00:01
In the following question, it is said that the addition of aluminium fin has been suggested to increase the rate of heat dissipations from one side of an electronic device, which is 1 meter wide and 1 meter tall.
00:15
The fins are to be rectangular in cross -section, 2 .5 cm long and 0 .2 cm thick.
00:23
There are to be 100 fins per meter.
00:26
The convection properties are 35 watt per meter square kelvin and 25 degrees centigrade.
00:32
Then determine the percent increase in the rate of heat transfer of the fined wall compared to the bare wall.
00:41
So for this, we have been given k equal to 204 watt per meter kelvin.
00:53
And the device area is given 1 meter into 1 which is equal.
00:58
To 1 meter square and the fins having rectangular cross section in x direction which is given by w into t equal to a fin so this is 1 into 0 .25 divided by 100 which is equal to 0 .0 025 meter square and number of fins given n equal to 100 fins and h is equal to 35 watt meter square kelvin and t is equal to 25 degree centigrade so first thing is exposed to convection at tip this is equal to t devised minus t surrounding and we know that q finn equal to k acme theta nought h cph hyperbolic m l plus m k sine hyperbolic m o divided by m k pos hyperbolic amyl plus h shine hyperbolic emm where m is equal to root under h p upon k ac then p is equal to two omega plus p so this is equal to two 1 plus 0 .25 divided by 100 then p is equal to 2 .05 meter meter so ac is equal to 0 .0025 square then m is equal to root under 35 into 2 .0 .05 divided by 0 .0025 into 204 which is equal to 11 .7 3 and l is given 0 .025 meters then m l equal to 0 .025 into 11 .73 so we get 0 .293 to 5.
04:20
So, cost hyperbolic ml equal to 1 .04 -3 and stein hyperbolic ml equal to 0 .2474 .74.
04:39
Now put this values in above equation.
04:42
So q fin equal to 204 into 0 .0025.
04:53
Into 11 .73 into peta not 351 .04 double three plus 11 .7...