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.001.00−1.00  z 1   0.000   309.09− 161.860.00 z 2  = − 1.506(4.41b)309.090.00805.20 z  3 3.051− 0.00170.1021{z} =0.0061Q{ } = 0.0204(4.42b)0.00440.0625One more iteration leads to 1.001.00−1.00  z 1   0.000   314.19− 127.010.00 z 2  = − 0.095(4.41c)314.190.00753.42 z  3 0.151− 0.00010.1022{z} = 0.0004Q{ } = 0.0200 m3/s(4.42c)0.00030.0622This solution is now sufficiently accurate!The solution of the H-equations by the Newton method uses basically the same equa-tions [D]{z} = {F} and {H}( m+1) = {H}( m) - {z}.These relations lead to a single update equationdFH(m+1 ) = H(m) − F1 )(4.43)111 / ( dH 1The derivative is© 2000 by CRC Press LLC1 /n −11 /n −11 /n −1dF12 31= − 1100 − H 185 − HH−11− 11 − 60dH 1n 1 K 1K1n 2 K 2K2n 3 K 3K3(4.44a)or−−−dF0.4930.4810.4931 = − 1100− H 1 − 185 − H 1 −1H 1 − 60 dH 12900  1469 4686  2432 11128  5646 (4.44b)If we initiate the solution procedure with the initial estimate of H( 0 ) = 84 , then the first1two iterative cycles producedFF1( 1 )1 = − 0.00415 ,= − 0.0136 , H= 84 − 0.00415 / 0.0136 = 83.70(4.45a)dH11anddFF1( 2 )1 = − 0.000247 ,= − 0.01251 , H= 83.70 − 0.000247 / 0.01251 = 83.68mdH11(4.45b)which will be regarded as adequate.Finally, we now solve the ∆ Q-equations by the Newton method using again the equa-tions [D] {z} = {F} and {∆ Q( m+1)} = {∆ Q( m)} - {z}.In this case we solve repeatedly the two-equation system for updated correction vectors {z} until it is declared to besufficiently small.Three cycles of computation yield these results:599307 z 1  − 6.97z  − 0.0234∆ Q 0.0234  = 1  = 1 =  (4.46a)307 1382 z 2 24.6z 2 0.0230∆ Q2 − 0.0230472309 z 1  −1.523z  − 0.0062∆ Q 0.0296  = 1  = 1 =  (4.46b)309 1115 z 2 3.13z 2 0.0045∆ Q2 − 0.0275441314 z 1  −0.0952z  − 0.0004∆ Q 0.0300  = 1  = 1 =  (4.46c)314 1068 z 2 0.1507z 2 0.0003∆ Q2 − 0.0278Now the discharges can be computed as Q1 = 0.1 + ∆ Q1 + ∆ Q2 = 0.1022 m3/s, Q2 =0.05 - ∆ Q1 = 0.0200 m3/s, and Q3 = 0.09 + ∆ Q2 = 0.0622 m3/s.Computer programs of differing complexity and generality can also be developed for thesolution of these equation systems by application of the Newton method.We will nowlook at two programs.The first program is relatively simple but must be recoded in partfor each application; it will be applied to the solution of the Q-equations for the three-reservoir problem.The second program is more versatile.Program 4.2, the FORTRAN program listed in Fig.4.21, is designed to solve three si-multaneous equations with the Newton method.It calls a matrix solver that has the coef-ficient matrix expanded by one column to contain the known vector, and it places theinverse in additional columns beyond the location of the known vector.The first part ofthe main program is currently written specifically to solve the Q-equations for the three-reservoir problem.However, the portion that numerically evaluates the derivatives in theJacobian matrix is written more generally, with N giving the size of the matrix problemto be solved.Careful study of this listing will clarify considerably how the various tasks© 2000 by CRC Press LLCare performed.The subroutine INVERM employs a common method in linear algebraproblems by using an expanded matrix.The coefficient matrix is square, here 3 rows by3 columns.The known vector is placed in the next column, in this case column 4.Thesubroutine solves the system of equations and provides the inverse matrix.The solution isreturned in the same column that initially contained the known vector, here column 4.************************************************************************** PROGRAM NO.4.2, NEWTON, FORTRAN* THIS PROGRAM HAS BEEN INCLUDED FOR THE CONVENIENCE OF THE READER.* THE AUTHOR ACCEPTS NO RESPONSIBILITY FOR ITS CORRECTNESS [ Pobierz całość w formacie PDF ]

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