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Disclaimer: I am not a physicist, but I am trying to unravel this problem and it is interesting me more and more the more I work and think about it.

I have the following problem: There is a Cryogenic tank with Liquified Natural Gas (LNG) in the tank. The tank has a volume $V_{tank}$ in cubic meters. In the tank is an amount of LNG still in the liquid phase with mass $m_{lng}$ in kilogrammes. The temperature of the LNG ranges between -160 and -120 degrees celcius, and constantly is in a boiling phase. In the 'ullage' of the tank is boil-off gas with a pressure between 3 and 10 bar.

In my calculations it's okay to ignore any influences from outside the tank.

From what I understand we have the following:

  • the LNG and boil-off gas are in thermal equilibrium (I don't know if this is the correct term): the amount of evaporation is equal to the amount of condensation.
  • The top layer of LNG is warmer than the LNG in the bottom of the tank.
  • The temperature of the top layer of the tank is equal to the temperature of the gas
  • For this reason we can calculate the temperature of the gas and the top layer of LNG using the ideal gas law.

I would like to know what happens to the pressure in the tank and the temperature of the LNG after new (colder) LNG is offloaded into the tank. I do not care about thermodynamics over time, I only want to calculate a equilibrium.

I assume the following:

  • a heat capacity $c$ of 2.1 kJ/kg.K
  • a latent heat of vaporization/condensation $L$ of 510 kJ/kg
  • a gas constant of 518.3 J/kg.K

Furthermore I have the following known variables from sensors:

  • The pressure in the tank $P_{old}$
  • The bottom temperature of the LNG $T_{old}^{bottom}$
  • The current mass in the tank $m_{old}^{lng}$

And the following known:

  • The volume of the tank in cubic meters. This is around 60 m3.
  • The current mass of LNG (rough estimate based on tank liquid level and density). (Between 5000-15000kg
  • The amount of new mass to be offloaded into the tank (this is agreed upon beforehand) between 5000-15000kg
  • The temperature of the new gas (this is known when the LNG is loaded in the offload trailer at the terminal, but could also be calculated using the tank trailer pressure sensor. Here I assume that due to the sloshing the LNG is well-mixed.)

Currently my idea of calculating the new equilibrium is as follows:

  • Calculate the temperature of the LNG in the top layer using a conversion table my company has for LNG temperature and tank pressure, giving: $T_{old}^{top}$. I realise that in order to compute this temperature using the gas law I would need the mass of the gas, which I do not have.

  • Calculate the mass of the gas using the gas law: $$m_{old}^{gas} = \frac{P_{\text{old}} \cdot V_{\text{tank}}}{R \cdot T_{old}^{top} }$$

  • Make a guesstimate of the average temperature of the LNG by assuming the top 1 tonne of the LNG has the top temperature and the rest has the bottom temperature, resulting in $T_{old}^{avg}$.

  • Calculate the amount of gas that will condensate: $$m_{new}^{condensation} = \frac{m_{new} \cdot c \cdot (T_{old}^{avg} - T_{\text{new}})}{L}$$

  • Calculate the new mass of the gas: $$m_{new}^{gas} = m_{old}^{gas} - m_{new}^{condensation}$$

  • Calculate the new mass of LNG: $$ m_{new}^{lng} = m_{old}^{lng} + m_{new}^{lng} + m_{new}^{condensation}$$

  • Calculate the new LNG average temperature:

$$T_{final} = \frac{m_{old}^{lng} \cdot c \cdot T_{old}^{avg} + m_{new}^{lng} \cdot c \cdot T_{new} + m_{new}^{condensation} \cdot L}{m_{new}^{lng} \cdot c}$$

  • Using this new average temperature, calculate the new tank pressure:

$$P_{new} = \frac{ m_{new}^{gas} \cdot R \cdot T_{final} }{ V_{tank} }$$

This model's estimate is not that far off from what I am seeing in practice. E.g.: 4000 kg in tank, 4000 kg to offload, -150 C temperature of new gas, current pressure 9 bar -> results in a pressure of 7 bar. Offloading 10000 kg results is around 4.5 bar.

