Page 279 - Modelling in Transport Phenomena A Conceptual Approach
P. 279
8.2. ENERGY TRANSPORT WITHOUT CONVECTION 259
Right-side of the wall
Using the physical properties evaluated at 0 "C, the Reynolds number becomes
The use of Eq. (E) in Table 4.2 gives
(Nu) = (0.037Re;" - 871) Pr1/3
= k.037(11.3 x 106)4/5 - 8711 (0.717)1/3 = 13,758 (18)
Therefore, the average heat transfer coeficient is
= (13,758) ( 24.071xg
= 33.1 W/ m2. K (19)
Substitution of the new values of the average heat transfer coeficients, Eqs. (16)
and (19), into Eq. (1) gives the heat transfer rate as
+-+-
-
21.5 20 33.1
The surface temperatures are
0
Ti = TA - ~
A (hA)
= 50- 20,756 18OC
(30) (21.5)
Q
T2 = TB + -
A (hB)
=-lo+ 20,756 lloc
(30) (33.1)
Since these values are almost equal to the previous values, then the rate of heat loss
is 20,756W.
Comment: The Biot numbers, i.e., (h)L,h/k, for this problem are calculated as
follows:
