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Internal combustion engine power plant

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1. Diesel Engine Power PlantFuel tankEngineGeneratorCoolingTowerFuelPumpCooling WaterPumpAir inAir out
2. Four-Stroke Cycle Engine: An engine that completes one cycle in tworevolutions of the camshaft.IntakeCompressionPowerExhaustintake compression power exhaust
3. Two-Stroke Cycle Engine: An engine that completes one cycle in onerevolution of the camshaft.Intake & CompressionPower & ExhaustExhaust port Exhaust portIntake port Intake portIntake & Compression Power & Exhaust
4. Engine Performance1. Heat Supplied by Fuel (QS)QS = mF x HV KJ/hrWhere: mf – fuel consumption in kg/hrHV – heating value of fuel in KJ/kgKW4(60)NnLDPIP2miπ=2. Indicated Power (IP)Where: Pmi – indicated mean effective pressure, KPaL – length of stroke, mD – diameter of bore, mN = (RPM)/2 For 4-stroke single actingN = (RPM) For 4-stroke double actingN = (RPM) For 2-stroke single actingN = 2(RPM) For 2-stroke double acting
5. KW4(60)NnLDPBP2mbπ=3. Brake Power (BP)KW60,000TN2BPπ=Where: Pmb – brake mean effective pressure, KPaL – length of stroke, mD – diameter of bore, mN = (RPM)/2 For 4-stroke single actingN = (RPM) For 4-stroke double actingN = (RPM) For 2-stroke single actingN = 2(RPM) For 2-stroke double actingWhere: T – brake torque in N-mN – no. of (RPM)
6. 4. Friction Power (FP)FP = IP - BP5. Indicated Mean Effective Pressure (Pmi)KPaLSAPmi =Where: A’ – area of indicator card, cm2S – spring scale, KPa/cmL’ – length of indicator card, cm6. Brake Torque (T)T = (P – tare)R N-mWhere: P – gross load on scale, Ntare – tare weight, NR – length of brake arm, m
7. 7. Piston Speed (PS)PS = 2LN m/min8. Displacement Volume (VD)secmPBPVsecmPIPVsecm4(60)NnLDV3mbD3miD32D==π=9. Specific Fuel Consumptiona. Indicated Specific fuel consumptionhr-KWkgIPmm Ffi =
8. b. Brake Specific fuel consumptionhr-KWkgBPmm Ffb =c. Combined Specific fuel consumptionhr-KWkgGPmm Ffc =Where: GP – Generator power10. Heat Rate (HR)a. Indicated Heat Rate (HRi)hr-KWKJIPQHR SI =
9. b. Brake Heat Rate (HRb)hr-KWKJBPQHR Sb =c. Combined Heat Rate (HRc)hr-KWKJGPQHR Sc =11. Generator Speed (N)RPMn120f=NWhere: n – number of generator poles (usually divisible by 4)
10. 12. Mechanical Efficiency (ηm)100%xBPGP=gη100%xIPBP=mη13. Generator Efficiency (ηg)14. Indicated Thermal Efficiency (ei)100%xQ3600(IP)eSi =15. BrakeThermal Efficiency (eb)100%xQ3600(BP)eSb =
11. 16. Combined Thermal Efficiency (ec)100%xQ3600(GP)eSc =17. Indicated Engine Efficiency (ηi)100%xeeii =η18. Brake Engine Efficiency (ηb)100%xeebb =η19. Combined Engine Efficiency (ηc)100%xeec=ηWhere: e – cycle thermal efficiency
12. 20. Volumetric Efficiency (ηv)100%xVolumentDisplacemendrawnairofvolumeActualηv =shhsshTTBBPP =21. Correction Factor for Non Standard Condition Considering Pressure and Temperature Effects Considering Temperature Effects aloneshshTTPP =
13.  Considering Pressure Effects alonehsshBBPP =Note: From US Standard AtmosphereK10006.5h-TTHgmm100083.312hBBshsh°=−=Where:P – power, KWB – pressure, mm HgT – temperature,°Kh – elevation, metersSubscript:s – refers to sea levelh – refers to the elevation
14. ENGINE HEAT BALANCEQs = Q1 + Q2 + Q3 + Q4Where:Q1 – heat converted to useful workQ2 – heat loss to cooling waterQ3 – heat loss due to exhaust gasesQ4 – heat loss due to friction, radiation and unaccounted forQ1 = 3600(BP) KJ/hrQ2 = mwCpw(tw0 – tw1) KJ/hrQ3 = Qa + QbQa = mgCpg(tg – ta) KJ/hrQb = mf(9H2)(2442) KJ/hrQ4 = Qs – (Q1 + Q2 + Q3) KJ/hr
15. EngineQsQ2 Q3Q1Q4Qs = Q1 + Q2 + Q3 + Q4 + Q5