### Given the resolution of the problem 5.102 below, develop a python program that shows the found temperature distribution PROBLEM 5.102 KNOWN: Fuel element of Example 5.8 is initially at a uniform temperature of 250°C with no internal generation; suddenly a uniform generation, q = 10³ W/m³, occurs when the element is inserted into the core while the surfaces experience convection (To,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC: PROBLEM 5.102 (Cont.) To be well within the stability limit, select At = 0.3s, which corresponds to Fo aAt 5x10 m²/sx0.3s 4x² t=pAt=0.3p(s). =0.375 (0.002m)² Fuel element- x=5x10 m²/s k-30W/m-K Coolant Too-250°C h-1100 W/m²-K رجا T(x,0)=250°C 9-10°W/m³ at +>0 Coolant Too,h L-10mm k+ax=2mm ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3) q=0, initially; at t> 0, q is uniform. ANALYSIS: As suggested, the explicit method with a space increment of 2mm will be used. Using the nodal network of Example 5.8, the same finite-difference equations may be used. Interior nodes, m=1, 2, 3, 4 TP+1 = Fo TP m-1 +TP 4(x) + m+1 2 -2 Fo)T (1) Midplane node, m = 0 Same as Eq. (1), but with TP Surface node, m = 5 T}+1=2 Fo|T{ +BiTot m-1 =TP 2k m+1' +(1-2F0-2Bi-Fo)T. The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: hAx 1100W/m² K×(0.002m) Bi= k 30W/m-K 1/2 Fo =0.466 (1+Bi) = 0.0733 Substituting numerical values with q = 108W/m³, the nodal equations become 18-0.375 2+10 W/m³ (0.002m)²/30W/mK]+(1-2x0.375)T TP+1=0.375 2TP +13.33 +0.25 T 3.33] -0.25 TP TP+1 = 0.375 [T+T+13.33] .33] +0.25 T=0.375[T+T+13.33] +0.25 T T¹-0.375 T+T+13.33+0.25 T = 13.33 2 TP+1=0.375[T+T+13.33] +0.25 T TP+1=2×0.375 T +0.0733×250+- +(1-2x0.375-2×0.0733×0.375) TP+1 =0.750 [T+24.99] +0.195 T The initial temperature distribution is T₁ = 250°C at all nodes. The marching solution, following the procedure of Example 5.8, is represented in the table below. (2) р t(s) To T₁ T2 T3 ΤΑ T5(°C) 0 0 250 250 250 250 250 1 0.3 255.00 255.00 255.00 255.00 255.00 250 254.99 2 0.6 260.00 260.00 260.00 260.00 260.00 259.72 3 0.9 265.00 265.00 4 1.2 270.00 270.00 265.00 265.00 264.89 264.39 270.00 269.96 269.74 268.97 5 1.5 275.00 275.00 274.98 274.89 274.53 273.50 ΔΙΣ =0.466 Fo(Ax)² (0.002m)² α = 0.373s. -6. 2 5x10 m/s Continued..... The desired temperature distribution T(x, 1.5s), corresponds to p = 5. COMMENTS: Note that the nodes near the midplane (0,1) do not feel any effect of the coolant during the first 1.5s time period. (4) (5) (7)