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18.2 LIQUEFACTION BY EXPANSION – METHOD (II)          431




               at B then the downstream temperature, T 2 , will always be less than T 1 . It is possible to analyse whether
               heating or cooling will occur by evaluating the sign of the derivative (vT/vp) h . This term is called the
               Joule–Thomson Coefficient, m.
                  If m < 0, then there can be either heating or cooling depending on whether the downstream pressure
               is between A and C, or to left of C. If m > 0, then the gas will be cooled on passing through the plug.
               (i.e. the upstream state point defined by (p 1 ,T 1 ) is to the left of B on the isenthalpic line.)
                  This situation can be analysed in the following way. From the Second Law of Thermodynamics
                                                 dh ¼ Tds þ vdp:                            (18.4)
                  But for this process dh ¼ 0, and thus, from Eqn (18.4),

                                                       vs

                                                 0 ¼ T      þ v:                            (18.5)
                                                       vp  h
                  If it is assumed that entropy is a continuous function of pressure and temperature, i.e. s ¼ s (p,T)then
                                                  vs         vs

                                            ds ¼       dp þ      dT                         (18.6)
                                                  vp  T      vT  p
               which can be rearranged to give
                                           vs       vs      vs    vT

                                                ¼       þ                                   (18.7)
                                           vp      vp       vT    vp
                                              h        T        p     h
                  Hence, substituting this expression into Eqn (18.5) gives

                                                vs       vs    vT
                                        0 ¼ T        þ               þ v                    (18.8)
                                                vp      vT     vp
                                                   T        P     h
                  Now, from thermodynamic relationships (Eqn (7.24)),

                                                     vs
                                                  T       ¼ c p
                                                     vT
                                                         p
               and, from the Maxwell relationships (Eqn (7.19(d)))

                                                 vs        vv
                                                      ¼         :
                                                 vp        vT
                                                     T         p
                  Thus

                                                   vv        vT
                                           0 ¼ T        þ c p     þ v                       (18.9)
                                                   vT        vp
                                                       p         h
               which may be rearranged to give
                                                     "          #
                                           vT      1     vv

                                                 ¼    T         v ¼ m:                     (18.10)
                                            vp     c p   vT
                                               h            p
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