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answered Sep 12, 2014 by ChemKB (8,210 points)

Heat loss in the wellbore never reaches steady state due to the large thermal mass in the formation around the well. It attains a quasi-steady state in which the rate of heat loss is a monotonically decreasing function of time. Because of this, transient effects will always be present and they must be considered also for steady state calculations. Some problems with this are:

What is the formation conductivity? Data for thermal conductivity for the earth (rock or soil) around the well is normally not available and they can vary a lot depending on the porosity and moisture content. As an example, heat conductivity for soil can vary from 0.3 to 3 W/m/K depending on the moisture content. With very wet soil, a value close to that of water of 0.6 W/m/K would be expected, but a common value used for thermal calculations is 2 W/m/K for both soil and rock. 

If OLGA is to be used, how far into the formation is it required to calculate the conduction. If very thick layer is required, the calculation time will drop. 

What is the influence of wells near to each other (heat transfer between the wells). 

How to model repeated shutdowns and restarts (with many wells near each other).

The recommended equation for calculation of the convection in the annulus is given in the chapter “recommended model for steady state applications.

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