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=108W/m³, occurs when the element is inserted into the core while the surfaces experience convection (Too,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC: Fuel element x=5x10 m³/s k-30W/m-K Coolant Too 250°C h=1100W/m² K xحا T(x,0)=250°C 9-10°W/m³ at t>0 2.3.4.5.h k+ax=2mm Coolant Too.h L-10mm PROBLEM 5.102 (Cont.) To be well within the stability limit, select At = 0.3s, which corresponds to At 5×10 m²/sx0.3s Fo= Ax² t=pAt=0.3p(s). (0.002m) = 0.375 Substituting numerical values with q =108W/m³, the nodal equations become T=0.375[21" +10°W/m²³ (0.002m)²/30W/m-K]+(1-2×0.375) Tf TP+1=0.375[2TP =0.375 21+13.33]+0.25 T TP+1 = 0.375[T+TE+13.33] +0.25 TP тр = T-0.375 1+1+13.33]+0.25 T 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 ġ(Ax)² + m+1 2 +(1-2 Fo)T (1) T=0.375 1+1+13.33] +0.25 T T=0.375 1+1+1333] +0.25 T T+1=2×0.375 +0.0733×250+ +(1-2×0.375-2×0.0733×0.375)T 13.33 2 TP+1=0.750 [T+24.99] +0.195 Tg. Midplane node, m = 0 Same as Eq. (1), but with TP TP Surface node, m = 5 T}+1=2 Fo| T} +BiTot m-1 = m+1' 9(Ax)2 2k +(1-2F0-2Bi-Fo)T. The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: Bi Fo≤ hAx 1100W/m² Kx(0.002m). k 1/2 (1+Bi) 30W/m-K = 0.466 Fo(Ax)² (0.002m)² ΔΙΣ α =0.466- = 0.0733 = 0.373s. 5x10 6m²/s 2 Continued..... 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) P t(s) Το Τι T₂ T3 T4 T5(°C) 0 0 250 250 250 250 250 1 0.3 255.00 255.00 2 0.6 3 0.9 260.00 260.00 265.00 265.00 255.00 255.00 255.00 260.00 260.00 260.00 259.72 250 254.99 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 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. (3) (4) (5) (6) (7) (8)

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=108W/m³, occurs when the element is inserted into the core while the surfaces experience convection (Too,h). FIND: Temperature distribution 1.5s after element is inserted into the core. SCHEMATIC: Fuel element x=5x10 m³/s k-30W/m-K Coolant Too 250°C h=1100W/m² K xحا T(x,0)=250°C 9-10°W/m³ at t>0 2.3.4.5.h k+ax=2mm Coolant Too.h L-10mm PROBLEM 5.102 (Cont.) To be well within the stability limit, select At = 0.3s, which corresponds to At 5×10 m²/sx0.3s Fo= Ax² t=pAt=0.3p(s). (0.002m) = 0.375 Substituting numerical values with q =108W/m³, the nodal equations become T=0.375[21" +10°W/m²³ (0.002m)²/30W/m-K]+(1-2×0.375) Tf TP+1=0.375[2TP =0.375 21+13.33]+0.25 T TP+1 = 0.375[T+TE+13.33] +0.25 TP тр = T-0.375 1+1+13.33]+0.25 T 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 ġ(Ax)² + m+1 2 +(1-2 Fo)T (1) T=0.375 1+1+13.33] +0.25 T T=0.375 1+1+1333] +0.25 T T+1=2×0.375 +0.0733×250+ +(1-2×0.375-2×0.0733×0.375)T 13.33 2 TP+1=0.750 [T+24.99] +0.195 Tg. Midplane node, m = 0 Same as Eq. (1), but with TP TP Surface node, m = 5 T}+1=2 Fo| T} +BiTot m-1 = m+1' 9(Ax)2 2k +(1-2F0-2Bi-Fo)T. The most restrictive stability criterion is associated with Eq. (2), Fo(1+Bi) ≤ 1/2. Consider the following parameters: Bi Fo≤ hAx 1100W/m² Kx(0.002m). k 1/2 (1+Bi) 30W/m-K = 0.466 Fo(Ax)² (0.002m)² ΔΙΣ α =0.466- = 0.0733 = 0.373s. 5x10 6m²/s 2 Continued..... 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) P t(s) Το Τι T₂ T3 T4 T5(°C) 0 0 250 250 250 250 250 1 0.3 255.00 255.00 2 0.6 3 0.9 260.00 260.00 265.00 265.00 255.00 255.00 255.00 260.00 260.00 260.00 259.72 250 254.99 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 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. (3) (4) (5) (6) (7) (8)

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter7: Arrays

Section7.5: Case Studies

Problem 15E

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