Fan Calculation 1

[proprot]

Hi

I am trying to calculate if the fan I have installed at a particular site is to large. The dimensions of the room are 4.7m x 3.0m x 3.1m that gives me a total volume of 43.71m^3 the inlet size of a single louvre vent into the room is 0.500m x 0.400m that gives me an inlet area of 0.2m^2 the amount of air changes required per hour is 1.9 the outlet size of the exhaust fan is 0.398m in diameter . The fan would draw air form the external atmosphere through the single louvre inlet vent into the room and expel it through the exhaust outlet where the fan is mounted to the external atmosphere. At 1.9 air changes per hour I calculated the required velocity using 43.71m^3 x (1.9/60) to be 1.3845 m^3/s the current fan uses the 1440 rpm curve but I see the value of 1.385 falls quite close to the origin of the chart. Furthermore ploting this value will omly give me an equivalent static pressure value in Pa, once I have this value what do I use it for or how do I use it to verify adequate fan sizing

 

[LittleInch]

"43.71m^3 x (1.9/60) to be 1.3845 m^3/s"

Errr, I think you'll find converting 1.9 changes per HOUR to m3/SECOND requires you to divide by 3600, not 60.

Some common sense should apply here. At an area of 0.2m2, this implies a velocity of 7 m/sec / 15 mph - That's a pretty strong breeze - gale force 4 (moderate wind)

Pressure will relate then to how much flow you can get out of your vent with that pressure. ASHRAE guidelines maybe?

Where does 1.9/hr come from? Seems very low according to this guide

http://www.engineeringtoolbox.com/air-change-rate-room-d_867.html

Equally 1.9 per minute or 114 per hour sounds rather big.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[proprot]

Hi

The application is for a battery room and the 1.9 comes from a calculation used to determined the changes per hour required in the room to keep the H2 (Hydrogen) concentration to below 0.8% this is when the battery's are gassing.

So on correcting my previous error the 43.71m^3 x (1.9/3600) will be 0.0231 m^3/s this falls even lower on the chart so this would give a velocity of 0.115 m/sec

But this in only considering the velocity flow through the inlet vent based on air change requirements and room size. Once I have this how would I select an adequate fan size ? would I opt for the 960rpm or 1440rpm fan. Given that this value is rather small 0.0231 m^3/s I would think both fans on the curve data sheet are to large

 

[LittleInch]

Everything is the wrong size (far far too big).

At 0.115m/sec your area is far too big - you won't find a fan to fit that and do that sort of flowrate / velocity

You need something like this

https://www.tlc-direct.co.uk/Main_Index/Ventilation_Index/5_Inch_Fans/

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[proprot]

Thanks

So essentially the fan size can be reduced dramatically. I looked at a the link you posted and plotted the 0.115 m/s on the curve it gives me a static pressure rating what would I use this for. The fan is venting into the open atmosphere essentially extracting the gas contents of the room and drawing in fresh air through the louvre inlet.

 

[ChasBean1]

0.023 m3/s gives you your 1.9 air changes per hour. All the velocity information you gave is superfluous. For the 1440 rpm curve, the lowest you get is 1.5 m3/s. So your fan is about 65 times too large.

 

[chicopee]

Since H2 is lighter than air, you may be overestimating the volume of the space that should be vented. I would also base my calculations on the allowable concentration of hydrogen which could reduce the 1.9 air change per hour and save on resources. The trick would be to find a H2 sensor that has limits on hydrogen gases and which could be connected to the exhaust fan.

 

[willard3]

You should hire an engineer who understands fans rather than trying to bootstrap this on a web forum