Problem 2. Ethane is fed to a dehydrogenation reactor at a rate of (NA)Fkg-mole/
ID: 718141 • Letter: P
Question
Problem 2. Ethane is fed to a dehydrogenation reactor at a rate of (NA)Fkg-mole/hr. The products are acetylene and hydrogen. A fractional conversion of ethane is fa%. ate of (NA)Fkg-mole/hr. The re C2Hs (g)CH2 2H2(g) Reactor (NAJF (NB)F (NA, (NB)P (Nc)p (a) Calculate the total molar flow rate (kg-mole/hr) of the product stream from the dehydrogenation reactor (b) Find the ratio of the kg-mole/hr hydrogen formed to the kg-mole acetylene formed (kg-mole/hr hydrogen formed/kg-mole/hr acetylene formed) in the product stream from the dehydrogenation reactor (c) Evaluate average molecular weight of the product stream from the dehydrogenation reactor (d) Determine the mass flow rate of acetylene in Kghr in the product stream from the dehydrogenation reactor Table. Operating conditions of a reactor Name (NAJr. kg-mole /hr | fa, % 230 82Explanation / Answer
Moles of ethane fed = 230 kmol/hr
Moles of ethane consumed = conversion x moles fed
= 0.82 x 230
= 188.6 kmol/hr
In the product stream
Moles of ethane in product = moles of ethane fed - ethane reacted
= 230 - 188.6 = 41.4 kmol/hr
Moles of acetylene = moles of ethane converted
= 188.6 kmol/hr
Moles of H2 = 2 x 188.6 = 377.2 kmol/hr
Part a
Total molar flow rate = moles of ethane + Moles of acetylene + Moles of H2
= 41.4 + 188.6 + 377.2
= 607.2 kmol/hr
Part b
In the product stream
Kmol H2 / kmol C2H2 = 377.2 / 188.6 = 2
Part c
Mol fraction of C2H6 = 41.4/607.2 = 0.068
Mol fraction of C2H2 = 188.6/607.2 = 0.31
Mol fraction of H2 = 377.2/607.2 = 0.62
Average molecular weight of product stream
Mavg = (Mole fraction x M) of ethane + (Mole fraction x M) of C2H2 + (Moles fraction x M) of H2
= 0.068 x 30 + 0.31 x 26 + 0.62 x 2
= 11.34 kg/kmol
Part d
Mass of acetylene formed = 188.6 kmol/hr x 26 kg/kmol
= 4903.6 kmol/hr
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