7.4 Acoustic TheoryNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
In the following we are going to derive the linear acoustic wave equation starting from the continuity and momentum equations on non-conservation differential form. The equations are repeated here for convenience.
D ρ D t + ρ ( ∇ ⋅ v ) = 0 \dfrac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0 D t D ρ + ρ ( ∇ ⋅ v ) = 0 ρ D v D t + ∇ p = 0 \rho\dfrac{D\mathbf{v}}{Dt}+\nabla p=0 ρ D t D v + ∇ p = 0 Remember that D / D t D/Dt D / D t
D D t = ∂ ∂ t + v ⋅ ∇ \dfrac{D}{Dt}=\dfrac{\partial }{\partial t}+\mathbf{v}\cdot\nabla D t D = ∂ t ∂ + v ⋅ ∇ where ∂ / ∂ t \partial/\partial t ∂ / ∂ t v ⋅ ∇ \mathbf{v}\cdot\nabla v ⋅ ∇
We are going to analyze acoustic waves in one dimension, which means that the equations above reduces to
∂ ρ ∂ t + u ∂ ρ ∂ x + ρ ∂ u ∂ x = 0 \dfrac{\partial \rho}{\partial t}+u\dfrac{\partial \rho}{\partial x}+\rho\dfrac{\partial u}{\partial x}=0 ∂ t ∂ ρ + u ∂ x ∂ ρ + ρ ∂ x ∂ u = 0 ρ ∂ u ∂ t + ρ u ∂ u ∂ x + ∂ p ∂ x = 0 \rho\dfrac{\partial u}{\partial t}+\rho u\dfrac{\partial u}{\partial x}+\dfrac{\partial p}{\partial x}=0 ρ ∂ t ∂ u + ρ u ∂ x ∂ u + ∂ x ∂ p = 0 Pressure is a thermodynamic property and thus it can be expressed as a function of two other thermodynamic properties. Let’s express pressure as a function of density (ρ \rho ρ s s s
p = p ( ρ , s ) ⇒ d p = ( ∂ p ∂ ρ ) s d ρ + ( ∂ p ∂ s ) ρ d s p=p(\rho,s)\Rightarrow dp=\left(\dfrac{\partial p}{\partial \rho}\right)_s d\rho+\left(\dfrac{\partial p}{\partial s}\right)_\rho ds p = p ( ρ , s ) ⇒ d p = ( ∂ ρ ∂ p ) s d ρ + ( ∂ s ∂ p ) ρ d s Since weak acoustic waves are considered, entropy will be constant and thus d s = 0 ds=0 d s = 0
d p = ( ∂ p ∂ ρ ) s d ρ = a 2 d ρ dp=\left(\dfrac{\partial p}{\partial \rho}\right)_s d\rho=a^2d\rho d p = ( ∂ ρ ∂ p ) s d ρ = a 2 d ρ ρ ∂ u ∂ t + ρ u ∂ u ∂ x + a 2 ∂ ρ ∂ x = 0 \rho\dfrac{\partial u}{\partial t}+\rho u\dfrac{\partial u}{\partial x}+a^2\dfrac{\partial \rho}{\partial x}=0 ρ ∂ t ∂ u + ρ u ∂ x ∂ u + a 2 ∂ x ∂ ρ = 0 The acoustic perturbations can be described as small deviations around a reference state
ρ = ρ ∞ + Δ ρ p = p ∞ + Δ p T = T ∞ + Δ T u = u ∞ + Δ u = { u ∞ = 0 } = Δ u \begin{aligned}
&\rho=\rho_\infty+\Delta \rho\\
&p=p_\infty+\Delta p\\
&T=T_\infty+\Delta T\\
