler Horsepower(RBHp)
a) For Water Tube Type
HS
RBHp =
0.91
b) For Fire Tube Type
RBHp =
HS
1.1
Where: HS – required heating surface, m2
3. Developed Boiler Horsepower (DBHp)
Dev. Bo. HP =
m s(hs − hf)
15.65(2257 )
Dev. Bo. HP =
m s(hs − hf)
35,322
50

51. One Boiler Horsepower is equivalent to the generation of 15.65 kg/hr of steam
from water at 100°C to saturated steam at 100°C. The latent heat of vaporization of
water at 100°C was taken at 2257 KJ/kg.
4. Percentage Rating
D e v .B o .H p
% R
x 1 0 0 %
= R a te d B o .H p
5. ASME Evaporation Units
ASME Evap. Units = ms(hs – hf) KJ/hr
)
)
(
(
6.Factor of Evaporation (FE)
h
h
F E
s
−5 7 f
2 2
=
7. Boiler Efficiency
ηB =
ms (hs − hf )
x 100%
mf (HHV)
8. Net Boiler Efficiency
ηN =
ms (hs − hf )- Auxiliaries
x 100%
mf (HHV)
9. Actual Specific Evaporation
Actual Sp. Evap. =
ms
mf
kg of steam
kg of fuel
10. Equivalent Evaporation
Equiv. Evap. = ms (FE)
11. Equivalent Specific Evaporation
Equiv. Sp. Evap. =
ms
(FE)
mf
51

52. BOILER HEAT BALANCE
Energy supplied to the boiler by 1 kg of fuel is distributed among the following items in
the ASME short-form heat balance, all expressed in units of KJ/kg of fuel.
1. Heat absorbed by steam generating unit
Q1 =
m s(hs - hf)
KJ/kg
mf
Where: ms – steam flow rate in kg/hr
mf – fuel consumption in kg/hr
hs – enthalpy of steam, KJ/kg
hf – enthalpy of fed water, KJ/kg
2. Heat loss due to Dry Flue Gas
Q2 = mdg(1.026)(tg – ta) KJ/kg
Where: mdg – mass of dry flue gas, Kggas/Kgfuel
3. Heat loss due to Moisture in Fuel
Q3 = M(h’- hf’) KJ/kg
Where: h’ – enthalpy of superheated steam at flue gas Temperature, KJ/kg
hf’ – enthalpy of liquid at temperature of fuel entering furnace, KJ/kg
Q3 = M(2493 + 1.926tg – 4.187tf) KJ/kg when tg < 302°C
Q3 = M(2482 + 2.094tg – 4.187tf) KJ/kg when tg > 302°C
4. Heat loss due to moisture from the combustion of hydrogen
Q4 = 9H2(h’- hf’) KJ/kg
Q4 = 9H2 (2493 + 1.926tg – 4.187tf) KJ/kg when
tg < 302°C
Q4 = 9H2 (2482 + 2.094tg – 4.187tf) KJ/kg when
tg > 302°C
52

53. 5. Heat loss due to moisture in air supplied
Q5 = W(1.926)maa(tg – ta) KJ/kg
Q5 = %age saturation(Ws)(1.926)maa(tg – ta) KJ/kg
6. Heat loss due to incomplete combustion
Q6 = 23516Ci KJ/kg
Q6 = 23516
CO
CO2 + CO
Cab KJ/kg
7. Heat loss due to unconsumed carbon in the refuse
Q7 = 33,820(C - Cab)
Wher: (C - Cab) = (Wr – A)
(Wr – A) = WrCr
Wr =
A
1 - Cr
C – carbon in fuel, kg/kg
Cab – carbon actually burned, kg/kg
Wr – weight of dry refuse kg/kg
Cr – weight of combustible in the refuse, kg/kg
8. Heat loss due radiation and unaccounted-for losses
Q8 = HHV –(Q1 + Q2 + Q3 + Q4 + Q5 + Q6 + Q7)
53

