Bhavik,

10 seconds of Googling:

http://www.motors-biz.com/item/70041/auto-radiator-fan-motor-12v-car-for-skoda-audi-seat-volkswagen-1K0959455DH-1K0959455Q-in-Guangzhou.html

James

Thanks for attempting my question. I saw you link.

12v 200w 3900r/min Fan (Radiator Fan Motors)

I believe it uses 200 w power to run motor of fan but my concern is: will it help to reduce the temp of air from 35 to 20. As i mentioned there should be a cooling energy that needs to dissipate. the power shown is above is just the energy which will consume to run the fan.

Thanks for the reply.

Bhavik,

I was utterly wrong. And I apologize - you are right - that is the electrical drawn power.

I should have realized this - as the cooling power will scale with the temperature difference between the incoming air and the circulating working fluid. So a cooling power must be presented *per degree of temperature difference*.

So.

If we look at this item:

https://www.qats.com/Products/Liquid-Cooling/Heat-Exchangers/ATS-HE-26

We see that it moves 250W with every degree of difference between the air and the liquid.

If the air is at 25degC, and you wish to dissipate 250W, then the fluid will warm by around 1 degree - if pumped sufficiently quickly around the loop.

(an important point!)

I hope that this helps.

1- If you want to reach 20°C then your air temperature has to be less which ever will be the air velocity you intend to obtain via fans

2- What is not mentioned in the data is the the transfer surface of the fins. So that you have to estimate it with the data available at the manufacturer site. What does it give ? The overall dimensions 394x120x39.5 [mm] and the fins density of 17/inch and that is all. You can of course with only those data not start any estimation but you have also a picture and with its help you can go further. What you have is to estimate the distance between two flat tubes and get this way the wave length of fins. Their width can be estimated from the 39.5 [mm] at about at least 30 [mm].

Also from the picture you can estimate the length of the finned zone which comes to be about 350 [mm].

Also from the picture you notice that there are 6 parallel fins on one side and 7 on the other. The overall width of 120 [mm] represents thus 13 fins width + 12 tubes and 2 walls. From the picture you get that the distance between 2 adjacent tubes is about 82% of the pitch. If you assume that the wall has about same thickness as a tube the fins are about 8.5 [mm].

The total convective are for a fin way will be 350/25.4*17*(8.5*30)*2=1.195E5 [mm²] = 0.1195 [m²].

On one side you obtain A(6)=0.717 [m²] and on the other A(7)=0.836 [m²].

The total transfer area will be the sum =1.55[m²]. Now you can estimate given an air velocity the heat exchange rate and if it is ok or not for your case. If you intend to use free convection then the convective coefficient is about 10 W/sqmK and (without computing the logarithmic temperature differences) you will need about 17°K of temperature difference (average) to obtain the transfer of the 169[W]. This indicates also that you air temperature has to be about 10°C if you intend to reach 20°C at the end.

This approach is not perfect but when all data are not available it brings you further. I hope it will help .

If you have other questions do not hesitate to ask.