Problem 13.211
[Difficulty: 4]
Given: The gas dynamic relations for compressible flow
Find: Exit Mach number and velocity
Solution: Begin with the 1-D gas dynamic relations for compressible flow
Governing equations:
k 1 2
V1 M 1 kRT1 ; T01 T1 1
M1
2
Assumption: The flow is compressible and supersonic
V1 M 1 kRTa 5 1.4 287 216.7 1475.4
m
s
Assuming M2 = 4.0, M3 = 3.295, and M4 = 1.26
A
4* 1.05
A
T04
and
1.317
T4
A5 A5 A4
5 1.05 5.25
A* A4 A*
M 5 3.23
With
k 1 2
M 5 3.11
T5
2
To find the temperature at state 5, we need to express the temperature in terms of the entrance
temperature and known temperature ratios:
T T T T0 T0 T
T5 T1 2 3 4 4 5 5
T1 T2 T3 T4 T04 T05
T05
1
Now since the stagnation temperatures at 4 and 5 are equal (isentropic flow through the nozzle):
1
T5 216.7 K 1.429 1.333 3.744 1.317 1
3.11
T5 654.5 K
Therefore, the exhaust velocity is:
m
V5 M 5 kRT5 3.23 1.4 287 654.5 1656
s
Problem 13.210
[Difficulty: 3]
Given: The gas dynamic relations for compressible flow
Find: The shock values and angles in each region
Solution: Begin with the 1-D gas dynamic relations for compressible flow
Governing equations:
k 1 2
V1 M 1 kRT1 ; T01 T1 1
M1
2
Assumption: The flow is compressible and supersonic
V1 M 1 kRTa 5 1.4 287 216.7 1475.4
V f V1 1475.4
m
s
m
s
k 1 2
T01 T1 1
M 1 1300 K
2
From (2) to (3) A second oblique shock with M 2 4.0 and 100
From the oblique shock tables
2 22.230
and
M 2 n M 2 sin 1.513
From normal shock tables
M 3n 0.698
M 3n
0.698
sin( ) sin12.23
M 3 3.295
M3
V1
10°
V1
10°
Problem 13.209
[Difficulty: 3]
Given: The gas dynamic relations for compressible flow
Find: The shock values and angles in each region
Solution: Begin with the 1-D gas dynamic relations for compressible flow
Governing equations:
k 1 2
V1 M 1 kRT1 ; T01 T1 1
M1
2
Assumption: The flow is compressible and supersonic
V1 M 1 kRTa 5 1.4 287 216.7 1475.4
V f V1 1475.4
m
s
m
s
k 1 2
T01 T1 1
M 1 1300 K
2
From (1) to (2) there is an oblique shock with M 1 =5 and 100
From the oblique shock figure (or tables)
1 19.38
M1n M1 sin( )
M1n 1.659
1
( )
M 1n
M1
M2
M 2n
From Normal Shock Tables
M 1n 1.659
M 2 n 0.65119
10
T2
1.429
T1
M2
M 2n
4.0
sin( )
M2
M1
Problem 13.208
[Difficulty: 4]
Given: Mach number and airfoil geometry
FU
1
Find:
Lift and Drag coefficients
FL
RU
RL
Solution:
R =
k =
p1 =
M1 =
The given or available data is:
286.9
1.4
95
2
J/kg.K
kPa
=
12
o
=
10
o
Equations and Computations:
Following the analysis of Example 13.14
the force component perpendicular to the major axis, per area, is
F V/sc = 1/2{(p FL + p RL) - (p FU + p RU)}
(1)
and the force component parallel to the major axis, per area, is
F H/sc = 1/2tan(/2){(p FU + p FL) - (p RU + p RL)}
(2)
using the notation of the figure above.
(s and c are the span and chord)
The lift force per area is
F L/sc = (F Vcos() - F Hsin())/sc
(3)
The drag force per area is
F D/sc = (F Vsin() + F Hcos())/sc
C L = F L/(1/2V 2A )
The lift coefficient is
(4)
(5)
But it can be shown that
V 2 = pkM 2
(6)
Hence, combining Eqs. 3, 4, 5 and 6
C L = (F V/sc cos() - F H/sc sin())/(1/2pkM 2)
(7)
Similarly, for the drag coefficient
C D = (F V/sc sin() + F H/sc cos())/(1/2pkM 2)
(8)
For surface FL (oblique shock):
We need to find M 1n
The deflection angle is
=
+ /2
=
17
o
From M 1 and , and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
For
=
17.0
o
=
48.2
o
(Use Goal Seek to vary so that = 17o)
From M 1 and
M 1n =
1.49
From M 1n and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
p2 =
230.6
p FL =
p2
p FL =
230.6
kPa
kPa
To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using built-in function NormM2fromM (M ,k ))
(13.48a)
M 2n =
0.704
The downstream Mach number is then obtained from
from M 2n, and , and Eq. 13.47b
M 2n = M 2sin( - )
M2 =
Hence
(13.47b)
1.36
For p 02 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
(13.7a)
p 02 =
693
kPa
For surface RL (isentropic expansion wave):
Treating as a new problem
Here:
M 1 is the Mach number after the shock
and M 2 is the Mach number after the expansion wave
p 01 is the stagnation pressure after the shock
and p 02 is the stagnation pressure after the expansion wave
M 1 = M 2 (shock)
M1 =
1.36
p 01 = p 02 (shock)
p 01 =
For isentropic flow
For the deflection
693
kPa
p 0 = constant
p 02 =
p 01
p 02 =
693
=
=
10.0
kPa
o
We use Eq. 13.55
(13.55)
and
Deflection =
2 - 1 = (M 2) - (M 1)
(3)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
Applying Eq. 3
1 =
7.8
2 =
1 +
2 =
17.8
o
o
From 2, and Eq. 13.55 (using built-in function Omega (M , k ))
2 =
M2 =
For
17.8
1.70
o
(Use Goal Seek to vary M 2 so that 2 = 17.8o)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 =
141
p RL =
p2
p RL =
141
kPa
kPa
For surface FU (isentropic expansion wave):
M1 =
2.0
p 0 = constant
For isentropic flow
p 02 =
p 01
p 01 =
p 02 =
743
743
For p 01 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
For the deflection
=
- /2
=
7.0
kPa
o
We use Eq. 13.55
and
Deflection =
2 - 1 = (M 2) - (M 1)
(3)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
Applying Eq. 3
1 =
26.4
2 =
1 +
2 =
33.4
o
o
From 2, and Eq. 13.55 (using built-in function Omega(M, k))
2 =
M2 =
For
33.4
2.27
o
(Use Goal Seek to vary M 2 so that 2 = 33.4o)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 =
62.8
p FU =
p2
p FU =
62.8
kPa
kPa
For surface RU (isentropic expansion wave):
Treat as a new problem.
Flow is isentropic so we could analyse from region FU to RU
but instead analyse from region 1 to region RU.
M1 =
For isentropic flow
TOTAL deflection
2.0
p 0 = constant
p 02 =
p 01
p 01 =
p 02 =
743
743
=
+ /2
=
17.0
kPa
kPa
o
We use Eq. 13.55
and
Deflection =
2 - 1 = (M 2) - (M 1)
(3)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
Applying Eq. 3
1 =
26.4
2 =
1 +
2 =
43.4
o
o
From 2, and Eq. 13.55 (using built-in function Omega(M, k))
2 =
M2 =
For
43.4
2.69
o
(Use Goal Seek to vary M 2 so that 2 = 43.4o)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 =
32.4
kPa
p RU =
p2
p RU =
32.4
kPa
p FL =
p RL =
p FU =
p RU =
230.6
140.5
62.8
32.4
kPa
kPa
kPa
kPa
The four pressures are:
From Eq 1
F V/sc =
138
kPa
From Eq 2
F H/sc =
5.3
kPa
From Eq 7
CL =
0.503
From Eq 8
CD =
0.127
Problem 13.207
Given:
Mach number and airfoil geometry
Find:
Plot of lift and drag and lift/drag versus angle of attack
Solution:
The given or available data is:
k =
p1 =
M1 =
1.4
50
1.75
=
c =
12
1
kPa
o
m
Equations and Computations:
The net force per unit span is
F = (p L - p U)c
Hence, the lift force per unit span is
L = (p L - p U)c cos()
(1)
The drag force per unit span is
D = (p L - p U)c sin()
(2)
For each angle of attack the following needs to be computed:
[Difficulty: 4]
For the lower surface (oblique shock):
We need to find M 1n
Deflection
=
From M 1 and , and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
find
(Use Goal Seek to vary so that is the correct value)
From M 1 and find M 1n
From M 1n and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
find
p2
and
pL =
p2
For the upper surface (isentropic expansion wave):
p 0 = constant
For isentropic flow
p 02 =
p 01
For p 01 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
(13.7a)
find
p 02 =
=
Deflection
266
kPa
we use Eq. 13.55
(13.55)
and
Deflection =
2 - 1 = (M 2) - (M 1)
(3)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
find
Applying Eq. 3
1 =
19.3
2 =
1 +
o
From 2, and Eq. 12.55 (using built-in function Omega (M , k ))
From 2
find
M2
(Use Goal Seek to vary M 2 so that 2 is the correct value)
(4)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
pU =
p2
Finally, from Eqs. 1 and 2, compute L and D
Computed results:
(o)
(o)
(o)
0.50
1.00
1.50
2.00
4.00
5.00
10.00
15.00
16.00
16.50
17.00
17.50
18.00
35.3
35.8
36.2
36.7
38.7
39.7
45.5
53.4
55.6
56.8
58.3
60.1
62.9
0.50
1.00
1.50
2.00
4.00
5.00
10.0
15.0
16.0
16.5
17.0
17.5
18.0
Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
Sum: 0.0%
M 1n
p L (kPa)
2 (o)
2 from M 2 (o)
1.01
1.02
1.03
1.05
1.09
1.12
1.25
1.41
1.44
1.47
1.49
1.52
1.56
51.3
52.7
54.0
55.4
61.4
64.5
82.6
106.9
113.3
116.9
121.0
125.9
133.4
19.8
20.3
20.8
21.3
23.3
24.3
29.3
34.3
35.3
35.8
36.3
36.8
37.3
19.8
20.3
20.8
21.3
23.3
24.3
29.3
34.3
35.3
35.8
36.3
36.8
37.3
Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
Sum: 0.0%
M2
p U (kPa)
L (kN/m)
D (kN/m)
1.77
1.78
1.80
1.82
1.89
1.92
2.11
2.30
2.34
2.36
2.38
2.40
2.42
48.7
47.4
46.2
45.0
40.4
38.3
28.8
21.3
20.0
19.4
18.8
18.2
17.6
2.61
5.21
7.82
10.4
20.9
26.1
53.0
82.7
89.6
93.5
97.7
102.7
110
0.0227
0.091
0.205
0.364
1.46
2.29
9.35
22.1
25.7
27.7
29.9
32.4
35.8
L/D
115
57.3
38.2
28.6
14.3
11.4
5.67
3.73
3.49
3.38
3.27
3.17
3.08
To compute this table:
1) Type the range of
2) Type in guess values for
3) Compute from Eq. 13.49
(using built-in function Theta (M ,, k )
4) Compute the absolute error between each and
5) Compute the sum of the errors
6) Use Solver to minimize the sum by varying the values
(Note: You may need to interactively type in new values
if Solver generates values that lead to no )
7) For each , M 1n is obtained from M 1, and Eq. 13.47a
8) For each , p L is obtained from p 1, M 1n, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
9) For each , compute 2 from Eq. 4
10) For each , compute 2 from M 2, and Eq. 13.55
(using built-in function Omega (M ,k ))
11) Compute the absolute error between the two values of 2
12) Compute the sum of the errors
13) Use Solver to minimize the sum by varying the M 2 values
(Note: You may need to interactively type in new M 2 values)
if Solver generates values that lead to no )
14) For each , p U is obtained from p 02, M 2, and Eq. 13.47a
(using built-in function Isenp (M , k ))
15) Compute L and D from Eqs. 1 and 2
Lift and Drag of an Airfoil
as a Function of Angle of Attack
L and D (kN/m)
120
100
80
Lift
60
Drag
40
20
0
0
2
4
6
8
10
12
14
16
18
20
o
()
Lift/Drag of an Airfoil
as a Function of Angle of Attack
140
120
L/D
100
80
60
40
20
0
0
2
4
6
8
10
(o)
12
14
16
18
20
Problem 13.206
[Difficulty: 4]
Given: Mach number and airfoil geometry
Find:
Drag coefficient
Solution:
The given or available data is:
R =
k =
p1 =
M1 =
286.9
1.4
95
2
J/kg.K
kPa
=
0
o
=
10
o
Equations and Computations:
The drag force is
D = (p F - p R)cs tan(/2)
(1)
(s and c are the span and chord)
This is obtained from the following analysis
Airfoil thickness (frontal area) = 2s (c /2tan(/2))
Pressure difference acting on frontal area = (p F - p R)
(p F and p R are the pressures on the front and rear surfaces)
The drag coefficient is
2
C D = D /(1/2V A )
But it can easily be shown that
V 2 = pkM 2
(2)
Hence, from Eqs. 1 and 2
C D = (p F - p R)tan(/2)/(1/2pkM 2)
(3)
For the frontal surfaces (oblique shocks):
We need to find M 1n
The deflection angle is
=
/2
=
5
o
From M 1 and , and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
=
5.0
o
=
34.3
o
M 1n =
1.13
For
(Use Goal Seek to vary so that = 5o)
From M 1 and
From M 1n and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
p2 =
125.0
pF =
p2
pF =
125.0
kPa
kPa
To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using built-in function NormM2fromM (M ,k ))
(13.48a)
M 2n =
0.891
The downstream Mach number is then obtained from
from M 2n, and , and Eq. 13.47b
M 2n = M 2sin( - )
M2 =
Hence
(13.47b)
1.82
For p 02 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
(13.7a)
p 02 =
742
kPa
For the rear surfaces (isentropic expansion waves):
Treating as a new problem
Here:
M 1 is the Mach number after the shock
and M 2 is the Mach number after the expansion wave
p 01 is the stagnation pressure after the shock
and p 02 is the stagnation pressure after the expansion wave
M 1 = M 2 (shock)
M1 =
1.82
p 01 = p 02 (shock)
p 01 =
For isentropic flow
For the deflection
742
kPa
p 0 = constant
p 02 =
p 01
p 02 =
742
=
=
10.0
kPa
o
We use Eq. 13.55
(13.55)
and
Deflection =
2 - 1 = (M 2) - (M 1)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
Applying Eq. 3
1 =
21.3
2 =
1 +
2 =
31.3
o
o
From 2, and Eq. 13.55 (using built-in function Omega(M, k))
For
2 =
M2 =
(Use Goal Seek to vary M 2 so that 2 = 31.3o)
31.3
2.18
o
(3)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
Finally, from Eq. 1
p2 =
71.2
pR =
p2
pR =
71.2
CD =
0.0177
kPa
kPa
Problem 13.205
[Difficulty: 3]
Given: Wedge-shaped airfoil
Find:
Lift per unit span assuming isentropic flow
Solution:
The given or available data is:
R
k
p
M
=
=
=
=
=
c =
286.9
1.4
70
2.75
7
1.5
J/kg.K
kPa
o
m
Equations and Computations:
The lift per unit span is
L = (p L - p U)c
(1)
(Note that p L acts on area c /cos(), but its
normal component is multiplied by cos())
For the upper surface:
pU =
p
pU =
70
kPa
For the lower surface:
=
=
-7.0
o
We use Eq. 13.55
(13.55)
and
Deflection =
L - = (M L) - (M )
(2)
From M and Eq. 13.55 (using built-in function Omega (M , k ))
=
44.7
=
L -
L =
+
L =
37.7
o
L =
ML =
37.7
2.44
o
o
Applying Eq. 2
From L, and Eq. 13.55
(using built-in function Omega (M , k ))
For
(Use Goal Seek to vary M L so that L is correct)
Hence for p L we use Eq. 13.7a
(13.7a)
The approach is to apply Eq. 13.7a twice, so that
(using built-in function Isenp (M , k ))
p L = p (p 0/p )/(p 0/p L)
From Eq 1
pL =
113
kPa
L =
64.7
kN/m
Problem 13.204
[Difficulty: 4]
Given: Mach number and airfoil geometry
Find:
Lift and drag per unit span
Solution:
The given or available data is:
R =
k =
p1 =
M1 =
=
c =
286.9
1.4
50
1.75
18
1
J/kg.K
kPa
o
m
Equations and Computations:
F = (p L - p U)c
The net force per unit span is
Hence, the lift force per unit span is
L = (p L - p U)c cos()
(1)
D = (p L - p U)c sin()
(2)
The drag force per unit span is
For the lower surface (oblique shock):
We need to find M 1n
The deflection angle is
=
=
18
o
From M 1 and , and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
For
=
18.0
o
=
62.9
o
(Use Goal Seek to vary so that is correct)
From M 1 and
M 1n =
1.56
From M 1n and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
p2 =
133.2
pL =
p2
pL =
133.2
kPa
kPa
For the upper surface (isentropic expansion wave):
p 0 = constant
For isentropic flow
p 02 =
p 01
For p 01 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
(13.7a)
For the deflection
p 01 =
266
kPa
p 02 =
266
kPa
=
=
18.0
(Compression )
o
We use Eq. 13.55
(13.55)
and
Deflection =
2 - 1 = (M 2) - (M 1)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
Applying Eq. 3
1 =
19.3
2 =
1 +
2 =
37.3
o
o
(3)
From 2, and Eq. 13.55 (using built-in function Omega (M , k ))
For
2 =
M2 =
37.3
2.42
o
(Use Goal Seek to vary M 2 so that 2 is correct)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 =
17.6
kPa
pU =
p2
pU =
17.6
kPa
From Eq. 1
L =
110.0
kN/m
From Eq. 2
D =
35.7
kN/m
Problem 13.203
[Difficulty: 3]
Given: Deflection of air flow
Find:
Mach numbers and pressures
Solution
The given or available data is:
R =
k =
p2 =
M2 =
286.9
1.4
10
4
1 =
15
o
2 =
15
o
J/kg.K
kPa
Equations and Computations:
We use Eq. 13.55
(13.55)
and
Deflection =
a - b = (M a) - (M b)
From M and Eq. 13.55 (using built-in function Omega (M , k ))
2 =
65.8
o
For the second deflection:
Applying Eq. 1
1 =
2 - 2
1 =
50.8
o
From 1, and Eq. 13.55
(using built-in function Omega (M , k ))
For
1 =
50.8
M1 =
3.05
o
(Use Goal Seek to vary M 1 so that 1 is correct)
(1)
Hence for p 1 we use Eq. 13.7a
(13.7a)
The approach is to apply Eq. 13.7a twice, so that
(using built-in function Isenp (M , k ))
p 1 = p 2(p 0/p 2)/(p 0/p 1)
p1 =
38.1
kPa
For the first deflection:
We repeat the analysis of the second deflection
Applying Eq. 1
2 + 1 =
2 -
= 2 - (2 + 1)
=
35.8
o
(Note that instead of working from state 2 to the initial state we could have
worked from state 1 to the initial state because the entire flow is isentropic)
From , and Eq. 13.55
(using built-in function Omega (M , k ))
For
=
35.8
M =
2.36
o
(Use Goal Seek to vary M so that is correct)
Hence for p we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p = p 2(p 0/p 2)/(p 0/p )
p =
110
kPa
Problem 13.202
[Difficulty: 4]
Given: Mach number and deflection angle
Find:
Static and stagnation pressures due to: oblique shock; compression wave
Solution:
The given or available data is:
R =
k =
p1 =
M1 =
286.9
1.4
50
3.5
J/kg.K
kPa
=
35
o
=
35
o
Equations and Computations:
For the oblique shock:
We need to find M 1n
The deflection angle is
From M 1 and , and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
For
=
35.0
o
=
57.2
o
(Use Goal Seek to vary so that = 35o)
From M 1 and
M 1n =
2.94
From M 1n and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
p2 =
496
kPa
To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using built-in function NormM2fromM (M ,k ))
(13.48a)
M 2n =
0.479
The downstream Mach number is then obtained from
from M 2n, and , and Eq. 13.47b
M 2n = M 2sin( - )
M2 =
Hence
(13.47b)
1.27
For p 02 we use Eq. 12.7a
(using built-in function Isenp (M , k ))
(13.7a)
p 02 = p 2/(p 02/p 2)
p 02 =
1316
kPa
For the isentropic compression wave:
p 0 = constant
For isentropic flow
p 02 =
p 01
p 01 =
3814
kPa
p 02 =
3814
kPa
For p 01 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
(Note that for the oblique shock, as required by Eq. 13.48b
(13.48b)
0.345
p 02/p 01 =
(using built-in function Normp0fromM (M ,k )
p 02/p 01 =
0.345
(using p 02 from the shock and p 01)
For the deflection
=
=
-35.0
(Compression )
o
We use Eq. 13.55
(13.55)
and
Deflection =
2 - 1 = (M 2) - (M 1)
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
Applying Eq. 1
1 =
58.5
2 =
1 +
2 =
23.5
o
2 =
M2 =
23.5
1.90
o
o
From 2, and Eq. 13.55
(using built-in function Omega (M , k ))
For
(Use Goal Seek to vary M 2 so that 2 = 23.5o)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 =
572
kPa
(1)
Problem 13.201
[Difficulty: 3]
Given: Air flow in a duct
Find:
Mach number and pressure at contraction and downstream;
Solution:
The given or available data is:
k =
M1 =
1.4
2.5
=
p1 =
30
50
o
kPa
Equations and Computations:
For the first oblique shock (1 to 2) we find from Eq. 13.49
(13.49)
Using built-in function theta (M, ,k )
=
7.99
o
Also, M 1n can be found from geometry (Eq. 13.47a)
M 1n =
1.250
Then M 2n can be found from Eq. 13.48a)
Using built-in function NormM2fromM (M,k )
(13.48a)
M 2n =
0.813
Then, from M 2n and geometry (Eq. 13.47b)
M2 =
2.17
From M 1n and Eq. 13.48d (using built-in function NormpfromM (M ,k ))
(13.48d)
p 2/p 1 =
p2 =
1.66
82.8
Pressure ratio
We repeat the analysis for states 1 to 2 for 2 to 3, for the second oblique shock
We choose for M 2 by iterating or by using Goal Seek to target (below) to equal
the previous , using built-in function theta (M, ,k )
=
7.99
o
=
34.3
o
Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a)
M 2n =
1.22
Then M 3n can be found from Eq. 13.48a)
Using built-in function NormM2fromM (M,k )
M 3n =
0.829
Then, from M 3n and geometry (Eq. 13.47b)
M3 =
1.87
From M 2n and Eq. 13.48d (using built-in function NormpfromM (M ,k ))
p 3/p 2 =
p3 =
1.58
130
Pressure ratio
Problem 13.200
[Difficulty: 3]
Given: Deflection of air flow
Find:
Pressure changes
Solution:
R
k
p
M
The given or available data is:
=
=
=
=
286.9
1.4
95
1.5
J/kg.K
kPa
1 =
15
o
2 =
15
o
Equations and Computations:
We use Eq. 13.55
(13.55)
and
Deflection =
a - b = (M a) - (M b)
From M and Eq. 13.55 (using built-in function Omega (M , k ))
=
11.9
1 =
1 -
1 =
1 +
1 =
26.9
o
For the first deflection:
Applying Eq. 1
o
From 1, and Eq. 13.55
(using built-in function Omega (M , k ))
For
1 =
26.9
o
(1)
M1 =
2.02
(Use Goal Seek to vary M 1 so that 1 is correct)
Hence for p 1 we use Eq. 13.7a
(13.7a)
The approach is to apply Eq. 13.7a twice, so that
(using built-in function Isenp (M , k ))
p 1 = p (p 0/p )/(p 0/p 1)
p1 =
43.3
kPa
For the second deflection:
We repeat the analysis of the first deflection
Applying Eq. 1
2 + 1 =
2 -
2 =
2 + 1 +
2 =
41.9
o
(Note that instead of working from the initial state to state 2 we could have
worked from state 1 to state 2 because the entire flow is isentropic)
From 2, and Eq. 13.55
(using built-in function Omega (M , k ))
For
2 =
41.9
M2 =
2.62
o
(Use Goal Seek to vary M 2 so that 2 is correct)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p (p 0/p )/(p 0/p 2)
p2 =
16.9
kPa
Problem 13.199
[Difficulty: 4]
Given: Air flow into engine
Find:
Pressure of air in engine; Compare to normal shock
Solution:
The given or available data is:
k =
p1 =
M1 =
1.4
50
3
=
7.5
kPa
o
Equations and Computations:
Assuming isentropic flow deflection
p 0 = constant
p 02 =
p 01
For p 01 we use Eq. 13.7a (using built-in function Isenp (M , k ))
(13.7a)
p 01 =
p 02 =
For the deflection
=
1837
1837
7.5
kPa
kPa
o
From M 1 and Eq. 13.55 (using built-in function Omega (M , k ))
(13.55)
1 =
Deflection =
Applying Eq. 1
49.8
2 - 1 = (M 2) - (M 1)
o
(1)
2 =
1 -
2 =
42.3
(Compression!)
o
From 2, and Eq. 13.55 (using built-in function Omega (M , k ))
For
2 =
M2 =
42.3
2.64
o
(Use Goal Seek to vary M 2 so that 2 is correct)
Hence for p 2 we use Eq. 13.7a
(using built-in function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
For the normal shock (2 to 3)
p2 =
86.8
M2 =
2.64
kPa
From M 2 and p 2, and Eq. 13.41d (using built-in function NormpfromM (M ,k ))
(13.41d)
p3 =
690
kPa
For slowing the flow down from M 1 with only a normal shock, using Eq. 13.41d
p =
517
kPa
Problem 13.198
[Difficulty: 3]
Given: Air flow in a duct
Find:
Mach number and pressure at contraction and downstream;
Solution:
The given or available data is:
k =
M1 =
1.4
2.5
=
p1 =
7.5
50
o
kPa
Equations and Computations:
For the first oblique shock (1 to 2) we need to find from Eq. 13.49
(13.49)
We choose by iterating or by using Goal Seek to target (below) to equal the given
Using built-in function theta (M, ,k )
=
7.50
o
=
29.6
o
Then M 1n can be found from geometry (Eq. 13.47a)
M 1n =
1.233
Then M 2n can be found from Eq. 13.48a)
Using built-in function NormM2fromM (M,k )
(13.48a)
M 2n =
0.822
Then, from M 2n and geometry (Eq. 13.47b)
M2 =
2.19
From M 1n and Eq. 13.48d (using built-in function NormpfromM (M ,k ))
(13.48d)
p 2/p 1 =
p2 =
1.61
80.40
Pressure ratio
We repeat the analysis of states 1 to 2 for states 2 to 3, to analyze the second oblique shock
We choose for M 2 by iterating or by using Goal Seek to target (below) to equal the given
Using built-in function theta (M, ,k )
=
7.50
o
=
33.5
o
Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a)
M 2n =
1.209
Then M 3n can be found from Eq. 13.48a)
Using built-in function NormM2fromM (M,k )
M 3n =
0.837
Then, from M 3n and geometry (Eq. 13.47b)
M3 =
1.91
From M 2n and Eq. 13.48d (using built-in function NormpfromM (M ,k ))
p 3/p 2 =
p3 =
1.54
124
Pressure ratio
Problem 13.197
[Difficulty: 4]
Given: Air passing through jet inlet
Find:
Pressure after one oblique shock; after two shocks totaling same overall turn, after
isentropic compression
Solution:
The given or available data is:
R =
k =
M1 =
p1 =
θ =
53.33
1.4
2
5
20
ft-lbf/lbm-°R
psia
°
Equations and Computations:
To find the shock angle, we have to iterate on the shock angle until we match the
deflection angle, which is a function of Mach number, specific heat ratio, and shock angle.
β =
53.423
°
θ =
20.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach number normal to the wave is:
1.6061
M 1n =
The pressure ratio across the shock wave is:
p 2/p 1 =
2.8429
Therefore, the post-shock pressure is:
p2 =
14.21
psia
Now if we use two 10-degree turns, we perform two oblique-shock calculations.
For the first turn:
39.314
°
β 1-2a =
θ =
10.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach number normal to the wave is:
1.2671
M 1n =
The post-shock Mach number normal to the wave is:
M 2an =
0.8032
The pressure ratio across the shock wave is:
p 2a/p 1 =
1.7066
Therefore, the post-shock pressure is:
p 2a =
8.5329
psia
So the Mach number after the first shock wave is:
M 2a =
1.6405
For the second turn:
49.384
°
β 2a-2b =
θ =
10.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach number normal to the wave is:
M 2an =
1.2453
The post-shock Mach number normal to the wave is:
0.8153
M 2bn =
The pressure ratio across the shock wave is:
p 2b/p 2a =
1.6426
Therefore, the post-shock pressure is:
14.02
psia
p 2b =
For the isentropic compression, we need to calculate the Prandtl-Meyer
function for the incident flow:
ω1 =
26.3798
°
The flow out of the compression will have a Prandtl-Meyer function of:
6.3798
°
ω 2i =
To find the exit Mach number, we need to iterate on the Mach number to
match the Prandtl-Meyer function:
M 2i =
1.3076
6.3798
°
ω 2i =
The pressure ratio across the compression wave is:
p 2i/p 1 =
2.7947
Therefore, the exit pressure is:
p 2i =
13.97
psia
Problem 13.196
[Difficulty: 4]
Given: Flow turned through an expansion followed by a oblique shock wave
Find:
Mach number and pressure downstream of the shock wave
Solution:
The given or available data is:
R =
k =
M1 =
p1 =
θ =
53.33
1.4
2
1
16
ft-lbf/lbm-°R
atm
°
Equations and Computations:
The Prandtl-Meyer function of the flow before the expansion is:
ω1 =
26.380
°
Since we know the turning angle of the flow, we know the Prandtl-Meyer function after
the expansion:
ω2 =
42.380
°
We can iterate to find the Mach number after the expansion:
M2 =
2.6433
42.380
°
ω2 =
The pressure ratio across the expansion wave is:
p 2/p 1 =
0.3668
Therefore the pressure after the expansion is:
p2 =
0.3668
atm
We can iterate on the shock angle to find the conditions after the oblique shock:
36.438
°
β 2-3 =
θ =
16.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach numbers normal and parallel to the wave are:
M 2n =
1.5700
2.1265
M 2t =
The post-shock Mach number normal to the wave is:
M 3n =
0.6777
The pressure and tempreature ratios across the shock are:
2.7090
p 3/p 2 =
1.3674
T 3/T 2 =
The pressure after the shock wave is:
p3 =
0.994
atm
We can get the post-shock Mach number parallel to the shock from the
temperature ratio:
M 3t =
1.8185
So the post-shock Mach number is:
1.941
M3 =
Problem 13.195
[Difficulty: 3]
Given: Wedge-shaped projectile
Find:
Speed at which projectile is traveling through the air
Solution:
The given or available data is:
R =
k =
p1 =
T1 =
T1 =
θ =
p2 =
53.33
1.4
1
10
470
10
3
ft-lbf/lbm-°R
psia
°F
°R
°
psia
Equations and Computations:
The pressure ratio across the shock wave is:
p 2/p 1 =
3.0000
For this pressure ratio, we can iterate to find the Mach number of the flow normal
to the shock wave:
M 1n =
1.6475
3.0000
p 2/p 1 =
We used Solver in Excel to iterate on the Mach number.
With the normal Mach number, we can iterate on the incident Mach number to
find the right combination of Mach number and shock angle to match the turning
angle of the flow and normal Mach number:
M1 =
4.9243
19.546
°
β 1-2 =
θ =
10.0000
°
The pre-shock Mach numbers normal and parallel to the wave are:
M 1n =
1.6475
4.6406
M 1t =
We used Solver in Excel to iterate on the Mach number and shock angle.
Now that we have the upstream Mach number, we can find the speed. The sound
speed upstream of the shock wave is:
c 1 = 1062.9839 ft/s
Therefore, the speed of the flow relative to the wedge is:
V1 =
5234
ft/s
Problem 13.194
[Difficulty: 4]
Given: Air turning through an incident and reflected shock wave
Find:
Pressure, temperature, and Mach number after each wave
Solution:
The given or available data is:
R =
k =
M1 =
p1 =
T1 =
T1 =
θ =
53.33
1.4
2.3
14.7
80
540
10
ft-lbf/lbm-°R
psia
°F
°R
°
Equations and Computations:
To find the shock angle, we have to iterate on the shock angle until we match the
deflection angle, which is a function of Mach number, specific heat ratio, and shock angle.
For the first turn:
β 1-2 =
34.326
°
θ =
10.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach numbers normal and parallel to the wave are:
M 1n =
1.2970
1.8994
M 1t =
The post-shock Mach number normal to the wave is:
0.7875
M 2n =
The pressure and temperature ratios across the shock wave are:
p 2/p 1 =
1.7959
1.1890
T 2/T 1 =
Therefore, the post-shock pressure and temperature are:
p2 =
26.4
psia
642
°R
T2 =
Since the parallel component of velocity is preserved across the shock and
the Mach number is related to the square root of temperature, the new parallel
component of Mach number is:
M 2t =
1.7420
So the Mach number after the first shock wave is:
M2 =
1.912
For the second turn:
β 2-3 =
41.218
°
θ =
10.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach numbers normal and parallel to the wave are:
M 1n =
1.2597
1.4380
M 1t =
The post-shock Mach number normal to the wave is:
0.8073
M 2an =
The pressure and temperature ratios across the shock wave are:
p 3/p 2 =
1.6845
1.1654
T 2/T 1 =
Therefore, the post-shock pressure is:
p3 =
44.5
psia
748
°R
T3 =
Since the parallel component of velocity is preserved across the shock and
the Mach number is related to the square root of temperature, the new parallel
component of Mach number is:
M 2t =
1.3320
So the Mach number after the second shock wave is:
M2 =
1.558
Problem 13.193
[Difficulty: 3]
Given: Air passing through jet inlet
Find:
Pressure after one oblique shock; pressure after two shocks totaling same overall turn
Solution:
The given or available data is:
R =
k =
M1 =
p1 =
θ =
53.33
1.4
4
8
8
ft-lbf/lbm-°R
psia
°
Equations and Computations:
To find the shock angle, we have to iterate on the shock angle until we match the
deflection angle, which is a function of Mach number, specific heat ratio, and shock angle.
β =
20.472
°
θ =
8.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach number normal to the wave is:
1.3990
M 1n =
The pressure ratio across the shock wave is:
p 2/p 1 =
2.1167
Therefore, the post-shock pressure is:
p2 =
16.93
psia
Now if we use two 4-degree turns, we perform two oblique-shock calculations.
For the first turn:
17.258
°
β 1-2a =
θ =
4.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach number normal to the wave is:
1.1867
M 1n =
The post-shock Mach number normal to the wave is:
M 2an =
0.8506
The pressure ratio across the shock wave is:
p 2a/p 1 =
1.4763
Therefore, the post-shock pressure is:
p 2a =
11.8100
psia
So the Mach number after the first shock wave is:
M 2a =
3.7089
For the second turn:
18.438
°
β 2a-2b =
θ =
4.0000
°
We used Solver in Excel to iterate on the shock angle.
The pre-shock Mach number normal to the wave is:
1.1731
M 2an =
The post-shock Mach number normal to the wave is:
M 2bn =
0.8594
The pressure ratio across the shock wave is:
p 2b/p 2a =
1.4388
Therefore, the post-shock pressure is:
16.99
psia
p 2b =
The pressure recovery is slightly better for two weaker shocks than a single
stronger one!
Problem 13.192
[Difficulty: 3]
Given: Air deflected at an angle, causing an oblique shock
Find:
Post shock pressure, temperature, and Mach number, deflection angle, strong or weak
Solution:
The given or available data is:
R =
k =
M1 =
T1 =
T1 =
p1 =
β =
53.33
1.4
3.3
100
560
20
45
ft-lbf/lbm-°R
°F
°R
psia
°
Equations and Computations:
The pre-shock Mach numbers normal and parallel to the wave are:
M 1n =
2.3335
2.3335
M 1t =
The sound speed upstream of the shock is:
c1 =
1160.30
ft/s
Therefore, the speed of the flow parallel to the wave is:
V 1t =
2707.51
ft/s
The post-shock Mach number normal to the wave is:
M 2n =
0.5305
The pressure and temperature ratios across the shock wave are:
p 2/p 1 =
6.1858
1.9777
T 2/T 1 =
Therefore, the post-shock temperature and pressure are:
p2 =
124
psia
1108
°R
T2 =
648
°F
T2 =
The sound speed downstream of the shock is:
c2 =
1631.74
ft/s
So the speed of the flow normal to wave is:
V 2n =
865.63
ft/s
The speed of the flow parallel to the wave is preserved through the shock:
V 2t =
2707.51
ft/s
Therefore the flow speed after the shock is:
V2 =
2842.52
ft/s
and the Mach number is:
M2 =
1.742
Based on the Mach number and shock angle, the deflection angle is:
θ =
27.3
°
Since the Mach number at 2 is supersonic, this is a weak wave. This can be
confirmed by inspecting Fig. 13.29 in the text.
Problem 13.191
[Difficulty: 3]
Given: Data on airfoil flight
Find:
Lift per unit span
Solution:
The given or available data is:
R =
k =
p1 =
M1 =
286.9
1.4
75
2.75
U =
5
o
L =
c =
15
2
o
J/kg.K
kPa
m
Equations and Computations:
The lift per unit span is
L = (p L - p U)c
(1)
(Note that each p acts on area c /cos(), but its
normal component is multiplied by cos())
For the upper surface:
We need to find M 1n(U)
The deflection angle is
U =
U
U =
5
o
From M 1 and U, and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
For
U =
5.00
o
U =
25.1
o
(Use Goal Seek to vary U so that U = U)
From M 1 and U
M 1n(U) =
1.16
From M 1n(U) and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
p2 =
106
kPa
pU =
p2
pU =
106
L =
L
L =
15
o
L =
15.00
o
L =
34.3
o
kPa
For the lower surface:
We need to find M 1n(L)
The deflection angle is
From M 1 and L, and Eq. 13.49
(using built-in function Theta (M , ,k ))
For
(Use Goal Seek to vary L so that L = L)
From M 1 and L
M 1n(L) =
1.55
From M 1n(L) and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
From Eq 1
p2 =
198
pL =
p2
pL =
198
kPa
L =
183
kN/m
kPa
Problem 13.190
[Difficulty: 3]
Given: Oblique shock Mach numbers
Find:
Deflection angle; Pressure after shock
Solution:
The given or available data is:
k =
p1 =
M1 =
1.4
75
4
M2 =
2.5
=
33.6
kPa
Equations and Computations:
We make a guess for :
o
From M 1 and , and Eq. 13.49 (using built-in function Theta (M , ,k ))
(13.49)
From M 1 and
From M 2, , and
=
21.0
M 1n =
M 2n =
2.211
0.546
o
(1)
We can also obtain M 2n from Eq. 13.48a (using built-in function normM2fromM (M ,k ))
(13.48a)
M 2n =
0.546
(2)
We need to manually change so that Eqs. 1 and 2 give the same answer.
Alternatively, we can compute the difference between 1 and 2, and use
Solver to vary to make the difference zero
Error in M 2n =
0.00%
Then p 2 is obtained from Eq. 13.48d (using built-in function normpfromm (M ,k ))
(13.48d)
p2 =
415
kPa
Problem 13.189
[Difficulty: 4]
Given: Airfoil with included angle of 60o
Find:
Angle of attack at which oblique shock becomes detached
Solution:
The given or available data is:
R =
k =
T1 =
p1 =
V1 =
286.9
1.4
276.5
75
1200
=
60
c1 =
333
M1 =
3.60
J/kg.K
K
kPa
m/s
o
Equations and Computations:
From T 1
Then
m/s
From Fig. 13.29, at this Mach number the smallest deflection angle for which
an oblique shock exists is approximately = 35o.
By using Solver , a more precise answer is
(using built-in function Theta (M ,, k )
M1 =
3.60
=
65.8
o
=
37.3
o
A suggested procedure is:
1) Type in a guess value for
2) Compute from Eq. 13.49
(using built-in function Theta (M ,, k ))
(13.49)
3) Use Solver to maximize by varying
For a deflection angle the angle of attack is
= - /2
=
7.31
o
Computed results:
(o)
(o)
(o) Needed
(o)
0.00
1.00
2.00
3.00
4.00
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.31
47.1
48.7
50.4
52.1
54.1
57.4
58.1
58.8
59.5
60.4
61.3
62.5
64.4
65.8
30.0
31.0
32.0
33.0
34.0
35.5
35.8
36.0
36.3
36.5
36.8
37.0
37.3
37.3
30.0
31.0
32.0
33.0
34.0
35.5
35.7
36.0
36.2
36.5
36.7
37.0
37.2
37.3
Sum:
Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
M 1n
p 2 (kPa)
T 2 (oC)
2.64
2.71
2.77
2.84
2.92
3.03
3.06
3.08
3.10
3.13
3.16
3.19
3.25
3.28
597
628
660
695
731
793
805
817
831
845
861
881
910
931
357
377
397
418
441
479
486
494
502
511
521
533
551
564
931
564
0.0%
Max:
To compute this table:
Type the range of
Type in guess values for
Compute Needed from = + /2
Compute from Eq. 13.49
(using built-in function Theta (M ,, k )
Compute the absolute error between each and Needed
Compute the sum of the errors
Use Solver to minimize the sum by varying the values
(Note: You may need to interactively type in new values
if Solver generates values that lead to no )
For each , M 1n is obtained from M 1, and Eq. 13.47a
For each , p 2 is obtained from p 1, M 1n, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
For each , T 2 is obtained from T 1, M 1n, and Eq. 13.48c
(using built-in function NormTfromM (M ,k ))
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
Pressure on an Airfoil Surface
as a Function of Angle of Attack
1000
p 2 (kPa)
900
800
700
600
500
0
2
4
6
8
6
8
o
( )
Temperature on an Airfoil Surface
as a Function of Angle of Attack
600
500
o
T 2 ( C)
550
450
400
350
300
0
2
4
(o)
Problem 13.188
[Difficulty: 3]
Given: Data on airfoil flight
Find:
Lift per unit span
Solution:
The given or available data is:
R =
k =
p1 =
M1 =
=
c =
286.9
1.4
70
2.75
7
1.5
J/kg.K
kPa
o
m
Equations and Computations:
The lift per unit span is
L = (p L - p U)c
(1)
(Note that p L acts on area c /cos(), but its
normal component is multiplied by cos())
For the upper surface:
pU =
p1
pU =
70.0
kPa
For the lower surface:
We need to find M 1n
=
The deflection angle is
=
7
o
From M 1 and , and Eq. 13.49
(using built-in function Theta (M , ,k ))
(13.49)
=
7.0
o
=
26.7
o
M 1n =
1.24
For
(Use Goal Seek to vary so that = )
From M 1 and
From M 1n and p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
From Eq 1
p2 =
113
kPa
pL =
p2
pL =
113
kPa
L =
64.7
kN/m
Problem 13.187
[Difficulty: 4]
Given: Airfoil with included angle of 20o
Find:
Mach number and speed at which oblique shock forms
Solution:
The given or available data is:
R =
k =
T1 =
=
286.9
1.4
288
10
J/kg.K
K
o
Equations and Computations:
From Fig. 13.29 the smallest Mach number for which an oblique shock exists
at a deflection = 10o is approximately M 1 = 1.4.
By trial and error, a more precise answer is
(using built-in function Theta (M ,, k )
M1 =
1.42
=
67.4
o
=
10.00
o
c1 =
V1 =
340
483
A suggested procedure is:
1) Type in a guess value for M 1
2) Type in a guess value for
m/s
m/s
3) Compute from Eq. 13.49
(using built-in function Theta (M ,, k ))
(13.49)
4) Use Solver to maximize by varying
5) If is not 10 o, make a new guess for M 1
o
6) Repeat steps 1 - 5 until = 10
Computed results:
M1
(o)
(o)
1.42
1.50
1.75
2.00
2.25
2.50
3.00
4.00
5.00
6.00
7.00
67.4
56.7
45.5
39.3
35.0
31.9
27.4
22.2
19.4
17.6
16.4
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
Sum:
0.0%
To compute this table:
1) Type the range of M 1
2) Type in guess values for
3) Compute from Eq. 13.49
(using built-in function Theta (M ,, k )
o
4) Compute the absolute error between each and = 10
5) Compute the sum of the errors
6) Use Solver to minimize the sum by varying the values
(Note: You may need to interactively type in new values
if Solver generates values that lead to no , or to
values that correspond to a strong rather than weak shock)
Oblique Shock Angle as a Function of
Aircraft Mach Number
90
75
60
(o) 45
30
15
0
1
2
3
4
M
5
6
7
Problem 13.186
[Difficulty: 4]
Given: Airfoil with included angle of 60o
Find:
Plot of temperature and pressure as functions of angle of attack
Solution:
R =
k =
T1 =
p1 =
V1 =
The given or available data is:
286.9
1.4
276.5
75
1200
=
60
c1 =
333
M1 =
3.60
J/kg.K
K
kPa
m/s
o
Equations and Computations:
From T 1
Then
m/s
Computed results:
( )
()
( ) Needed
()
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
20.00
22.00
24.00
26.00
28.00
30.00
47.1
44.2
41.5
38.9
36.4
34.1
31.9
29.7
27.7
25.7
23.9
22.1
20.5
18.9
17.5
16.1
30.0
28.0
26.0
24.0
22.0
20.0
18.0
16.0
14.0
12.0
10.0
8.0
6.0
4.0
2.0
0.0
30.0
28.0
26.0
24.0
22.0
20.0
18.0
16.0
14.0
12.0
10.0
8.0
6.0
4.0
2.0
-
o
o
o
o
Sum:
Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
M 1n
p 2 (kPa)
T 2 (oC)
2.64
2.51
2.38
2.26
2.14
2.02
1.90
1.79
1.67
1.56
1.46
1.36
1.26
1.17
1.08
1.00
597
539
485
435
388
344
304
267
233
202
174
149
126
107
90
75
357
321
287
255
226
198
172
148
125
104
84
66
49
33
18
3
597
357
Max:
To compute this table:
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
Type the range of
Type in guess values for
Compute Needed from = /2 -
Compute from Eq. 13.49
(using built-in function Theta (M ,, k )
Compute the absolute error between each and Needed
Compute the sum of the errors
Use Solver to minimize the sum by varying the values
(Note: You may need to interactively type in new values
if Solver generates values that lead to no )
For each , M 1n is obtained from M 1, and Eq. 13.47a
For each , p 2 is obtained from p 1, M 1n, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
For each , T 2 is obtained from T 1, M 1n, and Eq. 13.48c
(using built-in function NormTfromM (M ,k ))
Pressure on an Airfoil Surface
as a Function of Angle of Attack
700
600
p 2 (kPa)
500
400
300
200
100
0
0
5
10
15
20
25
30
25
30
( )
o
Temperature on an Airfoil Surface
as a Function of Angle of Attack
400
350
T 2 (oC)
300
250
200
150
100
50
0
0
5
10
15
( )
o
20
Problem 13.185
[Difficulty: 3]
Given: Velocities and deflection angle of an oblique shock
Find:
Shock angle ; pressure ratio across shock
Solution:
The given or available data is:
R =
k =
V1 =
V2 =
=
286.9
1.4
1250
650
35
J/kg.K
m/s
m/s
o
Equations and Computations:
From geometry we can write two equations for tangential velocity:
For V 1t
V 1t = V 1cos()
(1)
For V 2t
V 2t = V 2cos( - )
(2)
For an oblique shock V 2t = V 1t, so Eqs. 1 and 2 give
V 1cos() = V 2cos( - )
Solving for
(3)
= tan-1((V 1 - V 2cos())/(V 2sin()))
=
(Alternatively, solve Eq. 3 using Goal Seek !)
