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398    CHAPTER 17 GAS TURBINES





                                                   Maximum      3
                                                  temperature
                                    Temperature, T




                                              Q 1







                                          2’              p     4
                                                          i
                                                   2
                                                                       Q
                                                                 Q      3
                                                                  2
                                           1’       1
                                                  Intercooling
                                                                  Entropy, S
             FIGURE 17.16
             T–s diagram for gas turbine cycle with intercooling.
             17.2.3.1 Optimum pressure ratio for reheating of gas turbines with heat exchange
             In the same way as cycles were optimised previously to produce maximum work there must be an
             optimum pressure ratio at which reheating or intercooling should take place. This can be evaluated by
             consideration of the equations defining cycle efficiency.
              Reheat cycle
                Assume that reheat occurs at a pressure ratio given by
                                                    k 1      k 1

                                                 p R  k    p 4  k
                                           s R ¼        ¼                                (17.37)
                                                 p 1       p 1
                The efficiency of the cycle is given by Eqn (17.4):
                                                       w
                                                   h ¼
                                                       q in
                                                         vh
             and hence the maximum efficiency will occur when  ¼ 0:
                Now                                     vs R
                                            vh    1 vw     w vq in
                                               ¼           2                             (17.38)
                                            vs R  q in vs R  q vs R
                                                           in
             i.e. when
                                                  vq in  vw
                                                 h    ¼                                  (17.39)
                                                  vs R  vs R
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