Biphase Model for Saturated Tank¶

The tank holds a self pressurized oxidizer, which is in saturation condition between liquid and vapor.

Oxidizer Extraction Model¶

Suppose a tank is filled up to a certain point with liquid oxidizer and its vapor is saturated at a certain ambient temperature.

Parameters¶

The parameters of the tank before the extraction are:

\(V_T\) Internal volume of the tank
\(m_O\) Mass of oxidizer stored in the tank
\(T_0\) Initial extraction temperature

Other variables can thus be described, such as:

\(V_O\) Volume occupied by liquid oxidizer
\(f\) Ullage volume percentage

The following solver needs as initial conditions only:

\(T_0\) Starting saturation temperature
\(f\) Vapor volume percentage (ullage)

and the following data for the chosen oxidizer.

Oxidizer Data¶

Data for known properties of the oxidizer as a function of temperature are needed:

\(p_{Sat}(T)\) Saturation pressure for a given temperature
\(\rho_l(T)\) Density of the saturated liquid for a given temperature
\(\rho_v(T)\) Density of the saturated vapor for a given temperature
\(s_l(T)\) Specific entropy of the saturated liquid for a given temperature
\(s_v(T)\) Specific entropy of the saturated vapor for a given temperature

Extraction¶

The physical quantities are regulated by the following expressions:

\[
\begin{aligned}
v_l(T) &= \frac1{\rho_l(T)} \\
v_v(T) &= \frac1{\rho_l(T)} \\
x(v, T) &= \frac{v - v_l(T)}{v_v(T) - v_l(T)} \\
\end{aligned}
\]

The mapping of \((x,T) \to (s,v)\) is given by:

\[
\begin{aligned}
    s(x, T) &= (1 - x) \cdot s_l(T) + x \cdot s_v(T) \\
    v(x, T) &= (1 - x) \cdot v_l(T) + x \cdot v_v(T) \\
\end{aligned}
\]

The goal is to find the inverse map

\[
    (s,v) \to (x,T)
\]

since evaluating \(s\) and \(v\) for the mass extraction is trivial.

The functions \(s(x,T)\) and \(v(x,T)\) can be differentiated with respect to \(x\) and \(T\) as:

\[
\begin{aligned}
dv = \frac{\partial v}{\partial x}\, dx + \frac{\partial v}{\partial T}\,dT \\
ds = \frac{\partial s}{\partial x}\, dx + \frac{\partial s}{\partial T}\,dT
\end{aligned}
\]

which in matrix form becomes

\[
\begin{Bmatrix}
    dv \\ ds
\end{Bmatrix} =
\begin{bmatrix}
    \dfrac{\partial v}{\partial x} &
    \dfrac{\partial v}{\partial T}\\
    \dfrac{\partial s}{\partial x}&
    \dfrac{\partial s}{\partial T}
\end{bmatrix}
\begin{Bmatrix}
    dx \\ dT
\end{Bmatrix}
\]

and inverting algebraically

\[
\begin{Bmatrix}
    dx \\ dT
\end{Bmatrix} =
\begin{bmatrix}
    \dfrac{\partial v}{\partial x} &
    \dfrac{\partial v}{\partial T}\\
    \dfrac{\partial s}{\partial x}&
    \dfrac{\partial s}{\partial T}
\end{bmatrix}^{-1}
\begin{Bmatrix}
    dv \\ ds
\end{Bmatrix}
\]

The quantities sought after are then computed as:

\[
\begin{Bmatrix}
    x(s,v) \\
    T(s,v)
\end{Bmatrix} =
\begin{Bmatrix}
    x_0 \\
    T_0
\end{Bmatrix} +
\int_{(s_0,T_0)} [J]^{-1}
\begin{Bmatrix}
    dv \\
    ds
\end{Bmatrix},\quad
\text{until $x =1$}
\]

Numerical solution¶

Time constraints¶

The extraction time \(t_n\) can be considered adimensional

\[ t_n = 1\]

therefore the \(n\) computation steps are

\[n = \frac{t_n}{\Delta t}\]

where \(\Delta t\) is a reasonably low number (ex. \(\Delta t = 5\cdot10^{-4}\)).

Although the initial conditions are useful, some of them can be discarded by considering the problem adimensionalized with respect to the mass of oxidizer stored inside the tank.

Initial conditions¶

\[ m_0 = 1 \]

Therefore the liquid mass is

\[ m_{l,0} = \frac{m_0}{1 + f \cdot \frac{\rho_{v,0}}{\rho_{l,0}}} \]

And vapor mass is

\[ m_{v,0} = m_0 - m_{l,0}\]

The oxidizer liquid volume is simply

\[ V_{l,0} = \frac{m_{l,0}}{\rho_{l,0}} \]

which can be used to get the total volume

\[ V_0 = V_{l,0} \cdot (1+f) \]

and the vapor volume

\[ V_{v,0} = V_0 - V_{l,0}.\]

