### 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)

### 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)

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter2: Steady Heat Conduction

Section: Chapter Questions

Problem 2.11P

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