Sound transmission between a machine room and an operator's room: 22-Mec-B11 Acoustics and Noise Control, May 2017, Q7

Question 7: Sound transmission between a machine room and an operator's room (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 105 dB machine of directivity factor 2.0 stands 4 m from the party wall of a large machine room; a 30 dB, 16 m$^2$ wall separates it from an operator's room in which the operator stands 1.5 m from that wall.

Given data
Quantity	Machine room (1)	Operator's room (2)
Total surface area, $S$	900 m$^2$	100 m$^2$
Mean absorption coefficient, $\bar{\alpha}$	0.05	0.35
Receiver distance	$r_1 = 4$ m (machine to wall)	$r_2 = 1.5$ m (operator to wall)
Machine	$L_W = 105$ dB, $D_r = 2.0$	—
Separating wall	$TL = 30$ dB, $S_w = 16$ m$^2$

Find. General expressions for the two steady-state levels, their numerical values, and a discussion of every available noise-control option for the operator's room.

Figure 7.1 — Coupled rooms. Power incident on the party wall from the reverberant field of room 1 is transmitted into room 2, where the wall itself becomes the source.

Approach. Apply the room equation twice: once with the machine as the source in room 1, and once with the party wall as the source in room 2, the wall's radiated power following from the incident reverberant intensity and the transmission coefficient.

Part (i) — Derivation

Machine room. The operator-side wall receives the direct field of the machine plus the reverberant field of room 1. Applying the room equation with the machine's directivity factor $D_r$ and the room constant $R_1 = S_1\bar{\alpha}_1/(1-\bar{\alpha}_1)$,$$\boxed{L_{p1} = L_W + 10\log_{10}\!\left(\frac{D_r}{4\pi r_1^{2}} + \frac{4}{R_1}\right)}$$
Power incident on and transmitted through the wall. In a diffuse reverberant field the intensity incident on any boundary is $I_{\text{inc}} = W/R_1$, so the power striking the party wall is $W_{\text{in}} = S_w W/R_1$ and the fraction that gets through is$$W_2 = \tau\,W_{\text{in}} = \frac{\tau S_w W}{R_1}$$which is precisely the relation offered in the hint. Taking ten times the logarithm and using $\tau = 10^{-TL/10}$,$$\boxed{L_{W2} = L_W - TL + 10\log_{10}\!\left(\frac{S_w}{R_1}\right)}$$
Operator's room. The wall panel now acts as the source, radiating $W_2$ into room 2. Because the panel sits in the boundary it radiates into a half space, so its directivity factor is 2, and the operator at $r_2$ again receives direct plus reverberant contributions:$$\boxed{L_{p2} = L_W - TL + 10\log_{10}\!\left(\frac{S_w}{R_1}\right) + 10\log_{10}\!\left(\frac{2}{4\pi r_2^{2}}+\frac{4}{R_2}\right)}$$with $R_2 = S_2\bar{\alpha}_2/(1-\bar{\alpha}_2)$. Every quantity the question asked for — $L_W$, $R_1$, $R_2$, $TL$ and $S_w$ — appears explicitly.

Part (ii) — Numerical evaluation

Room constants. $$R_1=\frac{900(0.05)}{1-0.05}=\frac{45}{0.95}=47.4\ \text{m}^2,\qquad R_2=\frac{100(0.35)}{1-0.35}=\frac{35}{0.65}=53.8\ \text{m}^2$$Note that the small, well-treated operator's room has the larger room constant — absorption, not size, is what governs it.
Machine-room level. The direct term is $D_r/4\pi r_1^2 = 2/(4\pi\times16)=9.95\times10^{-3}$ and the reverberant term is $4/R_1 = 0.0844$, so the reverberant field dominates by a factor of 8.5:$$L_{p1}=105+10\log_{10}(9.95\times10^{-3}+0.0844)=105-10.25$$$$\boxed{L_{p1}=94.7\ \text{dB}}$$
Power level radiated into the operator's room.$$L_{W2}=105-30+10\log_{10}\!\left(\frac{16}{47.4}\right)=105-30-4.71$$$$L_{W2}=70.3\ \text{dB}$$
Operator's level. The direct term from the panel is $2/(4\pi\times1.5^2)=0.0707$ and the reverberant term is $4/R_2 = 0.0743$; here the two are nearly equal, so the operator stands almost exactly at the hall radius of room 2:$$L_{p2}=70.3+10\log_{10}(0.0707+0.0743)=70.3-8.39$$$$\boxed{L_{p2}=61.9\ \text{dB}}$$The overall noise reduction achieved between the two rooms is $94.7-61.9=32.8$ dB — slightly better than the wall's own 30 dB rating, because the operator's room is well absorbed.

Part (iii) — Options for further reduction

The levels computed above identify where the leverage is. Room 2 is close to direct-reverberant balance at the operator's position, and $L_{p2}$ depends on room 1 only through its reverberant field, so both source-side and receiver-side measures are available. The realistic options, in the order a noise-control engineer would rank them, are as follows.

