One hundred kmol/h of a binary feed mixture at 25 degrees celsius containing 40
ID: 1060304 • Letter: O
Question
One hundred kmol/h of a binary feed mixture at 25 degrees celsius containing 40 mol% n-hexane and 60mol% n-heptane are partially vaporized in an adiabatic flash drum at 111 degrees celsius and 1520 mmHg absolute pressure. Assuming Rault's Law applies and that the existing vapor and liquid streams are in equilibrium, determine the quantity of energy (Q) in kW that must be supplied to the flash drum.
1. One hundred kmol/hr of a binary feed mixture at 25eC containing 40 mol% n-hexane and 60 mol% n-heptane are partially vaporized in an adiabatic flash drum at 111 C and 1520 mm Hg absolute pressure. Assuming Raoult'sLaw applies and that the exiting vapor and liquid streams are in equilibrium, determine the quantity ofenergy (Q) in kW that must be supplied to the flash drum. Important data are provided below. Vapor nv (kmol/hr) F 100 kmolhr Liquid nL kmolhr) DATA: Hexane liquid Cp 189 J/mol C Hexane vapor CP 143 J/mol C Heptane liquid CP-212 JmolaC Heptane vapor CP 166 UmowC Hexane AH 28,900 J/mol at 69°C Heptane AHv 31,800 J/mol at 98.4°C HINT: First complete the material balances for the problem.Explanation / Answer
Assuming for per hour calculation:
No of moles (n-hexane ) = 40 % of 100 kmol = 40 Kmol (25 C)
No of moles (n-heptane) = 60 % of 100 kmol = 60 kmol (25 C)
Apply gas equation
n1 R T1 / P1 (at 250C) = n2 R T2 (1110C) / P2
Considering P1 = 1 atm (25 C), P2 = 2 atm (111 C)
moles of hexane at 111 = 62.08 kmol, moles of heptane at 111 = 93.125 kmol
moles in vapor state = 62 + 93 - 100 = 55 kmol
As per vapor liquid equilibrium curve mole fraction of hexane (vapor) is 0.7 and mole fraction of heptane (vapor) is 0.3
moles of hexane in vapor state = 38.5 kmol, moles of heptane in vapor state = 16.5 kmol
Q = heat supplied to hexane to achive boiling point + latent heat of vaporization of hexane + heat supplied to heptane to achieve boiling point + latent heat of vaporization of heptane + heat supplied to vaporized hexane to reach 1110C + heat supplied to vaporized heptane to reach 1110C
Q = 40 x 189 x (69-25) + 38.5 x (28,900) + 60 x 212 x (98.4-25) + 16.5 x (31,800) + 38.5 x 143 x (111 - 69) + 16.5 x 166 x (111 - 98.4)
Q = 3.48 x 109 J
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