Assignment 2 Solutions

Department of Civil Engineering and Applied Mechanics
McGill University

CIVE327 FLUID MECHANICS AND HYDRAULICS
Assignment No. 2

1. A 60-cm-wide belt moves as shown in Fig. P1.37. Calculate the horsepower requirement
assuming a linear velocity profile in the 10
o
C water? (Potter & Wiggert 1.37)

Figure P1.37

Solution:
Given: area of shear, A = 0.6 4 = 2.4 m
2
; strain rate,
dy
du
=
002 0
10
.

For water at T = 10
o
C,
water
= 1.308 10
3
N.s/m
2

Therefore,
Shear stress, = = =

002 0
10
10 308 1
3
water
.
.
dy
du
6.54 N/m
2

Force, F = A = 6.54 2.4 = 15.696 N
Power, P = work done over time = Fu = 15.696 10 = 156.96 W = 0.2105 hp
(where 1 hp = 745.7 W)

2. The velocity distribution in a 1.0-cm-diameter pipe is given by ( ) ( )
2
0
2
1 16 r r r u = m/s, where r
0

is the pipe radius. Calculate the shearing stress at the centerline, at r = 0.25 cm, and at the wall if
water at 20
o
C is flowing. (Potter & Wiggert 1.39)

Solution:

For water at 20
o
C, = 1.005 10
3
N.s/m
2

2 mm 10 m/s
4 m
( ) ( )
2
0
2
1 16 r r r u =
u
max

u(r)
r
y = r
0
r
r
0
= 0.005 m
Strain rate,
2
0
32
r
r
dr
du
=
Therefore,
Shear stress,
2
0
3
32
10 005 1
r
r
.
dr
du
= =

At r = 0.25 cm = 0.0025 m,
2
3
005 0
0025 0 32
10 005 1
.
.
.

=

= 3.216 Pa
At the pipe wall, r = r
0
= 0.005 m,
2
3
005 0
005 0 32
10 005 1
.
.
.

=

= 6.432 Pa

3. Calculate the maximum average velocity V with which 20
o
C water can flow in a pipe in the
laminar state if the critical Reynolds number (Re = VD/) at which transition occurs is 2000; the
pipe diameter is (Potter & Wiggert 7.1):
(a) 2 m (b) 2 cm (c) 2 mm

Solution:
For water at 20
o
C, =
= 1.007 10
6
m
2
/s
Reynolds number is given as
VD
Re =
Therefore,
For Re = 2000, maximum average velocity,
D
.
D
.
D
V
3 6
10 014 2 2000 10 007 1 Re

=

= =

(a) For D = 2 m, =
=

2
10 014 2 10 014 2
3 3
.
D
.
V 1.007 10
3
m/s
(b) For D = 2 cm = 0.02 m, =
02 0
10 014 2
3
.
.
V 0.1007 m/s
(c) For D = 2 mm = 0.002 m, =
002 0
10 014 2
3
.
.
V 1.007 m/s

4. A pressure drop of 0.07 psi occurs over a section of 0.8-in.-diameter pipe transporting water at
70
o
F. Determine the length of the horizontal section if the Reynolds number is 1600. Also, find
the shear stress at the wall and the friction factor. (Potter & Wiggert 7.18)

Solution:

Water
p
1
p
2

dp = p
2
p
1
= 0.07 psi
dx
At 70
o
F, kinematic viscosity of water, = 1.06 10
5
ft
2
/s
dynamic viscosity of water, = 2.05 10
5
lb.s/ft
2

density of water, = 1.94 slug/ft
3

For Re = 1600, the average velocity,
=

= =
12
8 0
1600 10 06 1 Re
5
.
.
D
V

0.2544 ft/s
For a pipe, pressure gradient is
2
0
8
r
V
dx
dz
g
dx
dp
dx
p d
= + =

For a horizontal pipe (z = 0), pressure gradient is given as

2
0
8
r
V
dx
dp
=
Therefore, the length of the horizontal section in consideration,
=

= =
2544 0 10 05 2 8
12
4 0
12 07 0
8
5
2
2
2
0
. .
.
.
V
r
dp dx
268.45 ft

Shearing stress, =
=
45 268
12 07 0
2
12
4 0
2
2
0
0
.
.
.
dx
p d r
6.258 10
4
lb/ft
2

