def omega_definition(Psat, Pc):
r'''Returns the acentric factor of a fluid according to its fundamental
definition using the vapor pressure at a reduced temperature of 0.7Tc.
.. math::
\omega \equiv -\log_{10}\left[\lim_{T/T_c=0.7}(P^{sat}/P_c)\right]-1.0
Parameters
----------
Psat : float
Vapor pressure of the fluid at a reduced temperature of 0.7 [Pa]
Pc : float
Critical pressure of the fluid [Pa]
Returns
-------
omega : float
Acentric factor of the fluid [-]
Notes
-----
Examples
--------
Water
>>> omega_definition(999542, 22048320.0)
0.3435744558761711
References
----------
.. [1] Poling, Bruce E. The Properties of Gases and Liquids. 5th edition.
New York: McGraw-Hill Professional, 2000.
'''
return -log10(Psat / Pc) - 1.0
def Stiel_polar_factor(Psat, Pc, omega):
r'''This function handles the calculation of a chemical's Stiel Polar
factor, directly through the definition of Stiel-polar factor.
Requires the vapor pressure `Psat` at a reduced temperature of 0.6,
the critical pressure `Pc`, and the acentric factor `omega`.
.. math::
x = \log P_r|_{T_r=0.6} + 1.70 \omega + 1.552
Parameters
----------
Psat : float
Vapor pressure of fluid at a reduced temperature of 0.6 [Pa]
Pc : float
Critical pressure of fluid [Pa]
omega : float
Acentric factor of the fluid [-]
Returns
-------
factor : float
Stiel polar factor of compound
Notes
-----
A few points have also been published in [2]_, which may be used for
comparison. Currently this is only used for a surface tension correlation.
Examples
--------
Calculating the factor for water:
>>> Stiel_polar_factor(Psat=169745, Pc=22048321.0, omega=0.344)
0.02322146744772713
References
----------
.. [1] Halm, Roland L., and Leonard I. Stiel. "A Fourth Parameter for the
Vapor Pressure and Entropy of Vaporization of Polar Fluids." AIChE
Journal 13, no. 2 (1967): 351-355. doi:10.1002/aic.690130228.
.. [2] D, Kukoljac Miloš, and Grozdanić Dušan K. "New Values of the
Polarity Factor." Journal of the Serbian Chemical Society 65, no. 12
(January 1, 2000). http://www.shd.org.rs/JSCS/Vol65/No12-Pdf/JSCS12-07.pdf
'''
Pr = Psat / Pc
factor = log10(Pr) + 1.70 * omega + 1.552
return factor
def nus(self):
T, P = self.T, self.P
Tcs, Pcs, omegas = self._Tcs, self._Pcs, self._omegas
regular, CASs = self.regular, self._CASs
nus = []
limit_Tr = self.version > 0
for i in range(self.N):
# TODO validate and take T, P derivatives; n derivatives are from regular solution only
Tr = T / Tcs[i]
Pr = P / Pcs[i]
if regular[i]:
coeffs = self.simple_coeffs
elif CASs[i] == '1333-74-0':
coeffs = self.hydrogen_coeffs
elif CASs[i] == '74-82-8':
coeffs = self.methane_coeffs
else:
raise ValueError("Fail")
A0, A1, A2, A3, A4, A5, A6, A7, A8, A9 = coeffs
log10_v0 = A0 + A1 / Tr + A2 * Tr + A3 * Tr**2 + A4 * Tr**3 + (
A5 + A6 * Tr + A7 * Tr**2) * Pr + (A8 +
A9 * Tr) * Pr**2 - log10(Pr)
log10_v1 = -4.23893 + 8.65808 * Tr - 1.2206 / Tr - 3.15224 * Tr**3 - 0.025 * (
Pr - 0.6)
if Tr > 1.0 and limit_Tr:
log10_v1 = 1.0
if regular[i]:
v = 10.0**(log10_v0 + omegas[i] * log10_v1)
else:
# Chao Seader mentions
v = 10.0**(log10_v0)
nus.append(v)
return nus
def balance_stoichiometry(matrix, rounding=9, allow_fractional=False):
r'''This function balances a chemical reaction.
Parameters
----------
matrix : list[list[float]]
Chemical reaction matrix for further processing; rows contain element
counts of each compound, and the columns represent each chemical, [-]
Returns
-------
coefficients : list[float]
Balanced coefficients; all numbers are positive, [-]
Notes
-----
Balance the reaction 4 NH3 + 5 O2 = 4 NO + 6 H2O, without knowing the
coefficients:
>>> matrix = stoichiometric_matrix([{'N': 1, 'H': 3}, {'O': 2}, {'N': 1, 'O': 1}, {'H': 2, 'O': 1}], [True, True, False, False])
>>> matrix
[[3, 0, 0, -2], [1, 0, -1, 0], [0, 2, -1, -1]]
>>> balance_stoichiometry(matrix)
[4.0, 5.0, 4.0, 6.0]
>>> balance_stoichiometry(matrix, allow_fractional=True)
[1.0, 1.25, 1.0, 1.5]
This algorithm relies on `scipy`.
The behavior of this function for inputs which do not have a unique
solution is undefined.
This algorithm may suffer from floating point issues. If you believe there
is an error in the result, please report your reaction to the developers.
References
----------
.. [1] Sen, S. K., Hans Agarwal, and Sagar Sen. "Chemical Equation
Balancing: An Integer Programming Approach." Mathematical and Computer
Modelling 44, no. 7 (October 1, 2006): 678-91.
https://doi.org/10.1016/j.mcm.2006.02.004.
.. [2] URAVNOTE, NOVOODKRITI PARADOKSI V. TEORIJI, and ENJA KEMIJSKIH
REAKCIJ. "New Discovered Paradoxes in Theory of Balancing Chemical
Reactions." Materiali in Tehnologije 45, no. 6 (2011): 503-22.
'''
import scipy.linalg
done = scipy.linalg.null_space(matrix)
if len(done[0]) > 1:
raise ValueError("No solution")
d = done[:, 0].tolist()
min_value_inv = 1.0 / min(d)
d = [i * min_value_inv for i in d]
if not allow_fractional:
from fractions import Fraction
max_denominator = 10**rounding
fs = [
Fraction(x).limit_denominator(max_denominator=max_denominator)
for x in d
]
all_denominators = set([i.denominator for i in fs])
if 1 in all_denominators:
all_denominators.remove(1)
for den in sorted(list(all_denominators), reverse=True):
fs = [num * den for num in fs]
if all(i.denominator == 1 for i in fs):
break
# May have gone too far
return [float(i) for i in fs]
# done = False
# for i in range(100):
# for c in d:
# ratio = c.as_integer_ratio()[1]
# if ratio != 1:
# d = [di*ratio for di in d]
# break
# done = True
# if done:
# break
#
# d_as_int = [int(i) for i in d]
# for i, j in zip(d, d_as_int):
# if i != j:
# raise ValueError("Could not find integer coefficients (%s, %s)" %(i, j))
# return d_as_int
else:
d = [round(i, rounding + int(ceil(log10(abs(i))))) for i in d]
return d