62.5
o
For p 2/p 1, we need M 1n for use in Eq. 13.48d
(13.48d)
We can compute M 1 from and , and Eq. 13.49
(using built-in function Theta (M ,, k ))
(13.49)
For
=
35.0
o
=
62.5
o
M1 =
3.19
This value of M 1 was obtained by using Goal Seek :
Vary M 1 so that becomes the required value.
(Alternatively, find M 1 from Eq. 13.49 by explicitly solving for it!)
We can now find M 1n from M 1. From M 1 and Eq. 13.47a
M 1n = M 1sin()
Hence
M 1n =
2.83
Finally, for p 2/p 1, we use M 1n in Eq. 13.48d
(using built-in function NormpfromM (M ,k )
p 2 /p 1 =
9.15
(13.47a)
Problem 13.184
[Difficulty: 3]
Given: Data on an oblique shock
Find:
Deflection angle ; shock angle ; Mach number after shock
Solution:
The given or available data is:
R =
k =
M1 =
T1 =
p 2 /p 1 =
286.9
1.4
3.25
283
5
J/kg.K
K
Equations and Computations:
From p 2/p 1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k )
and Goal Seek or Solver )
(13.48d)
For
p 2 /p 1 =
5.00
M 1n =
2.10
From M 1 and M 1n, and Eq 13.47a
M 1n = M 1sin()
=
40.4
(13.47a)
o
From M 1 and , and Eq. 13.49
(using built-in function Theta (M ,, k )
(13.49)
=
23.6
o
To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using built-in function NormM2fromM (M ,k ))
(13.48a)
M 2n =
0.561
The downstream Mach number is then obtained from
from M 2n, and , and Eq. 13.47b
M 2n = M 2sin( - )
Hence
M2 =
1.94
(13.47b)
Problem 13.183
[Difficulty: 3]
Given: Data on an oblique shock
Find:
Mach number and pressure downstream; compare to normal shock
Solution:
R =
k =
p1 =
M1 =
The given or available data is:
=
286.9
1.4
80
2.5
35
J/kg.K
kPa
o
Equations and Computations:
From M 1 and
M 1n =
M 1t =
1.43
2.05
From M1n and p1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
p2 =
178.6
V t1 =
V t2
The tangential velocity is unchanged
Hence
c t1 M t1 =
(T 1)
1/2
c t2 M t2
M t1 = (T 2)1/2 M t2
M 2t = (T 1/T 2)1/2 M t1
From M1n, and Eq. 13.48c
(using built-in function NormTfromM (M ,k ))
Hence
T 2/T 1 =
1.28
M 2t =
1.81
kPa
Also, from M1n, and Eq. 13.48a
(using built-in function NormM2fromM (M ,k ))
(13.48a)
M 2n =
0.726
The downstream Mach number is then
M 2 = (M 2t2 + M 2n2)1/2
M2 =
1.95
Finally, from geometry
V 2n = V 2sin( - )
Hence
= - sin-1(V 2n/V 2)
or
= - sin-1(M 2n/M 2)
=
13.2
o
570
kPa
For the normal shock:
From M1 and p1, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
p2 =
Also, from M1, and Eq. 13.48a
(using built-in function NormM2fromM (M ,k ))
M2 =
0.513
For the minimum :
The smallest value of is when the shock is a Mach wave (no deflection)
= sin-1(1/M 1)
=
23.6
o
Problem 13.182
[Difficulty: 3]
Given: Oblique shock in flow at M = 3
Find:
Minimum and maximum , plot of pressure rise across shock
Solution:
The given or available data is:
R =
k =
M1 =
286.9
1.4
3
J/kg.K
Equations and Computations:
The smallest value of is when the shock is a Mach wave (no deflection)
= sin-1(1/M 1)
The largest value is
=
19.5
o
=
90.0
o
The normal component of Mach number is
M 1n = M 1sin()
(13.47a)
For each , p2/p1 is obtained from M1n, and Eq. 13.48d
(using built-in function NormpfromM (M ,k ))
(13.48d)
Computed results:
(o)
M 1n
p 2/p 1
19.5
20
30
40
50
60
70
75
80
85
90
1.00
1.03
1.50
1.93
2.30
2.60
2.82
2.90
2.95
2.99
3.00
1.00
1.06
2.46
4.17
5.99
7.71
9.11
9.63
10.0
10.3
10.3
Pressure Change across an Oblique Shock
12.5
10.0
7.5
p 2/p 1
5.0
2.5
0.0
0
30
60
( )
o
90
Problem 13.181
[Difficulty: 3]
Given: Air deflected at an angle, causing an oblique shock
Find:
Possible shock angles; pressure and temperature corresponding to those angles
Solution:
The given or available data is:
R =
k =
M1 =
T1 =
p1 =
θ =
286.9
1.4
1.8
400
100
14
J/kg-K
K
kPa
°
Equations and Computations:
There are two possible shock angles for a given deflection, corresponding to the
weak and strong shock solutions. To find the shock angle, we have to iterate on the
shock angle until we match the deflection angle, which is a function of Mach number,
specific heat ratio, and shock angle.
The weak shock solution is:
β weak =
49.7
°
θ =
14.0000
°
The strong shock solution is:
β strong =
78.0
°
θ =
14.0000
°
We used Solver in Excel to iterate on the shock angles.
For the weak shock, the pre-shock Mach number normal to the wave is:
1.3720
M 1nweak =
The pressure and temperature ratios across the shock wave are:
p 2/p 1weak =
2.0295
1.2367
T 2/T 1weak =
Therefore, the post-shock temperature and pressure are:
p 2weak =
203
kPa
495
K
T 2weak =
For the weak shock, the pre-shock Mach number normal to the wave is:
M 1nstrong =
1.7608
The pressure and temperature ratios across the shock wave are:
3.4505
p 2/p 1strong =
1.5025
T 2/T 1strong =
Therefore, the post-shock temperature and pressure are:
p 2strong =
345
kPa
601
K
T 2strong =
Problem 13.180
[Difficulty: 3]
Given:
Normal shock
Find:
Approximation for downstream Mach number as upstream one approaches infinity
Solution:
2
Basic equations:
2
M 2n
M 1n
2
k1
(13.48a)
2 k
2
k 1 M1n 1
2
M 1n
Combining the two equations
M2
M 2n
sin( β θ)
1
M2
M 2n M 2 sin( β θ)
(13.47b)
2
k1
2 k M 2 1
k 1 1n
sin( β θ)
2
M 1n
2 k
2
k1
2
k 1 M1n 1 sin( β θ)
2
2
2
( k 1 ) M 1n
2 k 1 sin( β θ) 2
k 1 M1n2
As M1 goes to infinity, so does M1n, so
M2
1
2 k sin( β θ) 2
k 1
M2
k1
2 k sin( β θ)
2
Problem 13.179
[Difficulty: 4] Part 1/2
Problem 13.179
[Difficulty: 4] Part 2/2
Problem 13.178
[Difficulty: 4] Part 1/2
Problem 13.178
[Difficulty: 4] Part 2/2
Problem 13.177
[Difficulty: 3]
Problem 13.176
[Difficulty: 3]
Problem 13.175
[Difficulty: 3]
Given: Data on flow through gas turbine combustor
Find:
Maximum heat addition; Outlet conditions; Reduction in stagnation pressure; Plot of process
Solution:
R =
k =
cp =
T1 =
p1 =
M1 =
The given or available data is:
286.9
1.4
1004
773
1.5
0.5
p02
J/kg·K
T02
J/kg·K
K
MPa
p2
T2
p01
T
T01
T1
Equations and Computations:
p1
From
p1 1 RT1
1=
6.76
kg/m
From
V1 M 1 kRT1
V1 =
279
m/s
3
s
Using built-in function IsenT (M,k):
T 01 /T 1 =
1.05
T 01 =
812
K
Using built-in function Isenp (M,k):
p 01 /p 1 =
1.19
p 01 =
1.78
MPa
For maximum heat transfer:
M2 =
1
Using built-in function rayT0 (M,k), rayp0 (M,k), rayT (M,k), rayp (M,k), rayV (M,k):
T 01 /T 0* =
*
p 01 /p 0 =
*
T /T =
*
p /p =
/ =
*
0.691
T 0* =
1174
K
( = T 02)
1.114
*
1.60
MPa
( = p 02)
978
K
( = T 02)
p0 =
*
T =
0.790
*
1.778
p =
0.444
=
*
0.844
3
3.01
kg/m
-182
kPa
Note that at state 2 we have critical conditions!
Hence:
From the energy equation:
p 012 – p 01 =
Q
dm
-0.182
MPa
c p T02 T01
Q /dm =
364
kJ/kg
( = p 2)
MPa
( = 2)
Problem 13.174
[Difficulty: 3]
Problem 13.173
[Difficulty: 4]
Given: Air flow through a duct with heat transfer followed by converging duct, sonic at exit
Find:
Magnitude and direction of heat transfer
Solution:
The given or available data is:
R =
cp =
k =
M1 =
T1 =
p1 =
A 2/A 3 =
M3 =
53.33
0.2399
1.4
2
300
70
1.5
1
ft-lbf/lbm-°R
Btu/lbm-°R
°R
psia
Equations and Computations:
We can determine the stagnation temperature at the entrance:
T 01/T 1 =
1.8000
So the entrance stagnation temperature is:
°R
T 01 =
540.00
The reference stagnation temperature ratio at state 1 is:
T 01/T 0* =
0.7934
The reference conditions for Rayliegh flow can be calculated:
°R
T 0* =
680.6
Since the flow is sonic at state 3, we can find the Mach number at state 2:
M2 =
1.8541
We know that the flow must be supersonic at 2 since the flow at M 1 > 1.
The reference stagnation temperature ratio at state 2 is:
T 02/T 0* =
0.8241
Since the reference stagnation temperature at 1 and 2 are the same:
°R
560.92
T 02 =
The heat transfer is related to the change in stagnation temperature:
q 1-2 =
5.02
Btu/lbm
The heat is being added to the flow.
Problem 13.172
[Difficulty: 2]
Problem 13.171
[Difficulty: 2]
Problem 13.170
[Difficulty: 2]
Problem 13.169
[Difficulty: 3]
Given: Air flow through a duct with heat transfer
Find:
Heat transfer needed to choke the flow
Solution:
The given or available data is:
R =
cp =
k =
p1 =
T1 =
V1 =
286.9
1004
1.4
135
500
540
c1 =
448.1406
J/kg-K
J/kg-K
kPa
K
m/s
Equations and Computations:
The sonic velocity at state 1 is:
m/s
So the Mach number is:
1.2050
M1 =
We can determine the stagnation temperature at the entrance:
1.2904
T 01/T 1 =
So the entrance stagnation temperature is:
T 01 =
645.20
K
The reference conditions for Rayliegh flow can be calculated:
T 01/T 0* =
*
T0 =
0.9778
659.9
K
Since the flow is choked, state 2 is:
1.000
M2 =
659.85
K
T 02 =
The heat transfer is related to the change in stagnation temperature:
14.71
kJ/kg
q 1-2 =
To choke a flow, heat must always be added .
Problem 13.168
[Difficulty: 3]
Given: Air flow through a duct with heat transfer
Find:
Exit conditions
Solution:
The given or available data is:
R =
cp =
k =
m=
286.9
1004
1.4
20
J/kg-K
J/kg-K
A=
p1 =
T1 =
q 1-2 =
0.06
320
350
650
m2
kPa
K
kJ/kg
ρ1 =
3.1868
kg/m3
V1 =
104.5990
m/s
c1 =
374.9413
m/s
kg/s
Equations and Computations:
The density at the entrance is:
So the entrance velocity is:
The sonic velocity is:
So the Mach number is:
0.2790
M1 =
We can determine the stagnation temperature at the entrance:
T 01/T 1 =
1.0156
So the entrance stagnation temperature is:
355.45
K
T 01 =
The reference conditions for Rayliegh flow can be calculated:
T 01/T 0* =
0.3085
*
T0 =
1152.2
*
T 1/T =
0.3645
T* =
960.2
*
2.1642
p 1/p =
K
K
p* =
147.9
kPa
The heat transfer is related to the change in stagnation temperature:
T 02 =
1002.86
K
The stagnation temperature ratio at state 2 is:
T 02/T 0* =
We can now find the exit Mach number:
M2 =
0.8704
0.652
*
0.8704
T 02/T 0 =
(We used Solver to match the reference pressure ratio by varying M 2.)
We can now calculate the exit temperature and pressure:
T 2/T * =
T2 =
*
p 2/p =
T2 =
0.9625
924
K
1.5040
222
kPa
Problem 13.167
[Difficulty: 2]
Problem 13.166
[Difficulty: 3]
Problem 13.165
[Difficulty: 3]
Given:
Frictionless flow of air in a duct
Find:
Heat transfer without choking flow; change in stagnation pressure
k
Solution:
Basic equations:
T0
k1
1
T
M
2
mrate
p1 p2
A
p0
2
A
π
4
2
D
At state 1
From continuity
From momentum
A 78.54 cm
2
k 1.4
kg
mrate 0.5
s
D 10 cm
M2 1
cp 1004
J
R 286.9
kg K
kg K
kg
m
ρ1
ρ1 0.894
c 1 k R T1
c1 331
R T1
3
s
m
mrate
V1
m
then
V1
V1 71.2
M1
M 1 0.215
ρ1 A
c1
s
mrate
p
2
2
2
2 2
2
2
p1 p2
V2 V1 ρ2 V2 ρ1 V1 but
ρ V ρ c M
k R T M k p M
R T
A
2
2
From continuity
p1
p1
ρ1 V1
M 1 c1
M k R T1
R T1
R T1 1
p1 M1
T2
T02 T2 1
k1
or
1 k M 2
1
p2 p1
2
1 k M2
k p1 M1
ρ2 V2
R
T1
p2 M2
T2 T1
p1 M1
p2 M2
T1
2
M2
2
T02 1394 K
p 2 31.1 kPa
k p2 M2
R
T2
2
T2 1161 K
T01 T1 1
T2 888 °C
k1
2
M1
2
k
p 02 p 2 1
k1
Finally
J
p1
p1 p2 k p2 M2 k p1 M1
Then
2
cp T02 T01
Hence
Hence
k 1
p 1 70 kPa
dm
M
2
mrate ρ A V
δQ
Given or available data T1 ( 0 273 ) K
k1
p ρ R T
p
V2 V1
1
δQ
M2
2
2
p 02 58.8 kPa
MJ
cp T02 T01 1.12
kg
dm
(Using Rayleigh functions, at M 1 0.215
T01
T0crit
k
k 1
p 01 p 1 1
Δp0 p 02 p 01
T01
0.1975 T02
0.1975
T02
T01
T01 276 K
k1
2
M1
2
k 1
p 01 72.3 kPa
Δp0 13.5 kPa
T02 1395 K and ditto for p02 ...Check!)
Problem 13.164
[Difficulty: 3]
Problem 13.163
[Difficulty: 3]
Given: Nitrogen flow through a duct with heat transfer
Find:
Heat transfer
Solution:
The given or available data is:
R =
cp =
k =
M1 =
T 01 =
p1 =
p2 =
55.16
0.2481
1.4
0.75
500
24
40
ft-lbf/lbm-°R
Btu/lbm-°R
°R
psia
psia
Equations and Computations:
We can find the pressure and stagnation temperature at the reference state:
p 1/p * =
1.3427
*
T 01/T 0 =
0.9401
So the reference pressure and stagnation temperature are:
p* =
17.875
psia
°R
T0 =
531.9
We can now find the exit Mach number through the reference pressure:
*
p 2/p * =
M2 =
2.2378
0.2276
p 2/p * =
2.2378
(We used Solver to match the reference pressure ratio by varying M 2.)
Since the reference state is the same at stations 1 and 2, state 2 is:
T 02/T 0* =
0.2183
°R
116
T 02 =
The heat transfer is related to the change in stagnation temperature:
q 1-2 =
-95.2
Btu/lb
(The negative number indicates heat loss from the nitrogen)
Problem 13.162
[Difficulty: 3]
Problem 13.161
[Difficulty: 3]
Problem 13.160
Given:
Frictionless air flow in a pipe
Find:
Heat exchange per lb (or kg) at exit, where 500 kPa
[Difficulty: 2]
Solution:
Basic equations: mrate ρ V A
δQ
p ρ R T
Given or available data T1 ( 15 273 ) K
dm
cp T02 T01
p 1 1 MPa
M 1 0.35
D 5 cm
k 1.4
cp 1004
p1
ρ1
R T1
ρ1 12.1
V1 M 1 c1
m
V1 119
s
From momentum
p1 p2
V2
V1
ρ1 V1
m
V2 466
s
From continuity
ρ1 V1 ρ2 V2
V1
ρ2 ρ1
V2
ρ2 3.09
T2 564 K
T2 291 °C
At section 1
p2
T2
and
T02 T2 1
k1
T01 T1 1
k1
with
Then
ρ2 R
δQ
3
c1
p 2 500 kPa
J
kg K
k R T1
m
Hence
kg
2
2
M2
M1
2
2
p 1 p 2 ρ1 V1 V2 V1 (Momentum)
(Energy)
R 286.9
c1 340
M2 1
J
kg K
m
s
kg
3
m
T02 677 K
T02 403 °C
T01 295 K
T01 21.9 °C
Btu
kJ
cp T02 T01 164
383
lbm
kg
dm
T0
(Note: Using Rayleigh line functions, for M 1 0.35
0.4389
T0crit
so
T0crit
T01
0.4389
T0crit 672K close to T2 ... Check!)
Problem 13.159
Given:
Frictionless flow of Freon in a tube
Find:
Heat transfer; Pressure drop
[Difficulty: 2]
NOTE: ρ2 is NOT as stated; see below
Solution:
Basic equations: mrate ρ V A
p ρ R T
BTU
Given or available data h 1 25
lbm
lbm
ρ1 100
D 0.65 in
Then
Q mrate h 02 h 01
A
mrate
π
4
ft
3
2
V1 8.03
s
mrate
ft
V2 944
s
Q mrate h 02 h 01
The pressure drop is
Δp ρ1 V1 V2 V1
BTU
ρ2 0.850
lbm
h 01 h 1
h 02 h 2
Q 107
p 1 p 2 ρ1 V1 V2 V1
2
V1
2
h 01 25.0
2
V2
BTU
s
Δp 162 psi
lbm
ft
3
lbm
mrate 1.85
s
A 0.332 in
V1
ρ1 A
The heat transfer is
h 2 65
V
2
D
ft
V2
ρ2 A
2
h0 h
2
2
h 02 82.8
BTU
lbm
BTU
lbm
(74 Btu/s with the wrong ρ2!)
(-1 psi with the wrong ρ 2!)
Problem 13.158
[Difficulty: 3]
Given: Air flow through a duct with heat transfer
Find:
Heat addition needed to yield maximum static temperature and choked flow
Solution:
The given or available data is:
R =
cp =
k =
D=
V1 =
p1 =
T1 =
T1 =
53.33
0.2399
1.4
6
300
14.7
200
660
ft-lbf/lbm-°R
Btu/lbm-°R
c1 =
1259.65
ft/s
in
ft/s
psia
°F
°R
Equations and Computations:
The sound speed at station 1 is:
So the Mach number is:
0.2382
M1 =
We can determine the stagnation temperature at the entrance:
T 01/T 1 =
1.0113
So the entrance stagnation temperature is:
°R
T 01 =
667.49
The reference stagnation temperature for Rayliegh flow can be calculated:
T 01/T 0* =
0.2363
*
°R
T0 =
2824.4
For the maximum static temperature, the corresponding Mach number is:
M2 =
0.8452
Since the reference state is the same at stations 1 and 2, state 2 is:
T 02/T 0* =
0.9796
°R
2767
T 02 =
The heat transfer is related to the change in stagnation temperature:
q 1-2 =
504
Btu/lb
For acceleration to sonic flow the exit state is the * state:
q 1-* =
517
Btu/lb
Problem 13.157
[Difficulty: 2]
Problem 13.156
[Difficulty: 2]
Problem 13.155
[Difficulty: 4]
Given: Air flow from converging-diverging nozzle into heated pipe
Find:
Plot Ts diagram and pressure and speed curves
Solution:
The given or available data is:
R =
k =
53.33
1.4
ft·lbf/lbm·oR
cp =
0.2399
Btu/lbm·oR
187
ft·lbf/lbm·oR
R
o
T0 =
p0 =
pe =
710
25
2.5
Me =
2.16
Using built-in function IsenT (M ,k )
Te =
368
o
Using p e, M e, and function Rayp (M ,k )
p* =
7.83
psi
Using T e, M e, and function RayT (M ,k )
T* =
775
o
Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using built-in function IsenMfromp (M ,k ))
psi
psi
R
R
We can now use Rayleigh-line relations to compute values for a range of Mach numbers:
M
T /T *
2.157
2
1.99
1.98
1.97
1.96
1.95
1.94
1.93
1.92
1.91
1.9
1.89
1.88
1.87
1.86
1.85
1.84
1.83
1.82
1.81
1.8
1.79
1.78
1.77
1.76
1.75
1.74
1.73
1.72
1.71
0.475
0.529
0.533
0.536
0.540
0.544
0.548
0.552
0.555
0.559
0.563
0.567
0.571
0.575
0.579
0.584
0.588
0.592
0.596
0.600
0.605
0.609
0.613
0.618
0.622
0.626
0.631
0.635
0.640
0.645
0.649
T (oR)
368
410
413
416
418
421
800
424
750
427
430 700
433 650
436 600
440
T (oR) 550
443
500
446
449 450
452 400
455 350
459 300
462
0
465
468
472
475
479
482
485
489
492
496
499
503
c (ft/s)
940
993
996
1000
1003
1007
1010
1014
1017
1021
1024
1028
1032
1035
1039
1043
1046
1050
105410
1057
1061
1065
1069
1073
1076
1080
1084
1088
1092
1096
1100
V (ft/s)
p /p *
p (psi)
Δs
(ft·lbf/lbm·oR)
Eq. (12.11b)
2028
0.32
2.5
1985
0.36
2.8
1982
0.37
2.9
1979 Ts Curve
0.37 (Rayleigh)
2.9
1976
0.37
2.9
1973
0.38
2.9
1970
0.38
3.0
1966
0.38
3.0
1963
0.39
3.0
1960
0.39
3.0
1957
0.39
3.1
1953
0.40
3.1
1950
0.40
3.1
1946
0.40
3.2
1943
0.41
3.2
1939
0.41
3.2
1936
0.41
3.2
1932
0.42
3.3
1928
0.42
20
30
40 3.3
50
1925
0.43
3.3
.
o
s
(ft
lbf/lbm
R)
1921
0.43
3.4
1917
0.43
3.4
1913
0.44
3.4
1909
0.44
3.5
1905
0.45
3.5
1901
0.45
3.5
1897
0.45
3.6
1893
0.46
3.6
1889
0.46
3.6
1885
0.47
3.7
1880
0.47
3.7
0.00
13.30
14.15
14.99
15.84
16.69
17.54
18.39
19.24
20.09
20.93
21.78
22.63
23.48
24.32
25.17
26.01
26.86
27.70 60
28.54
29.38
30.22
31.06
31.90
32.73
33.57
34.40
35.23
36.06
36.89
37.72
70
80
1.7
1.69
1.68
1.67
1.66
1.65
1.64
1.63
1.62
1.61
1.6
1.59
1.58
1.57
1.56
1.55
1.54
1.53
1.52
1.51
1.5
1.49
1.48
1.47
1.46
1.45
1.44
1.43
1.42
1.41
1.4
1.39
1.38
1.37
1.36
1.35
1.34
1.33
1.32
1.31
1.3
1.29
1.28
1.27
1.26
1.25
1.24
1.23
1.22
1.21
1.2
1.19
1.18
1.17
1.16
1.15
1.14
1.13
1.12
1.11
1.1
1.09
1.08
1.07
1.06
1.05
1.04
1.03
1.02
1.01
1
0.654
0.658
0.663
0.668
0.673
0.677
0.682
0.687
0.692
0.697
0.702
0.707
0.712
0.717
0.722
0.727
0.732
0.737
0.742
0.747
0.753
0.758
0.763
0.768
0.773
0.779
0.784
0.789
0.795
0.800
0.805
0.811
0.816
0.822
0.827
0.832
0.838
0.843
0.848
0.854
0.859
0.865
0.870
0.875
0.881
0.886
0.891
0.896
0.902
0.907
0.912
0.917
0.922
0.927
0.932
0.937
0.942
0.946
0.951
0.956
0.960
0.965
0.969
0.973
0.978
0.982
0.986
0.989
0.993
0.997
1.000
507
1104
510
1107
514
1111
517
1115
521
1119
525
1123
529
1127
532
1131
536
1135
540
1139
544
1143
548
1147
2500
551
1151
555
1155
559 2000
1159
563
1164
567 1500
1168
571
1172
V (ft/s)
575
1176
579 1000
1180
583
1184
587 500
1188
591
1192
595
1196
0
599
1200
2.0
603
1204
607
1208
612
1213
616
1217
620
1221
624
1225
628
1229
632
1233
636
1237
9
641
1241
645 8
1245
649 7
1249
653
1253
6
657
1257
662 5
1261
p (psi)
666 4
1265
670
1269
3
674
1273
678 2
1277
682 1
1281
686
1285
0
690
1288
2.0
694
1292
699
1296
703
1300
706
1303
710
1307
714
1310
718
1314
722
1318
726
1321
730
1324
733
1328
737
1331
741
1334
744
1337
747
1341
751
1344
754
1347
757
1349
761
1352
764
1355
767
1358
769
1360
772
1362
775
1365
1876
0.48
3.7
38.54
1872
0.48
3.8
39.36
1867
0.48
3.8
40.18
1863
0.49
3.8
41.00
1858
0.49
3.9
41.81
1853
0.50
3.9
42.62
1849
0.50
3.9
43.43
1844
0.51
4.0
44.24
1839
0.51
4.0
45.04
1834
0.52
4.1
45.84
Velocity
V Versus
M (Rayleigh)
1829
0.52
4.1
46.64
1824
0.53
4.1
47.43
1819
0.53
4.2
48.22
1814
0.54
4.2
49.00
1809
0.54
4.3
49.78
1803
0.55
4.3
50.56
1798
0.56
4.3
51.33
1793
0.56
4.4
52.10
1787
0.57
4.4
52.86
1782
0.57
4.5
53.62
1776
0.58
4.5
54.37
1770
0.58
4.6
55.12
1764
0.59
4.6
55.86
1758
0.60
4.7
56.60
1752
0.60
4.7
57.33
1.8
1.6
1.4
1746
0.61
4.8
58.05
M 4.8
1740
0.61
58.77
1734
0.62
4.9
59.48
1728
0.63
4.9
60.18
1721
0.63
5.0
60.88
1715
0.64
5.0
61.56
1708
0.65
5.1
62.24
Pressure
p Versus
M (Rayleigh)
1701
0.65
5.1
62.91
1695
0.66
5.2
63.58
1688
0.67
5.2
64.23
1681
0.68
5.3
64.88
1674
0.68
5.3
65.51
1667
0.69
5.4
66.14
1659
0.70
5.5
66.76
1652
0.71
5.5
67.36
1645
0.71
5.6
67.96
1637
0.72
5.6
68.54
1629
0.73
5.7
69.11
1622
0.74
5.8
69.67
1614
0.74
5.8
70.22
1606
0.75
5.9
70.75
1598
0.76
6.0
71.27
1.8
1.6
1.4
1589
0.77
6.0
71.78
M 6.1
1581
0.78
72.27
1573
0.79
6.2
72.75
1564
0.80
6.2
73.21
1555
0.80
6.3
73.65
1546
0.81
6.4
74.08
1537
0.82
6.4
74.50
1528
0.83
6.5
74.89
1519
0.84
6.6
75.27
1510
0.85
6.7
75.63
1500
0.86
6.7
75.96
1491
0.87
6.8
76.28
1481
0.88
6.9
76.58
1471
0.89
7.0
76.86
1461
0.90
7.1
77.11
1451
0.91
7.1
77.34
1441
0.92
7.2
77.55
1430
0.93
7.3
77.73
1420
0.94
7.4
77.88
1409
0.95
7.5
78.01
1398
0.97
7.6
78.12
1387
0.98
7.6
78.19
1376
0.99
7.7
78.24
1365
1.00
7.8
78.25
1.2
1.2
1.0
1.0
Problem 13.154
[Difficulty: 2]
Given: Air flow through a duct with heat transfer
Find:
Exit static and stagnation temperatures; magnitude and direction of heat transfer
Solution:
The given or available data is:
R =
cp =
k =
M1 =
T1 =
M2 =
286.9
1004
1.4
3
250
1.6
J/kg-K
J/kg-K
K
Equations and Computations:
We can determine the stagnation temperature at the entrance:
T 01/T 1 =
2.8000
So the entrance stagnation temperature is:
T 01 =
700.00
K
The reference stagnation temperature for Rayliegh flow can be calculated:
T 01/T 0* =
0.6540
*
T0 =
1070.4
K
Since the reference state is the same at stations 1 and 2, state 2 is:
T 02/T 0* =
0.8842
946
K
T 02 =
1.5120
T 02/T 2 =
626
K
T2 =
The heat transfer is related to the change in stagnation temperature:
q 1-2 =
247
kJ/kg
Problem 13.153
[Difficulty: 4]
Given: Air flow from converging nozzle into heated pipe
Find:
Plot Ts diagram and pressure and speed curves
Solution:
The given or available data is:
R =
k =
53.33
1.4
ft·lbf/lbm·oR
cp =
0.2399
Btu/lbm·oR
187
ft·lbf/lbm·oR
R
psi
psi
o
T0 =
p0 =
pe=
710
25
24
Me =
0.242
Using built-in function IsenT (M ,k )
Te =
702
Using p e, M e, and function Rayp (M ,k )
p* =
10.82
psi
Using T e, M e, and function RayT (M ,k )
T* =
2432
o
Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using built-in function IsenMfromp (M ,k ))
o
R
R
We can now use Rayleigh-line relations to compute values for a range of Mach numbers:
M
T /T *
0.242
0.25
0.26
0.27
0.28
0.29
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.4
0.41
0.42
0.43
0.44
0.45
0.46
0.289
0.304
0.325
0.346
0.367
0.388
0.409
0.430
0.451
0.472
0.493
0.514
0.535
0.555
0.576
0.595
0.615
0.634
0.653
0.672
0.690
0.708
0.725
T (oR)
702
740
790
841
892
9433000
994
1046
2500
1097
1149
2000
1200
1250
T (oR) 1500
1301
1351
1000
1400
1448500
1496
1543 0
1589
0
1635
1679
1722
1764
c (ft/s)
1299
1334
1378
1422
1464
1506
1546
1586
1624
1662
1698
1734
1768
1802
1834
1866
1897
1926
1955
1982
2009
2035
2059
V (ft/s)
50
p /p *
p (psi)
315
2.22
24.0
334
2.21
23.9
358
2.19
23.7
384 Ts Curve
2.18(Rayleigh)
23.6
410
2.16
23.4
437
2.15
23.2
464
2.13
23.1
492
2.12
22.9
520
2.10
22.7
548
2.08
22.5
577
2.07
22.4
607
2.05
22.2
637
2.03
22.0
667
2.01
21.8
697
2.00
21.6
728
1.98
21.4
759
1.96
21.2
790
1.94
21.0
821
1.92
100
150 20.8
852
1.91
20.6
o
s (ft.lbf/lbm
884
1.89
20.4R)
916
1.87
20.2
947
1.85
20.0
Δs
(ft·lbf/lbm·oR)
Eq. (12.11b)
0.00
10.26
22.81
34.73
46.09
56.89
67.20
77.02
86.40
95.35
103.90
112.07
119.89
127.36
134.51
141.35
147.90
154.17
160.17
200
165.92
171.42
176.69
181.73
250
300
0.47
0.48
0.49
0.5
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.6
0.61
0.62
0.63
0.64
0.65
0.66
0.67
0.68
0.69
0.7
0.71
0.72
0.73
0.74
0.75
0.76
0.77
0.78
0.79
0.8
0.81
0.82
0.83
0.84
0.85
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1
0.742
0.759
0.775
0.790
0.805
0.820
0.834
0.847
0.860
0.872
0.884
0.896
0.906
0.917
0.927
0.936
0.945
0.953
0.961
0.968
0.975
0.981
0.987
0.993
0.998
1.003
1.007
1.011
1.014
1.017
1.020
1.022
1.024
1.025
1.027
1.028
1.028
1.029
1.029
1.028
1.028
1.027
1.026
1.025
1.023
1.021
1.019
1.017
1.015
1.012
1.009
1.006
1.003
1.000
1805
1845
1884
1922
1958
3000
1993
2027
2500
2060
2091
2000
2122
2150
V (ft/s) 1500
2178
2204
1000
2230
2253
500
2276
2298
0
2318
0.2
2337
2355
2371
2387
2401
2415
2427
2438
2449 30
2458
2466 25
2474
2480 20
2486
p 2490
(psi) 15
2494
2497 10
2499
2501 5
2502
2502 0
2501
0.2
2500
2498
2495
2492
2488
2484
2479
2474
2468
2461
2455
2448
2440
2432
2083
979
1.83
19.8
186.57
2106
1011
1.81
19.6
191.19
2128
1043
1.80
19.4
195.62
Velocity
V Versus
M (Rayleigh)
2149
1075
1.78
19.2
199.86
2170
1107
1.76
19.0
203.92
2189
1138
1.74
18.8
207.80
2208
1170
1.72
18.6
211.52
2225
1202
1.70
18.4
215.08
2242
1233
1.69
18.2
218.48
2258
1265
1.67
18.0
221.73
2274
1296
1.65
17.9
224.84
2288
1327
1.63
17.7
227.81
2302
1358
1.61
17.5
230.65
2315
1389
1.60
17.3
233.36
2328
1420
1.58
17.1
235.95
2339
1450
1.56
16.9
238.42
2350
1481
1.54
16.7
240.77
2361
1511
1.53
16.5
243.01
0.3
0.4
0.5
0.6
0.7
0.8
2370
1541
1.51
16.3
245.15
M 16.1
2379
1570
1.49
247.18
2388
1600
1.47
15.9
249.12
2396
1629
1.46
15.8
250.96
2403
1658
1.44
15.6
252.70
2409
1687
1.42
15.4
254.36
2416
1715
1.41
15.2
255.93
Pressure
p Versus
M (Rayleigh)
2421
1743
1.39
15.0
257.42
2426
1771
1.37
14.9
258.83
2431
1799
1.36
14.7
260.16
2435
1826
1.34
14.5
261.41
2439
1853
1.33
14.4
262.59
2442
1880
1.31
14.2
263.71
2445
1907
1.30
14.0
264.75
2447
1933
1.28
13.9
265.73
2449
1959
1.27
13.7
266.65
2450
1985
1.25
13.5
267.50
2451
2010
1.24
13.4
268.30
2452
2035
1.22
13.2
269.04
2452
2060
1.21
13.1
269.73
2452
2085
1.19
12.9
270.36
2452
2109
1.18
12.8
270.94
0.3
0.4
0.5
0.6
0.7
0.8
2451
2133
1.17
12.6
271.47
M
2450
2156
1.15
12.5
271.95
2449
2180
1.14
12.3
272.39
2448
2203
1.12
12.2
272.78
2446
2226
1.11
12.0
273.13
2444
2248
1.10
11.9
273.43
2441
2270
1.09
11.7
273.70
2439
2292
1.07
11.6
273.92
2436
2314
1.06
11.5
274.11
2433
2335
1.05
11.3
274.26
2429
2356
1.04
11.2
274.38
2426
2377
1.02
11.1
274.46
2422
2398
1.01
10.9
274.51
2418
2418
1.00
10.8
274.52
0.9
1.0
0.9
1.0
Problem 13.152
[Difficulty: 5] Part 1/2
Problem 13.152
[Difficulty: 5] Part 2/2
Problem 13.151
[Difficulty: 2]
Problem 13.150
[Difficulty: 4] Part 1/2
Problem 13.150
[Difficulty: 4] Part 2/2
Problem 13.149
[Difficulty: 2]
Given:
Isothermal air flow in a duct
Find:
Downstream Mach number; Direction of heat transfer; Plot of Ts diagram
Solution:
Basic equations:
h1
V1
2
2
δQ
h2
dm
V2
2
2
T0
T
1
k1
2
M
2
mrate ρ V A
Given or available data
T1 ( 20 273 ) K
p 1 350 kPa
M 1 0.1
From continuity
mrate ρ1 V1 A ρ2 V2 A
so
ρ1 V1 ρ2 V2
Also
p ρ R T
M
Hence continuity becomes
p1
R T1
T1 T2
Hence
M2
But at each state
p2
M 1 c1
Since
From energy
and
p1
p2
R T2
V
p 2 150 kPa
V M c
or
c
M 2 c2
c1 c2
p1 M1 p2 M2
so
M1
M 2 0.233
2
2
V2
V1
h
h 2 h 02 h 01 cp T02 T01
2 1
dm 2
δQ
T0
T
1
k1
2
M
2
or
T0 T 1
k1
2
M
2
p02
Since T = const, but M 2 > M 1, then T02 > T01, and
δQ
dm
0
T
p01
T02
T 01
so energy is ADDED to the system
p2
p1
s
Problem 13.148
Given:
Isothermal air flow in a pipe
Find:
Mach number and location at which pressure is 500 kPa
[Difficulty: 5]
Solution:
Basic equations:
Given or available data
From continuity
Since
Then
At M 1 0.176
At M 2 0.529
Hence
f Lmax
1 k M
mrate ρ V A
p ρ R T
T1 ( 15 273 ) K
p 1 1.5 MPa
m
V1 60
s
D 15 cm
k 1.4
R 286.9
ρ1 V1 ρ2 V2
or
T1 T2
and
c1
c1 340
p1
M2 M1
p2
D
D
D
1 k M1
f Lmax2
L12 18.2
D
D
f
2
2
m
s
ln k M 1
2
ln k M 2
2
f Lmax1
D
2
2
M1
18.819
0.614
18.819 0.614 18.2
L12 210 m
V1
c1
ln k M
2
f 0.013
p 2 500 kPa
J
kg K
p2
T2
V2
V M c M k R T
2
1 k M2
k M2
2
k M
V1
M 2 0.529
k M1
f Lmax2
f L12
p1
T1
k R T1
f Lmax1
D
p1
M2 M1
p2
M 1 0.176
Problem 13.147
[Difficulty: 4]
Given: Oxygen supplied to astronaut via umbilical
Find:
Required entrance pressure and power needed to pump gas through the tube
Solution:
The given or available data is:
R =
cp =
k =
Q=
D=
L=
f=
T1 =
T1 =
T2 =
p2 =
259.8
909.4
1.4
10
1
15
0.01
20
293
293
30
J/kg-K
J/kg-K
L/min
cm
m
°C
K
K
kPa
Equations and Computations:
At the exit of the pipe we can calculate the density:
kg/m3
ρ2 =
0.39411
m=
6.568E-05
kg/s
A=
7.854E-05
m2
V2 =
2.12
m/s
c2 =
326.5
m/s
so the mass flow rate is:
The pipe area is:
Therefore, the flow velocity is:
The local sound speed is:
So the Mach number is:
M 2 = 0.006500
From the exit Mach number we can calculate:
T 02/T 2 =
1.0000
16893.2
fL 2/D =
Given the length, diameter, and friction factor, we know:
fL 1-2/D =
15.0
16908.2
Therefore:
fL 1/D =
So from this information we can calculate the entrance Mach number:
M 1 = 0.006498
16908.2
fL 1/D =
(We use Solver to calculate the Mach number based on the friction length)
The entrance sound speed is the same as that at the exit:
c1 =
326.5
m/s
So the flow velocity is:
2.12
m/s
V1 =
We can calculate the pressure ratio from the velocity ratio:
30.0
kPa
p1 =
From the entrance Mach number we can calculate:
T 01/T 1 =
1.0000
So the entrance and exit stagnation temperatures are:
T 01 =
293.00
K
293.00
K
T 02 =
The work needed to pump the gas through the pipeline would be:
W = 1.3073E-07 W
W =
0.1307
microwatts
Problem 13.146
[Difficulty: 5]
Given: Air flowing through a tube
Find:
Mass flow rate assuming incompressible, adiabatic, and isothermal flow
Solution:
R =
k =
53.33
1.4
ν =
D=
L=
f=
p1 =
T1 =
p2 =
0.000163
1
10
0.03
15
530
14.7
ft2/s
in
ft
A=
0.005454
ft2
ρ1 =
0.07642
lbm/ft3
V1 =
100.56
ft/s
m incomp =
0.0419
lbm/s
The given or available data is:
ft-lbf/lbm-°R
psia
°R
psia
Equations and Computations:
The tube flow area is:
For incompressible flow, the density is:
The velocity of the flow is:
The mass flow rate is:
For Fanno flow, the duct friction length is:
3.600
fL 1-2/D =
and the pressure ratio across the duct is:
p 1/p 2 =
1.0204
To solve this problem, we have to guess M 1. Based on this and the friction length,
we can determine a corresponding M 2. The pressure ratios for M 1 and M 2 will be used
to check the validity of our guess.
M1
M2
fL 1/D
fL 2/D
fL 1-2/D
p 1/p 2
0.0800
0.0813
106.72
103.12
3.600
1.0167
0.0900
0.0919
83.50
79.90
3.600
1.0213
0.1000
0.1027
66.92
63.32
3.600
1.0266
0.1100
0.1136
54.69
51.09
3.600
1.0326
Here we used Solver to match the friction length. When both the friction length and
the pressure ratios match the constraints set above, we have our solution.
Therefore our entrance and exit Mach numbers are:
M1 =
0.0900
0.0919
M2 =
The density at 1 was already determined. The sound speed at 1 is:
1128.8
ft/s
c1 =
so the velocity at 1 is:
V1 =
101.59
ft/s
and the mass flow rate is:
0.0423
lbm/s
m Fanno =
To solve this problem for isothermal flow, we perform a calculation similar to that done
above for the Fanno flow. The only difference is that we use the friction length relation
and pressure ratio relation for isothermal flow:
M2
fL 1/D
fL 2/D
fL 1-2/D
p 1/p 2
M1
0.0800
0.0813
105.89216
102.29216
3.600
1.0167
0.0900
0.0919
82.70400
79.10400
3.600
1.0213
0.1000
0.1027
66.15987
62.55987
3.600
1.0266
0.1100
0.1136
53.95380
50.35380
3.600
1.0326
Here we used Solver to match the friction length. When both the friction length and
the pressure ratios match the constraints set above, we have our solution.
Therefore our entrance and exit Mach numbers are:
M1 =
0.0900
0.0919
M2 =
The density and sound speed at 1 were already determined. The velocity at 1 is:
V1 =
101.59
ft/s
and the mass flow rate is:
0.0423
lbm/s
m Isothermal =
Note that in this situation, since the Mach number was low, the assumption of
incompressible flow was a good one. Also, since the Fanno flow solution shows
a very small change in Mach number, the temperature does not change much, and so
the isothermal solution gives almost identical results.
