def BVirial_Tsonopoulos_extended(T, Tc, Pc, omega, a=0, b=0, species_type='', 
                                 dipole=0, order=0):
    r'''Calculates the second virial coefficient using the
    comprehensive model in [1]_. See the notes for the calculation of `a` and
    `b`.

    .. math::
        \frac{BP_c}{RT_c} = B^{(0)} + \omega B^{(1)} + a B^{(2)} + b B^{(3)}
    
        B^{(0)}=0.1445-0.33/T_r-0.1385/T_r^2-0.0121/T_r^3

        B^{(1)} = 0.0637+0.331/T_r^2-0.423/T_r^3 -0.423/T_r^3 - 0.008/T_r^8
        
        B^{(2)} = 1/T_r^6

        B^{(3)} = -1/T_r^8

    Parameters
    ----------
    T : float
        Temperature of fluid [K]
    Tc : float
        Critical temperature of fluid [K]
    Pc : float
        Critical pressure of the fluid [Pa]
    omega : float
        Acentric factor for fluid, [-]
    a : float, optional
        Fit parameter, calculated based on species_type if a is not given and
        species_type matches on of the supported chemical classes.
    b : float, optional
        Fit parameter, calculated based on species_type if a is not given and
        species_type matches on of the supported chemical classes.
    species_type : str, optional
        One of .
    dipole : float
        dipole moment, optional, [Debye]
    order : int, optional
        Order of the calculation. 0 for the calculation of B itself; for 1/2/3, 
        the first/second/third derivative of B with respect to temperature; and  
        for -1/-2, the first/second indefinite integral of B with respect to 
        temperature. No other integrals or derivatives are implemented, and an 
        exception will be raised if any other order is given.

    Returns
    -------
    B : float
        Second virial coefficient in density form or its integral/derivative if
        specified, [m^3/mol or m^3/mol/K^order]

    Notes
    -----
    Analytical models for derivatives and integrals are available for orders
    -2, -1, 1, 2, and 3, all obtained with SymPy.


    To calculate `a` or `b`, the following rules are used:

    For 'simple' or 'normal' fluids:
    
    .. math::
        a = 0
        
        b = 0

    For 'ketone', 'aldehyde', 'alkyl nitrile', 'ether', 'carboxylic acid',
    or 'ester' types of chemicals:
    
    .. math::
        a = -2.14\times 10^{-4} \mu_r - 4.308 \times 10^{-21} (\mu_r)^8
        
        b = 0
    
    For 'alkyl halide', 'mercaptan', 'sulfide', or 'disulfide' types of 
    chemicals:

    .. math::
        a = -2.188\times 10^{-4} (\mu_r)^4 - 7.831 \times 10^{-21} (\mu_r)^8
        
        b = 0

    For 'alkanol' types of chemicals (except methanol):
    
    .. math::
        a = 0.0878
    
        b = 0.00908 + 0.0006957 \mu_r
        
    For methanol:
    
    .. math::
        a = 0.0878
        
        b = 0.0525
    
    For water:
    
    .. math::
        a = -0.0109
        
        b = 0
    
    If required, the form of dipole moment used in the calculation of some
    types of `a` and `b` values is as follows:
    
    .. math::
        \mu_r = 100000\frac{\mu^2(Pc/101325.0)}{Tc^2}


    For first temperature derivative of B:
    
    .. math::
        \frac{d B^{(0)}}{dT} = \frac{33 Tc}{100 T^{2}} + \frac{277 Tc^{2}}{1000 T^{3}} + \frac{363 Tc^{3}}{10000 T^{4}} + \frac{607 Tc^{8}}{125000 T^{9}}

        \frac{d B^{(1)}}{dT} = - \frac{331 Tc^{2}}{500 T^{3}} + \frac{1269 Tc^{3}}{1000 T^{4}} + \frac{8 Tc^{8}}{125 T^{9}}

        \frac{d B^{(2)}}{dT} = - \frac{6 Tc^{6}}{T^{7}}

        \frac{d B^{(3)}}{dT} = \frac{8 Tc^{8}}{T^{9}}

    For the second temperature derivative of B:
    
