Problem 4: tuer eloi fash evaporator that separates a iquid feed stream containi

Question

Problem 4: tuer eloi fash evaporator that separates a iquid feed stream containing benzene (B) and toluene (T). (a) Use material balance equations together with Raoult's law to find the outlet flows and (b) Calculate the enthalpy compositions feed to the flash evaporator. The vapor pressures of benzene and toluene in mmHg are: of all streams in kJ/kg take as a reference the conditions of the 1203.531 T +219.888 log10 Pe = 6.89272- log10 6.95805-1346773 T+219.693 mol vapor product/s mole fraction of B mole fraction of T 100 mol feed /s 0.4 mole fraction of B 0.6 mole fraction of T T= 100 P 800 mmHg T= 50 mole fraction of B

Explanation / Answer

a) Vapor Pressure of Benzene at given condition = 10^(6.89-1203/(100+219)) = 1314 mmHg

Vapor Pressure of toluene at given condition = 10^(6.95-1346/(100+219)) = 537.7 mmHg

By Raoult's Law, Total Pressure = (Mole fraction of i in liquid Phase)*(Vapor Pressure of i)

Let Mole fraction of Benzene in liquid phase = x

Then, Mole fraction of toluene in liquid phase=1-x

=>800 = x*1314+(1-x)*537

Therefore, Mole fraction of bezene in liquid phase = 0.33

=>Mole fraction of toluene in liquid phase=1-x = 0.67

Partial Pressure of component i = Mole fraction of i in vapor phase *Total Pressure =(Mole fraction of i in liquid Phase)*(Vapor Pressure of i)

=>ybenzene*800=.33*1314

=>Mole fraction of benzene in vapor phase = 0.54

=>Mole fraction of toluene in vapor phase = 1-.54 = .46

Mole flowrate in, F = Mole flowrate in Vapor phase out, V+Mole flowrate in liquid phase out, l

=>F=100=V+l

Mole flowrate of component i in, F = Mole flowrate of component i in Vapor phase out, V+Mole flowrate of component i in liquid phase out, l

=>.4*100 = .54*V+.33*l

=>40=.54V+.33(100-V)

=>V=33.33 Mole/s

=>l=100-V=66.67 mole/s

b)Enthalphy of liquid phase w.r.t to Feed condition = Enthalpy required to rise temperature of Benzene from 50 to 100 C+Enthalpy required to rise temperature of Toluene from 50 to 100 C

Enthalpy required to rise temperature of Benzene from 50 to 100 C = Mole of benzene in liquid phase*Cp(specific heat of benzene)*(100-50)

Enthalpy required to rise temperature of Toluene from 50 to 100 C = Mole of toluene in liquid phase*Cp(specific heat of Toluen)*(100-50)

=>Enthalphy of liquid phase w.r.t to Feed condition = 66.67*.33*Cp(Benzene)*(100-50)+66.67*.67*Cp(Toluene)*(100-50)

Enthalphy of vapour phase w.r.t to Feed condition = Enthalpy required to rise temperature of Benzene from 50 to 100 C+Enthalpy required to rise temperature of Toluene from 50 to 100 C+Enthalpy required to Vaporise Benzene at 100C+Enthalpy required to Vaporise Toluene at 100C

=>Enthalphy of vapour phase w.r.t to Feed condition=33*.54*Cp(Benzene)*(100-50)+33*.46*Cp(Toluene)*(100-50)+33*.54*(Latent heat of Benzene)+33*.46*(Latent heat of Toluene)

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