Problems and Solutions Chapter 5
1. The owner of a 1000 ft3 swimming pool desires to fill it up during a 24 hour
period using one garden hose. Determine the following:
a. The necessary flow rate in gallons per minute (gpm) to fill the pool.
b. If two garden hoses flowing at 5 gpm are used, how long would it take to
fill up the pool?
Solution Part A
Start with Equation (5.1) and set the output term to zero since there are no leaks in
the pool.
(5.1)
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts 
 a c c u m u la tio n  =  in p u ts  - 0
Therefore, the accumulation term must be equal to the sum of all inputs.
 a c c u m u la tio n  =  in p u ts 
The accumulation term is determined as follows:
1 0 0 0 f t o f w a te r  7 .4 8 g a l 
7 .4 8  1 0 g a l

 
3
1d
ft
d


3
 a c c u m u la tio n  =
3

1 0 0 0 f t o f w a te r  7 .4 8 g a l   1 d   1 h




3
1d
ft

  2 4 h   6 0 m in 
3
 a c c u m u la tio n  =
 a c c u m u la tio n  =
5 .1 9 g a l
 in p u ts  =
gal
5 .1 9
m in
  in p u ts 
m in
Solution Part B
The input term for Part B consists of two hoses each with a flow rate of 5 gpm
and the accumulation term is equal to 4.48103gal/day.
 a c c u m u la tio n  =  in p u ts 
 7 .4 8  1 0 g a l  
5 g a l 6 0 m in
24 h



t

 = 2 

1d
m in
h
d


 
3
1
t  0 .5 1 9 d 
t  0 .5 1 9 d
24 h

1 2 .5 h
d
2. Four wastewater streams from a textile mill are blended in an equalization tank
prior to further treatment. The flow rate and pH of each stream are as follows:
grey water, 500 gallons at pH = 4.0; white water, 1000 gallons at pH = 7.3; dye
waste, 1500 gallons at pH =11.0, and kier waste, 500 gallons at pH = 11.8.
Perform a mass balance on the hydrogen ion concentration [H+] so that the pH of
the equalized flow can be determined. Recall that pH = - log [H+].
Solution
Must substitute the hydrogen ion concentration in moles/L into Equation (5.6)
rather than the pH value along with the appropriate flow rates.
C5 
C5 
5 0 0 g p m (1 0
4
Q1 C 1  Q 2 C 2  Q 3 C 3  Q 4 C 4
M )  1 5 0 0 g p m 1 0
 7 .3
500  1500 
C 5  1 .1 1  1 0
5
(5.6)
Q1  Q 2  Q 3  Q 4
M
  2 0 0 0 g p m  1 0   5 0 0 1 0
11
 1 1 .8

2000  500  gpm
M
The pH of the combined flow streams is calculated by substituting into the
following equation.
p H   lo g  H    lo g 1 .1 1  1 0



+
5


p H  4 .9 5
3. 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 [H+] concentration so that the pH of the three
combined streams may be determined. The pH of a solution is equal to the
negative logarithm of the hydrogen ion concentration (pH = -log [H+]).
Wastewater Stream
Flow (liters per minute)
pH
1
50
5.5
2
200
6.5
3
250
8.5
2
Solution
Must substitute the hydrogen ion concentration in moles/L into Equation (5.6)
rather than the pH value along with the appropriate flow rates.
C4 
C4 
Q1 C 1  Q 2 C 2  Q 3 C 3
(5.6)
Q1  Q 2  Q 3
5 0 L p m (1 0
 5 .5
M )  2 0 0 L p m 1 0
50 
C 4  1  4 .4 4  1 0
7
 6 .5
M

2 5 0 L p m 1 0
 8 .5

200  250  Lpm
M
The pH of the combined flow streams is calculated by substituting into the
following equation.
p H   lo g  H    lo g  4 .4 4  1 0



+
7


p H  6 .3 5
4. The recycle of various waste streams from unit operations and processes at a
wastewater treatment plant (WWTP) can have a significant impact on the influent
characteristics to the facility. Most recycle and waste streams from digesters and
sludge handling operations are typically blended with the incoming raw
wastewater at the headworks of the WWTP. Given the following schematic,
determine the actual influent BOD5 concentration (mg/L) that the facility must
actually process.
Sludge digester supernatant
Q = 41 m3/d
BOD5 = 5000 mg/L
Headworks
Headworks effluent
Q=?
BOD5 = ?
Influent
Q = 21,600 m3/d
BOD5 = 375 mg/L
Sludge dewatering filtrate
Q = 71 m3/d
BOD5 = 1500 mg/L
3
Solution
Must substitute the flows and BOD5 concentrations into Equation (5.6) to solve
for the flow weighted BOD5 in the effluent from the headworks.
C4 
C4 
Q1 C 1  Q 2 C 2  Q 3 C 3
(5.6)
Q1  Q 2  Q 3
 2 1 ,6 0 0 m

d

3
  375 m g   71 m

+ 
L
d
 

3
  1500 m g   41 m

+
L
d
 

 2 1, 6 0 0  7 1  4 1 
C4 
387
m
3
  5000 m g 


L


3
d
mg
L
5. Raw primary sludge at a solids concentration of 5% is mixed with waste activated
sludge (WAS) at a solids concentration of 1.0%. The primary and WAS flows are
29,000 gallons per day and 35,000 gallons per day, respectively. Assume that the
specific gravity of both sludges is 1.0 so that a 1.0% solids concentration is
equivalent to 10,000 mg/L. Determine the following:
a. The solids concentration (%) for the blended sludges.
b. The volume of thickened sludge and supernatant produced in gallons per
day if the blended sludge in Part A is thickened to 8.0% solids. Assume
that the thickener removes 100% of the solids that enter.
Solution Part A
Must substitute the flows and solids concentrations into Equation (5.6) to
solve for the flow weighted solids concentration in the combined flow. The
solids concentrations of the sludges may be expressed either as % solids or in
mg/L when substituted into Equation (5.6). It does not matter as long as
consistent units are used.
C3 
C3 
Q1 C 1  Q 2 C 2
Q1  Q 2

 2 9 , 0 0 0 g p d   5 %    3 5 , 0 0 0 g p d  1 % 
 29, 000
 3 5, 0 0 0  g p d
2 .8 1 % s o lid s
Alternatively, the solids concentrations in mg/L may be used. Recall that a
1% solids concentration is equivalent to 10,000 mg/L when the specific
gravity of the sludge is assumed to be equal to 1.0 i.e. the specific gravity of
water.
4
C3 
Q1 C 1  Q 2 C 2
Q1  Q 2
C 3  2 8 ,1 2 5
mg
L

 29, 000 gpd  50, 000 m g
L
 29, 000
C 3  2 8 ,1 2 5
 3 5 , 0 0 0 g p d  1 0 , 0 0 0 m g
L

 3 5, 0 0 0  g p d
m g  1 % s o lid s 

  2 .8 1 % s o lid s
L  1 0 ,0 0 0 m g L 
Solution Part B
A material balance must be performed around the thickener. Draw a
schematic of the process showing the flows and solids concentrations entering
and exiting the thickener.
Qsupernatant = ? gpd
Csupernatant = 0 % solids
=6.4105
Q
gpd
C = 2.81% solids
Thickener
Qunderflow = ? gpd
Cunderflow = 8.0% solids
Start with Equation (5.1) to perform a materials balance around the thickener.
Recall, we are assuming that the thickener is 100% efficient so there are no
solids in the supernatant. We are also assuming that the thickener is operating
at steady-state so the accumulation term may be set equal to zero. We will
make use of the following conversion factor:
8 .3 4
lb
MG
1
mg
L
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts 

 0 .6 4 5 M G   2 8 ,1 0 0 m g   8 .3 4 lb
0= 


mg
d
L



 MG
L




 8 0 ,0 0 0 m g   8 .3 4 lb
 - Q s u p e r n a ta n t  0 % s o lid s   Q t h ic k e n e d 

mg
L



 MG
L



 0 .6 4 5 M G   2 8 ,1 0 0 m g   8 .3 4 lb



d
L


 MG mg
L




 8 0 ,0 0 0 m g   8 .3 4 lb
  Q th ic k e n e d 

L

 MG mg

L


5










Q th ic k e n e d  0 .2 2 7
MG
d
M G  10 gal 

 
d  1MG 
6
Q th ic k e n e d  0 .2 2 7
227, 000
gal
d
6. Reverse osmosis (RO) is being used more widely for treating groundwater for
human consumption. RO treatment can produce water that essentially has no
hardness. Hardness in water is attributed to divalent metallic cations primarily,
calcium  C a 2 +  and magnesium  M g 2 +  . A 3.785 mega-liter (ML) RO water
treatment plant (WTP) treats groundwater containing 150 mg/L of hardness as
CaCO3. The product or permeate from the RO WTP will be blended with
untreated groundwater to achieve a final hardness of 50 mg/L as CaCO3. How
much of the untreated groundwater must be blended with permeate from the RO
WTP? See schematic diagram below to assist in the solution of the problem.
QRO = ?
Hardness = 0 mg/L
RO WTP
Qby-pass = ?
Hardness = 150 mg/L
Well
Solution
Start with Equation (5.1) and set the accumulation term to zero since steady-state
conditions are assumed.
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts 
(5.1)
 in p u ts    o u tp u ts 
Next, rewrite the above equation in terms of flows multiplied by concentration as
follows:
 Q RO   C RO  
Q
by  pass
 C 
by  pass