Control options and their expected benefit
Option	Mechanism	Expected effect
Reduce the machine's sound power at source	Balancing, bearing replacement, damping treatment, resilient mounts, quieter process parameters	dB-for-dB: every 1 dB off $L_W$ takes 1 dB off both rooms. Always the first choice in the hierarchy of controls.
Enclose the machine in room 1	An absorptively lined, sealed enclosure on vibration isolators intercepts the airborne path at source	15–25 dB is routine and it protects room 1 as well, where the 94.7 dB level is the real hazard.
Increase the wall transmission loss	Heavier or double-leaf construction with an isolated stud line and an absorbent cavity; seal all penetrations, and treat flanking through floor, ceiling and ducts	$L_{p2}$ falls dB-for-dB with $TL$: raising 30 dB to 40 dB gives exactly 10 dB. Only worth doing once flanking paths are controlled, or they will cap the gain.
Add absorption to room 1	Raising $\bar{\alpha}_1$ increases $R_1$ and so lowers the reverberant intensity incident on the party wall	$L_{p2}$ falls as $10\log_{10}R_1$. Going from $\bar{\alpha}_1=0.05$ to 0.20 raises $R_1$ from 47 to 225 m$^2$ and buys about 6.8 dB in both rooms.
Add absorption to room 2	Raising $R_2$ removes only the reverberant half of the operator's field	Limited: even $\bar{\alpha}_2=0.70$ buys only about 2.2 dB, because the direct field from the wall panel then dominates. Cheap, but not a solution on its own.
Move the operator away from the party wall	Reduces only the direct term $2/4\pi r_2^2$	Up to about 3 dB before the reverberant floor of room 2 is reached; combine with absorption in room 2 to go further.
Reduce the wall area $S_w$, or build an operator booth	$L_{p2}$ scales as $10\log_{10}S_w$; a small, heavy, well-sealed booth with a laminated viewing window replaces the whole room-to-room path	Halving $S_w$ gives 3 dB; a purpose-built booth routinely delivers 35–45 dB and is the standard answer where the machine cannot be quietened.
Administrative controls and hearing protection	Limiting occupancy time; Class A hearing protection	Last resort under Canadian OH&S practice. At 61.9 dB the operator's room is already below any hearing-conservation trigger — the issue there is speech interference and concentration, not hearing damage.

Ranking them, the machine-room level of 94.7 dB is the genuine hazard and it is unaffected by anything done to the wall, so source treatment or an enclosure should be specified first: it fixes both rooms at once. If the operator's room alone must be improved, raising the wall transmission loss returns the most decibels per dollar, provided flanking transmission through the floor slab, the ceiling void and any shared ductwork is controlled at the same time — otherwise the flanking paths, not the wall, will set the achievable limit.

Final results — Question 7
Part	Quantity	Result
i	Machine room	$L_{p1}=L_W+10\log_{10}\!\left(D_r/4\pi r_1^2+4/R_1\right)$
i	Operator's room	$L_{p2}=L_W-TL+10\log_{10}(S_w/R_1)+10\log_{10}\!\left(2/4\pi r_2^2+4/R_2\right)$
ii	Room constants $R_1$, $R_2$	47.4 m$^2$, 53.8 m$^2$
ii	Machine-room level $L_{p1}$	94.7 dB
ii	Power level radiated by the wall $L_{W2}$	70.3 dB
ii	Operator's level $L_{p2}$	61.9 dB
iii	Recommended first action	Source treatment or a lined machine enclosure (protects both rooms)

Topic: Coupled reverberant rooms — transmission through a common partition and the noise-reduction equation.

Key relations: $I_{\text{inc}} = W/R$ on any boundary of a diffuse field; $W_2 = \tau S_w W/R_1$; $L_{p2} = L_W - TL + 10\log_{10}(S_w/R_1)+10\log_{10}(2/4\pi r_2^2 + 4/R_2)$; the classical reverberant-to-reverberant form $NR = L_{p1}-L_{p2} = TL + 10\log_{10}(R_2/S_w)$; $R = S\bar{\alpha}/(1-\bar{\alpha})$.

Why this works: A diffuse field delivers the same intensity $W/R$ to every square metre of boundary, which is what lets a single number $R_1$ stand in for the whole complicated geometry of room 1 and turns the partition into an ordinary sound source of known power in room 2. The partition then radiates into a half space, so the receiving room is analysed with the same room equation used for any machine, with $Q=2$. The chain makes the design levers explicit and separable: source power, partition transmission coefficient, wall area, and the absorption in each room each enter as their own logarithmic term.

Common pitfall: Assuming the noise reduction between the rooms simply equals the wall's $TL$. The measured reduction also carries $10\log_{10}(R_2/S_w)$, so a large wall opening onto a live, hard receiving room can perform many decibels worse than the laboratory rating, while a small wall into a well-absorbed room performs better — as it does here, 32.8 dB against a 30 dB wall. The other classic pitfall is ignoring flanking: once the direct airborne path exceeds about 45–50 dB of isolation, structure-borne flanking usually governs and further upgrading the wall achieves nothing.

Source: Bies, Hansen & Howard, Engineering Noise Control, 5th ed., Ch. 7 (coupled rooms) and Ch. 8 (partitions, flanking transmission); Beranek & Vér, Noise and Vibration Control Engineering, Ch. 9 and Ch. 10.

Question 7 of 7: Sound transmission between a machine room and an operator's room

Question 7: Sound transmission between a machine room and an operator's room (20 marks)

Part (i) — Derivation

Part (ii) — Numerical evaluation

Part (iii) — Options for further reduction