Question

Ethane (C2H6) is combusted in 18.1% excess oxygen (O2) in an adiabatic reactor. The feed of ethane enters the reactor at 25.8 mol/hr and 93.7°C. The stream exiting the reactor contains ethane, oxygen, carbon dioxide (CO2), carbon monoxide (CO) and water (H2O), so ethane reacts by both complete and partial combustion reactions. The conversion of ethane is 15.8% and the conversion of oxygen is 12.3%. Assume that pressure effects are negligible.

a.Draw and label a process flow diagram. Number each stream and label the components (E=ethane; O2=oxygen; CO2=carbon dioxide, CO=carbon monoxide; W=water).

b.Calculate the component molar flow rates (mol/hr) for all of the exiting components.

c.Calculate the adiabatic flame temperature (°C) for the reactor.

d.If the reactor's insulation is damaged and the reactor could lose heat to the surroundings, would the exit temperature increase, decrease, or stay the same?

Answer 1

part b

Ethane combustion reaction is

C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(l)

C2H6 + 5/2 O2 - > 2CO + 3H2O

In the Feed line moles are

E = 25.8

25.8 moles theoretically need (7/2)* 25.8 moles of O2

but we have 18.1 % excess O2 so

moles of O2= 3.5 * 1.18 * 25.8 moles = 106.554

in the exiting streams

ethane conversion is 15.8 %

so

E reacted = .158 * 28.5 = 4.503

Also give that O2 conversion = 12.3 %

Moles of O2 reacted = .123 * 106.554 = 13.106 moles

let moles of E used in reaction 1 = a

let moles of E used in reaction 2 = b

a+b= 4.503

3.5 a+ 2.5 b = 13.106

a= 1.8485 b = 2.6545

so exit stream contain

CO2 = 1.8485 * 2 = 3.7 moles

CO= 2.6545 * 2 = 5.31 moles

H2O = 3*(a+b) = 13.51 moles

E= 25.8-4.503 = 21.297 moles

O2= 106.554- 13.106 = 93.45 moles

part c

ehthalpy of combustion of E is 1560.7 Kj/mol

specific heat vlaues of various comp are

E= 65.46 J/mol K

O2= 30 J/mol K

CO = 28.56 J/molK

CO2 = 38.01 J/(mol K)

H2O =75.3 J/mol K

.Let flame temp be T

let us make Heat balance

Heat in = Heat out
(25.8*65.46*93.7)+ (106.554*30*93.7)+ (1560.7*1000*4.503) = ((21.297*35.46)+(93.45*30)+(5.31*28.56)+(3.7*38.01)+(75.3*13.51) * T)

T = 1357 ^oC adiabatic temp flame temp

d

If insulation is damaged heat wwill be lost and this value will be subtracted in LHS of above equation so flame temp will decrease