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Old   August 5, 2002, 00:25
Default Reynold stress equation
  #1
mukkarumhussain
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Hi, My third question. after putting velocity u as a mean velocity U and fluctuating part u' we get Reynold tensor equation. I think derivative of mean velocity U w.r.t.t. should be equal to zero b/c mean velocity is not changing with time but it is present in reynold tensor equation. Why
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Old   August 5, 2002, 01:46
Default Re: Reynold stress equation
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S. S. Mudathir
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That is not true. Consider for example a flow in a converging nozzle. when you go along the nozzle the hole velocity increases due to the contraction in area and you cann't say that the increase is only in the fluctuating component in the velocity. due to this the mean velocity is not conistant and should be a function in the spatial coordinates and time for turbulent flow. by the word "mean" in this situation it is meant that the mean velocity in same spatial coordinates will be be constant with respect to time. The same approach could be extended if you have turbulent flow with heat transfer or phase change....

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Old   August 5, 2002, 04:39
Default Re: Reynold stress equation
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mukkarumhussain
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In nozzel also mean velocity w.r.t.t. is constant. i agree with you that in nozzle velocity increase due to the reduction in area but this change is not due to time but due to changing in x,y,z, coordinates. In nozzels at a certain point velocity doesn't change with time it remains constant. and if mean velocity changes then what is the meaning of mean velocity b/c i think mean velocity means a constant velocity across which velocity is fluctuating.waiting for your answer.
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Old   August 5, 2002, 08:31
Default Re: Reynold stress equation
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S. S. Mudathir
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It depends on what kind of problems you are trying to handle. I mean that when a low or a mathematical relation is driven it is put in it is general form. And when this assumption was made (the mean velocity and fluctuation one) it is not made for only steady flow, it could be used for flows which vary with time. Although you didn't mention whether you are trying to solve steady or unsteady problem butI understand now from your last message that you are handling a steady problem. If I didn't misunderstand you then you are right and the mean velocity doesn't vary with time and then the equation will be reduced to cover the variations with the spatial coordinates. will be pleased to hear your comment and I extremly welcome and cooperation between us. It looks like you are arabic, am I correct?
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Old   August 5, 2002, 23:56
Default Re: Reynold stress equation
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mukkarumhussain
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No, I am handling a unsteady flow, I am new in CFD that is why i am trying to get initial knowledge.I am a Research Assistant in NED University(Pakistan) My research project is revemping of Francis Turbine. We are purchasing Fluent software. I think all turbulent flows are unsteady ,am i right. and do you know some website from where i could download materials on Hydraulic or Francis Turbine. From which country you belong and what are you doing.
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Old   August 6, 2002, 02:55
Default Re: Reynold stress equation
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S. S. Mudathir
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No I think you are not right. Not all turbulent flows are unsteady and this depends on your understanding to what is steadiness. What I understand by steady flow is that some major physical quantities don't change with time, for example if a turbine Francis is developing 50 MW power output, carrying a fixed load and running at a fixed speed then the flow rate will be constant and the flow will be thought as steady although it is turbulent. Now consider a sudden decrease in the load for example from 50 MW to 45 MW, one of the possible methods of decreasing the developed power from the turbine is to decrease the flow rate by closing the gates partially and the flow rate will change from Q1 m/s (required to generate 50 MW) to Q2 m/s (required to generate 45 MW) during a specific time span and then after reaching 45 MW power generation the system is again stead. The time required to transfer the system from the first situation to the second is said to be the transience time and the problem is completely unsteady during this period and then the derivatives w.r.t.t could not be set to zero. During this time a mean velocity is imagined at each certain time step to facilitate modeling the situation. About me I am Sudanese Mechanical engineer working as a teaching assistant at the university of Khartoum and doing my master in the same university in the power engineering specialization (a course system) with a partial fulfillment titled "Comparison study of two commercial CFD packages". About the resources give me I think you have to get some materials about turbulence modeling. There also good sites in the Internet I will check them and send you again INSHA ALLAH.

S. S. mudathir
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Old   August 6, 2002, 03:37
Default Re: Reynold stress equation
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Lionel Larcheveque
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Hi,

I think the problem is related to the definition of the average you use:

Formally, Reynolds Averaged Navier-Stokes equations are related to an ensemble average (you perform measurements on several runs of the same experimental configuration and you average the measurements). The mean quantities can therefore be time dependent.

If you want to use time averaging, you have to invoke ergodicity. It postulates that time average and ensemble average are equivalent. But one of the hypothesis needed to use this relation is that the measurements should be statistically stationary.

Hope this help.
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Old   August 6, 2002, 23:47
Default Re: Reynold stress equation
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mukkarumhussain
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In your example of Francis Turbine flow is steady but velocity is changing in every point with time as i think, and if velocity is also not changing then flow would be laminar not turbulent.am i right.
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Old   August 6, 2002, 23:50
Default Re: Reynold stress equation
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mukkarumhussain
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Thanx for your kind help. Can you tell me what is the critaria to take characteristic length nd characteristic velocity.
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Old   August 7, 2002, 03:18
Default Re: Reynold stress equation
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S. S. Mudathir
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if you mean u` (the randomely varying component---> yes you are rigth, i you mean U (the mean velocity) no you are wrong.
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