54. Problems (Steam Generators)
1. A steam generator uses coal as fuel having the ultimate analysis as follows:
C = 72% ; H2 = 5%; O2 = 10%; N2= 1.2%; S = 3.3%; M = 0.1% & A = 8.4%
If this coal is burned with 20% excess air, Determine
a) the A/F ratio in kga/kgf
b) the volume of wet flue gas at101 KPa and 282°C per kg of coal
c) the %age of CO2 by volume in the dry flue gas
d) the dew point of the products
e) the fuel consumption in Metric tons per hour for a steaming capacity of 100 Metric
tons/hour, Factor of Evaporation of 1.15 and a steam generator efficiency of 73%.
2. A water tube boiler generates 7,300 kg of steam per hour at a pressure of1.4 MPa and a
quality of 98% when the feed-water is 24°C. Find
a) Factor of Evaporation
b) Equivalent Evaporation
c) Developed Boiler Horsepower
d) %rating developed if the heating surface is 190 m2
e) Overall efficiency if coal having a HHV of 5000 KCal/kg as fired is used at the rate of
3000 L/hr.
3. A water tube boiler generates 8,000 kg of steam per hour at a pressure of 1.4 MPa and a
quality of 985 when the feed-water is 24°C. Find
a) Factor of Evaporation
b) Equivalent Evaporation in kg/hr
c) Boiler horsepower developed
d) Percent rating developed if the heating surface is 185.9 m2
e) Overall efficiency if coal having a HHV of 20,940 KJ/kg as fired is used at a rate of
1500 kg/hr
4. At a load of 43,000 KW in a steam turbine generating set, 3600 RPM, the following data
appear in the log sheet.
Steam flow -190 Metric Tons/hour
Steam pressure - 8.93 MPaa
Steam temperature - 535C
Feed-water temperature - 230C
Fuel Flow:
Bunker Oil -3.4 Metric Tons/hr
HHV =10,000 KCal/hr
Local coal -18 Metric Tons/hr
HHV = 5350 KCal/hr
Determine thee overall boiler efficiency.
h at 8.93 MPa and 535°C - 3475.7 KJ/kg
hf at 230°C- 990.12 KJ/kg
5. A coal fired steam boiler uses 3000 kg of coal per hour. Air required for combustion is 15.5
kg/kg of coal at a barometric pressure of 98.2 KPa. The flue gas has a temperature of 285°C
and an average molecular weight of 30. Assuming an ash loss of 11% and allowable gas
velocity of 7.5 m/sec, find the diameter of the chimney. (D = 1.91 m)
6. Two boilers are operating steadily on 136,500 kg of coal contained in a bunker. One boiler is
producing 2386 kg of steam/hr at 1.15 FE and an efficiency of 75%, and the other boiler
produces 2047 kg of steam/hr at 1.10 FE and an efficiency of 70%. How many hours will the
coal in the bunker run the boilers if the heating value of the coal is 32,000 KJ/kg. (281.5 hrs)
54

55. 7. An industrial plant is to be designed based upon the following requirements; 5000 KW output
and generator efficiency of 98%. Steam is extracted at the rate of7.6 kg/sec at 0.2 MPa for
industrial use. Turbine inlet pressure is 1.2 MPa and temperature of 260C, exhaust at 0.014
MPa. Brake turbine efficiency is 75%. Extracted and exhaust steam are returned to the boiler
as liquid at 93C, respectively. Determine
a) Supplied steam to the turbine in kg/hr
b) Total heat supplied to the boiler in KJ/hr
At 1.2 MPa and 260C
h = 2957.6 KJ/kg
S = 6.8721 KJ/kg-K
At 93C; hf = 389.54 KJ/kg
At 0.014 MPa
Sf = 0.7366 KJ/kg-K ; sfg = 7.2959 KJ/kg-K
hf = 219.99 KJ/kg ; hfg = 2376.6 KJ/kg
At 0.2 MPa
sf = 1.55301 KJ/kg-K ; sfg = 5.5970 KJ/kg-K
hf = 504.7 KJ/kg ; hfg = 2201.9 KJ/kg
At S1 = S2 to 0.20 MPa ;
h2 = 2606.28 KJ/kg
At S3 = S4 to 0.014 MPa
h3 = 2218.596 KJ/kg
8. In a test of a Bobcock and Wilcox boiler with hand-fired furnace, the following date were
taken;
Rated HP - 350
Grate Surface - 2.323 m2
Duration of test - 24 hours
Steam pressure - 1.2 MPa
Feed-water temperature - 34C
Quality of steam formed - 99%
Total weight of coal fired (wet) - 7110 kg
Moisture in coal - 7.5%
Total weight of water fed to boiler - 54,000 kg
Determine:
a) Factor of Evaporation
b) Dry coal per m2 of grate surface per hour
c) Equivalent evaporation per hr - m2 of heating surface
d) Equivalent evaporation per hour
e) Boiler HP Developed
f) Percentage of Rated capacity developed
g) The equivalent evaporation per kg of dry coal
h) Combined efficiency of boiler, furnace and grate if the coal has a heating value of
28,590 KJ/kg
9. Coal with HHV = 6700 KCal/kg is consumed at the rate of 600 kg/hr in a steam generator
with a Rated Boiler HP of 200. The feed-water temperature is 82C and steam generator is
at 1.08 MPaa saturated. The Developed Boiler HP is equivalent to 305. Determine:
a) Heating Surface, m2
b) Rate of steam generated, kg/hr
c) Percentage Rating
d) ASME Evaporation units, J/hr
55

56. e) Factor of Evaporation
f) Overall thermal efficiency
g) Actual specific evaporation, kg steam/kg of coal
h) Equivalent specific evaporation
10. The boiler, furnace and grate efficiency of a steam generator is 82%. Coal with a moisture
content of 12% is burned at the rate of 10,000 kg per hour. The heating value per kg of dry
coal is 28,000 KJ. Steam is generated at 3.2 MPa and a temperature of 320C. Feed-water
temperature is 95C. Determine:
a) the kg of steam generated per hour
b) the Developed Boiler Hp.
c) the Equivalent evaporation in kg per kg of coal as fired
d) the cost to evaporate 500 kg of steam if coal costs P 150 per Metric Ton
56