Problem 13.145
[Difficulty: 4]
Given: Natural gas pumped through a pipe
Find:
Required entrance pressure and power needed to pump gas through the pipe
Solution:
The given or available data is:
R =
cp =
k =
D=
L=
f=
T1 =
T1 =
T2 =
m=
p2 =
96.32
0.5231
1.31
30
60
0.025
140
600
600
40
150
Equations and Computations:
At the exit of the pipe we can calculate the density:
p2 =
21.756
ft-lbf/lbm-°R
Btu/lbm-°R
in
mi
°F
°R
°R
lbm/s
kPa
psia
lbm/ft3
ρ2 =
0.05421
A=
4.909
ft2
V2 =
150.32
ft/s
c2 =
1561.3
ft/s
The pipe area is:
Therefore, the flow velocity is:
The local sound speed is:
So the Mach number is:
M2 =
0.09628
From the exit Mach number we can calculate:
T 02/T 2 =
1.0014
76.94219
fL 2/D =
Given the length, diameter, and friction factor, we know:
fL 1-2/D =
3168.0
3244.9
Therefore:
fL 1/D =
So from this information we can calculate the entrance Mach number:
M1 =
0.01532
3244.9
fL 1/D =
(We use Solver to calculate the Mach number based on the friction length)
The entrance sound speed is the same as that at the exit:
c1 =
1561.3
ft/s
So the flow velocity is:
23.91
ft/s
V1 =
We can calculate the pressure ratio from the velocity ratio:
136.8
psi
p1 =
From the entrance Mach number we can calculate:
T 01/T 1 =
1.0000
So the entrance and exit stagnation temperatures are:
°R
T 01 =
600.02
°R
600.86
T 02 =
The work needed to pump the gas through the pipeline would be:
W =
17.5810
Btu/s
W =
24.9
hp
Problem 13.144
[Difficulty: 3]
Problem 13.143
[Difficulty: 3]
Given:
Air flow in a CD nozzle and insulated duct
Find:
Duct length; Plot of M and p
Solution:
Basic equations:
Fanno-line flow equations, and friction factor
Given or available data T1 ( 100 460 ) R
p 1 18.5 psi
k 1.4
cp 0.2399
Then for Fanno-line flow at M 1 2
M1 2
BTU
Rair 53.33
lbm R
2
and at M 2 1
Also
p1
p crit
Dh
2
1 M2
lbm R
fave Lmax1
Dh
2
1 M1
k M1
2
( k 1) M 2
1
ln
0.305
k1
2 k
2
M1
2 1
2
k1
k M2
2
( k 1) M 2
2
ln
0
k1
2 k
2
M2
2 1
2
k1
p1
lbm
ρ1
ρ1 0.089
Rair T1
3
ft
V1 M 1 k Rair T1
7 lbf s
For air at T1 100 °F, from Table A.9
μ 3.96 10
ft
For commercial steel pipe (Table 8.1)
ft lbf
p crit 45.3 psi
0.4082
fave Lmax2
A 1 in
1
k 1
p1
p1
1
2
0.4082
k1
p crit
p2
M1
2
1
M1
2
so
2
M2 1
e 0.00015 ft
Hence at this Reynolds number and roughness (Eq. 8.37)
e
D
1.595 10
3
2
ft
V1 2320
s
so
and
4 A
D
Re1
π
D 1.13 in
ρ1 V1 D
μ
6
Re1 1.53 10
f .02222
1.13
Combining results
ft
fave Lmax2 fave Lmax1
12
L12
.02222 ( 0.3050 0 )
Dh
Dh
f
D
L12 1.29 ft
L12 15.5 in
These calculations are a LOT easier using the Excel Add-ins! The M and p plots are shown in the Excel spreadsheet on the next
page.
The given or available data is:
M
2.00
1.95
1.90
1.85
1.80
1.75
1.70
1.65
1.60
1.55
1.50
1.45
1.40
1.35
1.30
1.25
1.20
1.15
1.10
1.05
1.00
fL m ax/D ΔfL ma x/D
0.305
0.000
0.290
0.015
0.274
0.031
0.258
0.047
0.242
0.063
0.225
0.080
0.208
0.097
0.190
0.115
0.172
0.133
0.154
0.151
0.136
0.169
0.118
0.187
0.100
0.205
0.082
0.223
0.065
0.240
0.049
0.256
0.034
0.271
0.021
0.284
0.010
0.295
0.003
0.302
0.000
0.305
f = 0.0222
p * = 45.3 kPa
D =
1.13 in
Fanno Line Flow Curves(M and p )
x (in) p /p * p (psi)
0
0.8
1.6
2.4
3.2
4.1
4.9
5.8
6.7
7.7
8.6
9.5
10.4
11.3
12.2
13.0
13.8
14.5
15.0
15.4
15.5
0.408
0.423
0.439
0.456
0.474
0.493
0.513
0.534
0.557
0.581
0.606
0.634
0.663
0.695
0.728
0.765
0.804
0.847
0.894
0.944
1.000
18.49
19.18
19.90
20.67
21.48
22.33
23.24
24.20
25.22
26.31
27.47
28.71
30.04
31.47
33.00
34.65
36.44
38.37
40.48
42.78
45.30
2.0
45
1.9
40
1.8
1.7
35
1.6
M 1.5
30 p (psi)
1.4
25
1.3
M
1.2
20
Pressure
1.1
15
1.0
0
4
8
x (in)
12
16
Problem 13.142
Given:
Air flow through a CD nozzle and tube.
Find:
Average friction factor; Pressure drop in tube
[Difficulty: 2]
Solution:
Assumptions: 1) Isentropic flow in nozzle 2) Adiabatic flow in tube 3) Ideal gas 4) Uniform flow
Given or available data:
J
k 1.40
R 286.9
p 0 1.35 MPa
T0 550 K
kg K
p 1 15 kPa
where State 1 is the nozzle exit
D 2.5 cm
L 1.5 m
1
k 1
k
2 p0
From isentropic relations M 1
1
k 1 p1
2
M 1 3.617
Then for Fanno-line flow (for choking at the exit)
( k 1) M 2
1
ln
0.599
Dh
2
k1
2 k
2
k M1
M1
2 1
2
2
( k 1) M 2
D 1 M1
1
k1
fave
ln
fave 0.0100
2
k1
L
2 k
2
k
M
M
2
1
1
1
2
fave Lmax
Hence
1 M1
2
k1
1
2
k 1
p1
p1
1
2
0.159
k1
p crit
p2
M1
2
1
M1
2
p2
p1
1
2
k
1
1
2
M
k1
2
M1
1 1
2
Δp p 1 p 2
p 2 94.2 kPa
Δp 79.2 kPa
These calculations are a LOT easier
using the Excel Add-ins!
Problem 13.141
[Difficulty: 4] Part 1/2
Problem 13.141
[Difficulty: 4] Part 2/2
Problem 13.140
[Difficulty: 3]
Problem 13.139
Example 13.8
[Difficulty: 3]
Problem 13.138
[Difficulty: 3]
Problem 13.137
[Difficulty: 3] Part 1/2
Problem 13.137
[Difficulty: 3] Part 2/2
Problem 13.136
[Difficulty: 3]
Given: Air traveling through a cast iron pipe
Find:
Friction factor needed for sonic flow at exit; inlet pressure
Solution:
The given or available data is:
R =
k =
D=
L=
M1 =
T1 =
T1 =
M2 =
p2 =
53.33
1.4
3.068
10
0.5
70
530
1
14.7
ft-lbf/lbm-°R
in
ft
°F
°R
psia
Equations and Computations:
From the entrance Mach number we can calculate:
p 1/p * =
fL 1/D =
From the exit Mach number we can calculate:
2.1381
1.06906
p 2/p * =
1.0000
0.00000
fL 2/D =
To find the friction factor of the duct, first we need to friction length:
fL 1-2/D =
1.06906
Based on this, and the pipe length and diameter, the friction factor is:
f=
0.0273
We can calculate the critical pressure from the exit pressure:
p* =
14.7
Therefore, the static pressure at the duct entrance is:
p1 =
31.4
psia
psia
Problem 13.135
[Difficulty: 3]
Given: Air traveling through a square duct
Find:
Entrance static and stagnation conditions; friction factor
Solution:
The given or available data is:
R =
k =
s=
L=
M1 =
M2 =
T2 =
p2 =
53.33
1.4
2
40
3
1.7
500
110
ft-lbf/lbm-°R
ft
ft
°R
psia
Equations and Computations:
From the entrance Mach number we can calculate:
p 01/p 1 =
36.7327
2.8000
T 01/T 1 =
p 1/p * =
*
T 1/T =
fL 1/D =
From the exit Mach number we can calculate:
p 2/p * =
0.2182
0.4286
0.52216
0.5130
*
0.7605
T 2/T =
0.20780
fL 2/D =
Since we know static conditions at 2, we can find the critical pressure and temperature:
p* =
214.4
psia
*
T =
°R
657.5
Therefore, the static conditions at the duct entrance are:
p1 =
46.8
psia
°R
282
T1 =
and from the isentropic relations we can find stagnation conditions:
p 01 =
1719
psia
°R
789
T 01 =
To find the friction factor of the duct, first we need to friction length:
fL 1-2/D =
0.31436
The area and perimeter of the duct are:
ft2
A=
4.0
P=
8.0
ft
Therefore the hydraulic diameter of the duct is:
DH =
2.0
ft
From the hydraulic diameter, length, and friction length, the friction factor is:
f=
0.01572
Problem 13.134
[Difficulty: 2]
Problem 13.133
[Difficulty: 2]
Given:
Air flow in a converging nozzle and insulated duct
Find:
Length of pipe
Solution:
Basic equations:
Fanno-line flow equations, and friction factor
Given or available data
T0 ( 250 460 ) R
p 0 145 psi
p 1 125 psi
D 2 in
k 1.4
cp 0.2399
T2 ( 150 460 ) R
BTU
Rair 53.33
lbm R
ft lbf
lbm R
1
From isentropic relations
k 1
k
2 p0
M1
1
k 1 p1
T0
T1
Then for Fanno-line flow
1
k1
fave Lmax1
Dh
2
M 1 so
2
2
1 M1
k M1
2
1
2
M 1 0.465
T1
T0
T1 681 R
1 k 1 M 2
1
2
( k 1) M 2
1
ln
1.3923
k1
2 k
2
M1
2 1
2
k1
2
k 1
p1
p1
1
2
2.3044
k1
p crit
p2
M1
2
1
M1
2
p crit
p1
2.3044
k 1
T1
Tcrit
p crit 54.2 psi
2
1
k1
Also, for
Tcrit
Then
1.031
T2
Tcrit
2
1
fave Lmax2
Dh
k1
2
2
1.150
M1
2
Tcrit 592 R
k 1
T2
M2
k M2
2
leads
to
2
2
1 M2
T1 221 °F
M2
2
k1
T1
1.150
Tcrit 132 °F
k 1 Tcrit
2
( k 1) M 2
2
ln
0.01271
k1
2 k
2
M2
2 1
2
k1
Tcrit
T2
1
M 2 0.906
Also
p1
ρ1
Rair T1
ρ1 0.496
For air at T1 221 °F, from Table A.9 (approximately)
lbm
ft
3
7 lbf s
μ 4.48 10
ft
For commercial steel pipe (Table 8.1)
e 0.00015 ft
Hence at this Reynolds number and roughness (Eq. 8.37)
Combining results
e
D
4
9 10
so
2
and
Re1
ρ1 V1 D
μ
6
Re1 3.41 10
f 0.01924
2
ft
f
L
f
L
ave max1
D ave max2
12
L12
.01924 ( 1.3923 0.01271 )
Dh
Dh
f
These calculations are a LOT easier using the Excel Add-ins!
ft
V1 595
s
V1 M 1 k Rair T1
L12 12.0 ft
Problem 13.132
[Difficulty: 3]
Problem 13.131
[Difficulty: 4]
Problem 13.130
[Difficulty: 4]
Given: Air traveling through converging nozzle and constant-area duct with friction;
Find:
choked flow at duct exit.
Pressure at end of duct; exit conditions if 80% of duct were removed
Solution:
The given or available data is:
R =
k =
p1 =
T1 =
286.9
1.4
600
550
J/kg-K
kPa
K
Equations and Computations:
Station 1 is a stagnation state, station 2 is between the nozzle and friction duct,
and station 3 is at the duct exit.
For part (a) we know:
fL 2-3/D =
5.3
1
M3 =
Therefore, we can make the following statements:
fL 3/D =
0
5.300
fL 2/D =
So the Mach number at the duct entrance is:
0.300
M2 =
5.300
fL 2/D =
(we used Solver to find the correct Mach number to match the friction length)
The pressure at station 2 can be found from the Mach number and stagnation state:
1.0644
p 1/p 2 =
563.69
kPa
p2 =
Since state 3 is the critical state, we can find the pressure at state 3:
p 2/p * =
*
3.6193
p =
155.75
kPa
p3 =
155.7
kPa
For part (a) we know that if we remove 80% of the duct:
fL 2-3/D =
1.06
0.300
M2 =
5.300
fL 2/D =
563.69
kPa
p2 =
Since we know state 2 and the friction length of the duct, we can find state 3:
fL 3/D =
4.240
So the Mach number at the duct exit is:
M3 =
0.326
4.240
fL 2/D =
(we used Solver to find the correct Mach number to match the friction length)
To find the exit pressure:
p 2/p * =
*
3.6193
p =
155.75
p 3/p * =
3.3299
kPa
At state 3 the pressure ratio is:
So the pressure is:
p3 =
519
kPa
These processes are plotted in the Ts diagram below:
T
p1
T1
p2
p 3short
p*
*
s
Problem 13.129
[Difficulty: 4]
Problem 13.128
[Difficulty: 3]
Problem 13.127
[Difficulty: 3]
Given:
Air flow in a CD nozzle and insulated duct
Find:
Temperature at end of duct; Force on duct; Entropy increase
Solution:
Basic equations:
Given or available data
Fs p 1 A p 2 A Rx mrate V2 V1
T0
T1 ( 100 460 ) R
p 1 18.5 psi
k 1.4
cp 0.2399
T
k1
1
2
M
T2
p2
Δs cp ln
Rair ln
T1
p1
2
M1 2
2
M2 1
BTU
Rair 53.33
lbm R
A 1 in
ft lbf
lbm R
Assuming isentropic flow in the nozzle
k1
2
1
M1
T0 T2
2
so
k1
T1 T0
2
1
M2
2
Also c1
mrate ρ1 V1 A
ρ1 0.0892
lbm
ft
1
3
2
k1
2
M1
2
M2
2
c2
T2 840 R
k Rair T2 V2 M 2 c2
V1
ρ2 ρ1
V2
so
p 2 ρ2 Rair T2
Hence
Rx p 2 p 1 A mrate V2 V1
Finally
T2
p2
Δs cp ln
Rair ln
T1
p1
(Note: Using Fanno line relations, at M 1 2
k1
mrate ρ1 V1 A ρ2 V2 A2
lbm
mrate 1.44
s
T2 T1
ft
V1 2320
s
k Rair T1 V1 M 1 c1
p1
ρ1
Rair T1
1
T1
Tcrit
p1
p crit
T1
T2
p1
Rx 13.3 lbf
p2
0.6667
0.4083
ft
V2 1421
s
ρ2 0.146
(Force is to the right)
BTU
lbm R
T2
p2
p1
0.4083
T1
0.667
p 2 45.3 psi
lbm
ft
p 2 45.3 psi
Δs 0.0359
T2 380 °F
T2 840 R
Check!)
3
Problem 13.126
[Difficulty: 3]
Given: Nitrogen traveling through C-D nozzle and constant-area duct with friction
Find:
Exit temperature and pressure
Solution:
The given or available data is:
R =
k =
p 01 =
T 01 =
T 01 =
A e/A t =
fL /D =
55.16
1.4
105
100
560
4
0.355
ft-lbf/lbm-°R
psia
°F
°R
Equations and Computations:
Based on the area ratio of the nozzle, we can find the nozzle exit Mach number:
M1 =
2.940
The pressure and temperature at station 1 are therefore:
3.128
psia
p1 =
°R
205.2
T1 =
The critical temperature, pressure, and maximum friction length at 1 are:
p 1/p * =
0.2255
*
p =
13.867
*
0.4397
T 1/T =
psia
*
T =
°R
466.7
0.51293
fL 1/D =
Based on the maximum and actual friction lengths, the maximum friction
length at station 2 is:
fL 2/D =
0.15793
So the exit Mach number is:
M2 =
1.560
0.15793
fL 2/D =
(we used Solver to find the correct Mach number to match the friction length)
The critical pressure and temperature ratios at station 2 are:
p 2/p * =
*
T 2/T =
So the exit temperature and pressure are:
p2 =
T2 =
0.5759
0.8071
7.99
377
psia
°R
Problem 13.125
[Difficulty: 3]
Problem 13.124
[Difficulty: 3]
Problem 13.123
[Difficulty: 2]
Problem 13.122
[Difficulty: 3]
Given:
Air flow in a converging nozzle and insulated duct
Find:
Pressure at end of duct; Entropy increase
k
Solution:
T0
Basic
equations:
T
Given or available data
1
k1
2
M
2
p0
p
k1
1
2
T0 ( 250 460 ) R
p 0 145 psi
k 1.4
cp 0.2399
M
2
k 1
T2
p2
Δs cp ln
Rair ln
T1
p1
p 1 125 psi
BTU
Rair 53.33
lbm R
c
k R T
T2 ( 150 460 ) R
ft lbf
lbm R
Assuming isentropic flow in the nozzle
k 1
k
2 p 0
1
k1
p1
M1
M 1 0.465
In the duct T0 (a measure of total energy) is constant, soM 2
At each location
Then
k Rair T1
c1 1279
ft
c2
k Rair T2
c2 1211
ft
p 2 ρ2 Rair T2
p 2 60.8 psi
(Note: Using Fanno line relations, at M 1 0.465
Finall
y
Tcrit
p crit
T2
Tcrit
1.031
so
M 2 0.907
p2
p crit
2
M1
2
T1 681 R
T1 221 °F
M 2 0.905
V1 M 1 c1
ft
V1 595
s
V2 M 2 c2
ft
V2 1096
s
V1
ρ2 ρ1
V2
ρ2 0.269
lbm
so
p1
Then
s
ft
T1
1
k1
s
ρ1 0.4960
mrate ρ1 V1 A ρ2 V2 A
Hence
T0
T0
1
T2
k1
2
c1
p1
ρ1
Rair T1
Also
T1
1.150
3
lbm
ft
3
T2
p2
BTU
Δs cp ln
Rair ln Δs 0.0231
lbm R
T1
p1
T1
Tcrit
Tcrit 329 K
1.150
p1
2.306
p crit
1.119
p 2 1.119 p crit
2.3060
p crit 54.2 psi
p 2 60.7 psi
Check!)
Problem 13.121
[Difficulty: 3]
Given: Oxygen traveling through duct
Find:
Inlet and exit Mach numbers, exit stagnation conditions, friction factor and
absolute roughness
Solution:
The given or available data is:
R =
k =
D =
L =
m =
p1 =
T1 =
p2 =
259.8
1.4
35
5
40.0
200
450
160
cm
m
kg/s
kPa
K
kPa
A =
0.0962
m2
J/kg-K
Equations and Computations:
The area of the duct is:
The sound speed at station 1 is:
c1 =
404.57
m/s
The density at 1 can be calculated from the ideal gas equation of state:
kg/m3
ρ1 =
1.7107
So the velocity at 1 is:
V1 =
243.03
m/s
and the Mach number at 1 is:
0.601
M1 =
The critical temperature and pressure may then be calculated:
p 1/p * =
1.7611
*
p =
113.6
*
1.1192
T 1/T =
kPa
*
T =
402.1
K
Since the critical pressure is equal at 1 and 2, we can find the pressure ratio at 2:
1.4089
p 2/p * =
The static to critical pressure ratio is a function of Mach number. Therefore:
M2 =
0.738
p 2/p * =
1.4089
(we used Solver to find the correct Mach number to match the pressure ratio)
The exit temperature is:
T 2/T * =
T2 =
1.0820
435.0
K
Based on the exit Mach number, pressure, and temperature, stagnation conditions are:
p 02 =
230
kPa
482
K
T 02 =
The maximum friction lengths at stations 1 and 2 are:
0.48802
fL 1/D =
0.14357
fL 2/D =
So the friction length for this duct is:
fL /D =
0.34445
and the friction factor is:
f =
0.02411
Now to find the roughness of the pipe, we need the Reynolds number.
From the LMNO Engineering website, we can find the viscosities of oxygen:
2
μ 1 = 2.688E-05 N-s/m
2
μ 2 = 2.802E-05 N-s/m
Therefore the Reynolds number at station 1 is:
Re1 = 5.413E+06
At station 2, we will need to find density and velocity first. From ideal gas equation:
kg/m3
ρ2 =
1.4156
The sound speed at 2 is:
c2 =
397.79
m/s
So the velocity at 2 is:
V2 =
293.69
m/s
and the Reynolds number is:
Re2 = 5.193E+06
So the Reynolds number does not change significantly over the length of duct.
We will use an average of the two to find the relative roughness:
Re = 5.303E+06
The relative roughness for this pipe is:
e/D =
0.00222
f =
0.02411
(we used Solver to find the correct roughness to match the friction factor.)
Therefore, the roughness of the duct material is:
e =
0.0776
cm
Problem 13.120
[Difficulty: 4]
Given: Air flow from converging-diverging nozzle into pipe
Find:
Plot Ts diagram and pressure and speed curves
Solution:
The given or available data is:
R =
k =
53.33
1.4
ft·lbf/lbm·oR
cp =
0.2399
Btu/lbm·oR
187
T0 =
p0 =
pe =
710
25
2.5
Me =
2.16
Using built-in function IsenT (M ,k )
Te =
368
Using p e, M e, and function Fannop (M ,k )
p* =
6.84
Using T e, M e, and function FannoT (M ,k )
T* =
592
Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using built-in function IsenMfromp (M ,k ))
ft·lbf/lbm·oR
R
o
psi
psi
o
R
psi
o
R
We can now use Fanno-line relations to compute values for a range of Mach numbers:
M
T /T *
2.157
2
1.99
1.98
1.97
1.96
1.95
1.94
1.93
1.92
1.91
1.9
1.89
1.88
1.87
1.86
1.85
1.84
1.83
1.82
1.81
1.8
1.79
1.78
1.77
1.76
1.75
1.74
1.73
1.72
1.71
0.622
0.667
0.670
0.673
0.676
0.679
0.682
0.685
0.688
0.691
0.694
0.697
0.700
0.703
0.706
0.709
0.712
0.716
0.719
0.722
0.725
0.728
0.731
0.735
0.738
0.741
0.744
0.747
0.751
0.754
0.757
T (oR)
368
394
396
398
400
402
650
403
405 600
407
409 550
410
500
412
T (oR)
414 450
416
418 400
420
350
421
423 300
425
0
427
429
431
433
435
436
438
440
442
444
446
448
c (ft/s)
940
974
976
978
980
982
985
987
989
991
993
996
998
1000
1002
1004
1007
1009
1011 5
1013
1015
1018
1020
1022
1024
1027
1029
1031
1033
1036
1038
V (ft/s)
p /p *
p (psi)
s
(ft·lbf/lbm·oR)
Eq. (12.11b)
2028
0.37
2.5
1948
0.41
2.8
1942
0.41
2.8
1937 Ts Curve
0.41 (Fanno)
2.8
1931
0.42
2.9
1926
0.42
2.9
1920
0.42
2.9
1914
0.43
2.9
1909
0.43
2.9
1903
0.43
3.0
1897
0.44
3.0
1892
0.44
3.0
1886
0.44
3.0
1880
0.45
3.1
1874
0.45
3.1
1868
0.45
3.1
1862
0.46
3.1
1856
0.46
3.1
1850
0.46
10
15
20 3.2
25
1844
0.47
3.2
.
o
lbf/lbm
R)
s
(ft
1838
0.47
3.2
1832
0.47
3.2
1826
0.48
3.3
1819
0.48
3.3
1813
0.49
3.3
1807
0.49
3.3
1801
0.49
3.4
1794
0.50
3.4
1788
0.50
3.4
1781
0.50
3.5
1775
0.51
3.5
0.00
7.18
7.63
8.07
8.51
8.95
9.38
9.82
10.25
10.68
11.11
11.54
11.96
12.38
12.80
13.22
13.64
14.05
14.46 30
14.87
15.28
15.68
16.08
16.48
16.88
17.27
17.66
18.05
18.44
18.82
19.20
35
40
1.7
1.69
1.68
1.67
1.66
1.65
1.64
1.63
1.62
1.61
1.6
1.59
1.58
1.57
1.56
1.55
1.54
1.53
1.52
1.51
1.5
1.49
1.48
1.47
1.46
1.45
1.44
1.43
1.42
1.41
1.4
1.39
1.38
1.37
1.36
1.35
1.34
1.33
1.32
1.31
1.3
1.29
1.28
1.27
1.26
1.25
1.24
1.23
1.22
1.21
1.2
1.19
1.18
1.17
1.16
1.15
1.14
1.13
1.12
1.11
1.1
1.09
1.08
1.07
1.06
1.05
1.04
1.03
1.02
1.01
1
0.760
0.764
0.767
0.770
0.774
0.777
0.780
0.784
0.787
0.790
0.794
0.797
0.800
0.804
0.807
0.811
0.814
0.817
0.821
0.824
0.828
0.831
0.834
0.838
0.841
0.845
0.848
0.852
0.855
0.859
0.862
0.866
0.869
0.872
0.876
0.879
0.883
0.886
0.890
0.893
0.897
0.900
0.904
0.907
0.911
0.914
0.918
0.921
0.925
0.928
0.932
0.935
0.939
0.942
0.946
0.949
0.952
0.956
0.959
0.963
0.966
0.970
0.973
0.976
0.980
0.983
0.987
0.990
0.993
0.997
1.000
450
1040
452
1042
454
1045
456
1047
458
1049
460
1051
462
1054
464
1056
466
1058
468
1060
470
1063
472
1065
2500
474
1067
476
1069
478 2000
1072
480
1074
482 1500
1076
484
1078
V (ft/s)
486
1080
488 1000
1083
490
1085
492 500
1087
494
1089
496
1092
0
498
1094
2.0
500
1096
502
1098
504
1101
506
1103
508
1105
510
1107
512
1110
514
1112
516
1114
8
518
1116
520 7
1118
522
1121
6
524
1123
527 5
1125
529
1127
p (psi) 4
531
1129
533 3
1132
535
1134
2
537
1136
539 1
1138
541
1140
0
543
1143
2.0
545
1145
547
1147
549
1149
551
1151
553
1153
555
1155
557
1158
559
1160
561
1162
564
1164
566
1166
568
1168
570
1170
572
1172
574
1174
576
1176
578
1179
580
1181
582
1183
584
1185
586
1187
588
1189
590
1191
592
1193
1768
0.51
3.5
19.58
1761
0.52
3.5
19.95
1755
0.52
3.6
20.32
1748
0.53
3.6
20.69
1741
0.53
3.6
21.06
1735
0.53
3.7
21.42
1728
0.54
3.7
21.78
1721
0.54
3.7
22.14
1714
0.55
3.7
22.49
1707
Velocity
V 0.55
Versus M 3.8
(Fanno) 22.84
1700
0.56
3.8
23.18
1693
0.56
3.8
23.52
1686
0.57
3.9
23.86
1679
0.57
3.9
24.20
1672
0.58
3.9
24.53
1664
0.58
4.0
24.86
1657
0.59
4.0
25.18
1650
0.59
4.0
25.50
1642
0.60
4.1
25.82
1635
0.60
4.1
26.13
1627
0.61
4.1
26.44
1620
0.61
4.2
26.75
1612
0.62
4.2
27.05
1605
0.62
4.3
27.34
1597
0.63
4.3
27.63
1.8
1.6
1.4
1589
0.63
4.3
27.92
M 4.4
1582
0.64
28.21
1574
0.65
4.4
28.48
1566
0.65
4.5
28.76
1558
0.66
4.5
29.03
1550
0.66
4.5
29.29
1542
Pressure
p 0.67
Versus M 4.6
(Fanno) 29.55
1534
0.68
4.6
29.81
1526
0.68
4.7
30.06
1518
0.69
4.7
30.31
1510
0.69
4.8
30.55
1502
0.70
4.8
30.78
1493
0.71
4.8
31.01
1485
0.71
4.9
31.24
1477
0.72
4.9
31.46
1468
0.73
5.0
31.67
1460
0.74
5.0
31.88
1451
0.74
5.1
32.09
1443
0.75
5.1
32.28
1434
0.76
5.2
32.48
1426
0.76
5.2
32.66
1417
0.77
5.3
32.84
1.8
1.6
1.4
1408
0.78
5.3
33.01
M 5.4
1399
0.79
33.18
1390
0.80
5.4
33.34
1381
0.80
5.5
33.50
1372
0.81
5.6
33.65
1363
0.82
5.6
33.79
1354
0.83
5.7
33.93
1345
0.84
5.7
34.05
1336
0.85
5.8
34.18
1327
0.86
5.9
34.29
1318
0.87
5.9
34.40
1308
0.87
6.0
34.50
1299
0.88
6.0
34.59
1290
0.89
6.1
34.68
1280
0.90
6.2
34.76
1271
0.91
6.2
34.83
1261
0.92
6.3
34.89
1251
0.93
6.4
34.95
1242
0.94
6.5
34.99
1232
0.96
6.5
35.03
1222
0.97
6.6
35.06
1212
0.98
6.7
35.08
1203
0.99
6.8
35.10
1193
1.00
6.8
35.10
1.2
1.2
1.0
1.0
Problem 13.119
[Difficulty: 4]
Given: Air flow from converging nozzle into pipe
Find:
Plot Ts diagram and pressure and speed curves
Solution:
The given or available data is:
R =
k =
53.33
1.4
cp =
0.2399
187
T0 =
p0 =
pe =
710
25
24
Me=
0.242
Using built-in function IsenT (M ,k )
Te =
702
Using p e, M e, and function Fannop (M ,k )
p* =
5.34
Using T e, M e, and function FannoT (M ,k )
T* =
592
Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using built-in function IsenMfromp (M ,k ))
o
ft·lbf/lbm· R
o
Btu/lbm· R
o
ft·lbf/lbm· R
R
o
psi
psi
o
R
psi
o
R
We can now use Fanno-line relations to compute values for a range of Mach numbers:
M
T /T *
0.242
0.25
0.26
0.27
0.28
0.29
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.4
0.41
0.42
0.43
0.44
0.45
1.186
1.185
1.184
1.183
1.181
1.180
1.179
1.177
1.176
1.174
1.173
1.171
1.170
1.168
1.166
1.165
1.163
1.161
1.159
1.157
1.155
1.153
T (oR)
702
701
701
700
699
698720
697
697700
696
695680
694
660
o 693
T ( R)
692640
691
690620
689
600
688
687580
686
0
685
684
682
c (ft/s)
1299
1298
1298
1297
1296
1296
1295
1294
1293
1292
1292
1291
1290
1289
1288
1287
1286
1285
1284
1283
1282
1281
V (ft/s)
p /p *
p (psi)
315
4.50
24.0
325
4.35
23.2
337
4.19
22.3
350 Ts Curve
4.03 (Fanno)
21.5
363
3.88
20.7
376
3.75
20.0
388
3.62
19.3
401
3.50
18.7
414
3.39
18.1
427
3.28
17.5
439
3.19
17.0
452
3.09
16.5
464
3.00
16.0
477
2.92
15.6
489
2.84
15.2
502
2.77
14.8
514
2.70
14.4
527
2.63
14.0
539
2.56
13.7
10
20
30
552
2.50
13.4
.
o
s (ft lbf/lbm
564
2.44
13.0R)
576
2.39
12.7
s
o
(ft·lbf/lbm· R
) Eq. (12.11b)
0.00
1.57
3.50
5.35
7.11
8.80
10.43
11.98
13.48
14.92
16.30
17.63
18.91
20.14
21.33
22.48
23.58
24.65
25.68
26.67
27.63
28.55
40
50
0.46
0.47
0.48
0.49
0.5
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.6
0.61
0.62
0.63
0.64
0.65
0.66
0.67
0.68
0.69
0.7
0.71
0.72
0.73
0.74
0.75
0.76
0.77
0.78
0.79
0.8
0.81
0.82
0.83
0.84
0.85
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1
1.151
1.149
1.147
1.145
1.143
1.141
1.138
1.136
1.134
1.132
1.129
1.127
1.124
1.122
1.119
1.117
1.114
1.112
1.109
1.107
1.104
1.101
1.098
1.096
1.093
1.090
1.087
1.084
1.082
1.079
1.076
1.073
1.070
1.067
1.064
1.061
1.058
1.055
1.052
1.048
1.045
1.042
1.039
1.036
1.033
1.029
1.026
1.023
1.020
1.017
1.013
1.010
1.007
1.003
1.000
681
1280
589
2.33
12.4
29.44
680
1279
601
2.28
12.2
30.31
679
1277
613
2.23
11.9
31.14
677
1276
625
Velocity
V 2.18
Versus M11.7
(Fanno) 31.94
676
1275
638
2.14
11.4
32.72
675 1400
1274
650
2.09
11.2
33.46
674
1273
662
2.05
11.0
34.19
672 1200
1271
674
2.01
10.7
34.88
671
1270
686
1.97
10.5
35.56
669 1000
1269
698
1.93
10.3
36.21
668
1267
710
1.90
10.1
36.83
800
667
1266
722
1.86
9.9
37.44
V (ft/s)
665 600
1265
733
1.83
9.8
38.02
664
1263
745
1.80
9.6
38.58
662 400
1262
757
1.76
9.4
39.12
661
1260
769
1.73
9.2
39.64
659 200
1259
781
1.70
9.1
40.14
658
1258
792
1.67
8.9
40.62
0
656
1256
804
1.65
8.8
41.09
0.2
0.3
0.4
0.5
0.6
0.7
0.8
655
1255
815
1.62
8.6
41.53
M 8.5
653
1253
827
1.59
41.96
652
1252
839
1.57
8.4
42.37
650
1250
850
1.54
8.2
42.77
648
1248
861
1.52
8.1
43.15
647
1247
873
1.49
8.0
43.51
645
1245
884
Pressure
p 1.47
Versus M7.8
(Fanno) 43.85
643
1244
895
1.45
7.7
44.18
642 30
1242
907
1.43
7.6
44.50
640
1240
918
1.41
7.5
44.80
638 25
1239
929
1.38
7.4
45.09
636
1237
940
1.36
7.3
45.36
635
1235
951
1.35
7.2
45.62
20
633
1234
962
1.33
7.1
45.86
631
1232
973
1.31
7.0
46.10
p (psi) 15
629
1230
984
1.29
6.9
46.31
628
1228
995
1.27
6.8
46.52
10
626
1227
1006
1.25
6.7
46.71
624
1225
1017
1.24
6.6
46.90
5
622
1223
1027
1.22
6.5
47.07
620
1221
1038
1.20
6.4
47.22
0
619
1219
1049
1.19
6.3
47.37
0.2
0.3
0.4
0.5
0.6
0.7
0.8
617
1218
1059
1.17
6.3
47.50
M 6.2
615
1216
1070
1.16
47.63
613
1214
1080
1.14
6.1
47.74
611
1212
1091
1.13
6.0
47.84
609
1210
1101
1.11
6.0
47.94
607
1208
1112
1.10
5.9
48.02
605
1206
1122
1.09
5.8
48.09
603
1204
1132
1.07
5.7
48.15
601
1202
1142
1.06
5.7
48.20
600
1201
1153
1.05
5.6
48.24
598
1199
1163
1.04
5.5
48.27
596
1197
1173
1.02
5.5
48.30
594
1195
1183
1.01
5.4
48.31
592
1193
1193
1.00
5.3
48.31
0.9
1.0
0.9
1.0
Problem 13.118
[Difficulty: 2]
Given:
Air flow in an insulated duct
Find:
Mass flow rate; Range of choked exit pressures
k 1
Solution:
T0
Basic equations:
T
Given or available data
1
k1
2
M
2
c
2 ( k 1)
1 k 1 M2
A
1
2
k 1
Acrit
M
2
k R T
T0 ( 80 460 ) R
p 0 14.7 psi
k 1.4
Rair 53.33
p 1 13 psi
ft lbf
lbm R
D 1 in
2
A
π D
2
A 0.785 in
4
Assuming isentropic flow, stagnation conditions are constant. Hence
M1
k 1
k
2 p 0
1
k1
p1
c1
Also
M 1 0.423
k Rair T1
c1 341
p1
ρ1
Rair T1
m
s
ρ1 0.0673
mrate ρ1 V1 A
When flow is choked
M 2 1 hence
We also have
c2
From continuity
ρ1 V1 ρ2 V2
k Rair T2
1
k1
2
M1
2
T1 521 R
V1 M 1 c1
m
V1 144
s
T2 450 R
T2 9.7 °F
V2 c2
ft
V2 1040
s
T1 61.7 °F
3
lbm
mrate 0.174
s
T0
T2
k1
1
2
ft
c2 1040
s
V1
ρ2 ρ1
V2
p 2 ρ2 Rair T2
T0
lbm
ft
Hence
Hence
T1
ρ2 0.0306
lbm
ft
3
p 2 5.11 psi
The flow will therefore choke for any back pressure (pressure at the exit) less than or equal to this pressure
(From Fanno line function
p1
p crit
2.545
at
M 1 0.423
so
p crit
p1
2.545
p crit 5.11 psi Check!)
Problem 13.117
[Difficulty: 2]
Given: Nitrogen traveling through duct
Find:
Inlet pressure and mass flow rate
Solution:
The given or available data is:
R =
k =
D =
M2 =
T2 =
p2 =
T1 =
296.8
1.4
30
0.85
300
200
330
J/kg-K
cm
K
kPa
K
Equations and Computations:
We can find the critical temperature and pressure for choking at station 2:
T 2/T * =
1.0485
*
T =
286.1
*
1.2047
p 2/p =
K
*
p =
166.0
kPa
Knowing the critical state, the Mach number at station 1 can be found:
(we will use Goal Seek to match the Mach number)
T 1/T * =
M1 =
1.1533
0.4497
1.1533
T 1/T * =
The static to critical pressure ratio is a function of Mach number. Therefore:
p1 =
396
kPa
The sound speed at station 1 is:
c1 =
370.30
m/s
So the velocity at 1 is:
V1 =
166.54
m/s
The density at 1 can be calculated from the ideal gas equation of state:
kg/m3
ρ1 =
4.0476
The area of the duct is:
A =
0.0707
m2
m =
47.6
kg/s
So the mass flow rate is:
Problem 13.116
[Difficulty: 4]
Problem 13.115
[Difficulty: 5]
Problem 13.114
[Difficulty: 2]
Problem 13.113
[Difficulty: 4]
Given: Air flowing through a converging-diverging nozzle followed by diabatic duct
Find:
Mach number at duct exit and heat addition in duct
Solution:
The given or available data is:
R =
cp =
k =
p 0inlet =
T 0inlet =
A 1/A t =
Te =
286.9
1004
1.4
1
320
2.5
350
J/kg-K
J/kg-K
MPa
K
K
Equations and Computations:
The Mach number at the nozzle exit can be found based on the area ratio:
2.4428
M1 =
The static temperature is:
2.1934
T 0inlet/T 1 =
145.891
K
T1 =
The Rayliegh flow critical ratios at this condition are:
T 1/T 1* =
0.39282
*
T 01/T 01 =
0.71802
Since all we know is the static temperature at the exit, we need to iterate on
a solution. We can guess at a pre-shock Mach number at the duct exit, and
iterate on that value until we match the exit temperature:
M2 =
1.753
T 2/T 2* =
*
0.62964
T 02/T 02 =
0.84716
233.844
K
T2 =
377.553
K
T 02 =
0.6274
M3 =
1.4967
T 3/T 2 =
350.000
K
T3 =
In this case we used Solver to match the exit temperature.
Therefore, the exit Mach number is:
M3 =
0.627
The rate of heat addition is calculated from the rise in stagnation temperature:
57.78
kJ/kg
q 1-2 =
Problem 13.112
[Difficulty: 5]
Given: Air flowing through a converging-diverging nozzle followed by duct with friction
Find:
Back pressure needed for (a) normal shock at nozzle exit, (b) normal shock at duct exit,
(c) back pressure for shock-free flow
Solution:
The given or available data is:
R =
k =
p 0inlet =
T 0inlet =
A e/A t =
L/D =
f=
286.9
1.4
1
320
2.5
10
0.03
J/kg-K
MPa
K
Equations and Computations:
(a) For a shock wave at the nozzle exit:
The pre-shock Mach number can be found based on the area ratio:
M1 =
2.4428
The static pressure before the shock wave is:
p 0inlet/p 1 =
15.6288
63.984
kPa
p1 =
The Mach number and static pressure after the shock wave are:
M2 =
0.5187
6.7950
p 2/p 1 =
434.770
kPa
p2 =
The friction length and critical pressure ratio after the shock wave are:
fL/D 2 =
0.9269
p 2/p 2* =
2.0575
The friction length for the duct is:
fL/D 2-3 =
0.3000
Therefore, the friction length at the duct exit is:
0.6269
fL/D 3 =
Iterating on Mach number with Solver to match this friction length yields:
0.5692
M3 =
0.6269
fL/D 3 =
The critical pressure ratio for this condition is:
p 3/p 3* =
1.8652
Since the critical pressure at 2 and 3 are equal, the back pressure is:
pb = p3 =
394
kPa
(b) For a shock wave at the duct exit:
We use the same nozzle exit Mach number and pressure:
M1 =
2.443
63.984
kPa
p1 =
The friction length and critical pressure ratio at this condition are:
fL/D 1 =
0.4195
p 1/p 1* =
0.3028
The friction length for the duct is:
fL/D 1-2 =
0.3000
Therefore, the friction length at the duct exit is:
fL/D 2 =
0.1195
Iterating on Mach number with Solver to match this friction length yields:
1.4547
M2 =
0.1195
fL/D 2 =
The critical pressure ratio for this condition is:
p 2/p 2* =
0.6312
Since the critical pressure at 1 and 2 are equal, the pressure is:
p2 =
133.388
kPa
The Mach number and static pressure after the shock wave are:
M3 =
0.7178
2.3021
p 3/p 2 =
307
kPa
pb = p3 =
(c) For shock-free flow, we use the conditions from part b before the shock wave:
pb = p3 =
133.4
kPa
Problem 13.111
[Difficulty: 3]
Problem 13.110
[Difficulty: 4]
Problem 13.109
[Difficulty: 3]
Given: Air flowing through a converging-diverging nozzle with standing normal shock
Find:
Exit Mach number and static pressure; design point pressure
Solution:
The given or available data is:
R =
k =
p 0inlet =
T 01 =
T 01 =
A e/A t =
A 1/A t =
53.33
1.4
150
200
660
1.76
1.2
ft-lbf/lbm-°R
psia
°F
°R
Equations and Computations:
The pre-shock Mach number can be found based on the area ratio:
1.5341
M1 =
The static pressure before the shock wave is:
3.8580
p 0inlet/p 1 =
38.881
psia
p1 =
The Mach number and static pressure after the shock wave are:
0.689
M2 =
2.5792
p 2/p 1 =
100.282
psia
p2 =
The area ratio for the remainder of the nozzle is:
A e/A 2 =
1.4667
Based on this and the post-shock Mach number, we can determine
the exit Mach number:
A 2/A 2* =
*
A e/A 2 =
Me =
1.102
1.617
0.392
Therefore the exit pressure is:
p 02/p 2 =
1.374
1.112
p 02/p e =
123.9
psia
pe =
Based on the area ratio, the design Mach number is:
Md =
2.050
The pressure ratio for the third critical can be found from the design point Mach number:
p 0inlet/p b,3rd =
8.4583
0.1182
p b,3rd/p 0inlet =
So the design pressure is:
pd =
17.73
psia
Problem 13.108
[Difficulty: 4]
Problem 13.107
[Difficulty: 3]
Problem 13.106
[Difficulty: 3]
Problem 13.105
[Difficulty: 3]
Problem 13.104
[Difficulty: 3]
Problem 13.103
[Difficulty: 3]
Problem 13.102
[Difficulty: 3]
Problem 13.101
[Difficulty: 2]
Problem 13.100
[Difficulty: 3]
Given: Normal shock in CD nozzle
Find:
Exit pressure; Throat area; Mass flow rate
Solution:
The given or available data is:
R =
k =
T 01 =
p 01 =
M1 =
286.9
1.4
550
700
2.75
A1 =
25
cm2
Ae =
40
cm2
J/kg·K
K
kPa
Equations and Computations (assuming State 1 and 2 before and after the shock):
Using built-in function Isenp (M,k):
p 01 /p 1 =
25.14
p1 =
28
kPa
Using built-in function IsenT (M,k):
T 01 /T 1 =
2.51
T1 =
219
K
3.34
A 1* = A t =
7.49
cm2
284
kPa
Using built-in function IsenA (M,k):
A 1 /A 1* =
Then from the Ideal Gas equation:
Also:
So:
Then the mass flow rate is:
1 =
0.4433
c1 =
V1 =
297
815
m rate =
m rate =
kg/m3
m/s
m/s
1 V 1A 1
0.904
kg/s
For the normal shock:
Using built-in function NormM2fromM (M,k):
M2 =
0.492
Using built-in function Normp0fromM (M,k) at M 1:
0.41
p 02 /p 01 =
p 02 =
For isentropic flow after the shock:
Using built-in function IsenA (M,k):
But:
Hence:
A 2 /A 2* =
A2 =
1.356
A1
A 2* =
18.44
cm2
Using built-in function IsenAMsubfromA (Aratio,k):
A e /A 2* =
2.17
Me =
0.279
Using built-in function Isenp (M,k):
p 02 /p e =
1.06
pe =
269
For:
kPa
Problem 13.99
[Difficulty: 2]
Problem 13.98
[Difficulty: 3]
Given: Oxygen accelerating through a converging-diverging nozzle
Find:
Pressure ratios for critical points, show that a shock forms in the nozzle, pre- and postshock Mach numbers, exit Mach number
Solution:
The given or available data is:
R =
k =
p 0inlet =
pb =
A e/A t =
48.29
1.4
120
50
3
ft-lbf/lbm-°R
psia
psia
Equations and Computations:
Based on the area ratio, the design Mach number is:
Md =
2.637
The pressure ratio for the third critical can be found from the design point Mach number:
p 0inlet/p b,3rd =
21.1422
0.04730
p b,3rd/p 0inlet =
If a normal shock exists in the nozzle, the pressure ratio should be between the
first and second critical points. At the first critical point the exit Mach number is
M 1st =
0.197
Since at first critical the flow is isentropic, the pressure ratio is:
1.0276
p 0inlet/p b,1st =
0.9732
p b,1st/p 0inlet =
At second critical, the flow is isentropic to the exit, followed by a normal shock.
