Building Acoustics

Statistical energy analysis (SEA) 271

where p represents the acoustic pressure in the room and u is the velocity we are seeking.
The quantity w is the energy density in the room, for which we have used the simple
expression valid for plane waves. For the corresponding modal densities we shall write

2

(^123)
0

12

and.

(^24) L

V

nn

c ch

ω

π^2 π

⋅S

== (7.12)

The quantities h and cL are the plate thickness and longitudinal wave speed, respectively.
The equation for n 1 is derived from the classical expression giving the modal density in a
room of rectangular shape (see section 4.4.1), which is

2

32

000

4

().

2 8

Nf fL

nf V S

f ccc

Δ ππ

=≈ ⋅+⋅+

Δ

(7.13)

The quantity ΔN is the number of natural frequencies inside the bandwidth Δf, S and L
are here the total surface area of the room and the sum of all the edges in the room,
respectively. Going to higher frequencies, the first term will become dominant and is
therefore used in Equation (7.12) besides the transformation using the angular frequency.
It now remains an expression for η 21 , representing the radiated power from the
plate, i.e. the power radiated back into the room resulting from the movement of the
plate. This output power, radiated from both sides of the plate may be written as

WcSuEmSuout=⋅ ⋅ ⋅ ≡2.ρσωηωη0 0 ^22221 = ⋅⋅ 21  (7.14)

Hence, the relationship between the coupling loss factor η 21 and the radiation factor σ
will be

21 00

2

.

c

m

ρ

η σ

ω

= ⋅ (7.15)

Inserting Equations (7.11), (7.12) and (7.15) into Equation (7.10), we arrive at the
relation between the velocity of the plate and the acoustic pressure in the room:

(^22)
0
(^222)
00 L
00

12 1

.

(^21)
2
u c
p chc m m
c
π
ρω ηω
ρ σ

=⋅

+

⋅





(7.16)

Example To put in some numbers we shall use a steel plate of 5 mm thickness being
driven by a sound field where the sound pressure level in the room is 100 dB inside a
one-third-octave band centred on 1000 Hz. The loss factor η 2 will be quite small as the
plate is freely hanging in the room; we may estimate η 21 ≈ 10 -4. Using the following data:
h – 5 mm, m – 13.5 kg/m^3 , cL – 5200 m/s, ρ 0 c 0 – 415 kg/m^2 s and c 0 – 340 m/s we obtain
the term

2
00

0.01

.

2

m

c

ηω

ρ σσ

≈

⋅