7.5 Finite Nonlinear WavesNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
Starting point: the governing flow equations on partial differential form
Continuity equation:
∂ ρ ∂ t + u ∂ ρ ∂ x + ρ ∂ u ∂ x = 0 \frac{\partial \rho}{\partial t}+u\frac{\partial \rho}{\partial x}+\rho\frac{\partial u}{\partial x}=0 ∂ t ∂ ρ + u ∂ x ∂ ρ + ρ ∂ x ∂ u = 0 Momentum equation:
∂ u ∂ t + u ∂ u ∂ x + 1 ρ ∂ p ∂ x = 0 \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho}\frac{\partial p}{\partial x}=0 ∂ t ∂ u + u ∂ x ∂ u + ρ 1 ∂ x ∂ p = 0 Any thermodynamic property can be expressed as a function of two other thermodynamic properties. This means that we can get density as a function of pressure and entropy: ρ = ρ ( p , s ) \rho=\rho(p,s) ρ = ρ ( p , s )
d ρ = ( ∂ ρ ∂ p ) s d p + ( ∂ ρ ∂ s ) p d s d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp+\left(\frac{\partial \rho}{\partial s}\right)_p ds d ρ = ( ∂ p ∂ ρ ) s d p + ( ∂ s ∂ ρ ) p d s Assuming isentropic flow d s = 0 ds=0 d s = 0
d ρ = ( ∂ ρ ∂ p ) s d p d\rho=\left(\frac{\partial \rho}{\partial p}\right)_s dp d ρ = ( ∂ p ∂ ρ ) s d p ∂ ρ ∂ t = ( ∂ ρ ∂ p ) s ∂ p ∂ t = 1 a 2 ∂ p ∂ t ∂ ρ ∂ x = ( ∂ ρ ∂ p ) s ∂ p ∂ x = 1 a 2 ∂ p ∂ x \begin{aligned}
&\frac{\partial \rho}{\partial t}=\left(\frac{\partial \rho}{\partial p}\right)_s\frac{\partial p}{\partial t}=\frac{1}{a^2}\frac{\partial p}{\partial t}\\
& \\
&\frac{\partial \rho}{\partial x}=\left(\frac{\partial \rho}{\partial p}\right)_s\frac{\partial p}{\partial x}=\frac{1}{a^2}\frac{\partial p}{\partial x}
\end{aligned} ∂ t ∂ ρ = ( ∂ p ∂ ρ ) s ∂ t ∂ p = a 2 1 ∂ t ∂ p ∂ x ∂ ρ = ( ∂ p ∂ ρ ) s ∂ x ∂ p = a 2 1 ∂ x ∂ p Now, insert (7.112) (7.108)
∂ p ∂ t + u ∂ p ∂ x + ρ a 2 ∂ u ∂ x = 0 \frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}+\rho a^2\frac{\partial u}{\partial x}=0 ∂ t ∂ p + u ∂ x ∂ p + ρ a 2 ∂ x ∂ u = 0 Dividing (7.113) ρ a \rho a ρ a
1 ρ a ( ∂ p ∂ t + u ∂ p ∂ x ) + a ∂ u ∂ x = 0 \frac{1}{\rho a}\left(\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}\right)+a\frac{\partial u}{\partial x}=0 ρ a 1 ( ∂ t ∂ p + u ∂ x ∂ p ) + a ∂ x ∂ u = 0 A slightly modified form of the momentum equation is obtained by multiplying and dividing the last term by a a a
∂ u ∂ t + u ∂ u ∂ x + 1 ρ a ( a ∂ p ∂ x ) = 0 \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho a}\left(a\frac{\partial p}{\partial x}\right)=0 ∂ t ∂ u + u ∂ x ∂ u + ρ a 1 ( a ∂ x ∂ p ) = 0 If the continuity equation on the form (7.114) (7.115)
[ ∂ u ∂ t + ( u + a ) ∂ u ∂ x ] + 1 ρ a [ ∂ p ∂ t + ( u + a ) ∂ p ∂ x ] = 0 \left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]=0 [ ∂ t ∂ u + ( u + a ) ∂ x ∂ u ] + ρ a 1 [ ∂ t ∂ p + ( u + a ) ∂ x ∂ p ] = 0 If, instead, the continuity equation on the form (7.114) (7.115)
[ ∂ u ∂ t + ( u − a ) ∂ u ∂ x ] + 1 ρ a [ ∂ p ∂ t + ( u − a ) ∂ p ∂ x ] = 0 \left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]+\frac{1}{\rho a}\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]=0 [ ∂ t ∂ u + ( u − a ) ∂ x ∂ u ] + ρ a 1 [ ∂ t ∂ p + ( u − a ) ∂ x ∂ p ] = 0 Since u = u ( x , t ) u=u(x,t) u = u ( x , t )
