1.A wall between two rooms is 20 x 9 ft. The wall construction has Transmission Loss 50 dB at 1000 Hz. In the wall is a small window of size 4 x 2 ft. The window has a tested Transmission Loss of 20 dB at the same frequency. The receiving room has 200 ft^2 Sabine absorption. What is the Noise Reduction NR of the wall without the window?

Question

1.A wall between two rooms is 20 x 9 ft. The wall construction has Transmission Loss 50 dB at 1000 Hz. In the wall is a small window of size 4 x 2 ft. The window has a tested Transmission Loss of 20 dB at the same frequency. The receiving room has 200 ft^2 Sabine absorption. What is the Noise Reduction NR of the wall without the window?

2.Two identical offices are separated by a wall 15 x 8 ft. Sabine absorption in the offices is 150 ft2 . In one office the Sound Intensity Level is 72 dB. What transmission loss is required for the common wall if the SIL in the second office is to be 35 dB?

Answer 1

To calculate the Noise Reduction (NR) of the wall without the window, we need to understand a few concepts and equations used in acoustics calculations.

1. Transmission Loss (TL): TL is a measure of the reduction in sound energy as it travels through a barrier, such as a wall or window. It is typically measured in decibels (dB) and is dependent on the frequency of the sound.

2. Sabine absorption (A): Sabine absorption is a measure of the sound energy absorbed by a room's surfaces. It is typically measured in square feet (ft^2) and is used to calculate the reverberation time and the amount of sound energy reflected or absorbed in a room.

3. Noise Reduction (NR): NR is a measure of the reduction in sound intensity or sound pressure level offered by a particular barrier. It is calculated as the difference between the sound intensity level (SIL) in the source room and the overall SIL in the receiving room.

Now let's solve the first question:

1. Without the window, the wall itself has a Transmission Loss (TL_wall) of 50 dB at 1000 Hz.

2. We can calculate the total sound pressure level (SPL) reduction without the window using the formula:

TL_total = TL_wall + 10 * log10(A_receiving / A_source)

Where A_receiving is the Sabine absorption in the receiving room and A_source is the Sabine absorption in the source room.

3. In this case, the receiving room has 200 ft^2 Sabine absorption, so A_receiving = 200 ft^2.

4. As there is no mention of the source room's Sabine absorption, we assume it to be negligible or zero, so A_source = 0 ft^2.

5. Substituting the values into the formula:

TL_total = 50 + 10 * log10(200 / 0)
= 50 + 10 * log10(infinity)
= 50 + 10 * infinity
= infinity (As log10(infinity) approaches infinity)

Therefore, the total sound pressure level reduction without the window (NR_wall) is infinity dB.

Now let's move on to the second question:

1. We are given that two identical offices are separated by a wall, and the Sabine absorption in the offices is 150 ft^2.

2. One office has a Sound Intensity Level (SIL) of 72 dB, and we want to calculate the transmission loss required for the common wall if the SIL in the second office is to be 35 dB.

3. We can calculate the Transmission Loss (TL_required) using the formula:

TL_required = SIL_source - SIL_receiving

Where SIL_source is the SIL in the source room (72 dB) and SIL_receiving is the desired SIL in the receiving room (35 dB).

4. Substituting the values into the formula:

TL_required = 72 - 35
= 37 dB

Therefore, the transmission loss required for the common wall in this case is 37 dB.

Note: It's important to mention that windows and other openings in a barrier can significantly reduce its transmission loss, so it's crucial to consider them in acoustic calculations.