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H. DUCT SOUND BREAKOUT AND BREAKIN

# H. DUCT SOUND BREAKOUT AND BREAKIN

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SOUND

Equation 11-43

where Wj is the sound power (watts) in the duct, Wr
is the sound power (watts) radiated from the duct, Ai
is the cross sectional area (in2) of the inside of the
duct, and Ao is the sound radiation surface area (in2)
of the outside of the duct. Rearranging Equation 1143 yields

AND

where a is the larger duct cross-section dimension
(in), b is the smaller duct cross-section dimension
(in), and L is the exposed length (ft) of the duct [Figure 11-15(a)]. For rectangular ducts, the breakout
transmission loss curve shown in Figure 11-16 can
be divided into two regions: (1) a region where plane
mode transmission within the duct is dominant and
(2) a region where multi-mode transmission is dominant. The frequency, f,, that divides these two regions
is given by

Equation 11-44

where Lwr (dB) and Lwi (dB) are given by
Equation 11-45

VIBRATION

Equation 11-52

If f < f,, the plane mode predominates and TLout, is

given by
Equation 11-53

Equation 11-46

The breakin transmission loss, TLin (dB), associated
with ducts is given by
Equation 11-47

where Wi is the incident sound power (watts) on the
duct from the surrounding space and Wt is the sound
power (watts) that travels along the duct both upstream and downstream from the point where the
sound enters the duct. The sound power level of the
sound transmitted into the duct is obtained by rearranging Equation 11-47, or
Equation 11-48
where Lw, is given by equation 11-46 and L,, is given
by
Equation 11-49

where f is frequency (Hz), q is the mass/unit area (lb/
ft2) of the duct walls, and a and b are as described
above. If f >- f,, multi-mode transmission predominates and TLout is calculated from

Equation 11-54
where q and f are as specified above. The minimum
value of TLout occurs when Wi = Wr and is specified
by
Equation 11-55

Table 14-41 in Chapter 14 shows some values of TLout
calculated using the above equations.
The breakin transmission loss can be divided into
two region'swhich are separated by a cutoff frequency f1.The cutoff frequency is the frequency for
the lowest acoustic cross-mode in the duct. It is given
by
Equation 11-56

2. Rectangular Ducts
If the duct is a rectangular duct, Ai and Ao in Equations 11-43 and 11-44 are given by
Equation 11-50
Ai = a x b
Equation 11-51
Ao = 24 x L x (a + b)

11.24

CHAPTER 11

Figure 11-16 TLout ASSOCIATED WITH
RECTANGULAR DUCTS

Equation 11-58

The results are tabulated below.

TLin = TLout- 3

Table 14-42 shows some values of TLin calculated
using the above equations.

Example 11-13
Determine the breakout and breakin sound power for
a duct with the following dimensions: smaller duct
dimension-12 inches; larger duct dimension-24
inches; duct length-20 feet. The duct is constructed
of 24 gauge sheet metal. q = mass/unit area of 24
gauge sheet metal = 1.0 lb/ft2.

Solution

The results are tabulated below.

Sound breakout:
Ai = 24 x 12 = 288 in2
Ao = 24 x 20 x (24 + 12) = 17,280 in2

11.25

SOUND

3. Round Ducts
If the duct is round, Ai and Ao in Equations 11-43 and
11-44 are given by
Equation 11-59

AND

VIBRATION

where q is the mass/unit area (lb/ft2) of the duct wall,
f is frequency (Hz), d is the inside duct diameter
(inches), and
Co = 230.4 for long seam ducts
Co = 232.9 for spiral wound ducts
Equation 11-63
TLout

Equation 11-60
where d is the duct diameter (inches) and L is the
exposed length (feet) of the duct. Narrow band and
1/3 octave band breakout transmission loss values
for round ducts are very hard to predict and no simple
prediction techniques are available. However, if the
analysis is limited to 1/1 octave frequency bands,
TLout associated with round ducts can be approximated by a curve similar to the one shown in Figure
11-17. Table 14-43 shows experimentally obtained
TLout data for round ducts. If the breakout analysis is
limited to 1/1 octave band values, Equations 11-61
and 11-62 can be used to approximate the data in
Table 14-43.
Equation 11-61