At the design Mach number, the pressure ratio is:
p b,2nd/p b,3rd =
7.949
Therefore, the back pressure ratio at the second critical is:
p b,2nd/p 0inlet =
0.3760
The actual back pressure ratio is
0.4167
p b/p 0inlet =
This pressure ratio is between those for the first and second critical points, so a shock
exists in the nozzle. We need to use an iterative solution to find the exact location of
the shock wave. Specifically, we iterate on the pre-shock Mach number until we match
the exit pressure to the given back pressure:
2.55
M1 =
2.759
A 1/A t =
18.4233
p 0inlet/p 1 =
6.513
psia
p1 =
0.508
M2 =
7.4107
p 2/p 1 =
48.269
psia
p2 =
1.0873
A e/A 2 =
A 2/A 2* =
*
A e/A 2 =
Me =
p 02/p 2 =
p 02/p e =
pe =
(We used Goal Seek in Excel for this solution.)
1.324
1.440
0.454
1.193
1.152
50.000
psia
Problem 13.97
[Difficulty: 3]
Given: Air accelerating through a converging-diverging nozzle
Find:
Pressure ratios needed to operate with isentropic flow throughout, supersonic flow at
exit (third critical); isentropic flow throughout, subsonic flow at exit (first critical point);
and isentropic flow throughout, supersonic flow in the diverging portion, and a normal
shock at the exit (second critical point).
Solution:
The given or available data is:
k =
Md =
1.4
2.5
Equations and Computations:
The pressure ratio for the third critical can be found from the design point Mach number:
p 0inlet/p b,3rd =
17.0859
0.0585
p b,3rd/p 0inlet =
The area ratio for this nozzle is:
A /A * =
2.637
So to operate at first critical the exit Mach number would be:
M 1st =
0.226
Since at first critical the flow is isentropic, the pressure ratio is:
p 0inlet/p b,1st =
1.0363
0.9650
p b,1st/p 0inlet =
At second critical, the flow is isentropic to the exit, followed by a normal shock.
At the design Mach number, the pressure ratio is:
p b,2nd/p b,3rd =
7.125
Therefore, the back pressure ratio at the second critical is:
0.4170
p b,2nd/p 0inlet =
p b,1st/p 0inlet =
p b,2nd/p 0inlet =
p b,3rd/p 0inlet =
0.9650
0.4170
0.0585
Problem 13.96
[Difficulty: 2]
Problem 13.95
[Difficulty: 2]
Problem 13.94
[Difficulty: 2]
Problem 13.93
[Difficulty: 3]
Problem 13.92
[Difficulty: 2]
Problem 13.91
[Difficulty: 4]
Given: Air flowing through a wind tunnel, stagnation and test section conditions known
Find:
Throat area, mass flow rate, static conditions in test section, minumum diffuser area
Solution:
The given or available data is:
R =
k =
p 01 =
T 01 =
T 01 =
53.33
1.4
14.7
75
535
A1 =
M1 =
1
2.3
ft-lbf/lbm-°R
psia
°F
°R
ft2
A schematic of this wind tunnel is shown here:
Equations and Computations:
For the Mach number in the test section, the corresponding area ratio is:
A 1/A 1* =
2.193
So the throat area is:
ft2
0.456
At =
The mass flow rate can be calculated using the choked flow equation:
m=
22.2
lbm/s
The static conditions in the test section are:
p 01/p 1 =
12.5043
2.0580
T 01/T 1 =
1.176
psia
p1 =
°R
260
T1 =
The strongest possible shock that can occur downstream of the first throat is when
the shock wave is in the test section. The post-shock Mach number is then
M2 =
0.5344
The area ratio corresponding to this Mach number is:
A 2/A 2* =
Therefore, the minimum diffuser throat area is
A 2* =
1.2792
0.782
ft2
Problem 13.90
[Difficulty: 3]
Problem 13.89
[Difficulty: 4]
Problem 13.88
[Difficulty: 3]
Problem 13.87
[Difficulty: 3]
Problem 13.86
[Difficulty: 3]
Problem 13.85
Given:
Normal shock
Find:
Rankine-Hugoniot relation
[Difficulty: 4]
Solution:
Basic equations:
2
Momentum:
p 1 ρ1 V1 p 2 ρ2 V2
Energy:
h1
2
1
1
2
2
V1 h 2 V2
2
2
Mass:
ρ1 V1 ρ2 V2
Ideal Gas:
p ρ R T
2
2
From the energy equation
2 h 2 h 1 2 cp T2 T1 V1 V2 V1 V1 V1 V2
From the momentum equation
p 2 p 1 ρ1 V1 ρ2 V2 ρ1 V1 V1 V2
Hence
Using this in Eq 1
2
2
(1)
where we have used the mass equation
p2 p1
V1 V2
ρ1 V1
p2 p1
p2 p1
V2 p 2 p 1
ρ1
1
1
2 c p T2 T1
V1 V2
1
1
p2 p1
ρ1 V1
ρ1
ρ1
V1
ρ2
ρ1 ρ2
where we again used the mass equation
Using the ideal gas equation
p2
2 cp
ρ2 R
1
1
p 2 p 1
ρ1 R
ρ1 ρ2
p1
Dividing by p 1 and multiplying by ρ2, and using R = c p - cv, k = cp/cv
2
Collecting terms
cp
R
p2
p1
p2
p1
2 k
k 1
2 k
p2
p1
For an infinite pressure ratio
ρ2
ρ2 p 2
ρ2
k p2
2
1
1
ρ1
ρ1
k 1 p1
p1
ρ1
ρ2
ρ2
k 1 ρ1
ρ2
ρ1
2 k
1
k1
( k 1) ( k 1)
2 k
ρ2
ρ2
1
ρ1
k 1 ρ1 ρ1
1
ρ2
ρ1
1
ρ2
ρ1
0
( k 1 ) ρ2
1
( k 1 ) ρ1
( k 1)
( k 1)
or
ρ2
or
p2
p1
( k 1)
ρ1
ρ1
( k 1)
( k 1) ( k 1)
ρ1
ρ2
ρ2
k1
k1
ρ2
ρ1
(= 6 for air)
Problem 13.84
[Difficulty: 3]
Given: Stagnation pressure and temperature probes on the nose of the Hyper-X
Find:
Pressure and temperature read by those probes
Solution:
The given or available data is:
R =
k =
M1 =
z=
z=
p SL =
T SL =
53.33
1.4
9.68
110000
33528
14.696
518.76
ft-lbf/lbm-°R
ft
m
psia
°R
Equations and Computations:
At this altitude the local pressure and temperature are:
p 1/p SL =
0.008643
0.12702
psia
p1 =
°R
422.88
T1 =
The stagnation pressure and temperature at these conditions are:
p 01/p 1 = 34178.42
4341.36
psia
p 01 =
19.74
T 01/T 1 =
°R
8347.81
T 01 =
Downstream of the normal shock wave, the Mach number is:
M2 =
0.3882
The total pressure ratio across the normal shock is:
p 02/p 01 = 0.003543
So the pressure read by the probe is:
p 02 =
15.38
psia
Since stagnation temperature is constant across the shock, the probe reads:
°R
T 02 =
8348
Problem 13.83
[Difficulty: 2]
Problem 13.82
[Difficulty: 2]
Problem 13.81
[Difficulty: 2]
Problem 13.80
[Difficulty: 2]
Given: Normal shock
Find:
Speed; Change in pressure; Compare to shockless deceleration
Solution:
The given or available data is:
R =
k =
53.33
1.4
T1 =
p1 =
V1 =
452.5
14.7
1750
c1 =
M1 =
1043
2.46
ft·lbf/lbm·R
o
0.0685
Btu/lbm·R
R
psi
mph
2567
ft/s
p2 =
101
psi
p2 – p1 =
86.7
psi
781
ft/s
p2 =
197
psi
p2 – p1 =
182
psi
Equations and Computations:
From
Then
c1 kRT1
Using built-in function NormM2fromM (M,k):
M2 =
0.517
Using built-in function NormdfromM (M,k):
2 / 1 =
3.29
Using built-in function NormpfromM (M,k):
p 2 /p 1 =
6.90
V2
1
V
2 1
V2 =
532
Using built-in function Isenp (M,k) at M 1:
p 01 /p 1 =
16.1
Using built-in function Isenp (M,k) at M 2:
p 02 /p 2 =
1.20
Then
From above ratios and p 1, for isentropic flow (p 0 = const):
ft/s
mph
Problem 13.79
[Difficulty: 2]
Given: Normal shock
Find:
Speed and Mach number after shock; Change in stagnation pressure
Solution:
The given or available data is:
R =
k =
53.33
1.4
T1 =
p1 =
V1 =
445
5
2000
c1 =
M1 =
1034
2.84
ft·lbf/lbm·R
o
0.0685
Btu/lbm·R
R
psi
mph
2933
ft/s
793
ft/s
Equations and Computations:
From
Then
c1 kRT1
Using built-in function NormM2fromM (M,k):
M2 =
0.486
Using built-in function NormdfromM (M,k):
2 / 1 =
3.70
Using built-in function Normp0fromM (M,k):
p 02 /p 01 =
0.378
V2
1
V
2 1
V2 =
541
Using built-in function Isenp (M,k) at M 1:
p 01 /p 1 =
28.7
Then
ft/s
mph
From the above ratios and given p 1:
p 01 =
p 02 =
p 01 – p 02 =
143
54.2
89.2
psi
psi
psi
Problem 13.78
[Difficulty: 2]
Given: Normal shock
Find:
Pressure after shock; Compare to isentropic deceleration
Solution:
R =
k =
T 01 =
p 01 =
M1 =
286.9
1.4
550
650
2.5
Using built-in function Isenp (M,k):
p 01 /p 1 =
17.09
Using built-in function NormM2fromM (M,k):
M2 =
0.513
Using built-in function NormpfromM (M,k):
p 2 /p 1 =
7.13
Using built-in function Isenp (M,k) at M 2:
p 02 /p 2 =
1.20
The given or available data is:
J/kg·K
K
kPa
Equations and Computations:
But for the isentropic case:
Hence for isentropic deceleration:
p1 =
38
kPa
p2 =
271
kPa
p2 =
543
kPa
p 02 = p 01
Problem 13.77
[Difficulty: 2]
Problem 13.76
[Difficulty: 2]
Given: Normal shock
Find:
Speed and temperature after shock; Entropy change
Solution:
R =
k =
cp =
The given or available data is:
53.33
1.4
0.2399
T 01 =
p1 =
M1 =
1250
20
2.5
1 =
0.0432
V1 =
4334
T 01 /T 1 =
2.25
ft·lbf/lbm·R
0.0685
Btu/lbm·R
Btu/lbm·R
o
R
psi
Equations and Computations:
From
p1 1 RT1
slug/ft3
ft/s
Using built-in function IsenT (M,k):
T1 =
o
R
o
F
o
R
728
o
F
143
psi
556
96
Using built-in function NormM2fromM (M,k):
M2 =
0.513
Using built-in function NormTfromM (M,k):
T 2 /T 1 =
Using built-in function NormpfromM (M,k):
p 2 /p 1 =
From
V 2 M 2 kRT 2
From
T
s c p ln 2
T1
2.14
T2 =
7.13
V2 =
867
s =
0.0476
37.1
p2 =
ft/s
p
R ln 2
p1
Btu/lbm·R
ft·lbf/lbm·R
1188
Problem 13.75
[Difficulty: 3]
Given: Air accelerating through a converging-diverging nozzle, passes through a normal shock
Find:
Mach number before and after shock; entropy generation
Solution:
The given or available data is:
R =
k =
p 01 =
T 01 =
T 01 =
53.33
1.4
150
400
860
ft-lbf/lbm-°R
psia
°F
°R
At =
3
in2
A1 = A2 =
6
in2
Equations and Computations:
The isentropic area ratio at the station of interest is:
A 1/A 1* =
2.00
So the Mach number at 1 is:
2.20
M1 =
Downstream of the normal shock wave, the Mach number is:
0.547
M2 =
The total pressure ratio across the normal shock is:
0.6294
p 02/p 01 =
Since stagnation temperature does not change across a normal shock,
the increase in entropy is related to the stagnation pressure loss only:
ft-lbf/lbm-°R
Δs 1-2 =
24.7
Btu/lbm-°R
0.0317
Δs 1-2 =
Problem 13.74
[Difficulty: 3]
Given: Air approaching a normal shock
Find:
Pressure and velocity after the shock; pressure and velocity if flow were
decelerated isentropically
Solution:
The given or available data is:
R =
k =
V1 =
p1 =
T1 =
286.9
1.4
900
50
220
m/s
kPa
K
c1 =
297.26
m/s
J/kg-K
Equations and Computations:
The sonic velocity at station 1 is:
So the Mach number at 1 is:
3.028
M1 =
Downstream of the normal shock wave, the Mach number is:
0.4736
M2 =
The static pressure and temperature ratios are:
10.528
p 2/p 1 =
2.712
T 2/T 1 =
So the exit temperature and pressure are:
526
kPa
p2 =
596.6
K
T2 =
At station 2 the sound speed is:
c2 =
489.51
m/s
Therefore the flow velocity is:
232
m/s
V2 =
If we decelerate the flow isentropically to
M 2s =
0.4736
The isentropic pressure ratios at station 1 and 2s are:
38.285
p 0/p 1 =
1.166
p 0/p 2s =
32.834
p 2s/p 1 =
So the final pressure is:
1642
p 2s =
The temperature ratios are:
2.833
T 0/T 1 =
1.045
T 0/T 2s =
2.712
T 2s/T 1 =
So the final temperature is:
596.6
T 2s =
The sonic velocity at station 2s is:
489.51
c 2s =
Therefore the flow velocity is:
232
V 2s =
kPa
K
m/s
m/s
Problem 13.73
[Difficulty: 3]
Given: Pitot probe used in supersonic wind tunnel nozzle
Find:
Pressure measured by pitot probe; nozzle exit velocity
Solution:
The given or available data is:
R =
k =
M1 =
p1 =
T0 =
286.9
1.4
5
10
1450
J/kg-K
kPa
K
Equations and Computations:
Downstream of the normal shock wave, the Mach number is:
M2 =
0.4152
The static and stagnation pressure ratios are:
p 2/p 1 =
29.000
0.06172
p 02/p 01 =
So the static pressure after the shock is:
p2 =
290
kPa
The pitot pressure, however, is the stagnation pressure:
p 02/p 2 =
1.12598
327
kPa
p 02 =
The static temperature at the nozzle exit can be calculated:
T 01/T 1 =
6.000
241.67
K
T1 =
At the nozzle exit the sound speed is:
311.56
m/s
c2 =
Therefore the flow velocity at the nozzle exit is:
1558
m/s
V2 =
Problem 13.72
[Difficulty: 2]
Problem 13.71
[Difficulty: 2]
Problem 13.70
[Difficulty: 3]
Given:
C-D nozzle with normal shock
Find:
Mach numbers at the shock and at exit; Stagnation and static pressures before and after the shock
k 1
Solution:
1 k 1 M2
A
1
2
Basic equations: Isentropic flow
k 1
Acrit
M
2
2
M2
Normal shock
Given or available data
p2
k1
2 k M 2 1
k 1 1
k 1.4
Rair 53.33
2
At 1.5 in
k
p0
p1
k1
1
p
2
M
2
k 1
k
k 1 M 2
1
2
k1
2
1
M1
2
2
2
M1
2 ( k 1)
2 k
k1
2
M1
k1
p 02
k1
p 01
1
2 k M 2 k
1
k
k 1
ft lbf
lbm R
2
As 2.5 in
k 1
p 01 125 psi
T0 ( 175 460 ) R
(Shock area)
Ae 3.5 in
1
k 1
1
2
Because we have a normal shock the CD must be accelerating the flow to supersonic so the throat is at critical state.
Acrit At
At the shock we have
As
Acrit
k 1
1.667
At this area ratio we can find the Mach number before the shock from the
isentropic relation
1 k 1 M 2
1
1
2
Acrit
k 1
M1
2
2 ( k 1)
As
Solving iteratively (or using Excel's Solver, or even better the function isenMsupfromA from the Web site!)
M 1 1.985
The stagnation pressure before the shock was given:
p 01 125 psi
The static pressure is then
p1
p 01
k
k 1
1 k 1 M 2
1
2
p 1 16.4 psi
2
After the shock we have
M2
M1
2
k1
M 2 0.580
2 k
2
k 1 M1 1
k
Also
k 1 M 2
1
2
k1
2
1
M1
2
p 02 p 01
2 k M 2 k
1
k
k 1
and
k 1
1
p 02 91.0 psi
1
k 1
1
2 k
k 1
2
p 2 p 1
M1
k 1
k 1
p 2 72.4 psi
Finally, for the Mach number at the exit, we could find the critical area change across the shock; instead
we find the new critical area from isentropic conditions at state 2.
1 k 1 M 2
2
2
Acrit2 As M 2
k 1
2
At the exit we have
Ae
Acrit2
k 1
2 ( k 1)
2
Acrit2 2.06 in
1.698
At this area ratio we can find the Mach number before the shock from the
isentropic relation
k 1
2 ( k 1)
1 k 1 M 2
e
Ae
1
2
Acrit2
k 1
Me
2
Solving iteratively (or using Excel's Solver, or even better the function isenMsubfromA from the Web site!)
These calculations are obviously a LOT easier using the Excel functions available on the Web site!
M e 0.369
Problem 13.69
[Difficulty: 2]
Given:
Normal shock near pitot tube
Find:
Air speed
Solution:
Basic equations:
k
p 1 p 2 ρ1 V1 V2 V1
Given or available data T1 285 R
p
p 1 1.75 psi
k 1.4
Rair 53.33
k 1
k
2 p 02
1
k1
p 2
At state 2
M2
From momentum
p 1 p 2 ρ2 V2 ρ1 V1
2
2
M1
Also
c1
Then
2
p2
2
1 k M 2 1
k p1
1
k Rair T1
k1
2
M
2
p 02 10 psi
k 1
p 2 8 psi
ft lbf
lbm R
2
2
2
but
ρ V ρ c M
or
p1 1 k M1
p
2
R T
k R T M k p M
2
2
2
p2 1 k M2
M 1 2.01
c1 827
ft
s
ft
V1 1666
s
V1 M 1 c1
ft
V1 1822
s
Note: With p1 = 1.5 psi we obtain
(Using normal shock functions, for
1
M 2 0.574
2
p1 p2 k p2 M2 k p1 M1
Hence
p0
(Momentum)
p2
p1
4.571 we find
M 1 2.02
M 2 0.573 Check!)
Problem 13.68
[Difficulty: 3]
Given: Air flowing into converging duct, normal shock standing at duct exit
Find:
Mach number at duct entrance, duct area ratio
Solution:
The given or available data is:
R =
cp =
k =
M3 =
p 2/p 1 =
286.9
1004
1.4
0.54
2
J/kg-K
J/kg-K
Equations and Computations:
For the given post-shock Mach number, there can be only one Mach number
upstream of the shock wave:
M2 =
2.254
0.5400
M3 =
(We used Solver to match the post-shock Mach number by varying M 2.)
The stagnation pressure is constant in the duct:
11.643
p 0/p 2 =
23.285
p 0/p 1 =
So the duct entrance Mach number is:
2.70
M1 =
The isentropic area ratios at stations 1 and 2 are:
A 1/A * =
*
3.1832
A 2/A =
2.1047
A 1/A 2 =
1.512
So the duct area ratio is:
Problem 13.67
[Difficulty: 2]
Given: Standing normal shock
Find:
Pressure and temperature ratios; entropy increase
Solution:
R =
cp =
k =
M1 =
286.9
1004
1.4
1.75
p 2/p 1 =
3.41
T 2/T 1 =
The entropy increase across the shock is:
Δs =
1.495
The given or available data is:
J/kg-K
J/kg-K
Equations and Computations:
The pressure ratio is:
The tempeature ratio is:
51.8
J/kg-K
Problem 13.66
Given:
Normal shock due to explosion
Find:
Shock speed; temperature and speed after shock
[Difficulty: 3]
V
Shock speed Vs
Shift coordinates: (Vs – V)
(Vs)
Solution:
Basic equations:
2
M2
p2
p1
k1
V M c M k R T
2 k
2
k 1 M1 1
2 k
2
k1
Given or available data
k 1.4
From the pressure ratio
M1
Then we have
Shock at rest
2
2
M1
M1
1 k 1 M 2 k M 2 k 1
1
1
2
2
T1
2
k 1 M 2
1
2
T2
k1
k1
R 286.9
J
kg K
k 1 p2 k
2 k p1 k
1
1
1 k 1 M 2 k M 2 k
1
1
2
2
T2 T1
2
k 1 M 2
1
2
2
M2
M1
p 2 30 MPa
T2 14790 K
T2 14517 °C
1
2
k1
M 2 0.382
2 k M 2 1
k 1 1
V1 M 1 k R T1
m
V1 5475
s
After the shock (V2) the speed is
V2 M 2 k R T2
m
V2 930
s
V Vs V2
V 4545
V2 Vs V
T1 ( 20 273 ) K
M 1 16.0
Then the speed of the shock (Vs = V1) is
But we have
p 1 101 kPa
Vs V1
m
s
These results are unrealistic because at the very high post-shock temperatures experienced, the specific heat
ratio will NOT be constant! The extremely high initial air velocity and temperature will rapidly decrease as the
shock wave expands in a spherical manner and thus weakens.
m
Vs 5475
s
Problem 13.65
[Difficulty: 3]
Given: CO2 cartridge and convergent nozzle
Find:
Tank pressure to develop thrust of 15 N
Solution:
The given or available data is:
R =
k =
T0 =
pb =
Dt =
188.9
1.29
293
101
0.5
J/kg·K
At =
0.196
mm2
K
kPa
mm
Equations and Computations:
The momentum equation gives
R x = m flowV e
Hence, we need m flow and V e
For isentropic flow
pe =
pe =
pb
101
kPa
If we knew p 0 we could use it and p e, and Eq. 13.7a, to find M e.
Once M e is known, the other exit conditions can be found.
Make a guess for p 0, and eventually use Goal Seek (see below).
p0 =
44.6
MPa
From p 0 and p e, and Eq. 13.7a
(using built-in function IsenMfromp (M ,k )
(13.7a)
Me =
4.5
From M e and T 0 and Eq. 13.7b
(using built-in function IsenT (M ,k )
(13.7b)
Te =
74.5
From T e and Eq. 12.18
K
(12.18)
Then
ce =
134.8
m/s
Ve =
606
m/s
The mass flow rate is obtained from p 0, T 0, A t, and Eq. 13.10a
(13.10a)
m choked =
0.0248
kg/s
Finally, the momentum equation gives
R x = m flowV e
=
15.0
We need to set R x to 15 N. To do this use Goal Seek
to vary p 0 to obtain the result!
N
Problem 13.64
[Difficulty: 4]
Given: Rocket motor with converging-only nozzle
Find:
Nozzle exit pressure and thrust
Solution:
The given or available data is:
R =
k =
p0 =
T0 =
70.6
1.25
175
5400
ft-lbf/lbm-°R
At =
pb =
1
14.7
in2
psia
°R
psia
Equations and Computations:
If the diverging portion of the nozzle is removed, the exit Mach number is 1:
The exit Mach number can be calculated based on the pressure ratio:
Me =
1.0000
The isentropic area ratio at this Mach number is:
A e/A * =
1.0000
So the nozzle exit area is:
At =
1.00
in2
The exit temperature and pressure can be found from the Mach number:
Te =
4800.0
°R
97.1
psia
pe =
The sound speed at the exit is:
ce =
3693.2
ft/s
And so the exit flow speed is:
3693.2
ft/s
Ve =
The nozzle is choked, and we can use the choked flow equation:
From p 0, T 0, A t, and Eq. 13.9a
(13.9a)
m =
1.058
lbm/s
Based on the momentum equation, we can calculate the thrust generated:
F=
204
lbf
Problem 13.63
[Difficulty: 3]
Given: Rocket motor
Find:
Nozzle exit area, velocity, and thrust generated
Solution:
The given or available data is:
R =
k =
p0 =
T0 =
70.6
1.25
175
5400
ft-lbf/lbm-°R
At =
pe =
1
14.7
in2
psia
°R
psia
Equations and Computations:
The exit Mach number can be calculated based on the pressure ratio:
Me =
2.2647
The isentropic area ratio at this Mach number is:
A e/A * =
2.4151
So the nozzle exit area is:
Ae =
2.42
in2
The exit temperature can be found from the Mach number:
Te =
3290.4
°R
The sound speed at the exit is:
ce =
3057.8
ft/s
And so the exit flow speed is:
6925.2
ft/s
Ve =
The density can be calculated using the ideal gas equation of state:
3
ρ e = 0.009112 lbm/ft
The nozzle is choked, and we can use the choked flow equation:
From p 0, T 0, A t, and Eq. 13.10a
(13.10a)
m =
1.058
lbm/s
Based on the momentum equation, we can calculate the thrust generated:
Rx =
228
lbf
Note that since the flow expanded perfectly (the nozzle exit pressure is equal
to the ambient pressure), the pressure terms drop out of the thrust
calculation.
Problem 13.62
[Difficulty: 4]
Given:
Compressed CO 2 in a cartridge expanding through a nozzle
Find:
Throat pressure; Mass flow rate; Thrust; Thrust increase with diverging section; Exit area
Solution:
Basic equations:
Assumptions: 1) Isentropic flow 2) Stagnation in cartridge 3) Ideal gas 4) Uniform flow
Given or available data:
J
k 1.29
R 188.9
p 0 35 MPa
T0 ( 20 273 ) K
p0
From isentropic relations p crit
k
1 k
2
Since p b << pcrit, then
p t p crit
Throat is critical so
mrate ρt Vt At
Tt
Vt
At
k1
d t 0.5 mm
p crit 19.2 MPa
k 1
Tt 256 K
2
k R Tt
π d t
p atm 101 kPa
p t 19.2 MPa
T0
1
1
kg K
m
Vt 250
s
2
4
At 1.963 10
7
pt
ρt
R Tt
ρt 396
mrate ρt Vt At
kg
mrate 0.0194
s
kg
3
m
2
m
Rx p tgage At mrate Vt
For 1D flow with no body force the momentum equation reduces to
Rx mrate Vt p tgage At
p tgage p t p atm
Rx 8.60 N
When a diverging section is added the nozzle can exit to atmospheric pressure
p e p atm
1
k 1
k
2 p 0
Me
1
k 1 p e
Hence the Mach number at exit is
Te
ce
T0
1
k1
2
Me
2
k R Te
2
M e 4.334
Te 78.7 K
ce 138
m
s
m
Ve 600
s
Ve M e ce
The mass flow rate is unchanged (choked flow)
Rx mrate Ve
From the momentum equation
The percentage increase in thrust is
11.67 N 8.60 N
8.60 N
mrate ρe Ve Ae
The exit area is obtained from
mrate
Ae
ρe Ve
T
35.7 %
and
pe
ρe
R Te
ρe 6.79
6
T0
pt
Tt
Conv.
Nozzle
CD
Nozzle
Te
s
kg
3
m
Ae 4.77 10
p0
pb
Rx 11.67 N
2
m
2
Ae 4.77 mm
Problem 13.61
[Difficulty: 3]
Problem 13.60
[Difficulty: 3]
Problem 13.59
[Difficulty: 3]
Problem 13.58
Given:
Rocket motor on test stand
Find:
Mass flow rate; thrust force
[Difficulty: 3]
k
Solution:
Basic equations:
T0
T
1
k1
2
M
2
patm pe Ae Rx mrate Ve
Given or available data p e 75 kPa
k1
2
p 0 4 MPa
so the nozzle exit area is
T0
1 k 1 M 2
e
2
Then
mrate ρe Ae Ve
kg
mrate 19.3
s
p ρ R T
c
k R T
mrate ρ A V
T0 3250 K
k 1.25
π 2
Ae d
4
Ae 491 cm
k R Te
pe patm Ae MCV ax Ve mrate
Rx p e p atm Ae Ve mrate
R 300
J
kg K
2
M e 3.12
and
m
Ve 2313
s
k 1
ce
Ve M e ce
Hence
2
Te 1467 K
The exit speed is
The momentum equation (Eq. 4.33) simplifies to
M
Momentum for pressure pe and velocity Ve at exit; Rx is the reaction force
k 1
k
2 p 0
1
k1
p e
Me
The exit temperature is Te
p
1
p atm 101 kPa
d 25 cm
From the pressures
p0
Rx 43.5 kN
ce 742
pe
ρe
R Te
m
s
kg
ρe 0.170
3
m
Problem 13.57
[Difficulty: 3] Part 1/2
Problem 13.57
[Difficulty: 3] Part 2/2
Problem 13.56
[Difficulty: 2]
Problem 13.55
[Difficulty: 2]
Problem 13.54
[Difficulty: 3]
Given: Methane discharging from one tank to another via a converging nozzle
Find:
Mass flow rate at two different back pressures
Solution:
The given or available data is:
R =
k =
p0 =
T0 =
T0 =
96.32
1.31
75
80
540
Ae =
1
ft-lbf/lbm-°R
psia
°F
°R
in2
Equations and Computations:
If the nozzle were choked, the exit Mach number is 1 and the pressure would be:
p* =
40.79
psia
Therefore, in part a, when
pe =
15
psia
The nozzle is choked, and we can use the choked flow equation:
From p 0, T 0, A t, and Eq. 13.9a
(13.9a)
m =
1.249
lbm/s
In part b, when
pe =
60
psia
The nozzle is not choked. The exit Mach number is:
Me =
0.5915
The exit temperature can be found from the Mach number:
Te =
512.2
°R
The sound speed at the exit is:
ce =
1442.6
ft/s
And so the exit flow speed is:
853.3
ft/s
Ve =
The density can be calculated using the ideal gas equation of state:
lbm/ft3
ρe =
0.1751
The mass flow rate can then be calculated directly from continuity:
m=
1.038
lbm/s
Problem 13.53
[Difficulty: 2]
Problem 13.52
[Difficulty: 2]
Problem 13.51
[Difficulty: 3]
Given: Air flow through a converging-diverging nozzle equipped with pitot-static probe
Find:
Nozzle velocity and mass flow rate
Solution:
The given or available data is:
R =
k =
p1 =
p 01 =
T1 =
T1 =
286.9
1.4
75
100
20
293
A1 =
10
in2
A1 =
0.006452
m2
J/kg-K
kPa
kPa
°C
K
Equations and Computations:
At station 1 the local sound speed is:
c1 =
343.05
m/s
Based on the static and pitot pressures, the Mach number is:
0.6545
M1 =
Therefore the velocity is:
225
m/s
V1 =
The local density can be calculated using the ideal gas equation of state:
kg/m3
ρ1 =
0.8922
So the mass flow rate is:
m =
1.292
kg/s
Problem 13.50
[Difficulty: 3]
Given: Wind tunnel test section with blockage
Find:
Maximum blockage that can be tolerated; air speed given a fixed blockage
Solution:
The given or available data is:
R =
k =
M1 =
T1 =
T1 =
53.33
1.4
1.2
70
530
ft-lbf/lbm-°R
°F
°R
At =
1
ft
2
Equations and Computations:
*
The test section will choke if the blockage decreases the area to A . In the test section:
*
A 1/A =
1.0304
So the minimum area would be
*
ft2
A =
0.9705
And the blockage would be the difference between this and the test section area:
2
ft
A1 - A*=
0.0295
A1 - A*=
4.25
in
2
A1 - A =
3.0000
in
2
A actual =
The resulting isentropic area ratio is:
0.9792
ft
If we have a blockage of:
Then the actual area would be:
*
A actual/A =
2
1.0090
and the actual Mach number is:
1.1066
M actual =
(remember that since we're already supersonic, we should use the supersonic solution)
The stagnation temperature for the wind tunnel is (based on test section conditions)
682.64
°R
T0 =
So the actual static temperature in the tunnel is:
548.35
°R
T actual =
The sound speed would then be:
1148.17
ft/s
c actual =
And so the speed in the test section is:
V actual =
1270.5
ft/s
Problem 13.49
[Difficulty: 2]
Given: Design condition in a converging-diverging nozzle
Find:
Tank pressure; flow rate; throat area
Solution:
The given or available data is:
R =
k =
53.33
1.4
T0 =
560
Ae =
pb =
Me =
1
14.7
2
pe =
pb
pe =
14.7
o
ft.lbf/lbm. R
o
R
2
in
psia
Equations and Computations:
At design condition
psia
From M e and p e, and Eq. 13.7a
(using built-in function Isenp (M ,k )
(13.7a)
p0 =
115
psia
From M e and A e, and Eq. 13.7d
(using built-in function IsenA (M ,k )
(13.7d)
Hence
A* =
0.593
in2
At =
0.593
in
2
From p 0, T 0, A t, and Eq. 13.10a
(13.10a)
m choked =
1.53
lb/s
Problem 13.48
[Difficulty: 4] Part 1/2
Problem 13.48
[Difficulty: 4] Part 2/2
Problem 13.47
[Difficulty: 4]
Problem 13.46
Given:
CD nozzle attached to large tank
Find:
Flow rate
[Difficulty: 2]
k
Solution:
Basic equations:
T0
T
Given or available data
1
2
M
2
p0
p
1
k1
2
M
p 0 150 kPa
T0 ( 35 273 ) K
k 1.4
R 286.9
For isentropic flow
Me
Then
Te
Also
ce
Finally
k1
J
kg K
k 1
k
2 p 0
1
k1
p e
1 k 1 M 2
e
2
ce 332
k 1
mrate ρ V A
p e 101 kPa
D 2.75 cm
π 2
Ae D
4
Ae 5.94 cm
M e 0.773
T0
k R Te
2
m
s
pe
ρe
R Te
ρe 1.28
mrate ρe Ve Ae
kg
mrate 0.195
s
kg
3
m
Te 275 K
Te 1.94 °C
Ve M e ce
m
Ve 257
s
2
Problem 13.45
[Difficulty: 3]
Given: Air flow through a converging-diverging nozzle
Find:
Nozzle mass flow rate
Solution:
The given or available data is:
R =
k =
V1 =
p1 =
T1 =
T1 =
53.33
1.4
50
15
70
530
ft-lbf/lbm-°R
At =
1
ft2
c1 =
1128.80
ft/s
M1 =
0.0443
ft/s
psia
°F
°R
Equations and Computations:
At station 1 the local sound speed is:
So the upstream Mach number is:
So now we can calculate the stagnation temperature and pressure:
p0 =
T0 =
15.021
530.21
psia
°R
To find the mass flow rate, we will use the choked flow equation:
From p 0, T 0, A t, and Eq. 13.10a
(13.10a)
m =
50.0
lbm/s
Problem 13.44
[Difficulty: 4] Part 1/3
Problem 13.44
[Difficulty: 4] Part 2/3
Problem 13.44
[Difficulty: 4] Part 3/3
Problem 13.43
Given:
Ideal gas flow in a converging nozzle
Find:
Exit area and speed
[Difficulty: 4]
k 1
k
Solution:
T0
Basic
equations:
T
k1
1
2
p 1 35 psi
Given or available data
M
p0
2
p
ρ1 0.1
lbm
ft
c1
Check for choking:
Hence
M1
V1
k1
M2
2
2
M1
A1 1 ft
c1
2
k
p 2 25 psi
p1
c1 1424
ρ1
k 1.25
ft
s
p 0 37.8 psi
p crit 21.0 psi
k
1
k 1
k 1
k
2 p 0
1
k1
p2
k 1
2 ( k 1)
1 k 1 M 2
1
2
k 1
2
k
p ρ const
ρ A V const
Hence p2 > pcrit, so NOT choked
k 1
M 1 A1
Acrit
Finally from continuity
ft
V1 500
s
3
p0
k
2
For isentropic flow
2
k
The critical pressure is then p crit
From M1 we find
M
k R T1 or, replacing R using the ideal gas equation
p 0 p 1 1
Then we have
2
1 k 1 M2
A
1
2
k 1
Acrit
M
2
k 1
M 1 0.351
c1
Then
1
k1
2 ( k 1)
so
so
M 2 0.830
Acrit 0.557 ft
2
k 1
1 k 1 M 2
2
2
A2
M2
k 1
2
Acrit
1
p1
ρ2 ρ1
p2
k
A1 ρ1
V2 V1
A2 ρ2
ρ2 0.131
lbm
ft
ft
V2 667
s
3
2 ( k 1)
A2 0.573 ft
2
Problem 13.42
Given:
Spherical air tank
Find:
Air temperature after 30s; estimate throat area
[Difficulty: 4]
Solution:
Basic equations:
T0
T
1
k1
2
M
2
p
k
ρ dVCV
t
const
ρ
ρ V dACS 0
(4.12)
Assumptions: 1) Large tank (stagnation conditions) 2) isentropic 3) uniform flow
Given or available data
p atm 101 kPa
p 1 2.75 MPa
T1 450 K
D 2 m
V
ΔM 30 kg
Δt 30 s
k 1.4
R 286.9
J
p b p atm
The flow will be choked if p b/p1 < 0.528:
so
pb
p1
0.037
π
6
3
D
3
V 4.19 m
kg K
(Initially choked: Critical conditions)
We need to see if the flow is still choked after 30s
The initial (State 1) density and mass are
The final (State 2) mass and density are then
For an isentropic process
p
k
const
ρ
The final temperature is
T2
To estimate the throat area we use
p2
ρ2 R
so
p1
ρ1
R T1
ρ1 21.3
M 2 M 1 ΔM
M 2 59.2 kg
kg
3
M 1 ρ1 V
M 1 89.2 kg
M2
ρ2
V
ρ2 14.1
m
ρ2
p2 p1
ρ1
k
T2 382 K
p 2 1.55 MPa
pb
p2
0.0652
mtave ρtave At Vtave
Δt
or
The average stagnation temperature is
The average stagnation pressure is
T0ave
p 0ave
(Still choked)
ΔM
At
Δt ρtave Vtave
where we use average values of density and speed at the throat.
T1 T2
2
p1 p2
2
T0ave 416 K
p 0ave 2.15 MPa
3
m
T2 109 °C
ΔM
kg
Hence the average temperature and pressure (critical) at the throat are
Ttave
Hence
Finally
T0ave
1 k 1
2
Vtave
k R Ttave
ΔM
At
Δt ρtave Vtave
Ttave 347 K
and
p 0ave
p tave
k
1 k
2
p tave
ρtave
R Ttave
m
Vtave 373
s
4
At 2.35 10
2
m
2
At 235 mm
This corresponds to a diameter
Dt
4 At
π
Dt 0.0173 m
Dt 17.3 mm
The process is isentropic, followed by nonisentropic expansion to atmospheric pressure
1
p tave 1.14 MPa
k 1
ρtave 11.4
kg
3
m
Problem 13.41
[Difficulty: 3]
Problem 13.40
Given:
Gas cylinder with broken valve
Find:
Mass flow rate; acceleration of cylinder
[Difficulty: 3]
k
Solution:
T0
Basic
equations:
T
1
k1
2
M
2
p0
p
1
k1
2
M
2
k 1
p ρ R T
c
k R T
mrate ρ A V
(4.33)
Given or available data p atm 101 kPa
k 1.66
R 2077
p 0 20 MPa p atm 20.101 MPa
J
kg K
The exit temperature is Te
p b p atm
T0
Ve ce
The exit pressure is
pe
p0
k
1
3
5.025 10
M CV 65 kg
(Choked: Critical conditions)
Te 52.8 °C
ce
p e 9.8 MPa
and exit density is
pe
ρe
R Te
k 1
mrate ρe Ae Ve
ax
p0
The momentum equation (Eq. 4.33) simplifies to
Hence
pb
2
Ae 78.5 mm
k R Te
m
Ve 872
s
1 k
2
Then
so
Te 220 K
1 k 1
2
The exit speed is
π 2
Ae d
4
d 10 mm so the nozzle area is
The flow will be choked if p b/p0 < 0.528:
T0 ( 20 273 ) K
kg
mrate 1.468
s
pe patm Ae MCV ax Ve mrate
pe patm Ae Ve mrate
M CV
ax 31.4
m
2
s
The process is isentropic, followed by nonisentropic expansion to atmospheric pressure
ρe 21
kg
3
m
Problem 13.39
[Difficulty: 1]
Given: Hydrogen flow through a converging-diverging nozzle
Find:
Nozzle exit Mach number
Solution:
The given or available data is:
R =
k =
p0 =
T0 =
T0 =
pe =
766.5
1.41
100
540
1000
20
ft-lbf/lbm-°R
psia
°F
°R
psia
Equations and Computations:
Using the stagnation to exit static pressure ratio, we can find the exit Mach number:
(using built-in function Isenp (M ,k ))
Me =
1.706
Problem 13.38
[Difficulty: 3]
Given: Air flow through a converging-diverging nozzle
Find:
Nozzle exit area and mass flow rate
Solution:
The given or available data is:
R =
k =
p0 =
T0 =
pe =
286.9
1.4
2
313
200
MPa
K
kPa
At =
20
cm2
J/kg-K
Equations and Computations:
Using the stagnation to exit static pressure ratio, we can find the exit Mach number:
(using built-in function Isenp (M ,k ))
Me =
2.1572
A e/A * =
1.9307
From M e, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
At the throat the flow is sonic, so At = A*. Therefore:
Ae =
38.6
cm2
To find the mass flow rate at the exit, we will use the choked flow equation:
From p 0, T 0, A t, and Eq. 13.9a
(13.9a)
m =
17.646
kg/s
Problem 13.37
[Difficulty: 3]
Given: Air-driven rocket in space
Find:
Tank pressure; pressure, temperature and speed at exit; initial acceleration
Solution:
R =
k =
T0 =
286.9
1.4
398
K
At =
M =
m rate =
25
25
0.05
mm2
kg
kg/s
Because p b = 0
Hence the flow is choked!
pe =
p*
Hence
Te =
T*
The given or available data is:
J/kg.K
Equations and Computations:
From T 0, and Eq. 12.22b
(12.22b)
Also
Hence
T* =
332
Te =
332
K
58.7
o
Me =
Ve =
C
1
V* =
From T e and Eq. 12.18
Then
K
ce
(12.18)
ce =
365
m/s
Ve =
365
m/s
To find the exit pressure we use the ideal gas equation
after first finding the exit density.
The mass flow rate is m rate = eA eV e
Hence
e =
0.0548
pe =
5.21
kg/m3
From the ideal gas equation p e = eRT e
kPa
From p e = p * and Eq. 12.22a
(12.22a)
p0 =
9.87
kPa
We can check our results:
From p 0, T 0, A t, and Eq. 13.9a
(13.9a)
m choked =
m choked =
Then
0.050
m rate
kg/s
Correct!
The initial acceleration is given by:
(4.33)
which simplifies to:
pe At Max mrateV
ax =
or:
1.25
ax
m/s2
m rate V p e At
M
Problem 13.36
[Difficulty: 3]
Problem 13.35
[Difficulty: 3]
Problem 13.34
[Difficulty: 3]
Problem 13.33
Given:
Spherical cavity with valve
Find:
Time to reach desired pressure; Entropy change
[Difficulty: 3]
k
Solution:
Basic equations:
T0
T
1
k1
2
M
p0
2
p
k1
1
2
M
2
k 1
T2
p2
Δs cp ln R ln
T1
p1
k 1
Given or available data
Then the inlet area is
p ρ R T
c
k R T
mrate ρ A V
p 0 101 kPa
Tatm ( 20 273 ) K
p f 45 kPa
Tf Tatm
π 2
At d
4
At 0.785 mm
T0 Tatm
k 1.4
2
k 2
mchoked At p 0
R T0 k 1
R 286.9
pf
ρf
R Tf
ρf 0.535
kg
m
Since the mass flow rate is constant (flow is always choked)
k 1
k 2
We have choked flow so mrate At p 0
R T0 k 1
Δt
Hence
M
cp 1004
kg K
π
3
pb
so
p0
3
M mrate Δt
mrate 1.873 10
kg K
V 0.131 m
0.446
(Choked)
M 0.0701 kg
Δt
or
J
3
D
and final mass is M ρf V
2 ( k 1)
Δt 374 s
mrate
3
D 50 cm
J
and tank volume is V
The flow will be choked if p b/p0 < 0.528; the MAXIMUM back pressure is p b p f
The final density is
d 1 mm
2 ( k 1)
M
mrate
4 kg
s
Δt 6.23 min
The air in the tank will be cold when the valve is closed. Because ρ =M/V is constant, p = ρRT = const x T, so as the
temperature rises to ambient, the pressure will rise too.
T2
p2
For the entropy change during the charging process is given by Δs cp ln
R
ln
where
T1
p1
and
p1 p0
p2 pf
Hence
T2
p2
Δs cp ln
R ln
T1
p1
T1 Tatm T2 Tatm
Δs 232
J
kg K
Problem 13.32
[Difficulty: 2]
Given:
Isentropic air flow into a tank
Find:
Initial mass flow rate; Ts process; explain nonlinear mass flow rate
Solution:
Basic equations:
Given or available data
Then
k
T0
T
k1
1
2
M
p0
2
p
p 0 101 kPa
p b p 0 10 kPa
k 1.4
R 286.9
A
π
4
2
D
J
kg K
Avena 65 % A
pb
The flow will be choked if p b/p0 < 0.528
p0
0.901
1
k1
M
2
2
p b 91 kPa
k 1
mrate ρ A V
T0 ( 20 273 ) K
D 5 mm
2
Avena 12.8 mm
(Not choked)
k
Hence
p0
p vena
1
so
M vena
Then
Tvena
k1
2
M
2
k 1
wher
e
k 1
k
2 p 0
1
k1
pvena
T0
1
k1
2
M vena
2
p vena 91 kPa
M vena 0.389
Tvena 284 K
cvena 338
Tvena 11.3 °C
m
Then
cvena
and
Vvena M vena cvena
m
Vvena 131
s
Also
p vena
ρvena
R Tvena
ρvena 1.12
mrate ρvena Avena Vvena
mrate 1.87 10
Finally
k R Tvena
p vena p b
s
kg
3
m
3 kg
s
The Ts diagram will be a vertical line (T decreases and s = const). After entering the tank there will be turbulent mixing (s increases)
and the flow comes to rest (T increases). The mass flow rate versus time will look like the curved part of Fig. 13.6b; it is nonlinear
because V AND ρ vary
Problem 13.31
[Difficulty: 3]
Given: Temperature and pressure in a tank; nozzle with specific area
Find:
Mass flow rate of gas; maximum possible flow rate
Solution:
The given or available data is:
R =
k =
T0 =
p0 =
296.8
1.4
450
150
K
kPa
At =
30
cm2
At =
pb =
0.003
100
m2
J/kg.K
kPa
Equations and Computations:
Assuming that the nozzle exit pressure is the back pressure:
100
kPa
pe =
Then the nozzle exit Mach number is:
0.7837
Me =
This nozzle is not choked. The exit temperature is:
Te =
400.78
K
From T e and Eq. 12.18
Then
(12.18)
ce =
408.08
m/s
Ve =
319.80
m/s
From the ideal gas equation of state, we can calculate the density:
kg/m3
0.8407
e =
Therefore the mass flow rate is:
m =
0.807
kg/s
When the room pressure can be lowered, we can choke the nozzle.
p*
pe =
T*
Te =
From T 0, and Eq. 12.22b
(12.22b)
Also
Hence
T* =
p* =
375
79.24
Te =
375
Me =
Ve =
1
K
V* =
From T e and Eq. 12.18
Then
K
kPa
ce
(12.18)
ce =
395
m/s
Ve =
395
m/s
To find the mass flow rate we calculate the density from the ideal
gas equation of state:
Hence
e =
0.7120
kg/m3
m max =
0.843
kg/s
Therefore the mass flow rate is:
We can check our results:
From p 0, T 0, A t, and Eq. 13.9a
(13.9a)
Then
m choked =
m choked =
0.843
m rate
kg/s
Correct!
