ENGINEERING EVALUATION 343 



Appendix B 

 HEAT-LOSS CALCULATIONS-SEALAB II 



COEFFICIENTS OF THERMAL CONDUCTIVITY 



K(Btu/hr-ft-°F) 



Steel (1% C) 25 



Corkboard 0.025 



Concrete 0.47-0.81, 0.64 (avg.) 

 Air 0.015 



Helium 0.090 



Plexiglas 0.120 



HEAT-TRANSFER AREAS OF SEALAB II 



Two inches of cork insulation, sides and ends. 



Total side area = 2 [(7.5) (51.5)] = 772 ft^ 

 Total end area = 311 - 85 = 226 ft ^ 

 Total area, 2 in. cork = 998 ft " 



One inch cork insulation, overhead. 



A^ = 10 X 55.25 = 552 ft^ 

 Concrete deck (one foot average thickness). 



Ad = 376 ft^ 

 Port areas (Plexiglas, 1 in. thick). 

 "77X 2^ 



Ap = 11 



= 34 ft-^ 



4 



Uninsulated Hull Areas. 



Entry area (below deck) = 73 ft^ 



Surface Access Hatch = 5 ft^ 



Emergency Exit Hatch = 5 ft^ 



A, = 83 ft^ 



HEAT LOSS FROM SEALAB II 



One of the major problems in determining the heat loss in Sealab II is the effect the helium 

 atmosphere has on the insulation. It has been found previously that helium permeates most 

 materials. Since helium is six times as conductive as air, this seriously affects the thermal 

 conductivity of the insulation. 



It is known that if the gas within an insulating material is replaced by another gas having a 

 different conductivity, the conductivity of the insulation will be changed by an amount very nearly 

 equal to the difference in conductivity of the two gases. 



