# Opening velocity varies over time?

 User Name Remember Me Password
 Register Blogs Members List Search Today's Posts Mark Forums Read

 January 2, 2010, 05:24 Opening velocity varies over time? #1 Senior Member   Join Date: Jul 2009 Posts: 212 Rep Power: 9 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

 January 2, 2010, 06:24 #2 Member   Join Date: Nov 2009 Posts: 49 Rep Power: 7 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.

January 2, 2010, 07:29
#3
Senior Member

Join Date: Jul 2009
Posts: 212
Rep Power: 9
Quote:
 Originally Posted by Abou ali 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

 January 2, 2010, 10:37 #4 Member   Join Date: Nov 2009 Posts: 49 Rep Power: 7 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.

January 2, 2010, 18:52
#5
Senior Member

Join Date: Jul 2009
Posts: 212
Rep Power: 9
Quote:
 Originally Posted by Abou ali 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?

 January 3, 2010, 12:13 #6 Member   Join Date: Nov 2009 Posts: 49 Rep Power: 7 Hi, Can you display the error message sent by CFX here ?

 January 4, 2010, 05:37 #7 Senior Member   Join Date: Jul 2009 Posts: 212 Rep Power: 9 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. | |--------------------------------------------------------------------|

 January 4, 2010, 17:08 #8 Member   Join Date: Nov 2009 Posts: 49 Rep Power: 7 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.

January 4, 2010, 18:24
#9
Senior Member

Join Date: Jul 2009
Posts: 212
Rep Power: 9
Quote:
 Originally Posted by Abou ali 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

January 12, 2010, 19:37
#10
Senior Member

Join Date: Jul 2009
Posts: 212
Rep Power: 9
Quote:
 Originally Posted by Abou ali 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?
Attached Images
 nose3_Q.jpg (54.0 KB, 14 views)

 January 14, 2010, 07:19 #11 Member   Join Date: Nov 2009 Posts: 49 Rep Power: 7 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.

January 14, 2010, 07:39
#12
Senior Member

Join Date: Jul 2009
Posts: 212
Rep Power: 9
Quote:
 Originally Posted by Abou ali 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?
Attached Images
 nose3_Q_1.jpg (34.1 KB, 13 views)

 January 14, 2010, 18:05 #13 Member   Join Date: Nov 2009 Posts: 49 Rep Power: 7 Hi, I think that there is a problem in the simulation definitions!!!! If you attach the output file here, maybe I can help.

January 14, 2010, 18:21
#14
Senior Member

Join Date: Jul 2009
Posts: 212
Rep Power: 9
Quote:
 Originally Posted by Abou ali 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:

January 15, 2010, 04:05
#15
Member

Join Date: Nov 2009
Posts: 49
Rep Power: 7
Hi,
I have seen your output file and maybe the problem is in the symmetry condition.
I have run a test simulation of a nose (0.05*0.05*0.05 cm3) in a domain of 2*2*2 m3.
I use an opening boundary condition in one side of the nose with the same condition as yours, for the boundaries of the domain I set them as opening with 0 Pa.
You can see here the rusults.
Attached Images
 nose in room.PNG (17.5 KB, 7 views) mass flow.jpg (36.3 KB, 10 views)
Attached Files
 output.txt (7.8 KB, 3 views)

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post xiuying OpenFOAM Running, Solving & CFD 8 August 27, 2013 15:33 Alan OpenFOAM Running, Solving & CFD 10 April 6, 2012 14:02 Michelle FLUENT 2 February 18, 2008 13:38 R P CFX 2 October 26, 2004 02:13 Lam FLUENT 1 October 31, 2003 07:50

All times are GMT -4. The time now is 02:43.

 Contact Us - CFD Online - Top