## 2 Coupling Calculation of Temperature and Seepage Field
### Influence of seepage on Temperature Field  
Due to the water level difference between the internal and external rivers in embankment engineering, the seepage flows from the inside of the river to the outside. When the water passes through the dam, if there is a temperature difference between the two media, heat exchange will inevitably occur.  
In the case of one-dimensional heat conduction, when seepage exists, the heat flow includes two parts: one is caused by the conduction of the porous medium which is \(-\lambda(\partial T/\partial x)\); another is the heat reduction taken by the seepage water and is \(c_{w}\rho_{w}vT\). In conclusion, the heat flux inside the embankment is:  
\[q_{x}=c_{w}\rho_{w}vT-\lambda\frac{\partial T}{\partial x} \tag{7}\]  
where \(c_{w}\) is the specific heat capacity of water (J\(\cdot\)K\({}^{-1}\)); \(\rho_{w}\) is water density (kg\(\cdot\)m\({}^{-3}\)); \(v\) is seepage water velocity (m\(\cdot\)s\({}^{-1}\)); and \(\lambda\) is the heat conductivity of the soil (W\(\cdot\)m\({}^{-1}\cdot\)K\({}^{-1}\)).  
In the x-direction of unit time, the net heat flowing into the unit volume is:  
\[-\frac{\partial q_{x}}{\partial x}=-c_{w}\rho_{w}\frac{\partial(vT)}{\partial x }+\frac{\partial}{\partial x}\bigg{(}\lambda\frac{\partial T}{\partial x} \bigg{)} \tag{8}\]  
The net heat should be equal to the heat absorbed by the temperature rise of the unit volume medium in unit time, that is:  
\[c\rho\frac{\partial T}{\partial\tau}=-c_{w}\rho_{w}\frac{\partial(vT)}{ \partial x}+\frac{\partial}{\partial x}\bigg{(}\lambda\frac{\partial T}{ \partial x}\bigg{)} \tag{9}\]  
Expanding to a 3D condition and taking into account heat source \(Q_{T}\):  
\[\frac{\partial}{\partial x}\bigg{(}\lambda_{x}\frac{\partial T}{\partial x} \bigg{)}+\frac{\partial}{\partial y}\bigg{(}\lambda_{y}\frac{\partial T}{ \partial y}\bigg{)}+\frac{\partial}{\partial z}\bigg{(}\lambda_{z}\frac{ \partial T}{\partial z}\bigg{)}-c_{w}\rho_{w}\bigg{[}\frac{\partial(v_{x}T)}{ \partial x}+\frac{\partial(v_{y}T)}{\partial y}+\frac{\partial(v_{z}T)}{ \partial z}\bigg{]}+Q_{T}=c\rho\frac{\partial T}{\partial\tau} \tag{10}\]  
With [PERSON]'s law:  
\[v_{x}=-k_{x}\frac{\partial H}{\partial x};v_{y}=-k_{y}\frac{\partial H}{ \partial y};\,v_{z}=-k_{z}\frac{\partial H}{\partial z}\]  
The soil is considered to be isotropic and that is: \(k_{x}=k_{y}=k_{z}=k\). Substituting it into Equation (10):  
\[\lambda\
abla^{2}T+kc_{w}\rho_{w}\bigg{\{}\bigg{[}\frac{\partial H}{\partial x }\frac{\partial T}{\partial x}+\frac{\partial H}{\partial y}\frac{\partial T }{\partial y}+\frac{\partial H}{\partial z}\frac{\partial T}{\partial z} \bigg{]}+T\
abla^{2}H\bigg{\}}+Q_{T}=c\rho\frac{\partial T}{\partial\tau} \tag{11}\]  
which is simplified as:  
\[\lambda\
abla^{2}T+kc_{w}\rho_{w}\Big{\{}\
abla H\
abla T+T\
abla^{2}H\Big{\}} +Q_{T}=c\rho\frac{\partial T}{\partial\tau} \tag{12}\]The temperature field is steady during the long service process and no source term is included, which means \(\partial T/\partial\tau=0\) and \(Q_{T}=0\); substituting this into Equation (12):  
\[\lambda\
abla^{2}T+kc_{w}\rho_{w}\Big{\{}\
abla H\
abla T+T\
abla^{2}H\Big{\}}=0 \tag{13}\]  
Setting \(a=\lambda\) and \(b=c_{w}\rho_{w}\), Equation (12) can be transformed into:  
\[a\
abla^{2}T+kb\Big{\{}\
abla H\
abla T+T\
abla^{2}H\Big{\}}=0 \tag{14}\]