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  • OpenMollier Chart
  • OpenMollier Chart
  • OpenMollier Chart
  • Serves as standard of comparison for power cycles
  • Process:A-B: Liquid-Sat. Liquid,B-C: Sat. liquid-Two Phase,C-1: Two-phase-Sat Steam; 1-2: Sat. Steam-Two-Phase(Turbine), 2-3: Condenser Work
  • OpenMollier Chart
  • -Develop an understanding on equipment function, construction and related calculations
  • Magnetic field is rotating, armature is fixed
  • Magnetic field adjustments is normally done automatically in response to changes in loads.There are occasions when an operator must make adjustments manually
  • A drop in that temperature would result in condensation and the new cooler air would also be saturated. An increase in that temperature would make it unsaturated so that it could accept more water vapour.
  • Range is determined not by the cooling tower, but by the process it is serving. The range at the exchanger is determined entirely by the heat load and the water circulation rate through the exchanger and on to the cooling water. Range oC = Heat Load in kcals/hour / Water Circulation Rate in LPH the heat load and the water circulation rate through the exchanger and on to the cooling water. Range oC = Heat Load in kcals/hour / Water Circulation Rate in LPH
  • Saturated Air. This is air that can accept no more water vapour at its given temperature. A drop in that temperature would result in condensation and the new cooler air would also be saturated. An increase in that temperature would make it unsaturated so that it could accept more water vapour. This partial pressure of water vapour is saturated air equals the saturation pressure Psat (obtained from steam tables) at the air temperature. For example, saturated air at 60 deg F and 14.696 psia, would have: Partial Pressure of water vapour Psat = 0.256 psia Partial Pressure of dry air Pa = 14.440 psia Total Pressure = 14.696 psiaFor the previously given examples, air at 60degF, 14.696 psia, and 50 percent Φ would have ω = 0.005465 lbm water vapor/lbmda and ωsat = 0.0113.
  • Transcript

    • 1. POWER PLANT MODULEBenjie S. AndresPower Department
    • 2. 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
    • 3. I. Fundamentals of ThermodynamicsI. 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
    • 4. I. Fundamentals of ThermodynamicsPressure: 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
    • 5. I. Fundamentals of ThermodynamicsI. 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
    • 6. I. Fundamentals of Thermodynamics1.2 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
    • 7. I. Fundamentals of Thermodynamics1.3 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
    • 8. I. Fundamentals of ThermodynamicsI. Fundamentals of Thermodynamics1.3.1 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
    • 9. I. Fundamentals of Thermodynamics 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
    • 10. EXERCISE II. Fundamentals of Thermodynamics 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=50kPaI. Fundamentals of Thermodynamics B. Compute for the properties of steam given that the enthalpy and pressure is 2500 kJ and 120 kPa. 10
    • 11. MOLLIER CHART OR HS DIAGRAM 11
    • 12. I. Fundamentals of Thermodynamics1.6 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
    • 13. I. Fundamentals of Thermodynamics 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
    • 14. I. Fundamentals of Thermodynamics1.4 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
    • 15. I. Fundamentals of Thermodynamics1.5 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
    • 16. I. Fundamentals of Thermodynamics 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
    • 17. I. Fundamentals of Thermodynamics 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
    • 18. I. Fundamentals of Thermodynamics1.6 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
    • 19. EXERCISE III. Fundamentals of Thermodynamics • 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
    • 20. RECAP PART I 21
    • 21. 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
    • 22. THE FIRST LAW OF THERMODYNAMICS APPLIED:ENERGYCAN 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
    • 23. II. POWER PLANT COMPONENTS 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
    • 24. II. POWER PLANT COMPONENTSSTEAM 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
    • 25. II. POWER PLANT COMPONENTSSTEAM TURBINE COMPONENTS 2.1 Turbine Rotor Typically, higher pressure sections are impulse type and lower pressure stages are reaction type Date
    • 26. II. POWER PLANT COMPONENTSSTEAM TURBINE COMPONENTS 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
    • 27. STEAM TURBINE COMPONENTSSTEAM TURBINE COMPONENTS 2.3 Thrust and Journal Bearings 29
    • 28. II. POWER PLANT COMPONENTSSTEAM TURBINE COMPONENTS 2.4 Diaphragms 30
    • 29. II. POWER PLANT COMPONENTSSTEAM TURBINE COMPONENTS 2.5 Turning Gear A turbine component designed to turn the rotor slowly during startups and shutdowns, thus minimizing rotor eccentricity Turning gear
    • 30. II. POWER PLANT COMPONENTSSTEAM TURBINE COMPONENTS 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
    • 31. II. POWER PLANT COMPONENTS 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
    • 32. II. POWER PLANT COMPONENTS 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
    • 33. II. POWER PLANT COMPONENTS ELECTRIC GENERATOR ROTATING RECTIFIER EXCITER
    • 34. 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.
    • 35. ROTOR FIELD WINDING ROTATING RECTIFIER EXCITER PMG
    • 36. 138 kV Line 13.8 /138 kV AVR Main Transformer Thyristor Firing SENSING Circuit VOLTAGE 13.8 kV Line STATOR WINDING PMG EXCITERROTATING RECTIFIER ROTOR FIELD WINDING
    • 37. 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.
    • 38. 138 kV Line 13.8 /138 kV AVR Main TransformerThyristor Firing SENSING Circuit VOLTAGE 13.8 kV Line STATOR WINDINGPMGEXCITER ROTATING RECTIFIER ROTOR FIELD WINDING
    • 39. 138 kV Line 13.8 /138 kV AVR Main Transformer Thyristor Firing SENSING Circuit VOLTAGE 13.8 kV Line STATOR WINDING 13.8 kVac 120Vac, 42 0 Hz 100 Vdc 200 Vac 250 Vdc PMG EXCITE R ROTATING RECTIFIER ROTOR FIELD WINDING
    • 40. 60 80 60 80 40 40 100 100 20 GENERATOR 20 0 0 AIR COOLERPMG EXCITER
    • 41. 60 80 60 8040 40 100 100 20 20 0 0
    • 42. GENERATOR AIR COOLERPMG EXCITER 60 80 40 100 20 0 60 80 40 100 20 0 GAC FROM GENERATOR COOLING AIR MAKE-UP WATER
    • 43. 46
    • 44. II. POWER PLANT COMPONENTS • 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
    • 45. II. POWER PLANT COMPONENTS 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
    • 46. II. POWER PLANT COMPONENTS5.0 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
    • 47. Types of Cooling Towers Crossflow Tower 50
    • 48. 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
    • 49. 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
    • 50. Cooling Tower Calculations 53
    • 51. Cooling Tower Calculations • 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
    • 52. 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. 55
    • 53. Cooling Tower CalculationsEnergy Balance• Heat Loss of Water=Heat Gain of Air Hot Air, h2Lcpw(HWT-CWT)=G(h2-h1)(heat loss to evaporation is neglected) Hot WaterWhere:h = enthalpy of dry air, Btu/lbm or J/kgL = mass of cooling air, lb or kg/unit time Cold Air, h1G=mass of circulating water, lb or kg/unit timeHWT = Hot water temperature Cold WaterCWT=Cold water temperatureCpw= specific heat of water, 4.186kj/kg-C 56
    • 54. 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: Hot Air, H2(Mass of Water)in=Losses+ (Mass of Water)out(If heat loss to evaporation is considered) Hot WaterLcpw(HWT-CWT)+G(H2-H1)(hf)=G(h2-h1)Where: Cold Air, H1H=humidity ratio of inlet and outlet air, kg vapor/kg DAh = enthalpy of dry air, KJ/kgL = mass flow of cooling air, kg/unit time Cold WaterG=mass flow of circulating water, kg/unit timeHWT = Hot water temperature, ⁰CCWT=Cold water temperature, ⁰CCpw= specific heat of water, 4.186kj/kg-Chf= enthalpy of saturated liquid at CWT, kJ/kg 57
    • 55. 58
    • 56. 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
    • 57. Cooling Tower ExerciseSolution • 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
    • 58. Cooling Tower ExerciseSolution 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
    • 59. II. POWER PLANT COMPONENTS6.0 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
    • 60. 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
    • 61. II. POWER PLANT COMPONENTS  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 –
    • 62. Principle of OperationThe HEI* states “The operating principle of asteam ejector stage is that the pressureenergy in the motive steam isconverted into velocity energy in thenozzle, and, this high velocity jet of steamentrains the vapor or gas beingpumped. The resulting mixture, at theresulting velocity, enters the diffuser wherethis velocity energy is converted topressure energy so that the pressure of the 1. Diffuser 6. Suction 2. Suction Chamber 7. Dischargemixture at the ejector discharge is 3. Steam Nozzle 8. Steam Inletsubstantially higher than the pressure in 4. Steam Chest 9. Nozzle Throatthe suction chamber.” * See the HEI 5. Extension (if used) 10. Diffuser ThroatStandards for Steam Jet Vacuum Systems 65
    • 63. II. POWER PLANT COMPONENTSMalitbog 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
    • 64. End of PresentationThank You!
    • 65. 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
    • 66. 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
    • 67. • EFFECT of MOISTURE and SOLID PARTICLES to First Stage Diaphragms