Добавил:

ivanov666 Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Башкирский Государственный Аграрный Университет

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

акустика / lin_h_bengisu_t_mourelatos_zp_lecture_notes_on_acoustics_and.pdf

Скачиваний:

Добавлен:

04.05.2023

Размер:

12.68 Mб

Скачать

☆

<<< < Предыдущая 31 32 33 34 35 36 37 38 39 40 41 4243 / 7843 44 45 46 47 48 49 50 51 52 53 54 55 > Следующая >>>

184	7 Resonant Cavities

7.43D Boundary Conditions of Rectangular Cavities

It can be difﬁcult to plot and visualize 3D standing waves in a rectangular cavity on paper. However, it is possible to assume that there are no standing waves in an arbitrary dimension and draw the standing waves in other two dimensions as shown in Examples 7.3 and 7.4 in this section.

On the other hand, formulations for the 3D standing waves in a rectangular cavity can be extended from the formulas of 1D and 2D standing waves as the following (this is tedious but straightforward):

7.4.13D Standing Wave Solutions of the Wave Equation

psðx, y, z, tÞ ¼ TðtÞXðxÞYðyÞZðzÞ

¼ AtAxAyAz cos ðωt þ θtÞ cos ðkxxÞ cos kyy cos ðkzzÞ

From Euler’s force equation, the velocity of the 3D standing wave solution is given by:

usðx, y, z, tÞ ¼ uxðx, y, z, tÞbex þ uyðx, y, z, tÞbey þ uzðx, y, z, tÞbez

where:
	1		k				kyy cos ðkzzÞ
uxðx, y, z, tÞ ¼				x		AtAxAyAz sin ðωt þ θtÞ sin ðkxxÞ cos
	ρ0c			k
	1		k				kyy cos ðkzzÞ
uyðx, y, z, tÞ ¼				y		AtAxAyAz sin ðωt þ θtÞ cos ðkxxÞ sin
	ρ0c			k
	1	k					kyy sinsðkzzÞ
uzðx, y, z, tÞ ¼				z	AtAxAyAz sin ðωt þ θtÞ cos ðkxxÞ cos
	ρ0c			k

The procedures for deriving the 3D velocity from the 3D pressure are similar to the procedures for deriving the 2D velocity from the 2D pressure. It is repetitive and is not shown here.

7.4.23D Natural Frequencies and Mode Shapes

The mode shapes of a rectangular cavity with six rigid walls can be calculated from the above pressure and velocity formulas with boundary conditions imposed by the

7.4 3D Boundary Conditions of Rectangular Cavities

185

six rigid walls. Assume that the cavity is oriented at the origin of the coordinate, as shown in the ﬁgure below:

= = = 0	̂	̂
= = = 0

The boundary conditions on all rigid walls require that the ﬂow velocity is zero. By constraining the velocity at the ﬁxed boundaries will discretize the wavenumber similar to the 2D standing waves as:

	π			π			π
kxl ¼		l;	kym ¼		m;	kzn ¼		n;	l, m, n ¼ 1, 2, . . .
	Lx			Ly			Lz

Therefore, the discretized ﬂow velocity vector and pressure are shown below:

usðx, y, z, tÞ ¼ uxðx, y, z, tÞbex þ uyðx, y, z, tÞbey þ uzðx, y, z, tÞbez

where:
	1	k			kymy cos ðkznzÞ
uxðx, y, tÞ ¼		x		AtAxAyAz sin ðωt þ θtÞ sin ðkxlxÞ cos
	ρ0c	k
	1	k			kymy cos ðkznzÞ
uyðx, y, tÞ ¼		y		AtAxAyAz sin ðωt þ θtÞ cos ðkxlxÞ sin
	ρ0c	k
	1	k			kymy sin ðkznzÞ
uzðx, y, tÞ ¼		z	AtAxAyAz sin ðωt þ θtÞ cos ðkxlxÞ cos
	ρ0c	k
and:

186	7 Resonant Cavities
psðx, y, z, tÞ ¼ AtAxAyAz cos ðωt þ θtÞ cos ðkxlxÞ cos	kymy cos ðkznzÞ

The standing wave of the pressure is shown below. The derivation for these mode shapes is similar to the derivation for the 2D mode shapes.

