Quesiti esame Acoustic in building

Acoustics in building – Livio Mazzarella 2023/2024

Question 1

Derive the sound wave equation in air starting from basic conservation equations: assumptions and nature of the solution. Find out the general solution for the sound pressure. Derive the relationship between sound pressure and speed (impedance), the definition of acoustic intensity, the relationship between acoustic pressure, intensity and energy density (for plane waves). Definition of dB, and all possible sound levels.

To build a physical model of acoustic phenomena (vibrational motion in fluids), the needed tools are:

• Conservation principles 1. As mass, momentum and energy conservation laws, etc.
• Constitutive laws of matter 2. As ideal gas, ideal liquid or ideal solid laws, etc.
• Phenomenological relationships 3. (Cause-to-effect relations) as Fourier’s law for thermal diffusion, Fick’s law for mass diffusion, Hooke’s law for elasticity, etc.

Combining such tools with the aid of specific simplifying hypotheses, the phenomenon governing equation can be derived: wave equation. The useful conservation principles, expressed in local form (i.e. valid in any point of the fluid as continuum) are: )+ (⃗ = 0 t⃗ ���⃗)+ (⃗ ⃗ − − = 0 t∂ρ ����⃗�)+ (ρ ⃗ + (φ�����⃗) ⋅ ⃗ � − ρ⃗ ⋅ = 0 − � ∂t ����⃗�

With mass density [kg/m3]; velocity vector [m/s]; stress tensor [N/ m2]; body (field) force [N/ kg]; ⃗ total energy by mass [J/ kg]; heat flux density vector [W/ m2], divergence operator [1/m]; �����⃗ (⬞) net netadvective mass flow by volume [kg/(s m3)]; advective momentum flow by vol.) )(⃗ (⃗ ⃗ The first hypothesis is:[kg m/(s2 m3)]; net advective energy flow [W/m3] From)(⃗ = + . ⃗ ����⃗�����)⃗ ⋅ ( + (⃗ ⃗ − − ) = 0

It follows the mechanical energy balance equation: 2 2 /2 ����⃗�����+ ⃗ − ⃗ ⋅ − ⃗ ⋅ = 0� � 2t 1 2

The Energy conservation equation with become: = + 22 2 /2 ����⃗�)+ + ⃗ + (u⃗ + (�����⃗) ⋅ ⃗ � − ⃗ ⋅ = 0− �� � 2tt

Subtracting the mechanical energy balance equation from the energy conservation equation, the internal energy balance equation is derived as: ���]+ (u⃗ + (�����⃗) ⋅ ⃗ � − ⃗ ⋅ = 0− [�t � −� � ̅

The second Hypothesis is where thermodynamic pressure [N/m2]; identity tensor [N/m2]; = + ̅ �hydrostatic stress tensor [N/m2]; deviatoric stress tensor [N/ m2]. It follow:− ��� = −() + �D� Reversible power of compression /expansion�� ⋅ ⃗ � = −( ⋅ ⃗ ) + �D ⋅ ⃗ ����⃗ = −⃗ ⋅ () + ⃗ ⋅ �D⃗ ⋅ Local dissipation function�� ���� �����⃗ ����⃗�� )=∑ ∇ ⋅ ⃗ � − ⃗ ⋅ = −(⃗ + ∑ ∇ � � ���������⃗

Hypothesis (3): The next three hypothesis are the hypothesis of Lightweight and “Isentropic” Fluid. =̅ ���������⃗

Hypothesis (4): Hypothesis (5): �����⃗; − () ≅ , = ) ≅ and the ≅ ∀∆ with g( ≅ → μ��⃗ gravitational acceleration [m/s2]; z vertical coordinate [m]; gradient operator [1/m]; dynamic viscosity ∇ ����⃗� ����⃗ ,tensor [Ns/m2]; thermal conductivity tensor [W/(m K)]. It follows: ,≅ 0λ ∑ ∇ (φ) =≅ 0 ⃗ ⋅ �� .

Introducing Hypotheses (1) to (5) the useful equations reduce to: 0 )+ (⃗ = 0 t⃗ )+ (⃗ ⃗ + () = 0 t ) )+ ( ⃗ + (p ⋅ ⃗ = 0 t ) )+ (u⃗ + p (⃗ = 0 t2 2/2 + ⃗ − ⃗ ⋅ () = 0 ℎ � �2t

∆(�,) �(�,) ∆(�,) ∆(�,)∆

The sixth Hypothesis is: with∀�,≪ ; ≪ ; ≪ ; ≪ ρ(⃗, ) = ρ +0 (⃗,(⃗, + Δ⃗ ) ; (⃗, ) = + Δ(⃗, ); (⃗, ) = + Δ(⃗, ) where , , and Δρ(⃗, ); ⃗ ) = 0 0 0 0 0 0 0��������⃗ , and are the equilibrium value.