    .. math::
        \frac{d^2 B^{(0)}}{dT^2} = - \frac{3 Tc}{125000 T^{3}} \left(27500 + \frac{34625 Tc}{T} + \frac{6050 Tc^{2}}{T^{2}} + \frac{1821 Tc^{7}}{T^{7}}\right)

        \frac{d^2 B^{(1)}}{dT^2} = \frac{3 Tc^{2}}{500 T^{4}} \left(331 - \frac{846 Tc}{T} - \frac{96 Tc^{6}}{T^{6}}\right)

        \frac{d^2 B^{(2)}}{dT^2} = \frac{42 Tc^{6}}{T^{8}}

        \frac{d^2 B^{(3)}}{dT^2} = - \frac{72 Tc^{8}}{T^{10}}

    For the third temperature derivative of B:
    
    .. math::
        \frac{d^3 B^{(0)}}{dT^3} = \frac{3 Tc}{12500 T^{4}} \left(8250 + \frac{13850 Tc}{T} + \frac{3025 Tc^{2}}{T^{2}} + \frac{1821 Tc^{7}}{T^{7}}\right)

        \frac{d^3 B^{(1)}}{dT^3} = \frac{3 Tc^{2}}{250 T^{5}} \left(-662 + \frac{2115 Tc}{T} + \frac{480 Tc^{6}}{T^{6}}\right)

        \frac{d^3 B^{(2)}}{dT^3} = - \frac{336 Tc^{6}}{T^{9}}

        \frac{d^3 B^{(3)}}{dT^3} = \frac{720 Tc^{8}}{T^{11}}

    For the first indefinite integral of B:
    
    .. math::
        \int{B^{(0)}} dT = \frac{289 T}{2000} - \frac{33 Tc}{100} \log{\left (T \right )} + \frac{1}{7000000 T^{7}} \left(969500 T^{6} Tc^{2} + 42350 T^{5} Tc^{3} + 607 Tc^{8}\right)
        
        \int{B^{(1)}} dT = \frac{637 T}{10000} - \frac{1}{70000 T^{7}} \left(23170 T^{6} Tc^{2} - 14805 T^{5} Tc^{3} - 80 Tc^{8}\right)

        \int{B^{(2)}} dT = - \frac{Tc^{6}}{5 T^{5}}

        \int{B^{(3)}} dT = \frac{Tc^{8}}{7 T^{7}}

    For the second indefinite integral of B:

    .. math::
        \int\int B^{(0)} dT dT = \frac{289 T^{2}}{4000} - \frac{33 T}{100} Tc \log{\left (T \right )} + \frac{33 T}{100} Tc + \frac{277 Tc^{2}}{2000} \log{\left (T \right )} - \frac{1}{42000000 T^{6}} \left(254100 T^{5} Tc^{3} + 607 Tc^{8}\right)
    
        \int\int B^{(1)} dT dT = \frac{637 T^{2}}{20000} - \frac{331 Tc^{2}}{1000} \log{\left (T \right )} - \frac{1}{210000 T^{6}} \left(44415 T^{5} Tc^{3} + 40 Tc^{8}\right)

        \int\int B^{(2)} dT dT = \frac{Tc^{6}}{20 T^{4}}
        
        \int\int B^{(3)} dT dT = - \frac{Tc^{8}}{42 T^{6}}
        
    Examples
    --------
    Example from Perry's Handbook, 8E, p2-499. Matches to a decimal place.

    >>> BVirial_Tsonopoulos_extended(430., 405.65, 11.28E6, 0.252608, a=0, b=0, species_type='ketone', dipole=1.469)
    -9.679715056695323e-05