Friction factor, =

= =
2
4
2
0
2544 0 94 1
8
1
10 258 6
8
1
. .
.
V
f
0.0399

5. Find the angle of the 10-mm-diameter pipe of Fig. P7.19 in which water at 40
o
C is flowing
with Re = 1500 such that no pressure drop occurs. Also, find the flow rate. (Potter & Wiggert
7.19)

Figure P7.19

Solution:
At 40
o
C, kinematic viscosity of water, = 0.661 10
6
m
2
/s
For Re = 1500, the average velocity, =

= =
01 0
1500 10 661 0 Re
6
.
.
D
V

0.09915 m/s
p
1

p
2
= p
1

Water

Pressure gradient in a pipe is given as,

2
0
8
r
V
dx
dz
g
dx
dp
= +
Since dp = 0 and dz = dz, therefore

2
0
8
r
V
dx
dz
g

=
since sin
dx
dz
=
=

= =
2
6
2
0
005 0 81 9
09915 0 10 661 0 8 8
. .
. .
gr
V
sin
2.1378 10
3

= 0.1225
o

Flow rate, = = = = 09915 0 005 0
2 2
0
. . V r AV Q 7.787 10
6
m
3
/s

6. Water at 20
o
C flows in the 4-mm-diameter pipe of Fig. P7.26. The pressure rise over the 10-m
section is 6 kPa. Find the Reynolds number of the flow and the wall shear stress. Assume laminar
flow. (Potter & Wiggert 7.26)

Figure P7.26

p
1

p
2
= p
1

Water
dx

dz

10
o

?
Water at 20
o
C
10 m
Solution:

At 20
o
C, kinematic viscosity of water, = 1.007 10
6
m
2
/s
dynamic viscosity of water, = 1.005 10
3
N.s/m
2

density of water, = 998.2 kg/m
3

Pressure gradient in a pipe is given as,

2
0
8
r
V
dx
dz
g
dx
dp
= +
Therefore, ( ) ( )
o 3
3
2 2
0
10 10 81 9 2 998 10 6
10 10 005 1 8
002 0
8
sin . .
.
.
gdz dp
dx
r
V

= + =

= V 0.5475 m/s

Reynolds number, =
= =
6
10 007 1
004 0 5475 0
Re
.
. . VD
2175

Shear stress,
( ) ( )
10
10 10 81 9 2 998 10 6
2
002 0
2 2
o 3
0 0
0
sin . . .
dx
gdz dp r
dx
p d r
=
+
=
=

=
0
1.1004 Pa

7. Find the radius in a developed laminar flow in a pipe where (Potter & Wiggert 7.30):
(a) The velocity is equal to the average velocity.
(b) The shear stress is equal to one-half the wall shear stress.

Solution:
(a) For steady, laminar, developed flow in a circular pipe, the velocity distribution is
( )
2
0
2
4
1
r r
dx
p d
u
(1)

Since the maximum velocity occur at r = 0, Eq. (1) becomes

dx
p d r
u
max
=
4
2
0
(2)

10
o

dz

Water at 20
o
C
dx = 10 m

The average velocity,
max
u V
2
1
= , therefore

dx
p d r
u V
max
= =
4 2
1
2
1
2
0
(3)

For velocity in pipe to be equal to the average velocity, Eq. (3) is equal to Eq. (1),
( )
2
0
2
2
0
4
1
4 2
1
r r
dx
p d
dx
p d r

2
0
2 2
0
2
1
r r r =

2
0
2
2
1
r r =
r = 0.7071 r
0

(b) The shear stress is equal to one-half the wall shear stress.
Shear stress in pipe flow,

dx
p d r
=
2
(4)
At the pipe wall,
0
= at
0
r r =

dx
p d r
=
2
0
0
(5)

At point where the shear stress is equal to one-half of the wall shear stress,

0
2
1
=

dx
p d r
dx
p d r
=
2 2
1
2
0

0
2
1
r r = == =

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What are the main equations used to calculate pressure drop and shear stress in pipe flow?

What factors affect the velocity profile and shear stress distribution in fully developed laminar pipe flow?

Department of Civil Engineering and Applied Mechanics

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