Heat Transfer Coefficient
The Laplace equation for radial-symmetric heat transfer through
a cylindrical tube wall reads as follows, with $\,r$ = radius in polar coordinates:
$$
\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)=0
$$
So it has become an ordinary instead of a partial differential equation.
Solving step by step:
$$
r\frac{dT}{dr} = C \quad \Longrightarrow \quad T(r) = C\ln(r)+D
$$
Boundary conditions are $\,T(R_I)=T_I$ (inside tube) and $\,T(R_O)=T_O$ (outside tube) :
$$
T_I = C\ln(R_I)+D \quad ; \quad T_O = C\ln(R_O)+D \quad \Longrightarrow \\
C\ln(R_O/R_I) = T_O-T_I \quad ; \quad D = T_I-(T_O-T_I)\ln(R_I)/\ln(R_O/R_I) \quad \Longrightarrow \\
T(r) = \frac{(T_O-T_I)\ln(r)+T_I\ln(R_O/R_I)-(T_O-T_I)\ln(R_I)}{\ln(R_O/R_I)} \quad \Longrightarrow \\
T(r) = T_I + \frac{\ln(r/R_I)}{\ln(R_O/R_I)}(T_O-T_I)
$$
The heat flux $Q$ through all of the $N$ tube walls per unit length is described by:
$$
Q = N.2\pi r.\lambda\frac{d T}{d r} = N.2\pi r.\lambda\frac{T_O-T_I}{r.\ln(R_O/R_I)} =
\frac{N.2\pi\lambda}{\ln(D_O/D_I)}(T_O-T_I) = a_W\,(T_O-T_I)
$$
Where $D_I,D_O$ = inner/outer diameter of the tubes and $\lambda$ is the conductivity of the tube walls material.
We conclude that the heat transfer coefficient of the tube bundle walls is given by: $$a_W = \frac{N.2\pi\lambda}{\ln(D_O/D_I)}$$
The total heat transfer coefficient $\,a\,$ from primary medium to secondary medium actually consists of three contributions:
$$
\frac{1}{a} = \frac{1}{a_P} + \frac{1}{a_W} + \frac{1}{a_S} \approx \frac{1}{a_W}
$$
Where where $S,W,P$ = Secondary medium, Wall, Primary medium. But the tube wall contribution is by far the smallest of the three,
which means that $\,a\,$ is approximately equal to $\,a_W$ .