00:01
Hi, in this question the surface temperature is given as 130 degree celsius and the ambient temperature is given as 20 degree celsius.
00:08
A is given as 0 .5 meter square.
00:11
Then we are given with a value of so we are given with a value of cp.
00:16
So cp rho k and nu.
00:19
So now we have we have to find out prandtl number it is given by mu into cp divided by k.
00:25
So which can be written as rho into nu into cp divided by k.
00:29
So on substituting the values we have the rho value is given as 1 .07.
00:34
Nu value is given as 19 .1 into 10 power minus 6 into cp is 1 .007 into 10 power 3 divided by k value is 0 .029.
00:44
So prandtl number is 0 .709.
00:47
Then we have the so we have the grashof's number is given by l cube g beta into ts minus t infinity divided by nu square.
00:58
So on substituting the values we have 0 .5 cube into 9 .8 beta is 1 so beta is 1 by t.
01:07
So which is 273 plus 75 into ts minus t infinity is 130 minus 20 degree divided by nu is 19 .1 into 10 power minus 6 whole square.
01:18
So from this we have grashof's number is 1 .06 10 power 9.
01:24
Then we have the heat coefficient can be calculated by nul is given by c into gr into pr whole power m...