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Example 15: Mass Law (Acoustic Analysis)

General

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

  • An analysis that considers the weight of the wall is required. The acoustic analysis meets this condition and is performed here without accounting for the elastic wall. Refer to Example 6: Piezoelectric-Acoustic Couple Analysis for the analysis when the wall is an elastic material.

 

 

Analysis Space

Item

Settings

Analysis Space

2D

Model Unit

m

Thickness in Depth Direction

1.0 [m]

 

Analysis Conditions

Item

Settings

Solver

Acoustic Analysis [Mach]

Analysis Type

Harmonic Analysis

Options

Select [Calculate loss]

Tab

Setting Item

Settings

Harmonic Analysis

Frequency

Linear Step by Number of Divisions

 

Minimum: 100[Hz]

Maximum: 4000 [Hz]

Interval

Number of Divisions: 4

 

Model

A wall is placed between air bodies.

 

Body Attribute and Material Property 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]

Sound Speed

10000 [m/s]

 

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 resulting sound pressure distribution at 1.075e+03 [Hz] is shown below.

The contour diagram based on the setting below is shown.

 

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]