Question 5 of 7: Composite transmission loss of an enclosure wall

Question 5: Composite transmission loss of an enclosure wall (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 6 m × 3 m enclosure wall pierced by a 0.5 m square window of 28 dB transmission loss and a 1 m × 2 m door of 25 dB transmission loss, with an overall target of 30 dB.

Find. The transmission loss the plain wall must achieve, the ceiling that the window and door impose on the whole construction, and the composite transmission loss once a 25 mm gap is left under the door.

Approach. Convert every stated transmission loss to a transmission coefficient, area-weight them to form the composite coefficient, and invert; the same relation, solved for the unknown element, answers all three parts.

1. Write the composite relation. Transmitted power is the sum of the power passing each element, so the area-weighted transmission coefficients add:$$\bar{\tau}=\frac{\sum_i S_i \tau_i}{S_{\text{total}}},\qquad \tau_i = 10^{-TL_i/10},\qquad TL = -10\log_{10}\bar{\tau}$$Decibels themselves never add here — that is the single most important structural fact about composite partitions.
2. Evaluate the coefficients of the known elements. $\tau_{\text{win}}=10^{-2.8}=1.585\times10^{-3}$ and $\tau_{\text{door}}=10^{-2.5}=3.162\times10^{-3}$, so their contributions are$$S_{\text{win}}\tau_{\text{win}}=0.25(1.585\times10^{-3})=3.962\times10^{-4}$$$$S_{\text{door}}\tau_{\text{door}}=2.00(3.162\times10^{-3})=6.325\times10^{-3}$$summing to $6.721\times10^{-3}$. The door alone accounts for 94 % of that, even though it is only 11 % of the wall area.
3. Solve for the plain wall (i). The 30 dB target requires $\bar{\tau}=10^{-3}$, hence a total budget of $S_{\text{total}}\bar{\tau}=18\times10^{-3}=1.800\times10^{-2}$. Subtracting the openings leaves $1.800\times10^{-2}-6.721\times10^{-3}=1.128\times10^{-2}$ for the 15.75 m$^2$ of plain wall:$$\tau_{\text{wall}}=\frac{1.128\times10^{-2}}{15.75}=7.161\times10^{-4}$$$$\boxed{TL_{\text{wall}} = -10\log_{10}(7.161\times10^{-4}) = 31.5\ \text{dB}}$$Rebuilding the composite from this answer returns exactly 30.0 dB, which closes the calculation.
4. Find the theoretical ceiling (ii). Let the plain wall become perfect, $\tau_{\text{wall}}\to 0$. All the leakage is then through the window and door:$$\bar{\tau}_{\min}=\frac{6.721\times10^{-3}}{18}=3.734\times10^{-4}$$$$\boxed{TL_{\max}=34.3\ \text{dB}}$$No amount of investment in the wall itself can beat 34.3 dB while that door and window are in place, and the 30 dB design is already within 4.3 dB of that ceiling. Practically, this means the next design move must be a better door, not a heavier wall.
5. Add the crack under the door (iii). The gap is 25 mm high, but the floor image doubles it to an effective 50 mm, so over the 1 m door width the effective open area is $S_{\text{crack}}=1.0\times0.050 = 0.05\ \text{m}^2$. At 500 Hz the wavelength is 0.69 m, so the gap is acoustically small and behaves as a simple open aperture with $\tau = 1$ ($TL = 0$ dB). Taking that area out of the door leaf, the new sum is$$\sum S_i\tau_i = 15.75(7.161\times10^{-4}) + 0.25(1.585\times10^{-3}) + 1.95(3.162\times10^{-3}) + 0.05(1.0) = 6.784\times10^{-2}$$$$\bar{\tau}=\frac{6.784\times10^{-2}}{18}=3.769\times10^{-3}$$$$\boxed{TL_{\text{with crack}} = 24.2\ \text{dB}}$$The crack is 0.28 % of the wall area and yet carries 74 % of all transmitted power, costing 5.8 dB of the 30 dB design. It also drops the construction below the 34.3 dB ceiling of part (ii) by more than 10 dB.

The engineering conclusion is unambiguous: sealing is not a finishing detail but a primary acoustic element. A door that is specified at 25 dB but hung with a 25 mm undercut for ventilation delivers roughly 13 dB in service, and the whole partition follows it down. Fitting an automatic drop seal or a threshold gasket recovers the full 5.8 dB at negligible cost compared with upgrading 15.75 m$^2$ of wall.