CEE 357 Win 2015 HW#2 Solutions
1. (a) Because the system of interest involves a first-order reaction taking place in a CSTR at steadystate, we can write: cout cin
1
1 K CSTR
CSTR
VCSTR
1 cin
1
K cout
Q
1
0.1 h
75
CSTR
L s 1
50 1
490 h
490 h 3600
s h 1 m3
1000 L
132,300 m3
(b) If the same extent of conversion was accomplished in a PFR, the required reactor volume would be: cout cin
exp
PFR
1 cout ln K cin
PFR
V
K
Q
1 ln 50
0.1 h
L
75
16.1 h s s
3600
h
39.1 h
1 m3
1000 L
10,560 m3
Note that the exponential equation for the PFR is applicable because we know that the reaction is first order. The PFR is obviously a much more efficient reactor to accomplish the treatment goal.
(c) A PFR can be thought of as a batch reactor that is on a conveyor belt, or equivalently.
Therefore, if the initial concentration in the batch reactor is the same as the influent to the
PFR, the concentration in the batch reactor after time t will be the same as that in the PFR when the water exits (i.e., after it has been in the PFR for time ). We saw in part (b) that
PFR would have to be 16.1 h to achieve 98% destruction of the contaminant, so that is the time that the solution would have to be held in a batch reactor to achieve the same goal.
2. (a) Because the cyanide is undergoing a first-order reaction in steady-state CSTR, we can find in the CSTR as follows: cout cin
1
1 K CSTR
1
CSTR
1 cin
1
K cout
1
0.8 h
1
15 mg/L
1
1 mg/L
17.5 h
(b) If a baffle were used to convert the reactor to two, equal-sized CSTR’s in series, then the full flow (Q) would go through both reactors, but each reactor would have only one-half the residence time of the full reactor. Thus, the cyanide concentration in the effluent from the first reactor would be: cout,1 cin
1 K CSTR
15 mg/L
0.8 h 1 8.75 h
1
1.88
mg
L
The fluid with this concentration is then the influent to the second reactor, which has the same residence time as the first. The effluent concentration from the second reactor would therefore be:
cout,2
cin,2
1 K
CSTR
1.88 mg/L
0.8 h 1 8.75 h
1
0.23
mg
L
Thus, splitting the reactor into two smaller reactors in series is predicted to lead to a significant improvement in performance.
(c) This question describes a nonsteady-state situation in which the influent concentration to a
CSTR undergoes a ‘step change’. Assuming that the flows in and out of the reactor as constant and equal to one another, the mass balance for this situation is:
V
dc dt Qcin Qcout VKc
where c is the concentration inside the reactor. Because the reactor is well-mixed, we can substitute cout for c. Then, dividing through by Q and rearranging, we obtain: dcout dt
cin
cin
dcout
1 K cout
cout t
cout 0
1
1
K cout
dt
dcout
K cout cin
t
dt
0
2
dx
, with cout x , cin a , a bx and 1 K b . Therefore, integrating that expression according to the information given in the problem statement, we can write:
The integral on the left can be represented in the form
1
K
1
cin cin 1
1
cin cin ln
1
1
K cout t
K cout 0 cin cin
K cout t
K cout 0
exp
1
1
t 0
K
t
K cout 0 exp
cout t
1
1
K
t
K
Because Na+ is non-reactive, K = 0, so 1 + K = 1, and the expression simplifies to:
cout t
cin
cin cout 0 exp
t
Throughout the time of interest, cin = 75 mg/L. Before the increase in concentrations, the
Na+ concentration coming out of the reactor equals the value entering, so cout(0) is
15 mg/L. Finally, from part (a), we know that the hydraulic detention time is 17.5 h.
Therefore, at any time t, we know all the terms on the right side of the equation, and we can compute cout. Inserting t = 6 h, we find:
cout t
75
mg
L
32.4
75
mg mg 15 exp L
L
6h
17.5 h
mg
L
For CN , the approach is identical, but we have to include the terms that contain K, which has a value of 0.8 h 1. The expression 1 + K therefore equals {1 + (17.5 h)(0.8 h 1)}, or 15.
For CN , cout(0) is 1.0 mg/L, and we find:
