Solving the Convection-Diffusion Equation for this Pipe with a Heat Sink

[HumanistEngineer]

TL;DR Summary

Temperature propagation through an insulated pipe in time
The solution of convection-diffusion equation with a heat sink (heat loss from pipe to the ground)

Hi Again,

I try to solve the transient temperature propagation through a buried insulated pipe by means of solving the convection-diffusion equation with a heat sink that is the heat loss from the water mass to the ground. Below you can see the details of my calculation steps in my numerical analysis.
The problem is that my results are oscillating and/or abnormal. Would you please check my finite difference approximations, if they are correct or not?

Details:

 

[Chestermiller]

It looks OK. So??

 

[HumanistEngineer]

Here is the result (for different mesh numbers). The maximum (steady-state) temperature should be at around 70 °C.

 

[Chestermiller]

That last term in the differential equation should involve a heat transfer coefficient. Please provide the derivation of that last term in the equation. What is the inlet condition and the initial condition? What is the temperature of the ground?

I get $$c=\frac{h}{\rho C}\frac{4}{D}$$where h is the heat transfer coefficient

 

[HumanistEngineer]

Thank you Chestermiller for your time. You indicated c as c = h/(rho cp) (4/D) . Why is there this 4/D?

I use the thermal resistance, R [K/W], in the last term c, which means that h=1/R. Here how R was derived:

The initial condition for the pipe is that the water temperature is initially at 50 °C and in my numerical example the inlet water temperature is increased to 70 °C. Then the aim is to see how the temperature propagates through the pipe length in time. The ground temperature is constant at 10 °C.

 

[Chestermiller]

If U is the overall heat transfer coefficient based on the inside diameter of the pipe, then the balance equation over a length $\Delta x$ of pipe, in finite difference form, goes like:
$$\rho C_p\left(\frac{\pi d_i^2}{4}\Delta x\right)\frac{\partial T}{\partial t}+\ ... \ =\ ...\ +\pi d_i\Delta xU(T-T_{ground})$$
If we divide this by $\rho C_p\left(\frac{\pi d_i^2}{4}\Delta x\right)$ and take the limit as $\Delta x$ approaches zero, we obtain: $$\frac{\partial T}{\partial t}+\ ...\ =\ ...\ +\frac{U}{\rho C_p}\frac{4}{d_i}(T-T_{ground})$$
Here, $$\frac{1}{Ud_i}=\frac{1}{h_{flow}d_i}+\frac{1}{2\lambda}\ln{(d_{o}/d_{i})}+...$$

Note that $\Delta x$ is not present here.