10 - Conduction - general superposition

10 - Conduction - general superposition - Superposition...

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Unformatted text preview: Superposition IIT Kanpur • Nonhomogenities may be split into several simpler subproblems problems which can be added together (superimposed) to give solution to the original problem, • For superposition, the problem must be linear (both PDE and BCs should be linear) • Generally, for homogeneous PDE: the number of subproblems (to be solved by the method of separation of variables) is equal to the number of nonhomogeneous BCs plus one for the IC • Generally, for nonhomogeneous steady state PDE: the number of subproblems (to be solved by the method of separation of variables) is equal to number of BCs minus two • Always check that the superimposed solutions do indeed add up to satisfy the original problem (both PDE and BCs) 1 General Problem IIT Kanpur k, and are constants PDE BCs IC • • • • • , at x = 0 , at x = L , at y = 0 , at y = W , at z = 0 , at z = H , , , A1, A2, A3, A4, A5, A6, B1, B2, B3, B4, B5, B6 are constants [ref] N. Ozisik, 2012 , , at t = 0 Note: f1, f2, f3, f4, f5, f6 are not functions of time The PDE and the BCs are linear, the PDE and the BCs are nonhomogeneous Nonhomogeneous BCs = 6, IC = 1 Nonhomogensous PDE = 1 [6 – 2 = 4 subproblems] We can split this into (6 + 1 + 4 = 11 subproblems) 2 [ref] N. Ozisik, 2012 T(x,y,z,t) IIT Kanpur Superposition – step 1 • Split the original problem (T) into one steady state (TSS) and one transient problem (TH) : TH(x,y,z,t) 0 , , 0 TSS(x,y,z) T(x,y,z,t) = TH(x,y,z,t) + TSS(x,y,z) transient transient steady state PDE: NH PDE: NH PDE: H BCs: NH BCs: NH BCs: H IC: TH(t=0) = F – TSS : , , , , 0 • The transient problem (TH) will have homogeneous PDE (no generation) and all BCs in its homogeneous form, which can be obtained by equating the corresponding fi to zero (see general BCs) • Transient problem (TH) can be solved by the method of separation of variables with only one nonhomogenity (IC), note the initial condition depends on the solution of the steady state problem (TSS) • The steady state problem (TSS) will have nonhomogeneous PDE (with generation) and all 3 nonhomogeneous BCs, hence, it will require further splitting Superposition – step 2 [ref] N. Ozisik, 2012 IIT Kanpur TSS(x,y,z) • Split the steady state problem (TSS) into two sub problems (TA and TB) TSS(x,y,z) = TA(x,y,z) + TB(x,y,z) PDE: NH PDE: H PDE: NH BCs: NH BCs: NH BCs: H TA(x,y,z) TB(x,y,z) 0 • The first problem (TA) will have homogeneous PDE (no generation) and all nonhomogeneous BCs, this will require further splitting to solve • The second problem (TB) will have nonhomogeneous PDE (with generation) and all BCs in its homogeneous form, which can be obtained by equating the corresponding fi to zero (see general BCs), this will also require further splitting to solve 4 Superposition – step 3 [ref] N. Ozisik, 2012 TA(x,y,z) IIT Kanpur 0 (1) (2) T2(x,y,z) T1(x,y,z) 0 0 (3) (4) T3(x,y,z) T4(x,y,z) 0 0 • Split the problem (TA) into six sub problems (T1, T2, T3, T4, T5, and T6) TA(x,y,z) = T1(x,y,z) + T2(x,y,z) + T3(x,y,z) + T4(x,y,z) + T5(x,y,z) + T6(x,y,z) PDE: H PDE: H BCs: NH BCs: only one NH, other H for each subproblem • Each of (T1, T2, T3, T4, T5, and T6) will have homogeneous PDE (no generation) and only a single nonhomogeneous BC, all other BCs take the homogeneous form, which can be obtained by equating the corresponding fi to zero (see general BCs) Plus two problems for the third dimension (5) T5(x,y,z) (6) T6(x,y,z) 0 0 • Each nonhomogeneous BC is included once and only once such that all the BCs for TA are same as that of addition of T1, T2, T3, T4, T5, and T6 • The six subproblem (T1, T2, T3, T4, T5, and T6) can be solved by the method of separation of variables each with only 5 one nonhomogeneous BC Superposition – step 4 IIT Kanpur [ref] N. Ozisik, 2012 • Split the problem (TB) into two sub problems (TPDE and TODE) TB(x,y,z) = TPDE(x,y,z) + TODE(x) ODE: NH PDE: NH PDE: H BCs: H (2 along x) BCs: H BCs: H BCs: NH (#BCs – 2), coupled with TODE(x) TB(x,y,z) ODE: TPDE(x,y,z) 0 TODE(x) 0 BC1: 0 at x = 0 BC2: 0 at x = L C1, C2 can be obtained from BC1 and BC2 • While any spatial dimension will generally work (e.g. x for this case) for TODE, careful selection can minimize the coupling into the homogeneous PDE as nonhomogeneous BCs • Generally, choose dimension for TODE with most number of Neumann BCs in the original problem 6 Superposition – step 4 IIT Kanpur TB(x,y,z) = TPDE(x,y,z) + TODE(x) 0 PDE: BC1: 0 BC2: 0 at x = 0 • The PDE will have homogeneous BCs along the spatial dimension for at x = L TODE (e.g. x for this case) 0 BC3: Similarly: at y = 0 at y = 0 BC4: at y = W BC5: at z = 0 BC6: at z = H • Nonhomogeneous BCs • Coupled with TODE(x) • Needs splitting into 4 more subproblems (TPDE1, TPDE2, TPDE3, TPDE4) similar to that done in step 3 that can be solved by the method of separation of variables 7 Example – Superposition IIT Kanpur k, and are constants PDE , 0, , at x = L 1, BC2 at x = L , , IC , at x = 0 0, [ref] L. M. Jiji, 2009 at x = 0 1, BC1 , at t = 0 at t = 0 • The PDE and the BCs are linear • The PDE and the BCs are nonhomogeneous 8 [ref] L. M. Jiji, 2009 IIT Kanpur Example – Superposition • Split the original problem (T) into one steady state (TSS) and one transient problem (TH) T(x,t) = TH(x,t) + TSS(x) ODE: NH PDE: H BCs: NH BCs: H , IC: TH(t=0) = Ti – TSS • Split the steady state problem (TSS) into two sub problems (TA and TB) TH=0 TH=0 TSS(x) = TA(x) + TB(x) ODE: NH ODE: H ODE: NH BCs: NH BCs: NH BCs: H , : 0 0 TH=0 TH=0 9 ...
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• Winter '17
• Suchitra Mathur
• IIT Kanpur

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