1.1 power plant module1 batch3

POWER PLANT
MODULE

Benjie S. Andres
Power Department

Course Contents
• I. Fundamentals of Thermodynamics
1.1 Basics, concepts, terminologies, practices
1.2 Phases of water and Introduction of TS diagram
1.3 Steam Properties and Steam Tables
1.4 Thermodynamic Laws and Cycles,
1.5 Carnot and Rankine Cycle
• II. Power Plant Components
-Applications of Thermodynamics to power plant components
• III. Power Plant Facilities
-Familiarize with the various plant configurations
-Introduce the criteria for Technology selection
-Discussion on EDC Power Plant Facilities
• IV. Overview to Power Plant Operations

Date

I. Fundamentals of Thermodynamics

I. PR 1.1 Fluid Properties: INCIPLESRMODYNAMICS
Measurable or Quantifiable characteristic of a fluid
- Measurable: Pressure, Temperature, Specific Volume
- Quantifiable: Internal Energy, Enthalpy, Entropy

1.1.1 Measurable Properties
•Pressure: force per unit area (Which is greater, a PULL or a PUSH?)
•Temperature: measure of hotness or coldness
•Specific Volume: amount of space occupied/unit mass

1.1.2 Quantifiable Properties: Define

1.2.1 Internal Energy: Thermal energy within the substance itself
1.2.2 Enthalpy: Sum of internal and flow energy
1.2.3 Entropy: A measure of unavailable energy

Date


Pressure: Force per unit Area

Pressure
•ABSOLUTE PRESSURE
-Total pressure above a perfect Plenum
vacuum
-Patm+Pgage Local Atmospheric Pressure
-Patm-Pvac

Absolute
•GAGE PRESSURE Vacuum
-Pressure measured above
atmospheric Atmospheric
Absolute
•VACUUM PRESSURE
-Pressure below atmospheric or
negative gage pressure
Relationships Between Pressure Terms
•DIFFERENTIAL PRESSURE
-Pressure measured as
difference between two unknown
pressures
ATMOSPHERIC PRESSURE VARIES WITH LOCATION
STANDARD VALUES:1.01 bar, 1.03 ksc, 101.325kpa, 0.101mpa, 760 mmHg

Date


I. PRINCIPLES OF THERMODYNAMICS
Temperature: Measure of hotness or coldness
Conversion:
- ⁰C-⁰F: ⁰C =(⁰F-32)(5/9)
- ∆T: C⁰ =(F⁰)(5/9)

Degree Centigrade is the common scale but ⁰K is used for absolute
values (⁰C+273)

Date


1.2 PHASES OF WATER

T Triple Point
E
M Z Steam
P
E Water
R
A
T Mixture
U Steam & Water
R
E
,
Solid-Vapor Region
T

ENTROPY, s


1.3 STEAM RELATED TERMS AND DEFINITIONS
Saturated Vapor
• A vapor whose temperature and pressure are such that any compression of
its volume at constant temperature causes it to condense to liquid at a rate
sufficient to maintain a constant pressure.
Saturated Liquid
• A liquid whose temperature and pressure are such that any decrease in
pressure without change in temperature causes it to boil.
Wet Steam
• Mixture of steam and liquid.
Quality
• Percentage of steam in a two phase fluid.

7


1.3.1 STEAM PROPERTIES AND CALCULATIONS
Equation: m = x(mg - mf) + mf
Where:
• m = total mass
• mg= mass of vapor
• mf= mass of moisture
• x = steam quality

EQUATION IS ALSO APPLICABLE ALSO TO OTHER PROPERTIES
OF STEAM: ENTROPY(s), ENTHALPY(h), INTERNAL
ENERGY(u), AND SPECIFIC VOLUME(v)

