CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
-   CFX (https://www.cfd-online.com/Forums/cfx/)
-   -   Opening velocity varies over time? (https://www.cfd-online.com/Forums/cfx/71427-opening-velocity-varies-over-time.html)

 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.
Thank you for your quick reply.

(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:
 +--------------------------------------------------------------------+ | | | CFX Command Language for Run | | | +--------------------------------------------------------------------+ LIBRARY: CEL: EXPRESSIONS: Te = 60/17 Veln = 3[m/s]*sin(2*pi*t/Te/1[s]) END END ADDITIONAL VARIABLE: Contaminant Option = Definition Tensor Type = SCALAR Units = [m/m ] Variable Type = Specific END MATERIAL: Air Ideal Gas Material Description = Air Ideal Gas (constant Cp) Material Group = Air Data, Calorically Perfect Ideal Gases Option = Pure Substance Thermodynamic State = Gas PROPERTIES: Option = General Material ABSORPTION COEFFICIENT: Absorption Coefficient = 0.01 [m^-1] Option = Value END DYNAMIC VISCOSITY: Dynamic Viscosity = 1.831E-05 [kg m^-1 s^-1] Option = Value END EQUATION OF STATE: Molar Mass = 28.96 [kg kmol^-1] Option = Ideal Gas END REFERENCE STATE: Option = Specified Point Reference Pressure = 1 [atm] Reference Specific Enthalpy = 0. [J/kg] Reference Specific Entropy = 0. [J/kg/K] Reference Temperature = 25 [C] END REFRACTIVE INDEX: Option = Value Refractive Index = 1.0 [m m^-1] END SCATTERING COEFFICIENT: Option = Value Scattering Coefficient = 0.0 [m^-1] END SPECIFIC HEAT CAPACITY: Option = Value Specific Heat Capacity = 1.0044E+03 [J kg^-1 K^-1] Specific Heat Type = Constant Pressure END THERMAL CONDUCTIVITY: Option = Value Thermal Conductivity = 2.61E-2 [W m^-1 K^-1] END END END END FLOW: SOLUTION UNITS: Angle Units = [rad] Length Units = [m] Mass Units = [kg] Solid Angle Units = [sr] Temperature Units = [K] Time Units = [s] END SIMULATION TYPE: Option = Transient EXTERNAL SOLVER COUPLING: Option = None END INITIAL TIME: Option = Automatic with Value Time = 0 [s] END TIME DURATION: Option = Total Time Total Time = Te[s] END TIME STEPS: Option = Timesteps Timesteps = 0.05 [s] END END DOMAIN: Default Domain Coord Frame = Coord 0 Domain Type = Fluid Fluids List = Air Ideal Gas Location = B32 BOUNDARY: Default Domain Default Boundary Type = WALL Location = F28.32,F30.32,F31.32,F35.32,F36.32,F37.32,F38.32 BOUNDARY CONDITIONS: ADDITIONAL VARIABLE: Contaminant Option = Zero Flux END HEAT TRANSFER: Option = Adiabatic END WALL INFLUENCE ON FLOW: Option = No Slip END WALL ROUGHNESS: Option = Smooth Wall END END END BOUNDARY: Nose Boundary Type = OPENING Location = F29.32 BOUNDARY CONDITIONS: ADDITIONAL VARIABLE: Contaminant Additional Variable Value = 0 [m m^-1] Option = Value END FLOW REGIME: Option = Subsonic END HEAT TRANSFER: Option = Static Temperature Static Temperature = 34 [C] END MASS AND MOMENTUM: Option = Cartesian Velocity Components U = Veln V = 0 [m s^-1] W = 0 [m s^-1] END TURBULENCE: Option = Medium Intensity and Eddy Viscosity Ratio END END BOUNDARY SOURCE: SOURCES: EQUATION SOURCE: Contaminant Option = Total Source Total Source = 1e-15 [kg s^-1] END END END END BOUNDARY: Symm Boundary Type = SYMMETRY Location = F33.32,F34.32 END DOMAIN MODELS: BUOYANCY MODEL: Buoyancy Reference Density = 1.2 [kg m^-3] Gravity X Component = 0 [m s^-2] Gravity