SOLUTIONS CHAPTER 6 EXERCISES
1. Several reactor configurations are to be considered for reducing the influent
substrate concentration from 100 mg/L to 20 mg/L at a design flow rate of 1
million gallons per day. Assume that substrate removal follows first-order
kinetics and the first-order rate constant is 0.8 d-1. The following equation may be
used for estimating the removal efficiency for completely-mixed reactors
operating in series assuming first-order removal reactions.
Sn  1 

=
S o  1 + k θ 
n
Where So and Sn = Influent substrate concentration and effluent substrate
concentration from the nth completely-mixed reactor in series, mass/volume;
n = Number of completely-mixed reactors in series;
k = First-order removal rate coefficient, d-1, and
θ = Detention time in each of the completely-mixed reactors in series.
Determine and compare the volume required for the following reactor
configurations.
a. One continuous flow, ideal completely mixed reactor
From Table 6.3, use the following equation:
C t − 1) (100 mg L 20 mg L − 1)
=
= 5 days
k
0.8 d −1
 1 MG 
5
3
V = θ Q = 5 days 
 = 5 MG or 6.68 × 10 ft
d


θ=
(C o
b. One continuous flow, ideal plug flow reactor
From Table 6.4, use the following equation:
ln (C o ) − ln(C t ) ln(100 mg L ) − ln (20 mg L )
=
= 2.01 days
k
0.8 d −1
 1 MG 
5
3
V = θ Q = 2.01days 
 = 2.01 MG or 2.69 × 10 ft
 d 
θ=
c. Two continuous flow, ideal completely mixed reactors in series
Sn  1 

=
S o  1 + k θ 
n
Solutions Ch 6 July 3, 2008

20 mg L 
1

= 
−1 
100 mg L  1 + 0.8 d θ 
2
37
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 20 


 100 
( 12 )
=
1
1 + 0.8 θ
0.4472 (1 + 0.8 θ ) = 1
θ = 1.55 days
 1 MG 
5
3
V = θ Q = 1.55 days 
 = 1.55 MG or 2.07 × 10 ft
 d 
d. Four continuous flow, ideal completely mixed reactors in series
Sn  1 

