00:01
So, for a fully developed lumina flow with constant heat flow, with constant heat flow, na is given by 4 .36.
00:17
So which implies that h by k is equal to 4 .36.
00:23
So we will get the value of h as 94 .066.
00:28
So this we have to use in another equations.
00:31
So we have the equation for q which is equal to m cp into t2 minus t1.
00:40
So the values of t2 and t1 are given in the question.
00:44
So let us substitute and find the value of q which is equal to 1 .26 into 10 power minus 3 into 2 .09 into 10 power 3 into 75 .5 minus 20, which we will get the value of q as 146 .153 watt.
01:08
So we have to find the temperature here, right? when the, through the constant heat flow.
01:14
So we have q is equal to one more equation we have that is h into area into ts minus t2.
01:24
So we have to find the value of ts here.
01:28
So let us substitute in the given values, we will get the value 146 .15 that is q is equal to h into that is 94 .066 into area pi into 6 .35 into 10 power minus 3 into 1 .22 into ts minus t2...