PRESSURE INDICATED IN THE STEAM TABLES ARE
ABSOLUTE VALUES

8


1.3.2 STEAM TABLES
• Saturated Steam: Temperature Table
• a) Consists of columns for:
• 1) Temperature
• 2) Pressure - corresponds to temperature for saturation conditions.
• 3) Specific Volume
• 4) Enthalpy
• 5) Entropy
• b) The v, h, and s columns each have values for saturated liquid (vf)
• saturated vapor (vg), and the change (vfg) from liquid to vapor.
• Saturated steam: Pressure Table
• a) This table is set up the same as table above except the temperature and
• pressure columns are reversed.
• Superheated steam
• a) This table is set up differently. It consists of:
• 1) Abs pressure column with sat. temperature in parentheses.
• 2) Across the top is temperature - degrees Fahrenheit. This
• represents the actual temp of the steam.
• 3) Sh column represents the degrees super heat.
• 4) It then has columns for v, h, and s.

9

EXERCISE I

A. Using Steam Tables determine the steam quality at the following
conditions:
a. Steam quality at h = 2500 kJ/kg, P=50kPa
b. Steam quality at s=6 kJ/kg-C, P=50kPa


B. Compute for the properties of steam given that the enthalpy
and pressure is 2500 kJ and 120 kPa.

10

MOLLIER CHART OR HS DIAGRAM

11


1.6 THERMODYNAMIC CYCLES
•Thermodynamics: Study of energy conversion mainly of heat to work
•Thermodynamic Process: A system that undergoes energy change, associated with
changes in pressure, volume, internal energy, temperature, or heat transfer
•Thermodynamic Cycles: Repeating series of processes used for transforming
energy to useful effect

HOT THERMODYNAMIC COLD
SOURCE CYCLE
SINK

WORK: W= Fd http://www.bpreid.com/carnot.php

Date


4.1 Definitions:
Thermodynamics: Study of energy conversion mainly of heat to work
Thermodynamic Cycles: Repeating series of processes used for transforming
energy to useful effect

HOT COLD
SOURCE SINK

W=0
http://www.bpreid.com/carnot.php

Date


1.4 THERMODYNAMIC LAWS

1. Energy can be changed from one form to another, but it cannot be created or
destroyed.

Increase in internal energy of a system = heat supplied to the system -
work done by the system. U = Q - W

2. It is impossible to have a cyclic process that converts heat completely into
work. It is also impossible to have a process that transfers heat from cool
objects to warm objects without using work.

….the universe is constantly losing usable energy and never
gaining.

14


1.5 THERMODYNAMIC PROCESSES
•Adiabatic - a process with no heat transfer into or out of the system.(∆Q=0)

•Isochoric - a process with no change in volume, (∆V=0)

•Isobaric - a process with no change in pressure. (∆P=0)

•Isothermal- a process with no change in temperature. (∆T=0)

•Isentropic - a process with no change in entropy. (∆s=0)

It is possible to have multiple processes within a single process. The most obvious
example would be a case where volume and pressure change, resulting in no change
in temperature or heat transfer - such a process would be both adiabatic & isothermal.

15


A B
T T2
1. Heat addition: A-B
2. Isentropic expansion: B-C
3. Heat rejection: C-D
4. Isentropic compression: D-A WORK
QA=T2∆s
QR=T1∆s
T1 C
D
W=∆T∆s

s1 s2
S
CARNOT CYCLE: MOST EFFICIENT CYCLE BUT NOT PRACTICAL

16


Triple Point
T A
E P=C
M B
P Steam
E Water H=C
R D 1
A C
T
Mixture s=C
U
Steam & Water
R 3
E
,
2
RANKINE CYCLE: THE PRACTICAL CARNOT CYCLE AND
T
THE MOST COMMON CYCLE USED BY POWER PLANTS

ENTROPY, s


1.6 Computation for Turbine-Generator Power
EQUATION: (h1 h 2' )
Thermal Efficiency x100
(h1 h 2)
Where:
h1 = enthalpy at turbine inlet
h2’ = actual enthalpy at turbine outlet
h2 = isentropic enthalpy(enthalpy when entropy s1=s2)

P ms (h1 h2' ) t g

18

EXERCISE II

• The Malitbog Power Plant interface pressure and
temperature is 11.065 ksca and 185deg C. Steam
expands to the Turbine to a pressure of 69.01
mmHgabs. The generator output is 77.9 MW and the
steamflow is 471.00 TPH. Given these, compute for the
efficiency of the turbine.

