Liquefied natural gas (LNG) is being produced in the world in increasing amounts. It has a high ener — Physical Chemistry Chemistry Question
Liquefied natural gas
Liquefied natural gas (LNG) is being produced in the world in increasing amounts. It has a high energy density in comparison with the compressed natural gas, so that liquefaction is advantageous for transportation over long distances, especially by sea. The main component of LNG (> 95%) is methane. The dependence of the boiling point of methane on pressure is well described by the empirical equation:
log (p/bar) = 3.99 – 443 / (T/K – 0.49)
What is the boiling point of methane at atmospheric pressure?
Model Answer
T = 0.49 + 443 / (3.99 – log p) = 111.5 K
Estimate how many times larger is the energy density per unit volume of liquefied methane than that of gaseous methane in cylinders under 300 bar pressure at room temperature (298 K). (The compressibility factor of methane at these conditions is close to 1, and thus the ideal gas law can be applied.)
Model Answer
Under 300 bar pressure at 298 K, 40000 m3 have the mass m = pV M / RT = 7.75 × 10^6 kg, or 7750 tons. Thus, LNG has 16800 / 7750 = 2.17 times larger energy density.
A phase diagram of methane given above is plotted in the coordinates ‘logarithm of pressure in bars (log p) – internal energy (U)’. It is based on the experimental data [Setzmann and Wagner, 1991]. The area encircled by black dots (data points) corresponds to equilibrium coexistence of liquid and gaseous methane, while out of it methane is either completely liquid or completely gaseous. Using the diagram, answer the following questions:
What is the enthalpy of vaporization of methane under conditions of its transportation?
Model Answer
The pressure inside the tank is the saturated vapor pressure of methane at the given temperature: log p = 3.99 – 443 / (273.15 – 159 – 0.49) = 0.0924, p = 1.24 bar. Using the diagram, one can calculate the distance between two black points at log p = 0.1 to be about ∆U = 7.2 kJ mol–1. Thus, ∆H = ∆U + RT = 8.1 kJ mol–1.
What percent of methane will evaporate after 15 days of sailing, given that the total heat leakage through the cryogenic tank is 50 kW?
Model Answer
Total heat obtained by methane is Q = 50000 × 3600 × 24 × 15 = 6.48 × 10^10 J. It will lead to evaporation of m = Q / ∆H × M = 1.28 × 10^5 kg, or 128 tons, or 0.76 % of methane.
For long term storage of LNG, it was suggested not to discharge evaporating methane but to seal the tank. A pilot experiment has been conducted with the same tank initially filled with liquid methane (at temperature –159 °C) up to exactly a third of its volume. After 9 months of storage, the pressure inside the tank grew up to 16.4 bar. Which part of methane has evaporated inside the tank? Assume that the heat leakage is the same as in the previous question.
Model Answer
Total heat obtained by methane is Q = 50000 × 3600 × 24 × 9 × 30.5 = 1.19 × 10^12 J. It will lead to an increase of the internal energy per mole of methane by ∆U = (Q / m) × M = 1.19 × 10^12 / (16800000 / 3) × 0.016 = 3.39 kJ mol–1. From the diagram, the initial internal energy at –159 °C is 0.1 kJ·mol–1. The abscise of the point corresponding to the final state on the diagram will thus be approximately 0.1 + 3.4 = 3.5 kJ mol–1. The ordinate is log p = 1.2. The ratio of the lengths of the line segments from this point to the borders of the phase coexistence curve (blue and red line segments in the figure below) is equal to the ratio of the number of moles of methane in vapor and liquid phases. One can find that about 6 / 51 = 12 % of methane is in the gas phase.
What can be the maximum temperature of the liquefied methane? What would be the pressure in the reservoir containing it?
Model Answer
The maximum possible temperature is the critical temperature of methane, corresponding to the maximum of log p vs U curve. From the diagram we find log pc = 1.65, then pc= 44.7 bar and Tc = 0.49 + 443 / (3.99 – log p) = 190 K.