PHYSICS, CHEMISTRY AND ENGINEERING 181 



terior of the box by suitable circulation of the air in the box. 

 Direct current is supplied both coil and fan, and the total input 

 in watts is determined from the ammeter and voltmeter read- 

 ings. After the box has been under heat for at least 24 hours, 

 and a condition of equilibrium has been established, it is only- 

 necessary to make one set of readings and by substitution in 

 the equation US (t-t )=3.415 (Wi+W 2 ) find the value 

 of U. The right hand member is, of course, the heat equiva- 

 lent of the watts supplied, where W x and W 2 =watts per hour 

 supplied to fan and coil respectively. S=mean area of the six 

 sides of the box. 



9. Values of C and Kj are determined at the same time by 

 the use of thermo-couples, imbedded just in the surface of the 

 wall materials, and used for measuring the surface tempera- 

 tures tj and t 2 . Thus, for the determination of conductivity, 

 we have 



(t r t 2 ) =3.415 (Wj+W 2 ) 

 where the value of C is to be found per inch of thickness, and 

 X-f-thickness inches. In a similar manner we may find values 

 of K (still air) by using the outside surface temperature of 

 the box, which is standing in still air, or 



K 1 S(t 2 -t )=3.415 (W t +W 2 ). 



10. Since it is manifestly impossible to test all forms of 

 wall construction, it will, in general, be necessary to deter- 

 mine the value of U (the coefficient based on inside and out- 

 side air temperatures) by calculation. The equations already 

 derived provide the means of doing this if values for the con- 

 ductivity C and surface coefficients K x and K 2 are known. 

 Tests now in progress in the Mechanical Engineering Labora- 

 tory of the University of Illinois, have, as one of their objects, 

 determination of such data, and the figures numbered from 3 

 to 8 show application of this data to typical simple and com- 

 pound wall constructions in solving for proper values of U 

 for use in practice. Heat transmission tests on actual walls, 

 such as here shown, give results, which agree very closely with 

 the calculated values. 



11. In practical application it is only necessary to multiply 

 the coefficient U by the temperature range and then by the net 

 area of the wall through which the heat loss takes place. Thus, 

 for 70° inside air, and 0° outside air temperature, the total 

 transmission loss for 1000 square feet of wall, such as shown 

 in Fig. 3 is 0.291x(70— 01x1000=20370 B. t. U. per hour. 



