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2.9 EQUILIBRIUM       19




                  •  slow isothermal compression or expansion of gases (so that no temperature gradient is required
                     to transfer the heat into or out of the system)
                  •  electrolysis (with no resistance in the electrolyte)


               2.8.2 IRREVERSIBLE PROCESSES
               In reality all the processes have some losses, which might be friction, heat losses, hysteresis, etc.
               Hence, all real processes are irreversible. This does not mean that it is impossible for a system to
               perform a cycle made up of irreversible processes but simply that the surroundings will have been
               altered by the cycle having been performed. A number of common irreversible processes will be listed.

                  •  friction between solids, or in fluids
                  •  unresisted expansions
                  •  heat transfer across finite temperature differences
                  •  combustion – because the fuel and air are turned into carbon dioxide and water
                  •  mixing of unlike, and miscible fluids
                  •  hysteresis processes (internal friction in metals, hysteresis in electrical systems)
                  •  plastic deformation of materials
                                                                         2
                  •  electrical resistance to current flow (production of heat through I R losses)

               2.9 EQUILIBRIUM
               Most texts on thermodynamics restrict themselves to dealing exclusively with equilibrium thermo-
               dynamics. This book will also focus on equilibrium thermodynamics but the effects of making this
               assumption will be explicitly borne in mind. The majority of processes met by engineers are in
               thermodynamic equilibrium, but some important processes have to be considered by nonequilibrium
               thermodynamics. Most of the combustion processes that generate atmospheric pollution include
               nonequilibrium effects, and carbon monoxide (CO) and oxides of nitrogen (NO x ) are both the result of
               the inability of the system to reach thermodynamic equilibrium in the time available.
                  There are four kinds of equilibrium, and these are most easily understood by reference to simple
               mechanical systems shown in Fig. 2.2. Here a marble is placed on a number of different surfaces, and
               the effect of a perturbation is considered: the result is then related to systems in similar thermodynamic
               states.
               1. Stable equilibrium: marble in bowl (depicted in Fig. 2.2(a)):
                     For stable equilibrium DEÞ > 0 and DSÞ < 0, where D indicates the sum of terms in the
                                           S
                                                      E
                     Taylor’s series. 1
                     In this case any deflection results in motion back to the stable condition.
               2. Neutral equilibrium: marble in trough (depicted in Fig 2.2(b)):
                     DEÞ ¼ 0 and DSÞ ¼ 0 along the trough axis: The marble is in equilibrium at any point
                        S           E
                     along the axis.


                                                                       2
                                                                                            2
                                                                              3
               1                                                dS   1 d S  2  1 d S  3    1 d S  2
                The difference between DS and dS. Consider the Taylor’s series DS ¼  Dx þ  2 dx 2 Dx :
                                                                dx   2 dx 2 Dx þ 6 dx 3 Dx þ .ydS þ
               Thus dS is the first term only of the Taylor’s series, and dS would equal zero at the top of the bowl in Fig. 2.2(c),
                        2                                     2
                      1 d S  2                               d S
               but DS ¼  2 dx 2 Dx would not be zero because the second derivative  dx 2 is not zero.
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