Q
RO
 Q by  pass
 C 
fin is h e d
 3 .7 8 5 M L   0 m g 
 150 m g 
 3 .7 8 5 M L
  50 m g 
 
 Q b y  p a ss  


   Q b y  p a ss  


d
L
d
L

 L 

 b y  p a ss



6
fin is h e d
 150 m g 
0 
 Q by  pass

L

 by  pass
Q
b y  p a ss

1 .9

1 .8 9  1 0 M L  m g
2
d L

50 m g
L
Q by  pass
ML
d
7. During most of the year, the average concentration of suspended solids (SS) and
flow in a pristine stream near Fort Collins, Colorado is 5 mg/L and 3 m3/s,
respectively. During the spring, melting ice conveys 250 mg/L of SS at a rate of
0.6 m3/s into the stream. Estimate the concentration of SS in the stream during
spring assuming that complete-mixing of the two streams occurs.
Solution
Must substitute the flows and suspended solids concentrations into Equation (5.6)
to solve for the flow weighted suspended solids concentration of the combined
streams.
C3 
C3 
Q1 C 1  Q 2 C 2
Q1  Q 2
4 5 .8

 3 .0 m

s

3
  5 m g   0 .6 m

 
s
 L  
 3 .0 m

s

3

3
0 .6 m
s
   250 m g  


L


3



mg
L
8. Covington, Virginia which is located in southwest Virginia encounters inversions
on a regular basis since it is surrounded by mountains. Assume that the
inversions typically limit a mixing depth of approximately 2,000 feet above the
city. The city’s land area is approximately 6.0 square miles and particulate
emission rate is estimated at 10 lb of particulates per square foot per day.
Determine:
a. The concentration of particulates in the air above the city at the end of a
24 hour period when there is no wind. Assume that the length and width
of the land area is approximately 2.45 miles  6 m i 2 .
b. The concentration of particulates in the air above the city at the end of a
24 hour period when the wind speed is 20 miles per hour. Assume there is
no particulate matter in the air from outside of the city limits.
Solution Part A
We must perform a material balance on particulates using Equation (5.4) and
assuming complete-mixing of the air above the city. A box model is normally
used to model this type of air pollution event. The input and output terms in the
mass balance are set equal to zero since there is no wind. The reaction term is set
equal to the particulate emission rate.
7
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts    re a c tio n 
 a c c u m u la tio n  =
(5.4)
0 - 0   re a c tio n 
The particulate emission rate is calculated as follows.
Recall that 1 mile = 5,280 feet.
1 0 lb
r p a r tic u la te s 
ft  d
2
 6 .0 m i
 5 2 8 0 ft

2
mi

2
2
2

9 lb
  1 .6 7  1 0
d

The volume of the air space above the city is calculated as follows:
 5 2 8 0 ft 
2
V  A  H e ig h t  6 .0 m i 
  2 0 0 0 ft 
2
 1mi

2
V  3 .3 5  1 0
11
ft
 a c c u m u la tio n  =
C


p a r tic u la te s
1d
2
3
0 - 0   re a c tio n 

9 lb
 V = 1 .6 7 × 1 0
d

Recall the following conversion factors:
1 lb = 4 5 4 g
C
p a r tic u la te s
=
C
p a r tic u la te s
=
1 m = 3 .2 8 1 f t

9 lb 
 1 .6 7 × 1 0
 1 d 
3
3
d 
 4 5 4 g   3 .2 8 1 f t 




11
3
3
3 .3 5  1 0 f t
m
 lb  

80
g
m
3
Solution Part B
We must perform a material balance on particulates using Equation (5.4) and
assuming that complete-mixing of the air above the city. The input term will be
set equal to zero since the incoming air is assumed to contain no particulate
matter. The mass of particulates in the output term will be equal to the
concentration in the air above the city times the air flow rate. Steady-state
conditions are also assumed to take place. The reaction term is set equal to the
particulate emission rate calculated in Part A of the solution.
8
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts    re a c tio n 
0 = Q a ir
in
C
p a r t in

- Q a ir
out
C
p a rt o u t

(5.4)
9
1 .6 7 × 1 0 lb p a r tic u la te s
d
The air flow rate into and out of the imaginary box above the city is calculated as
follows, where U is the wind speed.
Q a ir in  Q a ir o u t 
 A c ro s s -s e c tio n a l   U 
5 2 8 0 ft   2 0 m i 5 2 8 0 ft 2 4 h 

Q a ir in  Q a ir o u t   2 .4 5 m i× 2 0 0 0 ft×
×
×


mi 
h
mi
d 

Q a ir
in
 Q a ir
out
0 = 6 .5 6  1 0
13
 6 .5 6  1 0
13
ft
d
C
C
part out

part out

3
ft
3
d
ft
3
d
6 .5 6  1 0
13
C
0 
- 6 .5 6  1 0
5
3
C
d
part out
2 .5 5 × 1 0
ft
13

p a rt o u t

9
1 .6 7 × 1 0 lb p a r tic u la te s
d
9

1 .6 7 × 1 0 lb p a r tic u la te s
d
lb p a r tic u la te s
d
2 .5 5 × 1 0
5
lb p a r tic u la te s  4 5 4 g   3 .2 8 1 f t 
 


3
3
ft
m
 lb  

3
3
0 .4 1
g
m
3
9. A company has been discharging a non-reactive pollutant at a concentration of 15
g/m3 to a lagoon for several years. The wastewater flow rate is 0.1 mega-liters
(ML) per day and the lagoon detention time is 10 days. The lagoon is assumed to
be completely-mixed. The overflow from the lagoon discharges into an adjacent
river. Determine:
a. The steady state concentration of the non-reactive pollutant in the effluent
from the lagoon that overflows into the river.
b. The concentration of the non-reactive pollutant in the effluent leaving the
lagoon 10 days after the influent concentration is suddenly increased to
150 g/m3.
9
Solution Part A
We first perform a material balance on the non-reactive pollutant around the
lagoon starting with Equation (5.4) and assuming steady-state conditions. Since
the pollutant is non-reactive or conservative, the reaction term is set equal to zero.
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts    re a c tio n 
(5.4)
0 =  in p u ts  -  o u tp u ts   0
Therefore, input must equal output.
 in p u ts    o u tp u ts 
Q in C in  Q o u t C o u t
 0 .1 M L   1 5 g   0 .1 M L 
 


 C out
3 
d
d

 m  


Recall: 1 m 3 = 1 0 0 0

C out

C   15

out
g
m
3
L
  1000 m g 
15 g  1 m
mg

  15
3 
1g
L
m  1000 L  

3

Solution Part B
Equation (5.23) was derived for non-steady state conditions in which a nonreactive pollutant was involved. We will use the same equation however, since
the pollutant in this case is non-reactive, the reaction rate constant k is set equal to
zero. C  represents the pollutant concentration at steady-state or equilibrium
which is 15 mg/L. The concentration in g/m3 is equivalent to mg/L as shown
above.
 
Q  
C t  C    C 0  C   exp    k 
 t
V  
 
(5.23)
The detention time is defined as the volume divided by the flow rate.
Recall that
V
   1 0 d s o th a t
Q
C 10 d  1 5
Q
V
mg
L
 1 5 0  1 5 

1


1
10 d
 

1 
exp    0 
 10 d 
L
10 d 
 

mg
10
mg
C 1 0 d  6 4 .7
L
10. A lagoon with a volume of 1,500 m3 receives an industrial wastewater flow of
100 m3/day containing a conservative pollutant at a concentration of 20 mg/L.
a. Assuming that complete mix conditions are achieved in the lagoon,
determine the concentration of the conservative pollutant in the effluent
from the lagoon at steady state.
b. If the influent pollutant concentration is suddenly increased to 60 mg/L,
what would the effluent concentration from the lagoon be 7 days later?
Solution Part A
We first perform a material balance on the conservative pollutant around the
lagoon starting with Equation (5.4) and assuming steady-state conditions. Since
the pollutant is conservative, the reaction term is set equal to zero and the
accumulation term is set to zero at steady-state.
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts    re a c tio n 
(5.4)
0 =  in p u ts  -  o u tp u ts   0
Therefore, input must equal output.
 in p u ts    o u tp u ts 
Q in C in  Q o u t C o u t
 100 m

d

3
  20 m g   100 m

  
L
d
 

C  
out
20
3

 C out



mg
L
Solution Part B
Equation (5.23) was derived for non-steady state conditions in which a nonreactive pollutant was involved. We will use the same equation, however, since
the pollutant in this case is conservative, the reaction rate constant k is set equal to
zero. C  represents the pollutant concentration at steady-state or equilibrium
which is 20 mg/L.
11
 