    References
    ----------
    .. [1] Tsonopoulos, C., and J. L. Heidman. "From the Virial to the Cubic
       Equation of State." Fluid Phase Equilibria 57, no. 3 (1990): 261-76.
       doi:10.1016/0378-3812(90)85126-U
    .. [2] Tsonopoulos, Constantine, and John H. Dymond. "Second Virial
       Coefficients of Normal Alkanes, Linear 1-Alkanols (and Water), Alkyl
       Ethers, and Their Mixtures." Fluid Phase Equilibria, International
       Workshop on Vapour-Liquid Equilibria and Related Properties in Binary
       and Ternary Mixtures of Ethers, Alkanes and Alkanols, 133, no. 1-2
       (June 1997): 11-34. doi:10.1016/S0378-3812(97)00058-7.
    '''
    Tr = T/Tc
    if order == 0:
        B0 = 0.1445 - 0.33/Tr - 0.1385/Tr**2 - 0.0121/Tr**3 - 0.000607/Tr**8
        B1 = 0.0637 + 0.331/Tr**2 - 0.423/Tr**3 - 0.008/Tr**8
        B2 = 1./Tr**6
        B3 = -1./Tr**8
    elif order == 1:
        B0 = 33*Tc/(100*T**2) + 277*Tc**2/(1000*T**3) + 363*Tc**3/(10000*T**4) + 607*Tc**8/(125000*T**9)
        B1 = -331*Tc**2/(500*T**3) + 1269*Tc**3/(1000*T**4) + 8*Tc**8/(125*T**9)
        B2 = -6.0*Tc**6/T**7
        B3 = 8.0*Tc**8/T**9
    elif order == 2:
        B0 = -3*Tc*(27500 + 34625*Tc/T + 6050*Tc**2/T**2 + 1821*Tc**7/T**7)/(125000*T**3)
        B1 = 3*Tc**2*(331 - 846*Tc/T - 96*Tc**6/T**6)/(500*T**4)
        B2 = 42.0*Tc**6/T**8
        B3 = -72.0*Tc**8/T**10
    elif order == 3:
        B0 = 3*Tc*(8250 + 13850*Tc/T + 3025*Tc**2/T**2 + 1821*Tc**7/T**7)/(12500*T**4)
        B1 = 3*Tc**2*(-662 + 2115*Tc/T + 480*Tc**6/T**6)/(250*T**5)
        B2 = -336.0*Tc**6/T**9
        B3 = 720.0*Tc**8/T**11
    elif order == -1:
        B0 = 289*T/2000. - 33*Tc*log(T)/100. + (969500*T**6*Tc**2 + 42350*T**5*Tc**3 + 607*Tc**8)/(7000000.*T**7)
        B1 = 637*T/10000. - (23170*T**6*Tc**2 - 14805*T**5*Tc**3 - 80*Tc**8)/(70000.*T**7)
        B2 = -Tc**6/(5*T**5)
        B3 = Tc**8/(7*T**7)
    elif order == -2:
        B0 = 289*T**2/4000. - 33*T*Tc*log(T)/100. + 33*T*Tc/100. + 277*Tc**2*log(T)/2000. - (254100*T**5*Tc**3 + 607*Tc**8)/(42000000.*T**6)
        B1 = 637*T**2/20000. - 331*Tc**2*log(T)/1000. - (44415*T**5*Tc**3 + 40*Tc**8)/(210000.*T**6)
        B2 = Tc**6/(20*T**4)
        B3 = -Tc**8/(42*T**6)
    else: 
        raise Exception('Only orders -2, -1, 0, 1, 2 and 3 are supported.')
    if a == 0 and b == 0 and species_type != '':
        if species_type == 'simple' or species_type == 'normal':
            a, b = 0, 0
        elif species_type == 'methyl alcohol':
            a, b = 0.0878, 0.0525
        elif species_type == 'water':
            a, b = -0.0109, 0
        elif dipole != 0 and Tc != 0 and Pc != 0:
            dipole_r = 1E5*dipole**2*(Pc/101325.0)/Tc**2

            if (species_type == 'ketone' or species_type == 'aldehyde'
            or species_type == 'alkyl nitrile' or species_type == 'ether'
            or species_type == 'carboxylic acid' or species_type == 'ester'):
                a, b = -2.14E-4*dipole_r-4.308E-21*dipole_r**8, 0
            elif (species_type == 'alkyl halide' or species_type == 'mercaptan'
            or species_type == 'sulfide' or species_type == 'disulfide'):
                a, b = -2.188E-4*dipole_r**4-7.831E-21*dipole_r**8, 0

            elif species_type == 'alkanol':
                a, b = 0.0878, 0.00908+0.0006957*dipole_r
    Br = B0 + omega*B1 + a*B2 + b*B3
    return Br*R*Tc/Pc
r'''Calculates the second virial coefficient using the
    comprehensive model in [1]_. See the notes for the calculation of `a` and
    `b`.

