Heat Transfer - Convection (Forced Air) 2

[ssmith]

I need to remove the heat (approx. 800 watts) that is generated inside an industrial enclosure that contains a matrix (8 x 20 ) of cylindrical electronic packages mounted through one side panel of the enclosure. I must remove the heat by means of convection only. My initial approach is to force air axially through the matrix of electronic packages. The forced air would exit the enclosure through a series of holes no larger than 1/2" in diameter that are placed between each of the electronic packages. (I hope this is making sense !?!?)

I need help in determining the equations needed to calculate the requirements to remove this heat (Volumetric Flow, Exit hole size , number of holes, air velocity, etc.)

Any and all help would be greatly appreciated- equations,references,websites,etc.

ssmith

 

[svanels]

Assumptions

External temperature T_e = 25^oC or 278 K

External Pressure p = 1 kgf/cm² or 98.066 kPa

Specific heat of ar C_p = 1005 joule/(kg*K)

Supposing a temperature differenitial (dt) of 3 degrees

Heat to be removed Q = 800 watt

1) Calculate mass flow of air needed

M_air := Q / (C_p * dT)

M_air = 955 kg/hr

2) Convert to Volume/time at normal conditions

R = 287 joule/( kg * K)

V_air := M_air * R * T_e / p

V_air = 13.9 m³/min or 490 CFM



A good source for enclosure fans is McMaster. You always can increase your dT to reduce the CFM needed.

I suggest to blow into the enclosure and make the total exit area 0.7 - 0.8 of the Fan Intake area

Hope this helps


Steven van Els
SAvanEls@cq-link.sr

 

[ssmith]

Thank you so much for time and your help. Your reply did help me. You suggested at least an exit area of .7 or .8 of the intake area, how can I calculate the effects of the heat removal with different exit areas.

Thank you

 

[svanels]

Calculating will be difficult, the only way is experimenting
If you can get hold of good infrared thermometer, mark some spots on the enclosure to make comparisons.

Regards
Steven van Els
SAvanEls@cq-link.sr

 

what is the temperature reduction you want ?

Cooling air volume would be decided by

a) the quantum of heat to be removed.
b) the temperater reduction - from __ Deg to ___ Deg C/K.
c) cooling air inlet temperature - ambient air temperature.
d) cooling air velocity permissible inside the enclosure.
e) cooling air exit temperature desired.

The volume of cooling air, and the velocity desired at exit determines the exit orifice area / cross section.

Generally an velocity of 6-10 meter / sec is sufficient for exit of cooling air.

You can give me the above data for an idea about the possible options.

vikas agrawal.

 

[ssmith]

Recuperator,

The allowable temperature rise is 5 C.
The inlet air temperature is ambient of 35 C
We are not all that concerned with the air velocity inside the enclosure.

I have determined the mass flow and the volumetric flow rate of the air; however, I need help determining the pressure drop that is created by the round exit holes (1/2"dia qty = 192 holes) inorder to properly size the blower. I would like to be able to calculate this on my own, I just am having trouble finding the equations that I need.

Thanks so much for all the help!!

ssmith

 

[subjectin]

You say that you can only use convection...have you considered using heat pipe technology? A good website for heat pipe info is

http://www.thermacore.com

Again just an idea...good luck.