Entrainment rate of wind over the top of a stack

[Befuddled]

Hi all,

I am newbie trying to work something out, so I'll try my best to explain.

I have a large warehouse (L x W x H) 9.5m x 9.6m x 46.3m in size, which has excellent ventilation coming in through large mesh screens at one end which are sheltered from wind.

There is old vertical stack at one end with a constant diameter of 2.4 m in the roof which is open to atmosphere and is approximately 37m in height (measured from the ground)

The stack has a regular circular flat top (no cowl etc), just straight out.

If i have a wind speed of 4.5m/s, is it possible to calculate an approximate value for how long it would take for the air inside the warehouse to be changed by just relying upon the stack effect (no forced ventilation from fans etc)

As the wind is a cross wind across the top of the stack, i'm not sure its as simple as saying the vertical velocity in the stack is 4.5 m/s (i feel it should be lower as its being entrained due to the moving air above the stack outlet)

There is also no temperature difference as there is no heating in the warehouse

Anyone got any pointers or useful formulas? i just need a rough rule of thumb way to determine a value.

Thanks in advance

Befuddled..

 

[LittleInch]

Where is the 4.5m/sec wind?

If its not blowing into the building and you have no temperature difference, there is no stack effect. Probably only a low noise like blowing across the top of a bottle.

Or at least in this very similar question https://www.eng-tips.com/threads/stack-effect.451536/

The best bet is probably one of these https://www.ventilation-alnor.co.uk/assets/files/technical-sheets/WD-TURBO-EN-Rotating-chimney.pdf

or similar. Biggest seem to be about 500 to 600mm so weld a plate to the top of your stack and fit a whole bunch of them?

Oh and are you sure about those dimensions? "(L x W x H) 9.5m x 9.6m x 46.3m" isn't W x H x L? Otherwise that's more like a silo.

 

[Snickster]

The flow of 4.5 m/sec across the top of the stack will not "entrain" air and cause it to flow through the stack. What it will do is cause a lower pressure than atmospheric at the stack exit due to the velocity relative to still air. Air at 4.5 m/sec produces a lower air pressure of 0.05 in. w.g. relative to still air (H =(V/4005)^2 in. wg. where V is in feet per minute). Air inside of the warehouse is still and therefore there will be a differential pressure of 0.05 in. wg available to push the air through the chimney. This assumes no temperature differences to further increase the driving differential pressure due to the density difference stack effect.

I don't have values for friction loss in a brick stack, but I have a Ductulator for friction loss in round smooth duct. If you assume that all the available pressure differential of 0.05 in wg is available for flow through the stack (no building inlet air losses) then the flow that will develop will be:

Total pressure loss = 0.05 in. wg = Head loss friction + velocity head of flow exiting chimney.

Most of the available pressure is used up in velocity head of flow H = (V/4005)^2

V = velocity in feet per minute
H = velocity head in inches water gage.

The actual flow will be such that the velocity head plus friction loss in stack = 0.05 inches wg.

I calculate this to be 36,000 CFM for 94.5 in. ID stack.

If you assume that about half of the available differential pressure (0.025" wg) is pressure loss across the inlet opening of the building (air must enter the warehouse somewhere or there will be no flow), then only 0.025" is available for flow through the stack. In this case I calculate 27,000CFM flow through the stack.

Actual flows will be somewhat lower due to friction factor of brick chimney versus smooth duct.

Note that without friction in the stack then all of the available pressure differential of 0.05" wg will go into velocity head inside the chimney since there is no friction loss. In this case the velocity attained in the chimney would be the same velocity as across the top of the stack since velocity across stack produces 0.05" wg which would be the total energy input to velocity in the stack without friction so H = 0.05" = (V/4005)^2 .

So V in chimney without friction comes out to same across top of chimney and flow would be V x A = 4.5 m/sec (Pi/4) (2.4)^2 = 20 m^3/sec = 43,102 CFM. However there will be friction which uses some of the available pressure differential so it needs to be taken into account as I did above.