57. CONDENSERS
Direct - contact or Open, condensers
This type of condenser are used in special cases, such as when dry cooling towers are used in
geothermal power plants and in power that use temperature differences in ocean waters
(OTEC). Modern direct contact condensers are of the spray type. Early designs were of the
barometric or jet type.
Turbine
exhaust
Dry cooling
tower
Noncondensables
to SJAE
2
5
Condenser
3
4
To plant feedwater
system
Pump
Schematic Diagram of a Direct - contact condenser
of the Spray type
By mass balance
m2 = m4
m3 = m2 + m5
By Energy balance
m2h2 + m5h5 = m3 = h3
And the ratio of circulating water to steam flow
m5 h2 − h 3
=
m 2 h 3 − h5
57

58. Surface Condenser
Turbine
exhaust h2
ms
Water in
mw twA
Water box
Water out
mw twB
Tubes
Support
Plate
ms
Condensate
h3
Let
Q = QR = Qw
QR – heat rejected by steam
Qw – heat absorbed by cooling water
ms – steam flow rate in kg/sec
mw – cooling water flow rate in kg/sec
twA – inlet temperature of cooling water in °C
twB – outlet temperature of cooling water in °C
Cpw = 4.187 KJ/kg-°C (specific heat of water)
QR = Qw
QR = ms(h2 – h3) KW
Qw = mw Cpw (twB – twA)
58

59. In terms of Overall coefficient of heat transfer U:
Q=
UA (LMTD)
KW
1000
where :
W
m - °C
LMTD − log mean temperature difference,°C
U - overall coefficient of heat transfer in
A − total heat transfer surface area, m2
A = πDL(N t )
D − outside diameter of tubes, m
L - length of tubes, m
Nt - total number of tubes
LMTD =
t wB − t wA
t − t wA
ln s
t s − t wB
t s − saturation temperature of steam,°C
TTD – Terminal Temperature difference
TTD = ts - twB
TEMPERATURE – AREA DIAGRAM
T
ts
θ1
θ2
twB
twA
A
θ2 = t s − t wA
θ1 = t s − t wB
LMTD =
θ2 − θ1
θ
ln 2
θ1
59
2
or
W
m -K
2

60. Problem
A 10,000 KW turbine generator uses 5 kg/KW-hr of steam at rated load. Steam supply pressure
is 4.5 MPa and 370°C and the pressure in the surface condenser is 3.4 KPa (tsat = .
Temperature of inlet circulating water is 16°C and outlet of 22°C. Combined efficiency of the
turbo-generator set is 92%. The condenser tubes are 2 mm; 1.2 mm thickness. Water velocity is
3.5 m/sec. Overall coefficient of heat transfer U = 4 W/m2-°C. Tube sheet thickness is 10 mm.
Determine:
a. Cooling water required in L/min
b. Number of tubes for 2-Pass design
c. Actual length of tubes
Other Data are as follows:
h1 = 3131.4 ;S1 = 6.5897
h2 = 1967.1 ;S2 = 6.5897 x2 = 76.17
h3 = 109.75 ;S3 = 0.3836
h4 = 114.27
GEOTHERMAL POWER PLANT
Geothermal energy is the power obtained by using heat from the Earth's interior. Most
geothermal resources are in regions of active volcanism. Hot springs, geysers, pools of
boiling mud, and fumaroles (vents of volcanic gases and heated groundwater) are the
most easily exploited sources of such energy
The most useful geothermal resources are hot water and steam trapped in subsurface
formations or reservoirs and having temperatures ranging from 176° to 662° F (80° to
350° C). Water and steam hotter than 356° F (180° C) are the most easily exploited for
electric-power generation and are utilized by most existing geothermal power plants. In
these plants hot underground water is drilled from wells and passes through a
separator-collector where the hot water is flashed to steam, which is then used to drive
a steam turbine whose mechanical energy is then converted to electricity by a
generator.
60

61. T
Well bottom pressure
well head pressure
0
Flasher – separator pressure
H
B
1
3
2
S
IDEAL TURBINE WORK
Wt = ms(h1 – h2) KW
ACTUAL TURBINE WORK
Wt’ = ηTms(h1 – h2) KW
61