This model is a first attempt and has some flaws that I do not know how to handle:

  • After the new LNG has entered the tank, we calculate the average new temperature of the LNG. However, after a while, the top layer will become warmer again and a new new gas-liquid equilibrium will form with a different pressure
  • Currently I use the total volume of the tank in the ideal gas law calculations. I think it might be better to calculate the volume of the ullage before the offload and after the offload. This can be done with an estimation of the density of the LNG, estimating the volume of the LNG and subtracting it from the volume of the tank. Naturally, the ullage volume will be smaller after the offload. Implementing this gives a large pressure (e.g. 21 bar after offload, while the tank is only made for max 10 bar). My guess would be that the increased pressure due to smaller ullage volume would lead to extra condensation? I find this difficult to model.
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  • $\begingroup$ "The temperature of the LNG ranges between -160 and -120 degrees celcius... The top layer of LNG is warmer than the LNG in the bottom of the tank." Why is the top layer warmer than the bottom layer? Is it because LNG far below the boiling point is added to the bottom routinely? In liquid helium cryostats usually the bottom is warmer than the top. Both the liquid helium on the top and the bottom are actively boiling, but the boiling temperature is higher on the bottom because the bottom is at higher pressure (pressure increases with depth in a fluid). $\endgroup$
    – AXensen
    Aug 4, 2023 at 12:47

1 Answer 1

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Your description is too long for me to go through it, so instead of commenting on your approach I will give an outline of the correct solution. This is a somewhat complicated problem because of its unsteady nature, but it is definitely solvable.

Say the tank at time $t$ contains mass $M$ at temperature $T$ and pressure $P$, and that we add the amount $dm$, which is at temperature $T_\text{in}$ and pressure $P_\text{in}$. The mass balance and the energy balance equations are $$\tag{1} dM = dm $$ $$\tag{2} dU^\text{tank} = H_\text{in} dm$$ The first equation says that the mass in the tank increases by the amount we add. The second equation says that the energy in the tank changes by the amount of enthalpy that enters the tank ($H_0$ is the specific enthalpy at the conditions in the inlet to the tank).

These two equations can be integrated from initial state to the state after an amount $\delta m$ has been added to the tank: $$\tag{3} U = H_\text{in}\frac{\delta m}{M_0+\delta m} + U_0\frac{M_0}{M_0+\delta m} $$ where: $$ \begin{array}{ll} U = U(T,P) & \text{specific internal energy in the tank at final $T$ and $P$}\\ M_0 & \text{initial mass in the tank} \\ U_0 = U(T_0,P_0) & \text{specific internal energy in the tank initially} \\ T_0 & \text{initial temperature in the tank}\\ P_0 & \text{initial pressure in the tank}\\ H_\text{in} = H(T_\text{in},P_\text{in}) & \text{specific enthalpy of LNG at inlet to the tank}\\ \end{array} $$ We also know the specific volume in the tank $$\tag{4} V = \frac{V^\text{tank}}{M+\delta m} $$ Equations (3) and (4) define the state completely, but it takes some tedious trial-and-error to calculate the pressure and temperature in the tank. It goes like this:

  1. Make a guess for the state in the tank. Since we know it is a V/L mixture, guess its temperature and the fraction of liquid. The pressure then is the saturation pressure at the guessed temperature.
  2. Using the above guess and with the help of thermodynamic tables (or equations of state or other similar equations), calculate the specific internal energy and specific volume.
  3. If the calculated $U$ and $V$ match these in Eqs. (3) and (4), we have the solution. Otherwise pick a different guess and continue.

Notice that we have two unknowns to guess, temperature and liquid fraction, but we also have two equations, Eq. (3) and (4), so the problem is mathematically sound, it just takes some tedious iterations to get the answer.

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  • $\begingroup$ Thanks a lot for this answer. This has actually helped me solve the question. Now I am struggling with the 'reverse' operation: i.e. we are taking vapour out of the tank in order to reduce pressure. The thing that is difficult is that there is only a limited mass of vapour in the tank (e.g. 400 kg) and in order to reduce a certain amount of pressure (e.g. 3 bar), we would need to take out more than 400kg of vapour, as the 400kg would only reduce around 1.5 bar of pressure. How would I best model this? $\endgroup$ Aug 31, 2023 at 6:15
  • $\begingroup$ This is also a standard problem in chemical engineering thermodynamics (see this book, p319). I'll be happy to walk you through the solution if you post as a new question. $\endgroup$
    – Themis
    Sep 1, 2023 at 22:58
  • $\begingroup$ Hi Themis, thank you I will take a look. I might have a different version. Do you mean problem 7.6? $\endgroup$ Sep 25, 2023 at 14:51
  • $\begingroup$ Or I guess you mean problem 6.43, as 6.42 is very similar to my previous question and your previous answer. $\endgroup$ Sep 25, 2023 at 15:14
  • $\begingroup$ I posted it as a new question: physics.stackexchange.com/q/781877 $\endgroup$ Sep 25, 2023 at 15:48

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