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