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# Optimization of Thermal Systems

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The project of Electronic Cooling System was presented for the Optimization of Thermal System class in April 2009.

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### Optimization of Thermal Systems

1. 1. Design and Evaluation of Electronic <br />Cooling Systems<br />By<br />SuabsakulGururatana<br />Prashant Ghadge<br />Serkan Ongun<br />Jatin Lad <br />Mechanical Engineering Department<br />Lamar University<br />
2. 2. Introduction<br />Problem Statement:<br />A piece of electronic equipment dissipates a total of 400 W. Its base dimension must not exceed 40 cm x 30 cm, and high must be less than 15 cm. Six boards are employed to mount the component. The temp. should not go beyond 120 ºC anywhere in the system.<br />The Electronic Equipment<br />
3. 3. Purpose and Overview<br /><ul><li>Ideas from our course work towards our recent project.
4. 4. Comparative study between two systems.
5. 5. Consideration of Design and Optimization criteria.
6. 6. Use of softwares for Modeling and Simulation.</li></li></ul><li>Design<br />
7. 7. Model - Forced Air-Cooled System <br />Forced Air-Cooled System <br />
8. 8. Model- Refrigeration System<br />Refrigeration System<br />
9. 9. Design Procedure<br /><ul><li>Gambit Software – Modeling
10. 10. Fluent software – Simulation
11. 11. Temperature of the board fixed to 120F
12. 12. Temperature of the incoming air 77F
13. 13. Velocity inlet is changed for several trials to get the rate of heat transfer (Q) = 400W
14. 14. We get the performance relation.
15. 15. Fan Selection</li></li></ul><li>Analysis-Forced Air-Cooled System <br />Temperature distribution over equipment for Forced Air Cooled System<br />
16. 16. Analysis- Refrigeration System<br />Temperature distribution over equipment for Refrigeration System<br />
17. 17. Optimization<br />
18. 18. Forced Air-Cooled System <br />Q-V diagram for forced air cooled system<br />
19. 19. Objective Function (I)<br /><ul><li>Q (w) = -0.22V2 + 25.20V + 34.37
20. 20. Constraint Q = AV
21. 21. Total Q = 0.86 m3/s
22. 22. For single fan Q = 0.29 m3/s
23. 23. Optimum heat transfer rate Q (w) = 744.22W</li></li></ul><li>1. Refrigeration System At Tevap = 30° F<br />Q-V diagram for Refrigeration System at Tevep = 30° F<br />
24. 24. Objective function (II)<br /><ul><li>Q (w) = -0.27V2 + 30.09V + 30.07
25. 25. Constraint Q = AV
26. 26. Total Q = 0.27 m3/s
27. 27. For single fan Q = 0.09 m3/s
28. 28. Optimum heat transfer rate Q = 484.21 W</li></li></ul><li>2. Refrigeration System At Tevap = 40° F<br />Q-V diagram for Refrigeration System at Tevep = 40° F<br />
29. 29. Objective function (III)<br /><ul><li>Q (w) = -0.27V2 + 29.9V + 28.67
30. 30. Constraint Q = AV
31. 31. Total Q = 0.83 m3/s
32. 32. For single fan Q = 0.27 m3/s,
33. 33. Rate of heat transfer Q = 855.34 W</li></li></ul><li>3. Refrigeration System AtTevap = 50° F<br />Q-V diagram for Refrigeration System at Tevep = 50° F<br />
34. 34. Objective Function (IV)<br /><ul><li>Q (w) = -0.27V2 + 29.7V + 28.34
35. 35. Constraint Q = AV
36. 36. Total Q = 0.84 m3/s
37. 37. For single fan Q = 0.28m3/s,
38. 38. Rate of heat transfer Q = 844.06 W</li></li></ul><li>4. Refrigeration System at V = 15 m/s<br />Q-T diagram for Refrigeration System at V = 15 m/s<br />
39. 39. Objective Function (V)<br /><ul><li>Q = 0.029t2 – 0.64t + 418
40. 40. Unconstraint
41. 41. Therefore, t = 11.03° C
42. 42. t =52.60° F and Q = 414 W</li></li></ul><li>Conclusions<br /><ul><li>Both systems are working systems
43. 43. Better system – Forced Air Cooled
44. 44. Special purpose – Refrigeration systems
45. 45. General purpose – Forced Air Cooled System </li></li></ul><li>Thank You<br />