Problem 13.30
[Difficulty: 2]
Problem 13.29
[Difficulty: 2]
Given: Temperature in and mass flow rate from a tank
Find:
Tank pressure; pressure, temperature and speed at exit
Solution:
The given or available data is:
R =
k =
T0 =
286.9
1.4
273
At =
m rate =
0.001
2
J/kg.K
K
m2
kg/s
Equations and Computations:
Because p b = 0
Hence the flow is choked!
pe =
p*
Hence
Te =
T*
From T 0, and Eq. 12.22b
(12.22b)
T* =
228
Te =
228
-45.5
K
K
o
C
Also
Hence
Me =
Ve =
1
V* =
From T e and Eq. 12.18
ce
(12.18)
Then
ce =
302
m/s
Ve =
302
m/s
To find the exit pressure we use the ideal gas equation
after first finding the exit density.
The mass flow rate is m rate = eA eV e
Hence
e =
6.62
kg/m3
pe =
432
kPa
From the ideal gas equation p e = eRT e
From p e = p * and Eq. 12.22a
(12.22a)
p0 =
817
kPa
We can check our results:
From p 0, T 0, A t, and Eq. 13.9a
(13.9a)
Then
m choked =
m choked =
2.00
m rate
kg/s
Correct!
Problem 13.28
[Difficulty: 3]
Given: Data on flow in a passage
Find:
Exit temperature and mass flow rate of air assuming isentropic flow
Solution:
The given or available data is:
R =
k =
T1 =
p1 =
p 01 =
53.33
1.4
450
45
51
A1 =
4
ft2
A2 =
3
ft2
ft-lbf/lbm-°R
°R
psia
psia
Equations and Computations:
From the static and stagnation pressures we can calculate M 1:
M1 =
0.427
T 01 =
466.38
°R
A *1 =
2.649
ft2
From the M 1 and T 1 we can get T 01:
From M 1, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
For isentropic flow (p 02 = p 01, T 02 = T 01, A *2 = A *1)
51
p 02 =
466.38
T 02 =
A *2 =
2.649
A 2/A *2 =
1.1325
psia
°R
ft2
Given subsonic flow in the duct, we can find the exit Mach number using
Equation 13.7d
M2 =
0.653
From the Mach number and stagnation state we can calculate the
static pressure and temperature:
p2 =
38.28
psia
430
°R
T2 =
From T 2 and Eq. 12.18
c2 =
V2 =
1016.38
664.11
ft/s
ft/s
Using the ideal gas law we calculate the density at station 2:
lbm/ft3
ρ2 =
0.2406
Now we can use the area, density, and velocity to calculate the mass flow rate:
m =
479
lbm/s
Problem 13.27
[Difficulty: 2]
Problem 13.26
[Difficulty: 2]
Problem 13.25
[Difficulty: 2]
Given: Data on converging nozzle; isentropic flow
Find:
Pressure and Mach number; throat area; mass flow rate
Solution:
The given or available data is:
R =
k =
286.9
1.4
J/kg.K
A1 =
T1 =
V1 =
p atm =
0.05
276.3
200
101
m2
K
m/s
kPa
Equations and Computations:
From T 1 and Eq. 12.18
Then
(12.18)
c1 =
333
M1 =
0.60
m/s
To find the pressure, we first need the stagnation pressure.
If the flow is just choked
pe =
p atm =
p* =
101
kPa
From p e = p * and Eq. 12.22a
(12.22a)
p0 =
191
kPa
From M 1 and p 0, and Eq. 13.7a
(using built-in function Isenp (M ,k )
(13.7a)
Then
p1 =
150
kPa
The mass flow rate is m rate = 1A 1V 1
Hence, we need 1 from the ideal gas equation.
1 =
1.89
kg/m3
m rate =
18.9
kg/s
The mass flow rate m rate is then
The throat area A t = A * because the flow is choked.
From M 1 and A 1, and Eq. 13.7d
(using built-in function IsenA (M ,k )
(13.7d)
Hence
A* =
0.0421
m2
At =
0.0421
m2
Problem 13.24
[Difficulty: 2]
Problem 13.23
[Difficulty: 2]
Problem 13.22
[Difficulty: 3]
Given: Data on three tanks
Find:
Mass flow rate; Pressure in second tank
Solution:
The given or available data is:
R =
k =
286.9
1.4
At =
1
J/kg.K
2
cm
We need to establish whether each nozzle is choked. There is a large total pressure drop so this is likely.
However, BOTH cannot be choked and have the same flow rate. This is because Eq. 13.9a, below
(13.9b)
indicates that the choked flow rate depends on stagnation temperature (which is constant) but also
stagnation pressure, which drops because of turbulent mixing in the middle chamber. Hence BOTH nozzles
cannot be choked. We assume the second one only is choked (why?) and verify later.
Temperature and pressure in tank 1:
T 01 =
308
650
p 01 =
527
We make a guess at the pressure at the first nozzle exit:
p e1 =
NOTE: The value shown is the final answer! It was obtained using Solver !
527
This will also be tank 2 stagnation pressure:
p 02 =
65
Pressure in tank 3:
p3 =
K
kPa
kPa
kPa
kPa
Equations and Computations:
From the p e1 guess and Eq. 13.17a:
Then at the first throat (Eq.13.7b):
M e1 =
T e1 =
0.556
290
K
The density at the first throat (Ideal Gas) is:
Then c at the first throat (Eq. 12.18) is:
Then V at the first throat is:
e1 =
6.33
341
190
0.120
kg/m
m/s
m/s
kg/s
Finally the mass flow rate is:
c e1 =
V e1 =
m rate =
3
First Nozzle!
For the presumed choked flow at the second nozzle we use Eq. 13.9a, with T 01 = T 02 and p 02:
m rate =
0.120
kg/s
For the guess value for p e1 we compute the error between the two flow rates:
m rate =
0.000
Use Solver to vary the guess value for p e1 to make this error zero!
Note that this could also be done manually.
kg/s
Second Nozzle!
Problem 13.21
[Difficulty: 2]
Given: Data on flow in a passage
Find:
Possible Mach numbers at downstream location
Solution:
The given or available data is:
R =
k =
M1 =
286.9
1.4
1
A1 =
0.2
m2
A2 =
0.5
m2
A *1 =
0.2
m2
A *2 =
0.2
m2
J/kg-K
Equations and Computations:
Since the flow is sonic at the entrance:
For isentropic flow (A *2 = A *1)
A 2/A *2 =
2.5
Now there are two Mach numbers which could result from this area change,
one subsonic and one supersonic.
From A 2/A * 2, and Eq. 13.7d
(using built-in functions)
0.2395
M 2sub =
2.4428
M 2sup =
Problem 13.20
Given:
Air flow in a converging nozzle
Find:
Mass flow rate
[Difficulty: 2]
Solution:
k
Basic equations:
mrate ρ V A
Given or available data p b 35 psi
pb
p0
T
p 0 60 psi
k 1.4
Since
T0
p ρ R T
Rair 53.33
ft lbf
lbm R
0.583 is greater than 0.528, the nozzle is not choked and
Hence
Mt
and
Tt
ct
k 1
k
2 p 0
1
k1
pt
T0
1
k1
2
Mt
k Rair Tt
2
1
k1
2
M
pt pb
ft
Vt 1166
s
mrate ρt At Vt
slug
mrate 0.528
s
At 0.0873 ft
Vt ct
3 slug
ft
k1
π 2
At Dt
4
Tt 106 °F
p
1
Dt 4 in
Tt 566 R
ρt 5.19 10
p0
T0 ( 200 460 ) R
M t 0.912
pt
ρt
Rair Tt
2
3
lbm
mrate 17.0
s
2
2
M
2
k 1
Problem 13.19
Given:
Isentropic air flow in converging nozzle
Find:
Pressure, speed and Mach number at throat
[Difficulty: 2]
Solution:
Basic equations:
k
T0
T
Given or available data
k1
1
2
M
p0
2
p
1
k1
2
p 1 350 kPa
m
V1 150
s
k 1.4
R 286.9
M
2
k 1
M 1 0.5
p b 250 kPa
J
kg K
The flow will be choked if p b/p0 < 0.528
k
k1
p 0 p 1 1
2
M1
2
k 1
pb
p 0 415 kPa
p0
0.602
(Not choked)
k
Hence
p0
pt
so
1
Mt
k1
2
Mt
2
k 1
where
k 1
k
2 p 0
1
k1
pt
Also
V1 M 1 c1 M 1 k R T1 or
Then
T0 T1 1
Hence
Then
Finally
k1
Tt
ct
2
T0
1
k1
2
k R Tt
Vt M t ct
Mt
2
M1
2
pt pb
p t 250 kPa
M t 0.883
V1
T1
k R M1
1
2
T1 224 K
T0 235 K
T0 37.9 °C
Tt 204 K
Tt 69.6 °C
ct 286
m
s
m
Vt 252
s
T1 49.1 °C
Problem 13.18
[Difficulty: 2]
Given: Flow in a converging-diverging nozzle to a pipe
Find:
Plot of mass flow rate
Solution:
The given or available data is
R =
k =
T0 =
p0 =
Dt =
286.9
1.4
293
101
1
J/kg·K
K
kPa
cm
2
At =
0.785
cm
p* =
53.4
kPa
De =
2.5
cm
Ae =
4.909
cm2
Equations and Computations:
The critical pressure is given by
This is the minimum throat pressure
For the CD nozzle, we can compute the pressure at the exit required for this to happen
2
A* = 0.785 cm
A e/A * = 6.25
or
M e = 0.0931
or
p e = 100.4
(= A t)
3.41
67.2
(Eq. 13.7d)
kPa (Eq. 13.7a)
Hence we conclude flow occurs in regimes iii down to v (Fig. 13.8); the flow is ALWAYS choked!
p*
M
T * (K)
c*
V * = c * = p /RT
(kPa) (Eq. 13.7a) (Eq. 13.7b) (m/s)
(m/s)
(kg/m3)
53.4
1.000
244
313
313
0.762
(Note: discrepancy in mass flow rate is due to round-off error)
Flow
Rate
(kg/s)
0.0187
0.0185
(Using Eq. 13.9)
Problem 13.17
[Difficulty: 3]
Given: Data on tank conditions; isentropic flow
Find:
Plot cross-section area and pressure distributions
Solution:
The given or available data is:
R =
k =
53.33
1.4
T0 =
p0 =
pe =
m rate =
500
45
14.7
2.25
ft·lbf/lbm·oR
o
R
psia
psia
lbm/s
Equations and Computations:
From p 0, p e and Eq. 13.7a (using built-in function IsenMfromp (M,k))
(13.7a)
Me =
1.37
Because the exit flow is supersonic, the passage must be a CD nozzle
We need a scale for the area.
From p 0, T 0, m flow, and Eq. 13.10c
(13.10c)
Then
At = A* =
0.0146
ft2
For each M , and A *, and Eq. 13.7d
(using built-in function IsenA (M ,k )
(13.7d)
we can compute each area A .
From each M , and p 0, and Eq. 13.7a
(using built-in function Isenp (M ,k )
we can compute each pressure p .
L (ft)
M
0.069
0.086
0.103
0.120
0.137
0.172
0.206
0.274
0.343
0.412
0.480
0.549
0.618
0.686
0.755
0.823
0.892
0.961
1.000
1.098
1.166
1.235
1.304
1.372
1.00
1.25
1.50
1.75
2.00
2.50
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
14.6
16.00
17.00
18.00
19.00
20.00
A (ft 2)
p (psia)
0.1234
0.0989
0.0826
0.0710
0.0622
0.0501
0.0421
0.0322
0.0264
0.0227
0.0201
0.0183
0.0171
0.0161
0.0155
0.0150
0.0147
0.0146
0.0146
0.0147
0.0149
0.0152
0.0156
0.0161
44.9
44.8
44.7
44.5
44.4
44.1
43.7
42.7
41.5
40.0
38.4
36.7
34.8
32.8
30.8
28.8
26.8
24.9
23.8
21.1
19.4
17.7
16.2
14.7
Area Variation in Passage
0.14
0.12
A (ft2)
0.10
0.08
0.06
0.04
0.02
0.00
0
5
10
15
20
L (ft)
p (psia)
Pressure Variation in Passage
50
45
40
35
30
25
20
15
10
5
0
0
2
4
6
8
10
L (ft)
12
14
16
18
20
Problem 13.16
[Difficulty: 2]
Given: Data on flow in a passage
Find:
Flow rate; area and pressure at downstream location; sketch passage shape
Solution:
The given or available data is:
R =
k =
286.9
1.4
J/kg.K
A1 =
T1 =
p1 =
V1 =
T2 =
M2 =
0.25
283
15
590
410
0.75
m2
K
kPa
m/s
Equations and Computations:
From T 1 and Eq. 12.18
Then
(12.18)
c1 =
337
M1 =
1.75
m/s
Because the flow decreases isentropically from supersonic to subsonic
the passage shape must be convergent-divergent
From p 1 and T 1 and the ideal gas equation
1 =
0.185
kg/m3
m rate =
27.2
kg/s
The mass flow rate is m rate = 1A 1V 1
From M 1 and A 1, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
(13.7d)
A* =
0.180
m2
A2 =
0.192
m2
From M 2 and A *, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
From M 1 and p 1, and Eq. 13.7a
(using built-in function Isenp (M ,k ))
(13.7a)
p 01 =
79.9
kPa
p 02 =
79.9
kPa
p2 =
55.0
kPa
For isentropic flow (p 01 = p 02)
From M 2 and p 02, and Eq. 13.7a
(using built-in function Isenp (M ,k ))
Problem 13.15
[Difficulty: 3]
Given: Flow in a converging nozzle to a pipe
Find:
Plot of mass flow rate
Solution:
The given or available data is
R =
k =
T0 =
p0 =
Dt =
287
1.4
293
101
1
J/kg·K
K
kPa
cm
2
A t = 0.785 cm
Equations and Computations:
The critical pressure is given by
p * = 53.4 kPa
Hence for p = 100 kPa down to this pressure the flow gradually increases; then it is constant
c V = M ·c = p /RT
(m/s) (m/s)
(kg/m3)
343
41
1.19
342
58
1.18
342
71
1.18
341
82
1.17
341
92
1.16
340
101
1.15
337
138
1.11
335
168
1.06
332
195
1.02
329
219
0.971
326
242
0.925
322
264
0.877
318
285
0.828
315
306
0.778
313
313
0.762
313
313
0.762
313
313
0.762
313
313
0.762
313
313
0.762
Flow
Rate
(kg/s)
0.00383
0.00539
0.00656
0.00753
0.00838
0.0091
0.0120
0.0140
0.0156
0.0167
0.0176
0.0182
0.0186
0.0187
0.0187
0.0187
0.0187
0.0187
0.0187
Flow Rate in a Converging Nozzle
0.020
0.018
0.016
0.014
Flow Rate (kg/s)
p
M
T (K)
(kPa) (Eq. 13.7a) (Eq. 13.7b)
100
0.119
292
99
0.169
291
98
0.208
290
97
0.241
290
96
0.270
289
95
0.297
288
90
0.409
284
85
0.503
279
80
0.587
274
75
0.666
269
70
0.743
264
65
0.819
258
60
0.896
252
55
0.974
246
53.4
1.000
244
53
1.000
244
52
1.000
244
51
1.000
244
50
1.000
244
0.012
0.010
0.008
0.006
0.004
0.002
0.000
Using critical conditions, and Eq. 13.9 for mass flow rate:
53.4
1.000
244
313
313
0.762
0.0185
(Note: discrepancy in mass flow rate is due to round-off error)
50
60
70
80
p (kPa)
90
100
Problem 13.14
[Difficulty: 3]
Given: Data on flow in a passage
Find:
Mach numbers at entrance and exit; area ratio of duct
Solution:
The given or available data is:
R =
k =
T1 =
p1 =
T2 =
T 02 =
p2 =
286.9
1.4
310
200
294
316
125
J/kg-K
K
kPa
K
K
kPa
Equations and Computations:
Since the flow is adiabatic, the stagnation temperature is constant:
316
K
T 01 =
Solving for the Mach numbers at 1 and 2 using Eq. 13.7b
(using built-in function IsenMfromT (Tratio ,k ))
M1 =
0.311
0.612
M2 =
Using the ideal gas equation of state, we can calculate the densities of the gas:
kg/m3
ρ1 =
2.249
Then
ρ2 =
1.482
kg/m3
c1 =
c2 =
V1 =
V2 =
352.9
343.6
109.8
210.2
m/s
m/s
m/s
m/s
From static temperatures and Eq. 12.18
Since flow is steady, the mass flow rate must be equal at 1 and 2.
So the area ratio may be calculated from the densities and velocities:
A 2/A 1 =
0.792
Note that we can not assume isentropic flow in this problem. While the flow is
adiabatic, it is not reversible. There is a drop in stagnation pressure from state 1 to 2
which would invalidate the assumption of isentropic flow.
Problem 13.13
[Difficulty: 2]
Problem 13.12
Given:
Air flow in a passage
Find:
Speed and area downstream; Sketch flow passage
[Difficulty: 3]
k 1
Solution:
Basic equations:
T0
T
Given or available data
1
k1
2
M
2
c
k R T
T1 ( 32 460 ) R
p 1 25 psi
M 1 1.75
T2 ( 225 460 ) R
k 1.4
Rair 53.33
D1 3 ft
Hence
2 ( k 1)
1 k 1 M2
A
1
2
k 1
Acrit
M
2
A1
T0 T1 1
k1
2
M1
2
π D1
2
4
T0 793 R
A1 7.07 ft
ft lbf
lbm R
2
T0 334 °F
For isentropic flow stagnation conditions are constant. Hence
We also have
2
T0
M2
k1
c2
k Rair T2
Hence
V2 M 2 c2
From state 1
Acrit
T2
1
M 2 0.889
c2 1283
ft
s
ft
V2 1141
s
A1 M 1
k 1
Acrit 5.10 ft
2 ( k 1)
1 k 1 M 2
1
2
k 1
2
k 1
Hence at state 2
2 ( k 1)
1 k 1 M 2
2
Acrit
2
A2
M2
k 1
2
A2 5.15 ft
Hence, as we go from supersonic to subsonic we must have a converging-diverging diffuser
2
2
Problem 13.
[
2]
Problem 13.10
[Difficulty: 3]
Given: Data on flow in a nozzle
Find:
Mass flow rate; Throat area; Mach numbers
Solution:
The given or available data is:
R =
k =
T0 =
p1 =
A =
286.9
1.4
523
200
J/kg·K
1
2
K
kPa
p2 =
50
kPa
cm
Equations and Computations:
We don't know the two Mach numbers. We do know for each that Eq. 13.7a applies:
Hence we can write two equations, but have three unknowns (M 1, M 2, and p 0)!
We also know that states 1 and 2 have the same area. Hence we can write Eq. 13.7d twice:
We now have four equations for four unknowns (A *, M 1, M 2, and p 0)!
We make guesses (using Solver) for M 1 and M 2, and make the errors in computed A * and p 0 zero.
M1 =
0.512
from Eq. 13.7a:
p0 =
239
and from Eq. 13.7d:
A* =
0.759
cm
For:
M2 =
1.68
kPa
p0 =
239
kPa
0.00%
2
A* =
0.759
cm2
0.00%
Note that the throat area is the critical area
Sum
The stagnation density is then obtained from the ideal gas equation
0 =
1.59
3
kg/m
The density at critical state is obtained from Eq. 13.7a (or 12.22c)
* =
Errors
1.01
kg/m3
The velocity at critical state can be obtained from Eq. 12.23)
V* =
418
m/s
m rate =
0.0321
kg/s
The mass flow rate is *V *A *
0.00%
Problem 13.9
[Difficulty: 3]
Given: Data on flow in a passage
Find:
Shape of flow passage; exit area provided the flow is reversible
Solution:
The given or available data is:
R =
k =
m=
p1 =
T1 =
T1 =
53.33
1.4
20
30
1200
1660
A1 =
M2 =
8
1.2
ft-lbf/lbm-°R
lbm/s
psia
°F
°R
in2
Equations and Computations:
Using the ideal gas law we calculate the density at station 1:
lbm/ft3
0.04880
ρ1 =
Now we can use the area and density to get the velocity from the mass flow rate:
V1 =
7377
ft/s
From T 1 and Eq. 12.18
Then
c1 =
1998
M1 =
3.69
ft/s
Since the flow is supersonic and the velocity is decreasing, this duct is converging.
From M 1, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
A *1 =
0.9857
in2
A *2 =
0.9857
in2
A 2/A *2 =
1.0304
A2 =
1.016
For isentropic flow ( A *2 = A *1)
Therefore the exit area is:
in2
Problem 13.8
[Difficulty: 2]
Given: Data on flow in a passage
Find:
Stagnation conditions; whether duct is a nozzle or diffuser; exit conditions
Solution:
The given or available data is:
R =
k =
p1 =
T1 =
V1 =
259.8
1.4
200
420
200
A1 =
0.6
m2
A2 =
0.5
m2
c1 =
391
m/s
M1 =
0.512
T 01 =
442
K
p 01 =
239
kPa
J/kg-K
kPa
K
m/s
Equations and Computations:
From T 1 and Eq. 12.18
Then
From M 1 and T 1, and Eq. 13.7b
(using built-in function Isent (M ,k ))
From M 1 and p 1, and Eq. 13.7a
(using built-in function Isenp (M ,k ))
Since the flow is subsonic and the area is decreasing, this duct is a nozzle.
From M 1, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
A *1 =
0.4552
For isentropic flow (p 01 = p 02, T 01 = T 02, A *2 = A *1)
p 02 =
239
T 02 =
442
A *2 =
0.4552
m2
kPa
K
m2
*
A 2/A 2 =
1.0984
From A 2/A * 2, and Eq. 13.7d
(using built-in function IsenMsubfromA (M ,k ))
Since there is no throat, the flow stays subsonic!
0.69
M2 =
From M 2 and stagnation conditions:
(using built-in functions)
p2 =
T2 =
173
403
kPa
K
Problem 13.7
[Difficulty: 2]
Given: Data on flow in a passage
Find:
Pressure at downstream location
Solution:
The given or available data is:
ft·lbf/lbm·oR
R =
k =
53.33
1.4
T1 =
p1 =
V1 =
M2 =
560
30
1750
2.5
c1 =
1160
M1 =
1.51
p 01 =
111
psi
p 02 =
111
psi
p2 =
6.52
psi
o
R
psi
ft/s
Equations and Computations:
From T 1 and Eq. 12.18
Then
ft/s
From M 1 and p 1, and Eq. 13.7a
(using built-in function Isenp (M ,k ))
For isentropic flow (p 01 = p 02)
From M 2 and p 02, and Eq. 13.7a
(using built-in function Isenp (M ,k ))
Problem 13.6
Given:
Air flow in a passage
Find:
Mach number; Sketch shape
[Difficulty: 2]
Solution:
k
Basic
equations:
Given or available data
The speed of sound at state 1 is
Hence
p0
p
1
k1
2
M
2
k 1
c
T1 ( 10 273 ) K
p 1 150 kPa
m
V1 120
s
p 2 50 kPa
k 1.4
R 286.9
c1
M1
k R T1
c1 337
V1
Solving for M2
M2
s
k
k
p 0 p 1 1
m
M 1 0.356
c1
For isentropic flow stagnation pressure is constant. Hence at state 2
Hence
k R T
k1
2
M1
2
p0
p2
1
k1
2
k 1
k 1
k
2 p 0
1
k1
p2
p 0 164 kPa
M 2 1.42
Hence, as we go from subsonic to supersonic we must have a converging-diverging nozzle
M2
2
k 1
J
kg K
Problem 13.5
[Difficulty: 2]
Given: Data on flow in a passage
Find:
Temperature and Mach number at downstream location
Solution:
R =
k =
T1 =
T1 =
M1 =
296.8
1.4
30
303
1.7
J/kg-K
A1 =
0.15
m2
A2 =
0.45
m2
T 01 =
478
K
A *1 =
0.1121
m2
478
K
=
0.1121
m2
A 2/A *2 =
4.0128
The given or available data is:
°C
K
Equations and Computations:
From M 1 and T 1, and Eq. 13.7b
(using built-in function Isent (M ,k ))
From M 1, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
For isentropic flow (T 01 = T 02, A *2 = A *1)
T 02 =
A
*
2
*
From A 2/A 2, and Eq. 13.7d
(using built-in function IsenMsupfromA (M ,k ))
Since there is no throat, the flow stays supersonic!
2.94
M2 =
From M 2 and T 02, and Eq. 13.7b
(using built-in function Isent (M ,k ))
T2 =
T2 =
175
-98
K
°C
Problem 13.4
[Difficulty: 2]
Given: Data on flow in a passage
Find:
Pressure and Mach number at downstream location
Solution:
The given or available data is:
R =
k =
p1 =
M1 =
296.8
1.4
450
0.7
J/kg-K
A1 =
0.15
m
A2 =
0.45
m2
p 01 =
624
kPa
0.1371
m2
624
kPa
kPa
2
Equations and Computations:
From M 1 and p 1, and Eq. 13.7a
(using built-in function Isenp (M ,k ))
From M 1, and Eq. 13.7d
(using built-in function IsenA (M ,k ))
A
*
1
=
For isentropic flow (p 01 = p 02, A *2 = A *1)
p 02 =
A
*
=
0.1371
A 2/A *2 =
3.2831
2
2
m
*
From A 2/A 2, and Eq. 13.7d
(using built-in function IsenMsubfromA (M ,k ))
Since there is no throat, the flow stays subsonic
0.1797
M2 =
From M 2 and p 02, and Eq. 13.7a
(using built-in function Isenp (M ,k ))
p2 =
610
kPa
Problem 13.3
[Difficulty: 2]
Given:
Steam flow through a nozzle
Find:
Speed and Mach number; Mass flow rate; Sketch the shape
Solution:
Basic
equations:
2
mrate ρ V A
h1
2
V1
h2
2
V2
2
Assumptions: 1) Steady flow 2) Isentropic 3) Uniform flow 4) Superheated steam can be treated as ideal gas
Given or available data
T0 ( 450 273 ) K
p 0 6 MPa
p 2 MPa
D 2 cm
k 1.30
R 461.4
J
(Table A.6)
kg K
From the steam tables (try finding interactive ones on the Web!), at stagnation conditions
Hence at the nozzle section
J
s0 6720
kg K
h 0 3.302 10
6 J
J
an
s s0 6720
kg K d
p 2 MPa
T 289 °C
From these values we find from the steam tables that
Hence the first law becomes
The mass flow rate is given by
Hence
For the Mach number we need
V
2 h0 h
mrate ρ A V
mrate
c
A V
v
k R T
v
6 J
h 2.997 10
s
2
A
kg
3
v 0.1225
m
V 781
A V
kg
π D
4
4
A 3.14 10
2
m
kg
mrate 2.00
s
c 581
The flow is supersonic starting from rest, so must be converging-diverging
m
s
M
V
c
M 1.35
m
kg
Problem 13.2
[Difficulty: 2]
Problem 13.1
Given:
Air extracted from a large tank
Find:
Mass flow rate
[Difficulty: 2]
Solution:
h1
V1
2
Basic
equations:
mrate ρ V A
h2
Given or available data
T0 ( 70 273 ) K
p 0 101 kPa
D 15 cm
cp 1004
mrate ρ A V
A
2
V2
( 1 k)
2
p
2
k
const
We need the density and velocity at the nozzle. In the tank
1
From the isentropic relation
p
ρ ρ0
p0
J
k 1.4
kg K
R 286.9
J
kg K
2
A 0.0177 m
4
p0
ρ0
R T0
ρ 0.379
const
p 25 kPa
π D
k
k
ρ
2
The mass flow rate is given by
T p
ρ0 1.026
kg
3
m
kg
3
m
We can apply the energy equation between the tank (stagnation conditions) and the point in the nozzle to find the velocity
2
h0 h
V
V
2
2 h0 h
2 c p T0 T
( 1 k)
p0
T T0
p
Fot T we again use insentropic relations
Then
The mass flow rate is
V
2 c p T0 T
mrate ρ A V
Note that the flow is supersonic at this point
Hence we must have a converging-diverging nozzle
V 476
k
T 230.167 K
T 43.0 °C
m
s
kg
mrate 3.18
s
c
k Rc T
304
m
s
M
V
c
M 1.57
Problem 12.92
Given:
Data on air flow in a ramjet combustor
Find:
Critical temperature and pressure at nozzle exit
[Difficulty: 1]
Solution:
k 1.4
The data provided, or available in the Appendices, is:
Stagnation conditions are:
k1
T02 T2 1
2
M2
2
M 2 0.9
T2 ( 1660 460) R
T02 2463.4R
p 2 1.6 psi
T02 2003.8°F
k
p 02 p 2 1
k1
2
2
M2
k 1
p 02 2.71 psi
The critical temperature and pressure are:
T02
Tcrit2
k1
Tcrit2
2
T02
Tcrit2 2052.9 R
k 1
2
k
p 02
p crit2
k
2
1
k 1
p crit2
p 02
k
k
2
1
k 1
p crit2 1.430 psi
Tcrit2 1593.2 °F
Problem 12.91
Given:
Data on air flow in a ramjet combustor
Find:
Critical temperature and pressure at nozzle exit
[Difficulty: 1]
Solution:
The data provided, or available in the Appendices, is:
k 1.4
p 0 1.7 MPa T0 1010 K
The critical temperature and pressure are:
T0
Tcrit
k1
Tcrit
2
T0
Tcrit 841.7 K
k 1
2
k
p0
p crit
k
2
1
k 1
p0
p crit
k
k
2
1
k 1
p crit 0.898 MPa
Problem 12.90
Given:
Data on hot gas stream
Find:
Critical conditions
[Difficulty: 1]
Solution:
The data provided, or available in the Appendices, is:
R 287
J
kg K
For critical conditions
k 1.4
T0
Tcrit
T0 ( 1500 273) K
k1
Tcrit
2
T0
T0 1773K
p 0 140 kPa
Tcrit 1478K
k 1
2
k
p0
p crit
k
2
1
k 1
p0
p crit
k
k
2
Vcrit
k R Tcrit
m
Vcrit 770
s
1
k 1
p crit 74.0 kPa
absolute
Problem 12.9
[
1]
Problem 12.88
Given:
Data on helium in reservoir
Find:
Critical conditions
[Difficulty: 1]
Solution:
The data provided, or available in the Appendices, is:
RHe 386.1
ft lbf
lbm R
For critical conditions
k 1.66
T0
Tcrit
T0 3600 R
k1
Tcrit
2
p 0 ( 725 14.7)psi
T0
p 0 740 psi
Tcrit 2707 R
k 1
2
k
p0
p crit
k 1
k 1
2
p0
p crit
k
k
2
Vcrit
k RHe Tcrit
1
ft
Vcrit 7471
s
k 1
p crit 361 psi
absolute
Problem 12.87
[Difficulty: 1]
Problem 12.86
Given:
Air leak in ISS
Find:
Mass flow rate
[Difficulty: 2]
1
Solution:
mrate ρ V A
Basic equations:
2 k
Vcrit
k1
ρ0
R T0
ρcrit
k 1
k 1
2
The interior conditions are the stagnation conditions for the flow
Given or available data
T0 ( 65 460 ) R
The density of air inside the ISS would be:
Then
ρ0
ρcrit
1
k
2
The mass flow rate is
1
p 0 14.7 psi
Rair 53.33
p0
ρ0
Rair T0
ρcrit 1.49 10
k 1
ft lbf
lbm R
ρ0 2.35 10
ft
3
2 k
Vcrit
2
A 0.001 in
3 slug
ft
3 slug
k 1.4
k1
3
Rair T0
ft
Vcrit 1025
s
mrate ρcrit Vcrit A
mrate 1.061 10
5 slug
s
4 lbm
mrate 3.41 10
s
Problem 12.85
Given:
Air flow leak in window of airplane
Find:
Mass flow rate
[Difficulty: 2]
1
Solution:
Basic equations:
mrate ρ V A
Vcrit
2 k
ρ0
R T0
k1
ρcrit
k 1
k 1
2
The interior conditions are the stagnation conditions for the flow
Given or available data
kg
ρSL 1.225
3
m
T0 271.9 K
ρ0 0.7812 ρSL
ρ0 0.957
kg
3
m
(Above data from Table A.3 at an altitude of 2500 m)
2
A 1 mm
Then
ρ0
ρcrit
1
k
2
The mass flow rate is
cp 1004
1
k 1
J
k 1.4
kg K
ρcrit 0.607
kg
Vcrit
3
m
mrate ρcrit Vcrit A
4 kg
mrate 1.83 10
s
R 286.9
2 k
k1
R T0
J
kg K
m
Vcrit 302
s
Problem 12.84
[Difficulty: 3]
Problem 12.83
[Difficulty: 2]
Given:
Air flow through turbine
Find:
Stagnation conditions at inlet and exit; change in specific entropy; Plot on Ts
diagram
k
Solution:
p0
Basic equations:
p
Given or available data
k1
1
2
2
k 1
T0
T
1
k1
2
M
2
M 1 0.4
p 1 625 kPa
T1 ( 1250 273) K
M 2 0.8
p 2 20 kPa
T2 ( 650 273) K
k 1.4
R 286.9
cp 1004
Then
M
J
kg K
T01 T1 1
k1
2
2
M1
T2
p2
Δs cp ln R ln
T1
p1
J
kg K
T01 1572K
T01 1299 °C
k
p 01 p 1 1
k1
T02 T2 1
k1
2
2
M1
k 1
2
M2
p 01 698 kPa
2
T02 1041 K
k
p 02 p 2 1
k1
2
M2
2
k 1
p 02 30 kPa
T2
p2
Δs cp ln
R ln
T1
p1
p01
T
T 01
p1
p 02
T1
T 02
p2
T2
s
Δs 485
J
kg K
T02 768 °C
Problem 12.82
[Difficulty: 2]
Problem 12.81
[Difficulty: 2]
Given:
Data on air flow in a ramjet combustor
Find:
Stagnation pressures and temperatures; isentropic or not?
Solution:
The data provided, or available in the Appendices, is:
Rair 53.33
M 1 0.2
ft lbf
lbm R
T1 ( 600 460 ) R
For stagnation temperatures:
cp 0.2399
k 1.4
lbm R
p 1 7 psi
T01 T1 1
k1
T02 T2 1
k1
The rate of heat addition is:
BTU
2
2
M rate 0.1
M 2 0.9
M1
M2
Q M rate cp T02 T01
2
2
lbm
s
T2 ( 1660 460 ) R
T01 1068.5 R
T01 608.8 °F
T02 2463.4 R
T02 2003.8 °F
Q 33.5
p 2 1.6 psi
BTU
s
k
For stagnation pressures:
p 01 p 1 1
k1
2
M1
2
k 1
p 01 7.20 psi
k
p 02 p 2 1
The entropy change is:
k1
2
M2
2
k 1
T2
p2
Δs cp ln
Rair ln
T1
p1
p 02 2.71 psi
Δs 0.267
The friction has increased the entropy increase across the duct, even though the heat addition has decreased.
BTU
lbm R
Problem 12.80
[Difficulty: 2]
Given:
Data on air flow in a ramjet combustor
Find:
Stagnation pressures and temperatures; isentropic or not?
Solution:
The data provided, or available in the Appendices, is:
Rair 53.33
M 1 0.2
ft lbf
lbm R
T1 ( 600 460 ) R
For stagnation temperatures:
cp 0.2399
BTU
k 1.4
lbm R
p 1 7 psi
T01 T1 1
k1
T02 T2 1
k1
2
2
M rate 0.1
M 2 0.9
M1
M2
2
2
lbm
s
T2 ( 1890 460 ) R
T01 1068.5 R
T01 608.8 °F
T02 2730.7 R
T02 2271 °F
p 2 4.1 psi
Since we are modeling heat addition, the stagnation temperature should increase.
The rate of heat addition is:
Q M rate cp T02 T01
Q 39.9
BTU
s
k
For stagnation pressures:
The entropy change is:
p 01 p 1 1
k1
2
M1
2
k
k 1
7.20 psi
T2
p2
BTU
Δs cp ln
Rair ln 0.228
lbm R
T1
p1
p 02 p 2 1
k1
2
M2
2
k 1
6.93 psi
T
p02
T02
p2
The entropy increases because heat is being added. Here is a Ts diagram
of the process:
p01
T01
T2
p1
T1
s
Problem 12.79
Given:
Air flow in duct with heat transfer and friction
Find:
Heat transfer; Stagnation pressure at location 2
[Difficulty: 3]
k
Solution:
Basic equations:
Given or available data
c
k R T
and from
V
p0
c
p
V1
h1
p 1 400 kPa
T1 325 K
p 2 275 kPa
T2 450 K
J
dm
h2
kg K
V2
k1
2
kg K
m
ρ V A const
ρ1
V2 V1
ρ2
m
V2 302
s
q c p T2 T1
c2
kg
k1
q 160
2
M2
2
m
s
s
o
M2
kJ
kg
V2
c2
k 1
p 02 385 kPa
3
m
2
V2 V1
c2 425
2
3
ρ2 2.13
2
k
kg
2
k R T2
p 02 p 2 1
J
p2
ρ2
R T2
q h2 h1
k 1
2
R 286.9
V2 V1
2
2
ρ1 4.29
δQ
M
m
V1 150
s
k 1.4
2
Hence
δQ
1
p1
ρ1
R T1
dm
We also have
2
2
Also
2
ρ V A const
cp 1004
Then
M
M 2 0.711
Problem 12.
[
3]
Problem 12.77
[Difficulty: 2]
Given:
Data on air flow in a duct
Find:
Stagnation temperatures; explain; rate of cooling; stagnation pressures; entropy change
Solution:
The data provided, or available in the Appendices, is: R 287
T1 ( 500 273 ) K
p 1 500 kPa
M 1 0.5
M 2 0.2
For stagnation temperatures:
J
kg K
M rate 0.05
T01 T1 1
k1
T02 T2 1
k1
2
2
M1
M2
cp 1004
2
2
J
kg K
k 1.4
T2 ( 18.57 273 ) K
p 2 639.2 kPa
T01 811.7 K
T01 539 C
T02 256.5 K
T02 16.5 C
kg
s
The fact that the stagnation temperature (a measure of total energy) decreases suggests cooling is taking place.
For the heat transfer:
Q M rate cp T02 T01
Q 27.9 kW
k
For stagnation pressures:
p 01 p 1 1
k1
2
M1
2
k 1
p 01 593 kPa
k
p 02 p 2 1
The entropy change is:
k1
2
M2
2
k 1
T2
p2
Δs cp ln
R ln
T1
p1
p 02 657 kPa
Δs 1186
J
kg K
The entropy decreases because the process is a cooling process (Q is negative).
δq
From the second law of thermodynamics: ds
becomes ds ve
T
Hence, if the process is reversible, the entropy must decrease; if it is irreversible, it may increase or decrease
Problem 12.76
[Difficulty: 2]
Given:
Data on air flow in a duct
Find:
Stagnation pressures and temperatures; explain velocity increase; isentropic or not?
Solution:
The data provided, or available in the Appendices, is:
R 287
M 1 0.1
J
cp 1004
kg K
T1 ( 20 273 ) K
For stagnation temperatures:
J
k 1.4
kg K
p 1 1000 kPa
T01 T1 1
k1
T02 T2 1
k1
2
2
M 2 0.7
M1
M2
2
2
T2 ( 5.62 273 ) K
p 2 136.5 kPa
T01 293.6 K
T01 20.6 C
T02 293.6 K
T02 20.6 C
(Because the stagnation temperature is constant, the process is adiabatic)
k
For stagnation pressures:
p 01 p 1 1
k1
2
M1
2
k 1
p 01 1.01 MPa
k
p 02 p 2 1
The entropy change is:
Note that
k1
2
M2
2
k 1
p 02 189 kPa
T2
p2
Δs cp ln
R ln
T1
p1
V1 M 1 k R T1
m
V1 34.3
s
Δs 480
V2 M 2 k R T2
Although there is friction, suggesting the flow should decelerate, because
the static pressure drops so much, the net effect is flow acceleration!
The entropy increases because the process is adiabatic but irreversible (friction).
δq
From the second law of thermodynamics ds
: becomes ds > 0
T
J
kg K
m
V2 229
s
Problem 12.75
[Difficulty: 2]
Given:
Data on air flow in a duct
Find:
Stagnation pressures and temperatures; explain velocity increase; isentropic or not?
Solution:
The data provided, or available in the Appendices, is:
Rair 287
At altitude:
J
kg K
cp 1004
J
k 1.4
kg K
M 9.68
p SL 101.3 kPa
kg
ρSL 1.225
3
m
33528 30000
z 110000 ft z 33528 m T1 226.5 K ( 250.4 K 226.5 K)
T 234.9 K
40000 30000 1
33528 30000
p 1 p SL 0.01181 ( 0.002834 0.01181 )
40000 30000
The sound speed is:
c
k Rair T1 307.239
So the stagnation temperature and pressure are:
m
p 1 0.8756 kPa
V M c 2974
so the flight speed is:
s
T01 T1 1
k1
2
M
2
m
V 9757
s
T01 4638 K
ft
s
T01 8348 R
k
p 01 p 1 1
k1
2
As the air passes through the shock wave, stagnation pressure decreases:
M
2
k 1
p 01 29.93 MPa
p 02 p 01 ( 1 0.996 )
Therefore, the total head probe sees a pressure of
Since there is no heat transfer through the shock wave, the stagnation temperature remains the same:
p 02 119.7 kPa
T02 T01
T02 8348 R
Problem 12.74
[Difficulty: 2]
Problem 12.73
[Difficulty: 2]
Problem 12.72
Given:
Wind tunnel test of supersonic transport
Find:
Lift and drag coefficients
[Difficulty: 3]
k
Solution:
Basic equations:
c
k R T
FL
CL
1
2
Given or available data
M
CD
2
ρ V A
M 1.8
Finally
2
k1
2
k Rair T
CD
2
ρ V A
FD
1
2
1
2
ρ V A
2
k 1
T0
T
1
k1
2
M
2
2
ρ V A
p 0 200 psi
Rair 53.33
k
FL 12000 lbf
ft lbf
lbm R
k 1
p 34.8 psi
T 123 °F
2
c 1183
ft
c 807 mph
s
ft
V 1452 mph
s
slug
ft
FL
2
2
M
FD
ρ 0.00501
Rair T
1
V 2129
p
CL
2
k1
T 583 R
M
V M c
ρ
M
T0
1
We also need
k1
p p 0 1
T
and
p
k 1.4
c
c
1
T0 ( 500 460 ) R
2
Then
p0
2
A 100 in
We need local conditions
V
CL 1.52
CD 0.203
3
FD 1600 lbf
Problem 12.71
[Difficulty: 2]
Problem 12.70
Given:
Wind tunnel at M = 2.5
Find:
Stagnation conditions; mass flow rate
[Difficulty: 2]
k
Solution:
Basic equations:
Given or available data
Then
c
k R T
M
V
p0
c
p
M 2.5
T ( 15 273 ) K
k 1.4
R 286.9
T0 T 1
k1
2
M
1
k1
2
M
2
k 1
T0
T
1
2
A 0.175 m
T0 648 K
T0 375 °C
J
kg K
2
p 0 p 1
k1
2
M
2
k 1
p 0 598 kPa
The mass flow rate is given by
mrate ρ A V
We need
c
k R T
c 340
ρ
p
ρ 0.424
and also
Then
R T
mrate ρ A V
m
V M c
s
kg
3
m
kg
mrate 63.0
s
2
p 35 kPa
k
Also
k1
V 850
m
s
M
2
Problem 12.69
[Difficulty: 2]
Given:
Flight altitude of high-speed aircraft
Find:
Mach number and aircraft speed errors assuming incompressible flow; plot
Solution:
The governing equation for pressure change is:
k
p0
p
Hence
1
k1
2
M
2
k 1
(12.20a)
k
k 1
k 1 2
Δp p 1
M
1
2
p0
Δp p 0 p p
1
p
(1)
For each Mach number the actual pressure change can be computed from Eq. 1
p
Assuming incompressibility, the Bernoulli equation applies in
the form
ρ
2
V
2
p0
V
so
ρ
2 Δp
and the Mach number based on this is
Using Eq. 1
M incomp
V
c
ρ
k R T
2 Δp
k ρ R T
k
k 1
2
k 1 2
M incomp
1
M
1
k
2
The error in using Bernoulli to estimate the Mach number is
ΔM
M
M incomp M
M
For errors in speed:
Actual speed:
V M c
Speed assuming incompressible flow:
The error in using Bernoulli to estimate the speed from the pressure difference is
V M k R T
Vinc M incomp k R T
ΔV
V
The computations and plots are shown below, generated using Excel:
Vincomp V
V
2 p0 p
ρ
2 Δp
ρ
The given or available data is:
R =
k =
T =
286.9
1.4
216.7
J/kg.K
K
(At 12 km, Table A.3)
Computed results:
c =
M
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
295
M in comp
0.100
0.201
0.303
0.408
0.516
0.627
0.744
0.865
0.994
m/s
ΔM/M
V (m/s)
0.13%
0.50%
1.1%
2.0%
3.2%
4.6%
6.2%
8.2%
10.4%
29.5
59.0
88.5
118
148
177
207
236
266
V incomp (m/s)
29.5
59.3
89.5
120
152
185
219
255
293
ΔV/V
0.13%
0.50%
1.1%
2.0%
3.2%
4.6%
6.2%
8.2%
10.4%
Error in Mach Number Using Bernoulli
12%
10%
ΔM/M
8%
6%
4%
2%
0%
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
M
Error in Speed Using Bernoulli
12%
10%
ΔV/V
8%
6%
4%
2%
0%
0
50
100
150
V (m/s)
200
250
300
Problem 12.68
[Difficulty: 1]
Problem 12.67
[Difficulty: 2]
Given:
Mach number of aircraft
Find:
Pressure difference; air speed based on a) compressible b) incompressible assumptions
Solution:
The data provided, or available in the Appendices, is:
R 287
J
cp 1004
kg K
J
kg K
T 223.3 K
From Table A.3, at 10 km altitude
k 1.4
M 0.65
p 0.2615 101 kPa
p 26.4 kPa
k
p0
The governing equation for pressure change is:
p
1
k1
2
M
2
k 1
(12.20a)
k
Hence
The pressure difference is
p 0 p 1
k1
2
M
2
k 1
p 0 35.1 kPa
p 0 p 8.67 kPa
a) Assuming compressibility c
k R T
c 300
m
V M c
s
V 195
m
s
b) Assuming incompressibility
Here the Bernoulli equation applies in the form
p
ρ
For the density
Hence
ρ
p
V
2
p0
ρ
so
V
2 p0 p
ρ 0.412
R T
V 205
2
ρ
kg
3
V
2 p0 p
m
ρ
m
s
In this case the error at M = 0.65 in computing the speed of the aircraft using Bernoulli equation is
205 195
195
5.13 %
Problem 12.66
[Difficulty: 1]
Problem 12.65
[Difficulty: 1]
Problem 12.64
[Difficulty: 1]
Problem 12.63
Given:
Aircraft flying at 12 km
Find:
Dynamic and stagnation pressures
[Difficulty: 2]
k
Solution:
Basic equations:
Given or available data
At h 12 km ,from Table A.3
c
k R T
M
V
p0
c
p
M 2
h 12 km
kg
ρSL 1.225
3
m
p SL 101.3 kPa
ρ 0.2546 ρSL
ρ 0.312
Also
Hence
p 0 p 1
k1
c
2
k R T
p dyn
1
2
M
2
2
2
M
2
k 1
p dyn
R 286.9
p 0.1915 p SL
p 19.4 kPa
p 0 152 kPa
m
s
p dyn 54.3 kPa
V M c
V 590
m
s
1
2
2
ρ V
J
k 1.4
k 1
c 295
ρ V
3
k1
m
k
Hence
kg
1
kg K
T 216.7 K
Problem 12.62
[Difficulty: 2]
Given:
Pressure data on aircraft in flight
Find:
Change in air density; whether flow can be considered incompressible
Solution:
The data provided, or available in the Appendices, is:
k 1.4
p 0 48 kPa
p 27.6 kPa
T ( 55 273 ) K
Governing equation (assuming isentropic flow):
p
k
constant
(12.12c)
ρ
1
Hence
ρ
ρ0
p
p
0
k
1
so
Δρ
ρ
ρ0 ρ
ρ
k
p0
1 1
ρ
p
ρ0
Δρ
ρ
48.5 %
NOT an incompressible flow!