&u=u_\infty+\Delta u=\{u_\infty=0\}=\Delta u\\
\end{aligned} ρ = ρ ∞ + Δ ρ p = p ∞ + Δ p T = T ∞ + Δ T u = u ∞ + Δ u = { u ∞ = 0 } = Δ u Inserted in Eqns. (7.68) (7.72)
∂ ∂ t ( Δ ρ ) + Δ u ∂ ∂ x ( Δ ρ ) + ( ρ ∞ + Δ ρ ) ∂ ∂ x ( Δ u ) = 0 \dfrac{\partial}{\partial t}(\Delta \rho)+\Delta u\dfrac{\partial}{\partial x}(\Delta \rho)+(\rho_\infty+\Delta \rho)\dfrac{\partial}{\partial x}(\Delta u)=0 ∂ t ∂ ( Δ ρ ) + Δ u ∂ x ∂ ( Δ ρ ) + ( ρ ∞ + Δ ρ ) ∂ x ∂ ( Δ u ) = 0 ( ρ ∞ + Δ ρ ) ∂ ∂ t ( Δ u ) + ( ρ ∞ + Δ ρ ) Δ u ∂ ∂ x ( Δ u ) + a 2 ∂ ∂ x ( Δ ρ ) = 0 (\rho_\infty+\Delta \rho)\dfrac{\partial}{\partial t}(\Delta u)+(\rho_\infty+\Delta \rho) \Delta u\dfrac{\partial}{\partial x}(\Delta u)+a^2\dfrac{\partial}{\partial x}(\Delta \rho)=0 ( ρ ∞ + Δ ρ ) ∂ t ∂ ( Δ u ) + ( ρ ∞ + Δ ρ ) Δ u ∂ x ∂ ( Δ u ) + a 2 ∂ x ∂ ( Δ ρ ) = 0 In the same way as pressure, being a thermodynamic variable, can be expressed as a function of two other thermodynamic varaibles, so can the speed of sound. Once again we will select density and entropy as the two thermodynamic variables
a 2 = a 2 ( ρ , s ) a^2=a^2(\rho,s) a 2 = a 2 ( ρ , s ) and since entropy is constant
a 2 = a 2 ( ρ ) a^2=a^2(\rho) a 2 = a 2 ( ρ ) Taylor expansion of a 2 a^2 a 2 a ∞ a_\infty a ∞ Δ ρ = ρ − ρ ∞ \Delta \rho=\rho-\rho_\infty Δ ρ = ρ − ρ ∞
a 2 = a ∞ 2 + ( ∂ ∂ ρ ( a 2 ) ) ∞ Δ ρ + ( ∂ 2 ∂ ρ 2 ( a 2 ) ) ∞ ( Δ ρ ) 2 + ⋯ a^2=a^2_\infty+\left(\dfrac{\partial}{\partial \rho}(a^2)\right)_\infty\Delta \rho+\left(\dfrac{\partial^2}{\partial \rho^2}(a^2)\right)_\infty(\Delta \rho)^2+\ \cdots a 2 = a ∞ 2 + ( ∂ ρ ∂ ( a 2 ) ) ∞ Δ ρ + ( ∂ ρ 2 ∂ 2 ( a 2 ) ) ∞ ( Δ ρ ) 2 + ⋯ Inserted in Eqn. (7.75)
( ρ ∞ + Δ ρ ) ∂ ∂ t ( Δ u ) + ( ρ ∞ + Δ ρ ) Δ u ∂ ∂ x ( Δ u ) + [ a ∞ 2 + ( ∂ ∂ ρ ( a 2 ) ) ∞ Δ ρ + ⋯ ] ∂ ∂ x ( Δ ρ ) = 0 (\rho_\infty+\Delta \rho)\dfrac{\partial}{\partial t}(\Delta u)+(\rho_\infty+\Delta \rho) \Delta u\dfrac{\partial}{\partial x}(\Delta u)+\left[a^2_\infty+\left(\dfrac{\partial}{\partial \rho}(a^2)\right)_\infty\Delta \rho+\ \cdots\right]\dfrac{\partial}{\partial x}(\Delta \rho)=0 ( ρ ∞ + Δ ρ ) ∂ t ∂ ( Δ u ) + ( ρ ∞ + Δ ρ ) Δ u ∂ x ∂ ( Δ u ) + [ a ∞ 2 + ( ∂ ρ ∂ ( a 2 ) ) ∞ Δ ρ + ⋯ ] ∂ x ∂ ( Δ ρ ) = 0 The perturbations Δ u \Delta u Δ u Δ ρ \Delta \rho Δ ρ Δ u ≪ a ∞ \Delta u \ll a_\infty Δ u ≪ a ∞ Δ ρ ≪ ρ ∞ \Delta \rho \ll \rho_\infty Δ ρ ≪ ρ ∞ a 2 a^2 a 2
