narrow area acoustics
Simulate in comsol5.5
- Suitable for waveguides with slowly changing (constant) cross-sections
- Suitable for all frequencies and boundary layer thicknesses
- 当 L > > R L>>R L>>R is the most accurate
- The homogenized boundary layer loss is the complex-valued wavenumber K c K_cKcand characteristic impedance Z c Z_cZc
- Based on analytical models or "cross-sectional analysis"
- Apply port boundary conditions
Pipe type
- slit
- Rectangular conduit
- Equilateral triangular conduit
- round catheter
- Very Narrow Circular Tube
Radius<<Thermal Boundary Layer - Wide pipe approximates
any shape as long as hydraulic diameter (=4 ∗ A/C) (=4*A/C)(=4∗A/C) - User-defined
Any shape, specifying complex wavenumbers and complex characteristic acoustic impedances
Eigenfrequency analysis
∇ ⋅ ( − 1 ρ 0 ∇ p ) + λ 2 p ρ 0 c 2 = 0 \displaystyle \drawing \cdot(-\frac{1}{ { { {\rho }_{0}}}} \ drawing p )+\frac{ { { {\lambda }^{2}}p}}{ { { {\rho }_{0}}{ {c}^{2}}}}=0∇⋅(−r01∇p)+r0c2l2p=0
- Suitable for finding resonant frequencies
- Assuming an interharmonic sound pressure field: p = p 0 exp [ j ( ω t + θ ) ] \displaystyle p={ {p}_{0}}\exp [j(\omega t+\theta )]p=p0exp[j(ωt+i ) ]
- Solve for one variable: j ω = − λ j\omega=-\lambdaj ω=− λ (lambda variable in Comsol)
- The characteristic frequency is f 0 = imag ( − λ ) / 2 π \displaystyle { {f}_{0}}=imag(-\lambda )/2\pif0=i m a g ( − λ ) / 2 π is stored in the variable acpr.freq
- Angular frequency: ω = 2 π f 0 \displaystyle \omega =2\pi { {f}_{0}}oh=2πf0Available through variable acpr.omega
The characteristic frequency selected in the study
can match the resonance peak.
Grid size settings
Spatially resolved wavelength
- At least 10 − 12 D o F 10-12DoF per wavelength10−12DoF
- Requires 5-6 units per wavelength (preset to second order L agrange LagrangeL a g r a n g e unit)
- By setting the maximum cell size hhh isl/5 l/5l / 5 orl/6 l/6l / 6 to achieve, generally takeh < l/5 h<l/5h<l/5
- If using linear L agrange LagrangeL a g r a n g e units, each wavelength requires>10 >10>10 units _
- In some areas with tip resonances, a more detailed mesh needs to be drawn.
- If an iterative solver is used, the resolution grid per wavelength is preferably > 6 > 6>6
If the mesh is not well divided, some frequencies will not be calculated, or the error will be large. The finer the mesh, the more precise the calculation results will be, but the calculation time will become longer.
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