346 ENGINEERING EVALUATION 



For the sides and ends (2 in. corkboard) 

 1 1 1 



U = 



1 1 0.559+1.67 2.23 



+ 



1.79 0.6 

 U = 0.448 Btu/hr-ft--°F 

 Area of sides and ends (less port area) = 998 - 34 = 964 ft". The heat loss is 

 Q = UA At = 0.443 X 964 x 40 



Qsides6.ends = 17,300 Btl,/hr. 



For the top (1 in. corkboard) 



= 0.746 Btu/hr-ft^-^F 



-^(t-tj = hp(t3 -t) Ap = 34ft2 



p 



0.12 X 12 



lip v,w.J . i/ 



2.88(71 - 48) = (88 - 71) 2.03 

 33.1 = 34.5 



Therefore 



hp = 2.02 



Then 



^p ~ _J_ J_ ~ 0.459 + 0.694 1.19 °-^'*° 

 2.02 ■^ 1.44 



Kp = 0.12 Btu/hr-ft-°F 

 t. = 48°F 

 t, = 88°F 



Xp = 1 in. 



