A stream of hot dry nitrogen flows through process unit that

A stream of hot dry nitrogen flows through process unit that contains liquid acetone. A substantively portion of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the recovery unit at 205degrees Celsius and 1.1 bar and enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at 10 degrees Celsius 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 1m^3 of gas fed to the condenser the mass of acetone condensed (kg) and the volume of gas leaving the condenser (m^3). (b) Suppose the volumetric flow rate of the gas leaving the condenser is 20.0m^3/h. Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.

Solution

solution:

Acetone = 58.08 g/mol
Nitrogen = 14.0067 g/mol

Condenser influent
---------------------------
205 C @ 1.1 bar (combined gas)

T = 205º C
T = 205º C + 273.15
T = 478.15 K

P = 1.1 bar
P = 110,000 Pa + 101,325 Pa
P = 211,325 Pa, absolute

Pi, partial pressure of acetone = 0.110 bar
Pi = 11,000 Pa + 101,325 Pa
Pi = 112,325 Pa, absolute

R = 8.314472 J/K·mol

n = PV/RT
n = ( 211,325 Pa )( 1 m³ )/( 8.314472 J/K·mol )( 478.15 K )
n = 53.156mol

Acetone mass in influent
-----------------------------------
(from acetone mole fraction within influent)
Xi = Pi/P
Xi = 112,325 Pa / 211,325 Pa
Xi = 0.53153

mi = n Xi Mi
mi = ( 53.156 mol )( 0.53153 mol acetone/total mol )( 58.08 g/mol )
mi = 1640 g
mi = 1.64 kg

Crosscheck...
mi = ( Mi Pi V ) / ( R T )
mi = ( 58.08 g/mol )( 112,325 Pa )( 1 m³ ) / ( 8314.472 J/K·mol )( 478.15 K )
mi = ( 1640 g )( 1 kg / 1000 g )
mi = 1.64 kg

Condenser effluent
--------------------------
10 C @ 40 bar (combined gas)

T = 10º C
T = 10º C + 273.15
T = 283.15 K

P = 40 bar
P = 4,000,000 Pa + 101,325 Pa
P = 4,101,325 Pa, absolute

Pi = 0.133 bar
Pi = 13,300 Pa + 101,325 Pa
Pi = 114,625 Pa, absolute (partial pressure of acetone)

Effluent acetone mole fraction
--------------------------------------...
Xi = Pi/P
Xi = 114,625 Pa / 4,101,325 Pa
Xi = 0.02795

Acetone condensate
------------------------------
Xi = influent - effluent
Xi = 0.53153 - 0.02795
Xi = 0.50358

mi = ( 53.156mol )( 0.50358 moles acetone/total moles )( 58.08 g/mol )
mi = 1554.7027 g
mi = 1.554kg

Moles Effluent
---------------------
n = influent - condensate
n =53.156mol - ( 53.156mol )( 0.50358 moles acetone/total moles )
n = 26.387mol

Effluent Volume
-----------------------
V = nRT/P = ( 26.387 mol )( 8.314472 J/K·mol )( 283.15º K ) / ( 4,101,325 Pa )
V = 0.01514 m³

Mass flow of Acetone
-------------------------------
Q = 20 m³/hr

From above, for every 1 m³, 1.554kg of acetone vaporized
into the condenser\'s influent stream

m\' = ( 1.154 kg/m³ )( 20 m³/hr )
m\' = 23.08 kg/hr

Answers
------------
acetone condensate = 1.554 kg

condenser discharge volume = 0.01514 M^3

acetone vaporization rate = 23.08kg/hr