# electronic cooling system : design of air system,effect of exhaust and inlet louvers, fan selection

Discussion in 'General Electronics Chat' started by Dhaval Trivedi, Jun 10, 2015.

1. ### Dhaval Trivedi Thread Starter Member

Jun 10, 2015
36
0
Hi,

I am working on designing the air system for the electronic cabinet.
It is a forced convection.
there are louvers at the inlet and the exhaust with axial fan at the inlet pulling the air inside thru the louvers.

I want ot understand he effect of louvers size
placement
angle of opening

Effect of turbulence
Air velocity
Area of opening at inleet v/s outlet

how to maintain good air velocity in the entire cabinet

etc.

Kindly reply me as early as possible.

2. ### #12 Expert

Nov 30, 2010
16,685
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What you have asked for requires several books, which I will not attempt to type into this website. I suggest you enroll in a 4 year college and try to select aerodynamics in the 3rd or 4th year of physics. I have worked air flow problems since I achieved my State Certification in 1985, and I still have to ask the manufacturers of louvers, turning vanes, dampers, etc. about their specifications.

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3. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
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With genuine respect:

This site is entitled All About Circuits -- as opposed to:

--OR--

Fora focusing upon either of the above topics are your better contacts!

With constructive intent
HP

Last edited: Jun 11, 2015
4. ### AnalogKid Distinguished Member

Aug 1, 2013
4,692
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OR, we can start by putting the problem into a familiar context, and analyze it.

Fan = battery
air = current
pressure = voltage
holes and obstructions = resistances

At a bare minimum, the total cross-sectional area of each louver set must be greater than the total cross-sectional area of the fan blade area, or you are choking the fan before it does anything. After that, it's all about pressure drops. Air flow through an enclosure from an inlet to an outlet is a series circuit like a battery and resistors. After all, in the common analogy of flowing water for electric current, voltage equates to water (or air) pressure. Anytime air does anything, bump into something, turn a corner, go through a hole, etc., it loses pressure, the equivalent of the voltage drop across a resistance. If you look at the fan datasheet, it will have a plot of the airflow through the fan versus pressure. Notice that as the pressure increases, the airflow decreases. If you are lucky, you can find the pressure drops of the louver patterns which gives you an estimate of the maximum air flow possible.

A louver is a relative low pressure-drop obstruction. The hole areas are much greater than the area of the metal webs between the holes. Also, even though it guides the air through a direction change, it does it with a curved surface. This equates to a coax cable or waveguide terminated in its characteristic impedance - it aids in maximum power transfer (air flow x pressure).

Turbulance is tough to estimate. More is better, because it brings a larger percentage of the total volume of air molecules in direct contact with the heat-emitting surfaces. Dig into this and you'll meet the Reynolds Number and other obscure goodies.

The combined back pressure of the components within the cabinet is impossible to calculate directly. If you need an accurate number, the most common method today is to create a 3-D model of everything in a CFD thermal analysis program, let it iterate while you go out to lunch, and then wait until it coughs up a number.

Fan manufacturers have app notes regarding fan selection that have more details about the math involved. Also, you probably can find pre-manufactured louver assemblies on the web that are similar to the ones in your cabinet, and their datasheets probably list the pressure drop.

ak

Last edited: Jun 11, 2015
5. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
2,845
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@AnalogKid = AnalogyKid

Best regards
HP

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6. ### #12 Expert

Nov 30, 2010
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It might be admirable to try to understand everything but, if you would ask how to cool a cabinet, you could probably get a useful answer. A dozen people here can tell you how to do the job at hand, but we can't give you a college education in one page.

7. ### Dhaval Trivedi Thread Starter Member

Jun 10, 2015
36
0
I only want to know the thumb rules ! I don't want to re-invent the wheel again.
If there is any theory or proved fact i will use the same.
I just want to understand the physics or basic engineering facts not the detail engineering analysis considering all the factors

8. ### #12 Expert

Nov 30, 2010
16,685
7,327
You still are not helping much. Shall I guess what size you are working with? A dozen watts? A thousand watts?
I will not guess.

For a closed box, 3.25P = CFM Dt (in Fahrenheit)
For a velocity against a heat sink, look up heat sink manufacturers like Thermalloy.
For air flow through louvers, look for static pressure maximum of the fan and the flow curves compared to pressure for the fan.

