pipe normal depth | Python Fiddle

import math
#import unit_conversion

def normal_depth_pipe(Q, D, Sb, n):
    """This function calculates normal depth for circular pipes 8/7/2013 created by t. wang"""    
    # Q = flow (gpm)
    # D = pipe diameter (in)
    # Sb = pipe slope (ft/1000ft)
    # n = pipe roughness
    nmax = 100
    epsilon = 0.001
    D = float(D / 12.0)   # convert the diameter unit from inch to ft
    Q = Q / 448.8
    # Calculate full pipe flow capacity:
    A = math.pi / 4 * D * D
    Q_full = 1.486 / 1.547 / n * A * (D / 4.0) ** (2.0 / 3) * Sb ** 0.5
    Q_full_gpm = Q_full * 448.8
    print'The full pipe capacity is %0.2f' %(Q_full_gpm), '(gpm),', ' ' *10,

if Q < Q_full:
        y1 = 0
        y2 = D
        for i in range(nmax):
            y_guess = 0.5 * (y1 + y2)
            theta = 2 * math.acos((D / 2 - y_guess) / (D / 2))
            A = (D / 2) ** 2 * (theta - math.sin(theta)) / 2
            Q_calc = 1.486 / 1.547 / n * A * (A / (theta * D / 2)) ** (2.0 / 3) * Sb ** 0.5
            if abs(Q_calc - Q) < epsilon:
                return y_guess *12
                break
            if Q_calc < Q:
                y1 = y_guess
            else:
                y2 = y_guess
    elif Q >= Q_full and Q < 1.01 * Q_full:
        return D * 12
    else:
        return "Pipe is surcharged"

""" Testing the function """
Q = 1200
D = 24
Sb = 0.001
n = 0.013
print' '*80, 'Input:'
print'-'*200
print'Flow = %d' %Q, '(gpm),',' '*4 ,
print'Diameter = %d' %D, '(inch),', ' '*4 ,
print'Sb = %.3f,' %Sb,' '*4 ,
print'n = %.3f' %n
print'-'*200
print' '*80, 'Results:'
print'-'*200
B = normal_depth_pipe(Q, D, Sb, n)    
if B == 'Pipe is surcharged':
    print(B)
else: print'The normal depth is %.2f' %B, '(inch),', ' '*10,
if B <> 'Pipe is surcharged':
    percent_full = B / D * 100
    print 'The pipe is %.2f percent full.' %percent_full
print'-'*200

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