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 ^{0}F = 285.93 K

T2 = 85 ^{0}F = 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/s^{2}

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

Design pressure exceeded.

CHEMICAL ENGINEERING QUESTION! I WILL RATE AND COMMENT!
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 the pressure be in the tank
when it is 85 F in the afternoon?
B) If the maxiumum design gauge pressure of the tank is 8 inches
of water,...

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