Solutions to Chemical and Engineering Thermodynamics, 3e
4
4.1
Using the Mollier diagram
=
(510 490) C
F T I = F T I =
H P K H P K c1.241 10 7.929 10 hPa
H
7
H
6
= 4.463 10 6 C Pa = 4.463 C MPa
S =
(510 490) C
F T I F T I =
H P K H P K c1.069 10 9.515 10
Preface
This manual contains more or less complete solutions for every problem in the
book. Should you find errors in any of the solutions, please bring them to my attention.
Over the years, I have tried to enrich my lectures by including historical
infor
Solutions to Chemical and Engineering Thermodynamics, 3e
5
5.1
(also available as a Mathcad worksheet)
$
(a) G = H TS at P = 2.5 MPa a nd T = 223.99 C = 497.14 K
equal with
$
$
$
G V = H V TS V = 28031 497.14 6.2575 = 307 .8 J g
.
the accuracy
$
$
$
G L =
Solutions to Chemical and Engineering Thermodynamics, 3e
6
6.1
(a) By Eqn. (6.2-3)
FG H IJ
HN K
i
= Gi ;
P , S , N j i
but Gi = Hi TSi . Thus
FG H IJ
HN K
i
f
F U IJ
dU = G
H S K
= Hi TSi
P , S , N j i
a
(b) Since U = U S ,V , N
FG U IJ dV + FG U IJ
H V K
Chapter 7
Solutions to Chemical and Engineering Thermodynamics, 3e
7
7.1
a
f
PV T , P, N1 , N2 K = Ni RT
a
f
Vi ( T , P, x) = V i (T , P )
U T , P, N 1 , N 2 K = N i U i ( T , P) U i ( T , P, x ) = U i ( T , P)
Also Si (T, P , x) = S i (T, P) R ln xi
U m
Section 8.1
Solutions to Chemical and Engineering Thermodynamics, 3e
8
8.1-1
fiV = fi L xi P vap i = yi P , since the pressure is low enough that fugacity coefficient corrections
i
will be small.
(a) For the ideal solution, i = 1 for all species; 1 = EB,
9
Note that many of the problems in this chapter can be solved relatively easily with two
programs. The first is CHEMEQ which makes the calculation of the chemical equilibrium
constant at any temperature very easy. The second is an equation solving progra
Polish Journal of Environmental Studies Vol. 14, No 6 (2005), 891-895
Letter to Editor
Growth and Activity of Sulphate-Reducing
Bacteria in Media Containing Phosphogypsum
and Different Sources of Carbon
M. Rzeczycka*, M. Baszczyk
Department of General Mic
ENVIRONMENTALIMPACTSONTHEHYDROLOGICCYCLE
(Ref Chapter 12, Environmental Hydrology by A.D.Ward and S.W.Trimble, LewisPub.2004)
Introduction
Anthropogenic (human) activities often result in a need to modify and manage runoffandstreamdischarge:(1)toprevent,
FUNDAMENTALSOFREMOTESENSINGANDGEOGRAPHIC
INFORMATIONSYSTEMSFORHYDROLGY
WhatareGeographicInformationSystems?
GIS are technologies in surveying, mapping, and information management.
They have the potential to both store and create maplike products. Data in
The Origin of the Solar System and Differentiation of Early Earth
The origin of the solar system was first explained in the Nebular Hypothesis by Kant and
Laplace in the 18th century. In this hypothesis a cloud of gas and dust (called Nebula)
collapsed by
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ENVE 309 FUNDAMENTALS OF BIOLOGICAL TREATMENT
2007 Fall Semester, Recitation V
1. Calculate the volume requirement for an extended aeration activated sludge plant designed
to treat an average wastewater flow of 1 MLD containing 220 mg/L BOD. Calculate als
Env e 302 Chemical Microbiology
Mid Term Exam II (2004)
Out of 100
1. Identify the problem and give possible reasons for the following observations in
activated sludge:
Industrial effluent, pH = 5.6, SVI=250
problem_cause_ remedy_
_
Domestic, turbid efflu
Env e 309 Chemical Microbiology
Mid Term Exam II (2004)
Out of 100
1. Identify the problem and give possible reasons for the following observations in
activated sludge:
Industrial effluent, pH = 5.6, SVI=250
problem_buking slduge_cause_fungi_ remedy resto
309 MT 2 Solutions 2009
1.