= the larger of TL1,2

The above equations yield good results except when
the diameter of the duct is equal to or greater than
26 inches and the 1/1 octave band center frequency
is equal to 4000 Hz. For this special case TLout is
given by
Equation 11-64
TLout = 176 log10[q] - 36.9 log10[d] + 90.6
The maximum allowable value for TL9out is 50 dB.
Thus, if the value for TL, obtained from equation 1163 exceeds 50 dB, the value should be set equal to
50 dB. Table 14-44 lists the calculated values for
TLout.
For calculating the breakin transmission loss for
round ducts, the cut-off frequency for the lowest
acoustic cross-mode is given by

TL, = 176 log10[q] - 49.8 log10[f]
- 55.3 log10[d] + Co

Equation 11-65
Equation 11-62

TL2 = 17.6 log10[q] - 6.6 log10[f]
- 36.9 log10[d] + 97.4

Figure 11-17 TLout

11.26

ASSOCIATED WITH ROUND
DUCTS

CHAPTER 11

Equation 11-66a and 11-66b

The results are tabulated below.

If f > f1,the breakin transmission loss is defined by
Equation 11-67
TLin = TLout - 3

Table 14-45 in Chapter 14 gives values for the breakin
transmission loss for various duct sizes obtained
from experimental data. Table 14-46 gives the corresponding values calculated using the above equations.

Example 11-14
Determine the breakout and breakin sound power of
a long seam round duct given the following information: diameter-14 inches; length-15 feet. The duct
is constructed of 24 gauge sheet metal.

Solution
q = mass/unit area of 24 ga sheet metal
= 1.0 lbm/ft2

Sound breakout:

4. Flat Oval Ducts
If the duct is a flat oval duct, Ai and Ao in Equations
11-43 and 11-44 are given by
Equation 11-68

Equation 11-69
Equation 11-70
where a is the length (inches) of the major duct axis,
b is the length (inches) of the minor duct axis, L is
the duct length (feet), Ai is the cross-section area
(in2), Ao is the surface area of the outside of the duct
(in2), and P is the perimeter of the duct in inches
(Figure 11-18). The fraction of the perimeter taken up
by the flat sides, o,is given by
Equation 11-71

The results are tabulated below.

Figure 11-18 FLAT OVAL DUCT

11.27

SOUND

The minimum breakout transmission loss, TLout (min)
(dB), or flat oval ducts is given by
Equation 11-72

AND

Solution
q = mass/unit area of 24 ga. sheet metal
= 1.0 Ibm/ft2

Sound breakout:
The low-to-mid frequency transmission loss, TLout
(dB), associated with flat oval ducts is specified by
Equation 11-73

The upper frequency limit, fL (Hz), of applicability of
Equation 11-73 is
Equation 11-74

Table 14-47 in Chapter 14 gives some values of TLout
for flat oval ducts of various sizes.
As was the case with rectangular and circular ducts,
TL,, can be written in terms of TLout. While there are
no exact solutions for the cut-off frequency for the
lowest acoustic cross-mode in flat oval ducts, Equation 11-75 gives an approximate solution.
Equation 11-75

where a and b are in inches. This equation is valid
when a/b >- 2. When a/b < 2, the accuracy of Equation 11-75 deteriorates progressively as a/b approaches unity. When f <- f1,TLin is given by
Equation 11-76a and 11-76b

When f > f1, TLin is given by
Equation 11-77
TLin= TLout - 3

Table 14-48 gives TL,, values for the duct sizes listed
in Table 14-47.

Example 11-15
Determine the breakout and breakin sound power of
a flat oval duct given the following information: major
axis-24 inches; minor axis-6 inches; length-20
feet. The duct is constructed of 24 gauge sheet metal.

11.28

VIBRATION

The results are tabulated below.

The results are tabulated below.