The combined wavenumber klmn and the combined frequency flmn of an eigenmode (l, m, n) are related to the discretized wavenumbers kxl and kym and kzn as:

k2lmn ¼ k2xl þ k2ym þ k2zn

f lmn ¼ 2π klmn

The boundary conditions imposed by six rigid walls require that the ﬂow velocity normal to the wall be zero on the six walls. A pattern of mode shape can be deﬁned by the nodal lines where the pressure is zero. As described for the 2D pressure in the previous section, the nodal lines can be determined by the standing waves of the sound pressure. An example of a mode shape and a detailed explanation of the nodal lines can be found at the end of the following example:

Example 7.3 (Cavity Resonant Frequencies)

A rectangular cavity has the dimensions (Lx, Ly, Lz) ¼ (5, 8, 10) [m].

= 8

̂ ̂

= = = 0

7.4 3D Boundary Conditions of Rectangular Cavities	187
(a) Calculate all resonant frequencies of	this cavity that are lower than

50 [Hz]. Indicate the associated mode number. For example, denote f213 for eigenmode (l, m, n) ¼ (2, 1, 3) corresponding to the dimensions (Lx, Ly, Lz).

(b)Plot the mode shape of the eigenmode (l, m, n) ¼ (0, 4, 7) using the pressure nodal lines, and indicate the peaks and valleys with “+” and “-” signs, respectively.

Example 7.3 Solution Part (a)

The frequency (combined) of standing waves in the cavities can be formulated using the following relationship of eigenmodes:

f lmn ¼ 2π

¼ 2π klmn ¼ 2

Lx þ

Ly þ

ωlmn

where the combined wavenumber klmn and combined frequency flmnof an eigenmode (l, m, n) are related to the discretized wavenumbers kxl and kym and kzn as:

klmn2 ¼ kxl2 þ kym2 þ kzn2
π	π	π

kxl ¼ Lx l; kym ¼ Ly m; kzn ¼ Lz n

The formula above shows that the larger the dimension (Lx,Ly or Lz) and the smaller the mode (l, m or n), the lower the modal frequency flmn.

As a result, the ﬁrst few lower modal frequencies can be calculated using combinations of the ﬁrst few modes. Lower modal frequencies correspond to larger dimensions. For this reason, if the dimensions of the cavity are deﬁned from short to long, the lower frequency mode will exist in the z-direction (the longest dimension).

The following procedures demonstrate a searching method for ﬁnding all the resonant frequencies of this cavity lower than 50 [Hz].

Step 1:

Set l¼0 and m¼0. Next, increase n by 1 from 1. Finally, calculate the combined frequency using the following formula until the calculated frequency is higher than the maximum prescribed frequency (50 Hz):

f lmn ¼ 2π	¼ 2		"	Lx þ			Ly þ		Lz		#
						2		2		2	1
ωlmn		c			l		m		n

Step 2:

Set l¼0 and m¼1. Next, increase n by 1 from 0. Finally, calculate the combined frequency using the above formula until the calculated frequency is higher than the prescribed frequency (50 [Hz]).

188	7 Resonant Cavities

Step 3:

Repeat Step 2 by increasing m by 1 until the calculated frequency is higher than the prescribed frequency maximum.

Step 4:

Increase l by 1 and set m¼0 and n¼0. Then, repeat Step 2 and Step 3 above until the calculated frequency is higher than the prescribed frequency maximum:

l	m	n	Frequency [Hz]	Note
0	0	1	17	< 50 (OK)

0	0	2	34	< 50 (OK)
0	0	3	51	> 50 (NOK)
Stop and increase m by 1
0	1	0	21.25	< 50 (OK)

0	1	1	27.21	< 50 (OK)
0	1	2	40.09	< 50 (OK)
0	1	3	55.25	> 50 (NOK)

Stop and increase m by 1

0	2	0	42.50	< 50 (OK)
0	2	1	45.77	< 50 (OK)
0	2	2	54.43	> 50 (NOK)