    .. math::
        \frac{BP_c}{RT_c} = B^{(0)} + \omega B^{(1)} + a B^{(2)} + b B^{(3)}
    
        B^{(0)}=0.1445-0.33/T_r-0.1385/T_r^2-0.0121/T_r^3

        B^{(1)} = 0.0637+0.331/T_r^2-0.423/T_r^3 -0.423/T_r^3 - 0.008/T_r^8
        
        B^{(2)} = 1/T_r^6

        B^{(3)} = -1/T_r^8

    Parameters
    ----------
    T : float
        Temperature of fluid [K]
    Tc : float
        Critical temperature of fluid [K]
    Pc : float
        Critical pressure of the fluid [Pa]
    omega : float
        Acentric factor for fluid, [-]
    a : float, optional
        Fit parameter, calculated based on species_type if a is not given and
        species_type matches on of the supported chemical classes.
    b : float, optional
        Fit parameter, calculated based on species_type if a is not given and
        species_type matches on of the supported chemical classes.
    species_type : str, optional
        One of .
    dipole : float
        dipole moment, optional, [Debye]
    order : int, optional
        Order of the calculation. 0 for the calculation of B itself; for 1/2/3, 
        the first/second/third derivative of B with respect to temperature; and  
        for -1/-2, the first/second indefinite integral of B with respect to 
        temperature. No other integrals or derivatives are implemented, and an 
        exception will be raised if any other order is given.

    Returns
    -------
    B : float
        Second virial coefficient in density form or its integral/derivative if
        specified, [m^3/mol or m^3/mol/K^order]

    Notes
    -----
    Analytical models for derivatives and integrals are available for orders
    -2, -1, 1, 2, and 3, all obtained with SymPy.


    To calculate `a` or `b`, the following rules are used:

    For 'simple' or 'normal' fluids:
    
    .. math::
        a = 0
        
        b = 0

    For 'ketone', 'aldehyde', 'alkyl nitrile', 'ether', 'carboxylic acid',
    or 'ester' types of chemicals:
    
    .. math::
        a = -2.14\times 10^{-4} \mu_r - 4.308 \times 10^{-21} (\mu_r)^8
        
        b = 0
    
    For 'alkyl halide', 'mercaptan', 'sulfide', or 'disulfide' types of 
    chemicals:

    .. math::
        a = -2.188\times 10^{-4} (\mu_r)^4 - 7.831 \times 10^{-21} (\mu_r)^8
        
        b = 0

    For 'alkanol' types of chemicals (except methanol):
    
    .. math::
        a = 0.0878
    
        b = 0.00908 + 0.0006957 \mu_r
        
    For methanol:
    
    .. math::
        a = 0.0878
        
        b = 0.0525
    
    For water:
    
    .. math::
        a = -0.0109
        
        b = 0
    
    If required, the form of dipole moment used in the calculation of some
    types of `a` and `b` values is as follows:
    
    .. math::
        \mu_r = 100000\frac{\mu^2(Pc/101325.0)}{Tc^2}


    For first temperature derivative of B:
    
    .. math::
        \frac{d B^{(0)}}{dT} = \frac{33 Tc}{100 T^{2}} + \frac{277 Tc^{2}}{1000 T^{3}} + \frac{363 Tc^{3}}{10000 T^{4}} + \frac{607 Tc^{8}}{125000 T^{9}}

        \frac{d B^{(1)}}{dT} = - \frac{331 Tc^{2}}{500 T^{3}} + \frac{1269 Tc^{3}}{1000 T^{4}} + \frac{8 Tc^{8}}{125 T^{9}}

        \frac{d B^{(2)}}{dT} = - \frac{6 Tc^{6}}{T^{7}}

        \frac{d B^{(3)}}{dT} = \frac{8 Tc^{8}}{T^{9}}

    For the second temperature derivative of B:
    
    .. math::
        \frac{d^2 B^{(0)}}{dT^2} = - \frac{3 Tc}{125000 T^{3}} \left(27500 + \frac{34625 Tc}{T} + \frac{6050 Tc^{2}}{T^{2}} + \frac{1821 Tc^{7}}{T^{7}}\right)

