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introduction to chemical engineering ch (8) .pdf



Original filename: introduction to chemical engineering ch (8).pdf
Title: Chapter 8
Author: Ken Solen

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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 result of random collisions between molecules. Replacement of A with a
larger molecule would make the collisions less effective in redistributing the molecules.
Therefore, the rate of diffusion would decrease.
c. The long branches would decrease the diffusion rate by making it more difficult for the
molecules to move past each other.

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 8.2:
a. The rate of mass transport by convection is increased due to an increase in the magnitude
of the mass transfer coefficient. The average concentration of A in the flowing stream
may also decrease.
b. As shown in Equation 8.4, the rate of transport by convection is proportional to the area
of the interface. Therefore, the transport rate increases with an increase in the interfacial
area.
c. With the increase in concentration at the surface, the driving force (
increased, leading to an increase in the transport rate.

) is

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 8.3:
a. The driving force is the difference between the concentration of salt at the surface of the
block and the concentration of salt in the stagnant pool.
b. The driving force is the difference in the concentration of the drug in the muscle and its
concentration in the blood.

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Reading Question 8.4:
a. Diffusion takes place in the polymer phase, and mass convection in the blood phase.
b. Option i): Increasing the flow rate of blood would have little impact on the transfer rate
because it does not decrease the limiting resistance. Therefore, the transfer rate would
increase only slightly, if at all.
Option ii) will increase the overall rate of mass transfer, as the rate of diffusion of the
smaller molecules through the polymer phase is likely to be faster than that of the larger
molecules. Therefore, Option B decreases the limiting resistance and will have a
significant effect.

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 8.1:
Reviewing Equations 8.1 and 8.4

c A − c A1
N˙ A, x = − DAB A 2
x 2 − x1

(8.1)

N˙ A = hm A(c A1 − c A2 )

(8.4)

For both types of mass transfer (molecular diffusion and mass convection), the rate of
transfer is linearly proportional to the concentration difference. In other words, doubling the
concentration difference results in a doubling of the transfer rate, etc. Thus, lowering the
concentration difference by 40% would decrease the transfer rate by 40%.

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 8.2:
a. Since the water is stagnant, transfer is occurring via molecular diffusion.
b. We begin by noting that we do not know the values of the diffusivity and the crosssectional area of transfer. It is actually the product of those two which is useful for this
calculation, so we will determine the value of that product from the initial conditions.
First, we designate locations “1” and “2” as follows

(near the waste)

(far from the waste)

solid waste

x

From Equation 8.1

c A − c A1
N˙ A, x = − DAB A 2
x 2 − x1
Thus, the product we want is
x − x1
gmol ⎞ ⎛⎜ 13.6 m − 0 ⎞
L

DAB A = - N˙ A,x 2
= - ⎝ 7.3
=
620.5
c A 2 − c A1
min ⎠ ⎝ 0 − 0.16 gmol L⎠
min
Now, under the new conditions
c A − c A1
L ⎞ ⎛⎜ 0 − 0.105 gmol L ⎞
gmol

N˙ A, x = − DAB A 2
= − ⎝ 620.5
=
6.6

x 2 − x1
min ⎠ ⎝
9.9 m − 0
min
c. The current of water flowing through the pipe produces mass convection, which is
described by
N˙ A = hm A(c A1 − c A2 )
Rearranging,

hm

=

N˙ A
A c A1 − c A 2

(

)

=

2(6.6gmol min )

(0.3 m2 )(0.105gmol L − 0)

= 419

L
m min
2

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 8.3:
First, we select locations “1” and “2” as follows:
Fast-moving Water
c=0
8 cm

Water

c = csat
Dissolving Solid

The relationship that was given was
N˙ salt ,stirred

= 4 N˙ salt,stationary

or
hm,stirred A(csalt,1 − c salt,2 )

=

Substituting given values,

hm,stirred A(csat − 0) =
so,
hm,stirred

=

4Dsalt
8 cm

=


c
−c

4⎜ − Dsalt A salt ,1 salt,2 ⎟

x1 − x 2 ⎠


c −0⎞
4⎜ −D salt A sat

0 − 8 cm ⎠

(

4 3.2x10 −4 cm 2 s
8 cm

)

= 1.6x10 −4 cm s

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 8.4:
From Equation 8.4, the rate of delivery (transfer) is

⎛ 1L ⎞
N˙ A = hm A(c A1 − c A 2 ) = (0.016cm /s)( 3.5cm 2 )(5.8x10−5 gmol /L − 0)⎜
3⎟
⎝ 1000cm ⎠
⎛ 357.79 g ⎞⎛ 3600 s ⎞
= 3.25x10−9 gmol /s⎜
⎟ = 0.0042 g /hr
⎟⎜
⎝ gmol ⎠⎝ hr ⎠

Thus, the time to deliver 50 mg is
t=

0.050 g
= 11.9 hr
0.0042 g /hr

Chapter 8 – Answer Key, Introduction to Chemical Engineering: Tools for Today and Tomorrow
Homework Problem 8.5:
The drawing for the system described is
cwater,air = ?

Air Flow

cwater,surface = 0.002 gmol/L

hm = 0.017 m/s

15 m
30 m

a. Because of the motion of the air blown across the paste, transfer is by convection.
b. N˙ water =

ρ water V˙liquid water

n˙ =

MWwater

=
c. N˙ water

(1g cm 3)(9.5 L min) ⎛⎜ 1000 cm 3 ⎞⎟
18g gmol



L



= 528gmol min

= hm A(cwater, surface − c water, air )

Rearranging,

c water, air

= c water, surface −

N˙ water
hm A

= 0.002 gmol L −

⎛ 1 m 3 ⎞⎛1 min ⎞
528 gmol min

⎟⎜

(0.017 m s)(30 m)(15 m) ⎝1000 L ⎠⎝ 60 s ⎠

= 0.002 gmol/L – 0.00115 gmol/L = 0.00085 gmol/L
d. Measures to increase transfer would be to
•increase the transfer area (by using a larger sheet of paste)
•increase the saturated water concentration at the surface (by raising the temperature)


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