Page 67 - Introduction to Computational Fluid Dynamics

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Fully Developed Heat Transfer 1D HEAT CONDUCTION
The equation governing laminar fully developed heat transfer in a tube is given by

∂ ∂T ∂T
k 2π r − 2π r ρ C p u fd = 0, (2.80)
∂r ∂r ∂z
2
2
where u fd = 2u (1 − r /R ) or can be taken from the numerical solution of Equa-
tion 2.77. Evaluation of ∂T /∂z can be carried out from the boundary conditions at
the tube wall as follows.
Constant Wall Heat Flux: From the overall heat balance and from the condition
of fully developed heat transfer [33], it can be shown that
∂T dT b 2q w
= = . (2.81)
∂z dz ρ C p uR

Therefore, Equation 2.80 can be written as

∂ ∂T r r 2
k2πr − 8π 1 − q w = 0. (2.82)
∂r ∂r R R 2
2
2
Thus, if ∂r is replaced by ∂x, A by 2πr, and q by − 4(1 − r /R )q w /R, Equa-

tion 2.82 is same as the steady-state form of Equation 2.5.
Constant Wall Temperature: In this case, the condition of fully developed heat
transfer implies that

∂T −1 dT b −1 2k ∂T /∂r | r=R
= (T w − T b ) = (T w − T b ) , (2.83)
∂z dz ρ C p uR

where T b is the mixed-mean or bulk temperature. Thus, by setting q =
2
2
−4k/R (1 − r /R )(T w − T b ) −1 ∂T /∂r | r=R , Equation 2.80 is same as Equa-
tion 2.5. However, T b and ∂T /∂r | r=R must be evaluated at each iteration. The
bulk temperature T b is evaluated as

R
ρ C p uT 2π rdr
0 . (2.84)
R
T b =
0 ρ C p u 2π rdr
Thermal Entry Length Solutions
Consider laminar flow between two parallel plates separated by distance 2b. When
Pr >> 1, it is possible to obtain the variation of the heat transfer coefficient h with
axial distance z by solving the following differential equation:

∂ ∂T ∂T
k = ρ C p u fd , (2.85)
∂y ∂y ∂z