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CE 452 – Introduction to Environmental Engineering Second Semester 2013 Problem Set 2 Solution Due: Thur Feb 28 SIMPLE and APPLIED KINETICS 1. You are provided the following rate constants for the hydrolysis of a variety of organic chemicals, Chloroform (7.13x10 -10 /min) and trichloroethane (1.24x10 -6 /min) a. Calculate the half-lives in years for the hydrolysis of these chemicals? From the units of the rate constant, observe that it is for a first order reaction years Trichlor t years Chloro t k t 06 . 1 850 , 1 693 . 0 5 . 0 5 . 0 5 . 0 = = = b. Which chemical will persist longest in the environment Chloroform because it has the larger half-life (Smaller k) 2. A treat ability study is run to investigate the rate at which a hazardous chemical can be treated. The following data are collected at 25 o C. Time(min) Concentration of chemical (μ M) 0 25 3 17.7 6 11.1 12 6.25 a. What is the order of the reaction? Plot the data as C versus t and ln C versus t. Inspect which plot results in a straight line. First order reaction b. Use the experimental data to evaluate the rate constant, k, of the reaction (Note: be sure to specify the units.) The rate constant equals the slope of the line c. What is the half-life for this reaction? d. Predict the chemical concentration at t=10 days 1 min 12 . 0 - = = SLope κ min 8 . 5 12 . 0 5 . 0 ln = = - t t L g C e L g C f f / 3 . 0 ] [ ) / 0 . 1 ( ] [ 10 * 12 . 0 μ μ = = -

Problem Set 2 Solution

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Page 1: Problem Set 2 Solution

CE 452 – Introduction to Environmental Engineering

Second Semester 2013 Problem Set 2 Solution Due: Thur Feb 28

SIMPLE and APPLIED KINETICS

1. You are provided the following rate constants for the hydrolysis of a variety of organic

chemicals, Chloroform (7.13x10-10

/min) and trichloroethane (1.24x10-6

/min)

a. Calculate the half-lives in years for the hydrolysis of these chemicals?

From the units of the rate constant, observe that it is for a first order reaction

yearsTrichlort

yearsChlorot

kt

06.1

850,1

693.0

5.0

5.0

5.0

=

=

=

b. Which chemical will persist longest in the environment

Chloroform because it has the larger half-life (Smaller k)

2. A treat ability study is run to investigate the rate at which a hazardous chemical can be

treated. The following data are collected at 25 oC.

Time(min) Concentration of chemical (µµµµ M)

0 25

3 17.7

6 11.1

12 6.25

a. What is the order of the reaction?

Plot the data as C versus t and ln C versus t. Inspect which plot results in a

straight line. First order reaction

b. Use the experimental data to evaluate the rate constant, k, of the reaction (Note: be

sure to specify the units.)

The rate constant equals the slope of the line

c. What is the half-life for this reaction?

d. Predict the chemical concentration at t=10 days

1min12.0 −== SLopeκ

min8.5

12.05.0ln

=

=−

t

t

LgC

eLgC

f

f

/3.0][

)/0.1(][ 10*12.0

µ

µ

=

= −

Page 2: Problem Set 2 Solution

3. Radon is a naturally occurring gas present in many houses. Radon decays into Polonium

which has a half-life of 5 minutes. If you inhaled Polonium, what % reduction of

Polonium would occur during one breath in your lungs if you were sleeping (assume 7

breaths per minute while sleeping). Radioactive decay is a first-order process.

MATERIALS BALANCE and REACTORS

4. A home washing machine removes grease and dirt from clothes in a first-order process in

which 12% of the grease on the clothes is removed per minute. The washing machine

holds 50.0L of water and has a wash cycle of 5.00 minutes before discharging the wash

water. What will be the grease concentration (in mg/L) in the discharge if the clothes

initially contain 0.500 g grease?

%9797.0

][

][

7

min1

min13.0

][

][5.0

7

1*23.0

int

1

min5*

2/1

==

=

=

=

eC

C

afterconcfindthen

K

eC

C

tFromKFind

f

K

o

o

Page 3: Problem Set 2 Solution

5. Plateau Creek carries 5.0 m3/s of water with a selenium (Se) concentration of 0.0015

mg/L. A farmer starts withdrawing 1.0 m3/s of the creek water to irrigate the land. During

irrigation, the water picks up selenium from the salts of the soil. One-half of the irrigation

water is lost to the ground and plants, and the other half is returned to Plateau Creek. The

irrigation run-off to the creek contain 1.00 mg/L of selenium. Selenium is a conservative,

nonreactive substance, and the stream does not pick up more selenium from any other

source.

a. If the farmer irrigates continuously, what will be the steady-state concentration of

selenium in the stream downstream from the farm (after the irrigation run-off

returns to the stream)?

b. Fish are sensitive to selenium levels over 0.04 mg/L. The farmer agrees not to use

more water than will keep the stream selenium level below this critical

concentration. How much water can the farmer withdraw from the stream to use

for irrigation?

Page 4: Problem Set 2 Solution

6. A CSTR is used to treat an industrial waste product, using a reaction that destroys the

pollutant according to first-order kinetics, with k=0.216 d-1

. The reactor volume is 500

m3, the volumetric flow rate of the single inlet and exit is 50 m

3 day

-1, and the inlet

pollutant concentration is 100 mg L-1

.

a. Assuming the CSTR has reached steady state, what is the outlet concentration

after treatment? State your answer in units of mg/L.

Lmg

mday

daym

daymLmg

kVQ

QCC

out

inin

out /6.31

500*216.0

/50

/50*/100*

33

3

=

+

=+

=

b. A plug-flow reactor (PFR) is used to model the chemical transformation of the

waste product presented in Part a. Assume that the flow rate and the first-order

decay rate constant are unchanged (Q = 50 m3 day

-1, k = 0.216 day

-1). Based on

this information answer the following. What is the volume required for the PFR to

obtain the same degree of pollutant reduction as the CSTR of Part a? Report your

answer in m3.

3

00432.0

)50

*216.0(

3.266

316.0

*/1006.31

mV

e

elmg

eCC

V

V

k

inout

=

=

=

=

− θ

c. What is the % reduction in volume as compared to the volume of the CSTR for

the same steady-state conditions?

Percent Reduction

%74.46100*500

3.266500=

CSTR Volume=500m3

PFR=266.3 m3