Problems with heat conduction in roadbed on permafrost layer
Problem Description
Laying the foundations of roads, airstrips and certain structures on permafrost soil. Analyzing the structure of this type of foundation, there are asphalt layers, concrete layers, dry sand layers, gravel layers, insulation material layers, and then there are wet sand layers and frozen soil layers (see Figure 1). It is known that the temperature of the outside world is a function of time, and the temperature of the permafrost layer is a constant subzero temperature. Please answer the following questions:
- Question 1: Analyze the rules of air temperature transmission into the roadbed and the temperature distribution of each material layer;
- Question 2: Since the brackets of some equipment cannot be fixed on the thawing soil layer and must be fixed in the permafrost layer, it is very important to determine the thawing position of the underground soil layer. Please give the dividing line between thawing sand soil and frozen sand soil;
- Question 3: Please give the optimal thickness of each layer of material (combining temperature distribution, cost and durability);
- Question 4: Please conduct simulation based on the permafrost foundation of my country's Qinghai-Tibet Railway and give reasonable suggestions to the construction unit.
Model solving
Analysis of question one
For question one, considering that the curvature at each point on the frozen soil surface is large enough, the foundation layer is regarded as a plane layer parallel to each other, and a one-dimensional parabolic partial differential equation model based on Fourier's law is established. Since the external temperature is about function of time, so we can collect the daily temperature data of Lhasa in four seasons and perform sinusoidal function fitting to obtain the fitting function of the change of external temperature with time. Taking this as the boundary condition of the heat conduction equation, according to the permafrost soil temperature is 0℃ and below, this group selects T 0 = 0 T_0=0T0=0 ℃ is used as the right boundary of the heat conduction equation. Under a given initial value condition, a higher-precision Crank-Nicolson difference format is given through the finite difference method. The differential quotient is used to replace the differential to obtain a discrete linear equation system, which is solved by the chasing method. Use a system of linear equations to obtain the temperature distribution in time and one-dimensional space, and perform stability analysis and error testing on the results.
One-dimensional coupled parabolic partial differential equation model
Heat conduction equation
According to Fourier's law of heat conduction, the microelement equation of heat obtained by a conductor is:
d Q = − λ ∇ u ⋅ d S → dt dQ=-\lambda \nabla u\cdot d\overrightarrow{S}dtdQ=−λ∇u⋅dSd t
considers the problem to be studied in one dimension, and after simplification through integration and Gaussian formula,t 1 t_1t1~ t 2 t_2 t2The heat flowing into the conductor during the time period is:
Q 1 = ∫ t 1 t 2 [ ∫ x 1 x 2 k ∂ 2 u ∂ x 2 dx ] dt Q_1=\int_{t_1}^{t_2}{\left[ \int_{ x_1}^{x_2}{k \frac{\partial ^2u}{\partial x^2}dx} \right] dt}Q1=∫t1t2[∫x1x2k∂x2∂2 anddx]d t
absorbs heat due to the increase in temperature in the conductor:
Q 2 = ∫ x 1 x 2 c ρ [ ∫ t 1 t 2 ∂ u ∂ tdt ] dx = ∫ t 1 t 2 [ ∫ t 1 t 2 c ρ ∂ u ∂ xdx ] dt Q_2=\int_{x_1}^{x_2}{c\rho \left[ \int_{t_1}^{t_2}{\frac{\partial u}{\partial t}dt} \right] dx}=\int_{t_1}^{t_2}{\left[ \int_{t_1}^{t_2}{c\rho \frac{\partial u}{\partial x}dx} \right] dt}Q2=∫x1x2cρ[∫t1t2∂t∂udt]dx=∫t1t2[∫t1t2cρ∂x∂udx]d tAccording
to the conservation of energy, the energy relationship equation is obtained:
Q 1 = Q 2 Q_1 =Q_2Q1=Q2
从而得到：
∂ u ( x , t ) ∂ t = k c ρ ∂ 2 u i ( x , t ) ∂ x 2 \displaystyle \frac{\partial u\left( x,t \right)}{\partial t} = \displaystyle \frac{k}{c\rho } \displaystyle \frac{\partial ^2u_i\left( x,t \right)}{\partial x^2} ∂t∂u(x,t)=cρk∂x2∂2 andi(x,t)
Among them: kkk is thermal conductivity,ccc is the specific heat of the conductor,ρ \rhoρ is the dry density of the conductor.
The schematic diagram of the micro-element method is shown in Figure 2.