∂ ∂ t ( Δ ρ ) + ρ ∞ ∂ ∂ x ( Δ u ) = 0 \dfrac{\partial}{\partial t}(\Delta \rho)+\rho_\infty\dfrac{\partial}{\partial x}(\Delta u)=0 ∂ t ∂ ( Δ ρ ) + ρ ∞ ∂ x ∂ ( Δ u ) = 0 ρ ∞ ∂ ∂ t ( Δ u ) + a ∞ 2 ∂ ∂ x ( Δ ρ ) = 0 \rho_\infty\dfrac{\partial}{\partial t}(\Delta u)+a_\infty^2\dfrac{\partial}{\partial x}(\Delta \rho)=0 ρ ∞ ∂ t ∂ ( Δ u ) + a ∞ 2 ∂ x ∂ ( Δ ρ ) = 0 Before making the assumption that the perturbations are small compared to the corresponding reference state flow quantities and thus justifying the cancelation of products of perturbations from the equations, the flow equations were still the exact fully non-linear equations. Eqns. (7.80) (7.81) (7.80) (7.81) (7.80) (7.81)
The temporal derivative of the continuity equation:
∂ 2 ∂ t 2 ( Δ ρ ) + ρ ∞ ∂ 2 ∂ x ∂ t ( Δ u ) = 0 \dfrac{\partial^2}{\partial t^2}(\Delta \rho)+\rho_\infty\dfrac{\partial^2}{\partial x\partial t}(\Delta u)=0 ∂ t 2 ∂ 2 ( Δ ρ ) + ρ ∞ ∂ x ∂ t ∂ 2 ( Δ u ) = 0 The divergence of the momentum equation:
ρ ∞ ∂ 2 ∂ x ∂ t ( Δ u ) + a ∞ 2 ∂ 2 ∂ x 2 ( Δ ρ ) = 0 \rho_\infty\dfrac{\partial^2}{\partial x \partial t}(\Delta u)+a_\infty^2\dfrac{\partial^2}{\partial x^2}(\Delta \rho)=0 ρ ∞ ∂ x ∂ t ∂ 2 ( Δ u ) + a ∞ 2 ∂ x 2 ∂ 2 ( Δ ρ ) = 0 The second term in the first equation is the same as the first term in the second equation. Substituting the term, the two equations reduces to one single equation
∂ 2 ∂ t 2 ( Δ ρ ) = a ∞ 2 ∂ 2 ∂ x 2 ( Δ ρ ) \frac{\partial^2}{\partial t^2}(\Delta \rho)=a^2_\infty\frac{\partial^2}{\partial x^2}(\Delta \rho) ∂ t 2 ∂ 2 ( Δ ρ ) = a ∞ 2 ∂ x 2 ∂ 2 ( Δ ρ ) which is a one-dimensional form of the classic wave equation with the general solution
Δ ρ = F ( x − a ∞ t ) + G ( x + a ∞ t ) \Delta \rho = F(x-a_\infty t)+G(x+a_\infty t) Δ ρ = F ( x − a ∞ t ) + G ( x + a ∞ t ) F F F G G G F F F x x x G G G x x x (7.84) Δ ρ \Delta \rho Δ ρ (7.85) (7.84)
∂ ∂ t ( Δ ρ ) = ∂ F ∂ ( x − a ∞ t ) ∂ ( x − a ∞ t ) ∂ t + ∂ G ∂ ( x + a ∞ t ) ∂ ( x + a ∞ t ) ∂ t \frac{\partial}{\partial t}(\Delta \rho)=\frac{\partial F}{\partial (x-a_\infty t)}\frac{\partial (x-a_\infty t)}{\partial t}+\frac{\partial G}{\partial (x+a_\infty t)}\frac{\partial (x+a_\infty t)}{\partial t} ∂ t ∂ ( Δ ρ ) = ∂ ( x − a ∞ t ) ∂ F ∂ t ∂ ( x − a ∞ t ) + ∂ ( x + a ∞ t ) ∂ G ∂ t ∂ ( x + a ∞ t ) ∂ ∂ t ( Δ ρ ) = − a ∞ F ′ + a ∞ G ′ \frac{\partial}{\partial t}(\Delta \rho)=-a_\infty F'+a_\infty G' ∂ t ∂ ( Δ ρ ) = − a ∞ F ′ + a ∞ G ′ ∂ 2 ∂ t 2 ( Δ ρ ) = − a ∞ ∂ F ′ ∂ ( x − a ∞ t ) ∂ ( x − a ∞ t ) ∂ t + a ∞ ∂ G ′ ∂ ( x + a ∞ t ) ∂ ( x + a ∞ t ) ∂ t \frac{\partial^2}{\partial t^2}(\Delta \rho)=-a_\infty\frac{\partial F'}{\partial (x-a_\infty t)}\frac{\partial (x-a_\infty t)}{\partial t}+a_\infty\frac{\partial G'}{\partial (x+a_\infty t)}\frac{\partial (x+a_\infty t)}{\partial t} ∂ t 2 ∂ 2 ( Δ ρ ) = − a ∞ ∂ ( x − a ∞ t ) ∂ F ′ ∂ t ∂ ( x − a ∞ t ) + a ∞ ∂ ( x + a ∞ t ) ∂ G ′ ∂ t ∂ ( x + a ∞ t ) ∂ 2 ∂ t 2 ( Δ ρ ) = a ∞ 2 F ′ ′ + a ∞ 2 G ′ ′ \frac{\partial^2}{\partial t^2}(\Delta \rho)=a_\infty^2 F''+a_\infty^2 G'' ∂ t 2 ∂ 2 ( Δ ρ ) = a ∞ 2 F ′′ + a ∞ 2 G ′′ ∂ ∂ x ( Δ ρ ) = ∂ F ∂ ( x − a ∞ t ) ∂ ( x − a ∞ t ) ∂ x + ∂ G ∂ ( x + a ∞ t ) ∂ ( x + a ∞ t ) ∂ x \frac{\partial}{\partial x}(\Delta \rho)=\frac{\partial F}{\partial (x-a_\infty t)}\frac{\partial (x-a_\infty t)}{\partial x}+\frac{\partial G}{\partial (x+a_\infty t)}\frac{\partial (x+a_\infty t)}{\partial x} ∂ x ∂ ( Δ ρ ) = ∂ ( x − a ∞ t ) ∂ F ∂ x ∂ ( x − a ∞ t ) + ∂ ( x + a ∞ t ) ∂ G ∂ x ∂ ( x + a ∞ t ) ∂ ∂ x ( Δ ρ ) = F ′ + G ′ \frac{\partial}{\partial x}(\Delta \rho)=F'+G' ∂ x ∂ ( Δ ρ ) = F ′ + G ′ ∂ 2 ∂ x 2 ( Δ ρ ) = ∂ F ′ ∂ ( x − a ∞ t ) ∂ ( x − a ∞ t ) ∂ x + ∂ G ′ ∂ ( x + a ∞ t ) ∂ ( x + a ∞ t ) ∂ x \frac{\partial^2}{\partial x^2}(\Delta \rho)=\frac{\partial F'}{\partial (x-a_\infty t)}\frac{\partial (x-a_\infty t)}{\partial x}+\frac{\partial G'}{\partial (x+a_\infty t)}\frac{\partial (x+a_\infty t)}{\partial x} ∂ x 2 ∂ 2 ( Δ ρ ) = ∂ ( x − a ∞ t ) ∂ F ′ ∂ x ∂ ( x − a ∞ t ) + ∂ ( x + a ∞ t ) ∂ G ′ ∂ x ∂ ( x + a ∞ t ) ∂ 2 ∂ x 2 ( Δ ρ ) = F ′ ′ + G ′ ′ \frac{\partial^2}{\partial x^2}(\Delta \rho)= F'' + G'' ∂ x 2 ∂ 2 ( Δ ρ ) = F ′′ + G ′′ Eqns. (7.89) (7.93) (7.84)
a ∞ 2 F ′ ′ + a ∞ 2 G ′ ′ = a ∞ 2 ( F ′ ′ + G ′ ′ ) a_\infty^2 F''+a_\infty^2 G''=a_\infty^2(F'' + G'') a ∞ 2 F ′′ + a ∞ 2 G ′′ = a ∞ 2 ( F ′′ + G ′′ ) which shows that Eqn. (7.85)
F F F G G G G = 0 G=0 G = 0
Δ ρ ( x , t ) = F ( x − a ∞ t ) \Delta \rho(x,t)=F(x-a_\infty t) Δ ρ ( x , t ) = F ( x − a ∞ t ) If Δ ρ \Delta \rho Δ ρ (7.95) ( x − a ∞ t ) (x-a_\infty t) ( x − a ∞ t )
x = a ∞ t + c ⇒ d x d t = a ∞ x=a_\infty t + c\Rightarrow \dfrac{dx}{dt}=a_\infty x = a ∞ t + c ⇒ d t d x = a ∞ From Eqn. (7.95)
