SANITARY AND STORM SEWER DESIGN A Direct Algebraic Solution by A. M. Saatçi
Partially Flowing Pipe D θ h
Manning’s Equation v = (1/n) R2/3S1/2 Q = (K/n) D8/3 S1/2
Given: D, D, n, h/D D, S, n, h/D Given θ = 2 cos-1(1-2h/D) R=D(θ-sinθ)/4θ K=0.0496 θ(1-sinθ/θ)5/3 Q = KD8/3S1/2 /n v = Q/A A= D2/8(θ-sinθ)
Given: D, S, n, Q D, S, n, Q Given Graphical Solutions Figure 1. h/D, Aflow, v θ-2/3 (θ - sin θ)5/3 - 20.161 n Q D-8/3 S-1/2 = 0 Trial & Error Solution:
Geometry θ = 2 cos-1 [(1-2h)/D] A = D2 (θ-sin θ)/8 From geometry, θ = 2 cos-1 [(1-2h)/D] A = D2 (θ-sin θ)/8 K = 0.04968θ-2/3 (θ -sin θ)5/3 θ-2/3 (θ - sin θ)5/3 - 20.161 n Q D-8/3 S-1/2 = 0 θ-2/3 (θ - sin θ)5/3 - 20.161 K = 0 Last two eqns can be solved for θ using trial and error methods.
Saatci Equation θ = (3π/2) 1- 1- πK K = (1/π) { 1- [1-(2θ/3π)2]2}2 It is suggested that an algebraic equation of the form: θ = (3π/2) 1- 1- πK which can also be written as: K = (1/π) { 1- [1-(2θ/3π)2]2}2 can give a quick solution of the water surface angle θ.
Trial and Error Solutions to solve for θ Q, D, S, n Given K = Qn/(D8/3S1/2) θ-2/3 (θ - sin θ)5/3 - 20.161 K = 0 Trial and Error Solutions to solve for θ Saatci's Eqn θ = (3π/2) 1- 1- πK Direct Solution Water Surface Angle θ A =( D2/8)(θ-sin θ) v = Q/A h/D=0.5(1-cos(θ/2)) R = (D/4)[(θ-sinθ) /θ]