Building The Formula For Calculating The Vapor Of The Lpg Liquid Generating In The Explosion Risk Of Lpg Tank

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BUILDING THE FORMULA FOR CALCULATING THE VAPOR OF THE LPG LIQUID GENERATING IN THE EXPLOSION RISK OF LPG TANK Ly Ngoc Minh1; Nguyen Van Quan2; Đinh Xuan Thang3 ABSTRACT: The paper introduces the way for building of the formula to calculate the vapor of the Liquefied Petroleum Gas (LPG) generating in the explosion risk of LPG tank. INTRODUCTION: Liquid stored under pressure above their normal boiling point temperature present substantial problems because of flashing. If the tank, pipe, or other containment device develops a leak or rupture, the liquid will partially flash in to vapor, sometimes explosively. When we assessment the risk of the gas explosion from the tank, pipe, equipment of the LPG, we need to calculate the vapor of the LPG generating from the LPG liquid. This paper will introduce the way to form the formula to calculating the LPG vapor generating from the LPG liquid in the explosion. I.CONDITIONS 

LPG equipment place in the atmosphere, pressure is 1 atm (1,01 bar);



Commercial LPG;



Assumes constant physical properties over the temperature range.



Maximum allow working temperature of the LPG tank is 50oC (according to the

Vietnam standard) II.DETERMINE THE RATE OF LPG LIQUID FLASHING Flashing occurs so rapidly that the process is assumed to be adiabatic. The excess energy contained in the superheated LPG liquid vaporizes the LPG liquid and lowers the temperature to the new boiling point. If m LLPG is the mass of LPG liquid, C pLPG , L ,1atm the heat capacity of the LPG liquid (energy/mass.deg.), To be the temperature of the LPG liquid before depressurization, and Tb is the depressurized boiling point of the LPG liquid, the excess energy contained in the superheated LPG liquid is given by

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Hochiminh University of Industry, Vietnam Nation Institute of Occupational Safety and Health, Vietnam 3 Hochiminh City Institute of Resource and Environment, Vietnam 2

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

LPG LPG Q  m LLPG  C pLPG , L ,1atm  To , 50 o C  TB ,1atm



(a)

This energy vaporizes the LPG liquid. If rLLPG V ,1atm is the heat of vaporization of the LPG liquid, the mass of LPG liquid vaporized mv is given by m LLPG V 

Q LPG L V ,1atm

r





LPG LPG m LLPG  C pLPG , L ,1atm  To , 50 o C  TB ,1atm

r

The fraction of the LPG liquid vaporized is f LLPG V 



LPG L V ,1atm



LPG LPG C pLPG m LLPG , L ,1atm  To , 50 o C  TB ,1atm V  m LLPG rLLPG V ,1atm



(b)

(c)

III.DETERMINE THE LPG VAPOR GENERATING The vapor of LPG generating is LPG V

V



LPG LPG 250  mLLPG  TB ,1atm LPG  V  C p , L .1atm   T0 LPG  LLPG ,1atm  r1atm

(d)

IV.CASE STUDY o The tank contents the Propane (C3H8), capacity is the 100 ton. C pLPG , L ,1atm  2.25kJ / kg C ;

T 

LPG 0 Max





 50 o C ; TBLPG  42.1o C ; rLLPG V ,1atm = 427.8 kJ/kg.

Substituting in to Eq. c gives

f LLPG V 





LPG C pLPG  TBLPG m LLPG 2.25  50   42.1 , L ,1atm  To V    0.484 LPG LPG 427.8 mL rL V ,1atm

That is, only 48.4% of the mass of the original LPG liquid (in this case, propane) is vaporized. So, the LPG vapor generating is LPG V

V



LPG LPG 250  mLLPG  TB ,1atm LPG  V  C p , L .1atm   T0 LPG  LLPG ,1atm  r1atm



250  0.48 100 103  23529 m3 510

V.CONCLUSION AND SUGGESTION 

There is only part of the mass of the original LPG liquid is vaporized.



The paper introduces the way to building the formula to can calculate the vapor

generating from the explosion risk of the LPG tank.

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

The result of this calculation can use to determine the dispersion of the LPG vapor

in the air when assess the risk of the gas explosion from the tank, pipe, equipment of the LPG. 

Research applying the result of the study to another liquid.