∂ ∂ t ( Δ ρ ) = − a ∞ F ′ \dfrac{\partial}{\partial t}(\Delta \rho)=-a_\infty F' ∂ t ∂ ( Δ ρ ) = − a ∞ F ′ ∂ ∂ x ( Δ ρ ) = F ′ \dfrac{\partial}{\partial x}(\Delta \rho)=F' ∂ x ∂ ( Δ ρ ) = F ′ and thus
∂ ∂ x ( Δ ρ ) = − 1 a ∞ ∂ ∂ t ( Δ ρ ) \dfrac{\partial}{\partial x}(\Delta \rho)=-\dfrac{1}{a_\infty}\dfrac{\partial}{\partial t}(\Delta \rho) ∂ x ∂ ( Δ ρ ) = − a ∞ 1 ∂ t ∂ ( Δ ρ ) which gives a relation between the temporal derivative of Δ ρ \Delta \rho Δ ρ Δ ρ \Delta \rho Δ ρ (7.99) (7.81)
∂ ∂ t ( Δ u ) = − a ∞ 2 ρ ∞ ∂ ∂ x ( Δ ρ ) = { ∂ ∂ x ( Δ ρ ) = − 1 a ∞ ∂ ∂ t ( Δ ρ ) } = a ∞ ρ ∞ ∂ ∂ t ( Δ ρ ) ⇒ \dfrac{\partial}{\partial t}(\Delta u)=-\dfrac{a_\infty^2}{\rho_\infty}\dfrac{\partial}{\partial x}(\Delta \rho)=\left\{\dfrac{\partial}{\partial x}(\Delta \rho)=-\dfrac{1}{a_\infty}\dfrac{\partial}{\partial t}(\Delta \rho)\right\}=\dfrac{a_\infty}{\rho_\infty}\dfrac{\partial}{\partial t}(\Delta \rho)\Rightarrow ∂ t ∂ ( Δ u ) = − ρ ∞ a ∞ 2 ∂ x ∂ ( Δ ρ ) = { ∂ x ∂ ( Δ ρ ) = − a ∞ 1 ∂ t ∂ ( Δ ρ ) } = ρ ∞ a ∞ ∂ t ∂ ( Δ ρ ) ⇒ ∂ ∂ t ( Δ u − a ∞ ρ ∞ Δ ρ ) = 0 ⇒ Δ u − a ∞ ρ ∞ Δ ρ = c o n s t \dfrac{\partial}{\partial t}\left(\Delta u-\dfrac{a_\infty}{\rho_\infty}\Delta \rho\right)=0\Rightarrow \Delta u-\dfrac{a_\infty}{\rho_\infty}\Delta \rho = const ∂ t ∂ ( Δ u − ρ ∞ a ∞ Δ ρ ) = 0 ⇒ Δ u − ρ ∞ a ∞ Δ ρ = co n s t In an undisturbed gas Δ u = Δ ρ = 0 \Delta u=\Delta \rho=0 Δ u = Δ ρ = 0
Δ u − a ∞ ρ ∞ Δ ρ = 0 \Delta u-\dfrac{a_\infty}{\rho_\infty}\Delta \rho=0 Δ u − ρ ∞ a ∞ Δ ρ = 0 or
Δ u = a ∞ ρ ∞ Δ ρ \Delta u=\dfrac{a_\infty}{\rho_\infty}\Delta \rho Δ u = ρ ∞ a ∞ Δ ρ If instead F F F G G G
Δ u = − a ∞ ρ ∞ Δ ρ \Delta u=-\dfrac{a_\infty}{\rho_\infty}\Delta \rho Δ u = − ρ ∞ a ∞ Δ ρ ( ∂ p ∂ ρ ) s = a 2 ⇒ Δ p = a ∞ 2 Δ ρ \left(\dfrac{\partial p}{\partial \rho}\right)_s=a^2\Rightarrow \Delta p=a_\infty^2 \Delta \rho ( ∂ ρ ∂ p ) s = a 2 ⇒ Δ p = a ∞ 2 Δ ρ Acoustic wave traveling in the positive x x x
Δ u = a ∞ ρ ∞ Δ ρ = 1 a ∞ ρ ∞ Δ p \Delta u=\dfrac{a_\infty}{\rho_\infty}\Delta \rho=\dfrac{1}{a_\infty \rho_\infty}\Delta p Δ u = ρ ∞ a ∞ Δ ρ = a ∞ ρ ∞ 1 Δ p Acoustic wave traveling in the negative x x x
Δ u = − a ∞ ρ ∞ Δ ρ = − 1 a ∞ ρ ∞ Δ p \Delta u=-\dfrac{a_\infty}{\rho_\infty}\Delta \rho=-\dfrac{1}{a_\infty \rho_\infty}\Delta p Δ u = − ρ ∞ a ∞ Δ ρ = − a ∞ ρ ∞ 1 Δ p