**2. Mathematical Model**

The heat transfer process for a granular bed with buried tubes is complex [14]. This process includes the convection heat transfer between high-temperature flue gas and filled particles, heat transfer within filled particles, heat transfer between gas films on a particle surface, heat transfer between contact particles and air film, heat transfer between particles and tube walls, heat transfer within the tube wall, and the convection heat transfer of the cooling water inside a tube wall. To simplify the calculation process, an equivalent heat transfer coe fficient method of a particle bed is used on the basis of logarithmic mean temperature di fference formula. Macroscopically, the heat in high-temperature gas is exchanged with cooling water through the filling particles and heat exchange tubes. This formula is expressed as follows [15]:

$$Q = h\_{ht} A\_{ht} \Delta T = \text{cinc}(t\_{w,out} - t\_{w,in}).\tag{1}$$

The total heat transfer coe fficient of the granular bed refers to the comprehensive heat transfer coe fficient between high-temperature flue gas and cooling water. The influencing factors include the convective heat transfer coe fficient of the inner and outer surfaces of the buried tube, the thermal conductivity of the buried tube, and the heat transfer characteristics.

The logarithmic mean temperature di fference [16] is defined as follows:

$$t\_{\text{max}} = t\_{\text{g,in}} - t\_{w,in} \tag{2}$$

$$t\_{\rm min} = t\_{\rm \%out} - t\_{\rm uv\beta wt} \tag{3}$$

$$
\Delta T = \frac{(t\_{\text{max}} - t\_{\text{min}})}{\ln \left(\frac{t\_{\text{max}}}{t\_{\text{min}}}\right)} \tag{4}
$$

The total heat exchange area is the sum of the total surface area of the heat exchange tubes and the surface area of the filled particles; this area can be expressed as follows:

$$A\_{ht} = n\pi dl + \frac{m}{m\_1} \times 4\pi r^2. \tag{5}$$

The formula for the total heat transfer coe fficient of the granular bed is presented as follows:

$$d\eta\_{lt} = \frac{\dot{\text{cm}} (t\_{w,out} - t\_{w,in})}{\left[\pi n dl + \frac{m}{m\_1} \times 4\pi r^2\right] \Delta T} \tag{6}$$
