Mathematical Physics - Volume II - Numerical Methods

Chapter 4. Finite volume method

128

where the quantity ε is determined by the relation ε = max h 0 , ( λ ( j ) − λ ( j ) L ) , ( λ ( j ) D − λ ( j ) ) i .

(4.41)

Zero, as the value of the function in the expression (4.41), eliminates introduced correction in case of shock wave. The alternative procedure [13] is reduced to relations

mod = ( |

λ ( j ) | ,

if | λ ( j ) | ≥ ε ,

| λ ( j ) |

(4.42)

( j ) 2

1 2 ( λ

/ ε + ε ) , if | λ ( j ) | < ε ,

thus ensuring the continuity of the first derivative of quantity | λ ( j ) | . When fluid flows through a pipe with variable cross area section S ( x ) , when cross area section change is small compared to the pipe length, the flow can be considered quasi one-dimensional and the first of the equations (4.30) takes the form ∂ ∂ t ( S q ) + ∂ ∂ x ( S F ) − H = 0 , (4.43) bearing in mind that vectors q and F are defined by the relations (4.31) and (4.32), re spectively. The vector H , present in equation (4.43), introduces the variation of the cross section area S ( x ) and is determined by the relation

dS dx  

0 p 0   .

H =

(4.44)

Implicit difference scheme applicable to Euler equations (4.43) is defined by an algorithm

q n + 1 − q n ∆ t

∂ ∂ x

+

( S F )

n + 1

− H n + 1 = 0 .

S

(4.45)

In the equation (4.43), bearing in mind implicit formulation, the second and the third term depend on the flow variables in the following moment, defined by index n + 1. If the flow variables vector increment in the time interval determined by the indices n and n + 1 is ∆ q = q n + 1 − q n , (4.46) applying local linearization with respect to known values of flow variables at time level n , it can be written for the flux vector F at time n + 1

∂ F n ∂ q

F n + 1 = F n +

∆ q + O ( ∆ t 2 ) .

(4.47)

From the equation (4.35), numerical flux at the boundary m is obtained by the first order accuracy flux vector linearization in the cell m − 1 / 2, ie. F n + 1 m = F n + 1 m − 1 / 2 + A − m ( q n + 1 m + 1 / 2 − q n + 1 m − 1 / 2 ) , (4.48)