00:01
Hello friends, let's move on this question.
00:02
So first of all what we have got in this question, we have given the alpha or we can say for this gray surface the alpha and the value of e is equal to 0 .8 and we have given the temperature as 100 degree celsius or we can say 100 plus 273.
00:21
So it will become in kelvin 373 kelvin.
00:24
Now moving on the g we have given g is equal to 1000 watt per meter square and area is 0 .1 meter square.
00:34
Now moving on the opaque surface, so we know for opaque surface gamma equal to 0.
00:39
So alpha plus p is equal to 1.
00:42
So p is equal to 1 minus alpha or we can say alpha is equal to 0 .8 we have given.
00:47
So this is equal to 0 .2.
00:49
So now moving on the radiosity has a formula of summation of sorry not summation it is e, eb plus pg.
00:58
So this is equal to 0 .8 multiply with the eb is equal to 5 .67 10 raised to power minus 8 plus 373 multiply with multiply with 373 power 4 plus 0 .2 into multiply with 1000 that is we have given.
01:16
So finally we will get 1078 watt per meter square radiosity.
01:21
Now moving on the net heat transfer...