Steam Turbine efficiency
Efficiency= Energy output/ Energy input
Heat input
= Steam inlet flow in TPH * Enthalpy of inlet steam (Kca/Kg)
= 168*811
= 136248 Kcal
Heat out
put = Power in MW* Heat equivalent of
Power
=30*860
=25,800 Kcal
Efficiency of Steam Turbine = Heat output/ Heat input
=
25,800/136248
= 18.93%
Rankine cycle efficiency
for given plant:
W1=30 MW
W2= 1MW (assumed)
Q1= Heat
input to a Boiler
=m*(h1-h4)
=336*(811-150) (where 336 is total steam generation in
Boilers)
. =222096
=246773 Kcal(assuming Boiler efficiency of
90%) .
In a power
plant that produces both Steam and power, we must give credit to Steam export
also
Useful
output as steam export= Steam export flow*Heat value (Kcal/Kg)
=150* 760
=114000 Kcal
Efficiency of cycle is =
((W1-W2)+( Heat value of export steam)) /Q1
=((60-2)*860)+114000)/ (246773)
=66.40%
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