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Home / Examples / Coupled Analysis / Piezoelectric-Acoustic Analysis [Rayleigh/Mach] / Example 6: Mass Law

Example 6: Mass Law (Piezoelectric-Acoustic Coupled Analysis)


General

  • The frequency response of the wall’s transmittance is solved.

 

 

Analysis Space

Item

Settings

Analysis Space

2D

Model Unit

m

Thickness in Depth direction

1.0 [m]

 

Analysis Conditions

The solvers are Rayleigh and Mach.

Item

Settings

Solver

Piezoelectric Analysis [Rayleigh]

Acoustic Analysis [Mach]

Analysis Type

Harmonic Analysis

Analysis Plane

2D Section

Variables to Constrain

Select [Electric Potential] and [Y Displacement] *

Analysis Options

Select [Fully-coupled Analysis]

Output Setting

Select [Calculate loss (Acoustic Analysis domain)]

* To exclude piezoelectricity from the analysis, constrain the electric potential.

 

Tab

Setting Item

Settings

Harmonic Analysis

Frequency

Minimum: 100 [Hz]

Maximum: 4000 [Hz]

Interval

Number of Divisions: 4

 

Model

A wall is placed between air bodies.

 

Body Attribute and Material Setting

Body Number/Type

Body Attribute Name

Material Name

0/Solid

AIR

000_Air *

1/Solid

WALL

Wall

2/Solid

AIR

000_Air *

* Available from the material DB

 

Material Name

Tab

Material Name

Wall

Density

1000 [kg/m3]

Piezoelectricity

Material Type: Perfect Conductor

Anisotropy: Isotropic

Young's Modulus: 206[GPa]

Poisson's Ratio: 0.28

 

 

Body Attribute Name

Analysis Domain (Solver)

AIR

Acoustic Analysis (Mach)

WALL

Piezoelectric Analysis (Rayleigh)

 

Cautions: Note for the piezoelectric acoustic coupled analysis

The body to be analyzed must be specified to be subject to the piezoelectric or acoustic analysis. That is, in the analysis domain tab, either Piezoelectric Analysis or Acoustic Analysis must be selected.

Boundary Conditions

Boundary Condition Name/Topology

Tab

Boundary Condition Type

Settings

v/Edge

Acoustic

Speed

 

1 [m/s]

Select [Specify the incident wave]

z/Edge

Acoustic

Acoustic Impedance

402.56[Pa・s/m] *

 

* Set the specific acoustic impedance of air.

Results

The resultant distribution of sound pressure at 1.075e+03 [Hz] is shown below. Confirm the analysis type of the result filed is the acoustic analysis.

The contour diagram is illustrated using the setting below.

 

Acoustic Analysis 1: 1.075000E+3 [Hz]
Sound pressure [Pa] Value
0 [deg] Linear

 

Table 1. Setting in Result Field for Fig. 1

 

Fig. 1 Sound Pressure Distribution 1.075 [kHz]

 

 

Calculate the transmittance (T). The equation below will give the transmittance.

 

T = It/Ii

 

It: Can be acquired on the [Loss at Boundary] tab in the result table.

Ii: Can be acquired on the [Input Power] tab in the result table.

 

Calculate using the results from Femtet.

The results at 100 [Hz] are shown below.

Ii = 3.25 [W]

It = 201.28 [W]

-10・Log(3.25/201.2) = 17.9 [dB]

 

Calculation of theoretical values

20・Log(f・m) - 42.5 = 17.5 [dB]

Mass: m = 10 [kg], calculated from the width of the wall, 1 [cm], the thickness in the depth direction, 1 [m], and the density, 1000 [kg/m3] of the analysis model.

Frequency: f = 100 [Hz]