 

[3DDave]

Counter to this effect is if the stack is colder than the factory. Lower temp = higher density and the pressure change may not be enough to overcome the additional weight. Alternatively, once the stack is heated enough by the sunshine, buoyancy may predominate.

 

[georgeverghese]

Google tells me atmospheric air cools down by an average of 0.65degC per 100metres rise in altitude, so this is the driving force for the natural convection effect.
There is an error in your warehouse height.
This should help

Stack effect - Wikipedia

en.wikipedia.org

 

[Befuddled]

Thanks for the responses. i think they have helped.

Good spot, my notation was incorrect, should have been:- I have a large warehouse (W x H x L) 9.5m x 9.6m x 46.3m

I hadn't considered the cooling affect of wind moving. i did some rough calculations this morning just based on wind chill and altitude, and i am getting some realistic type numbers of a few meters per second

I think i got too obsessed with Bernoulli's being the principle driving factor.

I hope to get up to the opening again early next week with a anemometer to get some data (didn't have it with me the last time) but the flow felt really low, so should be getting in the ball part now.

I'll update next week, it will be interesting to see if science and life agree.

Thanks again.

B

 

[LittleInch]

The flow of 4.5 m/sec across the top of the stack will not "entrain" air and cause it to flow through the stack. What it will do is cause a lower pressure than atmospheric at the stack exit due to the velocity relative to still air. Air at 4.5 m/sec produces a lower air pressure of 0.05 in. w.g. relative to still air (H =(V/4005)^2 in. wg. where V is in feet per minute). Air inside of the warehouse is still and therefore there will be a differential pressure of 0.05 in. wg available to push the air through the chimney. This assumes no temperature differences to further increase the driving differential pressure due to the density difference stack effect.

I don't have values for friction loss in a brick stack, but I have a Ductulator for friction loss in round smooth duct. If you assume that all the available pressure differential of 0.05 in wg is available for flow through the stack (no building inlet air losses) then the flow that will develop will be:

Total pressure loss = 0.05 in. wg = Head loss friction + velocity head of flow exiting chimney.

Most of the available pressure is used up in velocity head of flow H = (V/4005)^2

V = velocity in feet per minute
H = velocity head in inches water gage.

The actual flow will be such that the velocity head plus friction loss in stack = 0.05 inches wg.

I calculate this to be 36,000 CFM for 94.5 in. ID stack.

If you assume that about half of the available differential pressure (0.025" wg) is pressure loss across the inlet opening of the building (air must enter the warehouse somewhere or there will be no flow), then only 0.025" is available for flow through the stack. In this case I calculate 27,000CFM flow through the stack.

Actual flows will be somewhat lower due to friction factor of brick chimney versus smooth duct.

Note that without friction in the stack then all of the available pressure differential of 0.05" wg will go into velocity head inside the chimney since there is no friction loss. In this case the velocity attained in the chimney would be the same velocity as across the top of the stack since velocity across stack produces 0.05" wg which would be the total energy input to velocity in the stack without friction so H = 0.05" = (V/4005)^2 .

So V in chimney without friction comes out to same across top of chimney and flow would be V x A = 4.5 m/sec (Pi/4) (2.4)^2 = 20 m^3/sec = 43,102 CFM. However there will be friction which uses some of the available pressure differential so it needs to be taken into account as I did above.

I'm not convinced it is as simple as this. As soon as you get any air coming out of the stack it will disrupt the air flow horizontally creating turbulence and changing the pressure. At the sort of velocities you mention up the stack you will also have entry and exit losses which need to be accounted for. I think even in a strong wind like 4.5 m/sec, there will be only a very gentle flow up the stack if at all and any thermal effects could easily drive this into reverse and feed air into the building.

Hope to hear from you next week. Take temperature measurements also. I think they drive flow much more than the wind across the top of the vent.

 

[Befuddled]

Will do.

The end results don't have to be super accurate (just need to know its is minutes or hours)

Thanks for the help.

B.

 

[LittleInch]

If you stick those whirly things up there at least you have some kinetic energy involved. Anything else is so temperature dependant it would work one day and have cold air coming down the vent the next day. IMHO.