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July 3, 2014, 10:07 |
Sinusoidal Pressure inlet
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#1 |
New Member
Rafau
Join Date: Aug 2013
Posts: 20
Rep Power: 13 |
Hey,
I want to set boundary condition - sinusoidal pressure inlet at circle shape(2d). I think my UDF is not correct, can you take a look? #include "udf.h" DEFINE_PROFILE(unsteady_pressure, thread, position) { face_t f; real t = CURRENT_TIME; begin_f_loop(f, thread) { F_PROFILE(f, thread, position) = 100 + 200*sin(500.* t); } end_f_loop(f, thread) } |
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July 3, 2014, 11:03 |
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#2 |
Senior Member
Join Date: Nov 2013
Posts: 1,965
Rep Power: 27 |
I don't know if it is correct, because I don't know what you exactly want. "sinusoidal pressure inlet", what does that mean? Something has a sinus-shape, but what? The pressure as function of time? The pressure as function of position?
Your code looks like the pressure is a function of time. Is that what you want? And what makes you think that your UDF is not correct? Are the results not what you expect? |
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July 3, 2014, 11:13 |
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#3 |
New Member
Rafau
Join Date: Aug 2013
Posts: 20
Rep Power: 13 |
thank you for fast response ;-)
I want to model point (exactly it is the small circle in huge ambient) which is point-sound source. That's why I need to apply p(t) function. Sinusoidal would be the best. I've read some UDF tutorials, but most of them were focused on inlet at pipe like a straight line(2d). Some mistakes i have made were connected with timestep size. (but few question to left) 1. in eq. F_PROFILE(f, thread, position) = 500 + 300 * sin(500000.* t); 500 is some const 300 is an Amplitude 500000 would be "w" - omega (angular velo)? correct? |
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July 3, 2014, 11:21 |
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#4 |
Senior Member
Join Date: Nov 2013
Posts: 1,965
Rep Power: 27 |
If you use F_PROFILE(f, thread, position) = 500 + 300 * sin(500000.* t); the result is:
Average pressure is 500 Pa. (Gauge pressure, the absolute pressure depends on what you set as operating pressure.) Minimum pressure is 500 Pa - 300 Pa = 200 Pa. Maximum pressure is 500 Pa + 300 Pa = 800 Pa. Every 2*pi seconds (6.28... s), there are 500000 "sine shapes". In other words: your point has a frequency of 500000/2pi = 79.6 kHz. To resolve this, your time step needs to be in the order of microseconds. |
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July 3, 2014, 11:27 |
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#5 |
New Member
Rafau
Join Date: Aug 2013
Posts: 20
Rep Power: 13 |
I have realized that few minutes ago. Anyhow i really appreciate your response. That makes me sure of myself.
thank you Pakk |
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Tags |
sinusoidal udf |
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