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April 28, 2005, 01:41 |
different value in post - help needed
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#1 |
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Dear friends,
Can we belive totally on the values calcaulated in the post? because inspite of same area the area-average velocity@inlet & outlet are different and still the mass flow rate is same@inlet and outlet. Is it the problem of POST calculator? if then how can i check the correct ones? OR is it the problem of mesh at inlet and outlet? How much accurate will be the values calcuated by POST? Looking for +ve response. Thanking you in advance.. |
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April 28, 2005, 05:18 |
Re: different value in post - help needed
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#2 |
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Hi tuks, Your quess is correct, it is because of the mesh (surface area)in the outlet & inlet which are not 100% the same. (you know, depending on your mesh type, one should have a very fine mesh to capture the geometry 100%)
regards, houman |
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April 28, 2005, 06:48 |
Re: different value in post - help needed
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#3 |
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Thanks houman. But my one more question is, if area of inlet & outlet is same and avg velocity is different then why it gives the same mass flow rate@inlet & outlet?
Thanks once again tuks |
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April 28, 2005, 07:16 |
Re: different value in post - help needed
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#4 |
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Hi, to answer your question, pls let me know: In which scale have you inserted the input? by Normal speed (m/s or ...) or by mass flow rate (Kg/s or ...) if the latter is the case you will already get your answer. and If you have used normal speed; it is probable that you have't set the fluid output velocity, (which is also the usual way) so after the end of your simulation and when it is steady (the imbalance eq. is well converged) then "input=output" or do you have more that one input and one output?
regards, houman |
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April 28, 2005, 09:32 |
Re: different value in post - help needed
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#5 |
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Hi houman, I have defined the inlet condition by cartesian velocity component with u=Uprofile, v=w=0, the x,y,z directions. I have only one inlet and one outlet. Outlet condition is static Pressure=0. Please reply for more information.
Thanks for reply. Tuks |
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April 28, 2005, 09:40 |
Re: different value in post - help needed
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#6 |
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Hi Tuks, what I have in mind is that: since you have one input and one output, after you simulations becomes steady state, then your outflow rate (mass flow rate) will be equal to your inflow rate, regardless of the serface area of each of then.
regards, houman |
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April 28, 2005, 09:49 |
Re: different value in post - help needed
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#7 |
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Thanks houman. I got very good insight. Tuks
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April 29, 2005, 12:45 |
Re: different value in post - help needed
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#8 |
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Maybe your forget the variation of "density". Mass flow rate is function of density,velocity and Area.
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April 29, 2005, 17:47 |
Re: different value in post - help needed
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#9 |
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First of all, velocity is not a conserved quantity, whereas mass and momentum are. Assuming density is constant, then the area averaged **normal** velocity is equivalent to the average mass flow rate:
areaAve(v_norm)=sum_i(area_1*v_norm_i)/sum_i(area_i) If density is not constant, then this is not the case. Also, the same does not apply if you are computing the area averaged Velocity as opposed to the average of the normal component. Regards, Robin |
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