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 kingjewel1 January 2, 2010 05:24

Opening velocity varies over time?

Hi there,

How could I make an opening velocity vary over time. Eg someone breathing? We assume
the frequency of respiration under light physical work is
17 times per minute with a time-mean rate about
8.4 l/min:) How to implement in CEL?

Thank you

 Abou ali January 2, 2010 06:24

Hi,
First try to find a mathematical formulation of the velocity or the mass flowrat versus time.
I mean velocity=f(t) were f is a periodic function with a period T=60/17
And f(0s)=0m/s f(T)=0m/s
Finally: area/T*Integral[(velocity) between T and 0] =8.4/17 must be verified.

 kingjewel1 January 2, 2010 07:29

Quote:
 Originally Posted by Abou ali (Post 241279) Hi, First try to find a mathematical formulation of the velocity or the mass flowrat versus time. I mean velocity=f(t) were f is a periodic function with a period T=60/17 And f(0s)=0m/s f(T)=0m/s Finally: area/T*Integral[(velocity) between T and 0] =8.4/17 must be verified.

(1) so if f(t+p)=f(t), f(0)=f(T)=0 such a function could be sin(60/17+t).

(2) 1/T*Area*[F(T)-F(0)]=8.4/17, where F(t)=integral of f(t)

=> -cos(60/17+T)+cos(60/17)=8.4T/(17*Area)

My question is how to adjust the units so that CFX understands the CEL to be in m/s say instead of radians (from sin)?

Thank you and Happy New year:)

 Abou ali January 2, 2010 10:37

Hi,
In the first equation you must add Vmax, it be come Vmax*sin()
were sin is a dimensionless quantity and Vmax can be calculated from the second equation.

 kingjewel1 January 2, 2010 18:52

Quote:
 Originally Posted by Abou ali (Post 241295) Hi, In the first equation you must add Vmax, it be come Vmax*sin() were sin is a dimensionless quantity and Vmax can be calculated from the second equation.
Hi there and thank you for replying,

How can you make sin nondimensional as CFX tells me it is be radians?

 Abou ali January 3, 2010 12:13

Hi,
Can you display the error message sent by CFX here ?

 kingjewel1 January 4, 2010 05:37

I divided t/[s], now the error message disappears. So I have 1.7[ms^-1]*sin(60/17+t/[s]) and have made this a monitor point as well. However when I run a transient solution I'm given the error

ERROR #004100025 has occurred in subroutine SCL_RESCH. |
| Message: |
| A floating point overflow has occurred. Check solver |
| parameters, boundary conditions and mesh quality. The |
| calculation will be terminated. |
| Note: if this error occurs at the first iteration, that may |
| be caused by the initial guess, e.g., a zero initial |
| velocity and zero dynamic viscosity. |
|--------------------------------------------------------------------|

 Abou ali January 4, 2010 17:08

Hi,
It's earlier to speak about the solver error message. The formulation which you found doesn't describe exactly what you want. Try to plot the function in the Plot panel of the Expression window (use t=0s to 60s) and see what it gives.
I think that a right function can be: Vmax*sin(2pi*t/T).
The integration of this function in a period is 0, this signify that the mass flowrat at inspiration = the mass flowrat at expiration = 8.4/17 l/min then Vmax can be calculated from the following eq: area/T*Integral[(velocity) between T/2 and 0] =8.4/17.
Try to verify the new expression again.

 kingjewel1 January 4, 2010 18:24

Quote:
 Originally Posted by Abou ali (Post 241468) Hi, It's earlier to speak about the solver error message. The formulation which you found doesn't describe exactly what you want. Try to plot the function in the Plot panel of the Expression window (use t=0s to 60s) and see what it gives. I think that a right function can be: Vmax*sin(2pi*t/T). The integration of this function in a period is 0, this signify that the mass flowrat at inspiration = the mass flowrat at expiration = 8.4/17 l/min then Vmax can be calculated from the following eq: area/T*Integral[(velocity) between T/2 and 0] =8.4/17. Try to verify the new expression again.
Thank you. I'm wondering though, Sin(2pi*t/T) has a periodicity of 2pi but I'm looking for 60/17 hence the
sin(2pi*t+60/17). What am I missing?:)

2)area/T*int(Vmax*Sin(2pi*t/T))dt=8.4/17 between 0 and T/2
=> Vmax=8.4e-3/(Pi*17*Area)~0.621[m/s] which seems very reasonable.
presuming 8.4 litres is 8.4e-3 meters cubed

 kingjewel1 January 12, 2010 19:37

1 Attachment(s)
Quote:
 Originally Posted by Abou ali (Post 241468) Hi, It's earlier to speak about the solver error message. The formulation which you found doesn't describe exactly what you want. Try to plot the function in the Plot panel of the Expression window (use t=0s to 60s) and see what it gives. I think that a right function can be: Vmax*sin(2pi*t/T). The integration of this function in a period is 0, this signify that the mass flowrat at inspiration = the mass flowrat at expiration = 8.4/17 l/min then Vmax can be calculated from the following eq: area/T*Integral[(velocity) between T/2 and 0] =8.4/17. Try to verify the new expression again.
I'm just wondering how I can explain why CFX produces a much smaller flow rate on the exhalation than on inhalation (see Fig for 1 cycle). What do you think could be the reason for this?

 Abou ali January 14, 2010 07:19

Hi,
I think that you have used an inlet and outlet boundaries conditions. This is not consistent with this problem because this type of boundaries prevents the flow to change direction. The solution is to use an opening boundary condition in both side of the domain.

 kingjewel1 January 14, 2010 07:39

1 Attachment(s)
Quote:
 Originally Posted by Abou ali (Post 242562) Hi, I think that you have used an inlet and outlet boundaries conditions. This is not consistent with this problem because this type of boundaries prevents the flow to change direction. The solution is to use an opening boundary condition in both side of the domain.
No no, I used an opening boundary condition applied to a single face (0.05cm*0.05cm). You say 'both side of the domain', what do you mean by this?

Subsequently I tried using an outlet and obtained similar flow rates but biased towards the inhalation (logically). Picture is of the domain for reference purposes. small cube in the middle is cut out. One face is the opening.
Set up:
Domain temp: 21C
Nose opening temp: 34C
Fluid:Air Ideal gas
Buoyancy: ON
Buoyancy Ref density: 1.204kg/m^3
Opening Uvelocity: Vmax*sin(2*pi*t/T/1[s])

By the way thank you for your help. What do you think?:)

 Abou ali January 14, 2010 18:05

Hi,
I think that there is a problem in the simulation definitions!!!! If you attach the output file here, maybe I can help.

 kingjewel1 January 14, 2010 18:21

Quote:
 Originally Posted by Abou ali (Post 242665) Hi, I think that there is a problem in the simulation definitions!!!! If you attach the output file here, maybe I can help.
I tried to attach it in a number of formats without much success, so i've pasted the top bit here:

Quote: