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.5 ft3/s[2](1)(2)II[1](3)∆Q2(4)1.0 ft3/s0.8 ft3/s∆Q1I(5)90'[3]P i p eKn17.591.93629.631.901348.61.882439.71.768516.51.935Figure 4.24 A 5-pipe, 3-node network.© 2000 by CRC Press LLCexamined.The reader is encouraged to check numerically some of these steps.In the Q-equations the elements of the Jacobian will either be ∂ Fi / ∂ Qj = ±1 or zero in row i fora junction continuity equation row.The Jacobian terms for the energy loop equation rowsn −1will either be ∂ Fji / ∂ Qj = ± n j K jQor zero.The non-zero elements of the Jacobianj1 /n −1for the H-equations arem∂ Fi / ∂ H j = ± { 1 / (nmKm )}{(H j − Hk ) / Km }in whichthe sign is determined by the sign in front of this term in the equation and the sign beforeHj within the parentheses.Non-zero terms in the Jacobian for the ∆ Q-equations will ben −1of the form ∂ Fki / ∂∆ Qj = ± nk Kk (Qok ±∆ Qm∑ ).The Q-equations areF 1 = Q 1 − Q 2 − Q 4 − 1.0= 0F 2 = Q 2 + Q 3−1.5= 0F 3 = Q 4 − Q 3 + Q 5 − 0.8= 0(4.47)nnnF1454 = K 1 Q+ K− K−10 = 014 Q 45 Q 5nnnF2345 = K 2 Q− K− K= 023 Q 34 Q 4The Newton method is described by [ D]{ z}={ F} and { Q}( m+1) = { Q}( m) - { z} with 1−1−111[ D] = −111 (4.48)n −1n −1n −1n1451 K 1 Qn− n14 K 4 Q 45 K 5 Q 5n−1n −1n −1n2342 K 2 Q− n− n23 K 3 Q 34 K 4 Q 4If we choose the initial estimate of the solution vector to be2.00.9 Q{ } ( 0 ) = 0.6(4.49)0.11.3with which we have been careful to satisfy the junction continuity equations, so thesedischarges can be used in the ∆ Q-equations, the first evaluation of the Jacobian matrix andright-hand side leads to1.0000− 1.00000.0000− 1.00000.0000  z 1 0.0000  0.00001.00001.00000.00000.0000 z 2 0.0000  0.00000.0000−1.00001.00001.0000  z 3  =0.0000(4.50a)28.11330.00000.000011.9749− 40.8039    z 4 − 7.6935 0.0000 16.6487− 58.2888 − 11.97490.0000  z 5  − 11.3783© 2000 by CRC Press LLCwith the solution− 0.11692.1169−0.14701.0470{z} =0.1470 Q{ } ( 1 ) = 0.4530(4.51a) 0.0301 0.0699 0.11691.1831The Newton equations for the next cycle are1.0000− 1.00000.0000− 1.00000.0000  z 1 0.0000  0.00001.00001.00000.00000.0000 z 2 0.0000  0.00000.0000−1.00001.00001.0000  z 3  = 0.0000 (4.50b)29.64810.00000.00009.0910− 37.3636    z 4 − 0.0682 0.0000 19.0804− 45.4902 − 9.09100.0000  z 5  − 0.7993with the solution− 0.00222.1191−0.01111.0581{z} =0.0111 Q{ } ( 2 ) = 0.4419(4.51b) 0.0089 0.0610 0.00221.1809One more cycle would yield the final solution2.11911.0583Q{ } ( 3 ) = 0.4417(4.51c)0.06081.1809Referring again to Fig.4.24, since we have only three nodes, we must construct threeH-equations.They are1 /n 1 1 /n 2 1 /n 4F 1 = 100 − H 1− H 1 − H 2− H 1 − H 3−1.0 = 0K1K2K41 /n 2 1 /n 3F 2 = H 1 − H 2+ H 3 − H 2−1.5 = 0(4.52)K2K31 /n 4 1 /n 3 1 /n 5F 3 = H 1 − H 3− H 3 − H 2+ 90 − H 3− 0.8 = 0K4K3K5© 2000 by CRC Press LLCUsing [D]{ z} = { F} and { H}( m+1) = { H}( m) - { z} to implement the Newton method with an initial estimate of the nodal heads as93{ H} ( 0 ) = 85 (4.53)88 successive computational cycles produce− 0.166 0.0600.035  z 1   −1.258 15
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