q = dLA^t. = _x_Q_ (0.25)(8,850) 



x ' -K A (200)(0.378) ' or 29 ' 6 F (2) 



for a 1/4-inch-thick thimble. While not unreasonably high, it would 

 produce base temperatures well over 100°F with natural convection heat 

 transfer to 40°F seawater. Much better results would be obtained with 

 copper. For the Cu-Ki case, the temperature drop from the base would be 

 approximately as sho-wn schematically in Figure 2. This figure also shows 

 a first estimate of the full-scale cross section of idealized heavy fins. 

 The final shape and height, H, selected were eventually modified to reduce 

 manufacturing costs and to avoid interlocking of the fins of adjacent 

 modules on the prescribed base circle (21.25 inches diameter) for the 

 12 modules. The results are slightly nonconservative compared with the 

 ideal shape (Figure 2) with ample fillets at the base and parabolic cross 

 section. Th.?t is, the more expensive, longer fins would provide slightly 

 better cooling; the difference is marginal. 



The bare thimbles would have a base temperature well over 200°F, 

 which would be undesirably high. An extended surface (fins) to produce 

 additional heat transfer area is clearly indicated. Because of the high 

 heat loading, the optimum fins will require: (a) heavy cross section, 

 to avoid excessive temperature drop along their length, (b) a material 

 of high thermal conductivity for the same reason, and (c) as close spacing 

 as practical without reducing flow to the base. 



The actual value of the film coefficient, h, is determined by the 

 veil-known dimensionless equation as refined by Braun (1965) for seawater: 



Nu = f(Pr) (Ra) n (3) 



in which f is an experimental constant and the exponent n is typically 

 1/4 up to Grashoff's number of about 10 9 , and 1/3 above 10 9 . The lower 

 value is associated with laminar flow, the higher value with turbulence. 



For the fin length, L, dictated by the design and the physical 

 properties of seawater at pressures equivalent to 20,000-foot depth: 



c = 0.95 Btu/(lb)(°F) 



p 



c = 64 lb/ft 3 



u = 1.66 lb/(ft)(hr) 



K = 0.385 Btu/(hr)(ft 2 )(°F/ft) 



S = 1.15 x 10" 4 /°F 



I ■ = 0.5 ft 



g = 4.17 x 100 ft/(hr)(hr) 



Pr = c p U / K = 5 - 68 (dimensionless); at 1 atmosphere, this is 

 about 4.5 



