Question

one mole of a liquid solution containing 60.0 mole% benzene and the remainder toluene enters a flash column.The liquid product stream contains 47.0 mole% benzene and the vapor product stream contains 69.0 mole % benzene.The temperature of both output stream is at 93oC.If the operation is adiabatic,what is the temperature of the input stream that enters the flash column?

Answer 1

The overall material balance gives:

F = L + V

where:

F = Moles of Feed

L = Moles of Liquid in output stream

V = Moles of Vapour in output stream

Component Balance gives:

F.z = L.x + V.y

where:

z = molar composition of benzene in feed

x = molar composition of benzene in liquid stream

y = molar composition of benzene in vapor stream

We have;

F =1 mole

z = 60% = 0.6

x = 47% = 0.47

y = 69% = 0.69

Substituting these values in the two equations we get:

1 = L + V

0.6 = 0.47 L + 0.69 V

Substituting V = 1 - L in the above equation

0.6 = 0.47 L + 0.69 (1 - L)

0.6 = 0.47 L -0.69 L + 0.69

0.22 L = 0.09

L = 0.41 mol

V = 1 - L = 1 - 0.41 = 0.59 mol

Since the operation is adiabatic, no heat enters or leaves the column. Also; the process is at steady state; therefore there is no heat accumulation.

Therefore the heat balance equation gives:

Heat in = Heat out

mf.Cf.Tf = L.Cl.To + V.λ + V.Cv.To

where:

mf = moles of feed entering

Cf = heat capacity of feed

Tf = feed temperature

Cl = heat capacity of liquid stream

Cv = heat capacity of liquid stream

To = temperature of output stream

λ = latent heat of vaporisation of feed

For the mixtures:

Cf = z.Cbenzene + (1 - z).Ctoluene

Ctoluene in liquid = 157 J/mol oC

Cbenzene in liquid = 92.8 J/mol oC

Cf = 0.6 X 118 + 0.4 X 157 = 133.6 J/mol oC

Cl = x.Cbenzene + (1 - x).Ctoluene

Cl = 0.47 X 118 + 0.53 X 157 = 138.67 J/mol oC

Cf = z.Cbenzene + (1 - z).Ctoluene

Ctoluene in vapor = 103.7 J/mol oC

Cbenzene in vapor = 82.4 J/mol oC

Cv = y.Cbenzene + (1 - y).Ctoluene

Cv = 0.69 X 82.4 + 0.31 X 103.7 = 89 J/mol oC

λ = y.λbenzene + (1 - y).λtoluene

λbenzene = 33800 J/mol

λtoluene = 38000 J/mol

λ = 0.69 X 33800 + 0.31 X 38000 = 35102 J/mol

Subtituting all these values in the equation:

mf.Cf.Tf = L.Cl.To + V.λ + V.Cv.To

1 X 133.6 X Tf = 0.41 X 138.67 X 93 + 0.59 X 35102 + 0.59 X 89 X 93

133.6 Tf = 30881

i.e.; Tf = 30881 / 133.6 = 231 oC

which is the required feed temperature.