d u = ∂ u ∂ t d t + ∂ u ∂ x d x = ∂ u ∂ t d t + ∂ u ∂ x d x d t d t du=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}dx=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}\frac{dx}{dt}dt d u = ∂ t ∂ u d t + ∂ x ∂ u d x = ∂ t ∂ u d t + ∂ x ∂ u d t d x d t Now, let d x / d t = u + a dx/dt=u+a d x / d t = u + a
d u = ∂ u ∂ t d t + ( u + a ) ∂ u ∂ x d t = [ ∂ u ∂ t + ( u + a ) ∂ u ∂ x ] d t du=\frac{\partial u}{\partial t}dt+(u+a)\frac{\partial u}{\partial x}dt=\left[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}\right]dt d u = ∂ t ∂ u d t + ( u + a ) ∂ x ∂ u d t = [ ∂ t ∂ u + ( u + a ) ∂ x ∂ u ] d t which is the change of u u u d x / d t = u + a dx/dt=u+a d x / d t = u + a
In the same way
d p = ∂ p ∂ t d t + ∂ p ∂ x d x = ∂ p ∂ t d t + ∂ p ∂ x d x d t d t dp=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}dx=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}\frac{dx}{dt}dt d p = ∂ t ∂ p d t + ∂ x ∂ p d x = ∂ t ∂ p d t + ∂ x ∂ p d t d x d t and thus, in the direction d x / d t = u + a dx/dt=u+a d x / d t = u + a
d p = ∂ p ∂ t d t + ( u + a ) ∂ p ∂ x d t = [ ∂ p ∂ t + ( u + a ) ∂ p ∂ x ] d t dp=\frac{\partial p}{\partial t}dt+(u+a)\frac{\partial p}{\partial x}dt=\left[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}\right]dt d p = ∂ t ∂ p d t + ( u + a ) ∂ x ∂ p d t = [ ∂ t ∂ p + ( u + a ) ∂ x ∂ p ] d t If we go back and examine Eqn. (7.116) (7.119) (7.121)
d u d t + 1 ρ a d p d t = 0 ⇒ d u + d p ρ a = 0 \frac{du}{dt}+\frac{1}{\rho a}\frac{dp}{dt}=0\Rightarrow du+\frac{dp}{\rho a}=0 d t d u + ρ a 1 d t d p = 0 ⇒ d u + ρ a d p = 0 Eqn. (7.122) C + C^+ C + d x / d t = u + a dx/dt=u+a d x / d t = u + a x t xt x t C + C^+ C + C − C^- C − d x / d t = u − a dx/dt=u-a d x / d t = u − a x t xt x t
d u = [ ∂ u ∂ t + ( u − a ) ∂ u ∂ x ] d t du=\left[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}\right]dt d u = [ ∂ t ∂ u + ( u − a ) ∂ x ∂ u ] d t d p = [ ∂ p ∂ t + ( u − a ) ∂ p ∂ x ] d t dp=\left[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}\right]dt d p = [ ∂ t ∂ p + ( u − a ) ∂ x ∂ p ] d t which can be identified as subsets of Eqn. (7.117)
d u d t − 1 ρ a d p d t = 0 \frac{du}{dt}-\frac{1}{\rho a}\frac{dp}{dt}=0 d t d u − ρ a 1 d t d p = 0 In order to fulfill the relation above, either d u = d p = 0 du=dp=0 d u = d p = 0
d u − d p ρ a = 0 du-\frac{dp}{\rho a}=0 d u − ρ a d p = 0 Eqn. (7.126) C − C^- C − d x / d t = u − a dx/dt=u-a d x / d t = u − a x t xt x t C − C^- C −
Figure 7.2: Characteristic lines through a point (x 1 x_1 x 1 t 1 t_1 t 1
So, what we have done now is that we have have found paths through a point (x 1 x_1 x 1 t 1 t_1 t 1 (7.116) (7.117) (7.122) (7.126) C + C^+ C + C − C^- C − x t xt x t
Riemann Invariants ¶ If the compatibility equations are integrated along respective characteristic line, i.e., integrate (7.122) C + C^+ C + (7.126) C − C^- C − J + J^+ J + J − J^- J −
J + = u + ∫ d p ρ a = c o n s t J^+=u+\int\frac{dp}{\rho a}=const J + = u + ∫ ρ a d p = co n s t J − = u − ∫ d p ρ a = c o n s t J^-=u-\int\frac{dp}{\rho a}=const J − = u − ∫ ρ a d p = co n s t The Riemann invariants are constants along the associated characteristic line.
