Non-uniform flow analysis by Python: calculation of water surface frofile of settling basin

A program to calculate water surface profile of settling basin which has an open channel with varied width and varied invert level. Since the condition is subcritical flow, the calculation will be carried out from downstream section to upstream section.

Basic equation of non-uniform flow analysis is shown below.

$\begin{equation*} \left(\cfrac{Q^2}{2 g A_2{}^2}+h_2+z_2\right) - \left(\cfrac{Q^2}{2 g A_1{}^2}+h_1+z_1\right) = \cfrac{1}{2}\left(\cfrac{n_2{}^2 Q^2}{R_1{}^{4/3} A_1{}^2} + \cfrac{n_1{}^2 Q^2}{R_2{}^{4/3} A_2{}^2}\right)\cdot(x_2-x_1) \end{equation*}$

$Q$	Discharge (constant value)
$x_2, z_2, A_2, R_2, n_2, h_2$	Properties of upstream section (distance, invert level, flow section area, hydraulic radius, Manning's roughness coefficient, water depth)
$x_1, z_1, A_1, R_1, n_1, h_1$	Properties of downstream section (distance, invert level, flow section area, hydraulic radius, Manning's roughness coefficient, water depth)

Where, subscripu '1' is for downstream section and subscript '2' is for upstream section.

The equations for calculations of section area $A$ and hydraulic radius $R$ for rectangular cross section are shown below.

$\begin{equation*} A=b\cdot h \qquad R=\cfrac{b\cdot h}{b+2\cdot h} \end{equation*}$

where, $b$ means channel width, $h$ means water depth.

Since this case is a condition under subcritical flow, the conditions of downstream with subscript of '1' are all known values, and calculation is carried out to obtain the water depth of $h_2$ using the know conditions.

The section properties are defined as a function of x in def sec(x,h). Since the calculation model has many varied section, the part of definition of section properties (def sec(x,h)) is long. However, the core part for bi-section method is simple.

Two required initial values for bi-section method are defined as ha=3.0 and hb=7.0 in the function of def bisection(h1,x1,x2,q). These initial values should be changed appropriately, depending on the problem to be solved.

A function def sec(x,h) has a fuction to output section properties (z, A, R, n) from input data of distance (x) and water depth (h). When this function def sec(x,h) is changed, this program can be applied to a non-uniform flow analysis of open channel with various cross section.

A program is shown below.

# Non-Uniform Flow Analysis (Subcritical flow)
import numpy as np


def sec(x, h):
    # definition of section property
    # x  : distance
    # h  : water depth
    # zz : invert level
    # aa : secion area
    # rr : hydraulic radius
    # nn : Manning's roughness coefficient
    n0=0.014 # roughness coefficient (normal value)
    zz,aa,rr,nn=0,0,0,0
    if 0.0 <= x < 11.0:
        ds=11
        nn=n0
        z1=562.2; b1=4.0
        z2=560.5; b2=26.0
        zz=z1-(z1-z2)/ds*x
        bb=b1+(b2-b1)/ds*x
        aa=bb*h
        rr=bb*h/(bb+2*h)
    if 11.0 <= x < 19.0:
        ds=8.0
        nn=n0
        z1=560.5; b1=12
        z2=560.5; b2=12
        zz=z1-(z1-z2)/ds*(x-11)
        bb=b1+(b2-b1)/ds*(x-11)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 19.0 <= x < 47.0:
        ds=29.0
        nn=n0
        z1=560.5     ; b1=12.0
        z2=z1+ds*0.02; b2=12.0
        zz=z1-(z1-z2)/ds*(x-19)
        bb=b1+(b2-b1)/ds*(x-19)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 47.0 <= x < 62.0:
        ds=15.0
        nn=n0
        z1=561.06    ; b1=12.0
        z2=z1+ds*0.02; b2=6.0
        zz=z1-(z1-z2)/ds*(x-47)
        bb=b1+(b2-b1)/ds*(x-47)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 62.0 <= x < 69.0:
        ds=7.0
        nn=n0
        z1=561.36    ; b1=6.0
        z2=z1+ds*0.02; b2=6.0
        zz=z1-(z1-z2)/ds*(x-62)
        bb=b1+(b2-b1)/ds*(x-62)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 69.0 <= x < 69.5:
        ds=0.5
        nn=n0
        z1=561.5; b1=6.0
        z2=562.0; b2=6.0
        zz=z1-(z1-z2)/ds*(x-69)
        bb=b1+(b2-b1)/ds*(x-69)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 69.5 <= x < 73.5:
        ds=4.0
        nn=n0
        z1=562.0; b1=6.0
        z2=562.0; b2=6.0
        zz=z1-(z1-z2)/ds*(x-69.5)
        bb=b1+(b2-b1)/ds*(x-69.5)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 73.5 <= x < 74.0:
        ds=0.5
        nn=n0
        z1=562.0; b1=6.0
        z2=561.5; b2=6.5
        zz=z1-(z1-z2)/ds*(x-73.5)
        bb=b1+(b2-b1)/ds*(x-73.5)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 74.0 <= x < 77.0:
        ds=3.0
        nn=n0
        z1=561.5; b1=6.5
        z2=561.5; b2=6.5
        zz=z1-(z1-z2)/ds*(x-74.0)
        bb=b1+(b2-b1)/ds*(x-74.0)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 77.0 <= x < 96.625:
        ds=19.625
        nn=n0
        z1=561.5; b1=6.5
        z2=563.0; b2=6.5
        zz=z1-(z1-z2)/ds*(x-77.0)
        bb=b1+(b2-b1)/ds*(x-77.0)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    if 96.625 <= x <= 102.125:
        ds=5.5
        nn=n0
        z1=563.0; b1=6.5
        z2=563.0; b2=6.5
        zz=z1-(z1-z2)/ds*(x-96.625)
        bb=b1+(b2-b1)/ds*(x-96.625)       
        ah=bb*h
        rh=bb*h/(bb+2*h)
        aa=ah*2
        rr=rh*2
    return zz,aa,rr,nn


