Question

A steam of hot dry nitrogen flows through a process unit that contains liquid acetone. A substantial part of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the unit at 205 ◦C and 1.1 bar, are compressed and then enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at 10 ◦C and 40 bar. The partial pressure of acetone in the feed to the condenser is 0.100 bar, and that in the effluent gas from the condenser is 0.379 bar. Assume ideal gas behavior.

(a) Calculate, for a basis of m3 of gas fed to the condenser, the mass of ace- tone condensed (kg) and the volume of gas leaving the condenser (m3).
(b) Suppose the volumetric flow rate of the gas leaving the condenser is 20.0 m3/h. Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.

Answer 1

Answer:

a. V = 0.015 m³

b. ζ = 195.58 kg acetone / hr

Explanation:

a.

Assume ideal gas behavior can use the equation

n₁ = 1.1 bar / 1.013 bar = 1.085 * 1m³( 273 k / 478 k ) * ( 1 kmol / 274 m³)

n₁ = 0.0276 kmol

Pacetone = 0.1 bar ⇒ y₁ = 0.1 bar / 1.1 bar

y₁ = 0.09090

Pacetone = 0.379 bar ⇒ y₃ = 0.379 bar / 40 bar

y₃ = 9.475 x10 ⁻³

n₁ * y₁ = n₃ * y₃   ⇒ 0.0276 * ( 1 - 0.09090) = n₃ * ( 1 - 9.475 x 10⁻³)

n₃ = 0.0253 kmol

n₁ = n₂ + n₃  ⇒ n₂ = 0.0276 - 0.0253 = 2.30 x 10 ⁻³ kmol

Volume condesed

V = 0.0253 kmol * 22.4 m³ * ( 283 / 273 ) * ( 1013 bar / 40 bar )

V = 0.015 m³

b.

To calculate the rate vaporized

ζ = 20.0 m³ / hr *( 0.0276 / 0.014 kg * m³ ) *( 0.09090 kmol a / kmol  ) * ( 5808 kg a / kmol a  )

ζ = 195.58 kg acetone / hr