Fan modeling with HVAC

[Bertrand S.]

Hello,

We are now testing FDS on situations with interaction between fire and mechanical ventilation. I'm trying to use the HVAC feature, but I have some difficulties to do what I want.

As an illustration, I want to get a flow rate of 0.1 m3/s with a duct pressure difference of 6 Pa, which correspond to a head loss of 10, thanks to the Bernoulli relation. This relation is the same than the equation 9.3 in the verification guide and is a simplified version of the equation 10.3 in the Technical Reference Guide (version 6.1.2, September 2014), with a steady state assumption, without momentum source term.

I build a simple ventilation network (a vent, 2 nodes, a duct and a fan eventually) from outside to the fluid domain. I joined the fds file and the following results in the pdf file that I don't understand:

- Setting a pressure difference of 6 Pa in a duct with a head loss of 10 should give a flow rate of 0.1 m3/s. So I set the ambient node to the appropriate height to get 6 Pa but the flow rate remains at zero (fan losses in the pdf file).

- I try another way building a fan curve thanks to the same relation, but the flow still remains zero (fan curve in the pdf file).

- My last try is to shift the previous fan curve to set a flow rate of 0.1 m3/s at a zero pressure difference. I get a flow rate of 0.11 m3/s instead of 0.1 m3/s and a duct difference pressure of about 2.5 Pa instead of 0 Pa (fan curve shifted in the pdf file).

Am I missing something ? How can I set a simple vent knowing the nominal flow rate, the duct pressure difference and the head loss ?

Thanks for your help.

Best regards,
Bertrand

[dr_jfloyd]

LOSS is the loss coefficient, K not the pressure drop in the duct.  The pressure drop (head) is given by  1/2 rho K V^2. In your first case there is no driving force for there to be any flow. What you have is a  piece of duct at an angle. If you were to pick up a horizontal length of duct and rotate it vertically, you will not suddenly see flow.

In your second case you have defined a fan curve; however, your fan curve says that at zero pressure across the fan, that there  is zero flow.  At t=0, there is no net pressure at the fan. The difference in static pressure between the endpoints of the duct is  exactly balanced by the weight of the air in the duct. Your fan curve is also non-physical.  Look at the example in the user's guide. The volume flow should decrease with increasing pressure.  Your fan curve says the fan moves more air the harder it has to work against pressure.  

In the third case if you had a valid fan curve then the flow would be where dP_fan = 1/2 K rho v^2.  Since your duct is 0.1  m^2 this is dP~0.5 * 10 * 1.19 * vdot /0.1

[Bertrand S.]

In the two first cases, I tried to set a constant pressure to the node connected outside. I forgot that the buoyant term would be offset in the equation...

Following your recommendations, I define the good fan curve for my example. I expect a zero pressure difference since both sides of the duct are connected to the atmospheric pressure and a volume flow rate of 0.1 m3/s but I get a pressure difference of 0.8 Pa and a flow rate of 0.075 m3/s. In my mind, the node pressure is set to the zone pressure. Am I right ? Or the node pressure is set to the local mechanical pressure that would explain the results ?
In the last case, how can I configure a simplified network to get the desired flow rate knowing the head loss and the driving pressure ?

Thanks for the help.
Bertrand

[dr_jfloyd]

FDS uses the local pressure inside the domain (zone pressure + perturbation pressure).  You have flows of ~ 2 m/s at the face of your HVAC vent. This results in a slight perturbation pressure that you are seeing in the output.  If your VENT, for example, were 0.4 m x 0.4 m, then you would be near zero pressure difference. This is a real effect. Consider if you were blowing air at the wall from some other vent. You would raise the local pressure due to stagnation of the flow. The AMBIENT vent would still be at ambient pressure. The result would be a slight pressure difference.  

If you know the local pressure and the flow loss of the duct, then the volume flow is whatever is needed such the friction losses equals the pressure drop.

[Bertrand S.]

Thanks for the explanations.
Bertrand