Liu hard sphere equation of state - Revision history

Carl McBride: Slight tidy

2020-11-10T10:21:43Z

Slight tidy

← Older revision		Revision as of 12:21, 10 November 2020
Line 1:		Line 1:
	Hongqin Liu proposed a correction to the C-~~S EOS~~ which improved accuracy by almost two ~~order~~ of magnitude <ref>[https://arxiv.org/abs/2010.14357]</ref>:		Hongqin Liu proposed a correction to the [[Carnahan-Starling equation of state]] which improved accuracy by almost two orders of magnitude <ref>[https://arxiv.org/abs/2010.14357 Hongqin Liu "Carnahan Starling type equations of state for stable hard disk and hard sphere fluids", arXiv:2010.14357]</ref>:

	: <math>		: <math>
Line 5:		Line 5:
	</math>		</math>

	The conjugate virial coefficient correlation is given by:		The conjugate [[Virial equation of state \| virial coefficient]] correlation is given by:

	: <math>		: <math>
Line 11:		Line 11:
	</math>		</math>

	The excess Helmholtz ~~free~~ energy is given by:		The excess [[Helmholtz energy function]] is given by:

	: <math>		: <math>
Line 17:		Line 17:
	</math>		</math>

	The isothermal compressibility is given by:		The isothermal [[compressibility]] is given by:

	: <math>		: <math>
Line 28:		Line 28:
	\frac{ dZ}{d\eta} = \frac{ 4 + 4\eta - \frac {11}{13} \eta^2 - \frac{52}{13}\eta^3 + \frac {7}{2}\eta^4 - \eta^5 }{(1-\eta)^4 }.		\frac{ dZ}{d\eta} = \frac{ 4 + 4\eta - \frac {11}{13} \eta^2 - \frac{52}{13}\eta^3 + \frac {7}{2}\eta^4 - \eta^5 }{(1-\eta)^4 }.
	</math>		</math>
			== References ==
			<references/>
			[[Category: Equations of state]]
			[[category: hard sphere]]

Hqliu1018 at 23:23, 8 November 2020

2020-11-08T23:23:17Z

← Older revision		Revision as of 01:23, 9 November 2020
Line 14:		Line 14:

	: <math>		: <math>
	A^{ex} = \frac{ A - A^{id}}{~~Nk_b~~}= \frac{ 188\eta - 126\eta^2 - 13\eta^4 }{52(1-\eta)^2} - \frac{5}{13} ln(1-\eta).		A^{ex} = \frac{ A - A^{id}}{Nk_B}= \frac{ 188\eta - 126\eta^2 - 13\eta^4 }{52(1-\eta)^2} - \frac{5}{13} ln(1-\eta).
	</math>		</math>

Hqliu1018 at 23:22, 8 November 2020

2020-11-08T23:22:25Z

← Older revision		Revision as of 01:22, 9 November 2020
Line 20:		Line 20:

	: <math>		: <math>
	k_T = (\eta\frac{ dZ}{d\eta})^{-1} \rho^{-1}.		k_T = (\eta\frac{ dZ}{d\eta} + Z)^{-1} \rho^{-1}.
	</math>		</math>

Hqliu1018 at 23:21, 8 November 2020

2020-11-08T23:21:10Z

@@ Line 15: / Line 15: @@
 : <math>
 A^{ex} = \frac{ A - A^{id}}{Nk_b}= \frac{ 188\eta - 126\eta^2 - 13\eta^4 }{52(1-\eta)^2} - \frac{5}{13} ln(1-\eta).
 </math>

Hqliu1018 at 23:12, 8 November 2020

2020-11-08T23:12:01Z

← Older revision		Revision as of 01:12, 9 November 2020
Line 11:		Line 11:
	</math>		</math>

	The excess ~~entropy~~ is given by:		The excess Helmholtz free energy is given by:

	: <math>		: <math>
	S^{ex} = \frac{ S - S^{id}}{Nk_b}= \frac{ 188\eta - 126\eta^2 - 13\eta^4 }{52(1-\eta)^2} - \frac{5}{13} ln(1-\eta).		A^{ex} = \frac{ A - A^{id}}{Nk_b}= \frac{ 188\eta - 126\eta^2 - 13\eta^4 }{52(1-\eta)^2} - \frac{5}{13} ln(1-\eta).
	</math>		</math>

Hqliu1018 at 23:10, 8 November 2020

2020-11-08T23:10:42Z

← Older revision		Revision as of 01:10, 9 November 2020
Line 14:		Line 14:

	: <math>		: <math>
	S^{ex} = \frac{ S - S^{id}}{Nk_b}= \frac{ 188\eta - 126\eta^2 - 13~~\eta^4 -~~ \eta^4 }{52(1-\eta)^2 - \frac{5}{13} ln(1-\eta) }.		S^{ex} = \frac{ S - S^{id}}{Nk_b}= \frac{ 188\eta - 126\eta^2 - 13\eta^4 }{52(1-\eta)^2} - \frac{5}{13} ln(1-\eta).
	</math>		</math>

Hqliu1018 at 23:09, 8 November 2020

2020-11-08T23:09:24Z

← Older revision		Revision as of 01:09, 9 November 2020
Line 9:		Line 9:
	: <math>		: <math>
	B_n = 0.9423n^2 + 1.28846n - 1.84615, n > 3.		B_n = 0.9423n^2 + 1.28846n - 1.84615, n > 3.
			</math>

			The excess entropy is given by:

			: <math>
			S^{ex} = \frac{ S - S^{id}}{Nk_b}= \frac{ 188\eta - 126\eta^2 - 13\eta^4 - \eta^4 }{52(1-\eta)^2 - \frac{5}{13} ln(1-\eta) }.
	</math>		</math>

Hqliu1018 at 21:09, 28 October 2020

2020-10-28T21:09:07Z

← Older revision		Revision as of 23:09, 28 October 2020
Line 8:		Line 8:

	: <math>		: <math>
	B_n = 0.9423n^2 + 1.28846n - 1.84615, n > 3.		B_n = 0.9423n^2 + 1.28846n - 1.84615, n > 3.
	</math>		</math>

Hqliu1018 at 21:08, 28 October 2020

2020-10-28T21:08:28Z

← Older revision		Revision as of 23:08, 28 October 2020
Line 3:		Line 3:
	: <math>		: <math>
	Z = \frac{ 1 + \eta + \eta^2 - \frac{8}{13}\eta^3 - \eta^4 + \frac{1}{2}\eta^5 }{(1-\eta)^3 }.		Z = \frac{ 1 + \eta + \eta^2 - \frac{8}{13}\eta^3 - \eta^4 + \frac{1}{2}\eta^5 }{(1-\eta)^3 }.
			</math>

			The conjugate virial coefficient correlation is given by:

			: <math>
			B_n = 0.9423n^2 + 1.28846n - 1.84615, n > 3.
	</math>		</math>

144.172.191.71: Created page with "Hongqin Liu proposed a correction to the C-S EOS which improved accuracy by almost two order of magnitude [https://arxiv.org/abs/2010.14357]: : $Z = \frac{..."$

2020-10-28T14:09:58Z

Created page with "Hongqin Liu proposed a correction to the C-S EOS which improved accuracy by almost two order of magnitude <ref>[https://arxiv.org/abs/2010.14357]</ref>: : <math> Z = \frac{..."

New page

Hongqin Liu proposed a correction to the C-S EOS which improved accuracy by almost two order of magnitude <ref>[https://arxiv.org/abs/2010.14357]</ref>:

: <math>
Z = \frac{ 1 + \eta + \eta^2 - \frac{8}{13}\eta^3 - \eta^4 + \frac{1}{2}\eta^5 }{(1-\eta)^3 }.
</math>