The following are the initial conditions of the loop

\[
\begin{aligned}
T_0 &\\
p_0 &= p_{Sat}(T_0)\\
v_0 &= \frac{V_0}{m_0} \\
x_0 &= x(v_0,T_0) \\
s_0 &= s(x_0, T_0) \\
S_0 &= m_0 \cdot s_0 \\
\dot{m}_0 &= \frac{m_0}{t_n}
\end{aligned}
\]

Mass extraction¶

The mass that gets extracted at every computation step from the tank is computed as

\[
dm_i = \dot{m}_i \cdot dt
\]

and the remaining mass inside the tank is thus:

\[
m_{i+1} = m_i - dm_i
\]

Whether the extraction is done of liquid oxidizer or vapor oxidizer, the entropy \(S\) of the mass left in the tank is computed as:

\[
\begin{cases}
S_{i+1} = S_i - dm_i \cdot s_l(T_i),& \quad \text{liquid extraction}\\
S_{i+1} = S_i - dm_i \cdot s_v(T_i),& \quad \text{vapor extraction}
\end{cases}
\]

And thus the specific entropy \(s\) and specific volume \(v\) are

\[
\begin{aligned}
s_{i+1} &= \frac{S_{i+1}}{m_{i+1}} \\
v_{i+1} &= \frac{V_0}{m_{i+1}}
\end{aligned}
\]

Vapor Quality and Temperature Evaluation¶

The next goal is to compute the variation of vapor quality \(x\) and temperature \(T\) given the variation in specific volume \(v\) and specific enthalpy \(s\) after th extraction.

The problem is now finding the quantities inside the Jacobian matrix. The partial derivatives with respect to vapor quality \(x\) are immediate:

\[
\begin{aligned}
\frac{\partial v}{\partial x} &= v_v(T) - v_l(T) \\
\frac{\partial s}{\partial x} &= s_v(T) - s_l(T)
\end{aligned}
\]

but the derivatives with respect to temperature \(T\) must be computed numerically, in this case with central finite difference:

\[
\begin{aligned}
\frac{\partial v}{\partial T}
&= \frac1{2\varepsilon} [v(x,T+\varepsilon)-v(x,T-\varepsilon)] \\
\frac{\partial s}{\partial T}
&= \frac1{2\varepsilon} [s(x,T+\varepsilon)-s(x,T-\varepsilon)]
\end{aligned}
\]

Starting the evaluation from the current temperature \(T_i\), the target temperature \(T_{i+1}\) will be reached with successive approssimation of a progressively changing temperature \(T_j\) until the residue \(r\) is under the desired tolerance:

\[
r = \sqrt{\frac{dT_j}{T_j}^2+\frac{dx_j}{x_j}^2}\leq \text{tol}
\]

The guesses for temperature \(T_j\) and vapor quality \(x_j\) are updated as:

\[
\begin{aligned}
T_{j+1} = T_j + dT_j \\
x_{j+1} = x_j + dx_j
\end{aligned}
\]

until the final temperature \(T_{i+1}\) is the \(n\)-th guess temperature \(T_j\mid_{j=n}\).

In summary:

\[
\begin{aligned}
T_{i+1} - T_{i} = \sum_{\substack{T_j=T_i \\ x_j = x_i}}^{r<tol} dT_j(x_j,T_j) \\
x_{i+1} - x_{i} = \sum_{\substack{T_j=T_i \\ x_j = x_i}}^{r<tol} dx_j(x_j,T_j)
\end{aligned}
\]

where, using the formula for the inverse of a \(2 \times 2\) matrix:

\[
\begin{Bmatrix}
    dx_j \\
    dT_j
\end{Bmatrix} =
\frac{1}{\frac{\partial v}{\partial x}\frac{\partial s}{\partial T} - \frac{\partial s}{\partial x}\frac{\partial v}{\partial T}}
\begin{bmatrix}
    \frac{\partial s}{\partial x} &
    -\frac{\partial v}{\partial T}\\
    -\frac{\partial s}{\partial x}&
    \frac{\partial v}{\partial T}
\end{bmatrix}
\begin{Bmatrix}
    dv_j \\
    ds_j
\end{Bmatrix}
\]

with all partial derivatives as a function of the guess vapor quality and temperature \(f(x_j,T_j)\) and

\[
\begin{Bmatrix}
    dv_j \\
    ds_j
\end{Bmatrix} =
\begin{Bmatrix}
    v_{i+1} - v(x_j,T_j) \\
    s_{i+1} - s(x_j,T_j) \\
\end{Bmatrix}
\]

Ready for next step¶

Once the quantities for \(T_{i+1}\) and \(x_{i+1}\) are estimated, the new value for pressure \(p_i\) and mass flow \(\dot{m}_i\) are:

\[
\begin{aligned}
    p_{i+1} &= p_{Sat} (T_{i+1}) \\
    \dot{m}_{i+1} &= \frac{\dot{m}_0}{p_0} \cdot p_{i+1} \\
\end{aligned}
\]

The last equation assumes that the mass flow ratio and pressure is constant along the entire extraction.

Loop Termination¶

The extraction is over once the vapor quality \(x\) reaches the value of \(1\), which is when the tank contains only vapor and the system is not biphasic anymore.