Optimization of Thermal Systems
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Transcript of Optimization of Thermal Systems
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DESIGN AND EVALUATION OF ELECTRONIC
COOLING SYSTEMSBySuabsakul Gururatana
Prashant GhadgeSerkan Ongun
Jatin Lad
Mechanical Engineering DepartmentLamar University
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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.
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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.
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DESIGN
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MODEL - FORCED AIR-COOLED SYSTEM
Forced Air-Cooled System
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MODEL- REFRIGERATION SYSTEM
Refrigeration System
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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
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ANALYSIS-FORCED AIR-COOLED SYSTEM
Temperature distribution over equipment for Forced Air Cooled System
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ANALYSIS- REFRIGERATION SYSTEM
Temperature distribution over equipment for Refrigeration System
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OPTIMIZATION
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FORCED AIR-COOLED SYSTEM
y = -0.2203x2 + 25.204x + 34.741
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Velocity (m/s)
Q (
W)
Force air-cooled systems
Poly. (Force air-cooled systems)
Q-V diagram for forced air cooled system
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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
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1. REFRIGERATION SYSTEM At Tevap = 30° F
y = -0.272x2 + 30.09x + 30.06
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Q (
W)
Velovity (m/s)
Tevap. = 30 F
Poly. (Tevap. = 30 F)
Q-V diagram for Refrigeration System at Tevep = 30° F
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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
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y = -0.2726x2 + 29.994x + 28.674
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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
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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
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y = -0.2657x2 + 29.707x + 28.345
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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
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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
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4. Refrigeration System at V = 15 m/s
y = 0.0292x2 - 0.6446x + 418.29
415
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Tevap (C)
Q (
W)
V = 15 m/s
Poly. (V = 15 m/s)
Q-T diagram for Refrigeration System at V = 15 m/s
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OBJECTIVE FUNCTION (V)
Q = 0.029t2 – 0.64t + 418
Unconstraint
Therefore, t = 11.03° C
t = 52.60° F and Q = 414 W
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CONCLUSIONS
Both systems are working systems
Better system – Forced Air Cooled
Special purpose – Refrigeration systems
General purpose – Forced Air Cooled System
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THANK YOU