Chapter 14 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
SA 2000 kg/hr
AAH 5000 L/hr
PA 1250 L/hr
Case Study #1 - Aspirin Production
ultrapure water
22,000 L/hr
cooling
25C
water
Reactor
ASA
product
Acid
waste
70C
Mixer
Pu
Chapter 7 Haloalkanes
Multiple Choice
1. Which of the following structures have the correct IUPAC name? (Sec. 7.2)
Cl
H
C CH CH CH
2
2
3
CH3
CH3
F
R-2-chloropentane
I
R-4-fluoro-4-methylcyclohexene
II
H3CO
Br
H
CH3
Cl
Br
H
S-2-chloro-2-fluorobutane
IV
I,
Chapter 9 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 9.1:
a. The rate will increase. As temperature increases, the fraction of molecules having the
energy needed to react goes up.
b. The rate will incre
Chapter 7 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 7.1:
Gases are more compressible than liquids because the distance between the gas molecules is
much greater than the distance between molecules in a
Chapter 6 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 6.1:
You should have followed the example in the book and printed a spreadsheet page that looks
like the one illustrated in Figure 6.4.
Chapter 6 Ans
Chapter 12 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 12.1:
Feedback control is capable of correcting for any type of disturbance. However, it requires
that the system deviate from setpoint before a cor
Chapter 8 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 8.1:
a. An increase in temperature would increase the motion of the molecules, resulting in an
increase in the diffusion rate.
b. Diffusion is the re
Chapter 10 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 10.1:
All forms are operative in each of the scenarios. The dominant form changes with the
situation.
a. Convection is the main reason why the stude
Chapter 8 - Section B - Non-Numerical Solutions
8.12 (a) Because Eq. (8.7) for the efciency Diesel includes the expansion ratio, re VB / V A , we relate this quantity to the compression ratio, r VC / VD , and the Diesel cutoff ratio, rc V A / VD . Since V
Chapter 7 - Section B - Non-Numerical Solutions
7.2 (a) Apply the general equation given in the footnote on page 260 to the particular derivative of interest here: T S T = P S S P P T The two partial derivatives on the right are found from Eqs. (6.17) and
Chapter 6 - Section B - Non-Numerical Solutions
H S =T
P
6.1 By Eq. (6.8),
and isobars have positive slope T S
Differentiate the preceding equation: 2 H S2
2 H S2 =
P
=
P
P
Combine with Eq. (6.17):
T CP
and isobars have positive curvature.
6.2 (a) Applica
Chapter 5 - Section A - Mathcad Solutions
5.2 Let the symbols Q and Work represent rates in kJ/s. Then by Eq. (5.8) = TC Work = 1 QH TH TH := 798.15 K TC QH := 250 kJ s kJ s
TC := 323.15 K Work := Q H 1
TH
or
Work = 148.78
Work = 148.78 kW which is t
Chapter 11 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 11.1:
The emphasis on a great deal of strength will best be met by the high-carbon steel. This is
because the resulting affect of adding carbon to i
Chapter 13 Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 13.1:
The cost for the high-efficiency model is $1195
The cost for the standard model is $1198
Chapter 13 Answer Key, Introduction to Chemical Engin
9
. .
9.8
CH3
CH2
Ethyle benzene
H2
+
Styrene
Ethyl benzene ?
H+
H2 C
C
CH2
C
C
H+
OH
C
+
Cl
C
C
C
+
C
C
C
C+
C
H2O
C
C
C
AlCl3
+
C
C
O
O
C
C C
C
C
C
+
Cl
C
Zn/Cl
HCl
AlCl3
9.9
F-C Alkylation
C
C
C
+
Cl
AlCl3
C
C
C
Cl
AlCl3
C
CH3
+
C
C
C
Cl
AlCl3
F-C A
Transport numbers
The fraction of total current carried by the ions of a specified type.
I
t
I
t
I
I
The limiting transport number, t0, is defined for the limit of zero
concentration of the electrolyte solution.
The relationship between transportation n
Chapter 4 - Section B - Non-Numerical Solutions
4.5 For consistency with the problem statement, we rewrite Eq. (4.8) as: CP = A + B C T1 ( + 1) + T12 ( 2 + + 1) 2 3
where T2 / T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperatur
Chapter 3 - Section B - Non-Numerical Solutions
3.2 Differentiate Eq. (3.2) with respect to P and Eq. (3.3) with respect to T : P T =
T
1 V2
V P V T
T
V T V P
+
P
1 V
2V PT 2V T P
= +
2V PT 2V PT
=
P
1 V2
T
P
1 V
=
Addition of these two equations leads i
8.
1.
only,
Chapter 8
(a) Assume the average gas velocity is 20 m/s.
(See p. 189). The actual volume flow is
_ 10,000xlx293
3600x3x273
Crosssectional area required:
3 .= 0.994/20 = 0.0497m2 (0.535 ftz)
Schedule 40
standard pipe size,
(255mm ID).
q = 0.99
6.1
Re
3500 lb/h, and
3500/20.6 = 170 lb/h-ft: From Eq. (4.50),
51 z
I":
with p = 2.42 lb/ft-h:
3 x 2.42 2170
a = (
)U3
62.32 x 4.17 x 109x 0.7071
= 1.03 x 106 ft.
Total holdup is 1.03 x 104 x 29 = 0.03 ft3/ft3
Actual holdup is about twice that predicte