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22-Agric-B5 Power Units for Agricultural, Biosystems, and Food Industries · May 2015

Question 3 of 4: Occupational Noise Dose

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2015 — 04-Agric-B5, Power Units for Agricultural, Biosystems and Food Industries. Three-hour, open-book exam; any non-communicating calculator is permitted. Format: four questions constitute a complete paper, each of equal value; all questions require calculation.

Reference texts: Srivastava, Buckmaster & Hull, Engineering Principles of Agricultural Machines (2nd ed., ASABE) — sprocket/pulley speed ratios, indicated engine power, piston displacement and compression ratio; Kepner, Bainer & Barger, Principles of Farm Machinery (3rd ed.) — PTO power trains and tractor engine derating; OSHA 29 CFR 1910.95, Occupational Noise Exposure — noise-dose formula (5 dB exchange rate, as printed on this paper).

Problem 3: Occupational Noise Dose (equal value)

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. Noise dose $D(\%) = (C/T)\times100$, with reference duration $T(\text{h})=8/2^{(L-90)/5}$ (the paper's own 5 dB exchange-rate formula, criterion level 90 dBA / 8 h). (a) $L=55$ dBA, $C=1.25$ h. (b) $L=105$ dBA, $C=5$ min.

Find. The noise dose $D$ (%) for each exposure.

L=55, T=1024 h L=105, T=1 h 90 dBA, T=8 h (criterion) 11 9 7 5 3 1 log₂(T, hours) 50 70 90 110 Sound level, L (dBA)
Figure 3 — log₂(reference duration) vs. sound level is a straight line of slope −1/5: every 5 dB increase halves the permissible exposure time, per the source's own formula. The two evaluated points (55 dBA/1.25 h and 105 dBA/5 min) sit far apart on the T axis (1024 h vs. 1 h), which is why a log scale is used to show both at once.

Approach. Compute the reference duration $T$ at each sound level from the source's own formula, then form the dose $D=(C/T)\times100\%$.

  1. (a) Reference duration and dose at 55 dBA. $$T = \frac{8}{2^{(55-90)/5}} = \frac{8}{2^{-7}} = 8\times128 = 1024\ \text{h}$$ $$D = \frac{C}{T}\times100 = \frac{1.25}{1024}\times100 = \boxed{0.122\%}$$
  2. (b) Reference duration and dose at 105 dBA. $$T = \frac{8}{2^{(105-90)/5}} = \frac{8}{2^{3}} = 1\ \text{h}\qquad C=\frac{5}{60}=0.0833\ \text{h}$$ $$D = \frac{0.0833}{1}\times100 = \boxed{8.33\%}$$ Discussion: 55 dBA is far below the 90 dBA criterion level, so its permissible exposure balloons to 1024 h — 1.25 h there is a negligible (0.12%) dose. 105 dBA is well above the criterion, so its permissible exposure shrinks to just 1 h; even a brief 5-minute exposure already uses up 8.33% of the shift's allowable dose at that level, illustrating how sharply permissible time falls as sound level rises under a 5 dB exchange rate.
Case$T$ (h)Dose $D$
(a) 55 dBA, 1.25 h10240.122%
(b) 105 dBA, 5 min18.33%
Check: the source prints its own reference-duration formula $T=8/2^{(L-90)/5}$, which is the U.S. OSHA 5 dB exchange-rate convention. Most Canadian provincial OH&S jurisdictions instead use a 3 dB exchange rate ($T=8/2^{(L-85)/3}$) — the source's own formula governs this answer, per the "follow the question first" rule, since it is explicitly supplied.