The effect of perpendicular wind at channel inlet

  • #1
Gixxer99
6
0
TL;DR Summary
The impact of pressure drop on channel inlet when wind flow passing by perpendicular of channel opening.
Hello
I'm thinking of how to attack this problem.
Outside a channel-inlet, wind is passing perpendicular to the opening with a wind speed of 9 m/s and a fan sucks in a certain amount of air into this channel so that the velocity in the channel is 3 m/s.
When the outside air is 0 m/s I have no additional pressure drop that the fan needs to overcome, when entering the channel, when the outside air is a little bit windy = 9 m/s the I get a additional pressure drop of 40 Pa.

What I'm thinking of is how to estimate the additional pressure drop if the 2 velocitys are different, because I'm assuming that channel velocity also needs to be in this equation. If I could convince myself that channel velocity do not have an impact and that only the outdoor velocity has, then I should just derive a K factor from known conditions and do a minor pressure loss calcualtion to estimate new pressure drop to the changed conditions, but I think this is a to big shortcut.

Storm.png
 
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  • #2
Gixxer99 said:
TL;DR Summary: The impact of pressure drop on channel inlet when wind flow passing by perpendicular of channel opening.

Hello
I'm thinking of how to attack this problem.
Outside a channel-inlet, wind is passing perpendicular to the opening with a wind speed of 9 m/s and a fan sucks in a certain amount of air into this channel so that the velocity in the channel is 3 m/s.
When the outside air is 0 m/s I have no additional pressure drop that the fan needs to overcome, when entering the channel, when the outside air is a little bit windy = 9 m/s the I get a additional pressure drop of 40 Pa.

What I'm thinking of is how to estimate the additional pressure drop if the 2 velocitys are different, because I'm assuming that channel velocity also needs to be in this equation. If I could convince myself that channel velocity do not have an impact and that only the outdoor velocity has, then I should just derive a K factor from known conditions and do a minor pressure loss calcualtion to estimate new pressure drop to the changed conditions, but I think this is a to big shortcut.

View attachment 342159
By channel you mean a pipe? Also, place the fan in your diagram, and show a plausible pressure distribution from the fan to the inlet of the pipe (relative to atmospheric).
 
  • #3
Hello

Channel are in this case rectangualar ducting, here are some more pictures that maybe better explain what I want to solve.

No wind.png



Wind.png



To solve.png
 

Attachments

  • No wind.png
    No wind.png
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  • #4
Is the fan inside duct at the discharge, or just open to atmosphere right there? We need to find out what the pressure difference is across the fan in each scenario. The inlet pressure is only half the story.
 
  • #5
erobz said:
Is the fan inside duct at the discharge, or just open to atmosphere right there? We need to find out what the pressure difference is across the fan in each scenario. The inlet pressure is only half the story.
The fan is inside the duct
 
  • #6
Gixxer99 said:
The fan is inside the duct
To be clear I'm imagining some duct on the inlet, the fan, and some duct that takes the flow somewhere else?

Ok, so we need fan inlet pressure and outlet pressure (fan differential pressure) for the data to isolate fan performance from atmospheric noise. The wind blowing over a house is enough to slam doors shut, and in extreme cases tear roofs off. It could be that the fans differential pressure isn't changing as much as you believe (by only measuring one side of the fan)?
 
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  • #7
I'm thinking following:
Total pressure is constant meaning that the difference in static pressure is one part of the pressure drop that occurs during windy conditions, and this difference in static pressure comes from the two velcoitys, there is no change in static pressure inside the channel for the two first pictures.
Minor loss due to sudden contraction are assumed same for all conditions.
So we need to estimate the K factor for the redirection of the flow.

Example (9 m/s and 3 m/s, denisty 1,2):
Changes in outdoor static pressure = 1,2*9*9/2 = 48,6 Pa
Additional negative pressure the fan needs to overcome = 48,6 Pa (increased pressure drop)
But this implies that the redirection of air flow is no minor loss, more a minor gain because we only can see an increase of 40 Pa in gauge pressure before the fan, so what am I missing.

Maybe it's due that I actually has a wall surrounding the inlet and the drag reduce the dynamic pressure to zero at the wall?

1711202722726.png
 
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  • #8
Gixxer99 said:
I'm thinking following:
Total pressure is constant meaning that the difference in static pressure is one part of the pressure drop that occurs during windy conditions, and this difference in static pressure comes from the two velcoitys, there is no change in static pressure inside the channel for the two first pictures.
Minor loss due to sudden contraction are assumed same for all conditions.
So we need to estimate the K factor for the redirection of the flow.

Example (9 m/s and 3 m/s, denisty 1,2):
Changes in outdoor static pressure = 1,2*9*9/2 = 48,6 Pa
Additional negative pressure the fan needs to overcome = 48,6 Pa (increased pressure drop)
But this implies that the redirection of air flow is no minor loss, more a minor gain because we only can see an increase of 40 Pa in gauge pressure before the fan, so what am I missing.
I'm saying that because of the effects of flow over the exterior of a building the has an inlet outside, and a discharge somewhere in the building the pressures could be different from what you think because of the wind interacting with the structure. leave a door open on a windy day and crack a window. The door usually slams shut. Thats because the pressure inside does not match what is outside.

A fan or a pump supplies a differential pressure. On a calm day ( ## P_{inlet} = -100 kPa ##) tell us what is the measurement for ##P_{discharge}## on the other side of the fan under those conditions. Do the same when you measure on a windy day (## P'_{inlet} = -140kPa## )

I want to see if you find:

## \overbrace{P_{discharge} - P_{inlet}}^{not~windy} \approx \overbrace{P'_{discharge} - P'_{inlet}}^{windy} ##

Meaning you don't have a major change in performance that you think you have, because you are currently only measuring the pressure on one side of the fan. Thats not how fan/pump curves work. They tell you the differential pressure the fan/pump can supply. They don't really care about what absolute pressure is fed in (so long as it's not a structural problem for the fan). If you are trouble shooting the performance characteristics you need the differential pressure.
 
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  • #9
I understand, have following:

Pinlet = -100
Pdischarge = 0

P’inlet = -140
P’discharge = +5
 
  • #10
Gixxer99 said:
I understand, have following:

Pinlet = -100
Pdischarge = 0

P’inlet = -140
P’discharge = +5
1711204232948.png


You are saying its 0 Guage where I'm showing? Unless the entire duct terminates with the fan, thats a bad measurement.
 
  • #11
The fan terminate the long ducting and at it's blowout has only a very short ducting attached to it.
 
  • #12
Gixxer99 said:
The fan terminate the long ducting and at it's blowout has only a very short ducting attached to it.
Ok, then it seems like the differential change is isolated and distinct. So, I guess we do have to look for an explanation involving that 40 kPa difference. I would think it could be treated like a tee, but this is probably not trivial in a open flow like this. I'm not sure how to define the flow properties passing by the inlet from which you fan is drawing upon. They are not in a fixed geometry...
 
  • #13
Can you make the fan curve accessible?
 

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