Page 51 - DJJ20063- Thermodynamics 1
P. 51

DJJ20063- Thermodynamics 1



                            Feedback to Activity 2C



                                                                                                          o
               2.6    From  the  superheated  table  at  120  bar,  the  saturation  temperature  is  324.6  C.
                      Therefore, the steam is superheated.


                                                           o
                                                 o
                      Degree of superheat = 500  C – 324.6  C
                                                 o
                                                  = 175.4  C

                                             o
                      So, at 120 bar and 500  C, we have
                                    -2
                                        3
                      v = 2.677 x 10  m /kg
                      h = 3348 kJ/kg

                       From equation 8.6,

                      u = h – Pv

                                                                   -2
                                                                      3
                                                2
                                                      2
                        = 3348 kJ/kg – (120 x 10  kN/m )(2.677 x 10  m /kg)
                        = 3026.76 kJ/kg



               2.7    At 160 bar, hg = 2582 kJ/kg. This is less than the actual specific enthalpy of

                      3139 kJ/kg.  Hence, the steam is superheated.

                      From the superheated table at 160 bar, the specific enthalpy of 3139 kJ/kg is located at a

                                          o
                      temperature of 450  C.

                                                               o
                                                     o
                      The degree of superheat = 450  C – 347.3  C
                                                     o
                                                        = 102.7  C

                                                                       3
                                                                   -2
                                          o
                      At 160 bar and 450  C, we have v = 1.702 x 10  m /kg

                       From equation 2.6,

                      u = h – Pv

                                                                      3
                                                2
                                                      2
                                                                   -2
                        = 3139 kJ/kg – (160 x 10  kN/m )(1.702 x 10  m /kg)
                        = 2866.68 kJ/kg
                                                                                                 42 | P a g e
   46   47   48   49   50   51   52   53   54   55   56