Answer:
(a) the fractional conversion of pentane achieved in the furnace is 90% conversion
(b) the volumetric flow rates (Umin) of the feed air is 256 x 10³ 1/m
(c) the volumetric flow rates (Lmin) of the gas leaving the condenser is 404.9 x 103 l/min
Explanation:
a) Molecular weight of pentane = 72.15 g/mol
density of liquid pentane = 626 kg/m3
Flow rate of feed liquid nitrogen = 50.4 l/min
= 626*50.4*10-3
= 31.55 kg/min
= 31.55/72.15 kmol/min
= 0.4372 kmol/min
Pentane existng the burner = 3.175 kg/min
Fractional conversion = (31.55 - 3.175)/ 31.55
= 0.9 = 90% conversion
b)
C₂H₅ + 8O₂ -----------> 5CO₂ + 6H₂O
From the Stoichiometric reaction,
8 mol of O2 are used for combustion of 1 mol of pentane
for 0.4372 kmol/min of pentane = 8 * 0.4372 kmol/min of Oxygen will be required
= 3.49 kmol/min of O2
amount of air will be = 3.49/0.21 = 16.62 kmol/min
15% excess air = 16.62*1.15 = 19.12 kmol/min
assuming air to be ideal gas
V = nRT / P.........(1)
V = 19.12 X 8.314 X 336 / 208.6
= 256 m³ = 256 x 10³ 1/m
c) Oxygen:
Amount of pentane consumed = (31.55 - 3.175) = 28.375 kg/min = 28.375/72.15 = 0.3932 kmol/min......(2)
Amount of O2 consumed = 8*0.3932 = 3.146 kmol/min
Amount of O2 fed by air = 0.21*19.12 = 4.0215 kmol/mim
unused O2 left = 4.0215 - 3.146
= 0.8755 kmol/min = 19.75*103 l/min............ (using (1))
Carbon Dioxide:
1 mol of pentane = 5 mol of CO2
0.3932 kmol/min of pentane = 5*0.3932 kmol/min of CO2..................(from (2)
= 1.966 kmol/min
= 44.362*103 l/min..........................(using (1))
Nitrogen:
v = 0.79 x 19.12 x 8.314 x 275 / 100.325
= 340.8 m³ / min = 340.8 x 10³ 1/min
Total volumetric flow rate of gases leaving the condenser = (340.8 + 44.36 + 19.75) x 103 l/min
= 404.9 x 103 l/min
