def K_branch_diverging_Crane(D_run, D_branch, Q_run, Q_branch, angle=90):
    r'''Returns the loss coefficient for the branch of a diverging tee or wye
    according to the Crane method [1]_.
    
    .. math::
        K_{branch} = G\left[1 + H\left(\frac{Q_{branch}}{Q_{comb}
        \beta_{branch}^2}\right)^2 - J\left(\frac{Q_{branch}}{Q_{comb}
        \beta_{branch}^2}\right)\cos\theta\right]
            
    .. math::
        \beta_{branch} = \frac{D_{branch}}{D_{comb}}
    
    See the notes for definitions of H, J, and G.

    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 outlet 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 branch with respect to the velocity and inside 
        diameter of the combined flow inlet [-]

    Notes
    -----
    If :math:`\beta_{branch} = 1, \theta = 90^\circ`, H = 0.3 and J = 0. 
    Otherwise H = 1 and J = 2.
    
    G is determined according to the following pseudocode:
        
    .. code-block:: python
    
        if angle < 75:
            if beta2 <= 0.35:
                if Q_ratio <= 0.4:
                    G = 1.1 - 0.7*Q_ratio
                else:
                    G = 0.85
            else:
                if Q_ratio <= 0.6:
                    G = 1.0 - 0.6*Q_ratio
                else:
                    G = 0.6
        else:
            if beta2 <= 2/3.:
                G = 1
            else:
                G = 1 + 0.3*Q_ratio*Q_ratio

    Note that there are several errors in the text of [1]_; the errata can be  
    obtained here: http://www.flowoffluids.com/publications/tp-410-errata.aspx

    Examples
    --------
    Example 7-36 of [1]_. A DN150 schedule 80 wye has 1515 liters/minute of
    water exiting the straight leg, and 950 liters/minute of water
    exiting it through a 45° branch. Calculate the loss coefficient in
    the branch. The calculated value there is 0.4640.
    
    >>> K_branch_diverging_Crane(0.146, 0.146, 0.02525, 0.01583, angle=45)
    0.4639895627496694
    
    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 < 60 or beta <= 2/3.:
        H, J = 1., 2.
    else:
        H, J = 0.3, 0
    if angle < 75:
        if beta2 <= 0.35:
            if Q_ratio <= 0.4:
                G = 1.1 - 0.7*Q_ratio
            else:
                G = 0.85
        else:
            if Q_ratio <= 0.6:
                G = 1.0 - 0.6*Q_ratio
            else:
                G = 0.6
    else:
        if beta2 <= 2/3.:
            G = 1
        else:
            G = 1 + 0.3*Q_ratio*Q_ratio
    angle_rad = radians(angle)
    K_branch = G*(1 + H*(Q_ratio/beta2)**2 - J*(Q_ratio/beta2)*cos(angle_rad))
    return K_branch