62. GENERATOR POWER OUTPUT
W0 = ηGηTms(h1 – h2) KW
where
ms – steam flow rate in kg/sec
ηT - turbine efficiency
ηG – generator efficiency
o– underground water
H – well head condition
1 – saturated vapor condition leaving flasher – separator
B – saturated liquid condition leaving flasher – separator
2 – turbine exhaust
3 – saturated liquid leaving condenser
Example
A geothermal power plant has an output of 16,000 KW and mechanical - electrical efficiency of
80%. The pressurized groundwater at 17.0 MPa, 280°C leaves the well to enter the flash
chamber maintained at 1.4 MPa. the flash vapor passes through the separator - collector to
enter the turbine as saturated vapor at 1.4 MPa. the turbine exhaust at 0.1 MPa. The unflashed
water runs to waste. If one well discharges 195,000 kg/hr of hot water, how many wells are
required. ( 4 wells)
W0 = 0.80m s (h1 − h 2 )
From Steam Table
16000 = m s (0.80)(2786.4 − 2341.1)
At 17,000 KPa and 280°C
m s = 45 kg/sec
ho = 1231.7 KJ/kg
By mass and energy balance on the flasher - collector
At 1400 KPa Saturated Vapor
h1 = 2786.4 KJ/kg ; S1 = 6.4642 KJ/kg-K mo = ms + mB
mB = m o − m s → eq. 1
At S1 = S2 to 100 KPa
h2 = 2341.1 KJ/kg
moho = msh1 + (mo - ms )hB
At 1400 KPa saturated Liquid
m (h − hB )
mo = s 1
= 219 kg/sec = 788,400 kg/hr
hB = 829.6 KJ/kg
h −h )
o
Nwells
B
788,400
=
= 4 wells
195,000
62

63. The Diesel Power Plant
Air out
Cooling
Tower
Fuel Tank
Engine
Generator
Air in
Fuel
Pump
Cooling water
Pump
Two stroke cycle engine: An engine that completes one cycle in one revolution of the
crankshaft.
Four stroke cycle engine: An engine that completes one cycle in two revolution of the
crankshaft.
TERMS AND DEFINITIONS
Diesel engine is a type of internal combustion engine that uses low grade fuel oil and
which burns this fuel inside the cylinder by heat of compression. It is used chiefly for
heavy-duty work. Diesel engines drive huge freight trucks, large buses, tractors, and
heavy road-building equipment. They are also used to power submarines and ships, and
the generators of electric-power stations in small cities. Some motor cars are powered
by diesel engines.
Gasoline engine - is a type of internal combustion engine, which uses high grade of oil.
It uses electricity and spark plugs to ignite the fuel in the engine's cylinders.
Kinds of diesel engines. There are two main types of diesel engines. They differ
according to the number of piston strokes required to complete a cycle of air
compression, exhaust, and intake of fresh air. A stroke is an up or down movement of a
piston. These engines are (1) the four-stroke cycle engine and (2) the two-stroke cycle
engine.
Four Stroke Cycle Engine
1. Intake
2. Compression
3. Power
4. Exhaust
In a four-stroke engine, each piston moves down, up, down, and up to complete a cycle.
The first down stroke draws air into the cylinder. The first upstroke compresses the air.
63

64. The second down stroke is the power stroke. The second upstroke exhausts the gases
produced by combustion. A four-stroke engine requires exhaust and air-intake valves.
It completes one cycle in two revolutions of the crankshaft.
Two Stroke Cycle Engine
1. Intake-Compression stroke
2. Power-exhaust stroke
In a two-stroke engine, the exhaust and intake of fresh air occur through openings in
the cylinder near the end of the down stroke, or power stroke. The one upstroke is the
compression stroke. A two-stroke engine does not need valves. These engines have
twice as many power strokes per cycle as four-stroke engines, and are used where high
power is needed in a small engine. It completes one cycle in one revolution of the
crankshaft.
Governor - is a device used to govern or control the speed of an engine under varying
load conditions.
Purifier - a device used to purify fuel oil and lube oil.
Generator - a device used to convert mechanical energy.
Crank scavenging - is one that the crankcase is used as compressor.
Thermocouple - is made of rods of different metal that are welded together at one end.
Centrifuge - is the purification of oil for separation of water.
Unloader - is a device for automatically keeping pressure constant by controlling the
suction valve.
Planimeter - is a measuring device that traces the area of actual P-V diagram.
Tachometer - measures the speed of the engine.
Engine indicator - traces the actual P-V diagram.
Dynamometer - measures the torque of the engine.
Supercharging - admittance into the cylinder of an air charge with density higher than
that of the surrounding air.
Bridge Gauge - is an instrument used to find the radial position of crankshaft motor
shaft.
Piston - is made of cast iron or aluminum alloy having a cylinder form.
Atomizer - is used to atomize the fuel into tiny spray which completely fill the furnace in
the form of hollow cone.
Scavenging - is the process of cleaning the engine cylinder of exhaust gases by forcing
through it a pressure of fresh air.
Flare back - is due the explosion of a maximum fuel oil vapor and air in the furnace.
Single acting engine - is one in which work is done on one side of the piston.
Double acting engine - is an engine in which work is done on both sides of the piston.
Triple-expansion engine - is a three-cylinder engine in which there are three stages of
expansion.
The working pressure in power cylinder is from 50 psi to 500 psi.
The working temperature in the cylinder is from 800°F to 1000°F.
Air pressure used in air injection fuel system is from 600 psi to 1000 psi.
64

