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

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 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

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

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

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 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