The equation the describes steady convection-diffusion of thermal energy with constant fluid properties is \rho cp(v.\nabla T)=k\nabla^2T, where v is the velocity vector, and is T temperature.  When this equation is applied to axisymmetric flow in pipe , the velocity vector in general is v=u_r(r,z)e_r+u_z(r,z)e_z, were e_r and e_z are the unit vectors in the r and z coordinate directions. For such a flow \nablaT=\frac{\partial T}{\partial z}e_z+\frac{\partial T}{\partial r}e_r

Thus  (v.\nabla T)=u_z\frac{\partial T}{\partial z}+u_r\frac{\partial T}{\partial r}

Now if the flow is fully developed, then v=u_z(r) so that

(v.\nabla T)=u_z\frac{\partial T}{\partial z}

Thus the only way the convection of thermal energy can be zero is if the temperature T is a function of r only. However the temperature field depends on the boundary conditions imposed on T ( inlet , exit and pipe wall). In most physical circumstances T=T(r,z) and thus the convection of thermal energy is non zero ( except far downstream), even though the flow is fully developed. If the physical properties k, \rho, and Cp are functions of temperature all bets are off.

Thanks for the answers. I understand that  u_z\frac{\partial T}{\partial z} is not zero so that the flow can be cooled or heated in pipe. In literature, the nondimensional equation given in my question by the first derivative to axial direction is given zero  and therefore convection coefficient "h" is found constant for constant surface temperature or heat flux conditions. I just wonder the reason behind it.