def func(h2,h1,x1,x2,q):
    g=9.8
    z1,a1,r1,n1=sec(x1,h1)
    z2,a2,r2,n2=sec(x2,h2)
    e2=q**2/2/g/a2**2+h2+z2
    e1=q**2/2/g/a1**2+h1+z1
    e3=0.5*(n1**2*q**2/r1**(4/3)/a1**2 + n2**2*q**2/r2**(4/3)/a2**2)*(x2-x1)
    return e2-e1-e3


def bisection(h1,x1,x2,q):
    ha=3.0 # lower initial value for bisection method
    hb=7.0 # upper initial value for bisection method
    for k in range(100):
        hi=0.5*(ha+hb)
        fa=func(ha,h1,x1,x2,q)
        fb=func(hb,h1,x1,x2,q)
        fi=func(hi,h1,x1,x2,q)
        if fa*fi<0: hb=hi
        if fb*fi<0: ha=hi
        #print(fa,fi,fb)
        if np.abs(hb-ha)<1e-10: break
    return hi


def main():
    g=9.8    # gravity acceleration
    q=42.0   # discharge
    # starting point (sub-critical flow)
    h1=3.9 # water depth at starting point
    x1=0.0   # origin of distance coordinate
    z1,a1,r1,n1=sec(x1,h1) # section property
    v1=q/a1  # flow velocity
    print('{0:>8s}{1:>8s}{2:>8s}{3:>8s}{4:>10s}{5:>8s}'.format('x','z','h','z+h','Energy','v'))
    print('{0:8.3f}{1:8.3f}{2:8.3f}{3:8.3f}{4:10.5f}{5:8.3f}'.format(x1,z1,h1,z1+h1,z1+h1+v1**2/2/g,v1))

    # calculation point
    #xp=np.arange(1,103,1)
    xp=np.array([11,19,47,62,69,69.5,73.5,74,77,96.625,102.125])

    # water level calculation to upstream direction
    for x2 in xp:
        h2=bisection(h1,x1,x2,q)
        z2,a2,r2,n2=sec(x2,h2)
        v2=q/a2
        print('{0:8.3f}{1:8.3f}{2:8.3f}{3:8.3f}{4:10.5f}{5:8.3f}'.format(x2,z2,h2,z2+h2,z2+h2+v2**2/2/g,v2))
        x1=x2 # distance
        h1=h2 # water depth


#==============
# Execution
#==============
if __name__ == '__main__': main()

Calculated results are shown below. A water level drop is larger than expected.

       x       z       h     z+h    Energy       v
   0.000 562.200   3.900 566.100 566.46982   2.692
  11.000 560.500   5.971 566.471 566.47522   0.293
  19.000 560.500   5.971 566.471 566.47523   0.293
  47.000 561.060   5.410 566.470 566.47528   0.323
  62.000 561.360   5.091 566.451 566.47541   0.687
  69.000 561.500   4.950 566.450 566.47553   0.707
  69.500 562.000   4.444 566.444 566.47554   0.788
  73.500 562.000   4.444 566.444 566.47562   0.788
  74.000 561.500   4.954 566.454 566.47563   0.652
  77.000 561.500   4.954 566.454 566.47567   0.652
  96.625 563.000   3.431 566.431 566.47615   0.942
 102.125 563.000   3.431 566.431 566.47634   0.942

That's all. Thank you.

damyarou

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Non-uniform flow analysis by Python: calculation of water surface frofile of settling basin