validating in cyclone

masoodina · July 20, 2012, 09:50

hi . i have a problem in validating . im working on a cyclone .

as u see i used this paper for validation . in the tangential velocity the plot is kinda sinusoidal and the picture under it shows tangetial velocity for different length of outlets and its really very different from the other plot .

whats right ???

my plots are like the first (sinusoidal) .

it seems papers have exported data to excel or something and used the absolute value . because this is the only difference in my result and other papers . what do you think ???

please share your ideas with me . thanks

lxlylzl · July 20, 2012, 23:01

Fig. 9 has normalized coordinates, while fig. 10 shows the magnitude plot of the required parameters.

masoodina · July 21, 2012, 02:46

thanks , i get that velocity magnitude and normalization .

but why in fig 9 we have negative velocity ?

the air is rotating in one way so it has to be positive any where and zero at center !! what do you think ? im wrong ??

and u see in this picture tangential velocity is positive everywhere . (i think maybe one of them has made the data by himself!!)

and the second is my cyclones tangent_velocity .

are these contours tangetial velocity magnitude too ?? if so, how in fluent can we do this ?

flotus1 · July 21, 2012, 03:14

Obviously, the author took a cartesian coordinate system instead of a cylindrical one to define the "tangential" velocity in fig.9
I don't think that the data is made up, this is just a post-processing mistake.

lxlylzl · July 21, 2012, 09:19

I'm also new to this topic but I think that the first pic. isn't correct. Flow inside cyclone is almost symmetric about the geometrical axis. It has nothing to do with cartesian coordinates. We go for polar coordinates when working with 2D model. In fig. 9, one might have considered origin somewhere on the geometrical axis, and then might have considered the sign while going along -ve direction while normalising along the radial direction (this you have to check, and let me know too!). By the way what is your inlet velocity?

flotus1 · July 21, 2012, 10:09

Of course it can be explained with coordinates.
Take the absolute value of fig.9 and you get the actual tangential velocity in a cyclone.

This is exactly the difference you get when evaluating the tangential velocity as one of the velocity components in a cartesian coordinate system on the one hand or as the actual tangential component of a cylindrical coordinate system on the other hand.
If the coordinate systems are aligned to the axis of the cyclone, switching from cartesian to cylindrical equals taking the absolute value of the tangential velocity.

lxlylzl · July 21, 2012, 23:49

Yes flotus1, you are right. I should have made it more clear: My 1st four lines are in context with the tangential velocity contour (fig. with black background), and the rest part with reference to fig.9.

lxlylzl · July 22, 2012, 00:01

Hi masoodina, did you get time average tangential velocity as approx. 1.5 times (or above) that of inlet velocity. I'm facing problem with that. Although after convergence I'm getting that at any instantaneous time (not with time averaging).

masoodina · July 22, 2012, 06:52

Quote:

Originally Posted by lxlylzl

Hi masoodina, did you get time average tangential velocity as approx. 1.5 times (or above) that of inlet velocity. I'm facing problem with that. Although after convergence I'm getting that at any instantaneous time (not with time averaging).

hi . i have entered velocity as 20 m/s and obtained max tangential velocity about 30 m/s . (i used rsm steady solver) .

lxlylzl · July 22, 2012, 13:02

O.K. But you did not mention whether your result (i.e. 30 m/s) is instantaneous or time average? My instantaneous value turns out to be around 1.5 times that of the inlet velocity, while time averaging gives me around 1.3 times the inlet velocity! Did you notice this?

masoodina · July 22, 2012, 13:41

well im not sure which one it is . i just used plot>velocity>tangential velocity .

i think probably instantaneous and time averaging velocities are equal while solving steady .
and i have considered many papers they didint mention average or instantaeous velocities either and used just those data that is more like experiements . so dont worry, take it easy .

July 21, 2012, 10:09		#6
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,399 Rep Power: 46	Of course it can be explained with coordinates. Take the absolute value of fig.9 and you get the actual tangential velocity in a cyclone. This is exactly the difference you get when evaluating the tangential velocity as one of the velocity components in a cartesian coordinate system on the one hand or as the actual tangential component of a cylindrical coordinate system on the other hand. If the coordinate systems are aligned to the axis of the cyclone, switching from cartesian to cylindrical equals taking the absolute value of the tangential velocity. masoodina likes this.

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July 20, 2012, 23:01		#2
lxlylzl Member Join Date: Jul 2012 Posts: 49 Rep Power: 13	Fig. 9 has normalized coordinates, while fig. 10 shows the magnitude plot of the required parameters.

July 21, 2012, 03:14		#4
flotus1 Super Moderator Alex Join Date: Jun 2012 Location: Germany Posts: 3,399 Rep Power: 46	Obviously, the author took a cartesian coordinate system instead of a cylindrical one to define the "tangential" velocity in fig.9 I don't think that the data is made up, this is just a post-processing mistake.

July 21, 2012, 09:19		#5
lxlylzl Member Join Date: Jul 2012 Posts: 49 Rep Power: 13	I'm also new to this topic but I think that the first pic. isn't correct. Flow inside cyclone is almost symmetric about the geometrical axis. It has nothing to do with cartesian coordinates. We go for polar coordinates when working with 2D model. In fig. 9, one might have considered origin somewhere on the geometrical axis, and then might have considered the sign while going along -ve direction while normalising along the radial direction (this you have to check, and let me know too!). By the way what is your inlet velocity?

July 21, 2012, 23:49		#7
lxlylzl Member Join Date: Jul 2012 Posts: 49 Rep Power: 13	Yes flotus1, you are right. I should have made it more clear: My 1st four lines are in context with the tangential velocity contour (fig. with black background), and the rest part with reference to fig.9.

July 22, 2012, 00:01		#8
lxlylzl Member Join Date: Jul 2012 Posts: 49 Rep Power: 13	Hi masoodina, did you get time average tangential velocity as approx. 1.5 times (or above) that of inlet velocity. I'm facing problem with that. Although after convergence I'm getting that at any instantaneous time (not with time averaging).

July 22, 2012, 13:02		#10
lxlylzl Member Join Date: Jul 2012 Posts: 49 Rep Power: 13	O.K. But you did not mention whether your result (i.e. 30 m/s) is instantaneous or time average? My instantaneous value turns out to be around 1.5 times that of the inlet velocity, while time averaging gives me around 1.3 times the inlet velocity! Did you notice this?

July 22, 2012, 13:41		#11
masoodina New Member masood Join Date: May 2012 Posts: 18 Rep Power: 13	well im not sure which one it is . i just used plot>velocity>tangential velocity . i think probably instantaneous and time averaging velocities are equal while solving steady . and i have considered many papers they didint mention average or instantaeous velocities either and used just those data that is more like experiements . so dont worry, take it easy .