Coupled parabolic partial differential equations
Considering the differences in thermal conductivity, specific heat and dry density of the five different foundation structures, five one-dimensional heat conduction equations are established: ∂ ui ( x , t ) ∂ t
= kici ρ i ∂ 2 ui ( x , t ) ∂ x 2 , i = 1 ⋯ 5 , \begin{align} \displaystyle \frac{\partial u_i\left( x,t \right)}{\partial t} = \displaystyle \frac{k_i}{c_i\rho _i} \displaystyle \frac{\partial ^2u_i\left( x,t \right)}{\partial x^2},i=1 \cdots 5, \end{align}∂t∂ui(x,t)=ciriki∂x2∂2 andi(x,t),i=1⋯5,
According to Fourier's law of heat conduction, during the heat conduction process, the heat flux density q ⃗ \vec{q} at the critical surface of two adjacent foundation materialsq相同，得到热流量密度耦合条件:
k i ∂ u i ( x , t ) ∂ n i ∣ Γ i = k i + 1 ∂ u i + 1 ( x , t ) ∂ n i + 1 ∣ Γ i , i = 1 ⋯ 4 , \begin{align} k_i \displaystyle \frac{\partial u_i\left( x,t \right)}{\partial n_i}|_{\varGamma _i} = k_{i+1} \displaystyle \frac{\partial u_{i+1}\left( x,t \right)}{\partial n_{i+1}}|_{\varGamma _i},i=1 \cdots 4, \end{align} ki∂ni∂ui(x,t)∣Ci=ki+1∂ni+1∂ui+1(x,t)∣Ci,i=1⋯4,
Since the critical surface temperatures of adjacent conductors are equal at any time, the critical surface temperature coupling condition for conductor heat transfer is obtained:
ui ( x , t ) ∣ Γ i = ui + 1 ( x , t ) ∣ Γ i , \begin{align } u_i\left( x,t \right) |_{\varGamma _i}=u_{i+1}\left( x,t \right) |_{\varGamma _i}, \end{align}ui(x,t)∣Ci=ui+1(x,t)∣Ci,
By performing second-order fitting of the sine function on the searched daily temperature data of Lhasa in four seasons, we get:
a ( t ) = α i sin ( w 1 t + φ i ) + β i sin ( w 2 t + θ i ) , i = 1 ⋯ 4 , \begin{align} a(t)=\alpha _i\sin \left( w_1 t+\varphi _i \right) +\beta _i\sin \left( w_2 t+\theta _i \right) , i=1 \cdots 4, \end{align}a(t)=aisin(w1t+Phii)+bisin(w2t+ii),i=1⋯4,
Therefore, the fitting function is used as the change function of the external temperature as the boundary condition of the heat conduction equation:
u 0 ( 0 , t ) = a ( t ) , \begin{align} u_0\left( 0,t \right) =a\ left( t \right), \end{align}u0(0,t)=a(t),
According to the research, it is known that the temperature of the frozen soil layer is 0℃ and below. This group selects T 0 = 0 T_0 =0T0=0 , as the right boundary condition:
u 5 ( L 5 , t ) = T 0 , \begin{align} u_5\left( L_5,t \right) =T_0, \end{align}u5(L5,t)=T0,
Finally, the above formulas are sorted to obtain the heat conduction equation of the initial boundary value problem:
{ ∂ u i ( x , t ) ∂ t = k i c i ρ i ∂ 2 u i ( x , t ) ∂ x 2 , x ∈ ( L i − 1 , L i ) , k i ∂ u i ( x , t ) ∂ n i ∣ Γ i = k i + 1 ∂ u i + 1 ( x , t ) ∂ n i + 1 ∣ Γ i , u i ( x , t ) ∣ Γ i = u i + 1 ( x , t ) ∣ Γ i , u 0 ( 0 , t ) = a ( t ) , u 5 ( L 5 , t ) = T 0 , \begin{align} \begin{cases} \displaystyle \frac{\partial u_i\left( x,t \right)}{\partial t} = \displaystyle \frac{k_i}{c_i\rho _i} \displaystyle \frac{\partial ^2u_i\left( x,t \right)}{\partial x^2},x\in \left( L_{i-1},L_i \right),\\ k_i \displaystyle \frac{\partial u_i\left( x,t \right)}{\partial n_i}|_{\varGamma _i} = k_{i+1} \displaystyle \frac{\partial u_{i+1}\left( x,t \right)}{\partial n_{i+1}}|_{\varGamma _i},\\ u_i\left( x,t \right) |_{\varGamma _i}=u_{i+1}\left( x,t \right) |_{\varGamma _i},\\ u_0\left( 0,t \right) =a\left( t \right),\\ u_5\left( L_5,t \right) =T_0,\\ \end{cases} \end{align} ⎩
⎨
⎧∂t∂ui(x,t)=ciriki∂x2∂2 andi(x,t),x∈(Li−1,Li),ki∂ni∂ui(x,t)∣Ci=ki+1∂ni+1∂ui+1(x,t)∣Ci,ui(x,t)∣Ci=ui+1(x,t)∣Ci,u0(0,t)=a(t),u5(L5,t)=T0,
Processed in CN format
{ u j n + 1 − u j n τ = k i c i ρ i ⋅ [ ( u j + 1 n + 1 − 2 u j n + 1 + u j − 1 n + 1 ) + ( u j + 1 n − 2 u j n + u j − 1 n ) ] 2 h 2 k i u J n − u J − 1 n h = k i + 1 u J + 1 n − u J n h u i ( J , t ) = u i + 1 ( J , t ) u 1 ( 0 , t ) = a ( t ) u 5 ( N , t ) = T 0 J = L i L N \begin{align} \begin{cases} \displaystyle \frac{u_{j}^{n+1}-u_{j}^{n}}{\tau}=\displaystyle \frac{k_i}{c_i\rho _i}\cdot \displaystyle \frac{\left[ \left( u_{j+1}^{n+1}-2u_{j}^{n+1}+u_{j-1}^{n+1} \right) +\left( u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n} \right) \right]}{2h^2}\\ k_i\frac{u_{J}^{n}-u_{J-1}^{n}}{h}=k_{i+1}\frac{u_{J+1}^{n}-u_{J}^{n}}{h}\\ u_i\left( J,t \right) =u_{i+1}\left( J,t \right)\\ u_1\left( 0,t \right) =a\left( t \right)\\ u_5\left( N,t \right) =T_0\\ J=\frac{L_i}{L}N\\ \end{cases} \end{align} ⎩ ⎨ ⎧tujn+1−ujn=ciriki⋅2h _2[(uj+1n+1−2 andjn+1+uj−1n+1)+(uj+1n−2 andjn+uj−1n)]kihuJn−uJ−1n=ki+1huJ+1n−uJnui(J,t)=ui+1(J,t)u1(0,t)=a(t)u5(N,t)=T0J=LLiN
Question 1 result display
Follow-up questions
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