Question 4.6: For steady incompressible laminar flow through a long tube, ...

With T = T(r), Eq. (4.75) reduces for steady flow to
\rho c_p \frac{dT}{dt}=k \nabla^2 T +\Phi                                  (4.75)
\rho c_p v_r \frac{dT}{dr}=\frac{k}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right)+\mu \left(\frac{dv_z}{dr}\right)^2                          (1)
But since v_r = 0 for this flow, the convective term on the left vanishes. Introduce v_z into Eq. (1) to obtain
\frac{k}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right)=-\mu \left(\frac{dv_z}{dr}\right)^2=-\frac{4U^2 \mu r^2}{R^4}                          (2)
Multiply through by r/k and integrate once:
r\frac{dT}{dr}=-\frac{\mu U^2 r^4}{kR^4}+C_1                                         (3)
Divide through by r and integrate once again:
T=-\frac{\mu U^2 r^4}{4kR^4}+C_1 \ln r+C_2                                            (4)
Now we are in position to apply our boundary conditions to evaluate C_1  and  C_2.
First, since the logarithm of zero is -\infty, the temperature at r = 0 will be infinite unless
C_1=0                                        (5)
Thus we eliminate the possibility of a logarithmic singularity. The same thing will happen if we apply the symmetry condition dT/dr = 0 at r = 0 to Eq. (3). The constant C_2 is then found by the wall-temperature condition at r = R:
T=T_w=-\frac{\mu U^2}{4k}+C_2
or                    C_2=T_w+\frac{\mu U^2}{4k}                                       (6)
The correct solution is thus
T(r)=T_w+\frac{\mu U^2}{4k}\left(1-\frac{r^4}{R^4}\right)                                                       Ans. (7)
which is a fourth-order parabolic distribution with a maximum value T_0=T_w+\mu U^2/(4k) at the centerline.