Q  
C t  C    C 0  C   exp    k 
 t
V  
 
C7d 
20 m g
L
 60  20 
(5.23)
3
 

100 m d 
exp    0+
7
d


3
L
1500 m 
 

mg
C 7 d  4 5 .1
mg
L
11. A lagoon with a volume of 1,500 m3 receives an industrial wastewater flow of
100 m3/day containing a non-conservative or reactive pollutant at a concentration
of 20 mg/L.
a. Assuming that complete-mix conditions are achieved in the lagoon and a
reaction rate coefficient k = 0.20 d-1, determine the concentration of the
non-conservative pollutant in the effluent from the lagoon at steady state.
b. If the influent pollutant concentration is suddenly increased to 120 mg/L,
what would the effluent concentration from the lagoon be 7 days later?
Solution Part A
We first perform a material balance on the conservative pollutant around the
lagoon starting with Equation (5.4) and assuming steady-state conditions. Since
the pollutant is non-conservative, the reaction term is not set equal to zero.
 a c c u m u la tio n  =  in p u ts  -  o u tp u ts    re a c tio n 
(5.4)
0 =  in p u ts  -  o u tp u ts    re a c tio n 
At steady-state, the flow rate into the lagoon will equal the flow rate out of the
lagoon, Q in  Q o u t  Q .
0 = Q in C in - Q o u t C o u t  k C o u t V
Q C in - Q C o u t  k C o u t V
Q C in  C o u t  k V  Q

This mass balance yields Equation (5.10).
C out 
C out 
Q C in
k
V  Q

Q C in
k
V  Q



100 m
 0 .2 0 d
-1
-1
d  20 m g L 
3
×1500 m + 100 m
100 m
 0 .2 0 d
3
3
3
d
d  20 m g L 
3
×1500 m + 100 m
12
3
d
C  
out
5
mg
L
Solution Part B
Equation (5.23) was derived for non-steady state conditions in which a nonconservative pollutant was involved. C  represents the pollutant concentration at
steady-state or equilibrium which is 5 mg/L.
 
