Mr Wilson, On Convection of Heat. 423 



When r = a, we have 



o = cx+d = A + cpa- — ^~ > 



so that A =cx + d — %cpa?, 



hence the solution of the problem is 



n 7 q „ cpr 4 



v = ex + a — ^cpa- + cpr- — ±— % . 



When r = 0, 6 = ex + d — %cpar, 



so that the temperature at the axis of the pipe is lower than that 

 at the circumference by %cpa-. 



The mean temperature of the liquid flowing past the plane 

 x = q, is given by the equation 



- 2 f a 

 6 = rr- udrdr 

 Va 2 J 



= - j^ r K " -") [<* + *- -f + qpr- - ^) dr 



= cq + d - ^cpa 2 . 



Thus the mean temperature of the liquid flowing through the 

 pipe is less than the temperature of the pipe by ^cpa?. 



As an example, suppose a pipe 2 cms. in diameter with water 

 flowing through it with a mean velocity of 100 cms. per second. 

 Suppose also that the temperature of the pipe falls off in the 

 direction of the flow of the water one degree in 1000 cms. Then 

 we have c = 10 -3 , a = 1, and 



_spV_ 100 ^ 



p ~ 2k ~ 2 x 0-00136 " &1 X U ' 



Hence the mean temperature of the water in the pipe at any 

 plane is \\ x 10~ 3 x 3'7 x 10 4 = 17° above the temperature of the 

 pipe at that plane. In practice if the velocity of the liquid 

 exceeds a certain value the flow becomes turbulent, in which 

 case the mean temperature of the liquid would be more nearly 

 equal to that of the pipe. 



In conclusion I wish to say that my best thanks are due to 

 Prof. J. J. Thomson for several valuable suggestions with regard 

 to this paper. 



