PSI - Issue 18

Kostina A. et al. / Procedia Structural Integrity 18 (2019) 293–300 Author name / Structural Integrity Procedia 00 (2019) 000–000

296 4

,

(10)

p

p



0

0 f t 

p T T   ,

(11)

1

2 3 4 T         n , 0

(12)

,

(13)

v n

0

f   



1 2 3 4    

0   u ,

(14)

1

0  ,

(15)

u

u



r

z



 

3

2 4

where i  , ( 1, ..., 4) i  – boundaries of the computational domain, n – the outer normal to a boundary, p T – temperature of liquid circulating in the freezing well. The problem (1)–(8) with initial and boundary conditions (9)–(15) is solved numerically by the finite element method. The computational domain is divided on mapped elements of the first order for variables T , f p and the second order for variable u . The solution of the problem is used for an estimation of the external pressure acting on the freezing pipe. According to a technical standard adopted by Russia GOST 32388-2013 (2014) pipes should be used within the elastic range. For strength calculation the following criterion is proposed:

1 3          , 2 min , 3 UTS y

(16)



0,max

where y  – yield stress. In this work a mechanical behavior of the freezing pipe under pressure is determined by Hooke’s law. Stress and strain in the pipe is estimated by a numerical modeling. To simulate a position of initiation of a crack in the pipe and direction of its propagation the extended finite element method is applied. 3. Results of the numerical simulation The pressure acting on the freezing pipe during the artificial freezing and subsequent thawing is estimated within stratums of sandstone, sand and clay. Depth of each stratum is 3 m. Hydrodynamic and thermophysical properties of the soils are listed in Table 1, mechanical properties are listed in Table 2. The plot of the temperature p T of a liquid refrigerant is shown in Fig. 2 (a). 0,max  – maximal nominal stress, UTS  – ultimate tensile strength,

Table 1. Hydrodynamic properties and thermophysical properties of frozen and unfrozen soils at temperatures T= 263.15 K (index 1) and T =283.15 K (index 2). Soil n k, m 2 ,1 p c , J/(kgꞏK) ,2 p c , J/(kgꞏK) 1  , kg/m 3 2  , kg/m 3 1  , W/(m∙K) 2  , W/(m∙K) Sandstone 0.34 4.51ꞏ10 -12 1170 1910 2029 2058 6.15 5.30 Sand 0.35 5.07ꞏ10 -12 1180 1930 2037 2066 3.85 2.37 Clay 0.40 5.72ꞏ10 -12 1280 2140 1920 1954 1.5 1.29