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Powerhouse noise

Powerhouse noise

Powerhouse noise

(OP)
I have a hydro powerhouse with two 50MW generators. The generator housing is a fabricated steel structure approxamatly 30 ft dia, 12 feet high, inside a 50ft wide by 140 feet long, by 50 ft high building. There is a lot of resonate noise with a peak of 100dBA at 125 Hz. I am looking for a fire resitant noise solution. Any tips?

RE: Powerhouse noise

Your low frequency (125 Hertz) will not be easy to attenuate.  Your best bet is to solve the problem at the source i.e. enclosure or acoustic blankets.  Both solutions require lots of $$$.

RE: Powerhouse noise

You could use a fiberglass blanket with a 6 inch air gap behind it to obtain a high abosorption coefficient at low frequency.  The fiberglass should be 3 to 6 pounds per cubic foot and be encased in a 1mil vapor barrier.

RE: Powerhouse noise

I just joined eng-tips today, so this may be too late. A cellulose type insulation on some of the inside walls and ceiling would reduce reverberation. One type is K-13 from: http://www.internationalcellulose.com/ ;  A brochure gives 12 dB reduction at 125 Hz for a compressor house (should ask them how much of the building walls and ceiling were treated).

RE: Powerhouse noise

You can look up the absorption coefficient (alpha) for the fibreglass and space gap and then calculate the reduction in sound level from the extra absorption.  However consider the following points:

1. The building not including floors has 26,000 square feet of area.  If you acoustically insulate 50% that leaves 13,000 sq. ft. and at $7/sq ft. installed (my estimate), you are in the region of $70k+.  Just remember, this is a low estimate, try installing absorption panels on a 50 foot high ceiling with some steel work to provide the 6 inch gap and some sort of protection cover.  Fugitive fibres (loose fibreglass particles) are now becoming a concern, so you can't leave the fibreglass uncovered anymore.

2. 13,000 sq feet of uncovered plastic vapour barrier is a fire hazard!!!

3. Assuming that you don't have a strong standing wave (125 Hertz = approx 9 ft wavelength so I doubt it) , the direct sound propagation from the equipment may still be a problem.  Calculations will have to be done to estimate the noise reduction at various points in the building.  At any given point the total sound level is the sum of the reverberation sound and the direct sound.  Depending on location the direct sound may not change much and your problem is not solved.

4. The reduction in sound level is dependant on the total absorption, therefore the quote of 12 dB reduction can't be used because it may have assumed a 100% coverage in a small room.  In addition for a room as large as yours, air absorption becomes significant and adding additional absorption may require a high coverage and $$.

That is why I suggested earlier that you control the noise at source.  you really need to do a room reverberation calculation to estimate the noise reduction for various amounts of additional absorption.

Hope this helps.

RE: Powerhouse noise

K-13 is a spray-on insulation, also used for auditoriums. I have no idea what the installed cost/sq. ft. is; contact International Cellulose Corp. (1-800-444-1252).

RE: Powerhouse noise

so if 125 hertz = 9 foot wave -length would this be linear  and would 250 hertz = 36 foot wave-length

thanks
machine doctor

RE: Powerhouse noise

Wrong way around, the higher the frequency the smaller the wavelength.

The formula is:

wavelength (lambda) = (speed of sound)/frequency

speed of sound = 344 m/s
frequency = Hertz
Therefore units of lambda is metres.

For 125 Hertz 344/125=2.75 metres or 9 feet approx.
For 250 Hertz 344/250=1.38 metres or 4.5 feet approx.

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