Solutions, Chapter 11
_
11.1
f pipe R pipe = 16
See the discussion below Eq. 11.10, which shows that f pipe = 3 f pm and that
2
R pipe = R pm .
3
Substituting these, we find
3 R pm 3 fpm
= 72
f pmR pm = 16
2 R pipe f pipe
_
11.2 From Eq. 11.14 we see tha
Solutions, Chapter 13
_
dV n
dV
13.1 = K , for n = 1 this becomes = K
which is the same as Eq. 1.4 for
dy
dy
K = .
dV
dV
yield ,
= 0; yield , = yield + o
for yield this becomes
dy
dy
dV
0. = o
which is also the same as Eq. 1.4 if o =
dy
_
13.2
B A d
Solutions, Chapter 12
_
12.1 Following Ex. 12.1, we read the appropriate values from Fig 12.2, as shown below
for the values for various values of Qg/Ql and compute and then plot
Holdup
Ratio
Qg/Ql
0
10
20
30
40
(1-e)/e
avg/
e
water
1 infinite
3.5
0.35
4
Solutions, Chapter 10
_
AL
1
in 3
gal
60 s
2
-4 m
= 10 in 5 in = 50
= 13.0 gpm = 8.2 10
3
s
s 231 in min
t
s
_
10.1* Q =
gal lbf 231 in 3 ft
hp s
min
25 2
10.2 Po = Q P = 500
in
gal 12 in 550 ft lbf 60 s
min
= 7.29 hp = 5.44 kW
This answer is independ
Solutions Chapter 7
In the problems in Chapter 6 it says to assume pipes are Sch. 40. In most problems in
this chapter pipe inside dimensions are given, normally in integral inches. The solutions
here are based on those integral inch diameters, rather tha
Fluid Mechanics for Chemical Engineers, Third Edition
Noel de Nevers
Solutions Manual
This manual contains solutions to all the problems in the text.
Many of those are discussion problems; I have tried to present enough guidance so that
the instructor can
IE 3302 Homework
Chapter 5
Due Thursday, October 22, 2009 (start of class)
Chapter 5 suggested problems: 1, 3, 9, 13, 25, 51, 57, 59, 65, 77
Directions: This homework set should be turned in individually. Each problem may be presented
in typed or hand-wri
Solutions, Chapter 20
_
20.1 Example 20.1 shows the values for x = 0.25 . The others are done on a
spreadsheet. For x = 0.1 , the values of the function at x = 0.4 and 0.6 are 0.3744 and
0.4704. Thus we have
dy
y
0.4704 0.4375
= 0.329
forward =
dx a
Solutions, Chapter 19
_
mi gal
gal
= 2.4
hr 25 mi
mi
rev
min
rev
Engine speed =2500
60
= 150 000
so that
min
hrq
hr
Fuel flow
2.4
gal
=
= 0.000 016
This is divided by 4 to get fuel flow per
revolution 150 000
rev
cylinder per revolution, and then multipl
Solutions, Chapter 18
_
V 2 (V + v )
V 2 + 2 vV + v 2 V 2 + 2vV + v 2
=
=
=
2
2
2
2
2
2
v2
KE V + v
=
=1 +
but 2 vV = 2v V and v = 0 , so that
2
KEss
V
V
_
2
18,1
KE =
18.2
All of the x values in Ex. 18.1 are unchanged.
Vy = 0
cos t
1t
ft
ft
ft
ft
=0
,
Solutions, Chapter 17
17.1
(a) Use the force balance figure at the right,
for a constant density fluid. From F = ma we find
dVx
d (mVx )
dVx
A
= F=
dy
@ y = y + y dy @ y =y
dt
y = y+y
y=y
Here m = Ay so this becomes
dV
dVx
x
dy
@ y = y +
Solutions, Chapter 16
_
16.1
Figure Number
16.4
16.6
16.8
Electrostatic Field
Field between plates of a
condenser
Field around a charged wire
Charged particle inside a
charged condenser
Field between two wires
with opposite charges
16.16
Heat Conduction
S
Solutions, Chapter 15
_
15.1 The first three parts of this problem start with Eq. 15.7 and simply drop the
unnecessary parts
(Vx ) ( Vy )
0=
+
(a)
x
y
(Vx ) (Vy )
0=
+
(b)
Even though the flow is unsteady, is a constant, so
x
y
its time derivative is ze
Solutions, Chapter 14
_
14.1 They lower the surface tension of the final rinse, which makes the film more likely
to drain completely from the surface of the dishes, instead of breaking up into droplets,
which will leave marks on the dishes as they dry, an