## Time Step Planning (This Section is Under Construction)

During your simulation setup you will need at advanced stages of your projects is to study an unsteady case. Once you can verify that a steady simulation can be run and convergence of solution can be achieved. When an unsteady case is chosen the first question that comes to mind is what time step value to be taken. By default we start by considering static time where advancing the simulation is done through one second at a time. The problem comes up when we run the calculation and find out that convergence of the solution is not achieved with the progression of the time stepping process.

When the simulation dose crash try at first to refine your mesh. This done through changing your mesh from a coarse mesh to medium mesh, if the simulation still doesn’t work take fine mesh.

If the simulation still crashes where several time step advancements are achieved, try to increase the number of iteration loops for each time step. By doing this your are enforcing convergence more strictly from the first couple of time steps this would dampen the amplification of error that was being created previously.

This would mean that a dynamic time stepping has to be selected. The next problem that comes up to mind is what value should I select for the time stepping do I have to take a value greater than one or smaller than one.

When the simulation dose crash try at first to refine your mesh. This done through changing your mesh from a coarse mesh to medium mesh, if the simulation still doesn’t work take fine mesh.

If the simulation still crashes where several time step advancements are achieved, try to increase the number of iteration loops for each time step. By doing this your are enforcing convergence more strictly from the first couple of time steps this would dampen the amplification of error that was being created previously.

This would mean that a dynamic time stepping has to be selected. The next problem that comes up to mind is what value should I select for the time stepping do I have to take a value greater than one or smaller than one.

The researcher guide for the time problem, is to ask your him self what data will I compare against, that would usualy give some hints. Find in the publication the studied variable of intrest which is studied in relation to time. Then see the duration of the experiment

Duration_of_Experiment = Time-Time0, the range is shown on the time axis in the publication.

Number_of_Taken_Samples = This can be read from the plotted points in the Publication.

Time_Difference = Can be found from a series of two points plotted on the time axis of the experiment.

Number_of_Taken_Samples = This can be read from the plotted points in the Publication.

Time_Difference = Can be found from a series of two points plotted on the time axis of the experiment.

Duration_of_Experiment=Number_of_Taken_Samples*Time_Duration

The next point is to take the extracted values from the experimental data and chose the values based on your judgment on how much you have computer resources.

Duration_of_Simulation = The researcher choses.

Number_of_Time_Steps = The researcher choses.

Time_Step = The researcher choses.

Number_of_Time_Steps = The researcher choses.

Time_Step = The researcher choses.

Duration_of_Simulation=Number_of_Time_Steps*Time_Step

## Choosing the Initial Conditions

Initial conditions should start from what you specify as initial conditions values in the software and not the results from a previous simulation run. You will find that once you get to the validation stage of your project if you don’t use the proper initial conditions you will get the wrong results. You will find the generated output data in a complete mess and make no sense.

But sometimes (for a time stepped simulation) you stop your simulation at a certain time step and you intend to continue the calculation after launch or in a month’s time or …etc. Then you use the previous initial condition to finish the simulation run.

But sometimes (for a time stepped simulation) you stop your simulation at a certain time step and you intend to continue the calculation after launch or in a month’s time or …etc. Then you use the previous initial condition to finish the simulation run.

## Time Stepping for the Study of Porous Media

When studying porous media flows what you encounter at some instances very small velocities to the values of 10^-8 or 10^-14 this would introduce. As an initial assumption you will need to verify what is the time required for a practical to travel from one end of the mesh to another, based on the encountered velocities, then based on that you can asses. You will find that the required time would be very large making the simulation not that practical to do. But for a real life case as for water drainage in a porous media that would depend on the soils parameters such as porosity and permeability, so for a soil having a permeability of 10^-13 the water would require a long time to travel from one side to another and if you really decrease the time steps what will happen is you will find that the simulation changes are very small at each time step, you will have to curve fit your data with experimental one and what you find is that the drainage data was taken for a period of two weeks at that point you can change your time stepping and make it large.

## Example

If a partical wants to travel a distance of LX =1 m in a porous domain, having in mind that DX=0.001 m and the number of cells the partical has to travel through is M=1000, knowing that the velocity is V=1e-7 calculate the required time step for the simulation.

## Solution

Based on the velocity equation

V=LX/dt

The calculated time step value will be

dt = (M*DX)/V = 1e7 (sec)

Its is clear that the calculated number for time is big

dt = 115.7 (Days)

This shows that it would require 115.7 days for the partical to travel from one side of the porous domain to the other.

While for studying the required time for the partical to get from one end of the cell to the other side requires

dt = (DX)/V = 1e3 (sec)

## Time Stepping for Turbomachinary

When studying turbomachinary the occuring speeds are high. You will need to know what is the required time for a full rotaion of the turbine rotor. Then based on that you can think of how much detail you want to capture during one full rotation. Some times three time steps are required, it all depends on that can you curve fit the generated data with the experiemntal data. Then comes how much generated data for each saved time step and is it nesecary for that detail.

Duration_of_One_Full_Rotation = This is Calculated by Researcher.

Number_of_Time_Steps = This is estiamted by how much storage capacity is availabel and what type of experimetal data is available.

Time_Step = This is the unknow value.

Number_of_Rotations = This is estiamted by how much storage capacity is availabel and what time of experimetal data is available, you can take its value as 1 initially.

Number_of_Time_Steps = This is estiamted by how much storage capacity is availabel and what type of experimetal data is available.

Time_Step = This is the unknow value.

Number_of_Rotations = This is estiamted by how much storage capacity is availabel and what time of experimetal data is available, you can take its value as 1 initially.

Duration_of_Simulation=Number_of_Time_Steps*Time_Step*Number_of_Rotations*Duration_of_One_Full_Rotation

## Example

Consider having a 23 bladed high temperature turbine in a turbofan engine and you want to run a calculation run, knowing that the blade rotates at 100000 RPM calculate the required times steps knowning that we plan to take 5 time steps per blade rotation.

## Solution

Blade Rotates at and rpm= 100000 (RPM)

We Know that Omega= 2*pi*rpm where Omega has the units of (rad/min)

So the value of Omega in (rad/sec) , reminder that the

Conversion of 100000 (RPM) to (RPS) gives the following value 1666.6 (RPS). Meaning that

Omega = 10471.9 (rad/sec)

Omega= 2*pi*n = (theta-theta0)/ dt

Based on taking theta0 = 0 (rad)

Based on taking theta = 2*pi (rad)

dt= (theta-theta0) / Omega

dt = 2*pi / 3.35 =6e-4 (sec)

This is the required time for a full blade roation

If we say we want to capture 50 time frames during one roation.

We take M = 50

knowing that t0 = 0 (sec)

Time_Duration = t-t0=6e-4 (sec)

Time_Step =Time_Duration/M

Time_Step = 1.2e-5 (sec)

We Know that Omega= 2*pi*rpm where Omega has the units of (rad/min)

So the value of Omega in (rad/sec) , reminder that the

Conversion of 100000 (RPM) to (RPS) gives the following value 1666.6 (RPS). Meaning that

Omega = 10471.9 (rad/sec)

Omega= 2*pi*n = (theta-theta0)/ dt

Based on taking theta0 = 0 (rad)

Based on taking theta = 2*pi (rad)

dt= (theta-theta0) / Omega

dt = 2*pi / 3.35 =6e-4 (sec)

This is the required time for a full blade roation

If we say we want to capture 50 time frames during one roation.

We take M = 50

knowing that t0 = 0 (sec)

Time_Duration = t-t0=6e-4 (sec)

Time_Step =Time_Duration/M

Time_Step = 1.2e-5 (sec)

## Time Stepping for a Wind Turbine

You will need to know what is the required time for a full rotaion of the turbine rotor then based on that you can think of how much detail you want to capture during one full rotaion then how much generated data for each saved time step and is it nesecary for that detail.

Duration_of_One_Full_Rotation = This is Calculated by Researcher.

Number_of_Time_Steps = This is estiamted by how much storage capacity is availabel and what type of experimetal data is available.

Time_Step = This is the unknow value.

Number_of_Rotations = This is estiamted by how much storage capacity is availabel and what time of experimetal data is available, you can take its value as 1 initially.

Number_of_Time_Steps = This is estiamted by how much storage capacity is availabel and what type of experimetal data is available.

Time_Step = This is the unknow value.

Number_of_Rotations = This is estiamted by how much storage capacity is availabel and what time of experimetal data is available, you can take its value as 1 initially.

Duration_of_Simulation=Number_of_Time_Steps*Time_Step*Number_of_Rotations

## Example

Consider having a 3 bladed turbine and you want to run a calculation run, knowing that the blade rotates at 32 RPM calculate the required times steps knowning that we plan to take 5 time steps per blade rotation.

## Solution

Blade Rotates at and rpm= 32 (RPM)

We Know that Omega= 2*pi*rpm where Omega has the units of (rad/min)

So the value of Omega in (rad/sec) , reminder that the

Conversion of 32 (RPM) to (RPS) gives the following value 0.53333 (RPS). Meaning that

Omega = 3.35 (rad/sec)

Omega= 2*pi*n = (theta-theta0)/ dt

Based on taking theta0 = 0 (rad)

Based on taking theta = 2*pi (rad)

dt= (theta-theta0) / Omega

dt = 2*pi / 3.35 =1.87 (sec)

This is the required time for a full blade roation

If we say we want to capture 5 time frames during one roation.

We take M = 5

knowing that t0 = 0 (sec)

Time_Duration = t-t0=1.87 (sec)

Time_Step =Time_Duration/M

Time_Step = 0.375 (sec)

We Know that Omega= 2*pi*rpm where Omega has the units of (rad/min)

So the value of Omega in (rad/sec) , reminder that the

Conversion of 32 (RPM) to (RPS) gives the following value 0.53333 (RPS). Meaning that

Omega = 3.35 (rad/sec)

Omega= 2*pi*n = (theta-theta0)/ dt

Based on taking theta0 = 0 (rad)

Based on taking theta = 2*pi (rad)

dt= (theta-theta0) / Omega

dt = 2*pi / 3.35 =1.87 (sec)

This is the required time for a full blade roation

If we say we want to capture 5 time frames during one roation.

We take M = 5

knowing that t0 = 0 (sec)

Time_Duration = t-t0=1.87 (sec)

Time_Step =Time_Duration/M

Time_Step = 0.375 (sec)

I would recommned this site to help out through the units conversion stage, when converting from rpm to rps and visa versa:

http://www.convertunits.com/from/revolutions+per+second/to/RPM

http://www.convertunits.com/from/revolutions+per+second/to/RPM

## Time Stepping for Combustion

You will need to know what is the required for the study. At some instances your main intrest is the burning velocities of the fuel this would require that your time stepping will be very much related to the flow speed itself therefore you might stick to a time advancemnt scheme of 1 second a time , while

Duration_of_Simulation =

Number_of_Time_Steps =

Time_Step =

Number_of_Time_Steps =

Time_Step =

Duration_of_Simulation=Number_of_Time_Steps*Time_Step

## Time Stepping for Reciprocating Engines

Reciprocating engine is a case that a researcher would encounter,the reaction rates are of importnce with comparisson with the cell size

## Example

Consider having a 4 stroke engine and you want to run a calculation run, knowing that the shaft rotates at 2000-3000 RPM calculate the required times steps knowning that we plan to take 5 time steps per blade rotation.

## Solution

Blade Rotates at and rpm= 2500 (RPM)

We Know that Omega= 2*pi*rpm where Omega has the units of (rad/min)

So the value of Omega in (rad/sec) , reminder that the

Conversion of 2500 (RPM) to (RPS) gives the following value 41.6 (RPS). Meaning that

Omega = 261.7 (rad/sec)

Omega= 2*pi*n = (theta-theta0)/ dt

Based on taking theta0 = 0 (rad)

Based on taking theta = 2*pi (rad)

dt= (theta-theta0) / Omega

dt = 2*pi / 261.7 =0.024 (sec)

This is the required time for a full shaft roation

If we say we want to capture 5 time frames during one roation.

We take M = 5

knowing that t0 = 0 (sec)

Time_Duration = t-t0=0.024 (sec)

Time_Step =Time_Duration/M

Time_Step = 4.8e-3 (sec)

We Know that Omega= 2*pi*rpm where Omega has the units of (rad/min)

So the value of Omega in (rad/sec) , reminder that the

Conversion of 2500 (RPM) to (RPS) gives the following value 41.6 (RPS). Meaning that

Omega = 261.7 (rad/sec)

Omega= 2*pi*n = (theta-theta0)/ dt

Based on taking theta0 = 0 (rad)

Based on taking theta = 2*pi (rad)

dt= (theta-theta0) / Omega

dt = 2*pi / 261.7 =0.024 (sec)

This is the required time for a full shaft roation

If we say we want to capture 5 time frames during one roation.

We take M = 5

knowing that t0 = 0 (sec)

Time_Duration = t-t0=0.024 (sec)

Time_Step =Time_Duration/M

Time_Step = 4.8e-3 (sec)

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