Question

A stream of liquid n-pentane flows at a rate of 50.4 L/min into a heating chamber, where it evaporates into a stream of air 15% in excess of the amount needed to burn the pentane completely. The temperature and gauge pressure of the entering air are 336 K and 208.6 kPa. The pentane-laden heated gas flows into a combustion furnace in which a fraction of the pentane is burned. The product gas, which contains all of the unreacted pentane and no CO, goes to a condenser in which both the water formed in the furnace and the unreacted pentane are liquefied. The uncondensed gas leaves the condenser at 275 K and 1 atm absolute. The liquid condensate is separated into its components, and the flow rate of the pentane is measured and found to be 3.175 kg/min.
Calculate
(a) the fractional conversion of pentane achieved in the furnace
(b) the volumetric flow rates (Umin) of the feed air
(c) the volumetric flow rates (Lmin) of the gas leaving the condenser

Answer 1

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