Loss of Heat from Covered Steam Pipes 7 



Consider an infinitely thin annular element of the covering 

 at radius r, its thickness is dr, its area, 2*r, its conductivity for 

 one foot thickness is C' and the difference between the temper- 

 atures of its inner and outer surfaces is dt. Then treating it as 

 a flat plate we have M t_ 2 * r C' dt 



1VJ. - ~ 



dr 

 Integrating we have M'= \. - . . . . . (i) 



where N=2. 3 (log R' log R}. 



Now we have already shown that for a square foot of surface 

 M=(A + R) (/' 0) and remembering that M'=2itR' M and 

 calling A -f- R= Q we have 



M'=2KR' Q (/' 0) ....... (2) 



The amount of heat passing through the covering must equal 

 that escaping from the surface, so setting (i) equal to (2) we get 



i + QR' N 



C 



It is more convenient to use C the coefficient of conductivity 

 for one inch thickness instead of C the coefficient for one foot. 

 Since C= 1 2 C' we get 



QR' N ...... (4) 



C 



It is really more convenient to keep R' in feet. In working 

 up the value of N=2.^ (log R log R' ) from a table of loga- 

 rithms we can, of course, take R and R' in inches if we wish. 



For an example take a lo-inch pipe, covered with a thickness 

 of i -j^- inches of Keasbey's Magnesia; steam temperature 365.2 

 F.; air temp. 66 F.; C=.^. 



The difference between the logarithms of the inner and outer 

 radii is .087 X 2.3=.2=7V. 



We cannot yet compute the exact value of Q because we do 

 not know the temperature of the surface of the covering, but we 

 will assume it to be i .7. 



Then ^=3.44X1.7(365.2-66) _ 



* 



i+ /.. 7 X.548X.\ 

 \ -45 / 



45 



