Question

You are building a recording studio, and want to have a high level of sound isolation between the live room and the control room. This wall will be 3 m wide and 2.5 m tall. You decide on a construction that will have a TL of 60 at 1000 Hz. However, you need to install a window in this wall, and the best window you can afford has a TL of 45 at 1000 Hz. If the window is to be 1.5 m wide and .75 m high, what will the composite TL of the wall be? For the studio above, what would the NR of the wall be if the total absorption in the receiving room is 15 metric sabins?

Answer 1

Step-by-step explanation:

To calculate the composite TL (Transmission Loss) of the wall with the window, we can use the following formula:

TL_composite = 10 * log10(10^(TL1/10) + 10^(TL2/10))

Where TL1 is the TL of the wall without the window, and TL2 is the TL of the window.

Given:

TL1 = 60 (TL of the wall without the window)

TL2 = 45 (TL of the window)

Plugging in the values:

TL_composite = 10 * log10(10^(60/10) + 10^(45/10))

TL_composite = 10 * log10(10^6 + 10^4.5)

TL_composite = 10 * log10(1,000,000 + 31,622.7766)

TL_composite = 10 * log10(1,031,622.7766)

TL_composite ≈ 10 * 6.013

TL_composite ≈ 60.13

Therefore, the composite TL of the wall with the window is approximately 60.13 at 1000 Hz.

To calculate the Noise Reduction (NR) of the wall, we can use the following formula:

NR = TL + 10 * log10(A)

Where TL is the composite TL of the wall and A is the total absorption in the receiving room in metric sabins.

Given:

TL = 60.13 (composite TL of the wall with the window)

A = 15 metric sabins (total absorption in the receiving room)

Plugging in the values:

NR = 60.13 + 10 * log10(15)

NR ≈ 60.13 + 10 * 1.1761

NR ≈ 60.13 + 11.761

NR ≈ 71.891

Therefore, the Noise Reduction (NR) of the wall with the window, given the total absorption in the receiving room is 15 metric sabins, is approximately 71.891.