=
S o  1 + k θ 
 20 


 100 
( 14 )
=
n

20 mg L 
1

= 
−1 
100 mg L  1 + 0.8 d θ 
4
1
1 + 0.8 θ
0.6687 (1 + 0.8 θ ) = 1
θ = 0.619 days
 1 MG 
4
3
V = θ Q = 0.619 days 
 = 0.619 MG or 8.28 × 10 ft
 d 
2. During a chemical reaction, the concentration of Species A was measured as a
function of time. The observed concentration at various time intervals is
presented below. Determine the reaction order and rate constant, k. Is Species A
being removed or produced?
Time (min)
0
10
20
30
40
50
Concentration of A (mg/L)
100
80
60
40
20
0
A plot of concentration of A versus time on arithmetic paper produces a straight
line indicating a zero order reaction. The slop of the line is equal to k = 2
mg/L·min. Species A is being removed. See plot on next page.
Solutions Ch 6 July 3, 2008
38
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6.2
120
Concentration of A, mg/L
100
CA = -2 t + 100
2
R =1
80
Zero Order Removal Reaction
60
-1
-1
Slope = K = 2 mg L min
40
20
0
0
10
20
30
40
50
60
Time, minutes
3. The concentration of Species C was measured as a function of time during a
chemical reaction. The observed concentration of Species C at various time
intervals is presented below. Determine the reaction order and rate constant, k.
Is Species C being removed or produced?
Time (hr)
0
1.0
3.5
6.5
11.0
Concentration of C (mg/L)
100
80
50
26
10
A plot of the natural log of C versus time produces a straight line on arithmetic
paper indicating a first order reaction. The slop of the line is equal to k = 0.21h-1.
Species C is being removed. See plot below.
6.3
5.000
4.500
4.000
3.500
ln (C) = -0.2088 (t) + 4.6108
2
R = 0.9995
ln (C)
3.000
2.500
2.000
First Order Removal Reaction
1.500
Slope = K = 0.21 hr
1.000
-1
0.500
0.000
0
2
4
6
8
10
12
Time, hours
Solutions Ch 6 July 3, 2008
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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4. The concentration of Species D was measured as a function of time during a
chemical reaction. The observed concentration of Species D at various time
intervals is presented below. Determine the reaction order and rate constant, k. Is
Species D being removed or produced?
Time (hr)
0
1.0
2.0
3.0
4.0
5.0
Concentration of D (mg/L)
200
142
111
90
77
67
A plot of the reciprocal of D versus time produces a straight line on arithmetic paper
indicating a second order reaction. The slope of the line yields k = 0.002 L/mg·h.
Species D is being removed. See plot below.
Problem 6.4
0.016
1/(Concentration of D), L/mg
0.014
1/CD = 0.002(t) + 0.005
R2 = 0.9998
0.012
0.010
0.008
Slope = 0.002 L/mg-hr
0.006
0.004
Second Order Removal Reaction
0.002
0.000
0
1
2
3
4
5
6
Time, hours
5. During a chemical reaction, the concentration of Species B was measured as a
function of time. The observed concentration of Species B at various time
intervals is presented below. Determine the reaction order and rate constant, k. Is
Species B being removed or produced?
Time (hr)
0
5.0
10.0
15.0
20.0
Solutions Ch 6 July 3, 2008
Concentration of B (mg/L)
100
125
150
175
200
40
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A plot of the concentration of B versus time yields a straight line on arithmetic paper
indicating a zero order production reaction. The slope of the line is equal to k = 5
mg/L·h. Species B is being produced. See plot below.
Problem 6.5
250
CB = 5(t) + 100
R2 = 1
Concentration of B, mg/L
200
150
Slope = K = 5.0 mg/L-hr
100
Zero Order Production Reaction
50
0
0
5
10
15
20
25
Time, hours
6. If the half-life of a chemical compound is 30 days under anaerobic conditions,
determine the first-order removal rate constant, k.
C t = Co e− k t
0.5 C o = C o e
C t = 0.5 C o
when t = t 1 2
− k t1 2
0.5 C o
= e − k 30 d
Co
k = 0.023 h −1
7. Three wastewater streams are combined at a food processing facility to equalize
the pH prior to biological treatment. The flow rate and pH of each of the
wastewater streams is presented in the following table. Perform a mass balance
on flow and the hydrogen ion concentration (H+) so that the pH of the three
combined streams may be estimated. The pH of a solution is equal to the negative
logarithm of the hydrogen ion concentration (pH = -Log [H+]).
Wastewater Stream
1
2
3
Recall pH = - log[H+]
Solutions Ch 6 July 3, 2008
Flow (liters per minute)
5
20
25
pH
5.5
6.5
8.5
See Example 6.8.
41
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[H ]
average
[H ]
+
average
=
[H ] Q + [H ] Q + [H ]Q
=
+
1
+
+
2
1
2
+
3
3
Q1 + Q 2 + Q 3
10 −5.5 (5 Lpm) + 10 −6.5 (20 Lpm ) + 10 −8.5 [25 Lpm]
= 4.44 × 10 −7 moles L
5 + 20 + 25 Lpm
pH = - log[H+] = - log[4.44×10-7] = 6.35
8. A sanitary landfill receives 600 ft3 of municipal solid waste 5 days per week at a
density of 500 lb/yd3. If the solid waste is compacted to 1000 lb/yd3 and the
average depth of each cell is 10 feet, estimate the expected life of the landfill in
years if 25 acres of space are still available. Draw a materials balance diagram to
solve the problem. (1 acre = 43,560 ft2)
 600 ft 3   5 d   52 wk   1 yd 3   500 lb 
 

 = 2.89 × 10 6 lb yr
 
SW input = 
 
3 
3 
 d   wk   yr   27 ft   yd 
SWIN
SWOUT = 0
2.89x106 lb/yr
3
3
2.89x10 yd /yr
SW input = 2.89 × 10 6
ρ = 1000 lb/yd3
10 ft per lift
lb  1 yd 3 
yd 3

 = 2889
yr  1000 lb 
yr
 43,560 ft 2 
 = 1.089 × 10 6 ft 2
Area = 25 ac 
ac


3
3
yd  27 ft 3 
4 ft


SW input = 2889
7.8
10
=
×
yr  yd 3 
yr
Volume of one 10 ft cell = (1.089 × 10 6 ft 2 )(10 ft ) = 1.089 × 10 7 ft 3
Expected life =
Solutions Ch 6 July 3, 2008
1.089x10 7 ft 3
= 140 yr
7.8 × 10 4 ft 3 yr
42
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9. Perform a materials balance on substrate (S) around a chemostat (completely
mixed reactor without recycle) assuming a first-order removal (dS/dt = - kS) for
substrate with a rate coefficient k value of 0.5 days-1. The influent substrate
concentration is 150 mg/L and 90% removal is desired. Determine the detention
time in hours for the chemostat assuming steady-state conditions.
S = (1 − 0.90)150
mg
mg
= 15
L
L
Q
Q
So
S
S
V
 dS 
V = Q So − Q S + r V
 
