The purposes of this tutorial are to show that parameterised Q1s are easier than non-parameterised ones:
That tutorial explained how to introduce the necessary objects to the calculation domain, set their sizes and positions manually. You can examine your Q1 file if you completed all recommendations of the tutorial in question and kept the results of your work. However, you can also study the original Q1 file directly from the Input-File Library using the PHOENICS Commander environment.

It is there where the calculation domain, the grid and all the other objects are introduced and specified.
As you can see the sizes and positions are specified numerically, not with the help of parameters.



Variable 1 = P1 2 = U1 3 = W1 4 = KE 5 = EP
Minval= 1.000E-10 -2.974E-03 1.000E-10 5.668E-06 4.752E-08
Maxval= 5.361E-03 1.043E-02 5.892E-02 1.781E-04 1.656E-05
Cellav= 3.852E-03 3.696E-03 4.956E-02 1.029E-04 9.243E-06
Variable 6 = TEM1
Minval= 2.022E+01
Maxval= 2.278E+01
Cellav= 2.042E+01
1.00 6..2.+....+..3.+3.3.+3...+....+....+....+....+....5
. 1 3 3 3 4 5 2
0.90 + 3 3 3 4 5 1
. 3 3 3 5 1 .
0.80 + 1 5 3 3 3 3
. 3 1 4 1 1 .
0.70 + 2 1 5 +
. 1 1 1 1 4 .
0.60 + 1 1 5 +
. 5 .
0.50 + 5 4 +
. 5 4 5 5 4 5 5 .
0.40 + 5 5 5 5 2 +
. 5 .
0.30 + 4 2 +
. .
0.20 2 2 +
. 2 .
0.10 + 2 +
. 6 6 6 6 6 6 6 6 6 6
0.00 5..6.+....+....+2.6.+6..6+6..6+.6.6+.6..+....+....+
0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0
the abscissa is ISWP. min= 1.00E+00 max= 3.81E+02
Various variables at some specified monitor point, marked by numerals from 1 to 6, are plotted as a function of the iteration number, varying, as the graph indicates, from 1 to 381. The designation of variables and their limits are given above the plot.
Group 24. Dumps For Restartsup to the label
STOPat the very bottom of the file. This is the part of Q1 where concrete data on the objects sizes and positions are stored.

A new file will open.


Within familiar markers save25begin and save25end we have introduced a new PIL command incl(file_name) which loads the file named within the brackets into the instruction stack, i.e. in our case adding the required values to the list of instructions. You may read more about this command by clicking here.

xulast=2. yvlast=1. zwlast=1.above the line where the domain sizes are specified.

This was not evident when these positions and sizes were expressed numerically rather than via parameters.
It would convenient and reasonable to have all the objects depend on the calculation domain so that any changes introduced to the domain will result in corresponding changes of the whole scene; and we are about to find out how to achieve this.
Variable 6 = TEM1
Minval= 2.011E+01
Maxval= 2.140E+01
Cellav= 2.021E+01
i.e.half the previous values.
The reason for this is that we did not change the heat source in the L-BLOCK object. Doubling YVLAST doubled the area of the inlet and so caused twice as much air to enter; the temperature rise was therefore halved.
To rectify this, we need to double the heat flux, i.e. in the editor to change the line > OBJ, HEAT_FLUX, 0.000000E+00, 1.000000E+02
in the file gp25_1 to > OBJ, HEAT_FLUX, 0.000000E+00, 1.000000E+02*yvlast
When you run the Satellite and solver again, you should find that the earlier temperature distribution has been recovered.


xulast=2.to
xulast=4.
Variable 1 = P1 2 = U1 3 = W1 4 = KE 5 = EP
Minval= 1.000E-10 1.000E-10 -2.704E-02 5.652E-06 4.734E-08
Maxval= 1.395E-02 1.949E-02 1.000E-10 1.087E-05 2.689E-07
Cellav= 1.125E-02 1.773E-02 -2.253E-02 1.053E-05 2.533E-07
Variable 6 = TEM1
Minval= 2.001E+01
Maxval= 2.320E+01
Cellav= 2.017E+01
1.00 6....+....+4.4.+4.5.+5..5+5..5+.5.5+.5.5+.5..5.5..5
. 2 5 5 5 5 5 2 2 2 2 2 2 2 2 2 2 1 .
0.90 + 1 1 +
. 5 1 1 1 .
0.80 + 4 1 1 +
. 1 1 1 1 .
0.70 + 1 1 +
. .
0.60 + +
. .
0.50 + +
. .
0.40 + +
. .
0.30 + 3 3 +
. 3 3 .
0.20 + 3 3 +
. 3 3 .
0.10 + 3 3 3 +
. 3 3 3 3 .
0.00 5..6.6..6.+6.6.+6.6.+6..6+6..6+.6.6+.6.6+.6..6.6..6
0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0
the abscissa is ISWP. min= 1.00E+00 max= 3.81E+02
The numbers have been changed from their previous values; and the line-printer plot above confirms this.