Problem 12.61
Given:
Aircraft flying at 250 m/s
Find:
Stagnation pressure
[Difficulty: 1]
k
Solution:
Basic equations:
Given or available data
First we need
c
k R T
V 250
c
M
m
V
p0
c
p
T ( 50 273 ) K
s
k R T
c 299
m
s
then
1
k1
2
p 0 p 1
k1
2
M
2
2
M
k 1
p 0 44.2 kPa
V
c
k 1
p 28 kPa
k
Finally we solve for p0
M
k 1.4
M 0.835
R 286.9
J
kg K
Problem 12.60
[Difficulty: 2]
Given:
X-15 rocket plane traveling at fixed Mach number and altitude
Find:
Stagnation and dynamic pressures
k
Solution:
Basic equation:
Available data
At
c
k R T
R 286.9
M
J
kg K
z 58400 m
V
p0
c
p
k 1.4
1
V 7270
interpolating from Table A.3
k1
2
M
km
2
k 1
2
2
ρ V
kg
ρSL 1.225
3
m
p SL 101.3 kPa
hr
1
p dyn
T 270.7 K ( 255.8 K 270.7 K)
58400 50000
60000 50000
T 258 K
Hence
c
k R T
c 322
m
s
c 1159
km
M
and we have
hr
V
c
6.27
The static pressure and density can be found by interpolation:
58400 50000
p p SL 0.0007874 ( 0.0002217 0.0007874 )
60000 50000
p 0.0316 kPa
k
p 0 p 1
58400 50000
ρ ρSL 0.0008383 ( 0.0002497 0.0008383 )
60000 50000
k1
2
M
2
k 1
p 0 65.6 kPa
4 kg
ρ 4.21 10
3
m
p dyn
1
2
2
ρ V
p dyn 0.86 kPa
Problem 12.59
[Difficulty: 2]
Given:
Scramjet-powered missile traveling at fixed Mach number and altitude
Find:
Stagnation and dynamic pressures
k
Solution:
Basic equation:
Available data
At
c
k R T
R 286.9
M
J
kg K
z 85000 ft
V
p0
c
p
k 1.4
M 7
1
k1
2
M
p SL 14.696 psi
2
k 1
p dyn
ρSL 0.2377
1
2
2
ρ V
slug
3
ft
z 25908 m interpolating from Table A.3 T 220.6 K ( 222.5 K 220.6 K)
25908 24000
26000 24000
T 222 K
Hence
c
k R T
c 299
m
s
c 981
ft
and we have
s
V M c 6864
ft
s
The static pressure and density can be found by interpolation:
k
25908 24000
p p SL 0.02933 ( 0.02160 0.02933 )
p 0.323 psi
26000 24000
25908 24000
slug
ρ ρSL 0.03832 ( 0.02797 0.03832 )
ρ 0.00676 3
26000 24000
ft
p 0 p 1
p dyn
1
2
k1
2
2
ρ V
M
2
k 1
p 0 1336 psi
p dyn 1106 psi
Problem 12.58
[Difficulty: 1]
Given:
Car and F-1 race car traveling at sea level
Find:
Ratio of static to total pressure in each case; are compressiblilty effects experienced?
Solution:
k
Basic equations:
Given or available data
At sea level, from Table A.3
Hence
c
k R T
M
V
p0
c
p
Vcar 55 mph
ft
Vcar 80.7
s
k 1.4
Rair 53.33
T 288.2 K
or
c
k Rair T
p
p0
1
k1
c 1116
2
M car
M F1
VF1
2
p
p0
k 1
ft
VF1 323
s
VF1 220 mph
ft lbf
lbm R
ft
M car
s
ρ 0.002377
slug
ft
Vcar
3
p 14.696 psi
M car 0.0723
c
0.996
2
ρ Vcar
p
1
2 p
p0
1
0.996
M F1 0.289
c
1
2
k 1
The pressure ratio is
2
M
k
Note that the Bernoulli equation would give the same result!
For the Formula One car:
T 519 R
The pressure ratio is
k1
1
k1
2
M F1
2
k
k 1
0.944
Note that the Bernoulli equation would give almost the same result:
2
ρ VF1
p
1
2 p
p0
1
0.945
Incompressible flow can be assumed for both cases,
but the F1 car gets very close to the Mach 0.3 rule
of thumb for compressible vs. incompressible flow.
Problem 12.57
[Difficulty: 2]
Given:
X-15 rocket plane traveling at fixed Mach number and altitude
Find:
Stagnation temperature at the nose of the plane
Solution:
Basic equation:
Available data
At
T0 T 1
k1
R 286.9
J
2
kg K
z 58400 m
M
2
c
k 1.4
k R T
M
V 7270
interpolating from Table A.3
V
c
km
hr
T 270.7 K ( 255.8 K 270.7 K)
58400 50000
60000 50000
T 258 K
Hence
c
k R T
So the stagnation temperature is
c 322
m
s
T0 T 1
c 1159
k1
2
M
2
km
hr
and we have
M
V
c
6.27
T0 2289 K
Problem 12.56
Given:
Scramjet-powered missile traveling at fixed Mach number and altitude
Find:
Stagnation temperature at the nose of the missile
Solution:
T0 T 1
k1
Available data
R 286.9
J
At
z 85000 ft
Basic equation:
2
kg K
M
[Difficulty: 2]
2
k 1.4
M 7
z 25908 m interpolating from Table A.3 T 220.6 K ( 222.5 K 220.6 K)
T 222 K
So the stagnation temperature is
T0 T 1
k1
2
M
2
T0 2402 K
25908 24000
26000 24000
Problem 12.55
[Difficulty: 2]
Given:
Mach number range from 0.05 to 0.95
Find:
Plot of percentage density change; Mach number for 1%, 5% and 10% density change
Solution:
k 1.4
The given or available data is:
Basic equation:
1
ρ0
ρ
1
( k 1)
2
M
2
1
k 1
(12.20c)
Δρ
Hence
ρ0
ρ0 ρ
ρ0
1
ρ
so
ρ0
Δρ
ρ0
1 1
Here are the results, generated using Excel:
M
0.05
0.10
0.15
0.20
0.25
0.30
0.35
Δρ /ρ o
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
7.6%
9.4%
11%
14%
16%
18%
21%
23%
26%
29%
31%
34%
0.1%
0.5%
1.1%
2.0%
3.1%
4.4%
5.9%
To find M for specific density changes
use Goal Seek repeatedly
Δρ /ρ o
M
0.142
1%
0.322
5%
0.464
10%
Note: Based on ρ (not ρ o) the results are:
0.142
0.314
0.441
Density Variation with Mach Number
40%
Δρ/ρo
30%
20%
10%
0%
0.0
0.1
0.2
0.3
0.4
0.5
M
0.6
0.7
0.8
0.9
1.0
( k 1)
2
M
2
1 k
Problem 12.54
Given:
Supersonic transport aircraft
Find:
Explanation of sound wave refraction
[Difficulty: 5]
Solution:
A sound wave is refracted when the speed of sound varies with altitude in the atmosphere. (The variation in sound speed is caused by
temperature variations in the atmosphere, as shown in Fig. 3.3)
Imagine a plane wave front that initially is vertical. When the wave encounters a region where the temperature increase with altitude
(such as between 20.1 km and 47.3 km altitude in Fig. 3.3), the sound speed increases with elevation. Therefore the upper portion of
the wave travels faster than the lower portion. The wave front turns gradually and the sound wave follows a curved path through the
atmosphere. Thus a wave that initially is horizontal bends and follows a curved path, tending to reach the ground some distance from
the source.
The curvature and the path of the sound could be calculated for any specific temperature variation in the atmosphere. However, the
required analysis is beyond the scope of this text.
Problem 12.53
Given:
Speed of automobile
Find:
Whether flow can be considered incompressible
[Difficulty: 2]
Solution:
Consider the automobile at rest with 60 mph air flowing over it. Let state 1 be upstream, and point 2 the stagnation point
on the automobile
The data provided, or available in the Appendices, is:
R 287
J
k 1.4
kg K
V1 60 mph
p 1 101 kPa
T1 ( 20 273 ) K
1
The basic equation for the density change is
ρ0
ρ
1
( k 1)
2
M
2
k 1
(12.20c)
1
( k 1)
ρ0 ρ1 1
or
For the Mach number we need c
m
V1 26.8
s
ρ0 ρ1 1
k1
2
M1
2
k 1
ρ1 1.201
c1
c1 343
V1
3
m
s
M 1 0.0782
c1
ρ0 1.205
kg
m
k R T1
k 1
2
p1
ρ1
R T1
M1
1
2
M1
kg
The percentage change in density is
ρ0 ρ1
ρ0
3
m
0.305 %
This is an insignificant change, so the flow can be considered incompressible. Note that M < 0.3, the usual guideline for
incompressibility
V1 120 mph
For the maximum speed present
m
V1 53.6
s
M1
1
ρ0 ρ1 1
k1
2
M1
2
k 1
ρ0 1.216
kg
3
m
The percentage change in
density is
This is still an insignificant change, so the flow can be considered incompressible.
V1
c1
M 1 0.156
ρ0 ρ1
ρ0
1.21 %
Problem 12.52
[Difficulty: 4] Part 1/2
Problem 12.52
[Difficulty: 4] Part 2/2
Problem 12.51
[Difficulty: 3]
x
h
x = Vt
Given:
Supersonic aircraft flying overhead
Find:
Location at which first sound wave was emitted
Solution:
Basic equations:
c
k R T
Given or available data
V 1000
M
m
α asin
M
V
c
h 3 km
s
1
k 1.4
R 286.9
J
kg K
Δx h tan( α)
We need to find Δx as shown in the figure
The temperature is not constant so the Mach line will not be straight (α is not constant). We can find a range of
α and Δx by considering the temperature range
At h 3 km we find from Table A.3 that
Using this temperature
c
Hence
α asin
T 268.7 K
k R T
1
c 329
M
c
Hence
α asin
s
α 19.2 deg
V
an
d
M
Δx h tan( α)
Δx 1043 m
an
d
M
Δx h tan( α)
Δx 1085 m
c
M 3.04
T 288.2 K
At sea level we find from Table A.3 that
Using this temperature
m
k R T
M
1
c 340
m
s
α 19.9 deg
Thus we conclude that the distance is somwhere between 1043 and 1085 m. Taking an average
V
c
Δx 1064 m
M 2.94
Problem 12.50
[Difficulty: 3]
x
h
Given:
Supersonic aircraft flying overhead
Find:
Time at which airplane heard
Solution:
Basic equations:
c
k R T
Given or available data
V 1000
m
s
M
α asin
V
h 3 km
x
M
c
k 1.4
The time it takes to fly from directly overhead to where you hear it is Δt
If the temperature is constant then
1
R 286.9
J
kg K
x
V
h
tan ( α )
The temperature is not constant so the Mach line will not be straight. We can find a range of Δt by considering the temperature range
At h 3 km we find from Table A.3 that
Using this temperature
c
k R T
Hence
α asin
M
1
c
Hence
α asin
c 329
m
s
α 19.2 deg
M
and
x
h
tan ( α )
V
c
x 8625m
M 3.04
Δt
x
V
Δt 8.62s
T 288.2 K
At sea level we find from Table A.3 that
Using this temperature
T 268.7 K
k R T
M
1
c 340
m
s
α 19.9 deg
M
and
x
h
tan ( α )
V
c
x 8291m
Thus we conclude that the time is somwhere between 8.62 and 8.29 s. Taking an average
M 2.94
Δt
x
V
Δt 8.55 s
Δt 8.29s
Problem 12.49
[Difficulty: 2]
Problem 12.48
[Difficulty: 2]
x
h
Given:
High-speed jet flying overhead
Find:
Estimate speed and Mach number of jet
Solution:
Basic
equations:
c
k R T
Given or available data
T ( 25 273 ) K
M
α asin
V
1
M
c
h 3000 m
k 1.4
R 286.9
J
kg K
The time it takes to fly from directly overhead to where you hear it is Δt 7.5 s
The distance traveled, moving at speed V, is
x V Δt
tan( α)
The Mach angle is related to height h and distance x by
1
sin( α)
and also we have
M
c
cos( α)
h
x
h
(1)
V Δt
(2)
V
c V Δt
c Δt
V h
h
cos( α)
Dividing Eq. 2 by Eq 1
sin( α)
Note that we could have written this equation from geometry directly!
We have
Hence
Then the speed is
c
k R T
M
1
sin( α)
V M c
c 346
m
s
so
α acos
c Δt
h
M 1.99
V 689
m
s
Note that we assume the temperature of the air is uniform. In fact the temperature will vary over 3000 m, so the
Mach cone will be curved. This speed and Mach number are only rough estimates.
α 30.1 deg
Problem 12.47
[Difficulty: 2]
Problem 12.46
[Difficulty: 2]
Problem 12.45
[Difficulty: 1]
Problem 12.44
Given:
Projectile fired into a gas, Mach cone formed
Find:
Speed of projectile
[Difficulty: 3]
Solution:
Basic equations:
Given or available data
c
k R T
p 450 kPa
ρ 4.5
α asin
M
V
M
c
kg
3
k 1.625
m
Combining ideal gas equation of state and the sonic speed:
From the Mach cone angle:
M
1
sin( α)
M 4.62
c
k
p
ρ
α
1
25
2
p ρ R T
deg 12.5 deg
c 403.1
Therefore the speed is:
m
s
V M c
V 1862
m
s
Problem 12.43
[Difficulty: 3]
x
h
Given:
Hypersonic aircraft flying overhead
Find:
Time at which airplane is heard, how far aircraft travelled
Solution:
c
Basic equations:
k R T
M 7
Given or available data
M
α asin
V
R 286.9
J
kg K
The time it takes to fly from directly overhead to where you hear it is Δt
x
At
h 120000 ft h 36576 m
M
c
k 1.4
If the temperature is constant then
1
x
V
h
tan( α)
T 226.5 K ( 250.4 K 226.5 K)
interpolating from Table A.3
T 242.2 K
c
Using this temperature
Hence
M
α asin
1
k R T
α 8.2 deg
c 312
m
and
s
x
h
tan( α)
V M c
V 2183
m
s
x 253.4 km
Δt
x
V
Δt 116.06 s
36576 30000
40000 30000
Problem 12.42
Given:
Air flow at M = 1.9
Find:
Air speed; Mach angle
[Difficulty: 1]
Solution:
Basic equations:
c
k R T
M
T ( 77 460 ) R
M 1.9
Hence
c
c 1136
Then the air speed is
The Mach angle is given by
V M c
α asin
M
k 1.4
ft
V 2158
1
1
M
c
The given or available data is
k Rair T
α asin
V
s
ft
s
α 31.8 deg
V 1471 mph
Rair 53.33
ft lbf
lbm R
Problem 12.41
[Difficulty: 3]
Given:
Data on atmospheric temperature variation with altitude
Find:
Lapse rate; plot rate of change of sonic speed with altitude
Solution:
Rair 286.9
dz
c
dc
dz
z (km)
T (K)
-1
dc/dz (s )
0
1
2
3
4
5
6
7
8
9
10
288.2
281.7
275.2
268.7
262.2
255.8
249.3
242.8
236.3
229.8
223.3
-0.00383
-0.00387
-0.00392
-0.00397
-0.00402
-0.00407
-0.00412
-0.00417
-0.00423
-0.00429
-0.00435
T T0
m
m
Hence
T0 288.2 K T10k 223.3 K
z 10000 m
T T0 m z
dT
For an ideal gas
k 1.4
kg K
T10k T0
z
k R T
which can be evaluated at z = 10 km
z
m k R
2 c
3K
6.49 10
k R T0 m z
m
Here are the results, calculated using Excel:
Rate of Change of Sonic Speed
with Altitude
-0.0038
-0.0039
-1
For a linear temperature variation
J
dc/dz (s )
The given or available data is:
-0.0040
-0.0041
-0.0042
-0.0043
-0.0044
0
2
4
6
z (km)
8
10
Problem 12.40
[Difficulty: 2]
Given:
Data on atmospheric temperature variation with altitude
Find:
Sound of speed at sea level; plot speed as function of altitude
Solution
The given or available data is:
R =
k =
286.9
1.4
J/kg.K
Computing equation:
c kRT
Computed results:
(Only partial data is shown in table)
T (K)
c (m/s)
z (m)
288.2
284.9
281.7
278.4
275.2
271.9
268.7
265.4
262.2
258.9
255.7
249.2
242.7
236.2
229.7
223.3
340
338
336
334
332
330
329
326
325
322
320
316
312
308
304
299
Speed of Sound Variation with Altitude
350
325
c (m/s)
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
300
275
250
0
10000
20000
30000
40000
50000
z (m)
60000
70000
80000
90000 100000
Problem 12.39
Section 12-2
[Difficulty: 3]
Problem 12.38
Given:
Data on water specific volume
Find:
Speed of sound over temperature range
[Difficulty: 2]
Solution:
c
Basic equation:
As an approximation for a liquid c
ρ
at isentropic conditions
p
Δp
using available data.
Δρ
We use compressed liquid data at adjacent pressures of 5 MPa and 10 MPa, and estimate the change in density between these
pressures from the corresponding specific volume changes
Δp p 2 p 1
1
Δρ
v2
1
and
v1
c
Δp
Δρ
at each
temperature
Here are the results, calculated using Excel:
p2 =
p1 =
p =
10
5
5
MPa
MPa
MPa
Data on specific volume versus temperature can be obtained fro any good thermodynamics text (try the Web!)
p1
o
3
0
20
40
60
80
100
120
140
160
180
200
0.0009977
0.0009996
0.0010057
0.0010149
0.0010267
0.0010410
0.0010576
0.0010769
0.0010988
0.0011240
0.0011531
p2
Speed of Sound versus Temperature
3
3
T ( C) v (m /kg) v (m /kg) Δρ (kg/m ) c (m/s)
2.52
2.31
2.18
2.14
2.19
2.31
2.42
2.68
2.82
3.18
3.70
1409
1472
1514
1528
1512
1470
1437
1366
1330
1254
1162
1500
1400
c (m/s)
0.0009952
0.0009973
0.0010035
0.0010127
0.0010244
0.0010385
0.0010549
0.0010738
0.0010954
0.0011200
0.0011482
1600
1300
1200
1100
1000
0
50
100
o
T ( C)
150
200
Problem 12.37
[Difficulty: 2]
Given:
Echo heard while hammering near mountain lake, time delay of echo is known
Find:
How far away are the mountains
Solution:
Basic equation:
c
k R T
Assumption: Speed of light is essentially infinite (compared to speed of sound)
The given or available data is
T ( 25 273 ) K
Hence
c
The distance covered by the sound is:
k 1.4
k Rair T
L c Δt
Rair 287
c 346
L 1038 m
J
kg K
Δt 3 s
m
s
but the distance to the mountains is half that distance:
L
2
519 m
Problem 12.36
Given:
Shuttle launch
Find:
How long after seeing it do you hear it?
[Difficulty: 2]
Solution:
Basic equation:
c
k R T
Assumption: Speed of light is essentially infinite (compared to speed of sound)
The given or available data is
T ( 80 460 ) R
L 3.5 mi
Hence
c
k Rair T
c 1139
Δt
L
Δt 16.23 s
Then the time is
c
ft
s
In the winter:
T ( 50 460 ) R
Hence
c
k Rair T
c 1107
Δt
L
Δt 16.7 s
Then the time is
c
ft
s
k 1.4
Rair 53.33
ft lbf
lbm R
Problem 12.35
[Difficulty: 2]
Given:
Mach number and altitude of hypersonic aircraft
Find:
Speed assuming stratospheric temperature, actual speed, speed assuming sea level static temperature
Solution:
Basic equation:
c
k R T
M
J
Available data
Rair 286.9
Assuming
T 390 R 217 K
Hence
c
At
kg K
k Rair T
V
c
k 1.4
c 295
M 7
m
and we have
s
m
Vstrat M c 2065
s
z 120000 ft z 36576 m interpolating from Table A.3
T 226.5 K ( 250.4 K 226.5 K)
36576 30000
40000 30000
T 242 K
Hence
c
k Rair T
c 312
m
and we have
s
m
Vactual M c 2183
s
The error is:
Assuming
T 288.2 K
Hence
c
k Rair T
Vstrat Vactual
Vactual
c 340
m
s
and we have
5.42 %
m
Vsls M c 2382
s
The error is:
Vsls Vactual
Vactual
9.08 %
Problem 12.34
Given:
X-15 rocket plane speed and altitude
Find:
Mach number
[Difficulty: 2]
Solution:
Basic equation:
Available data
At
c
k R T
R 286.9
M
J
kg K
z 58400 m
V
c
k 1.4
V 7270
interpolating from Table A.3
km
hr
T 270.7 K ( 255.8 K 270.7 K)
T 258 K
Hence
c
k R T
c 322
m
s
c 1159
km
hr
and we have
M
V
c
6.27
58400 50000
60000 50000
Problem 12.33
Given:
Fireworks displays!
Find:
How long after seeing them do you hear them?
[Difficulty: 2]
Solution:
Basic equation:
c
k R T
Assumption: Speed of light is essentially infinite (compared to speed of sound)
The given or available data is
Hence
Then the time is
TJuly ( 75 460 ) R
cJuly
ΔtJuly
k Rair TJuly
L
cJuly
In January
TJan ( 5 460 ) R
Hence
cJan
Then the time is
ΔtJan
k Rair TJan
L
cJan
L 1 mi
k 1.4
cJuly 1134
ft
s
ΔtJuly 4.66 s
cJan 1057
ft
s
ΔtJan 5.00 s
Rair 53.33
ft lbf
lbm R
Problem 12.32
Given:
Airplane cruising at 550 mph
Find:
Mach number versus altitude
[Difficulty: 2]
Solution:
c
k R T
M
V
Here are the results, generated using Excel:
c
V = 500 mph
R = 286.90 J/kg-K
k = 1.40
(Table A.6)
Data on temperature versus height obtained from Table A.3
z (m)
T (K)
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
288.2
284.9
281.7
278.4
275.2
271.9
268.7
265.4
262.2
258.9
255.7
249.2
242.7
236.2
229.7
223.3
c (m/s) c (mph)
340
338
336
334
332
330
329
326
325
322
320
316
312
308
304
299
661
658
654
650
646
642
639
635
631
627
623
615
607
599
590
582
M
0.756
0.760
0.765
0.769
0.774
0.778
0.783
0.788
0.793
0.798
0.803
0.813
0.824
0.835
0.847
0.859
Mach Number versus Elevation
0.90
0.85
M
Basic equation:
0.80
0.75
0.70
0
1000
2000
3000
4000
5000
z (m)
6000
7000
8000
9000
10000
Problem 12.31
[Difficulty: 1]
Problem 12.30
[Difficulty: 1]
Problem 12.29
[Difficulty: 2]
Given:
Scramjet-powered missile traveling at fixed Mach number and altitude
Find:
Time necessary to cover specified range
Solution:
Basic equation:
c
k R T
M
J
Available data
R 286.9
At
z 85000 ft
kg K
V
c
k 1.4
M 7
6
Δx 600 nmi 3.65 10 ft
z 25908 m interpolating from Table A.3 T 220.6 K ( 222.5 K 220.6 K)
25908 24000
26000 24000
T 222 K
Hence
c
k R T
The time needed to cover the range is:
c 299
m
Δt
Δx
s
V
c 981
531 s
ft
and we have
s
Δt 8.85 min
V M c 6864
ft
s
This is about ten times as fast as the Tomahawk!
Problem 12.28
Given:
Airplane cruising at two different elevations
Find:
Mach numbers
[Difficulty: 1]
Solution:
Basic equation:
c
k R T
M
J
Available data
R 286.9
At
z 1500 m
Hence
c
kg K
k R T
Repeating at
z 15000 m
Hence
c
The Mach number is
k R T
c
k 1.4
T 278.4 K
c 334
M
The Mach number is
V
from Table A.3
m
s
V
c 1204
km
and we have
hr
V 550
km
hr
M 0.457
c
T 216.7 K
c 295
M
V
c
m
s
c 1062
M 1.13
km
hr
and we have
V 1200
km
hr
Problem 12.27
Given:
Submarine sonar
Find:
Separation between submarines
[Difficulty: 2]
Solution:
Basic equation:
Given (and Table A.2) data
For the seawater
c
Ev
ρ
Δt 3.25 s
c
Ev
SG ρw
SG 1.025
c 1537
Hence the distance sound travels in time Δt is
L c Δt
The distance between submarines is half of this
x
L
2
Ev 2.42
m
s
L 5 km
x 2.5 km
GN
2
m
kg
ρw 1000
3
m
Problem 12.26
Given:
Hunting dolphin
Find:
Time delay before it hears prey at 1/2 mile
[Difficulty: 2]
Solution:
Basic equation:
Given (and Table A.2) data
c
Ev
ρ
3
L 0.5 mi 2.64 10 ft
SG 1.025
5
Ev 3.20 10 psi
ρw 1.94
slug
ft
For the seawater
c
Ev
SG ρw
Hence the time for sound to travel distance L is
c 4814
Δt
L
c
ft
s
Δt 0.548 s
Δt 548 ms
3
Problem 12.25
Given:
Device for determining bulk modulus
Find:
Time delay; Bulk modulus of new material
[Difficulty: 2]
Solution:
Basic equation:
Hence for given data
c
Ev
ρ
Ev 200
GN
2
L 1 m
and for steel
SG 7.83
kg
ρw 1000
3
m
Δt 0.198 ms
Δt 198 μs
m
For the steel
c
Ev
SG ρw
Δt
Hence the time to travel distance L is
For the unknown material
M 0.25 kg
The density is then
ρ
M
2
L
The speed of sound in it is
Hence th bulk modulus is
c
c 5054
s
4
L
Δt 1.98 10
c
D 1 cm
ρ 3183
π D
Δt 0.5 ms
kg
3
m
4
L
c 2000
Δt
Ev ρ c
m
2
m
s
Ev 12.7
GN
2
m
s
Problem 12.24
[Difficulty: 3]
Given:
Sound wave
Find:
Estimate of change in density, temperature, and velocity after sound wave passes
Solution:
Basic
equations:
p ρ R T
Ev
dp
dρ
ρ
Assumptions: 1) Ideal gas 2) Constant specific heats 3) Infinitesimal changes
dp
To find the bulk modulus we need
in
dρ
Ev
dp
dρ
ρ
dp
dρ
ρ
p
For rapid compression (isentropic)
k
const
and so
ρ
Hence
Ev ρ k
p
ρ
dp
dρ
k
p
ρ
Ev k p
For gradual compression (isothermal) we can use the ideal gas equation
Hence
Ev ρ ( R T) p
p ρ R T
so
dp dρ R T
Ev p
We conclude that the "stiffness" (Ev) of air is equal to kp when rapidly compressed and p when gradually compressed. To give an
idea of values:
For water
Ev 2.24 GPa
For air ( k 1.4) at p 101 kPa
Rapid compression
Ev k p
Gradual compression Ev p
Ev 141 kPa
Ev 101 kPa
Problem 12.23
Given:
Five different gases at specified temperature
Find:
Sound speeds for each gas at that temperature
Solution:
Basic equation: c
k R T
The data provided, or available in the Appendices, is:
k H2 1.41
[Difficulty: 3]
RH2 4124
J
kg K
J
k CH4 1.31 RCH4 518.3
kg K
T ( 20 273 ) K
J
k He 1.66
RHe 2077
k N2 1.40
RN2 296.8
kg K
J
kg K
J
k CO2 1.29 RCO2 188.9
kg K
cH2
k H2 RH2 T
cH2 1305
m
cHe
k He RHe T
cHe 1005
m
cCH4
cN2
cCO2
k CH4 RCH4 T
k N2 RN2 T
k CO2 RCO2 T
s
s
cCH4 446
cN2 349
m
s
m
s
cCO2 267
m
s
Problem 12.22
[Difficulty: 3]
Given:
Sound wave
Find:
Estimate of change in density, temperature, and velocity after sound wave passes
Solution:
Basic equation:
p ρ R T
T2
p2
Δs cp ln
R ln
T1
p1
du cv dT
dh cp dT
Assumptions: 1) Ideal gas 2) Constant specific heats 3) Isentropic process 4) infinitesimal changes
Given or available data
T1 ( 20 273 ) K
c
k R T1
For small changes, from Section 11-2
p 1 100 kPa
c 343
dp 20 Pa
The air density is ρ1
R T1
Then
dVx
1
ρ1 c
2
dp c dρ
ρ1 1.19
so
dρ
dp
dρ 1.70 10
2
4 kg
Dividing by the ideal gas equation we find
m
kg
3
m
dVx 0.049
dp
p
dp
dρ
dT T1
p 1 ρ1
dρ
ρ
kg K
a very small change!
3
m
dp
J
s
This is the velocity of the air after the sound wave!
s
For the change in temperature we start with the ideal gas equation p ρ R T
Hence
R 286.9
m
c
p1
k 1.4
and differentiate dp dρ R T ρ R dT
dT
T
dT 0.017 K
dT 0.030 Δ°F
a very small change!
Problem 12.21
Given:
Data on flow rate and balloon properties
Find:
"Volumetric ratio" over time
[Difficulty: 3]
Solution:
The given or available data are:
Rair 53.3
ft lbf
Tatm 519 R
lbm R
p atm 14.7 psi
Standard air density
p atm
lbm
ρair
0.0765
Rair Tatm
3
ft
Mass flow rate
M rate Vrate ρair 1.275 10
From a force balance on each hemisphere
p patm π r2 σ 2 π r
Hence
p p atm
2 σ
The instantaneous volume is
4
3
Vball π r
3
The instantaneous mass is
M ball Vball ρ
The time to fill to radius r from r = 5 in is
t
4 lbm
s
2
or
p p atm 8 π k r
Rair Tair
M ball ( r) M ball ( r 5in)
M rate
ΔV Vball ( t Δt) Vball ( t)
The results, calculated using Excel, are shown on the next page:
3
σ k A k 4 π r
p
ρ
ft
Vrate 0.1
min
where
r
Density in balloon
3
lbf
ft
Basic equation:
The volume change between time steps t is
k 200
r (in)
p (psi)
ρ (lb/ft3 )
V ball (ft )
M ball (lb)
t (s)
ΔV/V rate
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
29.2
30.0
30.7
31.4
32.2
32.9
33.6
34.3
35.1
0.152
0.156
0.160
0.164
0.167
0.171
0.175
0.179
0.183
0.303
0.351
0.403
0.461
0.524
0.592
0.666
0.746
0.831
0.0461
0.0547
0.0645
0.0754
0.0876
0.101
0.116
0.133
0.152
0.00
67.4
144
229
325
433
551
683
828
0.00
42.5%
41.3%
40.2%
39.2%
38.2%
37.3%
36.4%
35.5%
3
Volume Increase of Balloon
as Percentage of Supplied Volume
44%
ΔV/V flow
42%
40%
38%
36%
34%
0
250
500
t (s)
750
1000
Problem 12.20
[Difficulty: 4]
Problem 12.19
[Difficulty: 3]
Given:
Data on performance degradation of turbine
Find:
Time necessary for power output to drop to 950 kW
Solution:
The data provided, or available in the Appendices, is:
3
p 1 10 bar 1 10 kPa
T1 1400 K
ηinitial 80 %
p 2 1 bar 100 kPa
cp 1004
J
kg K
Pinitial 1 MW
Pfinal 950 kW
Rgas 287
J
kg K
If the turbine expansion were isentropic, the actual output would be:
So when the power output drops to 950 kW, the new efficiency is:
Since the efficiency drops by 1% per year, the time elapsed is:
k 1.4
Pinitial
Pideal
1.25 MW
ηinitial
ηfinal
Pfinal
Pideal
Δt 4 yr
76 %
Problem 12.18
[Difficulty: 3]
Given:
Data on flow through compressor
Find:
Efficiency at which power required is 30 MW; plot required efficiency and exit temperature as functions of efficiency
Solution:
The data provided, or available in the Appendices, is:
R 518.3
J
kg K
cp 2190
J
cv cp R
kg K
cv 1672
J
k
kg K
T1 ( 13 273 ) K
p 1 0.5 MPa 101 kPa
m
V1 32
s
p 2 8 MPa 101 kPa
Wcomp 30 MW
D 0.6 m
cp
cv
k 1.31
The governing equation is the first law of thermodynamics for the compressor
2
2
V2
V1
M flow h 2
h 2 Wcomp
2 1
2
2
V2 V1
Wcomp M flow cp T2 T1
2
or
We need to find the mass flow rate and the temperature and velocity at the exit
p1 π 2
M flow ρ1 A1 V1
D V1
R T1 4
The exit velocity is then given by
M flow
p2
R T2
M flow
π
p1
R T1
π
2
D V1
4
2
D V2
4
V2
M flow 36.7
kg
s
4 M flow R T2
2
(1)
π p 2 D
The exit velocity cannot be computed until the exit temperature is determined!
Using Eq. 1 in the first law
Wcomp M flow cp T2 T1
2
4 Mflow R T2
V12
π p D2
2
2
In this complicated expression the only unknown is T2, the exit temperature. The equation is a quadratic, so
is solvable explicitly for T2, but instead we use Excel's Goal Seek to find the solution (the second solution
is mathematically correct but physically unrealistic - a very large negative absolute temperature). The exit
temperature is
T2 660 K
1 k
If the compressor was ideal (isentropic), the exit temperature would be given by
T p
k
constant
(12.12b)
1 k
Hence
k
p1
T2s T1
p2
T2s 529 K
For a compressor efficiency η, we have
η
h 2s h 1
η
or
h2 h1
with
V2
T2s T1
η
2
2
V2 V1
Wcomp M flow cp T2 T1
2
and
2
η 65.1 %
T2 T1
T2 T1
To plot the exit temperature and power as a function of efficiency we use
4 M flow R T2
T2s T1
π p 2 D
The dependencies of T2 and Wcomp on efficiency are plotted in Excel and shown here:
Required Compressor Power
as a Function of Efficiency
140
W comp (MW)
120
100
80
60
40
20
0
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
80%
90%
100%
η
Gas Exit Temperature
as a Function of Efficiency
2500
T (K)
2000
1500
1000
500
0
0%
10%
20%
30%
40%
50%
η
60%
70%
100%
Problem 12.17
[Difficulty: 3]
Problem 12.16
[Difficulty: 2]
Problem 12.15
[Difficulty: 3]
Problem 12.14
[Difficulty: 3]
Given:
Air is compressed from standard conditions to fill a tank
Find:
(a) Final temperature of air if tank is filled adiabatically and reversibly
(b) Heat lost if tank is filled isothermally
(c) Which process results in a greater mass of air in the tank
Solution:
The data provided, or available in the Appendices, is:
cp 1004
J
kg K
3
V 1 m
R 287
J
cv cp R
kg K
p 1 0.1 MPa
cv 717
T1 ( 20 273) K
J
kg K
k
cp
k 1.4
cv
p 2 2 MPa
k 1
p2
T2s T1
p1
Adiabatic, reversible process is isentropic:
For the isothermal process, we look at the first law:
The work is equal to: w p dv
v2
From Boyle's law: p 1 v 1 p 2 v 2
w 252
kJ
kg
v1
R T1
v
k
T2s 689.9 K
Δu q w cv ΔT
1
p1
p2
substituting this into the above equation:
kJ
kg
R T2s
(The negative sign indicates heat loss)
p V M R T
M
p2 V
R T1
23.8 kg
Q 5.99 10 kJ
The mass in the tank after compression isothermally is:
p2 V
p1
w R T1 ln
p2
3
Q M q
M
qw
v
The mass of the air can be calculated from the ideal gas equation of state:
For the isentropic compression:
Δu 0 and
2 1
v2
dv R T1 ln
dv R T1
v
v1
v
Therefore the heat transfer is q w 252
So the actual heat loss is equal to:
but ΔT = 0 so:
10.1 kg
M t 23.8 kg
Therefore the isothermal compression results in
more mass in the tank.
Problem 12.13
Given:
Data on diesel cycle
Find:
Plot of pV and Ts diagrams; efficiency
[Difficulty: 4]
Solution:
The data provided, or available in the Appendices, is:
cp 1004
J
kg K
R 287
J
cv cp R
kg K
kg K
k
cp
T1 ( 20 273 ) K
T3 ( 3000 273 ) K
V1 500 cc
V1
V2
12.5
V2 40 cc
V4 1.75 V3
V4 70 cc
M
V3 V2
p 1 V1
M 5.95 10
R T1
T v
For process 1-2 we have isentropic behavior
V1
T2 T1
V2
k 1
V1
p2 p1
V2
T2 805 K
V5 V1
4
k
constant (12.12a)
k 1
k 1.4
cv
p 1 100 kPa
Computed results:
Hence
J
cv 717
p v constant
kg
(12.12c)
k
p 2 3435 kPa
k
The process from 1 -2 is
V1
p ( V) p 1
V
The work is
p 1 V1 p 2 V2
2
W12
p ( V) dV
k1
V
and
s constant
V
1
For process 2 - 3 we have constant volume
Hence
T3
p3 p2
T2
V3 V2
V3 40 cc
p 3 13963 kPa
W12 218 J
Q12 0 J
(Isentropic)
The process from 2 -3 is
V V2 constant
T
Δs cv ln
T2
and
W23 0 J
(From Eq. 12.11a)
Q23 M Δu M cv dT
p4 p3
For process 3 - 4 we have constant pressure
The process from 3 - 4 is
p p 3 constant
Q23 M cv T3 T2
Q23 1052 J
p 4 13963 kPa
V4
T4 T3
V3
and
T
Δs cp ln
T3
T4 5728 K
(From Eq. 12.11b)
W34 p 3 V4 V3
W34 419 J
V4
T5 T4
V5
For process 4 - 5 we again have isentropic behavior
Q34 M cp T4 T3
Q34 1465 J
k 1
T5 2607 K
k
Hence
V4
p5 p4
V5
The process from 4 - 5 is
V4
p ( V) p 4
V
The work is
W45
p 5 890 kPa
k
and
p 4 V4 p 5 V5
s constant
W45 1330 J
k1
Q45 0 J
For process 5-1 we again have constant volume
The process from 5 -1 is
V V5 constant
and
T
Δs cv ln
T5
(From Eq. 12.11a)
Q51 M cv T1 T5
Q51 987 J
The net work is
Wnet W12 W23 W34 W45 W51
The heat added is
Qadded Q23 Q34
The efficiency is
η
Wnet
Qadded
Qadded 2517 J
η 60.8 %
W51 0 J
Wnet 1531 J
This is consistent with the expression from thermodynamics for the diesel efficiency
r k1
c
ηdiesel 1
k 1 k rc 1
r
1
where r is the compression ratio
r
V1
V2
V4
rc
V3
and rc is the cutoff ratio
r 12.5
rc 1.75
ηdiesel 58.8 %
The plots of the cycle in pV and Ts space, generated using Excel, are shown here:
p - V Diagram for Diesel Cycle
16000
p (kPa)
14000
12000
10000
8000
6000
4000
2000
0
0
100
200
300
400
500
V (cc)
T - s Diagram for Diesel Cycle
7000
T (K)
6000
5000
4000
3000
2000
1000
0
0
500
1000
s (J/kg.K)
1500
2000
Problem 12.12
Given:
Data on Otto cycle
Find:
Plot of pV and Ts diagrams; efficiency
[Difficulty: 4]
Solution:
The data provided, or available in the Appendices, is:
cp 1004
J
kg K
p 1 100 kPa
R 287
J
kg K
T1 ( 20 273 ) K
J
cv cp R
cv 717
T3 ( 2750 273 ) K
V1 500 cc
k
kg K
cp
cv
V1
V2
8.5
k 1.4
V2 58.8 cc
V4 V1
Computed results:
M
For process 1-2 we have isentropic behavior
T v
Hence
V1
T2 T1
V2
R T1
k 1
V1
p2 p1
V2
kg
(12.12 a and 12.12b)
k
p 2 2002 kPa
k
V2
p 1 V1 p 2 V2
W12
p ( V) dV
V
k1
1
T3
p3 p2
T2
p v constant
T2 690 K
V3 V2
4
k
constant
V1
p ( V) p 1
V
For process 2 - 3 we have constant volume
Hence
M 5.95 10
k 1
The process from 1 -2 is
The work is
p 1 V1
and
s constant
W12 169 J
V3 58.8 cc
p 3 8770 kPa
Q12 0 J
(Isentropic)
V V2 constant
The process from 2 -3 is
and
T
Δs cv ln
T2
W23 0 J
(From 12.11a)
Q23 M Δu M cv dT
Q23 M cv T3 T2
Q23 995 J
For process 3 - 4 we again have isentropic behavior
Hence
V3
T4 T3
V4
k 1
V3
p4 p3
V4
T4 1284 K
The process from 3 - 4 is
V3
p ( V) p 3
V
The work is
W34
k
p 4 438 kPa
k
and
p 3 V3 p 4 V4
k1
s constant
W34 742 J
Q34 0 J
T
Δs cv ln
T4
W41 0 J
For process 4-1 we again have constant volume
The process from 4 -1 is
V V4 constant
and
(From 12.11a)
Q41 M cv T1 T4
The net work is
Wnet W12 W23 W34 W41
The efficiency is
η
Wnet
Q23
This is consistent with the expression for the Otto efficiency
Q41 422 J
Wnet 572 J
η 57.5 %
ηOtto 1
1
k 1
r
where r is the compression ratio
r
V1
V2
r 8.5
ηOtto 57.5 %
Plots of the cycle in pV and Ts space, generated using Excel, are shown on the next page.
p - V Diagram for Otto Cycle
10000
p (kPa)
8000
6000
4000
2000
0
0
100
200
300
400
500
1000
1250
V (cc)
T - s Diagram for Otto Cycle
3500
T (K)
3000
2500
2000
1500
1000
500
0
0
250
500
750
s (J/kg.K)
Problem 12.11
[Difficulty: 3]
Given:
Air in a piston-cylinder
Find:
Heat to raise temperature to 1200oC at a) constant pressure and b) constant volume
Solution:
The data provided, or available in the Appendices, is:
T1 ( 100 273 ) K
T2 ( 1200 273 ) K
a) For a constant pressure process we start with
R 287
J
kg K
J
kg K
cv cp R
cv 717
J
kg K
T ds dh v dp
dh
dT
cp
T
T
Hence, for p = const.
ds
But
δq T ds
Hence
δq cp dT
b) For a constant volume process we start
cp 1004
q c p dT
q c p T2 T1
q c v dT
q c v T2 T1
q 1104
q 789
kJ
kg
T ds du p dv
du
dT
cv
T
T
Hence, for v = const.
ds
But
δq T ds
Hence
δq cv dT
Heating to a higher temperature at constant pressure requires more heat than at constant volume: some of the
heat is used to do work in expanding the gas; hence for constant pressure less of the heat is available for
raising the temperature.