 dt  accumulation
r = − kS
 dS 
V = Q So − Q S + − k S V
 
 dt  accumulation
At steady state, the accumulation term goes to zero; and the above equation reduces to the
following equation:
Q (S o − S) = k S V
θ=
V (S o − S) (150 − 15) mg L
=
=
= 18 h
Q
kS
0.5 h −1 (15 mg L )
10. Calculate the volume of an ideal plug flow reactor for the following scenario.
The volumetric flow rate is 6,500 m3/day and Species A is being removed or
converted according to a first-order reaction as follows: dCA/dt = - k CA, where
CA is the concentration of Species A and k = 9000 days-1. A 95% removal or
conversion of Species A is required.
C t = (1 − 0.95) C o = 0.05 C o
From Table 6.4, use the following equation:
Solutions Ch 6 July 3, 2008
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ln (C o ) − ln (C t )
θ=
=
k
C

ln 0

0.05
C
o

= 3.33 × 10 − 4 days
−1
9,000 d
 6500 m 3 
 = 2.16 m 3 or 2,164 L
V = θ Q = 3.33 × 10 days 
d


−4
A 1000 MW coal-burning power plant is burning West Virginia bituminous coal
with 8% ash content. The power plant is 33% efficient with 35% of the ash
settling out in the firing chamber as bottom ash. A simplified schematic diagram
is shown below. Assume 3.5 kwh per pound of coal.
11.
Exit Gases and
Fly Ash
Coal
Gases and
Fly Ash
Furnace
ESP
Bottom
Ash
a.
Stack
Fly
Ash
Determine the rate of coal input to the furnace in kg/day
Input Power =
Energy IN =
Output Power
Efficiency
Energy OUT 1000 MW
=
= 3030 MW
Efficiency
0.33
Energy Heat = 3030 − 1000 = 2030 MW
Solutions Ch 6 July 3, 2008
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 1000 kW   24 h   1 lb   454 g   1 kg 
 × 
 × 

Coal Input = 3030 MW × 
 × 
 × 
 1 MW   d   3.5 kWh   lb   1000 g 
Coal Input = 9.43 × 10 6 kg coal
day
a. Assuming that the electrostatic precipitator (ESP) is 99% efficient, calculate the
rate of fly ash emitted to the atmosphere in kg/day.
Ash Produced = 9.43 × 10 6
kg  0.08 kg ash 
kg

 = 7.55 × 10 5
d  kg coal 
d
Fly Ash Released to Atmosphere = (1 − 0.35)(1 − 0.990) × 7.55 × 10 5
kg
= 4.91 × 10 3 kg d
d
Energy Balance
b. Draw an energy diagram for the facility and calculate the rate of heat emitted to
the environment in kJ/s;
Energy IN = 3030 MW ×
1000 kW 1 kJ s
kJ
×
= 3.30 × 10 6
1 MW
1 kW
s
Energy OUT(Heat) = 2030 MW ×
1000 kW 1 kJ s
kJ
×
= 2.03 × 10 6
1 MW
1 kW
s
Energy OUT(Produced) = 1000 MW ×
EnergyIN
fuel
3.03x106 kJ/s
1000 kW 1 kJ s
kJ
×
= 1.10 × 10 6
1 MW
1 kW
s
EnergyOUT
Electricity
1.00x106 kJ/s
Power
Plant
EnergyOUT
heat
2.03x106 kJ/s
Solutions Ch 6 July 3, 2008
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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12. A pristine stream flowing at 100 cubic feet per second (cfs) in the Rocky
Mountains contains 5 mg/L of suspended solids (SS). During the spring, ice melt
conveys 250 mg/L of SS at a rate of 20 cfs into the stream. Determine the
concentration of SS in the stream during the spring.
SSaverage =
SSaverage =
SS1 Q1 + SS 2 Q 2
Q1 + Q 2
(5 mg L) (100 cfs) + (250 mg L) (20 cfs) = 46 mg
100 cfs + 20 cfs
L
13. Calculate the minimum rate at which 15°C make-up water from a river must be
pumped to evaporative cooling towers for a 1000 MW nuclear power plant. The
efficiency of the plant is 32% and all of the waste heat is assumed to be dissipated
through evaporative cooling with no direct heat lost to the atmosphere.
Energy IN =
Energy OUT 1000 MW
=
= 3125 MW
Efficiency
0.32
Energy Heat = 3125 − 1000 = 2125 MW
Using Equation (6.61) calculate the mass of water required.
[Rate of Change in Stored Energy] = m c ∆ T
× 4184
2125 MW = m
1 MW
J
× 15 D C × 6
D
kg ⋅ C
10 J s
= 3.38 × 10 4 kg s
m
Water Flow = 3.38 × 10 4
Solutions Ch 6 July 3, 2008
kg  1 m 3  3.9 m 3
=
×
s  1000 kg 
s
46
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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