def K_run_converging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90):
    r'''Returns the loss coefficient for the run of a converging tee or wye
    according to the Crane method [1]_.
    
    .. math::
        K_{branch} = C\left[1 + D\left(\frac{Q_{branch}}{Q_{comb}\cdot 
        \beta_{branch}^2}\right)^2 - E\left(1 - \frac{Q_{branch}}{Q_{comb}}
        \right)^2 - \frac{F}{\beta_{branch}^2} \left(\frac{Q_{branch}}
        {Q_{comb}}\right)^2\right]
            
    .. math::
        \beta_{branch} = \frac{D_{branch}}{D_{comb}}
    
    In the above equation, C=1, D=0, E=1. See the notes for definitions of F 
    and also the special case of 90°.

    Parameters
    ----------
    D_run : float
        Diameter of the straight-through inlet portion of the tee or wye
        [m]
    D_branch : float
        Diameter of the pipe attached at an angle to the straight-through, [m]
    Q_run : float
        Volumetric flow rate in the straight-through inlet of the tee or wye,
        [m^3/s]
    Q_branch : float
        Volumetric flow rate in the pipe attached at an angle to the straight-
        through, [m^3/s]
    angle : float, optional
        Angle the branch makes with the straight-through (tee=90, wye<90)
        [degrees]

    Returns
    -------
    K : float
        Loss coefficient of run with respect to the velocity and inside 
        diameter of the combined flow outlet [-]

    Notes
    -----
    F is linearly interpolated from the table of angles below. There is no 
    cutoff to prevent angles from being larger or smaller than 30 or 60
    degrees. The switch to the special 90° happens at 75°.

    +-----------+------+
    | Angle [°] |      |
    +===========+======+
    | 30        | 1.74 |
    +-----------+------+
    | 45        | 1.41 |
    +-----------+------+
    | 60        | 1    |
    +-----------+------+
    
    For the special case of 90°, the formula used is as follows. 
        
    .. math::
        K_{run} = 1.55\left(\frac{Q_{branch}}{Q_{comb}} \right)
        - \left(\frac{Q_{branch}}{Q_{comb}}\right)^2
    
    Examples
    --------
    Example 7-35 of [1]_. A DN100 schedule 40 tee has 1135 liters/minute of
    water passing through the straight leg, and 380 liters/minute of water
    converging with it through a 90° branch. Calculate the loss coefficient in
    the run. The calculated value there is 0.03258.
    
    >>> K_run_converging_Crane(0.1023, 0.1023, 0.018917, 0.00633)
    0.32575847854551254
    
    References
    ----------
    .. [1] Crane Co. Flow of Fluids Through Valves, Fittings, and Pipe. Crane,
       2009.
    '''
    beta = (D_branch/D_run)
    beta2 = beta*beta
    Q_comb = Q_run + Q_branch
    Q_ratio = Q_branch/Q_comb
    if angle < 75.0:
        C = 1.0
    else:
        return 1.55*(Q_ratio) - Q_ratio*Q_ratio

    D, E = 0.0, 1.0
    F = interp(angle, run_converging_Crane_angles, run_converging_Crane_Fs)
    K = C*(1. + D*(Q_ratio/beta2)**2 - E*(1. - Q_ratio)**2 - F/beta2*Q_ratio**2)
    return K