Q  
C t  C    C 0  C   exp    k 
 t
V  
 
C7d 
5mg
 1 2 0  5 
L
C 1 0 d  2 2 .8
(5.23)
3
 

100 m d 
1
e x p    0 .2 0 d +
7d
3 
L
1500 m 
 

mg
mg
L
12. Given the following reaction:
H 2+ D2  2 HD
where D is deuterium or heavy hydrogen and the rate law is given as follows:
r  k H 2 
0 .3 8
D 2 
0 .6 6
Ar
Determine:
a. The reaction order with respect to each reactant.
b. The total or overall reaction order.
Solution Part A
The reaction order with respect to H2 is 0.38
The reaction order with respect to D2 is 0.66
The reaction order with respect to Ar is 1.00
Solution Part B
The total reaction order is 0.38 + 0.66 + 1.00 = 2.04
13. Ammonia  N H 3  is often added to water treated with chlorine to form
chloramines that provide a longer-lasting chlorine residual than hypochlorous acid
13
 H O C l  . Monochloramine  N H 2 C l  is produced according to the following
reaction.
N H 3 + H O C l  N H 2 C l+ H 2 O
The proposed rate law is:
r   k  H O C l N H 3 
Determine:
a. The reaction order with respect to each reactant.
b. The total or overall reaction order.
Solution Part A
The reaction order with respect to HOCl is 1.0
The reaction order with respect to NH3 is 1.0
Solution Part B
The total reaction order is 1.0 + 1.0 = 2.0
14. Consider the irreversible conversion of a single reactant (A) to a single product
(P) for the following reaction: A  P . Evaluate the following data to determine
whether the reaction is zero-, first-, or second-order. Also determine the
magnitude of the rate constant k and list the appropriate units.
Time (minutes)
Concentration of A (g/L)
0
1.00
11
0.50
20
0.25
48
0.10
105
0.05
Solution
First, make an arithmetic plot of the concentration of A (g/L) plotted on the ordinate axis
versus time (minutes) on the abscissa axis. If the plot produces a straight line the reaction
is zero-order and the slope of the line is equal to the value of the rate constant k.
14
1.20
Concentration of A, g/L
1.00
0.80
0.60
0.40
0.20
0.00
0
20
40
60
80
100
120
Time, minutes
Since a plot of the concentration of A versus time produces a curve, the reaction is
probably not best modeled as zero-order. Therefore, make a semi-log plot of the natural
log (ln) of the concentration of A on the ordinate axis versus time (minutes) on the
abscissa axis. If the plot produces a straight line, the reaction is first-order. The natural
log of the concentration of A must be determined before plotting as shown below. It is
recommended to use Excel or similar spreadsheet software to complete such repetitive
calculations.
Time (minutes)
Concentration of A (g/L)
ln [A]
0
1.00
0.00
11
0.50
-0.69
20
0.25
-1.39
48
0.10
-2.30
105
0.05
-3.00
15
0.00
0
20
40
60
80
100
120
-0.50
ln(Concentration of A)
-1.00
-1.50
-2.00
-2.50
-3.00
-3.50
Time, minutes
A semi-log plot of the concentration of A versus time produces a curve indicating the
reaction is not first-order. Next, make a plot of the reciprocal of the concentration of A
(g/L) on the ordinate axis versus time (minutes) on the abscissa axis. If a straight line is
produced the reaction is second-order. The reciprocal of the concentration of A must be
determined before plotting as shown below.
Time (minutes)
Concentration of A (g/L)
1/A (L/g)
0
1.00
1.0
11
0.50
2.0
20
0.25
4.0
48
0.10
10.0
105
0.05
20.0
A plot of the reciprocal of A versus time yields a straight line so the reaction is secondorder. The slope of the line is equal to the magnitude of the reaction rate constant k,
which in this case is 0.19 with units are L/(g·min).
16
25
1/A = 0.1865 t + 0.5357
R2 = 0.9963
20
1/A, L/g
15
10
5
0
0
20
40
60
80
100
120
Time, minutes
If none of the above plots produces a straight line, the reaction order may be fractional or
variable. Other texts such as Levenspiel (1972) or Metcalf and Eddy (2003) should be
consulted.
15. During biological treatment of a synthetic wastewater, the concentration of
microorganisms denoted by X was measured as a function of time. The observed
microbe concentration at various time intervals is presented in the table below.
Determine the reaction order and rate constant k and list the appropriate units.
Are the microbes being produced or destroyed?
Time (hours)
Concentration of X (mg/L)
0
40
2.0
60
4.0
110
6.0
200
8.0
325
Solution
First, make an arithmetic plot of the concentration of X (mg/L) plotted on the ordinate
axis versus time (hours) on the abscissa axis. If the plot produces a straight line the
reaction is zero-order and the slope of the line is equal to the value of the rate constant k.
17
350
Microbe Concentration, mg/L
300
250
200
150
100
50
0
0
1
2
3
4
5
6
7
8
9
Time, hours
Since a plot of the concentration of X versus time produces a curve, the reaction is
probably not best modeled as zero-order. Therefore, make a semi-log plot of the natural
log (ln) of the concentration of X on the ordinate axis versus time (hours) on the abscissa
axis. If the plot produces a straight line, the reaction is first-order. The natural log of the
concentration of X must be determined before plotting as shown below. It is
recommended to use Excel or similar spreadsheet software to complete such repetitive
calculations.
Time (hours)
Concentration of X (mg/L)
ln [X]
0
40
3.689
2
60
4.094
4
110
4.700
6
200
5.298
8
325
5.784
18
7.00
6.00
5.00
ln (X)
4.00
3.00
ln(X) = 0.2697 t + 3.6344
R2 = 0.996
2.00
1.00
0.00
0.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
Time, hours
Since a straight line is produced with the semi-log plot, microbial growth appears to
follow first-order kinetics. The slope of the line of best fit is the value of the reaction
rate constant k which is equal to 0.27 h-1. The microbes are being produced i.e. they
are growing.
16. During biological treatment of a domestic wastewater, the concentration of
chemical oxygen demand (COD) was measured as a function of time. The
observed COD concentration at various time intervals is presented in the table
below. Determine the reaction order and rate constant k and list the appropriate
units. Is COD being produced or removed?
Time (hours)
COD Concentration (mg/L)
0
493
2.0
310
4.0
181
6.0
99
8.0
60
10.0
30
19
Solution
First, make an arithmetic plot of the concentration of COD (mg/L) plotted on the ordinate
axis versus time (hours) on the abscissa axis. If the plot produces a straight line the
reaction is zero-order and the slope of the line is equal to the value of the rate constant k.
600
COD Concentration, mg/L
500
400
300
200
100
0
0
2
4
6
8
10
12
Time, hours
Since a plot of the concentration of COD versus time produces a curve, the reaction is
probably not best modeled as zero-order. Therefore, make a semi-log plot of the
natural log (ln) of the COD concentration on the ordinate axis versus time (hours) on
the abscissa axis. If the plot produces a straight line, the reaction is first-order. The
natural log of the concentration of COD must be determined before plotting as shown
below. It is recommended to use Excel or similar spreadsheet software to complete
such repetitive calculations.
Time (hours)
COD Concentration (mg/L)
ln [COD]
0
493
6.201
2
310
5.737
4
181
5.198
6
99
4.595
8
60
4.094
10
30
3.401
20
7.00
6.00
ln(COD)
5.00
4.00
3.00
ln(COD) = -0.279 t + 6.2658
R2 = 0.9969
2.00
1.00
0.00
0
2
4
6
8
10
12
Time, hours
Since a straight line is produced with the semi-log plot, COD appears to follow firstorder removal kinetics. The slope of the line of best fit is the value of the reaction
rate constant k which is equal to 0.28 h-1. COD is being removed at an exponential
rate.
17. During a chemical reaction, the concentration of Species A was measured as a
function of time. The observed concentration of Species A at various time intervals is
presented in the table below. Determine the reaction order and rate constant k and list
the appropriate units. Is Species A being removed or produced?
Time (min)
Concentration of A (mg/L)
0
0
10
18
20
41
30
59
40
82
50
105
21
Solution
First, make an arithmetic plot of the concentration of A (mg/L) plotted on the ordinate
axis versus time (minutes) on the abscissa axis. If the plot produces a straight line the
reaction is zero-order and the slope of the line is equal to the value of the rate constant k.
120
[A] = 2.1 t - 1.6667
R2 = 0.9983
Concentration of A, mg/L
100
80
60
40
20
0
0
10
20
30
40
50
60
-20
Time, minutes
Since the plot produces a straight line on arithmetic paper, the kinetics appear to be
zero-order. The slope of the line of best fit is the reaction rate constant k which is
equal to 2.1 mg/(Lmin). Species A is being produced.
18. During a chemical reaction, the concentration of Species A was measured as a
function of time. The observed concentration of Species A at various time intervals is
presented in the table below. Determine the reaction order and rate constant k and list
the appropriate units. Is Species A being removed or produced?
Time (min)
Concentration of A (mg/L)
0
105
10
78
20
62
30
40
40
21
50
1
22
Solution
First, make an arithmetic plot of the concentration of A (mg/L) plotted on the ordinate
axis versus time (minutes) on the abscissa axis. If the plot produces a straight line the
reaction is zero-order and the slope of the line is equal to the value of the rate constant k.
120
Concentration of A, mg/L
100
[A] = - 2.0371 t + 102.1
R2 = 0.9966
80
60
40
20
0
0
10
20
30
40
50
60
Time, minutes
Since the plot produces a straight line on arithmetic paper, the kinetics appear to be zeroorder. The slope of the line of best fit is the reaction rate constant k which is equal to
2.04 mg/(Lmin). Species A is being removed.
19. 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
3.0
5.0
7.0
9.0
11.0
Concentration of D (mg/L)
200
160
105
65
45
30
20
23
Solution
First, make an arithmetic plot of the concentration of D (mg/L) plotted on the ordinate
axis versus time (hours) on the abscissa axis. If the plot produces a straight line the
reaction is zero-order and the slope of the line is equal to the value of the rate constant k.
250
Concentration of D, mg/L
200
150
100
50
0
0
2
4
6
8
10
12
Time, hours
Since a plot of the concentration of D versus time produces a curve, the reaction is
probably not best modeled as zero-order. Therefore, make a semi-log plot of the
natural log (ln) of the concentration of D on the ordinate axis versus time (hours) on
the abscissa axis. If the plot produces a straight line, the reaction is first-order. The
natural log of the concentration of D must be determined before plotting as shown
below. It is recommended to use Excel or similar spreadsheet software to complete
such repetitive calculations.
Time (hr)
0
1.0
3.0
5.0
7.0
9.0
11.0
Concentration of D (mg/L)
200
160
105
65
45
30
20
24
Ln [D]
5.30
5.08
4.65
4.17
3.81
3.40
3.00
6.00
5.00
ln[D]
4.00
3.00
2.00
ln[D] = -0.2093 t + 5.2771
R2 = 0.999
1.00
0.00
0
2
4
6
8
10
12
Time, hours
Since the semi-log plot produces a straight line, the kinetics appear to be first-order.
Slope of the line of best fit is the reaction rate constant k which is equal to 0.21h-1.
Species D is being removed at an exponential rate.
20. 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 (d)
0
2.0
4.0
6.0
8.0
10.0
12.0
Concentration of B (g/L)
50
60
75
90
110
140
170
25
Solution
First, make an arithmetic plot of the concentration of B (g/L) plotted on the ordinate axis
versus time (days) on the abscissa axis. If the plot produces a straight line the reaction is
zero-order and the slope of the line is equal to the value of the rate constant k.
180
160
Concentration of B, g/L
140
120
100
80
60
40
20
0
0
2
4
6
8
10
12
14
Time, days
Since a plot of the concentration of B versus time produces a curve, the reaction is
probably not best modeled as zero-order. Therefore, make a semi-log plot of the
natural log (ln) of the concentration of B on the ordinate axis versus time (days) on
the abscissa axis. If the plot produces a straight line, the reaction is first-order. The
natural log of the concentration of B must be determined before plotting as shown
below. It is recommended to use Excel or similar spreadsheet software to complete
such repetitive calculations.
Time (d)
0
2.0
4.0
6.0
8.0
10.0
12.0
Concentration of B
(g/L)
50
60
75
90
110
140
170
26
Ln [ B]
3.91
4.09
4.32
4.50
4.70
4.94
5.14
6.00
5.00
ln[B]
4.00
3.00
ln[B] = 0.1027 t + 3.8986
R2 = 0.999
2.00
1.00
0.00
0
2
4
6
8
10
12
14
Time, days
Since the semi-log plot produces a straight line, the kinetics appear to be first-order.
The slope of the line of best fit is the reaction rate constant k which is equal to 0.103
d-1. Species B is being produced at an exponential rate.
21. During a chemical reaction, the concentration of Species C was measured as a
function of time. 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 (h)
0
5.0
10.0
15.0
20.0
30.0
40.0
50.0
75.0
100
Concentration of C (mg/L)
500
20
9.5
6.5
5.0
3.5
2.5
2.0
1.5
1.0
27
Solution
First, make an arithmetic plot of the concentration of C (mg/L) plotted on the ordinate
axis versus time (hours) on the abscissa axis. If the plot produces a straight line the
reaction is zero-order and the slope of the line is equal to the value of the rate constant k.
600
Concentration of C, mg/L
500
400
300
200
100
0
0
20
40
60
80
100
120
-100
Time,hours
Since a plot of the concentration of C versus time produces a curve, the reaction is
probably not best modeled as zero-order. Therefore, make a semi-log plot of the
natural log (ln) of the concentration of C on the ordinate axis versus time (hours) on
the abscissa axis. If the plot produces a straight line, the reaction is first-order. The
natural log of the concentration of C must be determined before plotting as shown
below. It is recommended to use Excel or similar spreadsheet software to complete
such repetitive calculations.
Time (h)
0
5.0
10.0
15.0
20.0
30.0
40.0
50.0
75.0
100
Concentration of C (mg/L)
500
20
9.5
6.5
5.0
3.5
2.5
2.0
1.5
1.0
28
Ln[C]
6.21
3.00
2.25
1.87
1.61
1.25
0.92
0.69
0.41
0.00
7.00
6.00
5.00
ln[C]
4.00
3.00
2.00
1.00
0.00
0
20
40
60
80
100
120
Time, hours
A semi-log plot of the concentration of C versus time produces a curve indicating the
reaction is not first-order. Next, make a plot of the reciprocal of the concentration of C
(mg/L) on the ordinate axis versus time (hours) on the abscissa axis. If a straight line is
produced the reaction is second-order. The reciprocal of the concentration of C must be
determined before plotting as shown below.
Time (h)
0
5.0
10.0
15.0
20.0
30.0
40.0
50.0
75.0
100
Concentration of C (mg/L)
500
20
9.5
6.5
5.0
3.5
2.5
2.0
1.5
1.0
29
1/C (L/mg)
0.002
0.050
0.105
0.154
0.200
0.286
0.400
0.500
0.667
1.000
1.20
1/C = 0.0096 t + 0.0044
R2 = 0.9942
1/(Concentration of C), L/mg
1.00
0.80
0.60
0.40
0.20
0.00
0
20
40
60
80
100
120
Time, hours
A plot of the reciprocal of C versus time yields a straight line so the reaction is secondorder. The slope of the line is equal to the magnitude of the reaction rate constant k,
which in this case is 0.0096 with units of L/(g·h). Species C is being removed.
22. During a chemical reaction, the concentration of protein P was measured as a function
of time. The observed concentration of P at various time intervals is presented below.
Determine the reaction order and rate constant, k. Is P being removed or produced?
Time (h)
0
10.0
15.0
20.0
30.0
50.0
100.0
Concentration of P (mg/L)
0.50
0.55
0.58
0.62
0.70
0.95
10.00
Solution
First, make an arithmetic plot of the concentration of P (mg/L) plotted on the ordinate
axis versus time (hours) on the abscissa axis. If the plot produces a straight line the
reaction is zero-order and the slope of the line is equal to the value of the rate constant k.
30
12
Concentration of P, mg/L
10
8
6
4
2
0
0
20
40
60
80
100
120
Time, hours
Since a plot of the concentration of P versus time produces a curve, the reaction is
probably not best modeled as zero-order. Therefore, make a semi-log plot of the
natural log (ln) of the concentration of P on the ordinate axis versus time (hours) on
the abscissa axis. If the plot produces a straight line, the reaction is first-order. The
natural log of the concentration of P must be determined before plotting as shown
below. It is recommended to use Excel or similar spreadsheet software to complete
such repetitive calculations.
Time (h)
0
10.0
15.0
20.0
30.0
50.0
100.0
Concentration of P (mg/L)
0.50
0.55
0.58
0.62
0.70
0.95
10.00
31
Ln[P]
-0.693
-0.598
-0.545
-0.478
-0.357
-0.051
2.303
2.50
2.00
1.50
ln[P]
1.00
0.50
0.00
0
20
40
60
80
100
120
-0.50
-1.00
Time, hours
A semi-log plot of the concentration of P versus time produces a curve indicating the
reaction is not first-order. Next, make a plot of the reciprocal of the concentration of P
(mg/L) on the ordinate axis versus time (hours) on the abscissa axis. If a straight line is
produced the reaction is second-order. The reciprocal of the concentration of P must be
determined before plotting as shown below.
Time (h)
0
10.0
15.0
20.0
30.0
50.0
100.0
Concentration of P (mg/L)
0.50
0.55
0.58
0.62
0.70
0.95
10.00
32
1/P
2.000
1.818
1.724
1.613
1.429
1.053
0.100
2.50
1/(Concentration of P), L/g
2.00
1/P = - 0.019 t + 2.0026
R2 = 0.9999
1.50
1.00
0.50
0.00
0
20
40
60
80
100
120
Time,hours
A plot of the reciprocal of P versus time yields a straight line so the reaction is secondorder. The slope of the line is equal to the magnitude of the reaction rate constant k,
which in this case is 0.019 with units of L/(mg·h). Species P is being produced.
23. Many biological reactions can be modeled as a mixed-order function commonly
referred to as a Monod function named after the French microbiologist. Monod
found that microbial growth can be modeled with the follow equation:
 
