Question

A steam reformer operating at 650C and 1 atm uses propane as fuel for hydrogen production. At the given operating conditions, the reforming reaction can be approximated as a complete reaction with good accuracy, while the shifting reaction is at equilibrium. The equilibrium constant for this reaction is KP=2.03 at 650C and the overall reforming and shifting reactions are endothermic.
(a) Determine the concentrations of water ([H20]) and hydrogen ([H2]) if the concentrations of propane, carbon monoxide and carbon dioxide in the outlet of reformer are 0.005, 0.08 and 0.09 respectively.


(b) Without any calculations, how does increasing the operating temperature of the reformer affect the concentration of hydrogen in the outlet?


(c) What is the maximum hydrogen yield?

Answer 1

Answer:

Explanation:

a) for shifting reactions,

Kps =  ph2 pco2/pcoph20

=[h2] [co2]/[co] [h2o]

h2 + co2 + h2O + co + c3H8 = 1

it implies that

H2 + 0.09 + H2O + 0.08 + 0.05 = 1

solving the system of equation yields

H2 = 0.5308,

H2O = 0.2942

B)  according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.

c) to calculate the maximum hydrogen yield , both reaction must be complete

C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)

CO + H2O ⇒ CO2 + H2 ( SHIFTING)

C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)

SO,

Maximum hydrogen yield

= 10mol h2/3 molco2 + 10molh2

= 0.77

⇒ 77%