∂V ∂F+∂τ ∂ξ + S = 0 (1.1)
V = α Uα
F = Uα
! (1.2)
! (1.3)
and
U 2α + αF (α) − Γ
S = 0! (1.4)− ρc2 (ρg − ρcoα3/2
The computational domain is divided into equal shells with a mesh spacing Δξ in order to solve equation 4.7. The part of the tube for which ξj−1 ≤ ξ ≤ ξj is then the jth cell with interfaces at j-1/2 and j+1/2 so that ξj−1 = (j − )Δξ. if 4.7 is integrated, one gets
where the flux vectorFj−1/2 is the flux at the interface between cells j-1 and j. Using this, equation 4.11 can be written as;
d
(ξj − ξj−1)dτ Vj + Fj+ 1 − Fj− 1
= (ξj − ξj−1)Sj (1.6)
Figure 1.1: Discretization of the solution vector along the grid showing the discontinuities at each cell interface are the mean value of the source term and the mean value of V in the jth cell respectively. The equations 4.11 and 4.12 are the exact values of the conservation quantities in V, therefore a conservative scheme can be constructed by choosing a suitable approximations to the flux and the source terms and intergrating over the time step to get equation 4.15
(ξ − ξj−1)(Vjk+1 − Vjk) + ∫ τk+1 Fj+ 1 − Fj− 1 dτ = ξj − ξj−1 ∫ τk+1
Sjdτ (1.9)
the solution at the time step k in the cell j is denoted by Vjk. If Fj+ 1 k+ 1 and Sjk+ 1
The 2nd order Godunov scheme is an explicit second order finite volume scheme Roe and Krigbaum (1964), Pedley et al. (1996). The method captures the elastic jumps and is upwind biased. Steps involved are;
The grid is divided into N equal space cells horizontally ant T tines vertically.
2. First step order-its is assumed that the solution at time step k is known, the solution vector V and source terms S are uniform in each cell and constant over time step. These constant quantities are the ones that give rise to discontinuities at the cell interfaces. The first solution at time k + 1/2 is given by;
The fluxes are determined by ignoring the source terms (S=0) and solving the Riemann problem locally at the boundary of the cells. Equation 4.1 (with S = 0) is said to have initial value problem with discontinuous initial data of the form;
V (0) = VL = constanttotheleftoftheinterface.
Vr = Constanttotherightoftheinterface
The Riemann problem is like shock tube problem of gas dynamics and dam break problem of shallow water waves in open channels. The Riemann solver aims to calculate the position and type of the left and right waves resulting from the breakups and state at the interface which remains constant after the initial breakups. hence if the solution of the Riemann problem is V ∗(Vr, Vl) then the fluxes are found by F = F (V ∗) at all interfaces and it can be generalized as;
Fj− 1 k+ 1 = F (V ∗(Vj−1kVjk)) (1.15)
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