19

II. POWER PLANT COMPONENTS

• A geothermal power plant produces electricity by converting geothermal
steam into mechanical energy in the turbine; and by converting the
mechanical energy in the generator into electrical energy.

Date

THE FIRST LAW OF THERMODYNAMICS APPLIED:ENERGY
CAN NEITHER BE CREATED NOR DESTROYED

Mahanagdong 60 MW Unit POWER SUMMARY
GROSS POWER OUTPUT = 60,230 KW
COOLING WATER PUMPS = 1,740 KW
COOLING TOWER FANS = 1,065 KW
VACUUM PUMP = 872 KW
CONDENSATE PUMP = 52 KW
TRANSFORMER LOSSES = 301 KW
STEAM SUPPLY MISCELLANEOUS = 400 KW
Main NET POWER OUTPUT 55,800 KW
Transformer
5.94 P
2,755H
Steam Gas NCG VENT TO
444,666 G
157.8 T Turbine Generator EJECTOR GAS REMOVAL COOLING TOWER

SYSTEM
VACUUM
LEGEND PUMP
P: kg/cm² a INTER VACUUM
H: kJ/kg CONDENSER SEPARATOR
G: kg/hr Main
T: deg C Condenser

Hot Well
PUMP
DRIFT AND
EVAPORATION LOSS

L. O. COOLER
GEN AIR COOLER
BLOWDOWN
GLAND STEAM COND.

COOLING Cooling
CWR
Tower
WATER
COOLING WATER
SYSTEM PUMP
CWS
DESIGN BASIS:
NCG - 2.0%
WET BULB T - 230C

Date


1.0 Steam Gathering System
• Transport steam from the
point of delivery by FCRS to
the individual steam turbine
inlet (main steam) and the
gas removal system
(auxiliary steam);
• Scrub the steam and
remove impurities in the
steam before it is introduced
to steam turbines.

24


STEAM TURBINE COMPONENTS

 The heat from the geothermal fluid (steam and non-condensable gases) is converted
to mechanical work as it passes through the turbine rotor blades , causing the rotor
shaft to rotate and drive a generator to produce electricity;

CASING ROTOR
DIAPHRAGMS

TURNING GEAR
SEALS

THRUST BEARING

JOURNAL BEARING

JOURNAL BEARING

STEAM INLET

26



2.1 Turbine Rotor

Typically, higher pressure sections are
impulse type and lower pressure stages
are reaction type

Date


2.2 Turbine Casing
• Upper and Lower parts are
bolted at a horizontal joint with
the lower portion sitting on
levelling pads and grouted.

• Diaphragms are mounted and
supported by the casing

28



2.3 Thrust and Journal Bearings

29


2.4 Diaphragms

30


2.5 Turning Gear

A turbine component designed to
turn the rotor slowly during
startups and shutdowns, thus
minimizing rotor eccentricity

Turning gear


2.6 Control Valves
A butterfly valve at the turbine inlet that regulate the flow of steam supply during
normal operation and serve as a backup to the main stop valve during shutdown.
Commonly called governor valves.

Electro-
Hydraulic
Converter


Steam Turbine Related Terms
• What are the different types of Turbine at EDC?
– Back Pressure turbine
• Steam expands at plenum pressure
– Condensing Turbine
• Steam expands at vacuum pressure

33


3.0 ELECTRIC GENERATOR
Converts mechanical energy into electrical energy.
Three Elements Required to Induce a Voltage
• Relative Motion-provided by Turbine
• Magnetic Field
• Conductor


ELECTRIC GENERATOR
ROTATING
RECTIFIER EXCITER

ELECTRIC GENERATOR

Generator Construction
• The armature winding is more complex than the field and can be
constructed more easily on a stationary structure
• The armature winding can be braced more securely on a rigid
frame.
• It is easier to insulate and protect the high-voltage armature
windings.
• The armature winding is cooled more readily because the stator
core can be made large enough and with many air passages or
cooling duct for forced circulation.
• The low-voltage field can be constructed for efficient high-speed
operation.

