Partial Differential Equations - Modelling and ... - ResearchGate

116 R. Sanders and A.M. Tesdall
impinging on a thin wedge. The numerical solution appears to be in agreement
with what is found for the model problems from the previous sections.
2 The Weak Shock Thin Wedge Limit
The compressible Euler equations are given by
∂ρ
+ ∇·ρu =0,
∂t
∂ρu
+ ∇·ρu ⊗ u + ∇p =0,
(2)
∂t
∂ρe
∂t + ∇·(ρe + p)u =0,
where ρ is the fluid density, u =(u, v) isthex-y velocity vector, p is the
pressure and e is the total energy per unit mass. The internal energy per unit
mass ε = e − 1/2|u| 2 ,andwetakep =(γ − 1)ρε for a calorically perfect gas
with the constant ratio of specific heats γ>1.
Consider an incident planar shock with Mach number M =1+ε 2 striking
a thin wedge with half angle θ w = aε, whereε>0 is destined to vanish. Take
the undisturbed upstream state U r as ρ = ρ r , u = v =0andp = p r , yielding
an upstream speed of sound c r = √ γp r /ρ r . From (1), calculate that U l is
(
p l
=
p r
ρ l
ρ r
=
1+ 4γ )
γ +1 ε2 + O(ε 4 ),
(
1+ 4 )
γ +1 ε2 + O(ε 4 ),
u l
= 4
c r γ +1 ε2 + O(ε 4 ),
v l
=
−4
c r γ +1 aε3 + O(ε 5 ).
Hunter and Brio [HB00] observed the scales shown in (3) and proposed an
asymptotic model based on
p = p r (1 + ε 2 ˆp), u = c r ε 2 û,
ρ = ρ r (1 + ε 2 ˆρ), v = c r ε 3ˆv,
and the stretched independent variables
ˆx = x − p(t)
ε 2 , ŷ = y ε ,
where p(t) is the location where the incident shock would (neglecting possible
interactions) strike the wedge wall at time t,
p(t) =c r cos(θ w )(1 + ε 2 ) t = c r cos(aε)(1 + ε 2 ) t ≈ c r (1 − (1 − a 2 /2)ε 2 ) t,
(3)