From the first law:
Constant pressure:
Constant volume: q Δu
q Δu w
The two processes can be plotted using Eqs. 11.11b and 11.11a, simplified for the case of constant pressure
and constant volume.
a) For constant pressure
T2
p2
s2 s1 cp ln R ln
T1
p1
so
T2
Δs cp ln
T1
b) For constant volume
T2
v2
s2 s1 cv ln R ln
T1
v1
so
T2
Δs cv ln
T1
The processes are plotted in Excel and shown on the next page
kJ
kg
T-s Diagram for Constant Pressure and Constant Volume
Processes
1500
T (K)
1250
1000
750
500
a) Constant Pressure
250
b) Constant Volume
0
0
250
500
750
Δs (J/kg.K)
1000
1250
1500
Problem 12.10
[Difficulty: 2]
Given:
Cooling of air in a tank
Find:
Change in entropy, internal energy, and enthalpy
Solution:
Basic equation:
p ρ R T
T2
p2
Δs cp ln R ln
T1
p1
Δu cv ΔT
Δh cp ΔT
Assumptions: 1) Ideal gas 2) Constant specific heats
Given or available data M 5 kg
T1 ( 250 273) K
cp 1004
J
cv 717.4
kg K
For a constant volume process the ideal gas equation gives
J
p2
T2
p2
T1
T2
p2
Δs cp ln R ln
T1
p1
Δs 346
Δu cv T2 T1
cp
k 1.4
cv
T2
p 1 3 MPa
p
T1 1
R cp cv
R 287
p 2 1.85 MPa
J
kg K
Δu 143
Δh cp T2 T1
Δh 201
ΔS M Δs
ΔS 1729
ΔU M Δu
ΔU 717 kJ
ΔH M Δh
ΔH 1004 kJ
kJ
kg
kJ
kg
J
K
Here is a plot of the T-s diagram:
T-s Diagram for Constant Volume Cooling
750
1
T (K)
Total amounts are
k
kg K
p1
Then
T2 ( 50 273) K
500
250
0
-400
2
-350
-300
-250
-200
Δs (J/kg.K)
-150
-100
-50
0
J
kg K
Problem 12.9
[Difficulty: 2]
Given:
Supercharger
Find:
Pressure, temperature and flow rate at exit; power drawn
Solution:
Basic equation:
T2
p2
Δs cp ln R ln
T1
p1
p ρ Rair T
Δh q w
Δh cp ΔT
(First law - open system)
Assumptions: 1) Ideal gas 2) Adiabatic
In an ideal process (reversible and adiabatic) the first law becomes
Δh w
or for an ideal gas
wideal cp ΔT
k 1
For an isentropic process
T2
p2
Δs 0 cp ln
R ln
T1
p1
The given or available data is T1 ( 70 460 ) R
p2
T1
p1
T2
or
p 1 14.7 psi
p 2 ( 200 14.7) psi
k 1.4
cp 0.2399
T2 1140 R
T2 681 °F
p 2 215 psi
ρ1
Q2 Q1
ρ2
p 1 T2
Q2 Q1
p 2 T1
ft
Q2 0.0737
s
3
ft
Q1 0.5
s
k
Btu
lbm R
η 70 %
Rair 53.33
ft lbf
lbm R
k 1
Hence
p2
T2
p1
k
T1
We also have
mrate ρ1 Q1 ρ2 Q2
For the power we use
Pideal mrate wideal ρ1 Q1 cp Δ T
p1
From the ideal gas equation ρ1
Rair T1
Hence
ρ1 0.00233
ft
Pideal ρ1 Q1 cp T2 T1
The actual power needed is Pactual
Pideal
η
slug
or
3
3
ρ1 0.0749
lbm
ft
Pideal 5.78 kW
Pactual 8.26 kW
A supercharger is a pump that forces air into an engine, but generally refers to a pump that is driven directly by
the engine, as opposed to a turbocharger that is driven by the pressure of the exhaust gases.
3
Problem 12.8
Given:
Test chamber with two chambers
Find:
Pressure and temperature after expansion
[Difficulty: 2]
Solution:
Basic equation:
p ρ R T
Δu q w
(First law - closed system)
Δu cv ΔT
Δu 0
or for an Ideal gas
Vol 2 2 Vol 1
so
1
ρ2 ρ1
2
so
p2
Assumptions: 1) Ideal gas 2) Adiabatic 3) No work
For no work and adiabatic the first law becomes
We also have
From the ideal gas equation
M ρ Vol const
p2
p1
Hence
Note that
and
ρ2 T2
1
ρ1 T1
2
T2 20 °F
p2
200 kPa
2
T2
p2
1
Δs cp ln
R ln R ln 0.693 R
2
T1
p1
ΔT 0
T2 T1
1
2
p1
p 2 100 kPa
so entropy increases (irreversible adiabatic)
Problem 12.7
Given:
Data on turbine inlet and exhaust
Find:
Whether or not the vendor claim is feasible
[Difficulty: 2]
k 1
Solution:
Basic equations:
T2
p2
Δs cp ln
R ln
T1
p1
η
h1 h2
h 1 h 2s
p2
T1
p2
T2
Δh cp ΔT
k
when s = constant
T1 T2
T1 T2s
The data provided, or available in the Appendices, is:
3
p 1 10 bar 1 10 kPa
T1 1400 K
η 80 %
P 1 MW
p 2 1 bar 100 kPa
cp 1004
J
kg K
Rgas 287
J
kg K
k 1.4
k 1
p2
If the expansion were isentropic, the exit temperature would be: T2s T1
p1
Since the turbine is not isentropic, the final temperature is higher:
Then
kJ
Δh cp T1 T2 542.058
kg
k
725.126 K
T2 T1 η T1 T2s 860.101 K
T2
p2
Δs cp ln
Rgas ln
T1
p1
The mass flow rate is:
Δs 171.7157
m
P
Δh
J
kg K
1.845
kg
s
Problem 12.6
Given:
Adiabatic air compressor
Find:
Lowest delivery temperature; Sketch the process on a Ts diagram
[Difficulty: 2]
Solution:
Basic equation:
T2
p2
Δs cp ln
R ln
T1
p1
1 k
The lowest temperature implies an ideal (reversible) process; it is also adiabatic, so Δs = 0, and
The data provided, or available in the Appendices, is:
p 1 101 kPa p 2 ( 500 101 ) kPa
1 k
Hence
p1
T2 T1
p2
Temperature T
T2 864 R
p2
2
The process is
k
p1
1
Entropy s
p1
T2 T1
p2
T1 288.2 K
k
k 1.4
Problem 12.5
Given:
Air before and after expansion; process
Find:
Final temperature and change in entropy
[Difficulty: 2]
Solution:
Basic equations:
T2
p2
Δs cp ln
R ln
T1
p1
p V m R T
The data provided, or available in the Appendices, is:
p 1 50 psi
T1 660 R
p 2 1 atm 14.696 psi
cp 0.2399
From the process given:
p 1 V1
Btu
Rgas 53.33
lb R
1.3
p 2 V2
1.3
ft lbf
lb R
0.0685
From the ideal gas equation of state:
Btu
lb R
p 2 V2
p 1 V1
T2
V1
T1
V2
p 2 T1
p 1 T2
1
p2
When we combine these two equations we get:
p1
1
So the final temperature is:
Then
p1
T2 T1
p2
V1
V
2
1.3
p 2 T1
p 1 T2
1
1.3
T2
p2
Δs cp ln
Rgas ln
T1
p1
T2 497.5 R
Δs 0.0161
Btu
lb R
1.3
Solving for temperature ratio:
T1
T2
p2
p
1
1.3
1
Problem 12.4
Given:
Data on turbine inlet and exhaust
Find:
Whether or not the vendor claim is feasible
[Difficulty: 2]
Solution:
Basic equation:
T2
p2
Δs cp ln
R ln
T1
p1
The data provided, or available in the Appendices, is:
T1 ( 2200 460 ) R
T1 1.478 10 K
p 2 1 atm 14.696 psi
T2 ( 850 460 ) R
T2 727.778 K
BTU
Rgas 53.33
lb R
T2
p2
Δs cp ln
Rgas ln
T1
p1
Δs 0.0121
An example of this type of process is plotted in green on the graph.
Also plotted are an isentropic process (blue - 1-2s) and one with an
increase in entropy (red: 1-2i). All three processes expand to the same
pressure. The constant pressure curve is drawn in purple.The second
law of thermodynamics states that, for an adiabatic process
Δs 0
or for all real processes
Δs 0
ft lbf
lb R
0.0685
BTU
lb R
BTU
lb R
1
Temperature T
cp 0.2399
Then
3
p 1 10 atm 146.959 psi
2i
2
2s
Hence the process is NOT feasible!
Entropy s
Problem 12.3
Given:
Data on an air compressor
Find:
Whether or not the vendor claim is feasible
[Difficulty: 2]
Solution:
Basic equation:
T2
p2
Δs cp ln
R ln
T1
p1
The data provided, or available in the Appendices, is:
p 1 14.7 psi
T1 ( 50 460 ) R
p 2 ( 150 14.7) psi
T2 ( 200 460 ) R
Then
BTU
lb R
T2
p2
Δs cp ln
Rgas ln
T1
p1
Rgas 53.33
Δs 0.1037
ft lbf
lb R
0.0685
or for all real processes
Δs 0
lb R
lb R
We have plotted the actual process in red (1-2) on this temperature-entropy
diagram, and the ideal compression (isentropic) in blue (1-2s). The line of constant
pressure equal to 150 psig is shown in green. However, can this process actually
occur? The second law of thermodynamics states that, for an adiabatic process
Δs 0
BTU
BTU
2s
Temperature T
cp 0.2399
2
1
Hence the process is NOT feasible!
Entropy s
Problem 12.2
[Difficulty: 2]
Problem 12.1
Given:
Air flow through a filter
Find:
Change in p, T and ρ
[Difficulty: 2]
Solution:
Basic equations:
h 2 h 1 c p T2 T1
p ρ R T
Assumptions: 1) Ideal gas 2) Throttling process
In a throttling process enthalpy is constant. Hence
h2 h1 0
s
o
T2 T1 0
or
T constant
The filter acts as a resistance through which there is a pressure drop (otherwise there would be no flow. Hence p 2 p 1
From the ideal gas equation
p1
p2
ρ1 T1
ρ2 T2
The governing equation for entropy is
Hence
p2
p1
Δs R ln
so
T1 p 2
p2
ρ2 ρ1
ρ1
T2 p 1
p1
T2
p2
Δs cp ln
R ln
T1
p1
p2
and
1
p1
Entropy increases because throttling is an irreversible adiabatic process
Hence
ρ2 ρ1
so
Δs 0
Problem 11.75
Given:
Data on V-notch weir
Find:
Weir coefficient
[Difficulty: 1]
Solution:
5
Basic equation:
Q = Cw⋅ H
2
where
H = 180 ⋅ mm
Note that this is an "engineering" equation in which we ignore units!
Cw =
Q
5
H
2
Cw = 1.45
Q = 20⋅
L
s
Problem 11.74
Given:
Data on V-notch weir
Find:
Discharge
[Difficulty: 1]
Solution:
5
Basic equation:
Q = Cw⋅ H
2
where
H = 1.5⋅ ft
Cw = 2.50
Note that this is an "engineering" equation in which we ignore units!
5
Q = Cw⋅ H
2
Q = 6.89
ft
3
s
for
θ = 90⋅ deg
Problem 11.73
Given:
Data on V-notch weir
Find:
Flow head
[Difficulty: 1]
Solution:
5
Basic equation:
8
θ
2
Q = Cd ⋅ ⋅ 2⋅ g ⋅ tan ⎛⎜ ⎞ ⋅ H
15
⎝2⎠
where
Cd = 0.58
2
H =
5
Q
⎛
⎞
⎜
8
θ
⎜ Cd ⋅ ⋅ 2⋅ g ⋅ tan ⎛⎜ ⎞
15
⎝
⎝2⎠⎠
H = 0.514m
θ = 60⋅ deg
Q = 150⋅
L
s
Problem 11.72
Given:
Data on rectangular, sharp-crested weir
Find:
Required weir height
[Difficulty: 3]
Solution:
3
Basic equations:
2
2
Q = Cd ⋅ ⋅ 2 ⋅ g ⋅ b'⋅ H
3
where
Given data:
b = 1.5⋅ m
Q = 0.5⋅
Cd = 0.62 and
b' = b − 0.1⋅ n ⋅ H
with
n = 2
3
Hence we find
m
s
3
3
2
2
2
2
Q = Cd ⋅ ⋅ 2 ⋅ g ⋅ b'⋅ H = Cd ⋅ ⋅ 2 ⋅ g ⋅ ( b − 0.1⋅ n ⋅ H) ⋅ H
3
3
3
Rearranging
( b − 0.1⋅ n ⋅ H) ⋅ H
2
=
3⋅ Q
2 ⋅ 2 ⋅ g⋅ Cd
This is a nonlinear implicit equation for H and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
5
The right side evaluates to
For
H = 1⋅ m
3⋅ Q
2 ⋅ 2 ⋅ g⋅ Cd
= 0.273 ⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
3
2
5
= 1.30⋅ m
2
3
For
H = 0.3⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
H = 0.34⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
= 0.237 ⋅ m
H = 0.331 ⋅ m
But from the figure
( b − 0.1⋅ n ⋅ H) ⋅ H
H + P = 2.5⋅ m
2
H = 0.5⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
= 0.284 ⋅ m
2
For
H = 0.35⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
= 0.495 ⋅ m
2
For
H = 0.33⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
H = 0.331 m
P = 2.5⋅ m − H
P = 2.17 m
2
2
5
= 0.296 ⋅ m
3
5
= 0.273 ⋅ m
2
5
3
5
3
For
For
5
3
For
3
2
5
= 0.272 ⋅ m
2
Problem 11.71
Given:
Data on rectangular, sharp-crested weir
Find:
Discharge
[Difficulty: 1]
Solution:
3
Basic equation:
Q = Cw⋅ b ⋅ H
2
where
Cw = 3.33 and
b = 8 ⋅ ft
Note that this is an "engineering" equation, to be used without units!
3
Q = Cw⋅ b ⋅ H
2
Q = 26.6
ft
3
s
P = 2 ⋅ ft
H = 1 ⋅ ft
Problem 11.70
Given:
Data on rectangular, sharp-crested weir
Find:
Required weir height
[Difficulty: 3]
Solution:
3
Basic equations:
2
2
Q = Cd ⋅ ⋅ 2 ⋅ g ⋅ b'⋅ H
3
where
Given data:
b = 1.6⋅ m
Q = 0.5⋅
Cd = 0.62
and
b' = b − 0.1⋅ n ⋅ H
n = 2
with
3
Hence we find
m
s
3
3
2
2
2
2
Q = Cd ⋅ ⋅ 2 ⋅ g ⋅ b'⋅ H = Cd ⋅ ⋅ 2 ⋅ g ⋅ ( b − 0.1⋅ n ⋅ H) ⋅ H
3
3
3
Rearranging
( b − 0.1⋅ n ⋅ H) ⋅ H
2
=
3⋅ Q
2 ⋅ 2 ⋅ g⋅ Cd
This is a nonlinear implicit equation for H and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
5
The right side evaluates to
3⋅ Q
2 ⋅ 2 ⋅ g⋅ Cd
= 0.273 ⋅ m
2
3
For
H = 1⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
5
= 1.40⋅ m
2
3
For
H = 0.3⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
H = 0.31⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
= 0.253 ⋅ m
H = 0.316 ⋅ m
But from the figure
( b − 0.1⋅ n ⋅ H) ⋅ H
H + P = 2.5⋅ m
2
H = 0.5⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
= 0.265 ⋅ m
2
For
H = 0.35⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
2
= 0.530 ⋅ m
2
For
H = 0.315 ⋅ m
( b − 0.1⋅ n ⋅ H) ⋅ H
H = 0.316 m
P = 2.5⋅ m − H
P = 2.18 m
2
2
5
= 0.317 ⋅ m
3
5
= 0.273 ⋅ m
2
5
3
5
3
For
For
5
3
For
3
2
5
= 0.272 ⋅ m
2
Problem 11.69
Given:
Data on broad-crested wier
Find:
Maximum flow rate/width
Solution:
3
Basic equation:
Q = Cw⋅ b ⋅ H
Available data
H = 1 ⋅ ft
2
P = 8 ⋅ ft − 1 ⋅ ft
3
ft
3
Then
[Difficulty: 1]
Q
b
= q = Cw⋅ H
2
= 3.4⋅
s
ft
P = 7 ⋅ ft
Cw = 3.4
Problem 11.68
Given:
Data on optimum rectangular channel
Find:
Channel width and slope
[Difficulty: 2]
Solution:
Basic equations:
Q=
1.49
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
b = 2⋅ yn
and from Table 11.3, for optimum geometry
Note that the Q equation is an "engineering" equation, to be used without units!
Available data
Hence
Q = 100 ⋅
ft
3
n = 0.015
s
A = b⋅ yn = 2⋅ yn
2
Rh =
A
P
=
(Table 11.2)
2⋅ yn
2
yn + 2⋅ yn + yn
=
yn
2
We can write the Froude number in terms of Q
Fr =
V
g⋅ y
=
Q
Q
A⋅ g ⋅ y
=
or
1
2
2⋅ yn ⋅ g⋅ yn
Fr =
2
Q
5
2
2⋅ g⋅ yn
5
1=
Hence for critical flow, Fr = 1 and y n = y c, so
yc =
⎛ Q ⎞
⎜
⎝ 2⋅ g ⎠
or
Q = 2⋅ g⋅ yc
(ft)
and
5
2⋅ g⋅ yc
2
Hence
Q
2
2
5
y c = 2.39
b = 2⋅ yc
2
2
Then
Hence
1
3
Sc =
Using (from Table 11.2)
n⋅ Q
⎞
⎛
⎜
1
8
⎜
⎟
⎜ 1.49⋅ 2 3 ⋅ y 3
c ⎠
⎝
1.49
1
1
2 ⎛ yc ⎞
3
2
2
Q=
⋅ A⋅ Rh ⋅ Sb =
⋅ 2⋅ yc ⋅ ⎜
⋅ Sc
n
n
2
⎝ ⎠
1.49
b = 4.78
or
Q=
1.49⋅ 2
n
3
8
1
3
2
⋅ y c ⋅ Sc
(ft)
2
Sc = 0.00615
n = 0.013
Sc =
n⋅ Q
⎛
⎞
⎜
1
8
⎜
⎟
⎜ 1.49⋅ 2 3 ⋅ y 3
c ⎠
⎝
2
Sc = 0.00462
Problem 11.67
Given:
Data on wide channel
Find:
Critical slope
[Difficulty: 2]
Solution:
Basic equations:
Q=
1.49
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
A = b ⋅y
and
Rh = y
Note that the Q equation is an "engineering" equation, to be used without units!
3
ft
Available data
q = 20⋅
s
ft
From Table 11.2
n = 0.015
For critical flow
y = yc
Vc =
g⋅ yc
2
Q = A⋅ Vc = b ⋅ y c⋅ g ⋅ y c
so
Hence
⎛ Q ⎞
⎜
⎝ b⋅ g ⎠
3
yc =
3
⎛ q ⎞
⎜
⎝ g⎠
y c = 2.316 (ft)
Solving the basic equation for Sc
Sbcrit =
Q=
1.49
Sbcrit =
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
n⋅ Q
⎛
⎞
⎜
2
⎜
⎟
⎜ 1.49⋅ b⋅ y ⋅ y 3
c c ⎠
⎝
⎛ n⋅ q ⎞
⎜
5
⎜
⎟
⎜ 1.49⋅ y 3
c ⎠
⎝
=
1.49
2
1
3
2
⋅ b ⋅ y c⋅ y c ⋅ Sb
n
2
Sbcrit =
⎛ n⋅ q ⎞
⎜
5
⎜
⎟
⎜ 1.49⋅ y 3
c ⎠
⎝
n = 0.013
Note from Table 11.2 that a better roughness is
and then
yc =
or
2
2
Sbcrit = 0.00185
2
Sbcrit = 0.00247
Problem 11.66
Given:
Data on trapezoidal canal
Find:
Critical slope
[Difficulty: 3]
Solution:
Q=
Basic equations:
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
A = y ⋅ b + y ⋅ cot ( α )
and
y ⋅ ( b + y ⋅ cot( α) )
Rh =
b+
2⋅ y
sin( α)
Note that the Q equation is an "engineering" equation, to be used without units!
α = atan⎛⎜
b = 10⋅ ft
Available data
2⎞
α = 63.4⋅ deg
⎝1⎠
Q = 600⋅
ft
3
s
n = 0.015
For brick, a Google search gives
For critical flow
y = yc
Vc =
g⋅ yc
so
Q = A⋅ Vc = y c⋅ b + y c⋅ cot( α) ⋅ g ⋅ y c
(
)
(yc⋅ b + yc⋅ cot( α))⋅
g⋅ yc = Q
Q = 600⋅
with
ft
3
s
This is a nonlinear implicit equation for y c and must be solved numerically. We can use one of a number of numerical root finding
techniques, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with the given depth
For
yc = 5
( ft)
(yc⋅ b + yc⋅ cot( α))⋅
g ⋅ y c = 666
For
y c = 4.5
( ft)
(yc⋅ b + yc⋅ cot( α))⋅
g ⋅ y c = 569
For
y c = 4.7
( ft)
(yc⋅ b + yc⋅ cot( α))⋅
g ⋅ y c = 607
For
y c = 4.67 ( ft)
(yc⋅ b + yc⋅ cot( α))⋅
g ⋅ y c = 601
Hence
y c = 4.67
(ft)
and
Acrit = 49.0
(ft2)
Rhcrit = 2.818
(ft)
Acrit = y c⋅ b + y c⋅ cot( α)
Rhcrit =
(
y c⋅ b + y c⋅ cot( α)
b+
Solving the basic equation for Sc
Q=
1.49
2⋅ yc
)
sin( α)
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Sbcrit =
n⋅ Q
⎛
⎞
⎜
2
⎜
⎟
3
⎜ 1.49⋅ A ⋅ R
crit hcrit ⎠
⎝
2
Sbcrit = 0.00381
Problem 11.65
[Difficulty: 2]
Given:
Data on rectangular channel
Find:
Expressions valid for critical depth at optimum geometry
Solution:
Basic equations:
Q=
1
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
b = 2⋅ yn
and from Table 11.3, for optimum geometry
Note that the Q equation is an "engineering" equation, to be used without units!
Hence
A = b⋅ yn = 2⋅ yn
2
Rh =
A
P
=
2⋅ yn
yn + 2⋅ yn + yn
2
2
Then
1
1
1
3
2
1
1
2 ⎛ yn ⎞
3
2
2
Q = ⋅ A⋅ Rh ⋅ Sb = ⋅ 2 ⋅ y n ⋅ ⎜
⋅ Sb
n
n
⎝2⎠
or
Q=
or
Fr =
2
3
=
8
1
3
2
⋅ y ⋅ Sb
n n
yn
2
We can write the Froude number in terms of Q
V
Fr =
g⋅ y
=
Q
A⋅ g ⋅ y
Q
=
1
2
2⋅ yn ⋅ g⋅ yn
2
Q
5
2
2⋅ g⋅ yn
5
1=
Hence for critical flow, Fr = 1 and y n = y c, so
Q
or
5
2⋅ g⋅ yc
Q = 2⋅ g⋅ yc
5
2
Q = 6.26⋅ y c
2
24.7⋅ n
2
2
To find Sc, equate the expressions for Q and set Sb = Sc
1
Q=
2
3
8
1
3
2
⋅ y ⋅ Sc
n c
5
= 2⋅ g⋅ yc
2
4
or
3
−
2
Sc = 2 ⋅ g ⋅ n ⋅ y c
1
3
Sc =
1
yc
3
Problem 11.64
Given:
Data on rectangular flume
Find:
Optimum geometry
[Difficulty: 2]
Solution:
Basic equations:
Q=
1.49
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
and from Table 11.3, for optimum geometry
b = 2⋅ yn
Note that the Q equation is an "engineering" equation, to be used without units!
Available data
ft
Sb = 10⋅
mile
Sb = 0.00189
A = b⋅ yn = 2⋅ yn
ft
3
s
n = 0.013
For wood (unplaned), a Google seach gives
Hence
Q = 40⋅
2
Rh =
A
P
=
2⋅ y n
2
y n + 2⋅ y n + y n
=
yn
2
2
2
Then
1
3
1
1.49
1.49
2 ⎛ yn ⎞
3
2
2
Q=
⋅ A ⋅ Rh ⋅ Sb =
⋅ 2⋅ y n ⋅ ⎜
⋅ Sb
n
n
⎝ 2⎠
3
Solving for y n
2 ⎞
⎛⎜
⎜ Q⋅ n⋅ 2 3 ⎟
yn = ⎜
1⎟
⎜
⎟
⎜ 4 ⋅ 1.49⋅ Sb 2
⎝
⎠
5
y n = 2.00
(ft)
b = 2y n
b = 4.01
(ft)
Problem 11.63
Given:
Data on rectangular channel and weir
Find:
If a hydraulic jump forms upstream of the weir
[Difficulty: 4]
1
Solution:
Q=
Basic equations:
1
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
3
Note that the Q equation is an "engineering" equation, to be used without units!
For a rectangular channel of width b = 2.45⋅ m and depth y we find from Table 11.1
A = b ⋅ y = 2.45⋅ y
b⋅ y
Rh =
b + 2⋅ y
2.45⋅ y
=
2.45 + 2 ⋅ y
2
Q=
Hence
1
n
3
and also
n = 0.015
2
1
3
⋅ A⋅ Rh ⋅ Sb
2
1
=
0.015
⋅ 2.45⋅ y ⋅ ⎛⎜
2.45⋅ y
y
Q = 5.66⋅
⎞ ⋅ 0.0004 2 = 5.66
⎝ 2.45 + 2⋅ y ⎠
2
( 2.45 + 2 ⋅ y )
5.66⋅ 0.015
=
3
2
2
3
.0004 ⋅ 2.54⋅ 2.54
y
or
1
(Note that we don't use units!)
3
2
( 2.54 + 2 ⋅ y )
= 0.898
3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start with the given depth
5
For
y = 1.52
( m)
y
5
3
2
( 2.54 + 2 ⋅ y )
= 0.639
For
y = 2
( m)
3
y
y = 1.95
( m)
y
( 2.54 + 2 ⋅ y )
= 0.908
3
5
3
2
( 2.54 + 2 ⋅ y )
3
2
5
For
= 0.879
For
y = 1.98
3
( m)
y
3
2
( 2.54 + 2 ⋅ y )
= 0.896
3
1
y = 1.98
(m)
This is the normal depth.
s
1
5
3
m
3
5
Solving for y
Sb = 0.0004
and
We also have the critical depth:
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
3
y c = 0.816 m
Hence the given depth is 1.52 m > y c, but 1.52 m < y n, the normal depth. This implies the flow is subcritical (far enough upstream
it is depth 1.98 m), and that it draws down to 1.52 m as it gets close to the wier. There is no jump.
Problem 11.62
Given:
Rectangular channel flow
Find:
Critical depth
1
Solution:
Basic equations:
[Difficulty: 1]
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
3
Q=
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
For a rectangular channel of width b = 2 ⋅ m and depth y = 1.5⋅ m we find from Table 11.1
2
A = b⋅ y
A = 3.00⋅ m
n = 0.015
Manning's roughness coefficient is
Q=
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Rh =
and
1
Hence
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
3
y c = 0.637 m
b⋅ y
b + 2⋅ y
Sb = 0.0005
3
Q = 3.18⋅
m
s
Rh = 0.600 ⋅ m
Problem 11.61
Given:
Trapezoidal channel
Find:
Geometry for greatest hydraulic efficiency
[Difficulty: 5]
Solution:
From Table 11.1
A = y ⋅ ( b + y ⋅ cot ( α ) )
P=b+
2⋅ y
sin( α)
We need to vary b and y (and then α!) to obtain optimum conditions. These are when the area and perimeter are optimized. Instead
of two independent variables b and y, we eliminate b by doing the following
b =
Taking the derivative w.r.t. y
But at optimum conditions
Hence
∂
∂y
∂
∂y
A
y
P=0
But
∂
∂α
− cot( α) +
2
y
− y ⋅ cot ( α ) +
2⋅ y
sin ( α )
2
cot( α) + 1 =
−2 ⋅ cos( α) = −1
We can now evaluate A from Eq 1
A=
2
sin( α)
)=0
2
2⋅ y
2
2
− y ⋅ cot( α)
sin( α)
−
or
2
2
2
sin( α) + cos( α)
+1=
sin( α)
α = acos⎛⎜
2
1⎞
2⋅ y
2
2
−
3
=
2
(
2
1
sin( α)
1
2
3
⋅y =
2
⎛ 4 − 1 ⎞ ⋅ y2 = 3⋅ y2
⎜
3⎠
⎝ 3
2
But for a trapezoid
A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ⎛⎜ b +
Comparing the two A expressions
A = ⎛⎜ b +
⎝
⎝
1
3
⋅ y⎞ ⋅ y =
⎠
3⋅ y
2
1
3
⋅ y⎞
⎠
we find
b=
)
+ cot( α) + 1 = 0
α = 60 deg
⎝2⎠
− y ⋅ cot( α) =
2
(1)
2 ⋅ cos( α)
sin( α)
cos( α)
Hence
2⋅ y
− y ⋅ −1 − cot( α)
2
sin( α)
(
2
A =0
A=
or
sin( α)
2 ⋅ y ⋅ cos( α)
sin( α)
∂y
2
2
A=−
∂
and
A
0=−
A
P=
and so
2
1 ∂
A
− cot ( α ) +
⋅ A −
2
sin ( α )
y ∂y
y
P=
y
Now we optimize A w.r.t. α
− y ⋅ cot ( α )
⎛ 3 − 1 ⎞ ⋅y = 2 ⋅y
⎜
3
3⎠
⎝
2⋅ y
But the perimeter is
P=b+
In summary we have
α = 60 deg
and
b=
1
3
sin( α)
= b + 2⋅ y⋅
2
3
=b+
4
3
⋅ y = b + 2⋅ b = 3⋅ b
P−
⋅P
so each of the symmetric sides is
1
3
2
⋅P
=
1
3
⋅P
We have proved that the optimum shape is equal side and bottom lengths, with 60 angles i.e., half a hexagon!
Problem 11.60
Given:
Data on trapezoidal channel
Find:
Normal depth
Solution:
Q=
Basic equation:
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
[Difficulty: 3]
Note that this is an "engineering" equation, to be used without units!
b = 20⋅ ft
For the trapezoidal channel we have
⎞
⎝ 1.5 ⎠
α = atan⎛⎜
1
α = 33.7 deg
Q = 1000⋅
ft
3
s
S0 = 0.0002 n = 0.014
A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ( 20 + 1.5⋅ y )
Hence from Table 11.1
2
Hence
Q=
1.49
3
⋅ A⋅ Rh ⋅ Sb
n
2
1
2
Rh =
=
1.49
0.014
⋅ y ⋅ ( 20 + 1.5⋅ y ) ⋅ ⎡⎢
y ⋅ ( 20 + 1.5⋅ y ) ⎤
y ⋅ ( b + y ⋅ cot( α) )
b+
2⋅ y
=
y ⋅ ( 20 + 1.5⋅ y )
20 + 2 ⋅ y ⋅ 3.25
sin( α)
1
3
2
⎥ ⋅ 0.0002 = 1000 (Note that we don't use units!)
⎣ 20 + 2 ⋅ y⋅ 3.25⎦
5
Solving for y
[ ( 20 + 1.5⋅ y ) ⋅ y ]
3
2
( 20 + 2⋅ y⋅
3.25)
= 664
3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
5
For
y = 7.5
( ft)
[ ( 20 + 1.5⋅ y ) ⋅ y ]
3
2
( 20 + 2⋅ y⋅
5
3.25)
= 684
For
y = 7.4
( ft)
[ ( 20 + 1.5⋅ y ) ⋅ y ]
2
( 20 + 2⋅ y⋅
3
3.25)
5
For
y = 7.35
( ft)
[ ( 20 + 1.5⋅ y ) ⋅ y ]
( 20 + 2⋅ y⋅
The solution to three figures is
3.25)
= 667
3
5
3
2
3
= 658
For
y = 7.38
( ft)
[ ( 20 + 1.5⋅ y ) ⋅ y ]
2
( 20 + 2⋅ y⋅
3
y = 7.38
(ft)
3
3.25)
3
= 663
Problem 11.59
Given:
Data on trapezoidal channel
Find:
Geometry for greatest hydraulic efficiency
Solution:
Basic equation:
Q=
1
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
[Difficulty: 5]
Note that this is an "engineering" equation, to be used without units!
For the trapezoidal channel we have
From Table 11.1
α = atan⎛⎜
3
1⎞
α = 26.6⋅ deg
⎝2⎠
A = y ⋅ ( b + y ⋅ cot( α) )
Q = 250⋅
P=b +
m
Sb = 0.001
s
n = 0.020
2⋅ y
sin ( α )
We need to vary b and y to obtain optimum conditions. These are when the area and perimeter are optimized. Instead of two
independent variables b and y, we eliminate b by doing the following
b=
Taking the derivative w.r.t. y
But at optimum conditions
Hence
∂
∂y
∂
∂y
A
y
P=
0=−
Hence
Then
A
y
− y ⋅ cot ( α ) +
2⋅ y
sin ( α )
2
1 ∂
A
− cot ( α ) +
⋅ A −
2
sin ( α )
y ∂y
y
∂
and
A
− cot( α) +
2
2⋅ y
sin( α)
∂y
2
or
sin( α)
A = y ⋅ ( b + y ⋅ cot( α) )
b=
P=
and so
P=0
y
Comparing to
− y ⋅ cot( α)
we find
A=
2⋅ y
2
sin( α)
2
− y ⋅ cot( α)
A = y ⋅ ( b + y ⋅ cot( α) ) =
2⋅ y
2
sin( α)
2
− y ⋅ cot( α)
− 2 ⋅ y ⋅ cot( α)
A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ⎛⎜
2⋅ y
⎝ sin( α)
P=b+
A=0
2⋅ y
sin( α)
=
4⋅ y
sin( α)
− 2 ⋅ y ⋅ cot( α) + y ⋅ cot( α) ⎞ = y ⋅ ⎛⎜
2
⎠
− 2 ⋅ y ⋅ cot( α) = 2 ⋅ y ⋅ ⎛⎜
2
⎝ sin( α)
2
⎝ sin( α)
− cot( α) ⎞
⎠
− cot( α) ⎞
⎠
y ⋅ ⎛⎜
2
and
Rh =
A
P
=
⎝ sin( α)
⎠ = y
2
2
2 ⋅ y ⋅ ⎛⎜
− cot( α) ⎞
⎝ sin( α)
⎠
2
Hence
Q=
1
n
− cot( α) ⎞
2
3
⋅ A⋅ Rh ⋅ Sb
2
1
2
=
1
n
⋅ ⎡⎢y ⋅ ⎛⎜
2
2
⎣
⎝ sin( α)
8
− cot( α) ⎞⎤⎥ ⋅ ⎜
⎠⎦
⎛y⎞
⎝2⎠
1
3
⋅ Sb
2
1
3
2
y ⋅ Sb
2
⎛
⎞
Q= ⎜
− cot( α) ⋅
2
⎝ sin( α)
⎠
n⋅ 2
3
3
Solving for y
Finally
2
⎡
⎤
⎢
⎥
3
2 ⋅ n⋅ Q
⎢
⎥
y =
⎢
1⎥
⎢ 2
⎥
2
− cot( α) ⎞ ⋅ Sb ⎥
⎢ ⎛⎜
⎣ ⎝ sin( α)
⎠
⎦
b =
2⋅ y
sin( α)
− 2 ⋅ y ⋅ cot( α)
8
y = 5.66
(m)
b = 2.67
(m)
Problem 11.58
Given:
Data on trapezoidal channel
Find:
Normal depth and velocity
[Difficulty: 3]
Solution:
Basic equation:
Q=
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
b = 20⋅ ft
For the trapezoidal channel we have
α = atan( 2 )
α = 63.4 deg
A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ⎛⎜ 20 +
Hence from Table 11.2
⎝
1
2
Q = 400 ⋅
⋅ y⎞
⎠
Hence
3
s
Rh =
2
Sb = 0.0016
n = 0.025
y ⋅ ( b + y ⋅ cot( α) )
b+
2⋅ y
y ⋅ ⎛⎜ 20 +
⎝
=
1
2
⋅ y⎞
20 + y ⋅ 5
sin( α)
3
1
⎡ y ⋅ ⎛ 20 + 1 ⋅ y⎞ ⎤
⎜
⎢
⎥
1
1
1
2 ⎠
3
2
2
⎝
Q = ⋅ A⋅ Rh ⋅ Sb =
⋅ y ⋅ ⎛⎜ 20 + ⋅ y⎞ ⋅ ⎢
⎥ ⋅ 0.0016 = 400
n
0.025 ⎝
2 ⎠ ⎣ 20 + y ⋅ 5 ⎦
2
ft
1
(Note that we don't use units!)
5
Solving for y
3
⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤
⎢ ⎜
⎥
2 ⎠⎦
⎣ ⎝
= 250
This is a nonlinear implicit equation for y and must be solved numerically. We
can use one of a number of numerical root finding techniques, such as
Newton's method, or we can use Excel's Solver or Goal Seek, or we can
manually iterate, as below. We start with an arbitrary depth
2
( 20 + y⋅ 5) 3
5
For
y = 5
( ft)
5
3
⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤
⎢ ⎜
⎥
2 ⎠⎦
⎣ ⎝
= 265
For
2
y = 4.9
( ft)
( 20 + y⋅ 5) 3
3
⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤
⎢ ⎜
⎥
2 ⎠⎦
⎣ ⎝
= 256
2
( 20 + y⋅ 5) 3
5
For
y = 4.85
( ft)
⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤
⎢ ⎜
⎥
2 ⎠⎦
⎣ ⎝
2
5
3
= 252
For
y = 4.83
( ft)
( 20 + y⋅ 5) 3
The solution to three figures is y = 4.83⋅ ft
Finally, the normal velocity is V =
Q
A
3
⎡y⋅ ⎛ 20 + 1 ⋅ y⎞⎤
⎢ ⎜
⎥
2 ⎠⎦
⎣ ⎝
= 250
2
( 20 + y⋅ 5) 3
Then
A = ( b + y ⋅ cot( α) ) ⋅ y
V = 3.69⋅
ft
s
A = 108 ⋅ ft
2
⎠
Problem 11.57
[Difficulty: 3]
Given:
Triangular channel
Find:
Proof that wetted perimeter is minimized when sides meet at right angles
Solution:
From Table 11.1
2
A = y ⋅ cot( α)
P=
2⋅ y
sin( α)
y=
We need to vary z to minimize P while keeping A constant, which means that
Hence we eliminate y in the expression for P
For optimizing P
or
dP
dα
=−
2 ⋅ ( A⋅ cos( α) − A⋅ sin( α) ⋅ tan( α) )
sin( 2 ⋅ α) ⋅ A⋅ tan( α)
A⋅ cos( α) − A⋅ sin( α) ⋅ tan( α) = 0
P = 2⋅
A
⋅
A
cot( α)
with A = constant
1
cot( α) sin( α)
=0
1
tan( α)
= tan( α)
tan( α) = 1
α = 45⋅ deg
For α = 45o we find from the figure that we have the case where the sides meet at 90o. Note that we have only proved that this is
a minimum OR maximum of P! It makes sense that it's the minimum, as, for constant A, we get a huge P if we set α to a large
number (almost vertical walls); hence we can't have a maximum value at α = 45o.
Problem 11.56
Given:
Data on semicircular trough
Find:
New depth of flow
Solution:
Q=
Basic equation:
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
[Difficulty: 4]
Note that this is an "engineering" equation, to be used without units!
3
D = 1⋅ m
For the semicircular channel
Sb = 0.01
Q=
Hence
−
Solving for α
1
A=
α
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
2
3
8
2
⋅ ( α − sin( α) ) ⋅ D =
1
8
⋅ ( α − sin( α) )
Rh =
2
=
m
s
n = 0.022
For corrugated steel, a Google search leads to (Table 11.2)
From Table 11.1
Q = 0.5⋅
3
1
4
⋅ ⎛⎜ 1 −
⎝
sin( α) ⎞
α
⎠
⋅D =
1
4
⋅ ⎛⎜ 1 −
⎝
sin( α) ⎞
α
1
1
sin( α) ⎞⎤
1
2
⋅ ⎡⎢ ⋅ ( α − sin( α) )⎤⎥ ⋅ ⎡⎢ ⋅ ⎛⎜ 1 −
⎥ ⋅ 0.01 = 0.5 (Note that we don't use units!)
0.022 ⎣ 8
α ⎠⎦
⎦ ⎣4 ⎝
1
5
⋅ ( α − sin( α) )
3
= 2.21
This is a nonlinear implicit equation for α and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start
with a half-full channel
−
For
α = 180 ⋅ deg
α
−
For
α = 159 ⋅ deg
α
2
3
5
⋅ ( α − sin( α) )
2
3
3
−
= 3.14
For
α = 160 ⋅ deg
5
⋅ ( α − sin( α) )
3
y =
D
2
⋅ ⎛⎜ 1 − cos⎛⎜
⎝
α
−
= 2.20
For
α = 159.2 ⋅ deg
The solution to three figures is α = 159 ⋅ deg
From geometry
⎠
α ⎞⎞
⎝ 2 ⎠⎠
y = 0.410 m
α
2
3
5
⋅ ( α − sin( α) )
2
3
3
= 2.25
5
⋅ ( α − sin( α) )
3
= 2.212
Problem 11.55
Given:
Data on trapzoidal channel
Find:
New depth of flow
Solution:
Basic equation:
Q=
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
[Difficulty: 3]
Note that this is an "engineering" equation, to be used without units!
3
b = 2.4⋅ m
For the trapezoidal channel we have
α = 45⋅ deg
For bare soil (Table 11.2)
n = 0.010
Hence from Table 11.1
A = y ⋅ ( b + y ⋅ cot( α) ) = y ⋅ ( 2.4 + y )
2
Q=
Hence
1
n
3
2
Rh =
2
1
⋅ A⋅ Rh ⋅ Sb
Q = 7.1⋅
=
1
0.010
y ⋅ ( 2.4 + y )
m
Sb = 0.00193
s
y ⋅ ( b + y ⋅ cot( α) )
b+
=
2⋅ y
y ⋅ ( 2.4 + y )
2.4 + 2 ⋅ y ⋅ 2
sin( α)
1
3
⎤ ⋅ 0.00193 2 = 7.1 (Note that we don't use units!)
⎥
⎣ 2.4 + 2 ⋅ y⋅ 2⎦
⋅ y ⋅ ( 2.4 + y ) ⋅ ⎡⎢
5
Solving for y
[ y ⋅ ( 2.4 + y ) ]
3
2
= 1.62
( 2.4 + 2⋅ y⋅ 2) 3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start
with a shallower depth than that of Problem 11.49.
5
For
y = 1
( m)
[ y ⋅ ( 2.4 + y ) ]
5
3
2
= 2.55
For
y = 0.75
( m)
( 2.4 + 2⋅ y⋅ 2) 3
[ y ⋅ ( 2.4 + y ) ]
2
y = 0.77
( m)
[ y ⋅ ( 2.4 + y ) ]
5
3
2
= 1.60
For
y = 0.775
( m)
( 2.4 + 2⋅ y⋅ 2) 3
The solution to three figures is
= 1.53
( 2.4 + 2⋅ y⋅ 2) 3
5
For
3
[ y ⋅ ( 2.4 + y ) ]
3
2
( 2.4 + 2⋅ y⋅ 2) 3
y = 0.775
(m)
= 1.62
Problem 11.54
Given:
Data on trapzoidal channel
Find:
New depth of flow
Solution:
Basic equation:
Q=
1
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
[Difficulty: 3]
Note that this is an "engineering" equation, to be used without units!
3
b = 2.4⋅ m
For the trapezoidal channel we have
α = 45⋅ deg
For bare soil (Table 11.2)
n = 0.020
Hence from Table 11.1
A = y ⋅ ( b + cot( α) ⋅ y ) = y ⋅ ( 2.4 + y )
Q = 10⋅
R=
m
y ⋅ ( b + y ⋅ cot( α) )
b+
2
Q=
Hence
1
2
1
3
⋅ A⋅ Rh ⋅ Sb
n
2
=
1
0.020
y ⋅ ( 2.4 + y )
Sb = 0.00193
s
2⋅ y
y ⋅ ( 2.4 + y )
2.4 + 2 ⋅ y ⋅ 2
sin( α)
1
3
⎤ ⋅ 0.00193 2 = 10
⎥
⎣ 2.4 + 2 ⋅ y⋅ 2⎦
⋅ y ⋅ ( 2.4 + y ) ⋅ ⎡⎢
=
(Note that we don't use units!)
5
Solving for y
[ y ⋅ ( 2.4 + y ) ]
3
2
= 4.55
( 2.4 + 2⋅ y⋅ 2) 3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start
with a larger depth than Problem 11.49's.
