CAPE 3310: Reaction Engineering Paper

CAPE 3311 Reaction Engineering
Coursework 2
Name: Sharifah Irryanti Syd Hamdan
Student ID: 200802028
Date of submission : 4th December 2015
Supervisor: Dr Xiaojun Lai

Question 1

To determine the holding time to produce a desired product, which is 90% R, a graph of -1/rA against CA need to be plotted. For plug flow reactor, the area under the curve represents the holding time.
a) Plug flow reactor

Using integration method,

A = ∫_(C_A)^(C_A0)▒1/〖-r〗_A dCA = 1/k ∫_(C_A)^(C_A0)▒1/(C_A C_R ) dCA = 1/k ∫_(C_A)^(C_A0)▒1/(C_A (1-C_A)) dCA since CR =1 - CA

By partial fraction,
1/(C_A (1-C_A))= A/C_A + B/〖1-C〗_A
A

So, taking CA = 2.624 mol/L as the point where the reactor change, so size ratio of first reactor, V1 and second reactor, V2 = ([( 1.6437-0.01) (2.624-0.78)])/([(0.1452-0.01) (10-2.624)]) = 3.021

Conversion rate of first reactor, CA / CA0 = (2.624/10 ) x 100% = 26.24 %

Question 3

A R rR = k1CA
A + A S rS = k2CA2

a) Instantaneous fractional yield of R, φ (R/A)= (moles of R formed)/(moles of A reacted) = 〖dC〗_R/(-dC_A ) = r_R/〖-r〗_A = (k_1 C_A)/(k_1 C_A+2 k_2 C_A^2 ) Instantaneous fractional yield of R, φ (R/(R+S))= r_R/(r_R+r_S )= (k_1 C_A)/(k_1 C_A+ k_2 C_A^2 )

b) Since φ (R/A)= r_R/〖-r〗_A CR max = ∫_0^(C_AO)▒〖φ 〖dC〗_A 〗 = ∫_0^(C_AO)▒〖(k_1 C_A)/(k_1 C_A+2 k_2 C_A^2 ) 〖dC〗_A 〗 = ∫_0^(C_AO)▒〖k_1/(k_1+2 k_2 C_A ) 〖dC〗_A 〗 = ∫_0^(C_AO)▒〖1/([1+(2 k_2 C_A)/k_1 ]) 〖dC〗_A 〗

CR max = k_1/〖2k〗_2 [ln⁡〖(1+(2 k_2 C_A)/k_1 〗) ] 1¦0 = k_1/〖2k〗_2 [ln⁡〖(1+(2 k_2)/k_1 〗) ]
Let k1/k2 = K
CRmax = 1- 0.18 = 0.82 mol/L

CR = K/2 [ln⁡〖(1+(2 )/K〗) ]

Assume two values of K so that a graph of CR against K can be

The second reaction, which is decomposition of Y into Z is first order reaction, n =1.

X Y Z

- r Y = k CY where k2 = 1.5 hr-1
Since the first reaction is of zero order, k1 = C_Y/t = (100 mol/m^3)/(0.5 hr) = 200 mol/m3hr
Calculating K from k1 and k2 :
K = (k_2 C_XO)/k_1 = (1.5 ( 100 ))/200 = 0.75
CYmax/ CXO = (1 - e^(-K))/K = (1 - e^(-0.75))/0.75 = 0.70351
CYmax = 0.70351 x 100 = 70.351 mol/m3 tYmax = CXO / k1 = 100/200 tYmax = 0.5 hr

b) After half an hour, all X have been reacted into Y, which then dissolve into X. At t = 1 hr, only Y and Z exist.

When t > C_XO/k_1 CYmax/ CXO = (1 )/K (e^(K-k_2 t-e^(-k_2 t))) = (1 )/0.75 (e^(0.75-1.5(1) )- e^(-1.5(1) )) = 0.3323
Since all X has been converted into Y and Z, so X = Y + Z
CZ = [1- 0.3323] x 100 = 66.77 mol/m3

Question 5

First order A R

1. At equilibrium,

K = C_R/C_A = (C_AO X_Ae)/(C_AO (1-X_Ae))=X_Ae/(1-X_Ae )
X_Ae= K (1-X_Ae)
X_Ae= K – K