Page 186 - Design of Solar Thermal Power Plants
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3.3 THERMAL PERFORMANCE OF PARABOLIC TROUGH COLLECTOR  171

              In order to eliminate T b from Eq. (3.40), Eq. (3.42) and Eq. (3.43) are
           substituted into it, then the two energy balance equations can be com-
           bined as
                       2
                      d T f  C R þ C R þ C R  bf  dT f  T f   A a S
                             b ba
                                     f
                                       ba
                                            f
                 C C      þ                        þ       ¼
                  b f
                      ds 2         R bf R ba     ds  R bf R ba  R bf
                                                                       (3.44)

                  R ba þ R bf                  dT fo  dT fi   T a
                                           _
                           _ mc f ðT fo   T fi Þ  C b mc f     þ
                    R R ba                      ds    ds    R R ba
                                                              bf
                     bf
              Outlet temperature T fo of heat-transfer fluid within the parabolic
           trough collector is selected as the lumped temperature of heat-transfer
           fluid within the metal absorber tube. Thus Eq. (3.44) is changed into
                                       2
               1 þðR þ R Þ _ mc f     d T fo    1 þ _ mc R bf  R þ R ba  !  dT fo
                                                    f
                                                           bf
                     bf
                          ba
                               T fo ¼                   þ
                   C C R R ba          ds 2      C R bf   C R R ba   ds
                                                  f
                                                           b bf
                        bf
                    b f
                                       _ mc dT fi  ðR þ R Þ _ mc f
                                                       ba
                                         f
                                                  bf
                                     þ        þ             T fi
                                       C f  ds   C C R R ba
                                                  b f
                                                      bf
                                          1            A a
                                     þ          T a þ      S           (3.45)
                                      C C R R ba    C C R bf
                                                      b f
                                       b f
                                            bf
              Based on the above, a differential equation can be deduced
                        2
                       d T fo  dT fo         dT fi
                           þ A     þ BT fo ¼ C   þ DT fi þ ES þ FT a   (3.46)
                       ds 2     ds            ds
           in which
                                    1 þ _ mc f R bf  R bf þ R ba
                                A ¼           þ                        (3.47)
                                      C f R bf  C b R bf R ba
                                      1 þðR bf þ R ba Þ _ mc f
                                  B ¼                                  (3.48)
                                         C b C f R bf R ba
                                             _ mc f
                                        C ¼                            (3.49)
                                             C f
                                       ðR þ R Þ _ mc f
                                               ba
                                          bf
                                   D ¼                                 (3.50)
                                         C b C f R bf R ba
                                            A a
                                      E ¼                              (3.51)
                                          C C R bf
                                           b f
                                             1
                                     F ¼                               (3.52)
                                         C C R R ba
                                          b f
                                              bf
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