# Property calculator

A fluid properties calculator for liquid, vapor and super-critical states, based on the open source thermodynamic package CoolProp

# Fluid Info

Critical point, triple point, fluid limits, other fluid constants, data sources.

# Literature References

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# Saturation Data

Evaporation and condensation data

# P-H Diagram

Automatically generated log(P)-H diagrams for many fluids
The diagram is best displayed using the default ranges.

# 1.   Fluid property input

## 1.1.   Choice of state variables

For a pure fluid (fluid consisting of a single chemical species) two variables are necessary to completely describe its thermodynamic state. In theory, these could be any two from the following list:

• pressure
• temperature
• density
• specific enthalpy
• specific entropy

In reality, there are good choices and bad choices of state variables. A choice could be also good or bad depending on the particular state of the fluid. For an incompressible fluid, for example, choosing the density as a state variable is apparently not suitable. Pressure and temperature are always a good choice, except when properties are calculated in the two-phase region (boiling/condensation). Temperature and density are always a good choice for compressible fluids. Temperature and enthalpy are an unsuitable combination as they are strongly correlated, while pressure and enthalpy are a very good combination. How suitable the selected combination of state variables is depends also on the particular equations of state used.

## 1.2.   Two-phase region

If a liquid is heated beyond its boiling point at a pressure below its critical pressure, the fluid starts evaporating. During evaporation the temperature and pressure are linked to each other, and if, for example, the pressure is kept constant, so remains the temperature. This part of the fluid state diagram is called the two-phase region, because a partially evaporated fluid contains a liquid phase and a vapor phase. The mass fraction of the vapor phase is called vapor quality q. Thus $$q = 0$$ indicates an all-liquid (saturated liquid) state, and $$q = 1$$ indicates a saturaded vapor state. The property of the overall fluid in the two-phase region depends on the property of both phases and the q, e.g.

\begin{equation*} h = q \cdot h_v + (1 - q) \cdot h_l \end{equation*}
\begin{equation*} v = q \cdot v_v + (1 - q) \cdot v_l \end{equation*}

where subscript v indicates vapor properties and subscript l - liquid ones.

In order to obtain the boiling temperature at a given pressure, the user enters the pressure and an arbitrary value in the range 0-1 for q.

# 2.   CoolProp package

All the fluid properties on this page are calculated using the open-source property package CoolProp. CoolProp is an open-source, cross-platform, free property database based on C++ that includes pure fluids, pseudo-pure fluids, and humid air properties. A complete list of all the materials implemented in CoolProp can be found here

Caution: When using pseudo-pure fluids (like Air), there is a chance you may obtain funny values for properties under ceratain conditions (e.g. at a low temperature). Thus, better use pure fluids (e.g. Nitrogen) whenever possible.

# 3.   State diagrams

## 3.1.   P-H diagram

### 3.1.1.   Isentrops

A special algorithm was developed for tracing isentrops, so that most calculations can be expressed in terms of the fundamental variables of the equation of state: temperature and density.

1. Start from a seed poing $$f_0$$
2. Compute the fluid state based on $$s$$ and $$T$$ variables
3. Find $$\left(\frac{\partial\rho}{\partial T}\right)_{s}$$.
4. Select a step $$\Delta T$$ and find $$\Delta \rho$$ from
\begin{equation*} \frac{\mathrm{d}\rho}{\mathrm{d}T}=\left(\frac{\partial \rho}{\partial T}\right)_{s} \end{equation*}
1. Compute the fluid state at $$f_1$$ by $$T_0 + \Delta T$$ and $$\rho_0 + \Delta\rho$$
2. Go back to 3. using $$f_1$$

We use Maxwell relations to find the derivative:

\begin{equation*} \left(\frac{\partial v}{\partial T}\right)_{s}=-\left(\frac{\partial s}{\partial p}\right)_{v} \end{equation*}

Which can be expressed as:

\begin{equation*} \left(\frac{\partial\rho}{\partial T}\right)_{s}=\rho^2\cdot\left(\frac{\partial s}{\partial p}\right)_{v} \end{equation*}

Using the total derivative of $$s$$:

\begin{equation*} \mathrm{d}s=\left(\frac{\partial s}{\partial v}\right)_{T}\cdot \mathrm{d}v+\left(\frac{\partial s}{\partial T}\right)_{v}\cdot \mathrm{d}T \end{equation*}

And taking the partial derivative:

\begin{equation*} \left(\frac{\partial s}{\partial p}\right)_{v}=\left(\frac{\partial s}{\partial v}\right)_{T}\cdot\left(\frac{\partial v}{\partial p}\right)_{v}+\left(\frac{\partial s}{\partial T}\right)_{v}\cdot\left(\frac{\partial T}{\partial p}\right)_{v} \end{equation*}
\begin{equation*} \left(\frac{\partial s}{\partial p}\right)_{v}=\left(\frac{\partial s}{\partial T}\right)_{v}\cdot\left(\frac{\partial T}{\partial p}\right)_{v} \end{equation*}

Finally:

\begin{equation*} \left(\frac{\partial \rho}{\partial T}\right)_{s}=\rho^2\left(\frac{\partial s}{\partial T}\right)_{v}\cdot\left(\frac{\partial T}{\partial p}\right)_{v} \end{equation*}

In the single phase region the two partial derivatives on the right can be calculated directly. In the two-phase region, we can use:

\begin{equation*} s=q\cdot s_{V}+(1-q)\cdot s_{L} \end{equation*}

And therefore the total $$s$$ derivative can be expressed as:

\begin{equation*} \partial s=\left(\frac{\partial s}{\partial q}\right)_{T}\mathrm{d}q+\left(\frac{\partial s}{\partial T}\right)_{q}\mathrm{d}T \end{equation*}

Taking the partial derivative:

\begin{equation*} \left(\frac{\partial s}{\partial T}\right)_{v}=\left(\frac{\partial s}{\partial q}\right)_{T}\cdot\left(\frac{\partial q}{\partial T}\right)_{v}+\left(\frac{\partial s}{\partial T}\right)_{q}\cdot\left(\frac{\partial T}{\partial T}\right)_{v} \end{equation*}
\begin{equation*} \left(\frac{\partial s}{\partial T}\right)_{v}=\left(\frac{\partial s}{\partial q}\right)_{T}\cdot\left(\frac{\partial q}{\partial T}\right)_{v}+\left(\frac{\partial s}{\partial T}\right)_{q} \end{equation*}

The terms can be expressed as follows:

\begin{equation*} \left(\frac{\partial s}{\partial q}\right)_{T}=s_{V}-s_{L} \end{equation*}
\begin{equation*} \left(\frac{\partial q}{\partial T}\right)_{v}=-\frac{\left(\frac{\partial v}{\partial T}\right)_{q}}{\left(\frac{\partial v}{\partial q}\right)_{T}} \end{equation*}
\begin{equation*} \left(\frac{\partial s}{\partial T}\right)_{q}=q\cdot\left(\frac{\partial s}{\partial T}\right)_{SatV}+\left(1-q\right)\cdot\left(\frac{\partial s}{\partial T}\right)_{SatL} \end{equation*}

The right-hand side of the second equation contains two terms, which can be expressed as follows:

\begin{equation*} \left(\frac{\partial v}{\partial T}\right)_{q}=q\cdot\left(\frac{\partial v}{\partial T}\right)_{SatV}+\left(1-q\right)\cdot\left(\frac{\partial v}{\partial T}\right)_{SatL} \end{equation*}
\begin{equation*} \left(\frac{\partial v}{\partial q}\right)_{T}=v_{V}-v_{L} \end{equation*}

Therefore we have:

\begin{equation*} \left(\frac{\partial q}{\partial T}\right)_{v}=-\frac{q\cdot\left(\frac{\partial v}{\partial T}\right)_{SatV}+\left(1-q\right)\cdot\left(\frac{\partial v}{\partial T}\right)_{SatL}}{v_{v}-v_{L}} \end{equation*}
\begin{equation*} \left(\frac{\partial q}{\partial T}\right)_{v}=\frac{\frac{q}{\rho^{2}}\left(\frac{\partial\rho}{\partial T}\right)_{SatV}+\frac{1-q}{\rho^{2}}\left(\frac{\partial\rho}{\partial T}\right)_{SatL}}{\frac{1}{\rho_{v}}-\frac{1}{\rho_{L}}} \end{equation*}

### 3.1.2.   Isochores

Determine the appropriate ranges for $$T$$ and $$\rho$$ and compute the fluid state based on those variables

### 3.1.3.   Isotherms

Same as in the case of isochores