CIVE 3610 – Water Supply and Treatment
Fall Semester 2004
Homework #3 – Solution
1) A town requiring 1.0 m3/s of drinking water has two sources, a local well with
60 g/m3 nitrate and a distant reservoir with 10g/m3 nitrate.
a. Draw a diagram showing all the inputs, outputs, and variables.
b. What flow rates of well and reservoir are needed to meet the 45g/m3
drinking water standard and minimize the use of more expensive
reservoir water?
a. Draw a diagram showing all the inputs, outputs, and variables.
CW = 60g/m3
QW = ?
C = 45g/m3
Q = 1 m3/s
CR = 10g/m3
QR = ?
b. What flow rates of well and reservoir are needed to meet the 45g/m3 drinking
water standard and minimize the use of more expensive reservoir water?
QW * CW+ QR * CR= QF * CF
QW * (60g/m3) + QR * (10g/m3) = (1.0m3/s) * (45 g/m3)
60QW + 10QR = 45
Minimize Reservoir,
QW + QR = 1 m3/s
QR = 1-QW
60 QW +10(1- QW) = 45 g/s
60 QW + 10 –10 QW = 45 g/s
50 QW = 35
QW = 0.70 m3/s; QR = 0.3 m3/s
(2)
(1)
2) During a chemical reaction, the following concentrations of Species A at
various times were observed. Determine:
a. The reaction order and reaction rate constant, k
b. The concentration of A after 20hrs.
Time, hours
0
7.5
15
22.5
30
Concentration, mg/L
50.8
32
19.7
12.3
7.6
Solution:
a. The reaction order and reaction rate constant, k
Assume a first order reaction. The plot of ln (Ct) versus time
is shown in Figure 1. Since the plot is a straight line, the reaction is a first order
reaction. The slope of line gives the rate constant ‘k’.
Figure 1. Plot of ln (Ct) vs. Time
ln (Ct)
ln (Ct) vs. Time
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
7.5
15
22.5
30
Time
Rate constant K = [ln (50.8) –ln (7.6)] /30hrs] = [(3.92-2.03) / 30] = 0.063/hr
b. The concentration of A after 20hrs
Ct/ CO = e - (k * t)
Ct = (50.8 mg/L) e - (0.063 * 20)
=14.3 mg/L
3) A completely mixed chemical reactor has an influent flow with a
concentration of 150 mg/L of A and flowrate of 100 gal/ min. The reaction is
first order and the rate constant is 0.4/hr. Determine:
a. Draw a diagram showing all inputs, outputs, and variables.
b. The required detention time and the volume (gallons) of the reactor if
the effluent contains 20 mg/L of A.
c. Draw a diagram for a plug flow reactor with the same design
conditions.
d. How many times larger a completely mixed reactor must be than a
plug flow reactor for 80% removal or conversion.
e. How many times larger a completely mixed reactor must be than a
plug flow reactor for 90% removal or conversion.
f. Why does a plug flow reactor perform better than a completely mixed
reactor?
a. Draw a diagram showing all inputs, outputs, and variables.
150mg/L
CF
CF
100gal/min
100 gal/min
K=0.4/hr
b. The required detention time and the volume (gallons) of the reactor if the
effluent contains 20 mg/L of A.
θ = (CO- CF) / KCF = (150-20) / (0.40*20) = 16.3 hrs
Volume V= Q* θ = (100gal/min)* (60 min/hr) *16.3 hr = 97800 gals
c. Draw a diagram for a plug flow reactor with the same design conditions.
150mg/L
CF
K=0.4/hr
100 gal/min
100 gal/min
d. How many times larger a completely mixed reactor must be than a plug flow
reactor for 80% removal or conversion.
80% removal = (CO- CO * 80/100) = (150 -150* (80/100)) = 150-120 = 30
Therefore, CF = 30mg/L
PFR
CF/CO = e– (K * θ)
30/150 = e– (0.4 * θ)
θ = 4.02 hrs
CSTR
θ = (CO - CF)/ (K * CF )
= (150-30) / (0.4 * 30)
θ = 10 hrs
The ratio of CSTR to PFR
CSTR / PFR = 10 / 4 = 2.5
Therefore, CSTR = 2.5 times larger than PFR
e. How many times larger a completely mixed reactor must be than a plug flow
reactor for 90% removal or conversion.
