100 kmol/h of a gas stream containing 36.3 mole % n-hexane in nitrogen is fed to
ID: 989395 • Letter: 1
Question
100 kmol/h of a gas stream containing 36.3 mole % n-hexane in nitrogen is fed to a condenser at 75 degree C and 3 atm. The product gas which contains 2 mole% n-hexane in nitrogen leaves at 0 degree C and 3 atm. The condensate is liquid n-hexane at 0 degree C Calculate the rate of heat (kW) which must be transferred from the condenser (Show details of your calculations steps including the Hypothetical process paths). The boiling point of n-hexane is 68.74 degree C (at 1 atm) and its latent heat of vaporization is 28.85 kJ/molExplanation / Answer
MOLE BALANCE:
hexane inlet mole rate = 36.3 kmol/h
Nitrogen inlet mole rate = 63.7 kmol/h
All nitrogen is got out through gas stream.
Mole rate of gas stream = 63.7 kmol/h /(1-0.02) == 65 kmol/hr
mole rate of hexane in gas = 65-63.7 = 1.3 kmol/hr
mole rate of hexane in liquid = 36.3 kmol/hr - 1.3 kmol/hr = 35 kmol/hr
Calulate boiling point of Hexane at 3 atm using following equation:
ln(P1/ P2) = - delta H /R (1/T1 -1/T2 )
ln(3/1) = (28.85x 1000 J/mol ) / 8.314 x (1/T1 -1/341.89 K)
T1 =307.85 K == 34.7 oC
75 oC = 348.15 K
Heat required for hexane = heat lost by total hexane gas from (348.15 K to 307.85 K) +heat lost by 1.3 kmol gas from (307.85 K to 273.15 K ) + Latent heat of vapor + heat lost by liquid (307.85 K to 273.15 )
Integrate Cp values for the given temperatures and calculate mc dT
Heat lost by nitrogen = 63.7 kmol/hr x integrate (Cp from 348.15 to 273.15 )
Add total hexane + nitrogen heat
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