We have assumed isentropic flow and thus we may use the isentropic relations
p = C 1 T γ / ( γ − 1 ) = C 2 a 2 γ / ( γ − 1 ) p=C_1T^{\gamma/(\gamma-1)}=C_2a^{2\gamma/(\gamma-1)} p = C 1 T γ / ( γ − 1 ) = C 2 a 2 γ / ( γ − 1 ) where C 1 C_1 C 1 C 2 C_2 C 2 (7.129)
d p = C 2 ( 2 γ γ − 1 ) a [ 2 γ / ( γ − 1 ) − 1 ] d a dp=C_2\left(\frac{2\gamma}{\gamma-1}\right)a^{[2\gamma/(\gamma-1)-1]}da d p = C 2 ( γ − 1 2 γ ) a [ 2 γ / ( γ − 1 ) − 1 ] d a Now, if we further assume the gas to be calorically perfect
a 2 = γ R T = γ p ρ ⇒ ρ = γ p a 2 a^2=\gamma RT=\frac{\gamma p}{\rho}\Rightarrow \rho=\frac{\gamma p}{a^2} a 2 = γ RT = ρ γ p ⇒ ρ = a 2 γ p Eqn. (7.129) (7.131)
ρ = C 2 γ a [ 2 γ / ( γ − 1 ) − 2 ] \rho=C_2\gamma a^{[2\gamma/(\gamma-1)-2]} ρ = C 2 γ a [ 2 γ / ( γ − 1 ) − 2 ] and thus
J + = u + ∫ C 2 ( 2 γ γ − 1 ) a [ 2 γ / ( γ − 1 ) − 1 ] C 2 γ a [ 2 γ / ( γ − 1 ) − 2 ] a d a = u + ( 2 γ − 1 ) ∫ d a J^+=u+\int\frac{C_2\left(\frac{2\gamma}{\gamma-1}\right)a^{[2\gamma/(\gamma-1)-1]}}{C_2\gamma a^{[2\gamma/(\gamma-1)-2]}a}da=u+\left(\frac{2}{\gamma-1}\right)\int da J + = u + ∫ C 2 γ a [ 2 γ / ( γ − 1 ) − 2 ] a C 2 ( γ − 1 2 γ ) a [ 2 γ / ( γ − 1 ) − 1 ] d a = u + ( γ − 1 2 ) ∫ d a J + = u + 2 a γ − 1 J^+=u+\frac{2a}{\gamma-1} J + = u + γ − 1 2 a J − = u − 2 a γ − 1 J^-=u-\frac{2a}{\gamma-1} J − = u − γ − 1 2 a Eqns. (7.134) (7.135) C + C^+ C + C − C^- C − Figure 7.2 x 1 x_1 x 1 t 1 t_1 t 1
J + + J − = u + 2 a γ − 1 + u − 2 a γ − 1 = 2 u ⇒ u = 1 2 ( J + + J − ) J^++J^-=u+\frac{2a}{\gamma-1}+u-\frac{2a}{\gamma-1}=2u\Rightarrow u=\frac{1}{2}(J^++J^-) J + + J − = u + γ − 1 2 a + u − γ − 1 2 a = 2 u ⇒ u = 2 1 ( J + + J − ) J + = u + 2 a γ − 1 = 1 2 ( J + + J − ) + 2 a γ − 1 ⇒ a = γ − 1 4 ( J + − J − ) J^+=u+\frac{2a}{\gamma-1}=\frac{1}{2}(J^++J^-)+\frac{2a}{\gamma-1}\Rightarrow a=\frac{\gamma-1}{4}(J^+-J^-) J + = u + γ − 1 2 a = 2 1 ( J + + J − ) + γ − 1 2 a ⇒ a = 4 γ − 1 ( J + − J − )