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1. Steady State Analysis The fastest steady state simulation was given by successive over relaxation (SOR) method. It took 145 iterations and analysis time was only 0.8899 seconds. In SOR method we use relaxation parameter which helps in achieving the convergence faster. The value of T is increased more in each iteration…
Mohammad Saifuddin
updated on 08 Aug 2019
1. Steady State Analysis
The fastest steady state simulation was given by successive over relaxation (SOR) method. It took 145 iterations and analysis time was only 0.8899 seconds. In SOR method we use relaxation parameter which helps in achieving the convergence faster. The value of T is increased more in each iteration because of this relaxation parameter. The optimal value of relaxation parameter is very important in getting the results faster and correct solution. In 2D Steady state analysis in SOR method, Told values are updated inside the convergence loop and there is no time loop.
Another reason for getting faster results is because in steady state analysis there is no time derivative and thermal diffusivity. The solution do not depends on these two things. That is why we get such a fast convergence for our solution.
Formula for optimal value of relaxation parameter is given by: ω=21+sin(π⋅dx)ω=21+sin(π⋅dx)
Formula for SOR method:
Tn+1(i,j)=(1-ω)Tn+1(i,j)+ω(Tn+1(i-1,j)+Tn(i+1,j)+Tn+1(i,j-1)+Tn(i,j+1))4Tn+1(i,j)=(1−ω)Tn+1(i,j)+ω(Tn+1(i−1,j)+Tn(i+1,j)+Tn+1(i,j−1)+Tn(i,j+1))4
2. Transient State Analysis
The fastest transient state analysis was given by explicit method. Explicit method took 8.614 seconds to give the solution, in 1000 time steps which is same for all schemes in transient analysis. Transient state analysis by explicit method is slower than steady state SOR method, because here we have to take into consideration the time derivative and thermal diffusivity. Together, they both increases the simulation time.
k=αdtdx2k=αdtdx2
Formula for explicit method:
Tn+1(i,j)=Tn(i,j)+k(Tn(i-1,j)+Tn(i+1,j)+Tn(i,j-1)+Tn(i,j+1)-4⋅Tn(i,j))Tn+1(i,j)=Tn(i,j)+k(Tn(i−1,j)+Tn(i+1,j)+Tn(i,j−1)+Tn(i,j+1)−4⋅Tn(i,j))
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