65. Effect of over lubricating a diesel engine is:
Carbonization of oil on valve seats and possible explosive mixture is produced.
The average compression ratio of diesel engine is from 14:1 to 16:1.
Three types of piston:
1. barrel type
2. trunk type
3. closed head type
Three types of cam follower:
1. flat type
2. pivot type
3. roller type
Methods of mechanically operated starting valve:
1. the poppet
2. the disc type
Three classes of fuel pump:
1. continuous pressure
2. constant stroke
c. variable stroke
Type of pump used in transferring oil from the storage to the service tanks:
1. rotary pump
2. plunger pump
3. piston pump
4. centrifugal pump
Valve that is found in the cylinder head of a 4-stroke cycle engine:
1. fuel valve
2. air starting valve
3. relief valve
4. test valve
5. intake valve
6. exhaust valve
Four common type of governors used on a diesel engine:
1. constant speed governor
2. variable speed governor
3. speed limiting governor
4. load limiting governor
Kinds of piston rings used in an internal combustion engines:
1. compression ring
2. oil ring
3. firing ring
4. oil scraper ring
Reasons of smoky engine:
1. overload
2. injection not working
3. choked exhaust pipe
65

66. 4. fuel or water and leaky things
Methods of reversing diesel engines:
1. sliding camshaft
2. shifting roller
c. rotating camshaft
Arrangements of cylinders:
1. in-line
2. radial
3. opposed cylinder
4. V
5. opposed piston
Position of cylinders:
1. vertical
2. horizontal
3. inclined
Methods of starting:
1. manual, crank, rope, and kick
2. electric (battery)
3. compressed air
4. using another engine
Applications:
1. automotive
2. marine
3. industrial
4. stationary power
5. locomotive
6. aircraft
Types of internal combustion engine:
1. Gasoline engine
2. Diesel engine
3. Kerosene engine
4. Gas engine
5. Oil-diesel engine
Methods of ignition:
1. Spark
2. Heat of compression
Reasons for supercharging:
1. to reduce the weight to power
ratio
2. to compensate the power loss
due to high altitude
Types of superchargers:
1. engine-driven compressor
2. exhaust-driven compressor
66

67. 3. separately-driven compressor
Auxiliary systems of a diesel engine:
1. Fuel system
a. fuel storage tank
b. fuel filter
c. transfer pump
d. day tank
e. fuel pump
2. Cooling system
a. cooling water pump
b. heat exchanger
c. surge tank
d. cooling tower
e. raw water pump
3. Lubricating system:
a. lub oil tank
b. lub oil pump
c. oil filter
d. oil cooler
e. lubricators
4. Intake and exhaust system
a. air filter
b. intake pipe
c. exhaust pipe
d. silencer
5. Starting system
a. air compressor
b. air storage tank
Advantages of diesel engine over other internal combustion engines:
1. low fuel cost
2. high efficiency
3. needs no large water supply
4. no long warm-up period
5. simple plant layout
Types of scavenging:
1. direct scavenging
2. loop scavenging
3. uniflow scavenging
Color of the smoke:
1. efficient combustion - light brown baze
2. insufficient air - black smoke
3. excess air - white smoke
Causes of black smoke:
67

68. 1. fuel valve open too long
2. too low compression pressure
3. carbon in exhaust pipe
4. overload on engine
Causes of white smoke:
1. one or more cylinders not getting enough fuel
2. too low compression pressure
3. water inside the cylinder
ENGINE PERFORMANCE
1. Heat supplied by fuel (Qs): Total heat supplied by fuel.
Qs = mF (HV )
KJ
hr
Where:
mF – fuel consumption in kg/hr
HV – heating value of fuel in KJ/kg
2. Indicated Power (IP): Power developed within the working cylinders.
IP =
πPmiLD2Nn'
4(60)
KW
Where:
Pmi – indicated mean effective pressure in KPa
L – length of stroke in meters
D – diameter of bore in meters
N – no. of RPM
n’ – no. of cylinders
Note:
N = (RPM)
for 2-stroke, single acting
N = 2(RPM)
for 2-stroke, double acting
N = (RPM)
for 4-stroke, single acting
2
N = (RPM)
for 4-stroke, double acting
3. Brake or Shaft Power (BP): Power delivered by the engine to the shaft.
BP =
πPmbLD2Nn'
KW
4(60)
68

69. Where:
Pmb – brake mean effective pressure in KPa
Note:
N = (RPM)
for 2-stroke, single acting
N = 2(RPM)
for 2-stroke, double acting
N = (RPM)
for 4-stroke, single acting
2
N = (RPM)
for 4-stroke, double acting
Brake Power in Terms of torque:
BP =
2πTN
KW
60,000
Where:
T – brake torque in Newton – meter (N-m)
Note:
N - RPM
4. Friction Power (FP): Power due to friction.
FP = IP − BP
5. Brake Torque
T = (P − Tare)R N - m
Where:
P – Gross load on scales in Newton
Tare – tare weight, N
R – Length of brake arm in meters
6. Indicated Mean Effective Pressure (Pmi): Average pressure exerted by the
working substance (air-fuel mixture) on the piston to produce the indicated power.
Pmi =
A ' S'
KPa
L'
Where:
A’ – area of indicator card, cm2
S’ – spring scale in KPa/cm
L’ – length of indicator card, cm
69

