Page 55 - DJJ20063- Thermodynamics 1
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DJJ20063- Thermodynamics 1



                     ii.      Double Interpolation



                     In some cases a double interpolation is necessary, and it’s usually used in the Superheated

                     Steam  Table.    Double  interpolation  must  be  used  when  two  of  the  properties  (eg.
                     temperature and pressure) are not tabulated in the Steam Tables.  For example, to find the

                                                                     o
                     enthalpy of superheated steam at 25 bar and 320 C, an interpolation between 20 bar and
                                                                                                           o
                                                                                                o
                     30 bar is necessary (as shown in example 2.9).  An interpolation between 300 C and 350 C
                     is also necessary.


                     Example 2.9


                                                                                                      o
                               Determine the specific enthalpy of superheated steam at 25 bar and 320 C.


                     Solution to Example 2.9

                              An extract from the Superheated Steam Tables:

                                      o
                                         t( C)          300                320                 350
                                p(bar)

                                      20               3025                 h1                 3138

                                      25                                    h

                                      30               2995                 h2                 3117


                                                                                   o
                              Firstly, find the specific enthalpy (h1) at 20 bar and 320  C;

                              At 20 bar,

                                     T

                                   350                                    h  −  3025  3138 −  3025
                                                                           1       =
                                                                         320 − 300    350 − 300
                                   320

                                   300
                                                                                  h  =  3070  2 .  kJ/kg
                                                                                  1
                                                                  h
                                        3025    h1    3138



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