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24-Bld-A4 Building Environmental Control Systems · May 2017

Question 4 of 14: Reverberant-Field Sound Pressure Level

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

Notes on this paper

National Exams May 2017 — 07-Bld-A4, Building Environmental Control Systems (Building Engineering). 3-hour closed-book exam: Section 1 (four 20-mark essay/calculation questions) + Section 2 (ten 2-mark multiple-choice questions).

Reference texts: ASHRAE Handbook — Fundamentals (2021 ed.); ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality (2022 ed.); McQuiston, Parker & Spitler, Heating, Ventilating, and Air Conditioning: Analysis and Design (6th ed.); National Building Code of Canada 2020 (NBCC); National Energy Code of Canada for Buildings 2020 (NECB).

Question 1.4: Reverberant-Field Sound Pressure Level (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. Room 15 m × 8 m × 5 m (V = 600 m³). Sound power level Lw = 89 dB in the 500 Hz band. Reverberation time T = 0.9 s. Machine radiates omnidirectionally but sits ON THE FLOOR (a hard, reflecting boundary), so it radiates into a hemisphere — directivity factor Q = 2.

Find. The sound pressure level (SPL) in the 500 Hz band at r = 2 m and r = 4 m from the machine.

machiner = 2 mr = 4 mRoom plan: 15 m × 8 m (H = 5 m)
Fig. 1.4 — Plan view: floor-mounted machine at room center, evaluation radii 2 m and 4 m.

Approach. Use Sabine's equation to back out the room's total absorption (A, in metric sabins) from the given reverberation time, then combine the direct and reverberant sound fields with the standard room-acoustics SPL formula.

  1. Total room absorption from Sabine's equation. $$\boxed{A = \dfrac{0.16\,V}{T} = \dfrac{0.16\times600}{0.9} = 106.7\ \text{m}^2\ \text{sabins}}$$
  2. Combine direct and reverberant fields. The standard room-acoustics relation for SPL at distance r from a source of sound power level Lw, treating the Sabine absorption A directly as the room constant (a standard simplification when only T — not the room's surface area or mean absorption coefficient — is given, i.e. assuming the mean absorption coefficient is small enough that 1−ᾱ≈1): $$\text{SPL} = L_w + 10\log_{10}\!\left(\dfrac{Q}{4\pi r^2}+\dfrac{4}{A}\right)$$ with Q = 2 for the floor-mounted (hemispherical-radiation) source.
  3. Evaluate at r = 2 m. Direct term $Q/(4\pi r^2)=2/(4\pi\times4)=0.0398$; reverberant term $4/A=4/106.7=0.0375$; sum = 0.0773. $$\boxed{\text{SPL}_{2\,m} = 89+10\log_{10}(0.0773) = 77.9\ \text{dB}}$$
  4. Evaluate at r = 4 m. Direct term $2/(4\pi\times16)=0.00995$; reverberant term unchanged at 0.0375; sum = 0.0474. $$\boxed{\text{SPL}_{4\,m} = 89+10\log_{10}(0.0474) = 75.8\ \text{dB}}$$
Check: room constant R is approximated by the Sabine absorption A itself (i.e. 1−ᾱ≈1); the source is assumed to radiate hemispherically (Q=2) because it sits directly on the floor. Neither the surface area/mean absorption coefficient nor an explicit directivity factor was given in the source, so this is the standard textbook assumption for this class of problem.
Final Results — Question 1.4
QuantityValue
Total room absorption, A106.7 m² sabins
SPL at r = 2 m77.9 dB
SPL at r = 4 m75.8 dB

Section 2 — Multiple Choice (2 marks each)