70. 7. Displacement Volume (VD):
VD =
IP m3
Pmi sec
BP m3
VD =
Pmb sec
VD =
πLD2Nn' m3
4(60) sec
Note:
N = (RPM)
N = 2(RPM)
N = (RPM)
2
N = (RPM)
for 2-stroke, single acting
for 2-stroke, double acting
for 4-stroke, single acting
for 4-stroke, double acting
8. Specific Fuel Consumption
a. Indicated Specific Fuel consumption
mFi =
mF
kg
IP KW - hr
b. Brake Specific Fuel consumption
mFb =
mF
kg
BP KW - hr
c. Combined Specific Fuel Consumption
mF
kg
GP KW - hr
where :
mFc =
GP - generator power in KW
9. Heat Rate (HR): Heat rate is the amount of heat supplied divided by the KW
produced.
a. Indicated Heat Rate
HRi =
Qs mF (HV )
KJ
=
IP
IP
KW - hr
b. Brake Heat Rate
70

71. Qs mF (HV)
KJ
=
BP
BP
KW - hr
HRb =
c. Combined Heat Rate
HRc =
Qs mF (HV)
KJ
=
GP
GP KW - hr
10. Thermal Efficiency
a. Indicated Thermal Efficiency (ei)
ei =
3600(IP)
x 100%
Qs
b. Brake Thermal Efficiency
eb =
3600(BP)
x 100%
Qs
c. Combined Thermal Efficiency
3600(GP )
x 100%
Qs
eC =
11. Mechanical Efficiency
BP
x 100%
IP
P
ηm = mb x 100%
Pmi
ηm =
12. Generator Efficiency
ηg =
GP
x 100%
BP
13. Generator Speed
N=
120f
RPM
n
where
f - frequency in cps of Hertz
n - no. of generator poles
14. Volumetric Efficiency
ηv =
Actual Volume of air entering
x 100%
Displacement71
Volume

72. 15. Correction Factor for Non-Standard Condition
a. Considering Temperature and Pressure Effect
Ph = Ps
Bs
Bh
Th
Ts
b. Considering Temperature Effect alone
Ph = Ps
Th
Ts
c. Considering Pressure Effect alone
Ph = Ps
Bs
Bh
16. Engine Heat Balance
Q2
Engine
Qs
Q1
Q3
Q4
QS = Q1 + Q2 + Q3 + Q4
Q1 - heat converted to useful work
Q2 - heat loss to cooling water
Q3 - heat loss to exhaust gases
Q4 - heat loss due to friction, radiation and unaccounted for
Q1 = 3600(BP) KJ/hr
Q2 = mwCpw(two - twi) KJ/hr
Q3 = Qa + Qb KJ/hr
Qa = mgCpg(tg - ta) KJ/hr
Qb = mf(9H2)(2442.7) KJ/hr
Q4 = QS - (Q1 + Q2 + Q3) KJ/hr
H2 = 0.26 - 0.15S kgH/kgfuel
141.5
131.5 + °API
140
S=
130 + °Be
S=
72

73. where:
Qa - sensible heat of products of combustion
Qb - heat required to evaporate and superheat moisture formed from the
combustion of hydrogen in the fuel
tg - temperature of flue gas, °C
ta - temperature of air, °C
H2 - amount of hydrogen in the fuel kg H/kg fuel
Diesel Engine Maintenance
OPERATING A DIESEL ENGINE
Before starting:
There are several steps to be taken before starting a diesel engine, especially he first time,
and its good practice to work out a certain routine to be followed always:
1. All moving parts of the machine much be examined for proper adjustment, alignment, and
lubrication. This includes values, cams, value gear, fuel pumps, the fuel injection, the governor
lubricators, oil and water pumps, and the main driven machinery.
2. The whole engine and machinery must be examined for loose nuts, broken bolts, and loose
connection. And leaky jackets, joint or values. It well to remember that nothing must be tight.
3. All tools from the tool board should be checked to make sure none is missing. They may be
needed in a hurry when the engine is running or, is misplace and left on the engine, may drop off
from vibration and damage some moving parts.
4. All pipes and values for fuel, lubricating oil, water and air, as well as ducts, must be check for
clogging up, lack of adjustment, cleanliness, etc. Absence of foreign matter in the piping system
must be checked especially carefully, if the engine has been idle for sometime or is just being put
into service in the latter case it is advisable to blowout the entire piping system with compressed
air.
5. A complete check up must be given to the lubricating system to make sure that oil is present in
every placed required, that the lubricator and all bearings that are individually oiled have an ample
supply of clean oil, that all grease cups are filled. The lubricator should check for proper functioning
of the pumps and for the amount of oil delivery, and filled with oil to the proper level, the lubricator
should be turned by hands and the points to which its delivers oils should be lubricated. Make sure
that the engine well received proper lubrication the very moment its starts to run.
73

74. 6. The cooling system must checked, and if the pumps are driven by the electronic motors, they must
be started, the suction line opened to have water in the water engine before starting. The correct
amount of water circulation should be adjusted later, while the engine is being warm up. If the
engine has oil-cooled pistons with oil delivered by a especial pump, start the oil pump and adjust
the pressure to the amount stated in the name plate or given in the engine.
7. The fuel-oil system must be checked in every respect, to make sure that pipes are clean, pumps
are working, and a supply of fuel is in the tanks. The fuel-injection pumps should be primed and air
or water removed from the discharge line, valves or nozzles. One or two strokes on the fuelinjection pump in usually sufficient care should be taken not to force too much fuel the combustion
chamber or cylinder in order not to obtain and excessively high pressure with the first firing-causing
the safety valves to pop and not to get the fuel oil into the crankcase. However, the fuel pumps
must be primed sufficiently so that each discharge line in filled clear to the nozzles, the fuel
controlled level is set wide open so that injection will start at once. The fuel pump control is put in
the fuel on position.
8. The safety valve, usually installed on each cylinder head should be check. These valves are set to
pop off about 750 to 1250 psi, depending upon the maximum pressure allowed in the engine. The
values are exposed to high temperature gases and have a tendency to stick. The checking may be
done either by compressing the spring with crowbar or by unscrewing the cap and taking the valve
out of the inspections.
9.The engine should be turned over one or two times if it has not been operated for sometimes. To do
this it is necessary to open the indicator cocks or compressor-relief valves and to turn the engine
over, either by hands with a bar in the holes in the flywheel, or with a jack or air motor, as the case
maybe. Then the indicator cocks should be close with the same in proper position for starting-one
cylinder having the starting air valve open and the position about 100 past top center.
10. The air in the tanks must be checked to see that it is up to the required pressure. If, not it must be
pumped up the starting air system from the tanks to the starting air control valve must be opened,
either it has been checked that the main control valve is closed. With an air injection engine the
bottle within injection air must be checked and if necessary pumped up o the required pressure.
11. The engine load should be off, the switch should be open if the engine drives a generator, or the
clutch should be in neutral position. If the drive is through the friction clutch. If the engine drives a
pump or compressor, the by-pass should be open.
74

75. STARTING:
If all eleven points of the preparatory program have been observed starting with compressor
air is very simple.
First, the main starting – air valve is opened and the starting lever is manipulated according to
the instructions given in the engine instruction book.
Second, the engine is watched, no necessary air should be used. At the first indication of
combustion, air should be cut off and the ventilating valve opened, an in good condition usually
begin to between the second and fourth revolution of the crankshaft.
Third, if the engine fills to start after four or five revolution, there is something wrong. Useless
turning of the engine should be stopped, and the cost of trouble investigated.
Low air pressure, if the starting air is too low either from a slow loss of air through some leaky
joint or failure of the engine to start at the first attempt. And there is no air compressors to pump
up air several methods maybe used for securing the necessary starting pressure that never
should pure oxygen by used for starting purposes.
Flasks of compressed air may be obtained and the contents equalized into the engine
receivers, or a flack of carbon dioxide may be obtained from some local soda foundation and
piped to the starting battles. This gas is liquid at ordinary temperatures and about 800-psi
pressures. Therefore, it is necessary to apply some heat in order to evaporate this liquid carbon
dioxide. This heat may be applied by pouring hot water over the battle or by applying rags soaked
in hot water.
WARM UP:
After the engine is started, before putting on the load, its should be allowed to idle for a few
minutes (up to five minutes) and to warm up. During this five minutes the following observations must
be made.
1. Listen to find it out if combustion is regular and firing order and correct all. Cylinder for
combustion and note the working of the fuel injection pump to see whether they all operate
properly.
2. Observe the cooling water system throughout to see whether the pumps are working. There is
sufficient water, watch to see if the water temperature is building up properly, and regulate the
water flow accordingly.
75

76. 3. Observed lubrication pressure and the working of the lubrication and count the number of
drops for correct operation. Feel whether any of the cylinders is warming up too fast –
indicating an unlubricated piston and listen for unlubricated piston pin or crank pin bearing. If
any moving parts receive an insufficient amount of lubricating oil, serious trouble may result.
4. Observe the exhaust, color and sound, to note proper condition. These observations should
be repeated after the lead is put on. The color of the exhaust can tell many things.
The making of these observations during the first five minutes after starting should be
regular habit with the engine operator. This procedure is the best, the most reliable method of
preventing improper operation. It is based upon the fact that a diesel engine requires neither
much, but it requires proper attention at the proper time. It is also based on the known fact that
a diesel engine should be operating properly in five minutes or there is something wrong which
should be detected in these five minutes.
However, it should be noted that certain observations should be carried on even after
the 5-min. warming up period. Thus, if there are any leaky water jackets, injection valves, air
valves, etc… they may not show up until full expansion of the corresponding part has taken
place after the engine the has been in operation a longer time at normal load. No leaks of any
kind should be allowed, if they cannot be stopped while the engine is running the engine
should be stopped and not restarted until the trouble corrected.
RUNNING
In general the attention, which an operator must give to, the engine in
along the same lines as during the warm-up period.
regular operation is
The differences is that the corresponding
observations should be made periodically every 15 to 20 minutes and at least every half hour, even if
the engine is equipped with-a sufficient number of automatic danger-warming signal ad seconds, that
all observations must be entered in engine log.
THE ENTRANCES IN A COMPLETE ENGINE LOG ARE THE FOLLOWING:
1. Time of entering the readings, or rather the first reading in each series.
2. Engine load, or in the case of electric loads, volts and amperes reading.
3. Engine speed from the tachometer or if the engine has an adding revolution counter, the
counter reading, in this case it is essential to have in the engine room a large clock with a hand
indicating seconds, to enable the operator to read the revolution counter at exact intervals.
76