ROTOR FIELD
WINDING

ROTATING
RECTIFIER EXCITER PMG

138 kV
Line
13.8 /138 kV
AVR Main
Transformer
Thyristor
Firing SENSING
Circuit VOLTAGE
13.8 kV Line

STATOR
WINDING

PMG

EXCITER

ROTATING RECTIFIER
ROTOR FIELD
WINDING

Voltage Regulation

• Generator output is regulated by adjusting the strength
of the magnetic field around the generator rotor.
If the load of the generator decreases, the generator’s voltage output
increases.

To bring the voltage back down to normal, the system increase the
rheostat resistance in the exciter circuit. The increased resistance
reduces the current flow to the exciter stator windings, and weakens the
exciter magnetic field.

The reduced magnetic field restores the generator output voltage to
normal.

138 kV
Line
13.8 /138 kV
AVR Main
Transformer
Thyristor
Firing SENSING
Circuit VOLTAGE
13.8 kV Line

STATOR
WINDING

PMG

EXCITER
ROTATING
RECTIFIER
ROTOR FIELD
WINDING

138 kV
Line
13.8 /138 kV
AVR Main
Transformer
Thyristor
Firing SENSING
Circuit VOLTAGE
13.8 kV Line

STATOR
WINDING
13.8 kVac
120
Vac, 42
0 Hz

100
Vdc

200
Vac
250
Vdc

PMG

EXCITE
R
ROTATING
RECTIFIER
ROTOR FIELD
WINDING

60 80 60 80
40 40
100 100

20 GENERATOR 20
0 0
AIR COOLER

PMG
EXCITER

60 80 60 80
40 40
100 100

20 20
0 0

GENERATOR
AIR COOLER

PMG
EXCITER

60 80
40
100

20
0

60 80
40
100

20
0

GAC FROM GENERATOR
COOLING AIR MAKE-UP
WATER


• Cooling Water System
This system provides cold water to the main
condenser where it is used to absorb heat from steam
resulting to its condensation. The warm water from
the main condenser is then brought to a cooling tower
where ambient air is used to cool the warm water and
thus produce cold water for fresh supply to the
condensers.

47


4.0 Main Condenser
Maintains the turbine exhaust
at vacuum pressure to
maximize plant output; The
main condenser performs its
function by condensing steam
vapor leaving the turbine
exhaust to liquid which
results to a significant
reduction in the space the
vapor previously occupied
and this induces a void or a
vacuum.

48


5.0 Cooling Tower
• Cooling towers serve as the heat sink for condensers and auxiliary
equipment. Wet cooling towers dissipate heat rejected by the plant to the
environment through these mechanisms:
– Addition of sensible heat to the air
– Evaporation of a portion of the circulating water itself
• There are two types of wet-cooling tower, the natural draft and the
mechanical draft cooling towers.
EDCs cooling towers are all Wet Type, with the exception of the UMPP has
a Dry Type Cooling Tower.
• The mechanical draft type can be a crossflow or the counterflow type.

49

Types of Cooling Towers

Crossflow Tower

50

Common problems:

 High CT wet bulb temperature
 High ambient temperature
 Hot moist air recirculation
 Restriction of flow
 Instrument problem
 Clogging of sprinkler, fills and drift eliminator due to algae and scale build up
 Clogging of screen that result to false activation of level switch and false
indication from level transmitter

Cooling Tower Calculations
• Cooling Tower calculations is usually carried out with the aid
of the Psychometric Charts

– Approach. Difference between the cold-water temperature and
wet-bulb temperature
– Range (or cooling range). Difference between the hot-water
temperature and cold-water temperature.
– Saturated Air. This is air that cannot accept more water vapour at a
given temperature.
– Relative Humidity. the ratio of the actual vapor density to the
saturation vapor density at the temperature and barometric
pressure. Φ = 100% refers to saturated air.

