344 ENGINEERING EVALUATION 



The thermal conductivity of corkboard in a standard air atmosphere is 



K^ = 0.025 Btu/hr-ft-°F, 

 and 



Kheuu. -K,,, =0.09-0.015 = 0.075. 

 Then the new K,, would be 



Ke = 0.025 + 0.075 = 0.10 Btu/hr-ft-°F. 



The most difficult quantity to determine in any heat-transfer problem is the film coeffi- 

 cient of the filir next to the insulating material. This determines the heat transferred by natu- 

 ral convection. The simplest approach to solve for this quantity is by the following method. 



Consider a wall maintained at a constant temperature t„, coated with a layer of insulating 

 material of a thickness x and of thermal conductivity K. The outside of the insulation is in 

 contact with the atmosphere at temperature tg. Heat is transferred by conduction through the 

 insulation and by natural convection through the atmosphere. In the steady state, the rate at 

 which heat is conducted through a unit area of the insulation material is equal to the rate at 

 which it is supplied to the air by convection, or 



where t is the temperature of the outside surface of the insulation. Besides K and x, t„ and 

 tg are known. Since h varies as the fourth root of t - tg, the simplest way to solve for t is 

 by trial and error. Thus, assuming t to be any arbitrary value, h is calculated and then mul- 

 tiplied by t - tg. The value of (K/x) (t„ - t) is then calculated and compared with h(t - ta). 

 If these quantities are not equal, another value of t is chosen, and so on until the equation is 

 satisfied. To apply this to the present problem: 



^(t-t^) = h2(t3-t) t„ = 48° 



t. = 88° 

 0.10 Btu/hr-ft-°F x 12 in./ft 



(t - 48) = hi (88 - t) 



(corkboard) 



Xj = 2 in. 



Xj = 1 in. 

 (corkboard) 



0.6(70 - 48) = 2.05(88 - 70) 

 13.2 = 36.9 