90% removal = (CO- CO * 90/100) = (150 -150* (90/100)) = 150-135 = 15
Therefore, CF = 15 mg/L
PFR
CF/CO = e– (K * θ)
15 /150 = e– (0.4 * θ)
θ = 5.76 hrs
CSTR
θ = (CO - CF)/ (K*CF )
= (150-15) / (0.4*15)
θ = 22.5 hrs
The ratio of CSTR to PFR
CSTR / PFR = 22.5 / 5.76 = 3.91
Therefore, CSTR = 3.91 times larger than PFR
f. Why does a plug flow reactor perform better than a completely mixed reactor?
Plug flow reactors (PFRs) work better than completely mixed reactors
(CSTRs) because chemical conversion rates are generally first order. First order rates are
directly proportional to contaminant concentration. Since it is assumed there is no
dispersion in an idealized plug flow reactor, the average contaminant concentration is
greater than that in a CSTR, thus the reaction rate is faster, less detention time is needed
and the volume of the reactor can be smaller.
4) Three CSTRs are to be used in series. The second reactor has a volume twice that
of the first and third reactors. The influent flow has a concentration of 150 mg/L of
A, and the flowrate is 100 gal/min (380 l/min). The reaction is first order, the rate
constant is 0.4/hr.
a. Draw a diagram showing all inputs, outputs, and variables.
b. Determine the mean residence time and volume of each reactor if the
removal or conversion of A is 90%.
c. What is the concentration of A in each reactor?
d. What is the rate of conversion in each reactor?
a. Draw a diagram showing all inputs, outputs, and variables.
CF
150 mg/L
15 mg/L
Q = 100gpm
V1=V
V
V3=V
V1=V
V1=V
V1=V
V2=2V
K=0.4/h
r
100gpm
K=0.4/h
r
b. Determine the mean residence time and volume of each reactor if the removal or
conversion of A is 90%.
CF = CO / [(1+Kθ1)* (1+Kθ2)* (1+Kθ3)]
15 = 150/ [(1+0.4θ) *(1+0.4*2θ) *(1+0.4*θ)]
Rearrange equation,
1+1.6θ + 0.8θ2 +0.13θ3 = 10
V1= V3=V
V2 = 2V
We have, V= Q*θ
θ = V/Q
θ1= V1/Q = V/Q = θ
θ2= V2/Q = 2V/Q = 2θ
θ3= V3/Q = V/Q = θ
Trial and error solution to get the
Mean residence time
θ1 = θ3 = 2.25 hrs
θ2 =2 θ =2*2.25 = 4.5 hrs
θ
2
3
2.5
2.25
2.20
Volume of the reactors
V1= V3 = V = Qθ = (100gal/min) * (60 min/hr) * 2.25 hr =13500 gal
V2 = 2V = Qθ2 = (100gal/min) * (60 min/hr) * 4.5 hr = 27000 gal
c. What is the concentration of A in each reactor?
C1 = [CO/ (1+Kθ1)] = 150/ (1+0.4*(2.25)) = 79 mg/L
CO/CF =10
8.44
16.5
12.03
10.13
9.78
C2 = [C1/ (1+Kθ2)] = 79/ (1+0.4*(4.5)) = 28 mg/L
C3 = [C2/ (1+Kθ3)] = 28/ (1+0.4*(2.25)) = 15 mg/L
d. What is the rate of conversion in each reactor?
r1 = [(CO – C1) / θ1] = [(150-79) / 2.25] = 32 mg/L-h
r2= [(C1 – C2) / θ2] = [(79-28) / 4.5] = 32 mg/L-h
r3 = [(C2 – C3) / θ3] = [(28-15) / 2.25] = 5.8 mg/L-h