Page 36 - Advanced Gas Turbine Cycles
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Chapter 2


       REWERSIBILITY AND AVAILABILITY








       2.1.  Introduction

         In  Chapter  1, the  gas  turbine plant  was  considered briefly  in  relation  to  an  ideal
       plant based on the Carnot cycle. From the simple analysis in Section 1.4, it was explained
       that the closed cycle gas turbine failed to match the Carnot plant in thermal efficiency
       because of
       (a)  the ‘6 effect’ (that heat is not supplied at the maximum temperature and heat is not
           rejected at the minimum temperature) and
       (b)  the ‘u effect’ (related to any entropy increases within the plant, and the consequent
           ‘widening’ of the cycle on the T, s diagram).
         Since these were preliminary conclusions, further explanations of these disadvantages
       are  given  using  the  second  law  of  thermodynamics  in  this  chapter.  The  ideas  of
       reversibility, irreversibility, and the thermodynamic properties ‘steady-flow availability’
       and ‘exergy’ are also developed.
         In defining the thermal efficiency of the closed gas turbine cycle, such as the one shown
       in Fig.  1.2, we employed the first law of thermodynamics (in the form of the steady-flow
       energy equation round the cycle), which states that the heat supplied is equal to the work
       output plus the heat rejected, i.e.




       Here W is the net work output, i.e. the difference between the turbine work output ( WT) and
       the work required to drive the compressor (W,), W = WT - W,.
         For rhe open circuit gas turbine of Fig. 1.3, if the reactants (air Ma and fuel Mf) enter at
       temperature  To, and the exhaust products (Ma + Mf) leave at temperature T4, then the
       steady-flow energy equation yields




       where subscripts R  and P refer to reactants and products, respectively, and it has been
       assumed that there are no heat losses from the plant. If  we now consider unit air flow at
       entry with a fuel  flow .f  (=  Mf/Ma) then the  enthalpy flux HRo is equal to the sum of
       the  enthalpy  (hat)) and  the  enthalpy  of  the  fuel  flow ,f  supplied  to  the  combustion
      chamber  (fhfo), both  at  ambient temperature To, and  the  enthalpy of  the  exhaust gas

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