52


53

• Absolute Humidity. This is the air per unit mass of dry air (da).
Absolute humidity is given the symbol ω. Using PV=mRT for both
water vapor and dry air.

mv 53.3Pv 0.622 Pv
ma 85.7 Pa P Pv
Where
53.3 and 85.7 are the gas constants for dry air and water, respectively.
Pv=Partial Pressure of vapor=Psat from steam table at air temperature
P=Total Pressure=Atmospheric pressure
• Dry Bulb Temperature. This is the temperature of the air as commonly
measured and used. It is the temperature as measured by a
thermometer with a dry mercury bulb, and given the symbol DBT.

54

• Wet-bulb temperature. This is the temperature of the air as measured by a
psychrometer, in effect a thermometer with a wet gauze on its bulb. Air is
made to flow past the gauze. If the air is relatively dry, water would
evaporate from the gauze at a rapid rate, cooling the bulb and resulting in a
much lower reading than if the bulb were dry. If the air is humid, the
evaporation rate is slow and the wet-bulb temperature approaches the dry-
bulb temperature. Thus for a given ambient temperature T, the wet-bulb
temperature is lower the drier the air. The wet-bulb temperature is given the
symbol WBT
• Dew Point. The temperature below which water vapor in a given sample of
air begins to condense is called the dew point. It is equal to the saturation
temperature corresponding to the partial pressure of the water vapor in the
sample.

55

Energy Balance
• Heat Loss of Water=Heat Gain of Air
Hot Air, h2
Lcpw(HWT-CWT)=G(h2-h1)
(heat loss to evaporation is neglected) Hot Water
Where:
h = enthalpy of dry air, Btu/lbm or J/kg
L = mass of cooling air, lb or kg/unit time Cold Air, h1

G=mass of circulating water, lb or kg/unit time
HWT = Hot water temperature Cold Water
CWT=Cold water temperature
Cpw= specific heat of water, 4.186kj/kg-C

56

Mass Balance. The dry air goes through the tower
unchanged. The water vapor in the air gains mass due
to the evaporated water. Thus, based on a unit mass of
dry air:
Hot Air, H2
(Mass of Water)in=Losses+ (Mass of Water)out
(If heat loss to evaporation is considered) Hot Water

Lcpw(HWT-CWT)+G(H2-H1)(hf)=G(h2-h1)

Where: Cold Air, H1
H=humidity ratio of inlet and outlet air, kg vapor/kg DA
h = enthalpy of dry air, KJ/kg
L = mass flow of cooling air, kg/unit time Cold Water
G=mass flow of circulating water, kg/unit time
HWT = Hot water temperature, ⁰C
CWT=Cold water temperature, ⁰C
Cpw= specific heat of water, 4.186kj/kg-C
hf= enthalpy of saturated liquid at CWT, kJ/kg

57

Cooling Tower Exercises
• A cooling tower with a range of 20degF Hot Hot
receives 360,000 gal/min of 90degF circulating water water
water. The outside air is at In Out
60degF, 14.696psia, and 50% relative (A) (2)
humidity. The exit air is at 80degF saturated. hfA ha2
Calculate the mass of air required in ft3 /min. WA ω2
Neglect evaporation loss. hg2
• Conversion: 7.48 gal = 1 ft3
Density = 62.428 lb/ ft3 Cold
• Cp water = 1 Btu/(lbm deg F) Water h
In ωa1
(1) h 1 hfB
g1 WB
Cold
Water
Out
(B)

59

Cooling Tower Exercise
Solution
• TA = 90degF Hot Hot
water air
• TB = TA – Range = TA – 20degF = 70degF in out
• ha1 = h at 60degF and 50% RH (A) (2)
hfA ha2
• ha2 = h @ 80deg F and 100% RH WA ω2
hg2