It is obvious that the X-size of the domain as well as the labyrinth itself has been noticeably changed. The labyrinth now looks longer and thinner as compared to the previous picture.
These changes have been realized by changing only one parameter, the domain length, and all other objects being parametrically related to the domain, have been changed accordingly.
The temperature contours will look like these.

The velocity vectors and contours will be like these.

Having increased the labyrinth length while retaining its height, we thereby reduced its aspect ratio zwlast/xulast; and the effect of near-wall friction became a dominant factor leading to relative increase of the recirculation zone behind the IN-BLOCK.
The answer is positive, and all we have to do is to choose an object and cut off
its connections with the domain where necessary.
We are going to show all this, taking the IN-BLOCK as an example. And we shall start, as usual, with creating a new working directory for this test.
These varables are then inserted into the appropriate lines lower down in the data file, thus:
While, the X- and Z-sizes of the IN-BLOCK are expressed by the corresponding parameters, its Y-size still depends on the domain as the flow remains two-dimensional.
Variable 1 = P1 2 = U1 3 = W1 4 = KE 5 = EP
Minval= 1.000E-10 -2.974E-03 1.000E-10 5.668E-06 4.752E-08
Maxval= 5.361E-03 1.043E-02 5.892E-02 1.781E-04 1.656E-05
Cellav= 3.852E-03 3.696E-03 4.956E-02 1.029E-04 9.243E-06
Variable 6 = TEM1
Minval= 2.022E+01
Maxval= 2.278E+01
Cellav= 2.042E+01
1.00 6..2.+....+..3.+3.3.+3...+....+....+....+....+....5
. 1 3 3 3 4 5 2
0.90 + 3 3 3 4 5 1
. 3 3 3 5 1 .
0.80 + 1 5 3 3 3 3
. 3 1 4 1 1 .
0.70 + 2 1 5 +
. 1 1 1 1 4 .
0.60 + 1 1 5 +
. 5 .
0.50 + 5 4 +
. 5 4 5 5 4 5 5 .
0.40 + 5 5 5 5 2 +
. 5 .
0.30 + 4 2 +
. .
0.20 2 2 +
. 2 .
0.10 + 2 +
. 6 6 6 6 6 6 6 6 6 6
0.00 5..6.+....+....+2.6.+6..6+6..6+.6.6+.6..+....+....+
0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0
the abscissa is ISWP. min= 1.00E+00 max= 3.81E+02
If you now compare the obtained results with those for the runs made without changes in the initial data, you will find them identical. This is the proof that introduction of new parameters into the data file did not result in any changes so far.


You can see that the passage between the IN-BLOCK and the labyrinth wall is narrower now, which has resulted in the the corresponding increase of velocity.

The right-hand half of the picture confirms expectations: it shows that no fluid has penetrated there, and the velocity is zero everywhere.
The left-hand half is more puzzling; for how can there be any flow there, if there is no outlet to the right?
The explanation is that PHOENICS has been set an impossible problem: for it is still being told by the following lines in the file gp25_2:> OBJ, TYPE, INLET
> OBJ, PRESSURE, 0.000000E+00
> OBJ, VELOCITY, 5.000000E-02, 0.000000E+00, 0.000000E+00
that there is a finite X-direction velocity of 0.05 m/s at the INLET object.
This led to the following lines being printed near the bottom of the RESULT file: Nett source of R1 at patch named: OB7 (INLET ) = 2.675250E-02
Nett source of R1 at patch named: OB8 (OUTLET ) =-5.449870E-11
pos. sum=0.026752 neg. sum=-5.44987E-11
nett sum=0.026752
which indicates the PHOENICS could produce a solution only with a significant "nett sum" error, equal to the variable R1 which expresses the inlet mass flow rate.
What is to be done? The simplest solution is to introduce an additional parameter, say 'invel', to set invel equal to 0.05 as the default, but then to re-set it to zero when insbzz is greater than or equal to 0.9.
The following lines introduced to the top of gp25_2
and the statement specifying the inlet velocityin terms of invel > OBJ, VELOCITY, invel, 0.000000E+00, 0.000000E+00
accomplish this.
If you re-run the case with the above settings you will see that in the result file the 'nett sum' has become negligible.

However, if you wish to check your results in the viewer window, you might now find them to be rather strange, in the right-hand half.

However, we suggest that you pay attention to the velocity scale bar and recognise that the largest velocity at the outlet section, being 9.24E-11 m/s, could be nothing but a result of the inevitable round-off error.
In two cases, unexpected results (the lower temperatures 1 and continuity 2 errors, were explained as due to aspects of the input settings which had been over-looked, to which however PHOENICS responded faithfully.
In the second part of this tutorial,, accessed by clicking here, you will learn how to: