Answer:
DIFFERENTIAL EQUATIONS
d(Ca)/d(t) = - K*Ca*Cb + (Fao/V)*(1-(Ca/1.5)) # Concentration of A
d(Cb)/d(t) = - K*Ca*Cb - (Fao*Cb)/(V*1.5) # Concentration of B
d(Cc)/d(t) = K*Ca*Cb - (Fao*Cc)/(V*1.5) # Concentration of C
d(Cd)/d(t) = K*Ca*Cb - (Fao*Cd)/(V*1.5) # Concentration of D
d(Ce)/d(t) = K*Ca*Cb - (Fao*Ce)/(V*1.5) # Concentration of E
# AUXILIARY EQUATIONS
K = 5.1 # Reaction Rate Constant, [dm3/mol.h]
r = K*Ca*Cb # Reaction Rate, [mol/dm3.h]
V = Vo + (Fao/1.5)*t # Volume, [dm3]
X = 1 - (Cb*V)/(0.75*Vo) # Conversion
Nc = 0.75*Vo*X # Moles of C formed
# INITIAL CONDITIONS
Ca(0) = 0
Cb(0) = 0.75
Cc(0) = 0
Cd(0) = 0
Ce(0) = 0
Fao = 6 # Feed of A, [mol/h]
Vo = 1500 # Initial Volume, [dm3]
t(0) = 0 # Initial Time, [h]
t(f) = 250 # End Time, [h]
Explanation:
All the data was performed on polymath educational. If you have the educational version of plymath copy paste the following code to get the required values of moles of glycol formed. You can also get it for free from the polymath education free website which is free for 15 days.