 m ax S
Ks  S
where:
µ = specific growth rate of microorganism, d-1,
µmax= maximum specific growth rate of microorganism, d-1,
S = external substrate concentration, mg/L, and
Ks = half-velocity constant, mg/L.
During a biological reaction, the concentration of an amino acid denoted as C was
measured as a function of time, along with the rate of production denoted as r.
The observed concentration of C and rate of production of the amino acid at
various time intervals is presented below. The biological reaction can be modeled
by the following equation analogous to the equation proposed by Monod.
r 
kC
Ks  C
where:
r = rate of amino acid production, d-1,
k = maximum rate of amino acid production, d-1,
C = amino acid concentration, mg/L, and
Ks = half-velocity constant of the amino acid, mg/L.
33
The coefficient k and Ks can be determined by performing a double reciprocal plot of
1/r versus 1/C. Using the data in the table below, rearrange the equation presented
above to determine the slope and y-intercept of the line of best fit through the data so
that k and Ks
r (d-1)
C (mg/L)
0.85
10
1.50
20
2.50
50
3.00
75
3.75
150
4.00
200
Solution
First, rearrange the above equation to get it in the form of the equation of a straight line
where: Y = mX + B.
r 
kC
Ks  C
First invert both sides of the equation as follows and then separate variables.
1

Ks  C
r
1
kC

r
1
r
Ks

kC

kC
Ks 1
k
C
C

1
k
The equation is now in the form of a straight line with slope (m) equal to
K
s
and Y-
k
intercept equal to
1
. Before a plot of
k
1
r
vs
1
can be made, the reciprocal of the rate of
C
production and reciprocal of the concentrations must be determined as presented in the
table below.
r (d-1)
0.85
1.50
2.50
3.00
3.75
4.00
1/r (d)
1.176
0.667
0.400
0.333
0.267
0.250
C (mg/L)
10
20
50
75
150
200
34
1/C (L/mg)
0.1000
0.0500
0.0200
0.0133
0.0067
0.0050
1.40
1.20
1/r = 9.6811(1/C ) + 0.2009
R2 = 0.9993
1/r , days
1.00
0.80
0.60
Y -intercept = 0.2009=1/k
k = 4.98 d-1
0.40
Slope = K s /k = 9.6811
K s = 9.6811(4.98) = 48 mg/L
0.20
0.00
0.00
0.02
0.04
0.06
0.08
0.10
0.12
1/C , L/mg
A plot of
1
r
vs
1
produces a straight line which indicates the data follow a mixed-order.
C
The slope and y-intercept for the line of best fit are 9.68 (days·mg)/L and 0.20 days,
respectively. Therefore, k and Ks calculated as follows:
Y  in te r c e p t 
1
 0 .2 0 0 9 d a y s
k
1
k 

4 .9 8 d
-1
0 .2 0 0 9 d
S lo p e 
Ks
 9 .6 8 1 1
k
d mg
L
d mg 
d mg 


-1
K s   9 .6 8 1 1
  k   9 .6 8 1 1
  4 .9 8 d
L
L




Ks 
48 m g
L
35
24. If the half-life of a vinyl chloride in surface water is 5 days, estimate the
concentration of vinyl chloride remaining in a water sample after setting in a
laboratory for two weeks if the initial concentration was 20 mg/L.
Solution
Knowing the half-life of a compound allows one to calculate the reaction rate constant k.
The value of k is determined by substituting into Equation (5.49).
t1
2
 ln ( 0 .5 )


0 .6 9 3
k
k 
0 .6 9 3
t1
(5.49)
k

0 .6 9 3
 0 .1 4 d
-1
5 .0 d
2
Next, the concentration of vinyl chloride remaining in the sample after two weeks in the
laboratory can be determined from Equation (5.47).
Ct  Co e
C 14 d 
20 m g
e
 0 .1 4 d
-1
14 d
kt

(5.47)
2 .8
mg
L
L
25. Determine the half-life of a chemical compound if first-order removal kinetics were
observed to be followed during a laboratory study. The initial and final concentration
of the chemical at time 0 and 10 days later was 10 mg/L and 1.3 mg/L, respectively.
Solution
Equation (5.47) must be rearranged to solve for the reaction rate constant k.
k 
ln  C 0 C t

Ct  Co e
kt
ln  C t
ln  C o   k t


(5.47)
 10 m g L 
ln 

 1 .3 m g L 
t

0 .2 0 d
-1
10 d
Equation (5.49) then may be used to calculate the half-life of the chemical compound.
t1
2

 ln ( 0 .5 )
k

0 .6 9 3
k
36
(5.49)
t1
2

 ln ( 0 .5 )

0 .6 9 3
k

k
0 .6 9 3
0 .2 0 d
-1

3 .5 d
26. If the first-order removal rate constant k for a chemical compound is 0.5h-1, determine
the half-life in days.
Solution
Equation (5.49) then may be used to calculate the half-life of the chemical compound.
t1
2

t1
2

t1
2
 ln ( 0 .5 )

0 .6 9 3
k
(5.49)
k
 ln ( 0 .5 )

0 .6 9 3
k

k
0 .6 9 3
0 .5 h
 1d 
 1 .3 9 h 
 
 24 h 
-1
 1 .3 9 h
0 .0 5 8 d
27. If the half-life of DDT in surface water is 56 days determine the first-order removal
rate constant k in d-1.
Solution
Equation (5.49) then may be rearranged to calculate the reaction rate constant k given the
half-life of a chemical compound.
t1
2

k 
 ln ( 0 .5 )

0 .6 9 3
k
0 .6 9 3
t1
2
(5.49)
k

0 .6 9 3

1 .1 4  1 0
2
d
-1
56 d
28. Several reactor configurations are to be considered for reducing the influent substrate
concentration from 200 mg/L to 20 mg/L at a design flow rate of 38,000 m3/d.
Assume that substrate removal follows first-order kinetics and the first-order rate
constant k is 6.0 h-1. Determine the reactor volume required for the following
configurations operating at steady-state.
a. One ideal plug flow reactor.
b. One ideal complete-mix reactor.
c. Three ideal complete-mix reactors in series.
d. Ten ideal complete-mix reactors in series.
37
Solution Part A
We will start with Equation (5.89) which was derived for one-dimensional plug
flow and substitute a first-order removal reaction for the reaction term: r   k C .
Q C

 A  X

  r

Q d C

 A d X

   k C

1
k

Ct
1
C0
C
dC  
A
Q
ln  C t   ln  C 0

 
k
V  Q
V 


L
dX
0
AL
V
 
Q
ln  C 0   ln  C t

k
Q
 3 8 ,0 0 0 m
 
d

3
 l n  2 0 0   ln  2 0   1 d 



-1
6 .0 h

 24 h 
3
608 m
Solution Part B
Start with Equation (5.74) and substitute the appropriate influent and effluent
concentrations along with the reaction rate constant k into the equation.
 
