A 50/50 mixture of 1-propanol Vapor product and 2-propanol is separated nv mol b
ID: 524554 • Letter: A
Question
A 50/50 mixture of 1-propanol Vapor product and 2-propanol is separated nv mol by a continuous single-stage equilibrium flash tank. The mol 1-propanol/mo 80.00°C, 510.0 mmHg 1 mol basis liquid entering the tank is at 91.0°C. The tank operates at Liquid feed at 91.0°C 80.00°C and 510.0 mmHg and Liquid product product streams emerge 0.5 mol 1-propanollmol at these conditions. 0.5 mol 2-propanol/mol mol n x mol 1-propanoumol The vapor pressures of heat, Q Y x mol 2-propanollmol 1-propanol and 2-propanol are given by the Antoine equation and the mixture follows law. Within the process conditions, the specific enthalpies of 1-propanol given in the equations below. kJ propanol H 0.2 x T C +5.2 mol kJ 2-propanol H 0.281 x TI o C 7.0294 mol kJ 1 -propanol H 0.0849 x T o C 53.577 v mol kJ 2-propanol v H 0.0769 x T o C 34.778 mol Find the composition of the product streams, the moles in each product stream (using a 1 mole feed basis) and the required heat input, Q, by doing the following steps.Explanation / Answer
Solution:
By using dew and bubble point the moles of the liquid and vapour are computed as below,
T = 80 C = 353.15 K
1-Propanol --> P1sat = 400 mm Hg @ 353.15 K
2-Propanol --> P2sat = 692.017 mm Hg @ 353.15 K
moles of vapour liquid, (basis as 100 lt)
(400+692.017)(100*0.6455) = nL*62.36367*353.15
nL = 3.2 moles
moles of vapour product, (basis as 100 lt)
(400+692.017)(100*0.447)=nv*62.36367*353.15
nv = 2.216 moles
now converting to 1 mole basis so the answer will be,
nL = 3.2 / (3.2+2.216) = 0.59 moles
nv = 2.216 / (3.2+2.216) = 0.409 moles
Q = sum of H and by plugging T = 80 C in the given H equation and computing gives,
Q = 137.9496 kJ
Based on saturated pressure data,
minimal pressure = 400 mm Hg
maximum pressure = 692.017 mm Hg
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