Question

Assume 400 ft3/min (measured at STP) of a gas mixture (Stream F) at 150 °F and 1.70 atm is fed to a reactor. The mixture consists of 65 mol% methane and 35 mol% ethane. Air (Stream A) at 75 °F and 1.20 atm is fed to the reactor in 20% excess. Aside from the excess air, the flue gas from the reactor contains only CO2 and H2O(vapor), and comes out at 1300 °F and 1.0 atm. Assume all pressures are observed by reading a pressure gauge. Keep the units used in your solution consistent with those in the problem statement.

a). Draw a labeled flowchart. Be sure to show units for all rates and compositions. Use the names (F, A, etc.) given in the description.

b). Calculate the molar flow rate (lbmol/min) of the gas mixture (Stream F) fed to the reactor.

c). Solve for the required volumetric flow rate of the air stream. Express your answer as ft3/min of air at STP conditions.

d). Calculate the volumetric flow rate of the furnace stack gas in units of ft3/min.

Answer 1

The schematic diagram is shown below

B) Molar flowrate of gas mixture

V = 400 ft3/min = 11.335 m3/min

The gas is measured at STP

STP conditions

T = 32°F =273 K

P = 1 atm = 1.013×105 Pa

R = 8.314 J/gmol K

(PV/RT) = n

(1.013×105×11.335) /(8.314×273) =

505.892 gmol/min = 0.50589 kgmol/min =

= 1.12 lbmol/min

Molar flowrate = 1.12 lbmol/min

C)

Amount of methane in feed = 1.12(0.65) =

0.728 lbmol/min

Amount of ethane in feed = 1.12(0.35) =

0.392 lbmol/min

Complete combustion occurs

The reactions taking place are

From stiochiometry

0.728 lbmol methane requires 2(0.728) =

1.456 lbmol O2

0.392 lbmol ethane requires (7/2) (0.392) = 1.372 lbmol O2

Total required O2 = 1.372 + 1.456 = 2.828 lbmol

O2 is 20% excess

So total O2 = 2.828(1.2) = 3.3936 lbmol/min

Total air supplied = 3.3936/0.21 =

16.16 lbmol/min

At STP conditions

T = 32°F = 273 K

P = 1 atm

n = 16.16 lbmol/min = 7.33 kmol/min

V = nRT/P

V = 7.33(1000) (8.314) (273) /(1.013×105)

V = 164.235 m3/min

V = 5795.470 ft3/min at STP

D)

Exit gas analysis

Component	lbmol/min	mol%
O2	3.3936-2.828= 0.5656	3.23
N2	16.16(0.79)= 12.766	73.05
CO2	0.728(1) +(0.392) (2) = 1.512	8.65
H2O	0.728(2) +(0.392) (3) = 2.632	15.060
Total	17.4756	100

Total lbmoles = 17.4756

n = 17.4756 lbmoles = 7.9290 kgmol/min

Outlet temperature = 1300°F = 977.4 K

P = 1 atm

V = RTn/P

V = (8.314×977.4×7.9290(1000)) /(1.013×105)

V = 636.050 m3/min

V = 22444.65 ft3/min at outlet conditions

Please upvote if helpful