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9. ### Dhaval Trivedi Thread Starter Member

Jun 10, 2015
36
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Now I have few points :

Turbulence : actually what i found was that turbulence surely increase the air velocity and air volume in a area but sometimes it leads to flow separation which creates problem of some hot spots in the region.

Q : Also, even though it guides the air through a direction change, it does it with a curved surface. This equates to a coax cable or waveguide terminated in its characteristic impedance - it aids in maximum power transfer (air flow x pressure) : Can u describe more about this. little confused !

10. ### #12 Expert

Nov 30, 2010
16,685
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Turbulence in a waveguide model is bad because it is merely a waste of power. Turbulence at the interface where heat is being moved into the air stream is good.

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11. ### Dhaval Trivedi Thread Starter Member

Jun 10, 2015
36
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I have total power loss of 919 watt inside the unit. With 2 axial fan at the inlet : each 83 CFM. Size of the Unit is 590*418*212 mm3.
I have Inlet louvers with opening area of around 7000 mm2 and exhaust stamped louvers with opening area of 6500 mm2.

Now i am trying to optimize my exhaust louvers. Such that i get got air velocity accross the entire unit.

I used your formulae : 3.25P=CFM Dt
3.35 P = 166 * 60
P = 3064 F3/Min * Fahrenheit.

For 2nd Point : can u explain me more That velocity at heat sink ; howe manufacturer will describe the velocity.

For 3rd Point : How to make this selection. i want to fudamentally understand the max static pressure of fan. and flow curves comapred to pressure of fan.

12. ### AnalogKid Distinguished Member

Aug 1, 2013
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Flow separation is exactly what you want, because laminar air flow has an insulating effect, the opposite of what you want.

ak

13. ### Dhaval Trivedi Thread Starter Member

Jun 10, 2015
36
0
true : but it is hard to control the turbulence and make it happen where we want. What i have usually seen is air has the tendency to move towards low resistance then when it is having turbulence then mosot of air goes along the sides of hte component rather then passing thru hte component.

14. ### Dhaval Trivedi Thread Starter Member

Jun 10, 2015
36
0
flow separation is actually occuring exactly near to my component so all the air passes away from the component rather going to hte component

15. ### #12 Expert

Nov 30, 2010
16,685
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Can't follow your math. 3.25 x 919 Watts = (166 CFM) Dt
(3.25 x 919)/166 CFM = 18 degrees F rise in air temperature at the outlet of the box.

ps, you must add the watts of the fan motors if placing them on the input side of the air flow.

16. ### #12 Expert

Nov 30, 2010
16,685
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You must force the location of the air stream with sheet metal guides or fins or shapes.

17. ### AnalogKid Distinguished Member

Aug 1, 2013
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Fans: 2 x 83 CFM
Power: 1000 W
Maximum ambient (air intake) temperature: ???
Maximum allowable temperature rise inside the cabinet: ???

That last one is the biggest problem. Semiconductor datasheets sometimes list the maximum junction temperature, the maximum package temperature, the maximum ambient temperature, or some combinations. For something like relays in a control panel that data can be hard to find. Cooling system design depends on the information about the devices. So, 919 watts of what? Power supplies, relays, motors, industrial controllers, etc.? The first question is How much air do I need? To know that you have to know the environmental limits of the devices you are cooling. Based on experience I'd guess that 166 CFM gross is not enough to cool 1000 W. But if the allowed temperature inside the cabinet is 100C, then no problem.

ak

Nov 30, 2010
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19. ### #12 Expert

Nov 30, 2010
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For fan curves, you must look to the manufacturer of the fan. Every fan has a stall pressure where no air flows. Every fan has a maximum, unrestricted flow. Every fan has a highest efficiency point where x pressure and y flow rate use the motor most efficiently.

20. ### Dhaval Trivedi Thread Starter Member

Jun 10, 2015
36
0
Currently : Instead of creating a baffle to direct the air, I runned the short flow analysis on solidworks to check the airflow across the unit . and selected the case of the louver which was having the least turbulence induced inside the system. due to that i have somewhat less velocity at the exhaust but almost similar velocity across the unit.

In another case there was lot of turbulence due to that exhaust louver design. which gave me good velocity at exhaust but only little more veclity