# Interpolation of $H_{mix}$

Given two points $$(P,T, H_{mix})$$, how can I interpolate $$H_{mix}$$ between them?

In other words, if I had two points $$(H_1, P_1, T_1)$$ and $$(H_2, P_2, T_2)$$, where $$H_{mix}$$ is the enthalpy of mixing, $$P$$ is the pressure and $$T$$ is the temperature, how would I interpolate between them? I started with the differential $$dH = \frac{\partial H_{mix}}{\partial P}dP + \frac{\partial H_{mix}}{\partial T}dT$$ To interpolate, I would need to integrate this from $$P_1,T_1$$ to $$P_2,T_2$$. $$\Delta H_{mix} =\int _{P_{1}}^{P_{2}}\left(\frac{\partial H_{mix}}{\partial P}\right)_{T} dP+\int _{T_{1}}^{T_{2}}\left(\frac{\partial H_{mix}}{\partial T}\right)_{P} dT$$ This simplifies to $$\Delta H_{mix} =\int _{P_{1}}^{P_{2}} -C_{P} \mu _{J} dP+\int _{T_{1}}^{T_{2}} C_{P} dT=-C_P\mu_J(P_2-P_1) +C_P(T_2-T_1)$$ $$C_P$$ is the isobaric heat capacity and $$\mu_J$$ is the Joule-Thomson Coefficient. However, I do not know if this is correct or what my next step would be.

• Are the pressures outside the range of ideal gas behavior? Jan 19 at 12:08
• @ChetMiller They are in the range of $0-10 GPa$ Jan 19 at 12:53
• OK. Is the mixture of constant composition? If so, you are going to need to know (or estimate) the PVT behavior of the gas. Please also be aware that Cp is a function of pressure as well as temperature (beyond the ideal gas recon). You, at least, are going to need to know Cp vs T in the limit of ideal gas pressure range. Jan 19 at 13:55
• The mixture is not constant composition. It is $x:64$ O2:H2O, where $x$ can be anything from $25$ to $35$. Jan 19 at 14:13
• Is it constant between T1,P1 and T2,P2? Jan 19 at 16:11

You have the right idea, but if you want to interpolate you don't need heat capacities etc, just use the interpolation formula. In your case you need a double interpolation, which means you need four tabulated enthalpies, at two temperatures and two pressures that hopefully bracket the temperature and pressure of your calculation. The double interpolation is schematically shown below on the right.

In yous case $$x$$ is $$T$$, $$y$$ is $$P$$ and $$f$$ is $$H$$. The double interpolation formula is

$$H(T,P) = (1-a)(1-b) H_{11} + b(1-a) H_{12} + a(1-b) H_{21} + a b H_{22}$$ with $$H_{ij}=H(T_i,P_j)$$ and $$a = \frac{T-T_1}{T_2-T_1},\quad b = \frac{P-P_1}{P_2-P_1},\quad$$ For this to work you need four points, not just two.

(The figure is from this book)

• Thank you for the answer. I have a perhaps naive question: as I have 3 variables $(x,y,f(x,y))$, would the interpolation not result in a 3D plot? Jan 19 at 22:02
• Well, the interpolation itself will not yield a plot, you will get the enthalpy at point $(T,P)$. Maybe I don't understand your question? Jan 20 at 10:37
• Right, I am referring to the fact that I have three variables $(P,T,H)$, so I would need to plot everything in 3D, correct? Jan 20 at 14:16
• @DarkRunner If $H$ is a function of $T$ and $P$, you have a 2D surface. You don't have a triplet of points (i.e., $(P,\,T,\,H)$), you have a function $H(P,\,T)$. Jan 20 at 15:19
• @KyleKanos I see, thank you. Jan 20 at 16:12