It follows that high order deviations as: , ; ΔΔ ΔΔ⃗ΔΔ⃗ Δ⃗ Δ⃗ ΔΔΔ⃗ are negligible compared to first order deviations , , and Introducing Hypothesis (6) theΔ Δ Δ Δ.linearized equations are obtained:

Mass conservation

+ Δ)(0 )+ (( + Δ)⃗ = 00t][( + Δ)Δ⃗0 ]+ [( + Δ)Δ⃗ Δ⃗ + ( + Δ) = 00 0t ) )+ (Δ⃗ + (ΔΔ⃗ ) = 0 → + (Δ⃗ = 00 0tt

Momentum conservation

[( + Δ)( + Δ)]0 0 ] (+ [( + Δ)( + Δ)Δ⃗ + + Δ)(Δ⃗ ) = 00 0 0t 2 2Δ+ Δ)(Δ /2)][(0 + �( + Δ) Δ⃗ � + Δ⃗ ( + Δ) = 00 0t 2ΔΔ⃗Δ⃗ ) + + (Δ⃗ Δ⃗ + (ΔΔ⃗ Δ⃗ ) + (Δ) = 00 0tt

Internal energy balance

ΔΔuΔu Δ+ + + (Δ⃗ ) + (ΔuΔ⃗ ) + (ΔΔ⃗ ) + (ΔΔΔ⃗) + (0 0 0 0 0 0 0tt t+ Δ)(Δ⃗ ) = 0 = 0 for mass conservationΔu )] (+ [ + (Δ⃗ + + Δ)(Δ⃗ ) = 00 0 0 0tt

Mechanical energy balance

2 2 2 2[Δ + Δ Δ Δ /2 /2] + + div( Δ⃗ ) + div(Δ Δ⃗ ) + Δ⃗ ⋅ ( + Δ) = 00 0 02 2t t

It is a set of 3 independent equations in 4 unknowns, ∆, ∆, ∆⃗ , ∆, if we considering that the mechanical balance equation is linearly dependent on the momentum equation but We need 1 more equation to solve them.

Summing the internal and mechanical energy balance equations, the following acoustic energy conservation equation attains: 2Δu /2 [ + ] + [( + Δ)(Δ⃗ ) + Δ⃗ ⋅ ( + Δ)] = 00 0 0t t 1 2

Applying the definitions of acoustic energy density and the acoustic intensity[/3] ≡ ρ �Δ + Δ �0 2⃗ vector it follows:(≡ + Δ)Δ⃗ [/2]0 w (⃗+ ) = 0t

To be consistent with Hypothesis (4) and (5), the fluid cannot exchange energy through irreversible processes, assuming , where s is specific entropy by mass and expanding in Tailor’s series keeping = = (, ) ̅ (isentropic processes), it follows: 2∂∂ρ ρ1 2( ) ( )ρ − ρ = � � + ⋯− + − � �0 0 02∂ ∂20,̅ 0,̅

And neglecting the expansion terms higher than first order, it follows: ( )= � � � ∆∆ = − − = �0 0 0,̅ 0,̅1 ∂ρ

By definition of isentropic compressibility coefficient it follows the linear relationshipκ = � � > 0,0 ρ ∂0 0,̅ between pressure and mass density (Linear Isentropic Constitutive Equation):1 2∆ = ∆∆ = 0 ,0

Introducing the Linear Isentropic Constitutive Equation in the mass conservation equation, it turns into a pressure equation: ∆ 2 )+ (Δ⃗ = 0 ()0t

Which can be combined with the momentum conservation equation:Δ⃗ + (Δ) = 0 ()0 t

Applying the partial time derivative to the first and the gradient operator to the second, commuting the differential operators and substituting:2∂ ∂Δ⃗ −1Δ ∂2 2 2)� = −⃗ = −= − ρ ρ ρ � (Δ)����(Δ ��0 0 02 ∂t ρ∂t ∂t 0

Wave equation in sound pressure

The following results: 1 ()= ℎ =2t �0 ,0

Using the same approach but applying the partial time derivative to the second and the gradient operator to Wave Equation in vibrational the first, commuting the differential operators and substituting, a vector velocity results: ∂Δ ∂2 2 2 [(Δ)])] (Δ⃗ )= − � � = − ρ [(Δ⃗ ρ ∇ =0 0∂t ∂t2 ∂∂ Δ⃗ ⃗2 2 [(Δ)] (Δ ) (⃗ )= ρ = − ρ ∇ = ⃗ →0 02 2∂∂t t

Wave equation

Substituting the Linear Isentropic Constitutive Equation into the pressure Wave Equation, ain mass density is derived:2 Δ 2 2 ( Δ) 2 2 (Δ)= ∇ 2 2 2 (()= ∇ Δ) → = →� 2t 2 2t t2 ΔΔ =

The acoustic pressure wave equation can be solved using the Variables Separation Technique, that is assuming: it follows a set of two ordinary differential equations:Δ(⃗, ) = Δρ (⃗)()0 2 () 2+ ω () = 0 (1)22 2[Δ (⃗)] (⃗)+ ∇ Δ = 0 (2)0 0

Where is the coupling constant to be determined by the boundary conditions. The time-dependent ordinary 2 2 differential equation (1) is easily solved thought its characteristic algebraic equation: which has + = 0 two imaginary roots the solution is then:2 = √− = ±,1,2 ) () = c cos ω + c sin ω = c cos(ω + φ 1,ω 2,ω ω ω

Where , are constants to be determined by the boundary conditions as well as the parameter Thec .ω space-dependent ordinary