V
Q
V 

Co
 Ct


k Ct
Q  C o C t  1
Co
C t  1
3 8, 0 0 0 m

k
V 
2375 m
(5.74)
k
3
d
 200 m g
6 .0 h
-1
L
20 m g L
1

1d 


 24 h 
3
The volume of one complete-mix flow reactor is much larger than one plug flow reactor
i.e. 3.9 times larger in volume.
Solution Part C
We start with Equation (5.83) and substitute the appropriate influent and effluent
concentrations along with the reaction rate constant k and number of complete38
mix reactors into the equation. Then the equation must be rearranged to solve for
the detention time . The detention time in Equation (5.83) represents the
detention time in each of the complete-mix reactors. The total detention time is
determined by multiplying  by n, the number of reactors in series.
n
Cn


1
 C0 

1 k 
C3


1
 C0 

-1
 1  6 .0 h   
C3
C0
(5.83)


1
 

-1
 1  6 .0 h   
3
3


1
 

-1
200 m g L
 1  6 .0 h   
3
20 m g L
 20 m g L 


 200 m g L 
 0 .1 0 
 13 
 0 .1 0 
 13 

 13 


1
 

-1
 1  6 .0 h   
1
1  6 .0 h
1 
6 .0 h
0 .4 6 4  1  6 .0 h
-1
-1
 
-1

 


1
1
0 .4 6 4  2 .0 7 8   1
  0 .1 9 h p e r r e a c to r
The total detention time in the three reactors is:
n    3  0 .1 9 h = 0 .5 7 h
Recall, that detention time is calculated as follows:
 
V
.
Q
Therefore,
V    Q  0 .5 7 h  3 8 , 0 0 0
 1d 
3
 
  903 m
d
 24 h 
m
3
39
The total volume of three complete-mix flow reactor in series is closer in volume to one
plug flow reactor. Three complete-mix reactors are approximately 1.5 times the volume
of the single plug flow reactor volume.
Solution Part D
We start with Equation (5.83) and substitute the appropriate influent and effluent
concentrations along with the reaction rate constant k and number of complete-mix
reactors into the equation. Then the equation must be rearranged to solve for the
detention time . The detention time in Equation (5.83) represents the detention time in
each of the complete-mix reactors. The total detention time is determined by multiplying
 by n, the number of reactors in series.
Cn
C 10
C 10
C0


1
 C0 

1 k 
n
(5.83)


1
 C0 

-1
 1  6 .0 h   


1
 

-1
 1  6 .0 h   
10
10


1
 

-1
200 m g L
 1  6 .0 h   
10
20 m g L
 20 m g L 


 200 m g L 
 0 .1 0 
 11 0 
 0 .1 0 
 11 0 

 11 0 


1
 

-1
 1  6 .0 h   
1
1  6 .0 h
 1  6 .0 h
0 .7 9 4  1  6 .0 h
-1
 
-1
-1

 


1
1
0 .7 9 4  4 .7 7   1
  0 .0 4 3 h p e r r e a c to r
The total detention time in the three reactors is:
40
n    1 0  0 .0 4 3 h = 0 .4 3 h
Recall, that detention time is calculated as follows: 

V
.
Q
Therefore,
V    Q  0 .4 3 h  3 8 , 0 0 0
 1d 
3
 
  681 m
d
 24 h 
m
3
The total volume of ten complete-mix flow reactors in series is approaching the volume
of one plug flow reactor. Ten complete-mix reactors are approximately 1.12 times the
volume of the single plug flow reactor volume.
29. Several reactor configurations are to be considered for reducing the influent substrate
concentration from 100 mg/L to 15 mg/L at a design flow rate of 5 million gallons per
day (MGD). Assume that substrate removal follows first-order kinetics and the firstorder rate constant k is 8.0 d-1. Determine the reactor volume required for the
following configurations operating at steady-state.
a. One ideal plug flow reactor.
b. One ideal complete-mix reactor.
c. Three ideal complete-mix reactors in series.
d. Ten ideal complete-mix reactors in series.
Solution Part A
We will start with Equation (5.89) which was derived for one-dimensional plug
flow and substitute a first-order removal reaction for the reaction term: r   k C .
Q C

 A  X

  r

Q d C

 A d X

   k C

1
k

Ct
1
C0
C
A
dC  
Q
ln  C t   ln  C 0

 
k
V  Q
V 

ln  C 0
1 .5 9  1 0

ft
L
dX
0
AL
Q
ln  C t
k
5


 
V
Q
3

 5 ,0 0 0 ,0 0 0 g a l  l n  1 0 0   ln  1 5   1 f t
 



-1
d
8 .0 d


 7 .4 8 g a l 
3
41
Solution Part B
Start with Equation (5.74) and substitute the appropriate influent and effluent
concentrations along with the reaction rate constant k into the equation.
 
V

Co
Q
V 
 Ct

Co

k Ct
C t  1
Q  C o C t  1
5, 0 0 0, 0 0 0

k
V 
4 .7 3  1 0
5
(5.74)
k
ft
 100 m g L

 1

3
d

 15 m g L
  1 ft


-1
8 .0 d
 7 .4 8 g a l 
gal
3
The volume of one complete-mix flow reactor is much larger than one plug flow reactor
i.e. approximately 3 times larger in volume.
Solution Part C
We start with Equation (5.83) and substitute the appropriate influent and effluent
concentrations along with the reaction rate constant k and number of completemix reactors into the equation. Then the equation must be rearranged to solve for
the detention time . The detention time in Equation (5.83) represents the
detention time in each of the complete-mix reactors. The total detention time is
determined by multiplying  by n, the number of reactors in series.
n
Cn


1
 C0 

1 k 
C3


1
 C0 

-1
 1  8 .0 d   
C3
C0
(5.83)


1
 

-1
 1  8 .0 d   
3
3


1
 

-1
100 m g L
 1  8 .0 d   
3
15 m g L
 15 m g L 


 100 m g L 
 13 


1
 

-1
 1  8 .0 d   
42
 0 .1 5 
 13 
 0 .1 5 
 13 

1
1  8 .0 d
 1  8 .0 d
0 .5 3 1  1  8 .0 d
-1
-1
 
-1

 


1
1
0 .5 3 1  4 .2 4 8   1
  0 .1 1 d p e r r e a c to r
The total detention time in the three reactors is:
n    3  0 .1 1 d = 0 .3 3 d
Recall, that detention time is calculated as follows:
V
 
.
Q
Therefore,
V    Q  0 .3 3 d  5 , 0 0 0 , 0 0 0
 1 ft

5
3
 
  2 .2 1  1 0 ft
d
 7 .4 8 g a l 
3
gal
The total volume of three complete-mix flow reactor in series is closer in volume to one
plug flow reactor. Three complete-mix reactors are approximately 1.4 times the volume
of the single plug flow reactor volume.
Solution Part D
We start with Equation (5.83) and substitute the appropriate influent and effluent
concentrations along with the reaction rate constant k and number of complete-mix
reactors into the equation. Then the equation must be rearranged to solve for the
detention time . The detention time in Equation (5.83) represents the detention time in
each of the complete-mix reactors. The total detention time is determined by multiplying
 by n, the number of reactors in series.
Cn
C 10
C 10
C0


1
 C0 

1 k 
n
(5.83)


1
 C0 

-1
 1  8 .0 d   


1
 

-1
 1  8 .0 d   
10
43
10


1
 

-1
100 m g L
 1  8 .0 d   
10
15 m g L
 15 m g L 


 100 m g L 
 0 .1 5 
 11 0 
 0 .1 5 
 11 0 

 11 0 


1
 

-1
 1  8 .0 d   
1
1  8 .0 d
 1  8 .0 d
0 .8 2 7  1  8 .0 d
-1
-1
 
-1


 

1
1
0 .8 2 7  6 .6 1 6   1
  0 .0 2 6 d p e r r e a c to r
The total detention time in the three reactors is:
n    1 0  0 .0 2 6 d = 0 .2 6 d
Recall, that detention time is calculated as follows: 

V
.
Q
Therefore,
V    Q  0 .2 6 d  5 , 0 0 0 , 0 0 0
 1 ft

5
3
 
  1 .7 4  1 0 ft
d
7
.4
8
g
a
l


gal
3
The total volume of ten complete-mix flow reactors in series is approaching the volume
of one plug flow reactor. Ten complete-mix reactors are approximately 1.1 times the
volume of the single plug flow reactor volume.
30. A complete-mix flow reactor is designed to treat an influent waste stream containing
130 mg/L of casein at a flow rate of 380 liters per minute (Lpm). Assume that casein
removal follows first-order removal kinetics with a rate constant k of 0.5 h-1 and that
the effluent should contain 13 mg/L of casein at steady state. Determine:
a. The detention time in hours.
b. The volume of the reactor in cubic meters.
Solution Part A
Substitute the influent and effluent casein concentration and the rate constant k into
Equation (5.73) to calculate the detention time.
44
 
V

Co
Q
 
V
 Ct


Co
C t  1
k Ct

Co
Q
(5.73)
k
 Ct


k Ct
1 3 0 - 1 3  m g
0 .5 h
-1
1 3 m g
L
L
 18 h
Solution Part B
The volume of the complete-mix reactor is calculated by multiplying the detention time
time by the flow rate as follows:
V    Q  18 h× 380
L
×
6 0 m in
m
×
h
m
3
 410 m
3
1000 L
31. A complete-mix flow reactor is designed to treat an influent stream containing 150
mg/L of total organic carbon (TOC) at a flow rate of 150 gallons per minute (gpm).
Assume that TOC removal follows first-order removal kinetics with a rate constant k
of 0.4 h-1. The volume of the CMFR is 13,500 ft3and steady-state conditions exist.
Determine:
a. The detention time in hours.
b. The effluent TOC concentration.
Solution Part A
The detention time is defined as the volume divided by the flow rate, Equation (5.63)
 
V

Q
 
3
 7 .4 8 g a l

3
1 5 0 g a l m in 
ft
1 3 ,5 0 0 f t
 1d 
 1h



  6 0 m in   2 4 h 
(5.63)
0 .4 7 d
Solution Part B
The effluent TOC concentration from a complete-mix reactor with a first-order removal
reaction can be determined by rearranging Equation (5.73).
 