Stop and increase m by 1

0	3	0	63.75	> 50 (NOK)
Stop and increase l by 1
1	0	0	34	< 50 (OK)

1	0	1	38.01	< 50 (OK)

1	0	2	48.08	< 50 (OK)
1	0	3	61.30	> 50 (NOK)
Stop and increase m by 1

1	1	0	40.09	< 50 (OK)

1	1	1	43.55	< 50 (OK)
1	1	2	52.57	> 50 (NOK)
Stop and increase m by 1

1	2	0	54.43	> 50 (NOK)

Stop and increase l by 1
2	0	0	68.00	> 50 (NOK)
Stop

Frequencies lower than 50 [Hz] can be calculated by hand or in MATLAB or Excel as follows (partial list):

			2			2			2	1
	340	0	2		0	2		1	2	2	∙	1	½Hz& ¼ 17½Hz&
f lmn ¼ f 001 ¼	2	5		þ	8		þ	10		½Hz& ¼ 170	∙	10	½Hz& ¼ 17½Hz&

7.4 3D Boundary Conditions of Rectangular Cavities

189

																			1
	340	0	2				0				2			2			2		2							2
f lmn ¼ f 002 ¼	2	5			þ		8				þ			10					½Hz& ¼ 170 ∙							10		½Hz& ¼ 34	½Hz&
																			1
	340	0	2				0				2			3			2		2							3
f lmn ¼ f 003 ¼	2	5			þ		8				þ			10					½Hz& ¼ 170 ∙							10		½Hz& ¼ 51	½Hz&
																		1
340		0	2		1					2			0			2		2						1
f lmn ¼ f 010 ¼	2	5		þ		8					þ		10				½Hz& ¼							170 ∙	8		½Hz& ¼ 21:25 ½Hz&
									2						2						2		1
		340			0				2			1			2				1		2		2
f lmn ¼ f 011 ¼		2			5						þ	8				þ		10						½Hz& ¼ 27:21½Hz&
																						1
f lmn ¼ f 012 ¼		340			0			2		þ		1		2		þ			2	2		2		½Hz& ¼ 40:09 ½Hz&
f lmn ¼ f 012 ¼		2			5					þ		8				þ		10						½Hz& ¼ 40:09 ½Hz&
																						1
f lmn ¼ f 111 ¼		340			1			2		þ		1		2		þ			1	2		2		½Hz& ¼ 43:55 ½Hz&
f lmn ¼ f 111 ¼		2			5					þ		8				þ		10						½Hz& ¼ 43:55 ½Hz&

Part (b)

The pressure nodal lines can be determined by the following three properties of sound pressure:

Property #1: The relationship between wavelength and length of the cavity

The formulas of the discretized wavelengths shown below can be used for plotting the mode shapes:

λym

;

λzn

kym

kzn

where:

kym ¼

kzn ¼

m, n ¼ 1, 2, . . .

In the x-direction, the half wavelength of m ¼ 4 is:

λym ¼ Ly ¼ Ly

2 m 4

In the y-direction, the half wavelength of n ¼ 7 is:

λ2zn ¼ Lnz ¼ L7z

Property #2: Boundary conditions of ﬁxed surface

Pressure is maximum at ﬁxed boundaries because the ﬂow velocity on a ﬁxed surface is zero. We can use this property to align the harmonic function (cosine or sine function) to the boundaries.

Property #3: Multiplication of spatial functions

190 7 Resonant Cavities

The peaks and valleys can be determined by the multiplication of signs of ps(y, 0, t0) in y-direction and ps(0, z, t0) in z-direction because standing wave solutions are the result of multiplication of temporal function T(t) and spatial functions X(x), Y( y), and Z(z) as:

pðx, y, z, tÞ ¼ TðtÞXðxÞ YðyÞ ZðzÞ

¼ Plmn cos ðωt þ θtÞ cos ðkxlxÞ cos kymy cos ðkznzÞ

Therefore, the pressure nodal lines and valleys of eigenmode (l, m, n) ¼ (0, 4, 7) can be drawn as follows:

= 4	, 0,

̂ −1 × −1 = +1

-		+	-	+	-	+	-	′+ : Peak
								′+ : Peak
+		-	+	-	+	-	+	′− : Valley
								′− : Valley
-		+	-	+	-	+	-

