One hundred kmol/hr of a binary feed mixture at 25 o C containing 40 mol% n-hexa

Question

One hundred kmol/hr of a binary feed mixture at 25oC containing 40 mol% n-hexane and 60 mol% n-heptane are partially vaporized in an adiabatic flash drum at 111oC and 1520 mm Hg absolute pressure. Assuming Raoult’s Law applies and that the exiting vapor and liquid streams are in equilibrium, determine the quantity of energy (Q) in kW that must be supplied to the flash drum. Important data are provided below.

DATA:            Hexane liquid Cp = 189 J/mol/oC

                        Hexane vapor Cp = 143 J/mol/oC

                        Heptane liquid Cp = 212 J/mol/oC

                        Heptane vapor Cp = 166 J/mol/oC

                        Hexane ?Hvap = 28,900 J/mol at 69oC

                        Heptane ?Hvap = 31,800 J/mol at 98.4oC

HINT: First complete the material balances for the problem.

Explanation / Answer

Writing overall balance, nF= nV+nL, 100= nV+nL   , nV= 100-nL (1)

Writing C6 balance, F*C6=nV*yC6+ L*xc6 , 100*0.4= nV*yC6+ L*xc6    (2)

Since for solutions obeying Raoults law yC6P= xc6Pc6sat   and ycP= xcPcsat

Addition of the two equations gives P= xc6Pc6sat+ xcPcsat

and xc6 =1-xc

P= xc6Pc6sat+ (1-xc6)*Pcsat

Hexane : Log(PC6sat) = 7.01051- 1246.33/(t+232.988)

Log(PCsat) = 6.89677- 1264.90(t+216.544)

Both equations are valid for the temperature of 111 deg.c

where C6sat is in mm Hg and t in deg.c

log(PC6sat)= 7.1051- 1246.33/(111+ 232.988)

log(Pcsat)= 6.89677- 1264.90(111+216.544)

PC6sat : 2440 mm Hg, Pcsat= 1084mm Hg

Given P=1520 mm Hg= 2440*xc6+(1-xc6)*1084

1520 = 2440xc6-1084xc6+1084

1520-1084= xc6*(2440-1084)

Hence xc6= .32 and xc = 1-0.32= .68

And yc6= 0.3217*2440/1520=0.52 and yc= 1-0.52= 0.48

Eq.2 now becomes 40 = nV*0.52+ nL*0.3213(2A)

From Eq.1, 40= (100-nL)*0.52+nL*0.3213

Hence nL*(0.52-0.3213)= 52-40= 12

nL = 12/(0.52-0.3213)= 60 , nV= 100-60= 40

hence

Input : Hexane = 40 and heptane = 60

Out put : Vapor : Hexane = 40*0.52=21 and heptane= 40*0.48= 19

Liquid : Hexane : 40-21= 19 and heptane = 60-19=41

Heat input

Hexane need to be supplied with sensible heat up to 68.4 deg.c ,heat of vaporization   and then super heat from 68.4 to 111deg.c.

Sensible heat of hexane = 21*189*(69- 25)=174636 Joules/hr, super heat of vapor = 21*143*(111-68.4)=127928 joules /hr and enthalpy of vaporization= 21*28900 =606900 Joules/hr

Sensible heat of liquid remaining = 19*189*(111-25)= 155849 joules/hr

For heptane, sensible heat = 19*212*(98.4-25) =295655 joules/hr

Super heat= 19*166*(111-98.4)= 39740 joules/hr, enthalpy of vaporization = 19*31800= 606100 joules/hr

Sensilbe heat of remaining liquid = 41*212*(111-98.4)=109519/hr

Total heat= 174636+127928+606900+155849 +2956655 +39740+606100+109519 joules/hr=47779327 joules/hr =1.327KW

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