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 timemean rate about 8.4 l/min:) How to implement in CEL? Thank you 
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. 
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(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:) 
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. 
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How can you make sin nondimensional as CFX tells me it is be radians? 
Hi,
Can you display the error message sent by CFX here ? 
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.   
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. 
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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=8.4e3/(Pi*17*Area)~0.621[m/s] which seems very reasonable. presuming 8.4 litres is 8.4e3 meters cubed 
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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. 
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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?:) 
Hi, I think that there is a problem in the simulation definitions!!!! If you attach the output file here, maybe I can help. 
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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. 
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