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5.2 RATIONAL EFFICIENCY       103




                  In the case of the irreversible engine even if T 0 ¼ T 3 the rational efficiency is still less than unity
               because of the increase in entropy caused by the irreversible expansion from 2 to 3. The loss of
               available energy, in this irreversibility, is depicted by area D in Fig 5.1(b). In the case of T 0 ¼ T 3 ,
                                                       T 3 ðs 3   s 3s Þ
                                              h ¼ 1              :                          (5.16)
                                               R
                                                       dE 12   dF 12
                  If the dead-state temperature is less than T 3 then the rational efficiency is even lower because of the
               loss of available energy shown as area C on Fig 5.1(b).
                  Up till now it has been assumed that the cycle is similar to a Carnot cycle, with isothermal heat supply
               and rejection.Such a cycle is typical of one in which theworking fluid is a vapour which can change phase.
               However, many cycles use air as a working fluid (e.g. Otto, Diesel, Joule cycles, etc. – see Chapter 3), and
               inthiscaseitisnot possibletosupplyand rejectheatatconstanttemperature.A general cycleofthistypeis
               shown in Fig 5.1(c), and it can be seen that the heat is supplied over a range of temperatures from T 1 to
               T 2 , and rejected over a range of temperatures from T 3 to T 4 . If only the heat engine is considered then it is
               possible to neglect the temperature difference of the heat supply: this is an external irreversibility.
                  However, it is not possible to neglect the varying temperature of heat rejection, because the engine
               is rejecting available energy to the surroundings. If the cycle is reversible, i.e. 1-2-3s-4-1, then the
               rational efficiency of the cycle is
                                                         w net
                                                   h ¼                                      (5.17)
                                                    R
                                                       b 2   b 1
                  If the cycle shown is a Joule (gas turbine) cycle (see Chapter 3) then
                    w net ¼ h 2   h 1  ðh 3s   h 4 Þ¼ b 2   b 1 þ T 0 ðs 2   s 1 Þ fb 3s   b 4 þ T 0 ðs 3s   s 4 Þg
                                                                                            (5.18)
                                            ¼ b 2   b 1   fb 3s   b 4 g
                  Hence, the net work is made up of the maximum net work supplied and the maximum net work
               rejected, i.e.

                                          w net ¼½ b w net Š supplied   ½ b w net Š rejected  (5.19)
                  These terms are defined by areas 1-2-3s-5-1 and 4-3s-5-4 respectively in Fig 5.1(c). Thus the
               rational efficiency of an engine operating on a Joule cycle is

                                            b 2   b 1  ðb 3s   b 4 Þ  b 3s   b 4
                                       h ¼                   ¼ 1                            (5.20)
                                        R
                                                 b 2   b 1         b 2   b 1
                  Equation (5.20) shows that it is never possible for an engine operating on a cycle in which the
               temperature of energy rejection varies to achieve a rational efficiency of 100%. This is simply because
               there will always be energy available to produce work in the rejected heat.
                  If the cycle is not reversible, e.g. if the expansion is irreversible, then the cycle is defined by 1-2-3-
               4-1, and there is an increase in entropy from 2 to 3. Eqn (5.18) then becomes
                     w net ¼ h 2   h 1  ðh 3   h 4 Þ¼ b 2   b 1 þ T 0 ðs 2   s 1 Þ  fb 3   b 4 þ T 0 ðs 3   s 4 Þg
                         ¼ b 2   b 1   ðb 3   b 4 Þ   T 0 ðs 3   s 3s Þ
                             |fflfflfflffl{zfflfflfflffl}  |fflfflfflfflfflffl{zfflfflfflfflfflffl}  |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}           (5.21)
                              area       area        area
                            1-2-3s-5-4-1  4-3-6-4  5-6-7-8-5
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