	-		+	-	+	-	+
	-		+	-	+	-	+

= 7

The index l is zero in the eigenmode (l, m, n) ¼ (0, 4, 7) which means that this eigenmode has no mode shape in the x-direction. In other words, the sound pressure of this eigenmode is the same everywhere in the x-direction. Therefore, the standing wave of the sound pressure can be reduced to:

pðx, y, z, tÞ ¼ AtAxAyAz cos ðωt þ θtÞ cos ðkxlxÞ cos kymy cos ðkzmzÞ

! pðy, z, tÞ ¼ AtAxAyAz cos ðωt þ θtÞ cos kymy cos ðkzmzÞ

The pressure at y and z directions can be presented as:

pðy, z ¼ 0, tÞ ¼ AtAxAyAz cos

ωt

θt

cos

kymy

0, z, t

Þ ¼

t x y z

ð þ

tÞ

A A A A cos

ðωt

Þ cos

where:

7.4 3D Boundary Conditions of Rectangular Cavities								191
		π				π		4
kym ¼			m	¼
	Ly					Ly
		π			π		7
kzn ¼			n ¼
		Lz			Lz
Example 7.4 (Standing Wave Amplitude)								are given as (Lx, Ly, Lz) ¼
The dimensions of a rectangular cavity

(10, 5, 2) [m]. Assume that this rectangular cavity is positioned at the origin.

If the RMS pressure of mode (1,1,0) is 1 [Pa] at x ¼ 2 [m] and y ¼ 1 [m], what is the pressure magnitude of this mode?

Example 7.4 Solutions

The standing wave pressure solution of the wave equation is:

pðx, y, z, tÞ ¼ TðtÞXðxÞ YðyÞ ZðzÞ

¼ Plmn cos (ωt + θt) cos (kxlx) cos (kymy) cos (kznz)

[REP] where kxl, kym , and kzn are the discretized values that satisfy the boundary conditions of the rigid walls:

rad

kxl ¼

! kx1

kym ¼

rad

m ! ky1

rad

kzn ¼

! kz0

Therefore, the pressure at the given

location (x ¼ 2

[m],

¼ 1 [m],

and z ¼ 0 [m]) is:

ð π þ Þ

ð Þ

Þ ¼

p 1, 1, 0, t

Plmn cos

ωt θt cos

cos

¼ Plmn cos

2 cos 5 1 cos ðωt þ θtÞ

¼ A cos ðωt þ θtÞ

Note that in the above equation, A is the amplitude of cos(ωt + θt):

A ¼ Plmn cos

2 cos 5 1

Amplitude A is related to the RMS pressure of the standing wave (from the section of one-dimensional plane waves):

<<< < Предыдущая 31 32 33 34 35 36 37 38 39 40 41 4243 / 7843 44 45 46 47 48 49 50 51 52 53 54 55 > Следующая >>>

Соседние файлы в папке акустика

#
05.05.202343.36 Mб107beranek_l_akusticheskie_izmereniia.doc
#
04.05.20232.74 Mб81bergman_dr_computational_acoustics_theory_and_implementation.pdf
#
04.05.20235.27 Mб83gan_ws_acoustical_imaging_techniques_and_applications_for_en.pdf
#
04.05.20234.71 Mб80gan_ws_nonlinear_acoustical_imaging.pdf
#
04.05.202312.68 Mб84lin_h_bengisu_t_mourelatos_zp_lecture_notes_on_acoustics_and.pdf
#
04.05.20231.95 Mб96nagaenko_av_pezoelektricheskie_preobrazovateli.pdf
#
04.05.202360.48 Mб114self_douglas_the_design_of_active_crossovers.pdf
#
04.05.20234.58 Mб89speaks_ce_introduction_to_sound_acoustics_for_the_hearing_an.pdf
#
04.05.20234.46 Mб76tsvikker_k_kosten_k_zvukopogloshchaiushchie_materialy.doc
#
04.05.202311.15 Mб74tsvikker_k_kosten_k_zvukopogloshchaiushchie_materialy.pdf