Y Component = -9.81 [m s^-2] Gravity Z Component = 0 [m s^-2] Option = Buoyant BUOYANCY REFERENCE LOCATION: Option = Automatic END END DOMAIN MOTION: Option = Stationary END MESH DEFORMATION: Option = None END REFERENCE PRESSURE: Reference Pressure = 1 [atm] END END FLUID MODELS: ADDITIONAL VARIABLE: Contaminant Kinematic Diffusivity = 1e-5 [m^2 s^-1] Option = Transport Equation END COMBUSTION MODEL: Option = None END HEAT TRANSFER MODEL: Option = Thermal Energy END THERMAL RADIATION MODEL: Option = None END TURBULENCE MODEL: Option = k epsilon BUOYANCY TURBULENCE: Option = None END END TURBULENT WALL FUNCTIONS: Option = Scalable END END END INITIALISATION: Option = Automatic INITIAL CONDITIONS: Velocity Type = Cartesian ADDITIONAL VARIABLE: Contaminant Additional Variable Value = 0 [m m^-1] Option = Automatic with Value END CARTESIAN VELOCITY COMPONENTS: Option = Automatic with Value U = 0 [m s^-1] V = 0 [m s^-1] W = 0 [m s^-1] END EPSILON: Option = Automatic END K: Fractional Intensity = 0.05 Option = Automatic with Value END STATIC PRESSURE: Option = Automatic with Value Relative Pressure = 0 [Pa] END TEMPERATURE: Option = Automatic with Value Temperature = 21 [C] END END END OUTPUT CONTROL: MONITOR OBJECTS: MONITOR BALANCES: Option = Full END MONITOR FORCES: Option = Full END MONITOR PARTICLES: Option = Full END MONITOR POINT: Ve Expression Value = Veln Option = Expression END MONITOR RESIDUALS: Option = Full END MONITOR TOTALS: Option = Full END END RESULTS: File Compression Level = Default Option = Standard END TRANSIENT RESULTS: Transient Results 1 File Compression Level = Default Option = Standard OUTPUT FREQUENCY: Option = Every Timestep END END END SOLVER CONTROL: ADVECTION SCHEME: Option = High Resolution END CONVERGENCE CONTROL: Maximum Number of Coefficient Loops = 3 Timescale Control = Coefficient Loops END CONVERGENCE CRITERIA: Residual Target = 0.000001 Residual Type = RMS END TRANSIENT SCHEME: Option = Second Order Backward Euler TIMESTEP INITIALISATION: Option = Automatic END END END END COMMAND FILE: Version = 11.0 Results Version = 11.0 END EXECUTION CONTROL: INTERPOLATOR STEP CONTROL: Runtime Priority = Standard EXECUTABLE SELECTION: Double Precision = Off END MEMORY CONTROL: Memory Allocation Factor = 1.0 END END PARALLEL HOST LIBRARY: HOST DEFINITION: phipc Remote Host Name = PHI-PC Host Architecture String = winnt Installation Root = C:\Program Files\ANSYS Inc\v%v\CFX END END PARTITIONER STEP CONTROL: Multidomain Option = Independent Partitioning Runtime Priority = Standard EXECUTABLE SELECTION: Use Large Problem Partitioner = Off END MEMORY CONTROL: Memory Allocation Factor = 1.0 END PARTITIONING TYPE: MeTiS Type = k-way Option = MeTiS Partition Size Rule = Automatic Partition Weight Factors = 0.250, 0.250, 0.250, 0.250 END END RUN DEFINITION: Definition File = C:/Users/Phi/Documents/New \ folder/cylinder/nose3_Q_fast.def Interpolate Initial Values = Off Run Mode = Full END SOLVER STEP CONTROL: Runtime Priority = Standard EXECUTABLE SELECTION: Double Precision = Off END MEMORY CONTROL: Memory Allocation Factor = 1.0 END PARALLEL ENVIRONMENT: Number of Processes = 4 Start Method = PVM Local Parallel Parallel Host List = phipc*4 END END END

 Abou ali January 15, 2010 04:05

3 Attachment(s)
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.

 All times are GMT -4. The time now is 14:16.