Cold
air h
in ωa1
(1) h 1 hfB
g1 WB
Cold
water
out
(B)

60

Cooling Tower Exercise
Solution
Hot Hot
• Heat Gained (Air)=Heat Loss (water) water air
• Wa(ha2-ha1)=Wcpw(TA-TB ) in out
(A) (2)
hfA ha2
ω2
• Dry Air Required: WA
hg2
=9019440 lbDA/min

Cold
air h
in ωa1
(1) h 1 hfB
g1 WB
Cold
water
out
(B)

61


6.0 Gas Removal System
• The Gas Removal System removes the NCG from the
condenser. The purpose of this is to maintain the pressure
in the condenser since the NCG occupies space in the
condenser, increasing the condenser pressure.
• Steam ejectors are designed to convert the pressure
energy of a motivating fluid to velocity energy to entrain
suction fluid and then to recompress the mixed fluids by
converting velocity energy back into pressure energy. This
is based on the theory that a properly designed nozzle
followed by a properly designed throat or venturi will
economically make use of high pressure fluid to compress
from a low pressure region to a higher pressure. This
change from pressure head to velocity head is the basis of
the jet vacuum principle.

62

Gas Removal System Components

• Components from the condenser to the GRS
– NCG, Air, Water Vapor
• The capacity of the GRS is determined by the NCG Load. An
approximation of the NCG Load can be computed as follows:
– The total mass flow of the NCG, Air and water vapor will
determine the needed suction capacity of the GRS. The
computed total mass flow is termed as the Dry Air Equivalent
(DAE)
• The GRS can be either of the following configurations:
– Steam Gas Ejectors (SGE) for all stages
– Hybrid (SGE for the first or first two stages and Liquid Ring
Vacuum Pump for the last stage)
– Mechanical Extractor

63


 Gas Removal
System
 This system acts like
a pump to extract
non-condensable
gases from the main
condenser so it does
not degrade the
vacuum by
occupying space; Its
extraction capacity
must be equal to the
non-condensable gas
flow entering the
turbine and the main
condenser.

Power Generation Sector –

Principle of Operation
The HEI* states “The operating principle of a
steam ejector stage is that the pressure
energy in the motive steam is
converted into velocity energy in the
nozzle, and, this high velocity jet of steam
entrains the vapor or gas being
pumped. The resulting mixture, at the
resulting velocity, enters the diffuser where
this velocity energy is converted to
pressure energy so that the pressure of the
1. Diffuser 6. Suction
2. Suction Chamber 7. Discharge
mixture at the ejector discharge is
3. Steam Nozzle 8. Steam Inlet
substantially higher than the pressure in
4. Steam Chest 9. Nozzle Throat
the suction chamber.” * See the HEI
5. Extension (if used) 10. Diffuser Throat
Standards for Steam Jet Vacuum Systems

65


Malitbog Power Plant Gas Removal System
• There are three
ejector stages with 3
ejector trains each.
Each has a
condenser in each
stage.

Motive Steam
Steam from
condenser
Cooling Water

66

End of Presentation
Thank You!

EXAM I

A Geothermal Power Plant receives steam at an interface pressure
and temperature of 10.4 bara and 185deg C. The steam expands to
the Turbine to a pressure of 1.3 bara. The generator output is
20,310 kW and the steamflow is 254,750 kg/hr. Given
these, compute for the efficiency of the turbine and indicate what
type of turbine this is.

68

EXAM II

A cooling tower with a range of 10degF receives 400,000 gal/min of
100degF circulating water. The outside air is at 60
degF, 14.696psia, and 70% relative humidity. The exit air is at 85degF
saturated. Calculate the outside air required in lbmDA. Include all heat
and mass balance drawings.

69

• EFFECT of MOISTURE and SOLID
PARTICLES to First Stage Diaphragms

1.1 power plant module1 batch3

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