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Airflow Effect on Fire Sprinklers Exhaust Duct

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hvacnerd

Mechanical
Mar 25, 2005
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We are designing an exhaust system that requires fire sprinklers in the duct including the discharge stack. The fire marshall is concerned that the high velocity in the duct will blow the water up the stack and out. Does anyone know what velocity it would take to blow water up the stack? I have searched the internet, talked to fan sales reps and sprinkler reps and nobody has ever had a problem like this. Any input would be greatly appreciated.
 
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I'm not sure why they don't want the fans off. We have some calculations that should take care of it. Thanks for the input.
 
Just as a data point, 500 ft/min is a general rule-of-thumb for evaporator coil face velocity maximum. Above that the condensing water on the fins gets entrained in the airstream.
 
NFPA 45 explanatory material would have you maintain exhaust flow and not shut them off. Keeping the exhaust during a fire is better than shutting off exhaust, based on data from past fires. There is obviously not an easy answer because if you think about it a bit, it would take a very high air velocity to carry a water stream upward and a very low air velocity to carry a water mist upward. Here, you would have a combination of both, in no particular location or order.

I think your best reference for the fire authority would be ACGIH Industrial Ventilation. ACGIH recommends for exhaust stacks 2,600 feet per minute to keep rain water out of stacks and 3,000 fpm for a good exhaust plume (e.g., over building structures and away from intakes).

To me, the maximum rain water into an exhaust system would pale in comparison to sprinkler heads discharging within an exhaust duct. Based on this, I would think a much higher air velocity than 2,600 fpm would be required to prevent downwash of sprinkler water (i.e., to actually reverse the bulk of the water's direction and blow it upward).

The ACGIH data should allow you to demonstrate that because your exhaust velocity is below 2,600 fpm (which I hope it is), that the vast majority of the water will drain in its proper direction (downward).

-CB
 


Just an idea...

if the problem is simplified as force balance (downward by the mass of droplet and upward direction by airstrem force), below equation can be written.

we can assume the water droplets have spherical shapes (dia,R) ,lifted by air stream and hanged by zero velocity, airstream has velocity V and dynamic pressure V²/2g ,
simply balancing , F(air stream)=F(mass of droplet) ;


mass of droplet ,having force downward,F=ma
F=(4/3)x(3,14)xR³x density x 9,81 m/s²...(1)

pressure is force by unit surface (cross-section of sphere surface faced with airstream ,3,14*R²)

P=F/A
F=3,14 x R² x P ...................(2)

P= V²/(2x3,14) .....................(3)

3,14 x R² x V²/2x(3,14) = 4/3)x(3,14)xR³x density x 9,81 m/s²

V²=(8/3)x (3,14)² x density(1000 kg/m³)x R
V= 162,1 x R^(1/2)

assuming the droplets of water have 3 mm diameter;

R=0,003 m ; V =8,88 m/s ; V = 1574,8 ft/min

is the min velocity of air that can prevent water droplet to fall down
 
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