V
Q

Co
 Ct


Co
k Ct
C t  1
k
 k Ct  Co  Ct
C t 1   k   C o
45
(5.73)
Ct 
Co
1  
 k


150 m g L
 1 + 1 8 h × 0 .4 h 
-1
=
1 8 .3
mg
L
32. A plug flow reactor (PFR) is designed to treat an influent stream containing 200 mg/L
of acetic acid at a flow rate of 400 liters per minute (Lpm). The reactor has been in
operation for several months and acetic acid removal is observed to follow secondorder removal kinetics with a rate constant k of 0.0085 L/(mgh). Ninety percent
acetic acid removal is required. Determine:
a. The detention time in hours.
b. The volume of the plug flow reactor in m3.
Solution Part A
The detention time can be calculated by substituting the influent and effluent acetic acid
concentrations and reaction rate constant into Equation (5.93). First though, the effluent
acetic acid concentration must be determined so that 90% removal is achieved. Equation
(5.94) presented below must be rearranged to solve for the effluent acetic acid
concentration.
 C in  C o u t  1 0 0
p e rc e n t re m o v a l  %  
(5.94)
C in
90 =
 200
m g L - C out  1 0 0
200 m g L
mg 

0 .9 0  2 0 0
 = 2 0 0 m g L - C out
L 

C out  2 0
 
0 .0 0 8 5
 
L
 1 
1  1 

  

k   C t 
 C o  
1
 
mg
L
(5.93)




1
1

  

 200 m g L 
 20 m g L 
mg h
5 .3 h
Solution Part B
The volume of the plug flow reactor can be calculated by multiplying the detention time
by the flow rate.
46
 
V
(5.63)
Q

 6 0 m in   m



m in 
h
  1000 L 
V  127 m
3
L
V    Q  5 .3 h  4 0 0
3
33. A plug flow reactor (PFR) is designed to treat 10 million gallons per day (MGD) of
industrial wastewater containing contaminant A. Bench-scale studies indicate that
contaminant A removal follows first-order removal kinetics with a reaction rate
constant k of 9.0 d-1. Steady-state conditions exist and 95% removal of contaminant A
is required. Determine:
a. The detention time in hours.
b. The volume of the plug flow reactor in ft3.
Solution Part A
To solve for the detention time, we start out with the one dimensional flow equation for
plug flow, Equation (5.89) and substitute a first-order removal reaction for the reaction
term as follows:
Q C

 A  X
Q d C

  r

(5.89)
 kC
A d X
The equation above is integrated, with integration limits for C going from C0, the initial
concentration at the front of the reactor, where length is equal to zero to Ct, the
concentration at the end of the length of the reactor, L.


Ct
1
C0
C
Ct
1
C0
C
dC   k
A
Q
dC   k
A
Q


L
dX
0
L
dX
0
 Ct 
AL
V
ln 
 k
 k
 k
Q
Q
 C0 
 
C 
ln  0 
 Ct 
k

 0 .0 5 C 0 
ln 

C0


 9 .0 d
1
47
 0 .3 3 d
Solution Part B
The volume of the plug flow reactor can be calculated by multiplying the detention time
by the flow rate.
 
V
(5.63)
Q

gal 
ft


d  7 .4 8 g a l 
3
V    Q  0 .3 3 d  1 0  1 0
V  4 .4 1  1 0 ft
5
34.
5
3
A 1,500 megawatt (MW) coal-fired power plant has a 33% efficiency of
converting the energy of coal into electrical energy. The coal has an energy
content of 24 kJ/ g and contains 55% carbon, 2.0% sulfur, and 7% ash. Also
assume that 65% of the ash in the coal is released as fly ash with 35% of it settling
outside of the firing chamber where it is collected as bottom ash. Approximately
15% of the waste heat is assumed to exit in the stack gases, and the cooling water
dissipates the remaining heat. Assume that air emission standards restrict sulfur
and particulate quantities to 260 g SO2 per 106 kJ of heat input and 13 g
particulates per 106 kJ of heat input into the coal-fired power plant. Perform a
materials and energy balance around the coal-fired power plant to answer the
following questions and draw a simplified schematic of the process.
Determine:
a. The quantity of heat loss to the cooling water (MW).
b. The quantity of cooling water (kg/s) and flow (m3/s) assuming a 10ºC
increase in the temperature of the cooling water.
c. The efficiency of the sulfur dioxide removal system to meet air
emission standards.
d. The efficiency of the particulate removal system to meet air emission
standards.
Solution Part A
Perform the energy balance around the coal-fired power plant using Equation (5.26) as
the basis.
 th e r a te o f c h a n g e o f 
 th e r a te a t w h ic h 
 th e r a te a t w h ic h 






to ta l e n e r g y

e n e r g y e n te r s

e n e r g y le a v e s












in C V
CV
CV
(5.26)
Recall that at steady state, the energy accumulated is zero and the equation reduces to the
following form.
 th e ra te a t w h ic h 
 th e ra te a t w h ic h 




e n e rg y e n te rs

e n e rg y le a v e s








CV
CV
48
e n e rg y o u t
e n e rg y o u t
 e n e rg y in 
 e n e rg y o u t  
 


  
  
  

coal


 in s ta c k g a s e s   in c o o lin g w a te r   u s e fu l e le c tria l p o w e r 
We estimate the energy in the coal by dividing the useful energy produced as electrical
power by the efficiency of the coal-fired power plant as follows:
in p u t p o w e r 
o u tp u t p o w e r

1500 M W e
e ffic ie n c y
0 .3 3
= 4545 M W t
Determine the total energy losses in the system as follows:
total losses = energy input – energy output = 4545 – 1500 = 3045 MWt
Estimate the stack losses assuming 15% of the total energy losses as follows:
stack losses = 0.15 (3045 MWt) = 457 MWt
Calculate the energy loss in the cooling water.
e n e rg y o u t
e n e rg y o u t
 e n e rg y in 
 e n e rg y o u t  
 


  
  
  

coal


 in s ta c k g a s e s   in c o o lin g w a te r   u s e fu l e le c tria l p o w e r 
4545
MWt
 = 457 M W t  
e n e rg y o u t



  1 5 0 0 M W e
 in c o o lin g w a te r 

e n e rg y o u t


2588 M W t

 
 in c o o lin g w a te r 
Solution Part B
The mass flow rate  m  of water required for cooling is calculated from Equation (5.25)
and a specific heat (c) equal to 4.18 kJ/(kg·ºC).
rate
of change
in stored
energy

m c  T


kJ
o
2 5 8 8 M W t  m  4 .1 8
 1 0 C
o
k
g

C


(5.25)
 1 M W   1000 J 


6
 10 J s   kJ 

m  6 .1 9 × 1 0 k g s
4
The volumetric flow rate of the cooling water is determined by dividing the mass flow
rate by the density of the water (1000 kg/m3).
49
Q 
m

4

6 .1 9 × 1 0 k g s
1000 kg m
3
=
6 1 .9 m
3
s
Solution Part C
Calculate the quantity of SO2 produced and that can be emitted per unit of heat input into
the coal-fired power plant. First calculate the quantity of heat input into the coal-fired
plant as follows:
 10 W   1 kW   24 h   1 kJ s   3600 s 
11 k J
4545 M W 




  3 .9 3  1 0
h
d

 1 M W   1000 W   d   kW  
6
Next, calculate the quantity of coal input into the coal-fired plant as follows:
q u a n tity o f c o a l b u r n e d d a ily = 3 .9 3 × 1 0
11
kJ  g   1kg 
7 kg
  1 .6 4  1 0


d  24kJ   1000 g 
d
Next determine the quantity of SO2 that is produced daily knowing that the molecular
weight of sulfur dioxide is 32 + 2(16) = 64.
S O 2 p r o d u c e d = 1 .6 4 × 1 0
7
kg coal 
kg S   64 kg SO 2 
5 kg SO 2
 0 .0 2

  6 .5 6  1 0
d
kg coal   32 kg S 
d

The quantity of SO2 that is permitted to be discharged daily to the atmosphere is
calculated as:
 260 g SO 2  
 1 kg 
11 k J 
5 kg
S O 2 d is c h a r g e d = 
  3 .9 3  1 0
  1 .0 2  1 0

6
d   1000 g 
d
 10 kJ  
The efficiency of air pollution control equipment can be calculated from the following
equations:
r e moval
efficiency

C in
 C out  100

C in
S O 2 rem o v a l e ffic ie n c y  
M
in
 M
M
 6 .5 6 × 1 0
5
- 1 .0 2 × 1 0
5
out
 100
(5.94)
in
5
 kg
d ×100
=
8 4 .4 %
6 .5 6 × 1 0 k g d
Solution Part D
Determine the quantity of particulates or fly ash that enters the air pollution control
equipment.
f ly a s h p r o d u c e d = 1 .6 4 × 1 0
7
kg coal 
kg ash  
k g f ly a s h 
5 kg
 0 .0 7
  0 .6 5
  7 .4 6  1 0
d
kg coal  
kg ash 
d

50
The quantity of fly ash or particulate matter that is permitted to be discharged daily is
calculated as:
 1 3 g p a r tic u la te s  
 1 kg 
11 k J 
3 k g p a r tic u la te s

