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Hydraulics Prof. B.S. Thandaveswara Solution of manning equation by Newton Raphson Method There is no general analytical solution to manning equation for determining the flow depth given the flow rate because the area A and hydraulic radius R may be complicated functions of the depth. Newton Raphson method can be applied iteratively to obtain a numerical solution. Suppose that at iteration k the depth yk is selected and the flow rate Qn, is computed using manning formula using the area and hydraulic radius corresponding to yk. This Qk is compared with actual flow Qn; then the objective is to chose y such that the error. f (yk) = Qk - Qn is within the tolerance limit. The gradient of f with respect to y is df(yk ) dQk = dyk dyk because Qn is constant. Hence, assuming manning roughness is constant, 2 ⎞ ⎛ df ⎞ ⎛1 1 3 ⎜ ⎟ = ⎜ So 2 A k R k ⎟ dy n ⎝ ⎠ ⎝ ⎠k ⎛ ⎞ -1 2 ⎜ 1 1 2A R 3 dR dA ⎟⎟ = So 2 ⎜ +R 3 n 3 dy dy ⎟ ⎜ ⎜ ⎟ ⎝ ⎠k 2 ⎛ 2 dR 1 1 1 dA ⎞ = So 2 A k R k 3 ⎜ + ⎟ n ⎝ 3R dy A dy ⎠k ⎛ 2 dR ⎛ df ⎞ 1 dA ⎞ + ⎜ ⎟ = Qk ⎜ A dy ⎟⎠k ⎝ dy ⎠k ⎝ 3R dy in which the subscript k out side the bracket indicates that the quantities in the bracket computed for y = yk. In Newton's method, given a choice of yk , yk+1 is chosen to satisfy 0- f (y)k ⎛ df ⎞ = ⎜ ⎟ ⎝ dy ⎠k yk + yk+1 This yk+1 is the value of yk , f (yk ) yk+1 = yk ( df / dy ) k Indian Institute of Technology Madras Hydraulics Prof. B.S. Thandaveswara Which is the fundamental equation of the Newton's method. Iterations are continued until there is no significant change in yn; this will happen when the error is nearly zero or an acceptable prescribed tolerance. Thus for manning equation it may be written as y k+1 = y k - 1 - Q / Qk ⎛ 2 dR 1 dA ⎞ + ⎜ ⎟ ⎝ 3R dy A dy ⎠ k For rectangular channel A = bo y and R = bo y / ( bo + 2y ) where bo is the channel width; The quantity in denominator can be for rectangular channel ( ) d d (R ) = A P dy dy 1 dA A dP = − P dy P 2 dy ⎡ T R dP ⎤ =⎢ − ⎥ ⎣ P P dy ⎦ consider 2 dR 1 dA + 3R dy A dy 2 P ⎡ T R dP ⎤ T − + 3 A ⎢⎣ P P dy ⎥⎦ A dP ⎤ T 21⎡ T −R ⎥+ ⎢ dy ⎦ A 3 A⎣ 2 T 2 R dP T − + 3 A 3 A dy A ⎡ 5 T 2 1 dP ⎤ ⎢ 3 A − 3 P dy ⎥ ⎣ ⎦ For rectangular channel 5 bo 2 1 − 2 3 bo y 3 ( bo + 2 y ) 51 4 1 − 3 y 3 ( bo + 2 y ) 5 ( bo + 2 y ) − 12 y 3 y ( bo + 2 y ) = = 5bo + 10 y − 4 y 3 y ( bo + 2 y ) 5bo + 6 y 3 y ( bo + 2 y ) y k+1 = y k - 1 - Q /Qk 5 bo + 6 y k ⎛ ⎞ ⎜ ⎟ ⎜ 3y b + 2 y ⎟ k ⎠ ⎝ k o Similarly the channel shape function ⎡⎣( 2/3R )( dR/dy ) + ( 1/A ) (dA/dy)⎤⎦ ( for other cross sections can be derived. Indian Institute of Technology Madras ) Hydraulics Prof. B.S. Thandaveswara (1 + m2 ) + 4my 2 (1 + m2 ) 3y ( bo + my ) ⎛⎜ bo + 2 y (1 + m 2 ) ⎞⎟ ⎝ ⎠ ( bo + 2my ) + 6 y Trapezoided Channel 8 3y Triangular Channel Circular Conduit 4 ( 2sinθ + 3θ − 5θ cos θ ) ⎛θ ⎞ 3do (θ )(θ − sin θ ) sin ⎜ ⎟ ⎝2⎠ in which ⎛ θ = 2 cos −1 ⎜1 − ⎝ 2y ⎞ ⎟ do ⎠ Example: Compute the flow depth in a 0.6 m wide rectangular channel having n= 0.015, S0 = 0.025, and Q = 0.25 m3s-1. y B Solution: Let wide bo = 0.6m Manning coefficient n = 0.015 bed slope Sο = 0.025 discharge Q = 0.25m3 s −1 normal depth y = ? Hyraulic mean radius R = 2 1 1 Q = AR 3 Sο 2 n 2 ⎞3 1 ⎛ byk 1 Q = bo y ⎜ ⎟ Sο 2 + 2 n b y k ⎠ ⎝ Indian Institute of Technology Madras bo A = p bo + 2 y Hydraulics Prof. B.S. Thandaveswara 5 ⎤ ⎡ 1 by ( ) 1⎢ k 3 ⎥ Q= ⎢ S 2 2⎥ ο n ⎢⎣ ( b + 2 yk ) 3 ⎥⎦ 5 ⎤ ⎡ 1 1 ⎢ ( 0.6 × yk ) 3 ⎥ ×⎢ × 0.025 Qk = 2 ( ) 2⎥ 0.015 ⎢⎣ ( 0.6 + 2 yk ) 3 ⎥⎦ 5 53 53 0.6 3 y 4.4993 y k k = Qk = 10.5409* 23 23 0.6 + 2 yk 0.6 + 2 yk ( ) Shape function = = ( (1) ) 5bo + 6 yk 3 yk ( b + 2 yk ) 5 ( 0.6 ) + 6 yk 3 yk ( 0.6 + 2 yk ) = 3 + 6 yk 1 + 2 yk = 3 yk ( 0.6 + 2 yk ) yk ( 0.6 + 2 yk ) ⎛ 0.25 ⎞ ⎜1 − ⎟ yk ( 0.6 + 2 yk ) Qk ⎠ ⎝ y k+1 = yk − (1 + 2 yk ) (2) Iteration (k) yk ( m ) 1 0.100 2 0.1815 3 0.1727 Q(m3s -1 ) 0.1125 0.2684 0.2488 Froude number F = F= 0.2488 / ( 0.6* 0.1727 ) ( 9.81*0.1727 ) ∴ super critical flow Indian Institute of Technology Madras Q/ A V = gy gy = 1.844