77. 4. Fuel consumption enter the instantaneous reading of s rotameter or the reading of a fuel meter
in which case it is also important to make the reading at exact intervals.
5. Exhaust:
a.) Reading of the temperature of exhaust from each cylinder;
b.) Exhaust temperature in the exhaust line close to the exhaust manifold;
c.) Color of exhaust either by simple description such as clear, little haze, light gray, gray,
dark gray and very dark gray or better, by a number according to a standardized smoke
scale, such as Ringleman’s scale.
6. Lubricating oil:
a.) Pressure as discharged from the oil pressure pump.
b.) Temperature of the oil before the oil cooler.
c.) Temperature of the oil after the oil cooler.
7. Cooling water:
a.) Temperature of the water delivered to the water-cooling manifold.
b.) Temperature as discharge from each cylinder, or in the water outlet line.
c.) Flow, gallon per minute, either from the rotameter or a water meter.
8. Scavenge air:
a.) Temperature after blower
b.) Pressure after blower, usually in inches of mercury.
9. Super charger conditions:
a.) Temperature of air after booster pump.
b.) Pressure of the air after booster pump, Psi or inches of mercury.
10. Barometric pressure, inches of mercury.
11. Temperature of the air intake, before the air filter.
12. Remarks about what happened at e certain moment during operation of the engine, such as,
put second engine online or stopped it, found lubricating oil filter clogged by dirt as indicated by
excessive pressure drop, switched to the second filter, or by-passed filter and exchanged filter
element, etc. Between taking readings and entering them in the engine log, the operator
should listen to find out if the engine is running uniformly, without unusual sounds or knocks.
He should feel whether the bearing are running warmer than usual and particularly watch that
the engine as a whole doest not become overloaded or some of the cylinders become
77

78. overload. Because in the combustion in one or two cylinders doest not proceed correctly, as
indicated by a considerable lower or higher temperature from that exhaust particular cylinders.
Naturally, the operator must also see that the day fuel tank is not depleted and if the engine
has hand lubricated places that they are oiled at regular intervals. Should be oiled every two
hours the exhaust valve stems should receive a few drops of kerosene instead of oil every
three or four hours in order to keep them in good working condition. The circular groove
around the valves and the whole top of the cylinder head must be wiped clean at all times. Oil
must be allowed to accumulate on the cylinder head and run down the side of the engine, as it
could easily work into the joints between the cylinder and heads and decompose the rubber
gaskets with form the water joint.
If the flow of the cooling water or oil should stop for any reason, the engine or any of the
cylinder will become overheat. The engine must be stopped at once and permitted to cool
gradually. It is extremely dangerous to admit water to a hot engine as a sudden change in
temperature nay cause the pistons or one of the cylinder heads, liners or the exhaust manifold
to crack.
The exhaust from the engine should be perfectly clean. However, if the engine is
operating under an overload, the exhaust may become visible, with a light grayish smoke. If
the engine is visible under over than overload conditions, the cause should be found
immediately reminded. An engine under no condition is operated for any length of time with a
visible smoky exhaust.
If the pyrometer with thermocouples is installed on the engine cylinder that yields a
smoky exhaust may be found by nothing the exhaust temperatures of the various cylinders. If
abnormal condition exist in any of the cylinders, this condition will usually be accompanied by
an increase in the temperature of the exhaust from the cylinders, do not get their share of fuel,
and a result, the other cylinder are overloaded. If possible the engine should be stopped and
the cause rounds and reminded.
STOPPING THE ENGINE:
To stop the engine, proceed as follows: move the fuel pump controls to stop position
and shut the fuel supply valve.
78

79. The cooling water and piston cooling oil should be left running after the engine is shut
down until the outlet temperature are not more than 5 to 10 of higher than the inlet
temperature. This prevents local overheating which would cause scale deposits.
The jackets, if hard water used ad the engine is supplied with direct connect pumps, it
will be necessary to start the auxiliary pumps to cool the engine as indicated above.
If the engine is to be shut down for a considerable length of time the water jackets must
be completely drained so as to prevent rust and in cold weather also protect the jackets from
bursting if the watering the engine room should freeze. Naturally, all drops oiliest must be
stopped. All switches cut-out, and friction clutches put in neutral position.
79