... cosh nπ Thus -1 Smax ⎡2 ⎤ ⎡4 ⎤ q′′s Wd (-1 ) n+1 + (-1 )(n-1)/2 (n-1)/2 = = ⎢ ∑ (-1 ) nπ = ta nπ nh ⎥ ⎢ ⎥ ∑ 2 k(T2,max - T1 ) ⎢⎣ d n odd n π ⎥⎦ ⎢⎣ d n odd n π ⎥⎦ -1 where d is the depth of the rectangle... θ(y = W) = T2 - T1 = ∑ Cn W n=1 W ∫ sin nπx dx sinh nπ = W ∞ ∑ Cn n=1 - (-1 )n sinh nπ nπ Thus ⎡ ∞ [ (-1 ) n+1 + 1][ 1- (-1 )n ] ⎤ q′′s Wd S= = ⎢ ∑ nπ ⎥ 3 k(T2 - T1 ) ⎣⎢ d n=1 n π ⎦⎥ -1 ⎡8 ⎤ = ⎢ ∑... surface areas and heat transfer rates for the objects of this problem are half as much as for the objects of Table 4.1 q′′ As q* A q* (4πAs )1/2 q S= = = ss s = ss k(T1 - T2 ) k(T1 - T2 ) 2Lc where