optimization of thermal systems
DESCRIPTION
The project of Electronic Cooling System was presented for the Optimization of Thermal System class in April 2009.TRANSCRIPT
DESIGN AND EVALUATION OF ELECTRONIC
COOLING SYSTEMSBySuabsakul Gururatana
Prashant GhadgeSerkan Ongun
Jatin Lad
Mechanical Engineering DepartmentLamar University
INTRODUCTION
The Electronic Equipment
Problem Statement: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.
PURPOSE AND OVERVIEW
Ideas from our course work towards our recent project.
Comparative study between two systems.
Consideration of Design and Optimization criteria.
Use of softwares for Modeling and Simulation.
DESIGN
MODEL - FORCED AIR-COOLED SYSTEM
Forced Air-Cooled System
MODEL- REFRIGERATION SYSTEM
Refrigeration System
DESIGN PROCEDURE
Gambit Software – Modeling
Fluent software – Simulation
Temperature of the board fixed to 120F
Temperature of the incoming air 77F
Velocity inlet is changed for several trials to get the rate of heat transfer (Q) = 400W
We get the performance relation.
Fan Selection
ANALYSIS-FORCED AIR-COOLED SYSTEM
Temperature distribution over equipment for Forced Air Cooled System
ANALYSIS- REFRIGERATION SYSTEM
Temperature distribution over equipment for Refrigeration System
OPTIMIZATION
FORCED AIR-COOLED SYSTEM
y = -0.2203x2 + 25.204x + 34.741
0
100
200
300
400
500
0 5 10 15 20 25
Velocity (m/s)
Q (
W)
Force air-cooled systems
Poly. (Force air-cooled systems)
Q-V diagram for forced air cooled system
OBJECTIVE FUNCTION (I)
Q (w) = -0.22V2 + 25.20V + 34.37
Constraint Q = AV
Total Q = 0.86 m3/s
For single fan Q = 0.29 m3/s
Optimum heat transfer rate Q (w) = 744.22W
1. REFRIGERATION SYSTEM At Tevap = 30° F
y = -0.272x2 + 30.09x + 30.06
0
100
200
300
400
500
600
0 4 8 12 16 20
Q (
W)
Velovity (m/s)
Tevap. = 30 F
Poly. (Tevap. = 30 F)
Q-V diagram for Refrigeration System at Tevep = 30° F
OBJECTIVE FUNCTION (II)
Q (w) = -0.27V2 + 30.09V + 30.07
Constraint Q = AV
Total Q = 0.27 m3/s
For single fan Q = 0.09 m3/s
Optimum heat transfer rate Q = 484.21 W
y = -0.2726x2 + 29.994x + 28.674
0
100
200
300
400
500
600
0 4 8 12 16 20
Velovity (m/s)
Q (
W)
Tevap. = 40 F
Poly. (Tevap. = 40 F)
Q-V diagram for Refrigeration System at Tevep = 40° F
2. REFRIGERATION SYSTEM At Tevap = 40° F
OBJECTIVE FUNCTION (III)
Q (w) = -0.27V2 + 29.9V + 28.67
Constraint Q = AV
Total Q = 0.83 m3/s
For single fan Q = 0.27 m3/s,
Rate of heat transfer Q = 855.34 W
y = -0.2657x2 + 29.707x + 28.345
0
100
200
300
400
500
600
0 4 8 12 16 20
Velovity (m/s)
Q (
W)
Tevap. = 50 F
Poly. (Tevap. = 50 F)
3. REFRIGERATION SYSTEM At Tevap = 50° F
Q-V diagram for Refrigeration System at Tevep = 50° F
OBJECTIVE FUNCTION (IV)
Q (w) = -0.27V2 + 29.7V + 28.34
Constraint Q = AV
Total Q = 0.84 m3/s
For single fan Q = 0.28m3/s,
Rate of heat transfer Q = 844.06 W
4. Refrigeration System at V = 15 m/s
y = 0.0292x2 - 0.6446x + 418.29
415
416
417
418
419
420
421
-3 -2 -1 0 1 2 3 4 5 6
Tevap (C)
Q (
W)
V = 15 m/s
Poly. (V = 15 m/s)
Q-T diagram for Refrigeration System at V = 15 m/s
OBJECTIVE FUNCTION (V)
Q = 0.029t2 – 0.64t + 418
Unconstraint
Therefore, t = 11.03° C
t = 52.60° F and Q = 414 W
CONCLUSIONS
Both systems are working systems
Better system – Forced Air Cooled
Special purpose – Refrigeration systems
General purpose – Forced Air Cooled System
THANK YOU