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Moving grids and time-dependent co-ordinate systems  187

                           7.5 Application to a moving boundary problem


                        The ‘in-cylinder’ calculation for an internal combustion engine gives us an application
                        of a time-dependent co-ordinate system. The physical domain in this case is the volume
                        between cylinder head and piston face. The piston moves between the B.D.C. (bottom
                        dead centre) position and the T.D.C. (top dead centre) position, and so the physical
                        domain has a moving boundary. Here we take a two-dimensional model of the situation,
                        as shown in Fig. 7.1.
                          Our objective is to obtain a set of co-ordinates such that the irregularly-shaped time-
                        dependent physical domain is mapped to a fixed rectangular computational domain at
                        all times. In Fig. 7.1 the shape of the cylinder head is given by the time-independent
                        function y = y b (x), while the shape of the moving piston is given by y = y a (x, t).
                          A possible time-dependent co-ordinate transformation is

                                                   ξ = x/l
                                                         y − y a (x, t)
                                                   η =               .                     (7.40)
                                                       y b (x) − y a (x, t)
                        where l is the width of the cylinder head. This clearly maps the physical domain
                        onto a unit square in the computational plane, in which the transport equations are
                        to be solved, once they have been transformed to ξ, η co-ordinates. To carry out the
                        transformation of transport equations, we require the various components of the metric
                        tensors and Christoffel symbols.
                          We begin with
                                                          1
                                                     ξ x =  ,  ξ y = 0.
                                                          l
                        Exercise 3. Show that

                                             ∂y a   dy b        −1                −1
                              η x =− (1 − η)    + η     (y b − y a )  ,  η y = (y b − y a )  .  (7.41)
                                             ∂x     dx
                        From eqns (1.160) and (1.162) the Jacobian of the transformation is
                                     √                                  g        −1
                                       g = x ξ y η − x η y ξ = g(ξ x η y − ξ y η x ) =  (y b − y a )  .
                                                                        l


                                          y
                                                                   h
                                                    y =y (x)
                                                       b
                                          D         C              1
                                                                     D   C
                                          A         B
                                                    y =y (x,t)     0  A  B
                                                       a
                                                                          1   x
                                          0
                                                   L     x
                        Fig. 7.1 Piston-cylinder assembly with irregular cylinder head and piston face.
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