Acentric factor

Definition of acentric factor: $$\omega = -\left(\log_{10}\frac{p_s}{p_c}\right)_{T_r =0.7}-1$$ We introduce a new scaled variable $$\Theta = \frac{T_c}{T}-1 $$ So if $T_r=0.7$, $\Theta=0.4285714285714286$

At $\Theta=1/0.7-1$, $$p_s = p_c\cdot 10^{1-\omega}$$ In general, $\log_{10}\frac{p_s}{p_c}$ is close to a linear function of $\theta$

At the critical point, $\Theta = 1$, and $\log_{10}(p_s/p_c)=0$



In [8]:

    
import CoolProp, numpy as np, matplotlib.pyplot as plt
%matplotlib inline
PropsSI = CoolProp.CoolProp.PropsSI
for fluid in ['Ethane','Water','R125']:
    Tc = PropsSI('Tcrit',fluid)
    pc = PropsSI('pcrit',fluid)
    Ts = np.linspace(PropsSI('Ttriple',fluid),Tc)
    ps = PropsSI('P','T',Ts,'Q',0,fluid)
    plt.plot(Tc/Ts-1,-np.log10(ps/pc),label = fluid)
plt.axvline(1/0.7-1,dashes=[2,2])
plt.xlabel('$\Theta=T_c/T-1$')
plt.ylabel('$-\log_{10}(p/p_c)$')
plt.show()

Therefore, we can express the slope of the line as: $$ m = \frac{(\omega+1)-0}{(1/0.7-1)-0}$$ or $$ -\log_{10}(p/p_c) = \frac{\omega+1}{1/0.7-1}\Theta $$



In [12]:

    
import CoolProp, numpy as np, matplotlib.pyplot as plt
%matplotlib inline
PropsSI = CoolProp.CoolProp.PropsSI
for fluid in ['Ethane','Water','R125']:
    Tc = PropsSI('Tcrit',fluid)
    pc = PropsSI('pcrit',fluid)
    acentric = PropsSI('acentric',fluid)
    Ts = np.linspace(PropsSI('Ttriple',fluid),Tc)
    ps = PropsSI('P','T',Ts,'Q',0,fluid)
    line, = plt.plot(Tc/Ts-1,-np.log10(ps/pc),label = fluid)
    plt.plot(Tc/Ts-1,(acentric+1)/(1/0.7-1)*(Tc/Ts-1),label = fluid,color = line.get_color(),dashes = [2,2])
    
plt.axvline(1/0.7-1,dashes=[2,2])
plt.xlabel('$\Theta=T_c/T-1$')
plt.ylabel('$-\log_{10}(p/p_c)$')
plt.show()

So we can clearly see that while this formulation is not perfect, it's not at all bad