96 Applications of the Formulas 



per hour is 5.98, * the cooling in question would take place in 

 =6 t 



5.98 



This supposes the interior temperature to be maintained at 

 59 , and the cooling of the wall to take place in the way that we 

 have assumed; but in reality the cooling would be much less 

 rapid, because the temperatures of the outer surface would be 

 much lower than we have admitted and because the temperatures 

 of the different layers of the wall would succeed each other 

 according to a different law which would also aid in retarding 

 the cooling. 



905. We see from this that if the cooling of a room took 

 place through the walls alone, the variations of the exterior tem- 

 perature would manifest themselves within very slowly, and 

 feebly. But rooms have always glazed windows, and as the glass 

 almost instantaneously assumes the mean of the inner and outer 

 temperatures, we must in order to maintain a constant tempera- 

 ture, supply an increased quantity of heat which will vary with 

 the exterior temperature and is in general very much greater than 

 that which would result from the transmission of heat through 

 the walls. 



*When the loss was 5.98 B. T. U. the exterior temperature was 42.8 and the temper- 

 ature of the outer surface was 48.19 ; this is an excess of temperature of 5.39 not 16.19. 



fThis figure is just twice that given in the original text for the reason explained 

 above. 



The reasoning inlthis paragraph is certainly difficult to follow. We may consider 

 the problem in another way. P6clet appears to admit that during the cooling of the wall 

 the mean temperature of the outer surface is a mean between its temperatures at the 

 beginning and end of the process. Then we may assume the same to be true for the 

 mean temperature of the inner surface. Then the difference between the emission from 

 the outer surface and the absorption by the inner surface per hour will be the mean 

 amount of the hourly cooling of the wall. Then the total amount of cooling, which we 

 have already calculated, divided by this quantity, will give the time of cooling. 



The mean temperature of the outer surface is- ' =43.81 



and its excess over the outer temperature is 43.81 32= 11.81. 



Taking Q as in 858 =1.14 we have ; B. T. U. emitted per hour per square foot = 1.14 

 X 11.81 = 13.47. 



The mean temperature of the inner surface is 53-O9 



The excess is 59 53.09 = 5.91 and; B. T. U. absorbed per hour per square foot = 1.14 

 X 5-91 = 6-74- 



Then the mean hourly loss of heat by wall = 13.476.74=6.73 and the duration of 



cooling= =42 hours. 