From plot of q vs. Se slope is K which is 0.05 l/mg.d see Xcel sheet
From plot of q vs. 1/c slope is 1/YT intspt is Kd/YT see Excel sheet
Q(S0-Se)
q = -Va X
YT = 0.4
Kd = 0.07 d-1
Q=0.5 MLD
Se= 10 mg/L
q = K Se
q = 0.05 (10) = 0
07.12.2009
ENVE-309 FUNDAMENTALS OF BIOLOGICAL TREATMENT
HOMEWORK-3a (Due date: 14.12.2009 till 17:00 )
1. The BODu of wastewater from an industrial plant is observed to vary between 100 and
300 mg/L. The influent N and P concentrations are 6 and 8 mg/L r
ENV E. 309 Fundamentals of Biological Treatment
MT II (2009)
Out of 80
1.
Suppose you have been in charge of operating an activated sludge plant with 0.25 ML
tank volume and handling a wastewater flow of 1 MLD. Average influent BODu is
250 mg/L. During th
ENV E. 309 Fundamentals of Biological Treatment
MT II (20032)
Out of 80
1.
An extended aeration A.S. plant is to be designed for a flow of 0.5 MLD. Calculate
aeration tank volume and oxygen requirement. Take that all the nitrogen is
oxidized.The following
ENV E. 309 Fundamentals of Biological Treatment
MT II (20032)
Out of 80
1.
An extended aeration A.S. plant is to be designed for a flow of 0.5 MLD. Calculate
aeration tank volume and oxygen requirement. Take that all the nitrogen is
oxidized.The following
Design an anaerobic process to treat WW flow of 0.2 MLD from a meat packing plant. The
raw WW temperature is 20C ans a COD of 3000 mg/L. It is desired that the fermentation
temperature be maintained at 30 C and the MLVSS conc at 3500 mg/L. A safety factor
ENVE-309 FUNDAMENTALS OF BIOLOGICAL TREATMENT
HOMEWORK-4a
1. Design an SBR for a flow of 2 MLD and an average BODu concentration of 200 mg/L.
The following design criteria are applicable
Se < 15 mg/L BODu
xE = 10 days (effective SRT)
K = 0.1 L/mg.d
YT = 0
ENVE 309 FUNDAMENTALS OF BIOLOGICAL TREATMENT
2007 Fall Semester, Recitation IV Answers
1. Design criteria: Q = 5 MLD
S0 = 200 mg/l BOD Se = 20 mg/L BOD
X = 2000 mg/L kd = 0.1 d-1
YT = 0.5
K = 17 d-1
Unit of K = d-1 Grau-Grady model
S
20mg / L
q = K e = 1
ENVE 309 FUNDAMENTALS OF BIOLOGICAL TREATMENT
2007 Fall Semester, Recitation IV
1. Effluent from an industrial plant is variable between 100 400 mg/L BODu. If a BODu of
200 mg/L is selected for design, calculate required aeration tank volume for 5 MLD and
ENVE 309 FUNDAMENTALS OF BIOLOGICAL TREATMENT
2007 Fall Semester, Recitation V
1. Design criteria:
Q = 1 MLD
X = 2500 mg/L
YT = 0.5
S0 = 220 mg/L BOD
kd = 0.05 d-1
K = 0.15 L/mgd
k d 0.05d 1
=
= 0.1d 1
YT
0. 5
Unit of K is L/mg.d Discontinuous model
q
0.1
ENVE 309 FUNDAMENTALS OF BIOLOGICAL TREATMENT
2007 Fall Semester, Recitation VI
1. Calculate the volume requirement for a 3-stage step activated sludge plant designed to
treat an average wastewater flow of 15 MLD containing 200 mg/L BOD.
Design criteria: