## 3 Test Result and Discussion  
### Variation Law of Freezing Temperature Field  
(1) Brine temperature: According to Figure 5, the change curve of brine temperature to the circuit can be roughly divided into three stages: 1 Rapid cooling section. After freezing for 9.6 h, the temperature of desalted brine decreased rapidly to \(-18\)\({}^{\circ}\)C, and the average cooling rate reaches \(1.88\)\({}^{\circ}\)C/h. At the initial stage of freezing, the temperature difference of de-loop brine is small, and it rapidly increases to \(7.5\)\({}^{\circ}\)C with the increase in freezing time. 2 Slow cooling section. After freezing for 18.2 h, the temperature of the desalted brine reaches the freezing design temperature of \(-20.2\)\({}^{\circ}\)C. This process lasts for about 8.6 h, and the average cooling rate is \(0.26\)\({}^{\circ}\)C/d. The temperature difference of de-loop brine decreases gradually. 3 Stable section. The temperature of the desalted brine continuously fluctuates at \(-20\)\({}^{\circ}\)C in this stage, with the fluctuation range of \(\pm 1.5\)\({}^{\circ}\)C. The temperature difference of de-loop brine is gradually stable, indicating that the frozen wall is in good condition.  
Figure 4: Vertical section drawing. 1—sand layer; 2—clay layer; 3—freezing pipe; 4—strain gauge; 5—T-type thermocouple; 6—soil layer interface.  
(2) Frozen temperature field: It can be seen from Figures 6-8 that the temperature field of the sand layer and clay layer changes in a similar way. The temperature rises from the freezing pipe to both sides; the curve shape of temperature field is 'W'; and the temperature at the interface of the outer ring pipe is slightly higher than that of the inner ring pipe.  
Figure 5: Temperature and temperature difference of brine.  
Figure 6: Temperature contour drawing of sand layer.  
Figure 7: Temperature contour drawing of clay layer.  
The change rate of the freezing temperature field can also be divided into three stages: fast decreasing stage, slow decreasing stage, and stable stage. At the initial stage of freezing, the temperature drop rate of the soil around the freezing pipe is large due to the large temperature difference between the freezing brine and the soil around the freezing pipe. The farther away from the freezing pipe, the less cold the soil absorbs from the freezing pipe. With the increase in freezing time, the temperature difference between the freezing brine and the frozen soil around the freezing pipe continues to decrease, and the temperature change rate also gradually decreases until it becomes stable. However, in the whole process, the temperature decreasing rate of sand decreases faster than that of clay.  
(3) It can be seen from Figures 6-8 that the closure time and average temperature of the frozen wall in the sand layer are \(14.8\) h and \(-10.3\)\({}^{\circ}\)C, while those in the clay layer are \(24.5\) h and \(-9.0\)\({}^{\circ}\)C. It is mainly because the clay contains more bound water, while the sand contains less. In addition, the freezing of bound water requires more cooling than free water, and the specific heat of clay particle skeleton is larger than that of sand. Therefore, the cold absorbed by clay freezing is more than that of sand, which makes the freezing efficiency of clay less than that of sandy soil, resulting in the closure time and average temperature of frozen wall in sand layer being lower than those in clay layer.