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

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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 itself1.2.2 Enthalpy: Sum of internal and flow energy1.2.3 Entropy: A measure of unavailable energy

I. PR 1.1 Fluid Properties: INCIPLESRMODYNAMICS

I. Fundamentals of Thermodynamics

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Pressure: Force per unit Area• ABSOLUTE PRESSURE-Total pressure above a perfect vacuum-Patm+Pgage-Patm-Pvac

• GAGE PRESSURE-Pressure measured above atmospheric

• VACUUM PRESSURE-Pressure below atmospheric or negative gage pressure

• DIFFERENTIAL PRESSURE-Pressure measured as difference between two unknown pressures

Pres

sure

Abso

lute

Absolute

Plenum

Atmospheric

Vacuum

Relationships Between Pressure Terms

Local Atmospheric Pressure

ATMOSPHERIC PRESSURE VARIES WITH LOCATIONSTANDARD VALUES:1.01 bar, 1.03 ksc, 101.325kpa, 0.101mpa, 760 mmHg


Date

Conversion:- C- F: ⁰ ⁰ C =( F-32)(5/9)⁰ ⁰- ∆T: C =(F )(5/9)⁰ ⁰

I. PRINCIPLES OF THERMODYNAMICSTemperature: Measure of hotness or coldness

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


TEMPERATURE,T

ENTROPY, s

Triple Point

MixtureSteam & Water

Water

SteamZ

Solid-Vapor Region

1.2 PHASES OF WATER


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.

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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)

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PRESSURE INDICATED IN THE STEAM TABLES ARE ABSOLUTE VALUES



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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.


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

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


I. Fundamentals of Thermodynamics EXERCISE I

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MOLLIER CHART OR HS DIAGRAM

1.6 THERMODYNAMIC CYCLES

Date

• 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

http://www.bpreid.com/carnot.php

HOT SOURCE

COLD SINK

WORK: W= Fd

THERMODYNAMIC CYCLE



Date

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


HOT SOURCE

COLD SINK

W=0



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1.4 THERMODYNAMIC LAWS

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

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.

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

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


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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.


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CARNOT CYCLE: MOST EFFICIENT CYCLE BUT NOT PRACTICAL

TA B

s1s2

S

CDT1

T21. Heat addition: A-B2. Isentropic expansion: B-C3. Heat rejection: C-D4. Isentropic compression: D-A WORK

QA=T2∆s

QR=T1∆s

W=∆T∆s


TEMPERATURE,T

ENTROPY, s

P=C

H=C

Triple Point

MixtureSteam & Water

WaterSteam

s=C

A

2

B

1

C

D

RANKINE CYCLE: THE PRACTICAL CARNOT CYCLE AND THE MOST COMMON CYCLE USED BY POWER PLANTS

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1.6 Computation for Turbine-Generator Power

EQUATION:

Where:

h1 = enthalpy at turbine inlet

h2’ = actual enthalpy at turbine outlet

h2 = isentropic enthalpy(enthalpy when entropy s1=s2)

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100)(

)'(Efficiency Thermal

21

21x

hh

hh

gthhmP s )( '21


• 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.

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I. Fundamentals of Thermodynamics EXERCISE II

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RECAP PART I

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.

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Date

Turbine Generator

Main Condenser

Cooling Tower

STEAM SUPPLY

VACUUM PUMP

NCG VENT TO COOLING TOWER

BLOWDOWN

CWR

CWS

Steam Gas EJECTOR

INTER CONDENSER

VACUUM SEPARATOR

L. O. COOLER GEN AIR COOLER GLAND

STEAM COND.

DRIFT AND EVAPORATION LOSS

Hot Well PUMP

COOLING WATER PUMP

LEGENDP: kg/cm² a

H: kJ/kg G: kg/hr

T: deg C

5.94 P 2,755H

444,666 G

157.8 T

POWER SUMMARYGROSS POWER OUTPUT = 60,230 KWCOOLING WATER PUMPS = 1,740 KWCOOLING TOWER FANS = 1,065 KWVACUUM PUMP = 872 KWCONDENSATE PUMP = 52 KWTRANSFORMER LOSSES = 301 KWMISCELLANEOUS = 400 KWNET POWER OUTPUT 55,800 KW

DESIGN BASIS:NCG - 2.0%

WET BULB T - 230C

Main Transformer

GAS REMOVAL SYSTEM

COOLING WATER SYSTEM

Mahanagdong 60 MW Unit

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

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.

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TURNING GEAR

CASING

THRUST BEARING

JOURNAL BEARING

ROTOR

DIAPHRAGMS

SEALS

JOURNAL BEARING

STEAM INLET

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;


STEAM TURBINE COMPONENTS

2.1 Turbine Rotor

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

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• 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

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2.2 Turbine Casing




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2.3 Thrust and Journal Bearings



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2.4 Diaphragms


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

Turning gear

2.5 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

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


EXCITERROTATING RECTIFIER

ELECTRIC GENERATOR


• 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.