CHAPTER 11

5. Insertion Loss of External
Duct Lagging
External acoustic lagging is often applied to rectangular ductwork to reduce the transmission of sound
energy from within the duct to surrounding areas. The
lagging usually consists of a layer of soft, flexible,
porous material, such as fiberglass, covered with an
outer impervious layer (Figure 11-19). A relatively rigid
material, such as sheet metal or gypsum board, or a
limp material, such as sheet lead or loaded vinyl, can
be used for the outer covering.
With respect to the insertion loss of externally lagged
rectangular ducts, different techniques must be used
for rigid and limp outer coverings. When rigid materials are used for the outer covering, a pronounced
resonance effect between the duct walls and the
outer covering usually occurs. With limp materials the
variation in the separation between the duct and its
outer covering dampens the resonance so that it no
longer occurs. For both techniques, it is necessary to
determine the low frequency insertion loss, IL(lf) (dB).
It is given by
Equation 11-78

mass per unit area of the duct (lb/ft2), and M2 is the
mass per unit area of the outer covering (lb/ft2). P1
and P2 are specified by
Equation 11-79
P1 = 2 (a + b)

Equation 11-80
P2 = 2 (a + b + 4 h)

where a is the duct width (inches), b is the duct height
(inches), and h is the thickness (inches) of the soft,
flexible, porous material between the duct wall and
the outer covering.
If a rigid outer covering is used, it is necessary to
determine the resonance frequency, fr (Hz), associated with the interaction between the duct wall and
outer covering. fr is given by

Equation 11-81

where M1,M2, P1, and P2 are as previously defined.
S is the cross-section area (in2) of the absorbent
material and is given by
Equation 11-82
S = 2 h x (a + b + 2 h)

where P1 is the perimeter of the duct (inches), P2 is
the perimeter of the outer covering (inches), M, is the

The following procedures for determining the insertion loss for external duct lagging should be used for
rigid and limp outer coverings.

Figure 11-19 EXTERNAL DUCT LAGGING ON
RECTANGULAR DUCTS

11.29

SOUND

a. RIGID COVERING MATERIALS
If 1/3 octave band values are desired, draw a line
from point B (0.71 fr) to point A (fr) on Figure 11-

20(a). The difference in IL (dB) between points B and
A is 10 dB. The equation for this line is
Equation 11-83

Next draw a line from point A (fr) to point C (1.41 fr)
on Figure 11-20(a). The equation for this line is
Equation 11-84

From point C (1.41 fr), draw a line with a slope of 9
dB/octave. The equation for this line is
Equation 11-85

If 1/1 octave band values are desired, use Equation
11-78 for the 1/1 octave bands below the one that
contains fr. For the 1/1 octave band that contains fr,

subtract 5 dB from IL(lf) obtained from Equation 1178. For the 1/1 octave bands above the one that
contains fr, use Equation 11-85.

b. LIMP COVERING MATERIALS
Since there is no pronounced resonance with limp
covering materials, the low frequency insertion loss,

AND VIBRATION

IL(lf), is valid up to fr,after which the insertion loss
increases at a rate of 9 dB per octave [Figure 1120(b)]. For frequencies above fr, the equation for insertion loss is
Equation 11-86

The insertion loss of duct lagging probably does not
exceed 25 dB.
The insertion loss predictions using the procedures
described above should be fairly accurate up to about
1,000 Hz for most ducts. Duct lagging may not be a
particularly effective method for reducing low frequency (<100 Hz) duct sound breakout. A more effective method for reducing duct breakout is the use
of round ductwork, which has a high transmission
loss at low frequencies.
Example 11-16
Determine the 1/1 octave band insertion loss associated with the external lagging of a rectangular sheet
metal duct with the following characteristics: duct dimensions-8 in x 8 in; duct constructed of 18 gauge
sheet metal; thickness of absorbent material-1 inch;
outer covering-1/2 inch gypsum board.

Solution
Ml = mass/unit area of 18 gauge sheet metal
= 2.0 lb/ft2

Figure 11-20 INSERTION LOSS ASSOCIATED WITH RECTANGULAR EXTERNAL DUCT LAGGING

11.30

CHAPTER 11

M2 = mass/using area of one sheet of 1/2 inch
gypsum board = 2.1 lb/ft2
P1 = 2(8 + 8) = 32 inches
P2 = 2(8 + 8 + 4 x 1) = 40 inches

usually large rectangular enclosures with an inlet and
one or more outlet sections. The transmission loss
associated with a plenum chamber can be expressed
as

Equation 11-87

Thus,

Referring to Figure 11-21, Sout is the area (ft2) of the
output section of the plenum, S is the total inside
surface area (ft2) of the plenum minus the inlet and
outlet areas, r is the distance (feet) between the centers of the inlet and outlet sections of the plenum, and
aA is the average absorption coefficient of the plenum
lining. aA is given by

Equation 11-88
The results are summarized below.

J

DUCT ELEMENT SOUND
ATTENUATION

The duct elements covered in this section include:
sound plenums, unlined rectangular ducts, acoustically lined rectangular ducts, unlined circular ducts,
acoustically lined circular ducts, elbows, acoustically
lined circular radiused elbows, duct silencers, duct
branch power division, and duct end reflection loss.
O

where a, and S1 are the sound absorption coefficient
and corresponding surface area (ft2) of any bare or
unlined inside surfaces of the plenum chamber and
a2 and S2 are the sound absorption coefficient and
corresponding surface area (ft2) of the acoustically
lined inside surfaces of the plenum chamber. In many
situations, 100 percent of the inside surfaces of a
plenum chamber are lined with a sound absorbing
material. For these situations, aA = a2.
Q in Equation 11-87 is the directivity factor which
equals 2 if the inlet section is near the center of the
side of the plenum on which it is located. This corresponds to the situation where sound from the inlet
section of the plenum chamber is radiating into half
space. Q equals 4 if the inlet section is located in the
corner where two sides of the plenum come together.
This corresponds to the situation where sound from
the inlet section is radiating into quarter space.
in Equation 11-87 is the angle of the vector representing r relative to the horizontal plane. cos 0 and r
can be written
Equation 11-89
Equation 11-90

1. Plenum Chambers
The plenum chamber is usually placed between the
discharge section of a fan and the main duct of the
air distribution system. These chambers are usually
lined with acoustically absorbent material to reduce
fan and other types of noise. Plenum chambers are

where rh and rv are the horizontal and vertical distances (ft), respectively, between the inlet and outlet
sections of the plenum (Figure 11-21).
Equation 11-87 treats a plenum as if it is a large
enclosure. Thus, Equation 11-87 is valid only for the

11.31

SOUND

AND

VIBRATION

Figure 11-21 SCHEMATIC OF A PLENUM
CHAMBER
case where the wavelength of sound is small compared to the characteristic dimensions of the plenum.
For frequencies which correspond to plane wave
propagation in the duct, the results predicted by
Equation 11-87 are usually not valid. Plane wave
propagation in a duct exists at frequencies below
Equation 11-91

where o is sound attenuation per unit length in the
chamber (dB/ft), I is the horizontal length of the
plenum chamber (feet), co is the speed of sound in
air (ft/sec), f is frequency (Hz), and m is the ratio of
the cross-sectional area of the plenum divided by the
cross-sectional area of the inlet section of the plenum.
m is given by
Equation 11-94

where co is the speed of sound in air (ft/sec) and a
is the larger cross-section dimension (feet) of a rectangular duct, or below
Equation 11-92

where d is the diameter (feet) of a circular duct. The
cutoff frequency, fco, is the frequency above which
plane waves no longer propagate in a duct. At these
higher frequencies the waves that propagate in the
duct are referred to as cross or spinning modes. At
frequencies below fco, the plenum chamber can be
treated as an acoustically lined expansion chamber.
The equation for the transmission loss of an acoustically lined expansion chamber is
Equation 11-93

For frequencies less than fco, the transmission loss
of a plenum is given by Equation 11-93. For frequencies greater than or equal to fco, the transmission loss
of a plenum is given by Equation 11-87 fcoassociated
with Equations 11-91 and 11-92 is calculated on the
bases of the inlet section of the plenum. Table 14-49
gives the absorption coefficients of typical plenum
materials.
Equations for olfor the 1/1 octave frequency bands
from 63 Hz to 500 Hz are:
Equation 11-95
63 Hz: ol= [0.00306 x (P/A)1.959 x t0.917] x I
Equation 11-96
125 Hz: ol
= [0.01323 x (P/A)'4'0 x t0.941] x l
Equation 11-97
250 Hz: ol = [0.06244 x (P/A)0.824 x t1.079] x j
Equation 11-98
500 Hz: ol= [0.23380 x (P/A)0.500 x t1.087] x l
where P/A is the perimeter (P) of the cross-section
of the plenum chamber (feet) divided by the area (A

11.32