  3 .9 3  1 0
  5 .1 1  1 0

6
10 kJ
d   1000 g 
d


Estimate the particulate removal efficiency to meet air standards.
 7 .4 6 × 1 0
p a rtic u la te rem o v a l e ffic ie n c y 
5
k g d - 5 .1 1 × 1 0
5
3
 100
=
9 9 .3 %
7 .4 6 × 1 0 k g d
A simplified schematic of the materials and energy flow through the coal-fired power
plant is presented below.
457 MW t
Energy loss
Particulates:
5.11x10
x10 3 kg/d
SO 2: 1.02x10 5 kg/d
4545 MW e
Energy input
3.93x10 11kJ/d
1.64x10 7 kg/d coal
1.15x10 6 kg/d ash
3.28x10 5 kg/d S
2588 MW t
Energy loss
to cooling
water
1/3%%
33.5
33%
Efficient
Power
power
Plant
plant
Fly ash
4.02x10 5 kg/d
SO2
6.56x10 5 kg/d
Air
pollution
Pollution
Control
control
System
system
Stack
1500 MW e
Electrical
output
7.46x10 5 kg/d
Bottom ash
35. A solar collector panel with a surface area of 32 ft2 receives energy from the sun
at a rate of 160 BTU per hour per ft2 of surface area. If 40% of the incoming
energy is lost to the surroundings with the remainder going to heat water from 100
to 150°F, determine how many gallons of water at 150°F can be produced by
eight solar collector panels in a 30 minute time period. There is a negligible
pressure drop through the panel. Neglect potential and kinetic energy effects.
51
Solution
Start with Equation (5.130) assuming that the Rate of Energy Adsorbed is equal to the
rate of Energy Stored.
 ra te o f
c h a n g e in s to re d e n e rg y   m c  T   ra te o f e n e rg y a d s o rb e d 
 r a te o f e n e r g y a d s o r b e d  
160 BT U
 r a te
24, 576 B T U
o f e n e rg y a d so rb e d  
h  ft
2
3 2 ft

c o lle c to r
1 BTU
lb 
m  4 9 1 .5 2
1 5 0
 8 c o lle c to r s   1  0 .4 0 
h
m c  T   r a te o f e n e r g y a d s o r b e d  
m 
2

F  100 F 
24, 576 B T U
h
24, 576 B T U
h
lb
h
V o lu m e tr ic f lo w r a te 
m

G a llo n s p r o d u c e d = 5 8 .9

4 9 1 .5 2 lb h  7 .4 8 g a l 
gal
  5 8 .9
3 
3
6 2 .4 lb f t 
ft
h


gal  1 h

  3 0 m in
h  6 0 m in 

 2 9 .5 g a l
36. A bomb calorimeter is a device used to determine the heat energy value of
materials when they are combusted. Typically, a sample is place into a stainless
steel ball to which oxygen under high pressure is added. The bomb is placed into
an adiabatic water bath so that heat cannot be transferred to the surroundings.
Wires lead from the bomb to an electrical source so that a spark can be
administered to combust the material. Determine the energy content of refuse
derived fuel (RDF) if a 10 gram sample is placed into a bomb calorimeter that
holds 5 L of water. During combustion, the temperature in the water bath
increases by 15°C. Ignore the mass of the bomb.
Solution
The rate at which energy enters the system is equal to the amount of thermal energy that
is generated within the bomb calorimeter that is dissipated by the water bath. We will
use a modification of Equation (5.20) since mass is not flowing into or out of the control
volume.
(5.20)
 c h a n g e in s to re d e n e rg y   m c  T
52
 c h a n g e in
change
 1m

 1 0 0 0 k g   4 .1 8 k J 
s to r e d e n e r g y    5 L   

 15 C 

3
o
m

  kg  C 
 1000 L 


3
in s to re d e n e rg y   3 1 4 k J
The energy of the refuse derived fuel (RDF) is estimated by dividing the change in stored
energy by the grams of RDF.
e n e rg y o f R D F 
314 kJ

3 1 .4
10 g
kJ
g
37. Three complete-mix reactors in series treat a municipal wastewater flow rate of
1.0 million gallons per day (MGD) containing 200 mg/L of BOD5. If the volume
of each reactor is 0.5 million gallons and the BOD removal rate coefficient, k =
0.21 h-1, calculate the effluent BOD5 concentration from the third reactor.
Soution
The effluent from the third complete-mix reactor in series can be determined from
Equation (5.83). First, the detention time in each CMFR must be determined using
Equation (5.63).
 
V
(5.63)
Q
 
V
Q
0 .5 M G

= 0 .5 d
1 .0 M G D
 24 h 
  12 h
 d 
  0 .5 d  
Cn
C3


1
 C0 

1 k 
n

mg BOD 5 
1
 200


-1
L
 1 + 0 .2 1 h × 1 2 h 
(5.83)
3
C3 
4 .5 9
mg BOD 5
L
38. A 5 ft  10 ft solar collector is used to heat water flowing at 2.0 gallons per
minute. Assume that 50% of the sunlight is captured by the collector and the
intensity of sunlight is 434 BTU/(ft2h). Determine the temperature of the water
exiting the solar collector if the feed water temperature is 55°F.
53
Solution
Perform the energy balance around the solar collector using Equation (5.26) as the basis.
 th e r a te o f c h a n g e o f 
 th e r a te a t w h ic h 
 th e r a te a t w h ic h 






to ta l e n e r g y

e n e r g y e n te r s

e n e r g y le a v e s












in C V
CV
CV
(5.26)
Recall that at steady state, the energy accumulated is zero and the equation reduces to the
following form.
 th e ra te a t w h ic h 
 th e ra te a t w h ic h 




e n e rg y e n te rs

e n e rg y le a v e s








CV
CV
Therefore, the rate at which energy is adsorbed by the solar collector will be equal to the
rate of change in stored energy. The rate of change in stored energy is determined from
Equation (5.20).
(5.20)
 c h a n g e in s to re d e n e rg y   m c  T
change
 r a te
 2 g a l   8 .3 4 lb   1 B T U   6 0 m in 
in s to r e d e n e r g y   
 

 
T
o
h
 m in   g a l   lb × F  

 434 BT U 
BTU
o f e n e rg y a d so rb e d   
  5 f t× 1 0 f t   0 .5 0   1 0 , 8 5 0
2
h
 ft  h 
 ra te
o f e n e rg y a d s o rb e d    c h a n g e in s to re d e n e rg y 
10, 850
BTU
h
 2 g a l   8 .3 4 lb   1 B T U   6 0 m in 
 



T
o
h
 m in   g a l   lb × F  

 T  1 0 .8 F
o
Therefore, the temperature of the incoming water is increased from 55°F to
65.8°F.
o
o
T e m p e ra tu re o f H e a te d W a te r = 5 5 F + 1 0 .8 F =
o
6 5 .8 F
39. Approximately, two-thirds of the energy content of uranium fuel entering a 2,000
MW nuclear power plant is removed by condenser cooling water which is
withdrawn from an adjacent stream. The flow upstream of the nuclear power
plant is 200 m3/s and the temperature in the stream is 19°C. Perform an energy
balance on the nuclear power plant to answer the following questions and draw a
simplified schematic of the process. Determine:
a. The energy input to the nuclear power plant.
54
b. The necessary flow rate in the stream if the temperature in the cooling
water is only allowed to rise 10°C in temperature.
c. The temperature in the stream after the heated cooling water is released
back into the stream.
Solution
First, calculate the energy input to the power plant by dividing by the energy output by
efficiency.
e n e rg y in p u t =
e n e rg y o u tp u t

2000 M W e
e ffic ie n c y
1 3
= 6000 M W t
Total energy losses that must be dissipated by the cooling water are calculated as follows:
6000 M W  2000 M W = 4000 M W
Next, substitute into Equation (5.130) to determine the mass flow rate.
 ra te o f
c h a n g e in s to re d e n e rg y   m c  T

1J
6
 10 W  

s
4000 M W 


1
M
W
1
W








kJ
  1 kJ 
 m  4 .1 8

o
  1000 J 
kg  C 





m  9 .5 7  1 0
4
1 0
o
C
kg
s
The necessary stream flow rate is calculated as follows:
Q 
m

4

9 .5 7 × 1 0 k g s
1000 kg m
3
= 9 5 .7 m
3
s
The temperature rise in the stream can be estimated as follows assuming that the total
flow in the stream remains 200 m3/s i.e. 95.7 m3/s is withdrawn for cooling and 104.3
m3/s remains before the heated cooling water is discharged back into the river.
 ra te o f
c h a n g e in s to re d e n e rg y   m c  T
55
(5.130)
T 
 r a te
o f c h a n g e in s to r e d e n e r g y 


1J
6
 10 W  

s
4000 M W 


1
M
W
1
W






  1 kJ 
  1000 J 





m   1 0 0 0 k g   4 .1 8 k J 
 200



3
o
s 
m
  kg  C 

3
m c
 T  4 .8 C
The actual temperature in the stream will be 4 .8  1 9 .0

2 3 .8 C
Alternatively, the temperature in the stream can be calculated using Equation (5.6).
Tn 
Q 1T1  Q 2 T 2 
Q nTn
Q1  Q 2 
Qn

m 
o
 2 0 0  9 5 .7
 1 9 C
s 

3
T 

200

m 
o
+  9 5 .7
 1 9 + 1 0 C
s 

3
m

3
=
s
The schematic below shows the energy balance for the nuclear power plant.
Electrical output
2000 MWe
Uranium 6000 MW
1/3% efficient
nuclear plant
Cooling water 2000MWt
Qc = 95.7 m3/s T =29°C
Cooling water
Qc = 95.7 m3/s T =19°C
Q = 100 m3/s
T = 23.8
22.8°C
Stream
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Q = 200
100 m3/s
18°C
T = 19
o
2 3 .8 C