Billboard Analysis - Part 5

(Continuation of Part 4)

The following picture is the graph of relative total pressure.

Like in the case 3, we can predict that the pressure in the left side of the billboard is lower than the pressure in the right side. This is the table that shows it.

J	relative static pressure left (Pa)	relative static pressure right (Pa)
21	-1.18E+00	-4.83E+01
22	1.09E+01	-4.80E+01
23	1.65E+01	-4.75E+01
24	2.00E+01	-4.69E+01
25	2.25E+01	-4.64E+01
26	2.42E+01	-4.60E+01
27	2.54E+01	-4.56E+01
28	2.62E+01	-4.54E+01
29	2.67E+01	-4.52E+01
30	2.70E+01	-4.51E+01
31	2.70E+01	-4.51E+01
32	2.67E+01	-4.52E+01
33	2.62E+01	-4.54E+01
34	2.54E+01	-4.56E+01
35	2.42E+01	-4.59E+01
36	2.25E+01	-4.64E+01
37	2.01E+01	-4.69E+01
38	1.67E+01	-4.74E+01
39	1.12E+01	-4.80E+01
40	-1.12E+00	-4.82E+01
Average	1.99E+01	-4.64E+01

We can see that the average relative static pressure in the case 4 is higher than in the case 1 (or 2 or 3). This is due to the cells setting.

We want to know the force acting on the billboard. We can do this by two ways, i.e. by using integration of the pressure and integration of momentum difference. While for the case 1 we can’t use the integration of momentum difference due to the cell settings.

For the integration of the pressure difference, the rule is like in the case 3. In each case the calculated pressure is relative static pressure. The following table is the calculation result using the integration of pressure difference.

J	relative static pressure left (Pa)	relative static pressure right (Pa)	Area (m^2)	Force left (N)	Force right (N)
21	-1.18E+00	-4.83E+01	1.03E-01	-1.22E-01	-5.00E+00
22	1.09E+01	-4.80E+01	1.03E-01	1.13E+00	-4.97E+00
23	1.65E+01	-4.75E+01	1.03E-01	1.71E+00	-4.91E+00
24	2.00E+01	-4.69E+01	1.03E-01	2.07E+00	-4.85E+00
25	2.25E+01	-4.64E+01	1.03E-01	2.33E+00	-4.80E+00
26	2.42E+01	-4.60E+01	1.03E-01	2.50E+00	-4.76E+00
27	2.54E+01	-4.56E+01	1.03E-01	2.63E+00	-4.72E+00
28	2.62E+01	-4.54E+01	1.03E-01	2.71E+00	-4.70E+00
29	2.67E+01	-4.52E+01	1.03E-01	2.76E+00	-4.68E+00
30	2.70E+01	-4.51E+01	1.03E-01	2.79E+00	-4.67E+00
31	2.70E+01	-4.51E+01	1.03E-01	2.79E+00	-4.67E+00
32	2.67E+01	-4.52E+01	1.03E-01	2.76E+00	-4.68E+00
33	2.62E+01	-4.54E+01	1.03E-01	2.71E+00	-4.70E+00
34	2.54E+01	-4.56E+01	1.03E-01	2.63E+00	-4.72E+00
35	2.42E+01	-4.59E+01	1.03E-01	2.50E+00	-4.75E+00
36	2.25E+01	-4.64E+01	1.03E-01	2.33E+00	-4.80E+00
37	2.01E+01	-4.69E+01	1.03E-01	2.08E+00	-4.85E+00
38	1.67E+01	-4.74E+01	1.03E-01	1.73E+00	-4.90E+00
39	1.12E+01	-4.80E+01	1.03E-01	1.16E+00	-4.97E+00
40	-1.12E+00	-4.82E+01	1.03E-01	-1.16E-01	-4.99E+00
Total			2.07E+00	4.11E+01	-9.61E+01
Average	1.99E+01	-4.64E+01		2.05E+00	-4.80E+00

As the relative pressure is higher the force acting is also higher. We can see that the total force left sode on case 3 is just 34.8 N while for the case 4 is 41.1 N. So that with the total of force right side.

Now, we will calculate the force acting on the billboard by using integration of momentum difference. The flow of the wind inside the domain is influenced by the billboard. While the mass flow rate is constant but the velocity changes. So that calculation can be done by determining the velocity magnitude at a certain location before and after the wind collide with the billboard. The question is where is the proper location?

It’s easy to determine the location for determining velocity magnitude before the wind collide to the billboard, i.e. the inlet velocity. For velocity magnitude after the wind collide, I’ve decided to choose the location where there is no wind velocity vector whose direction is following the –x axis from the center of the billboard, i.e. J= 30 and J = 31 nodes. We know that the location of the billboard is in I = 21 and J = 21-40. The location I choose is in I = 185 nodes. It can be known by looking at the picture of velocity vector, then zooming it. The following picture is an example of velocity vector I meant.

 

From the color, it can be predicted that the magnitude of the velocity is about 4 m/s. With using “lihat lafa” menu, we can know the exact location. Choose “lihat alfa” menu then choose “Batas Tayangan” for determining the I and J range. The chosen I range are 23 until 200. We know that the billboard location is in I = 21. The chosen J is in the center of the billboard, i.e. J 30 and 31. The following picture shows that the velocity magnitude of about 4 m/s is at I = 185 (shown by oval-red line).

 

Choosing I = 185 to calculate the force by using integration of momentum is incorrect. It showed uas that the inlet average velocity is almost the same with average velocity in I = 185. The following table shows this.

So now we will delay this calculation first. While I’m trying to find the right location to determine the total momentum after the wind pass the billboard. Let’s go to Case 5.

1.       Case 5

Like Mr. Ahmad Indra said that try to change the upper and lower cell of the domain from symmetry to wall. So, in this case I try to do that. The domain and condition are the same with in case 1. Just the upper and lower cells is different. Here is the “grid permukaan” of this case.

And the result for stream function is shown below.

We can see that the stream function is almost the same with the result of case 1. So, I think there’s no need for us to calculate the force acting on the billboard with this case. It’s enough with the case 4.