A steam turbine has extractions to provide heating steam for

A steam turbine has extractions to provide heating steam for the feedwater heaters in a power plant. The following parameters are known: a. Isentropic Efficiency: 0.85 b. Inlet flow: 500 kg/s c. Inlet pressure: 4500 kPa d. Inlet Temperature: 330 C e. Extraction 1: Pressure=3300 kPa, flow rate=50 kg/s f. Extraction 2: Pressure =2200 kPa, flow rate=40 kg/s g. Extraction 3: Pressure=1600 kPa, flow rate=30 kg/s h. Extraction 4: Pressure=1200 kPa, flow rate=20 kg/s i. Turbine Exhaust: Pressure=100 kPa Determine the power, in MW, of the turbine

Solution

ENTHALPY AT EXTRACTION 1 AT 3300KPA= 2802KJ/Kg

ENTHALPY AT EXTRACTION 2 AT 2200KPA= 2799KJ/Kg

ENTHALPY AT EXTRACTION 3 AT 1600KPA= 2791KJ/Kg

ENTHALPY AT EXTRACTION 4 AT 1200KPA= 2782KJ/Kg

amount of heat loss in extraction 1=2802*50=140100 KW

amount of heat loss in extraction 2=2799*40=111960KW

amount of heat loss in extraction 3=2797*30=83910KW

amount of heat loss in extraction 4=2782*20=55640KW

TOTAL HEAT LOSS IN EXTRACTION=391610 KW

HEAT LOSS IN TURBINE EXHAUST=ENTHALPY AT 100KPA*FLOWRATE

=2675KJ/ Kg *140= 374500 KW

TOTAL HEAT LOSS=HEAT LOSS IN ALL EXTRACTION+HEAT LOSS IN TURBINE EXHAUST

=391610KW + 374500KW= 766110 KW

ENERGY AVAILABLE AT 4500 KPA =2796 KJ/KG*500=1398000 KW

POWER= ENERGY AVAILABLE- TOTAL ENERGY LOST

=1398000-766110=631890 KW=631.890 MW