What is the specific gravity of LNG at 5 bar and -134 deg C?

  • Thread starter Thread starter nikcosmo
  • Start date Start date
  • Tags Tags
    Gravity Specific
AI Thread Summary
The discussion focuses on calculating the specific gravity of liquefied natural gas (LNG) at 5 bar and -134 degrees Celsius using a composition certificate. The user notes that the expected density is around 0.4 kg/m^3 but struggles to achieve this value using the ideal gas equation, which is inappropriate for liquids. Suggestions include consulting density tables and engineering handbooks, as theoretical equations for liquid density often rely on experimentally determined coefficients. The consensus emphasizes the need for empirical data rather than theoretical calculations for accurate density assessment. Accurate density values are essential for determining LNG charging rates per liter.
nikcosmo
Messages
2
Reaction score
0
Hi Everyone...

I'm trying to calculate the specific gravity of a sample of LNG using a compostion certificate and temperature and pressure variables of 5 bar at -134 deg C saturated. Its do perform a check for the value used to calculate how much is charged per litre by converting density to litres. The density should be around 0.4 kg/m^3 but i can't seem to get a close answer to that using the ideal gas equation, pv = nrt.

Infor on composition certificate;
Gas composition;
Nitrogen ; 1.18%
Methane; 96.66%

Gas density at 15 deg C = 0.699 k/m^3
Ave MW = 16.487 g/mol

Can someone please help? thanks.
 
Physics news on Phys.org
pV=nRT is and ideal GAS equation, LNG is a LIQUEFIED natural gas. You can't use theory that describes gaseous state to liquid state.

--
 
Any suggestions to which way i should approach it then? cheers.
 
Only approach I can think of is to find density tables for LNG, engineering handbooks should have these. There are no theoretical equations that would allow calculation of density of a liquid (and even when such equations exist, they make heavy use of experimentally determined coefficients - so we are back to the handbook and tables).

--
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
View full post »

Hot Threads

Recent Insights

Back
Top