Question 6 of 7: Room absorption design for a factory machine
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2017 — 16-Mec-B11 Acoustics and Noise Control; 3 hours; CLOSED BOOK (one approved Casio or Sharp calculator). Seven questions of equal value (20 marks each); FIVE questions constitute a complete paper. All seven are solved here so the set works as a study resource.
Reference texts. D. A. Bies, C. H. Hansen and C. Q. Howard, Engineering Noise Control, 5th ed. (CRC Press) — the standard reference for this exam code; L. L. Beranek and I. L. Vér, Noise and Vibration Control Engineering, 2nd ed. (Wiley); L. E. Kinsler, A. R. Frey, A. B. Coppens and J. V. Sanders, Fundamentals of Acoustics, 4th ed. (Wiley); CSA Z107 series and the provincial OH&S noise regulations for the Canadian occupational-exposure context.
Units and constants used throughout. Reference pressure $p_{\text{ref}}=20\ \mu\text{Pa}$; reference power $W_{\text{ref}}=10^{-12}\ \text{W}$; reference intensity $I_{\text{ref}}=10^{-12}\ \text{W/m}^2$. Where a question does not state the air temperature, air at $20\ {}^\circ\text{C}$ is assumed: $c=343\ \text{m/s}$, $\rho_0 c = 413\ \text{rayl}$. Question 3 states its own values and they are used there.
Question 6: Room absorption design for a factory machine (20 marks)
Given. A 0.01 W machine stands on the concrete floor at the centre of a 10 m × 10 m × 4 m room; the floor ($\alpha = 0.01$) cannot be treated, but the ceiling and all four walls can be lined.
Given data
Machine acoustic power, $W$
0.01 W $\Rightarrow L_W = 100$ dB
Room dimensions
10 m × 10 m × 4 m
Floor area (untreatable), $S_f$
100 m$^2$ at $\alpha_f = 0.01$
Ceiling + walls (treatable), $S_\ell$
$100 + 4(10\times4) = 260$ m$^2$
Total surface area, $S$
360 m$^2$
Machine position
on the floor, mid-room $\Rightarrow Q = 2$
Find. The absorption coefficient the lining must provide for an 83 dB reverberant field, the radius of the 90 dB contour, and the room constant that would correspond to an 85 dB reverberant field.
Figure 6.1 — Section through the factory bay. Near the machine the direct field dominates; away from it the level flattens out at the reverberant value.
Approach. Use the standard room equation, which splits the field into a direct term that falls off with distance and a reverberant term set by the room constant; inverting the reverberant term gives the required absorption, and equating the total to 90 dB gives the contour radius.
Set up the room equation and the sound power level. With $W = 0.01$ W, $L_W = 10\log_{10}(0.01/10^{-12}) = 100$ dB. The steady-state level at distance $r$ from a source of directivity factor $Q$ in a room of constant $R$ is$$L_p = L_W + 10\log_{10}\!\left(\frac{Q}{4\pi r^2}+\frac{4}{R}\right),\qquad R = \frac{S\bar{\alpha}}{1-\bar{\alpha}}$$Far from the source the first term vanishes and the reverberant level is $L_{p,\text{rev}} = L_W + 10\log_{10}(4/R)$.
Invert the reverberant term for the required room constant (i). Setting $L_{p,\text{rev}} = 83$ dB,$$10\log_{10}\!\left(\frac{4}{R}\right)=83-100=-17\ \text{dB}\quad\Longrightarrow\quad \frac{4}{R}=10^{-1.7}=0.01995$$$$R = 200.5\ \text{m}^2$$
Convert the room constant to a mean absorption coefficient and a total absorption. Solving $R = S\bar{\alpha}/(1-\bar{\alpha})$ for $\bar{\alpha}$ gives $\bar{\alpha}=R/(S+R)$:$$\bar{\alpha}=\frac{200.5}{360+200.5}=0.3577\quad\Longrightarrow\quad A = S\bar{\alpha}=360(0.3577)=128.8\ \text{m}^2\ \text{sabin}$$
Allocate the absorption between the untreatable floor and the lining (i). The concrete floor supplies only $S_f\alpha_f = 100(0.01) = 1.0\ \text{m}^2$ sabin, so the 260 m$^2$ of ceiling and walls must supply the remaining 127.8:$$\alpha_\ell = \frac{A - S_f\alpha_f}{S_\ell}=\frac{128.8-1.0}{260}$$$$\boxed{\alpha_\ell = 0.49}$$That is a routine specification — a 50 mm mineral-wool blanket behind a perforated facing comfortably exceeds $\alpha = 0.5$ across the mid frequencies — so the design is achievable with standard proprietary lining.
Locate the 90 dB contour (ii). Now the direct field matters. With $Q = 2$ (the machine sits on the floor, so it radiates into a hemisphere) and $R = 200.5\ \text{m}^2$, set $L_p = 90$ dB:$$10^{(90-100)/10}=0.1=\frac{2}{4\pi r^2}+0.01995$$so the direct term must supply $0.1-0.01995=0.08005$, giving $4\pi r^2 = 2/0.08005 = 24.99\ \text{m}^2$ and$$\boxed{r = 1.41\ \text{m}}$$Beyond about 1.4 m from the machine the level is below 90 dB, and it keeps falling only until it flattens at the 83 dB reverberant floor. The hall radius, where direct and reverberant contributions are equal, is $r_h=\sqrt{QR/16\pi}=2.82$ m — so the 90 dB contour lies inside the direct-field-dominated zone, as expected.
Find the room constant for an 85 dB reverberant field (iii). Repeating Step 2 with the higher target,$$\frac{4}{R}=10^{(85-100)/10}=10^{-1.5}=0.03162$$$$\boxed{R = 126.5\ \text{m}^2}$$Accepting 2 dB more reverberant noise cuts the required room constant by 37 %, which is why the last few decibels of reverberant control are always the expensive ones: the reverberant level falls only as $10\log_{10}R$.