        \frac{d^2 B^{(1)}}{dT^2} = \frac{3 Tc^{2}}{500 T^{4}} \left(331 - \frac{846 Tc}{T} - \frac{96 Tc^{6}}{T^{6}}\right)

        \frac{d^2 B^{(2)}}{dT^2} = \frac{42 Tc^{6}}{T^{8}}

        \frac{d^2 B^{(3)}}{dT^2} = - \frac{72 Tc^{8}}{T^{10}}

    For the third temperature derivative of B:
    
    .. math::
        \frac{d^3 B^{(0)}}{dT^3} = \frac{3 Tc}{12500 T^{4}} \left(8250 + \frac{13850 Tc}{T} + \frac{3025 Tc^{2}}{T^{2}} + \frac{1821 Tc^{7}}{T^{7}}\right)

        \frac{d^3 B^{(1)}}{dT^3} = \frac{3 Tc^{2}}{250 T^{5}} \left(-662 + \frac{2115 Tc}{T} + \frac{480 Tc^{6}}{T^{6}}\right)

        \frac{d^3 B^{(2)}}{dT^3} = - \frac{336 Tc^{6}}{T^{9}}

        \frac{d^3 B^{(3)}}{dT^3} = \frac{720 Tc^{8}}{T^{11}}

    For the first indefinite integral of B:
    
    .. math::
        \int{B^{(0)}} dT = \frac{289 T}{2000} - \frac{33 Tc}{100} \log{\left (T \right )} + \frac{1}{7000000 T^{7}} \left(969500 T^{6} Tc^{2} + 42350 T^{5} Tc^{3} + 607 Tc^{8}\right)
        
        \int{B^{(1)}} dT = \frac{637 T}{10000} - \frac{1}{70000 T^{7}} \left(23170 T^{6} Tc^{2} - 14805 T^{5} Tc^{3} - 80 Tc^{8}\right)

        \int{B^{(2)}} dT = - \frac{Tc^{6}}{5 T^{5}}

        \int{B^{(3)}} dT = \frac{Tc^{8}}{7 T^{7}}

    For the second indefinite integral of B:

    .. math::
        \int\int B^{(0)} dT dT = \frac{289 T^{2}}{4000} - \frac{33 T}{100} Tc \log{\left (T \right )} + \frac{33 T}{100} Tc + \frac{277 Tc^{2}}{2000} \log{\left (T \right )} - \frac{1}{42000000 T^{6}} \left(254100 T^{5} Tc^{3} + 607 Tc^{8}\right)
    
        \int\int B^{(1)} dT dT = \frac{637 T^{2}}{20000} - \frac{331 Tc^{2}}{1000} \log{\left (T \right )} - \frac{1}{210000 T^{6}} \left(44415 T^{5} Tc^{3} + 40 Tc^{8}\right)

        \int\int B^{(2)} dT dT = \frac{Tc^{6}}{20 T^{4}}
        
        \int\int B^{(3)} dT dT = - \frac{Tc^{8}}{42 T^{6}}
        
    Examples
    --------
    Example from Perry's Handbook, 8E, p2-499. Matches to a decimal place.

    >>> BVirial_Tsonopoulos_extended(430., 405.65, 11.28E6, 0.252608, a=0, b=0, species_type='ketone', dipole=1.469)
    -9.679715056695323e-05

    References
    ----------
    .. [1] Tsonopoulos, C., and J. L. Heidman. "From the Virial to the Cubic
       Equation of State." Fluid Phase Equilibria 57, no. 3 (1990): 261-76.
       doi:10.1016/0378-3812(90)85126-U
    .. [2] Tsonopoulos, Constantine, and John H. Dymond. "Second Virial
       Coefficients of Normal Alkanes, Linear 1-Alkanols (and Water), Alkyl
       Ethers, and Their Mixtures." Fluid Phase Equilibria, International
       Workshop on Vapour-Liquid Equilibria and Related Properties in Binary
       and Ternary Mixtures of Ethers, Alkanes and Alkanols, 133, no. 1-2
       (June 1997): 11-34. doi:10.1016/S0378-3812(97)00058-7.