Generator Construction

ELECTRIC GENERATOR

GENERATOR STATOR

GENERATOR ROTOR

PMGEXCITERROTATING RECTIFIER

ROTOR FIELD WINDING

GENERATOR COMPONENTSAVR

Thyristor Firing Circuit

PMG

EXCITER

ROTATING RECTIFIER

STATOR WINDING

ROTOR FIELD WINDING

13.8 /138 kV Main

Transformer

138 kV Line

13.8 kV Line

SENSING VOLTAGE

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.

GENERATOR EXCITATIONAVR


PMG

EXCITER

ROTATING RECTIFIER

STATOR WINDING

ROTOR FIELD WINDING

13.8 /138 kV Main

Transformer

138 kV Line

13.8 kV Line

SENSING VOLTAGE

GENERATOR EXCITATIONAVR

PMG

EXCITER

ROTATING RECTIFIER

STATOR WINDING

ROTOR FIELD WINDING

13.8 /138 kV Main

Transformer

138 kV Line

13.8 kV Line

SENSING VOLTAGE

120 Vac, 420 Hz

100 Vdc

200 Vac

13.8 kVac

250 Vdc


Temp Inlet

8060

40

20

100

0

Temp Outlet

8060

40

20

100

0

COOLING SYSTEM

PMG

EXCITER

GENERATOR AIR COOLER

TOP VIEW

Temp Inlet

8060

40

20

100

0

Temp Outlet

8060

40

20

100

0

TOP VIEW

Temp Outlet

8060

40

20

100

0

Temp Inlet

8060

40

20

100

0

SIDE VIEW

PMG

EXCITER

GENERATOR AIR COOLER

GACCOOLING

WATER

FROM GENERATOR AIR MAKE-UP

• 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.

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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.

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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.

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Types of Cooling Towers

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Crossflow Tower

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.

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• 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.

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.

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53.3 0.622

85.7

v vva

a v

P Pmm P P P

Cooling Tower Calculations• 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.

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Cooling Tower CalculationsEnergy Balance• Heat Loss of Water=Heat Gain of Air

Lcpw(HWT-CWT)=G(h2-h1)(heat loss to evaporation is neglected)

Where:

h = enthalpy of dry air, Btu/lbm or J/kg

L = mass of cooling air, lb or kg/unit time

G=mass of circulating water, lb or kg/unit time

HWT = Hot water temperature

CWT=Cold water temperature

Cpw= specific heat of water, 4.186kj/kg-C

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Hot Air, h2

Cold Air, h1

Cold Water

Hot Water

Cooling Tower CalculationsMass 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:

(Mass of Water)in=Losses+ (Mass of Water)out

(If heat loss to evaporation is considered)

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

Where:

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

G=mass flow of circulating water, kg/unit time

HWT = Hot water temperature, ⁰CCWT=Cold water temperature, ⁰CCpw= specific heat of water, 4.186kj/kg-C

hf= enthalpy of saturated liquid at CWT, kJ/kg

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Hot Air, H2

Cold Air, H1

Cold Water

Hot Water

Cooling Tower Exercises• A cooling tower with a range of 20degF

receives 360,000 gal/min of 90degF circulating water. The outside air is at 60degF, 14.696psia, and 50% relative humidity. The exit air is at 80degF saturated. Calculate the mass of air required in ft3 /min. Neglect evaporation loss.

• Conversion: 7.48 gal = 1 ft3

Density = 62.428 lb/ ft3 • Cp water = 1 Btu/(lbm deg F)

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Hot water

In(A)

Hot waterOut(2)

ColdWaterOut(B)

ColdWater

In(1)

ha1 ω 1

hg1 hfB W B

hfA W A

ha2 ω 2

hg2

Cooling Tower Exercise Solution

• TA = 90degF

• TB = TA – Range = TA – 20degF = 70degF

• ha1 = h at 60degF and 50% RH

• ha2 = h @ 80deg F and 100% RH

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Hot water

in(A)

Hot airout(2)

Coldwaterout(B)

Coldairin(1)

ha1 ω 1

hg1 hfB W B

hfA W A

ha2 ω 2

hg2

Cooling Tower Exercise Solution• Heat Gained (Air)=Heat Loss (water)• Wa(ha2-ha1)=Wcpw(TA-TB )

• Dry Air Required:=9019440 lbDA/min

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Hot water

in(A)

Hot airout(2)

Coldwaterout(B)

Coldairin(1)

ha1 ω 1

hg1 hfB W B

hfA W A

ha2 ω 2

hg2

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.

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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

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Gas Removal SystemThis 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 – Geothermal 101


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Principle of OperationThe HEI* states “The operating principle of a steam ejector stage is that the pressure energy in the motive steam isconverted into velocity energy in the nozzle, and, this high velocity jet of steam entrains the vapor or gas beingpumped. The resulting mixture, at the resulting velocity, enters the diffuser where this velocity energy is converted topressure energy so that the pressure of the mixture at the ejector discharge is substantially higher than the pressure inthe suction chamber.” * See the HEI Standards for Steam Jet Vacuum Systems

1. Diffuser 6. Suction2. Suction Chamber 7. Discharge3. Steam Nozzle 8. Steam Inlet4. Steam Chest 9. Nozzle Throat5. Extension (if used) 10. Diffuser Throat

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

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End of PresentationThank You!

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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.

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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.

• EFFECT of MOISTURE and SOLID PARTICLES to First Stage Diaphragms