Question

After being purged with nitrogen, a low-pressure tank used to store flammable liquids is at a total pressure of 0.03 psig.

a. If the purging process is done in the morning when the tank and its contents are at 55°F, what will be the pressure in the tank when it is at 85°F in the afternoon?

b. If the maximum design gauge pressure of the tank is 8 inches of water, has the design pressure been exceeded?

Answer 1

a. If the purging process is done in the morning when the tank and its contents are at 55°F, what will be the pressure in the tank when it is at 85°F in the afternoon?

Answer: The volume of tank is constant and the moles of nitrogen gas purged is also constant. so

T_(K) = (T_(°F) + 459.67)× 5/9

T1 = 55 ⁰F = 285.93 K

T2 = 85 ⁰F = 302.59 K

P1 = 0.00204 atm

P2 = 0.00216 atm = 0.0317 Psig

b. If the maximum design gauge pressure of the tank is 8 inches of water, has the design pressure been exceeded?

maximum design gauge pressure =

h = 8 inches = 8 X 0.0254 m = 0.2032 meter

g = 9.8 m/s²

Pressure = 0.2032 X 1000 X 9.8 = 1991.36 pascal = 0.144 psig

Design pressure exceeded.