5
For
y = 1.5
( m)
[ y ⋅ ( 2.4 + y ) ]
5
3
2
= 5.37
For
y = 1.4
( m)
( 2.4 + 2⋅ y⋅ 2) 3
[ y ⋅ ( 2.4 + y ) ]
2
y = 1.35
( m)
[ y ⋅ ( 2.4 + y ) ]
5
3
2
= 4.41
For
y = 1.37
( m)
( 2.4 + 2⋅ y⋅ 2) 3
The solution to three figures is
= 4.72
( 2.4 + 2⋅ y⋅ 2) 3
5
For
3
[ y ⋅ ( 2.4 + y ) ]
3
2
( 2.4 + 2⋅ y⋅ 2) 3
y = 1.37
(m)
= 4.536
Problem 11.53
Given:
Data on flume with plastic liner
Find:
Depth of flow
[Difficulty: 3]
Solution:
Basic equation:
Q=
1.49
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width b = 6 ⋅ ft and depth y we find from Table 11.1
A = b⋅ y = 6⋅ y
n = 0.010
and also
R=
1 ⋅ ft
Sb =
1000⋅ ft
and
2
Q=
Hence
1.49
n
3
2
b + 2⋅ y
2
=
1.49
0.010
6⋅ y
⋅ 6 ⋅ y ⋅ ⎛⎜
y
6 + 2⋅ y
1
⎞ ⋅ 0.001 2 = 85.5 (Note that we don't use units!)
⎝ 6 + 2⋅ y ⎠
5
3
2
( 6 + 2⋅ y)
6⋅ y
3
5
Solving for y
=
Sb = 0.001
1
⋅ A⋅ Rh ⋅ Sb
b⋅ y
3
85.5⋅ 0.010
=
or
1
2
2
3
1.49⋅ .001 ⋅ 6 ⋅ 6
y
3
2
( 6 + 2⋅ y)
= 0.916
3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start
with Problem 11.46's depth
5
For
y = 3
( feet)
y
5
3
2
( 6 + 2⋅ y)
= 1.191
For
y = 2
( feet)
3
y
2
( 6 + 2⋅ y)
5
For
y = 2.5
( feet)
y
2
= 0.931
For
y = 2.45
( feet)
3
y = 2.47
( feet)
y
3
2
( 6 + 2⋅ y)
3
y
3
3
2
( 6 + 2⋅ y)
5
For
= 0.684
5
3
( 6 + 2⋅ y)
3
= 0.916
y = 2.47
(feet)
3
= 0.906
Problem 11.52
Given:
Data on semicircular trough
Find:
Discharge
[Difficulty: 1]
Solution:
Basic equation:
Q=
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
For the semicircular channel
D = 1⋅ m
α = 180 ⋅ deg
For corrugated steel, a Google search leads to (Table 11.2)
Hence from Table 11.1
1
A =
8
Rh =
4
⋅ ⎛⎜ 1 −
⎝
2
Then the discharge is
Q=
1
2
A = 0.393 m
⋅D
Rh = 0.25 m
⋅ ( α − sin( α) ) ⋅ D
1
sin( α) ⎞
α
n = 0.022
⎠
2
1
3
3
2 m
⋅ A⋅ Rh ⋅ Sb ⋅
s
n
3
Q = 0.708
m
s
Sb = 0.01
Problem 11.51
Given:
Data on semicircular trough
Find:
Discharge
[Difficulty: 2]
Solution:
Basic equation:
Q=
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
For the semicircular channel
D = 1⋅ m
y = 0.25⋅ m
Hence, from geometry
⎛y− D⎞
⎜
2
α = 2 ⋅ asin⎜
⎟ + 180 ⋅ deg
⎜ D
⎝ 2 ⎠
α = 120 ⋅ deg
n = 0.022
For corrugated steel, a Google search leads to
Hence from Table 11.1
1
A =
8
Rh =
4
⋅ ⎛⎜ 1 −
⎝
2
Then the discharge is
Q=
1
2
A = 0.154 m
⋅D
Rh = 0.147 m
⋅ ( α − sin( α) ) ⋅ D
1
Sb = 0.01
sin( α) ⎞
α
⎠
2
1
3
3
2 m
⋅ A⋅ Rh ⋅ Sb ⋅
s
n
3
Q = 0.194
m
s
Problem 11.50
Given:
Data on triangular channel
Find:
Required dimensions
[Difficulty: 1]
Solution:
Basic equation:
Q=
1
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
Note that this is an "engineering" equation, to be used without units!
3
α = 45⋅ deg
Sb = 0.001
For concrete (Table 11.2)
n = 0.013
(assuming y > 60 cm: verify later)
Hence from Table 11.1
A = y ⋅ cot( α) = y
For the triangular channel we have
2
2
Hence
Q=
1
3
⋅ A⋅ Rh ⋅ Sb
n
2
Rh =
y ⋅ cos( α)
2
2
1
2
=
1
n
=
m
s
y
2⋅ 2
8
1
1
8
1
3
⎞ ⋅S = 1 ⋅y 3 ⋅⎛ 1 ⎞ ⋅S 2 = 1 ⋅y 3 ⋅S 2
⎜
b n
b
b
2⋅ n
⎝8⎠
⎝ 2⋅ 2 ⎠
⋅ y ⋅ ⎛⎜
2
3
Q = 10⋅
y
3
Solving for y
y=
⎛ 2⋅ n⋅ Q ⎞
⎜ S
⎝ b⎠
8
y = 2.20 m
(The assumption that y > 60 cm is verified)
Problem 11.49
Given:
Data on trapezoidal channel
Find:
Bed slope
[Difficulty: 1]
Solution:
Basic equation:
Q=
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
3
For the trapezoidal channel we have
b = 2.4⋅ m
α = 45⋅ deg
For bare soil (Table 11.2)
n = 0.020
Hence from Table 11.1
A = y ⋅ ( b + cot( α) ⋅ y )
Hence
Sb =
⎛ Q⋅ n ⎞
⎜
2
⎜
⎟
⎜ A⋅ R 3
h ⎠
⎝
y = 1.2⋅ m
2
A = 4.32 m
2
Sb = 1.60 × 10
Rh =
−3
Q = 7.1⋅
y ⋅ ( b + y ⋅ cot( α) )
b+
2⋅ y
sin( α)
m
s
Rh = 0.746 m
Problem 11.48
Given:
Data on square channel
Find:
Dimensions for concrete and soil cement
[Difficulty: 2]
Solution:
Basic equation:
Q=
1
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
Note that this is an "engineering" equation, to be used without units!
A= b
For a square channel of width b we find
2
R=
b⋅ y
b + 2⋅ y
=
b
b + 2⋅ b
3
2
Hence
1
1
8
2
3
Sb
b⎞
2
3
⎛
Q = ⋅b ⋅⎜
⋅ Sb =
⋅b
2
n
3
⎝ ⎠
1
2
n⋅ 3
3
or
⎛⎜ 2
⎞
3
⎜ 3 ⋅Q ⎟
b=⎜
⋅n
1 ⎟
⎜
⎟
⎜ Sb 2
⎝
⎠
3
The given data is
Q = 20⋅
m
s
For concrete, from Table 11.2 (assuming large depth)
Sb = 0.003
n = .013
b = 2.36 m
For soil cement from Table 11.2 (assuming large depth)
n = .020
b = 2.77 m
2
8
=
b
3
Problem 11.47
Given:
Data on flume
Find:
Slope
[Difficulty: 1]
Solution:
Basic equation:
Q=
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width b = 3 ⋅ ft and depth y = 6 ⋅ ft we find
A = b⋅ y
A = 18⋅ ft
2
Rh =
b⋅ y
Rh = 1.20⋅ ft
b + 2⋅ y
n = 0.0145
For wood (not in Table 11.2) a Google search finds n = 0.012 to 0.017; we use
Sb =
n⋅ Q
⎛
⎞
⎜
2
⎜
⎟
⎜ 1.49⋅ A⋅ R 3
h ⎠
⎝
2
Sb = 1.86 × 10
−3
with
Q = 90⋅
ft
3
s
Problem 11.46
Given:
Data on flume
Find:
Discharge
[Difficulty: 1]
Solution:
Basic equation:
Q=
1.49
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width b = 6 ⋅ ft and depth y = 3 ⋅ ft we find from Table 11.1
A = b⋅ y
A = 18⋅ ft
n = 0.013
For concrete (Table 11.2)
Q=
2
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Rh =
and
b⋅ y
b + 2⋅ y
1 ⋅ ft
Sb =
1000⋅ ft
Q = 85.5⋅
ft
3
s
Rh = 1.50⋅ ft
Sb = 0.001
Problem 11.45
Given:
Data on trapezoidal channel
Find:
Depth of flow
[Difficulty: 3]
Solution:
Basic equation:
Q=
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
b = 2.5⋅ m
For the trapezoidal channel we have
α = atan⎛⎜
1⎞
⎝2⎠
3
α = 26.6 deg
Q = 3⋅
m
S0 = 0.0004
s
n = 0.015
A = y ⋅ ( b + cot( α) ⋅ y ) = y ⋅ ( 8 + 2 ⋅ y )
Hence from Table 11.1
R=
y ⋅ ( b + y ⋅ cot( α) )
b+
2
Q=
Hence
1
n
2
1
3
⋅ A⋅ Rh ⋅ Sb
2
=
1
0.015
⎡ ( 2.5 + 2 ⋅ y) ⋅ y⎤
2⋅ y
=
y ⋅ ( 2.5 + 2 ⋅ y )
2.5 + 2 ⋅ y ⋅ 5
cot( α)
1
3
2
⎥ ⋅ 0.0004 = 3
⎣ 2.5 + 2⋅ y ⋅ 5 ⎦
⋅ y ⋅ ( 2.5 + 2 ⋅ y ) ⋅ ⎢
(Note that we don't use units!)
5
Solving for y
[ y ⋅ ( 2.5 + 2 ⋅ y ) ]
3
2
= 2.25
( 2.5 + 2⋅ y⋅ 5) 3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
5
For
For
y = 1
y = 0.81
( m)
( m)
[ y ⋅ ( 2.5 + 2 ⋅ y ) ]
5
3
2
= 3.36
For
y = 0.8
( m)
3
2
( 2.5 + 2⋅ y⋅ 5) 3
( 2.5 + 2⋅ y⋅ 5) 3
5
5
[ y ⋅ ( 2.5 + 2 ⋅ y ) ]
3
2
= 2.23
For
y = 0.815
( m)
( 2.5 + 2⋅ y⋅ 5) 3
The solution to three figures is
[ y ⋅ ( 2.5 + 2 ⋅ y ) ]
[ y ⋅ ( 2.5 + 2 ⋅ y ) ]
3
2
( 2.5 + 2⋅ y⋅ 5) 3
y = 0.815
(m)
= 2.17
= 2.25
Problem 11.44
Given:
Data on trapzoidal channel
Find:
Depth of flow
[Difficulty: 3]
Solution:
Basic equation:
Q=
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
α = atan ⎛⎜
b = 8 ⋅ ft
For the trapezoidal channel we have
1⎞
⎝ 2⎠
α = 26.6deg
Q = 100⋅
ft
3
S0 = 0.0004
s
n = 0.015
A = y ⋅ ( b + y ⋅ cot ( α ) ) = y ⋅ ( 8 + 2⋅ y )
Hence from Table 11.1
2
Q=
Hence
1.49
n
3
⋅ A⋅ Rh ⋅ Sb
Rh =
2
1
2
=
1.49
0.015
⎡ y⋅ ( 8 + 2⋅ y)⎤
y ⋅ ( b + y ⋅ cot(α))
b +
2⋅ y
=
y ⋅ ( 8 + 2⋅ y )
8 + 2⋅ y ⋅ 5
sin ( α )
1
3
2
⎥ ⋅ 0.0004 = 100(Note that we don't use units!)
⎣ 8 + 2⋅ y⋅ 5 ⎦
⋅ y⋅ ( 8 + 2⋅ y) ⋅ y⋅ ⎢
5
Solving for y
[ y⋅ ( 8 + 2⋅ y) ]
3
2
= 50.3
( 8 + 2⋅ y⋅ 5) 3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below.
5
For
For
y = 2
y = 2.6
( ft)
( ft)
[ y⋅ ( 8 + 2⋅ y) ]
5
3
2
= 30.27
For
y = 3
( ft)
3
2
( 8 + 2⋅ y⋅ 5) 3
( 8 + 2⋅ y⋅ 5) 3
5
5
[ y⋅ ( 8 + 2⋅ y) ]
3
2
= 49.81
For
y = 2.61
( ft)
( 8 + 2⋅ y⋅ 5) 3
The solution to three figures is
[ y⋅ ( 8 + 2⋅ y) ]
[ y⋅ ( 8 + 2⋅ y) ]
3
2
( 8 + 2⋅ y⋅ 5) 3
y = 2.61
(ft)
= 65.8
= 50.18
Problem 11.43
Given:
Data on rectangular channel
Find:
Depth of flow
[Difficulty: 3]
Solution:
Basic equation:
Q=
1
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Note that this is an "engineering" equation, to be used without units!
3
For a rectangular channel of width b = 2.5⋅ m and flow rate Q = 3 ⋅
Manning's roughness coefficient is
n = 0.015
Q=
Hence the basic equation becomes
n
b⋅ y
⋅ b ⋅ y ⋅ ⎛⎜
s
we find from Table 11.1
A = b⋅ y
R=
b⋅ y
b + 2⋅ y
Sb = 0.0004
and
2
1
m
1
3
⎞ ⋅S 2
b
⎝ b + 2⋅ y ⎠
2
3
⎞ = Q⋅ n
1
⎝ b + 2⋅ y ⎠
y ⋅ ⎛⎜
Solving for y
b⋅ y
b ⋅ Sb
2
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below, to make the
Q⋅ n
left side evaluate to
= 0.900 .
1
b ⋅ Sb
2
2
For
y = 1
( m)
y ⋅ ⎛⎜
b⋅ y
2
3
⎞ = 0.676
⎝ b + 2⋅ y ⎠
For
y = 1.2
( m)
y ⋅ ⎛⎜
b⋅ y
2
For
y = 1.23
( m)
The solution to three figures is
y ⋅ ⎛⎜
b⋅ y
2
3
⎞ = 0.894
⎝ b + 2⋅ y ⎠
3
⎞ = 0.865
⎝ b + 2⋅ y ⎠
For
y = 1.24
( m)
y = 1.24
(m)
y ⋅ ⎛⎜
b⋅ y
3
⎞ = 0.904
⎝ b + 2⋅ y ⎠
Problem 11.42
Given:
Rectangular channel flow
Find:
Discharge
[Difficulty: 1]
Solution:
Basic equation:
Q=
1
n
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
Note that this is an "engineering" equation, to be used without units!
For a rectangular channel of width b = 2 ⋅ m and depth y = 1.5⋅ m we find from Table 11.1
2
A = b⋅ y
A = 3.00⋅ m
n = 0.015
Manning's roughness coefficient is
Q=
1.49
2
1
3
2
⋅ A⋅ Rh ⋅ Sb
n
Rh =
and
b⋅ y
b + 2⋅ y
Sb = 0.0005
3
Q = 3.18⋅
m
s
Rh = 0.600 ⋅ m
Problem 11.41
Given:
Tidal bore
Find:
Speed of undisturbed river
[Difficulty: 3]
At rest
V1 = Vr + Vbore
Solution:
2
Basic equations:
V2 ⋅ y 2
g
+
y2
2
2
2
V1 ⋅ y 1
=
g
+
y1
y2
y1
2
2
(This is the basic momentum equation for the flow)
V2
V2 ⋅ y 2 = V1 ⋅ y 1
or
Given data
Vbore = 18⋅ mph
ft
Vbore = 26.4⋅
s
Then
2
2
y 1 − y 2 = ⋅ ⎛ V2 ⋅ y 2 − V1 ⋅ y 1⎞ =
⎝
⎠
g
2
2
2
2
2
2 ⋅ V1
y1 − y2 =
Dividing by (y 2 - y 1)
2
y2
y 1 = 8 ⋅ ft
y 2 = y 1 + 12⋅ ft
y 2 = 20⋅ ft
⎡⎛ V ⎞ 2
⎤ 2 ⋅ V 2 ⎡⎛ y ⎞ 2
⎤
1 ⎢ 1
⎢ 2
⎥
⎥
⋅ ⎜
⎢ V ⋅ y2 − y1⎥ = g ⋅ ⎢⎜ y ⋅ y 2 − y 1⎥
g
⎣⎝ 1 ⎠
⎦
⎣⎝ 2 ⎠
⎦
2 ⋅ V1
2
2
2
y1
y1 + y2 = 2⋅
⋅
g y2
g⋅ y2
y1
⎞ 2⋅ V 2⋅ y ( y − y )
⎛y 2
1
2
1 1
⎜ 1
⋅
− y1 =
⋅
⎜
g
g
y2
⎝ y2
⎠
V1
V1 =
But
V1
=
or
⎛
y2 ⎞
⎝
y1
⋅⎜1 +
V1 = Vr + Vbore
⎠
or
(
y1 + y2
g
2
V1 = ⋅ y 2 ⋅
y1
2
)
ft
V1 = 33.6⋅
s
V1 = 22.9⋅ mph
Vr = V1 − Vbore
ft
Vr = 7.16⋅
s
Vr = 4.88⋅ mph
Problem 11.40
Given:
Surge wave
Find:
Surge speed
[Difficulty: 3]
V2
At rest
y1
Solution:
2
Basic equations:
V1 ⋅ y 1
g
+
y1
2
2
=
2
V2 ⋅ y 2
g
+
y2
V 2 = VSurge
2
2
(This is the basic momentum equation for the flow)
V1 ⋅ y 1 = V2 ⋅ y 2
Then
2
2
y 2 − y 1 = ⋅ ⎛ V1 ⋅ y 1 − V2 ⋅ y 2⎞ =
⎠
g ⎝
2
2
2
2
2 ⋅ V2
2
y2
y1
⎡⎛ V ⎞ 2
⎤ 2 ⋅ V 2 ⎡⎛ y ⎞ 2
⎤
2 ⎢ 2
⎢ 1
⎥
⎥
⋅ ⎜
⋅ y1 − y2 =
⋅ ⎜
⋅ y1 − y2
⎢ V
⎥
⎢ y
⎥
g
g
⎣⎝ 2 ⎠
⎦
⎣⎝ 1 ⎠
⎦
2 ⋅ V2
2
2
2
y2
y2 + y1 = 2⋅
⋅
g y1
g⋅ y1
=
⎛y 2
⎞ 2⋅ V 2⋅ y ( y − y )
2
1
2 2
⎜ 2
⋅
− y2 =
⋅
⎜y
g
g
y1
⎝ 1
⎠
V2
V2 =
But
V2
2
y2 − y1 =
Dividing by (y 2 - y 1)
V1
or
⎛
y1 ⎞
⎝
y2
⋅⎜1 +
V2 = VSurge
(
or
y2 + y1
g
2
V2 = ⋅ y 1 ⋅
y2
2
so
VSurge =
)
⎠
g⋅ y1
2
⎛
y1 ⎞
⎝
y2
⋅⎜1 +
⎠
y2
Problem 11.39
Given:
Data on sluice gate
Find:
Water depth before and after the jump
[Difficulty: 3]
Solution:
E1 =
Basic equation:
y3
y2
=
V1
2
2⋅ g
1
2
V2
p2
+ y1 =
+
= E2
2⋅ g
ρ⋅ g
⎛
⎝
⋅ −1 +
1 + 8 ⋅ Fr 2
For the gate
2⎞
For the jump (state 2 before, state 3 after)
⎠
m
V1 = 0.2⋅
s
y 1 = 1.5⋅ m
The given data is
2
2
q = y 1 ⋅ V1
Hence
Then we need to solve
V2
q = 0.3
m
E1 =
s
2
+ y 2 = E1
2⋅ g
q
or
V1
2
2⋅ g
+ y1
E1 = 1.50 m
2
2⋅ g⋅ y2
2
+ y 2 = E1
with
E1 = 1.50 m
We can solve this equation iteratively (or use Excel's Goal Seek or Solver)
2
y 2 = 0.5⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 0.518 m
2
2⋅ g
y 2 = 0.055 ⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.57 m
2
2⋅ g
For
y 2 = 0.0563⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.50 m
2
2⋅ g
Then
q
V2 =
y2
m
V2 = 5.33
s
For
2
For
y 2 = 0.05⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.89 m
2
2⋅ g
y 2 = 0.057 ⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.47 m
2
2⋅ g
2
For
2
For
2
For the jump (States 2 to 3)
y3 =
⎛
2 ⎝
y2
⋅ −1 +
Hence
Note that
1 + 8 ⋅ Fr 2
2⎞
⎠
y 2 = 0.056 m
Fr 2 =
y 3 = 0.544 m
is the closest to three figs.
V2
g⋅ y2
Fr 2 = 7.17
Problem 11.38
[Difficulty: 2]
Given:
Data on rectangular channel flow
Find:
Depth after hydraulic jump; Specific energy change
Solution:
2⎞
⎛
= ⋅ −1 + 1 + 8 ⋅ Fr 1
⎝
⎠
y1
2
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = E1 − E2 = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2
2⋅ g ⎠
⎝
y 1 = 0.4⋅ m
b = 1⋅ m
Then
Q = V1 ⋅ b ⋅ y 1 = V2 ⋅ b ⋅ y 2
Q
V1 =
b⋅ y1
Then Fr 1 is
Fr 1 =
Hence
y2 =
and
Q
V2 =
b⋅ y2
Basic equations:
y2
1
3
The given data is
For the specific energies
V1
y1
2
⎛
⎝
⋅ −1 +
Note that we could use
1 + 8 ⋅ Fr 1
2⎞
⎠
y 2 = 4.45 m
m
V2 = 1.46
s
V1
2
E1 = y 1 +
2⋅ g
V2
The energy loss is
Fr 1 = 8.20
g⋅ y1
E1 = 13.9 m
2
E2 = y 2 +
2⋅ g
E2 = 4.55 m
Hl = E1 − E2
Hl = 9.31 m
Hl =
( y2 − y1)3
4⋅ y1⋅ y2
Hl = 9.31⋅ m
Q = 6.5
m
s
m
V1 = 16.3
s
Problem 11.37
[Difficulty: 2]
Given:
Data on wide spillway flow
Find:
Depth after hydraulic jump; Specific energy change
Solution:
Basic equations:
2⎞
⎛
= ⋅ −1 + 1 + 8 ⋅ Fr 1
⎠
y1
2 ⎝
y2
1
m
V1 = 25
s
The given data is
y 1 = 0.9⋅ m
Then Fr 1 is
Fr 1 =
Hence
y2 =
Then
Q = V1 ⋅ b ⋅ y 1 = V2 ⋅ b ⋅ y 2
For the specific energies
V1
⎛
2 ⎝
⋅ −1 +
V1
Note that we could use
1 + 8 ⋅ Fr 1
2⎞
⎠
y 2 = 10.3 m
y1
V2 = V1 ⋅
y2
m
V2 = 2.19
s
2
E1 = y 1 +
2⋅ g
V2
The energy loss is
Fr 1 = 8.42
g⋅ y1
y1
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = E1 − E2 = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2
2⋅ g ⎠
⎝
E1 = 32.8 m
2
E2 = y 2 +
2⋅ g
E2 = 10.5 m
Hl = E1 − E2
Hl = 22.3 m
Hl =
( y2 − y1)3
4⋅ y1⋅ y2
E2
E1
= 0.321
Hl = 22.3⋅ m
Problem 11.36
Given:
Data on wide channel and hydraulic jump
Find:
Flow rate; Head loss
[Difficulty: 2]
Solution:
Basic equations:
The given data is
2⎞
⎛
= ⋅ −1 + 1 + 8 ⋅ Fr 1
⎝
⎠
y1
2
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = E1 − E2 = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2
2⋅ g ⎠
⎝
b = 5 ⋅ ft
y 1 = 0.66⋅ ft
y2
1
We can solve for Fr 1 from the basic equation
2
1 + 8 ⋅ Fr 1 = 1 + 2 ⋅
y 2 = 3.0⋅ ft
y2
y1
2
Fr 1 =
y2 ⎞
⎛
−1
⎜ 1 + 2⋅
y1
⎝
⎠
8
Fr 1 = 3.55
Hence
V1 = Fr 1 ⋅ g ⋅ y 1
ft
V1 = 16.4⋅
s
Then
Q = V1 ⋅ b ⋅ y 1
Q = 54.0⋅
Q
V2 =
b⋅ y2
ft
V2 = 3.60⋅
s
Also
⎛⎜
V1
The energy loss is Hl = ⎜ y 1 +
2⋅ g
⎝
Fr 1 =
and
ft
s
2
⎛⎜
V2 ⎞
− ⎜ y2 +
2⋅ g ⎠
⎠ ⎝
Fr 2 =
V2
g⋅ y2
Hl = 1.62⋅ ft
Hl =
g⋅ y1
3
2⎞
Note that we could use
V1
( y2 − y1)3
4⋅ y1⋅ y2
Hl = 1.62⋅ ft
Fr 2 = 0.366
Problem 11.35
Given:
Data on wide channel and hydraulic jump
Find:
Jump depth; Head loss
[Difficulty: 2]
Solution:
Basic equations:
The given data is
1
Q = 200 ⋅
ft
3
s
Also
Q = V⋅ A = V⋅ b ⋅ y
Hence
Q
V1 =
b⋅ y1
Then
y2 =
⎛
2 ⎝
y1
⋅ −1 +
Q
V2 =
b⋅ y2
The energy loss is
Note that we could use
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = E1 − E2 = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2 2⋅ g ⎠
⎝
2⎞
⎛
= ⋅ −1 + 1 + 8⋅ Fr 1
⎝
⎠
y1
2
y2
1 + 8 ⋅ Fr 1
b = 10⋅ ft
y 1 = 1.2⋅ ft
ft
V1 = 16.7⋅
s
Fr 1 =
2⎞
V1
g⋅ y1
Fr 1 = 2.68
y 2 = 3.99⋅ ft
⎠
ft
V2 = 5.01⋅
s
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2
2⋅ g ⎠
⎝
Hl =
Fr 2 =
V2
g⋅ y2
Hl = 1.14⋅ ft
( y2 − y1)3
4⋅ y1⋅ y2
Hl = 1.14⋅ ft
Fr 2 = 0.442
Problem 11.34
Given:
Data on wide channel and hydraulic jump
Find:
Jump depth
[Difficulty: 1]
Solution:
y2
Basic equations:
y1
⎛
2 ⎝
1
=
⋅ −1 +
1 + 8 ⋅ Fr 1
2⎞
⎠
3
m
The given data is
Q
b
= 2⋅
s
y 1 = 500 ⋅ mm
m
Also
Q = V⋅ A = V⋅ b ⋅ y
Hence
Q
V1 =
b⋅ y1
Then
y2 =
Note:
Q
V2 =
b⋅ y2
⎛
2 ⎝
y1
⋅ −1 +
m
V1 = 4.00
s
1 + 8 ⋅ Fr 1
2⎞
Fr 1 =
V1
g⋅ y1
Fr 1 = 1.806
y 2 = 1.05⋅ m
⎠
ft
V2 = 6.24⋅
s
Fr 2 =
V2
g⋅ y2
Fr 2 = 0.592
Problem 11.33
Given:
Data on wide channel and hydraulic jump
Find:
Jump depth; Head loss
[Difficulty: 2]
Solution:
Basic equations:
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = E1 − E2 = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2 2⋅ g ⎠
⎝
2⎞
⎛
= ⋅ −1 + 1 + 8⋅ Fr 1
⎝
⎠
y1
2
y2
1
3
m
The given data is
Q
b
= 10⋅
s
m
Also
Q = V⋅ A = V⋅ b ⋅ y
Hence
Q
V1 =
b⋅ y1
Then
y2 =
⎛
2 ⎝
y1
⋅ −1 +
y 1 = 1⋅ m
m
V1 = 10.0
s
1 + 8 ⋅ Fr 1
2⎞
The energy loss is
Note that we could use
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2
2⋅ g ⎠
⎝
Hl =
( y2 − y1)3
4⋅ y1⋅ y2
V1
g⋅ y1
Fr 1 = 3.19
y 2 = 4.04 m
⎠
Q
V2 =
b⋅ y2
Fr 1 =
m
V2 = 2.47
s
Fr 2 =
V2
g⋅ y2
Hl = 1.74 m
Hl = 1.74 m
Fr 2 = 0.393
Problem 11.32
Given:
Data on rectangular channel and hydraulic jump
Find:
Flow rate; Critical depth; Head loss
[Difficulty: 2]
1
Solution:
Basic equations:
The given data is
2⎞
⎛
= ⋅ −1 + 1 + 8⋅ Fr 1
⎝
⎠
y1
2
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = E1 − E2 = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2 2⋅ g ⎠
⎝
b = 4⋅ m
y 1 = 0.4⋅ m
y2
1
y 2 = 1.7⋅ m
2
1 + 8⋅ Fr 1 = 1 + 2⋅
We can solve for Fr 1 from the basic equation
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b 2
⎝
⎠
y2
y1
2
Fr 1 =
y2 ⎞
⎛
−1
⎜ 1 + 2⋅
y1
⎝
⎠
8
Fr 1 = 3.34
Fr 1 =
and
Hence
V1 = Fr 1 ⋅ g ⋅ y 1
m
V1 = 6.62
s
Then
Q = V1 ⋅ b ⋅ y 1
Q = 10.6⋅
V1
g⋅ y1
3
m
s
1
3
The critical depth is
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
Also
Q
V2 =
b⋅ y2
The energy loss is
2
2
⎛⎜
V1 ⎞ ⎛⎜
V2 ⎞
Hl = ⎜ y 1 +
− ⎜y +
2⋅ g ⎠ ⎝ 2
2⋅ g ⎠
⎝
Note that we could used
y c = 0.894 m
m
V2 = 1.56
s
Hl =
Fr 2 =
V2
g⋅ y2
Hl = 0.808 m
( y2 − y1)3
4⋅ y1⋅ y2
Hl = 0.808 m
Fr 2 = 0.381
3
Problem 11.31
Given:
Hydaulic jump data
Find:
Energy consumption; temperature rise
[Difficulty: 2]
Solution:
Basic equations:
P = ρ⋅ g ⋅ Hl⋅ Q
(1)
Hl is the head loss in m of fluid); multiplying by ρg produces energy/vol; multiplying by Q produces energy/time, or power
Urate = ρ⋅ Q⋅ cH2O⋅ ∆T
(2)
Urate is the rate of increase of internal energy of the flow; cH20∆T is the energy increase per unit mass due to a ∆T temperature rise;
multiplying by ρQ converts to energy rise of the entire flow/time
3
Given data:
From Eq. 1
From Example 11.5
P = ρ⋅ g ⋅ Hl⋅ Q
Equating Eqs. 1 and 2
Q = 9.65⋅
m
Hl = 0.258 ⋅ m
s
P = 24.4 kW
kg
ρ = 999 ⋅
and
3
m
cH2O = 1 ⋅
kg⋅ K
a significant energy consumption
ρ⋅ g ⋅ Hl⋅ Q = ρ⋅ Q⋅ cH2O⋅ ∆T
or
∆T =
g ⋅ Hl
∆T = 6.043 × 10
cH2O
The power consumed by friction is quite large, but the flow is very large, so the rise in temperature is insignificant.
In English units:
P = 32.7 hp
kcal
5
Q = 1.53 × 10 gpm
∆T = 1.088 × 10
−3
∆°F
−4
∆°C
Problem 11.30
[Difficulty: 2]
Given:
Rectangular channel flow with hump and/or side wall restriction
Find:
Whether critical flow occurs
1
Solution:
Basic equations:
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
3
2
Q
E=y+
2
3
A = b⋅ y
Emin =
h = 350 ⋅ mm
Q = 2.4⋅
2⋅ g⋅ A
2
⋅ yc
(From Example 11.4)
3
Given data:
b = 2⋅ m
y = 1⋅ m
h = 35⋅ cm
E1 = y +
2
(a) For a hump with
Then for the bump
Ebump = E1 − h
Q
2⋅ g⋅ b
2
⋅
1
y
m
s
E1 = 1.07 m
2
Ebump = 0.723 m
(1)
1
⎡⎢ ⎛ Q ⎞ 2⎤⎥
⎢ ⎜⎝ b ⎠ ⎥
yc = ⎢
⎣ g ⎥⎦
For the minimum specific energy
3
y c = 0.528 m
Emin =
3
2
⋅ yc
Emin = 0.791 m (2)
Comparing Eqs. 1 and 2 we see that the bump IS sufficient for critical flow
(b) For the sidewall restriction with
b const = 1.5⋅ m
as in Example 11.4 we have
Econst = E1
Econst = 1.073 m (3)
1
With b const:
⎡ ⎛ Q 2⎤
⎞ ⎥
⎢
⎜
⎢ ⎝ b const ⎠ ⎥
yc = ⎢
⎥
g
⎣
⎦
3
y c = 0.639 m
Eminconst =
3
2
⋅ yc
Eminconst = 0.959 m (4)
Comparing Eqs. 3 and 4 we see that the constriction is NOT sufficient for critical flow
(c) For both, following Example 11.4
Eboth = E1 − h
Eboth = 0.723 m
(5)
Eminboth = Eminconst
Eminboth = 0.959 m
(6)
Comparing Eqs. 5 and 6 we see that the bump AND constriction ARE sufficient for critical flow (not surprising, as the bump alone is
sufficient!)
Problem 11.29
Given:
Data on sluice gate
Find:
Water depth and velocity after gate
[Difficulty: 2]
Solution:
E1 =
Basic equation:
y3
y2
=
V1
2
2⋅ g
1
2
V2
p2
+ y1 =
+
= E2
2⋅ g
ρ⋅ g
⎛
⎝
⋅ −1 +
1 + 8 ⋅ Fr 2
For the gate
2⎞
For the jump (state 2 before, state 3 after)
⎠
m
V1 = 0.2⋅
s
y 1 = 1.5⋅ m
The given data is
2
2
q = y 1 ⋅ V1
Hence
Then we need to solve
V2
q = 0.3
m
s
E1 =
2
+ y 2 = E1
2⋅ g
q
or
V1
2
2⋅ g
+ y1
E1 = 1.50 m
2
2⋅ g⋅ y2
2
+ y 2 = E1
with
E1 = 1.50 m
We can solve this equation iteratively (or use Excel's Goal Seek or Solver)
2
y 2 = 0.5⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 0.518 m
2
2⋅ g
y 2 = 0.055 ⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.57 m
2
2⋅ g
For
y 2 = 0.0563⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.50 m
2
2⋅ g
Then
q
V2 =
y2
m
V2 = 5.33
s
For
2
For
y 2 = 0.05⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.89 m
2
2⋅ g
y 2 = 0.057 ⋅ m
⎛q⎞
⎜y
⎝ 2 ⎠ + y = 1.47 m
2
2⋅ g
2
For
2
For
2
Hence
Note that
y 2 = 0.056 m
Fr 2 =
is the closest to three figs.
V2
g⋅ y2
Fr 2 = 7.17
Problem 11.28
Given:
Data on sluice gate
Find:
Flow rate
[Difficulty: 2]
Solution:
Basic equation:
p1
ρ⋅ g
2
+
V1
p2
2
V2
+ y1 =
+
+ y2
2⋅ g
ρ⋅ g
2⋅ g
The Bernoulli equation applies because we have steady,
incompressible, frictionless flow.
Noting that p 1 = p 2 = p atm, (1 = upstream, 2 = downstream) the Bernoulli equation becomes
2
V1
2⋅ g
2
V2
+ y1 =
+ y2
2⋅ g
The given data is
b = 3 ⋅ ft
y 1 = 6⋅ ft
y 2 = 0.9⋅ ft
Also
Q = V⋅ A
so
Q
V1 =
b ⋅ y1
2
⎛ Q ⎞
⎜ b⋅ y
⎝ 1⎠ + y =
1
2⋅ g
Using these in the Bernoulli equation
2
Solving for Q
Note that
Q =
2
2⋅ g⋅ b ⋅ y1 ⋅ y2
y1 + y2
2
Q = 49.5⋅
ft
and
Q
V2 =
b ⋅ y2
2
⎛ Q ⎞
⎜ b⋅ y
⎝ 2⎠ + y
2
2⋅ g
3
s
Q
V1 =
b⋅ y1
ft
V1 = 2.75⋅
s
Fr 1 =
Q
V2 =
b⋅ y2
ft
V2 = 18.3⋅
s
Fr 2 =
V1
g⋅ y1
V2
g⋅ y2
Fr 1 = 0.198
Fr 2 = 3.41
Problem 11.27
Given:
Data on sluice gate
Find:
Water level upstream; Maximum flow rate
[Difficulty: 2]
Solution:
Basic equation:
p1
ρ⋅ g
+
V1
2
p2
V2
2
+ y1 =
+
+ y2 + h
2⋅ g
ρ⋅ g
2⋅ g
The Bernoulli equation applies because we have steady,
incompressible, frictionless flow.
Noting that p 1 = p 2 = p atm, and V 1 is approximately zero (1 = upstream, 2 = downstream) the Bernoulli equation becomes
y1 =
The given data is
Q
b
V2
2
+ y2
2⋅ g
2
= 10⋅
m
y 2 = 1.25⋅ m
s
Hence
Q = V2 ⋅ A2 = V2 ⋅ b ⋅ y 2
Then upstream
⎛⎜ V 2
⎞
2
y1 = ⎜
+ y2
⎝ 2⋅ g
⎠
Q
V2 =
b⋅ y2
or
m
V2 = 8
s
y 1 = 4.51 m
The maximum flow rate occurs at critical conditions (see Section 11-2), for constant specific energy
In this case
V2 = Vc =
Hence we find
y1 =
Hence
yc =
g⋅ yc
2
g⋅ yc
3
+ yc =
+ yc = ⋅ yc
2⋅ g
2⋅ g
2
Vc
2
3
⋅ y1
y c = 3.01 m
Vc =
3
m
Q
b
= Vc⋅ y c
Q
b
= 16.3⋅
s
m
(Maximum flow rate)
g⋅ yc
m
Vc = 5.43
s
Problem 11.26
Given:
Data on wide channel
Find:
Stream depth after rise
[Difficulty: 3]
Solution:
p1
Basic equation:
ρ⋅ g
V1
+
2
2⋅ g
V2
p2
2
+ y1 =
+
+ y2 + h
2⋅ g
ρ⋅ g
The Bernoulli equation applies because we have steady,
incompressible, frictionless flow. Note that at location 2 (the
bump), the potential is y 2 + h, where h is the bump height
2
Recalling the specific energy E =
V
+y
2⋅ g
At each section
Q = V⋅ A = V1 ⋅ b ⋅ y 1 = V2 ⋅ b ⋅ y 2
The given data is
y 1 = 2 ⋅ ft
Hence
Then
E1 =
V1
E1 = E2 + h
and noting that p1 = p 2 = p atm, the Bernoulli equation becomes
y1
V2 = V1 ⋅
y2
ft
V1 = 3 ⋅
s
h = 0.5⋅ ft
2
2⋅ g
+ y1
V2
E1 = 2.14⋅ ft
2
2
E1 = E2 + h =
+ y2 + h =
2⋅ g
V1 ⋅ y 1
2
2⋅ g
2
⋅
1
y2
+ y2 + h
2
or
V1 ⋅ y 1
2
2⋅ g
1
⋅
y2
2
+ y 2 = E1 − h
This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select
y 2 so the left side of the equation equals E1 − h = 1.64⋅ ft
2
For
y 2 = 2 ⋅ ft
V1 ⋅ y 1
y 2 = 1.3⋅ ft
Hence
y 2 = 1.31⋅ ft
Note that
y1
V2 = V1 ⋅
y2
so we have
Fr 1 =
V1
g⋅ y1
V1 ⋅ y 1
2⋅ g
2
1
⋅
2⋅ g
2
For
2
y2
+ y 2 = 2.14⋅ ft
2
y 2 = 1.5⋅ ft
For
2
V1 ⋅ y 1
⋅
y2
2
+ y 2 = 1.63⋅ ft
y 2 = 1.31⋅ ft
For
ft
V2 = 4.58⋅
s
Fr 1 = 0.37
and
Fr 2 =
V2
g⋅ y2
Fr 2 = 0.71
⋅
2⋅ g
2
1
2
V1 ⋅ y 1
2⋅ g
1
y2
2
+ y 2 = 1.75⋅ ft
2
+ y 2 = 1.64⋅ ft
2
⋅
1
y2
Problem 11.25
Given:
Data on rectangular channel and a bump
Find:
Local change in flow depth caused by the bump
[Difficulty: 3]
Solution:
Basic equation:
p1
ρ⋅ g
+
V1
2
2⋅ g
V2
p2
2
+ y1 =
+
+ y2 + h
2⋅ g
ρ⋅ g
The Bernoulli equation applies because we have steady,
incompressible, frictionless flow. Note that at location 2 (the
bump), the potential is y 2 + h, where h is the bump height
2
Recalling the specific energy E =
V
2⋅ g
+y
Q
At each section
Q = V⋅ A = V⋅ b ⋅ y
or
V=
The given data is
b = 10⋅ ft
y 1 = 0.3⋅ ft
h = 0.1⋅ ft
Q
V1 =
b⋅ y1
ft
V1 = 6.67⋅
s
Hence we find
E1 =
and
V1
b⋅ y
Q = 20⋅
ft
3
s
2
2⋅ g
+ y1
E1 = 0.991 ⋅ ft
V2
2
2
E1 = E2 + h =
+ y2 + h =
2⋅ g
Hence
E1 = E2 + h
and noting that p1 = p 2 = p atm, the Bernoulli equation becomes
2
Q
2
2⋅ g⋅ b ⋅ y2
+ y2 + h
2
or
Q
2
2⋅ g⋅ b ⋅ y2
2
+ y 2 = E1 − h
This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select y 2
so the left side of the equation equals E1 − h = 0.891 ⋅ ft
2
For
y 2 = 0.3⋅ ft
Q
2
2⋅ g⋅ b ⋅ y2
2
+ y 2 = 0.991 ⋅ ft
2
For
+ y 2 = 0.901 ⋅ ft
2
For
y 2 = 0.35⋅ ft
2
For
Hence
y 2 = 0.33⋅ ft
Q
2
2⋅ g⋅ b ⋅ y2
Q
V2 =
b⋅ y2
so we have
Fr 1 =
V1
g⋅ y1
2
2⋅ g⋅ b ⋅ y2
2
+ y 2 = 0.857 ⋅ ft
2
+ y 2 = 0.891 ⋅ ft
2
y 2 = 0.334 ⋅ ft
Note that
Q
y 2 = 0.334 ⋅ ft
y2 − y1
and
y1
= 11.3⋅ %
ft
V2 = 5.99⋅
s
Fr 1 = 2.15
and
Fr 2 =
V2
g⋅ y2
Fr 2 = 1.83
Q
2
2⋅ g⋅ b ⋅ y2
Problem 11.24
Given:
Data on rectangular channel and a bump
Find:
Local change in flow depth caused by the bump
[Difficulty: 3]
Solution:
Basic equation:
p1
ρ⋅ g
V1
+
2
2⋅ g
V2
p2
2
+ y1 =
+
+ y 2 + h The Bernoulli equation applies because we have steady,
2⋅ g
ρ⋅ g
incompressible, frictionless flow. Note that at location 2 (the
bump), the potential is y 2 + h, where h is the bump height
2
E=
Recalling the specific energy
V
+y
2⋅ g
Q
At each section
Q = V⋅ A = V⋅ b ⋅ y
or
V=
The given data is
b = 10⋅ ft
y 1 = 1 ⋅ ft
h = 0.25⋅ ft
Q = 20⋅
Q
ft
+ y2 + h
2
or
Hence we find
and
Hence
V1 =
b⋅ y1
E1 =
V1
E1 = E2 + h
and noting that p1 = p 2 = p atm, the Bernoulli equation becomes
b⋅ y
ft
3
s
V1 = 2 ⋅
s
2
2⋅ g
+ y1
E1 = 1.062 ⋅ ft
2
V2
2
E1 = E2 + h =
+ y2 + h =
2⋅ g
Q
2
2⋅ g⋅ b ⋅ y2
2
Q
2
2⋅ g⋅ b ⋅ y2
+ y 2 = E1 − h
2
This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select
y 2 so the left side of the equation equals E1 − h = 0.812 ⋅ ft
2
For
y 2 = 0.75⋅ ft
2
Q
2
2⋅ g⋅ b ⋅ y2
+ y 2 = 0.861 ⋅ ft
2
For
+ y 2 = 0.797 ⋅ ft
2
For
y 2 = 0.7⋅ ft
Q
2
2⋅ g⋅ b ⋅ y2
2
For
Hence
y 2 = 0.65⋅ ft
2⋅ g⋅ b ⋅ y2
y 2 = 0.676 ⋅ ft
Note that
Q
V2 =
b⋅ y2
so we have
Fr 1 =
V1
g⋅ y1
+ y 2 = 0.827 ⋅ ft
2
+ y 2 = 0.812 ⋅ ft
2
Q
2
2
and
y 2 = 0.676 ⋅ ft
y2 − y1
y1
Q
2
2⋅ g⋅ b ⋅ y2
= −32.4⋅ %
ft
V2 = 2.96⋅
s
Fr 1 = 0.353
and
Fr 2 =
V2
g⋅ y2
Fr 2 = 0.634
Problem 11.23
Given:
Data on rectangular channel and a bump
Find:
Elevation of free surface above the bump
[Difficulty: 3]
Solution:
p1
Basic
equation:
ρ⋅ g
+
V1
2
2⋅ g
V2
p2
2
+ y1 =
+
+ y 2 + h The Bernoulli equation applies because we have steady,
2⋅ g
ρ⋅ g
incompressible, frictionless flow. Note that at location 2 (the
bump), the potential is y 2 + h, where h is the bump height
2
E=
Recalling the specific energy
V
2⋅ g
+y
and noting that p1 = p 2 = p atm, the Bernoulli equation becomes
Q
At each section
Q = V⋅ A = V⋅ b ⋅ y
or
V=
The given data is
b = 10⋅ ft
y 1 = 1 ⋅ ft
h = 4 ⋅ in
Q
V1 =
b⋅ y1
ft
V1 = 10⋅
s
Hence we find
E1 =
and
V1
b⋅ y
Q = 100 ⋅
ft
3
s
2
2⋅ g
+ y1
V2
E1 = 2.554 ⋅ ft
2
2
E1 = E2 + h =
+ y2 + h =
2⋅ g
Hence
E1 = E2 + h
2
Q
2
2⋅ g⋅ b ⋅ y2
+ y2 + h
2
Q
or
2
2⋅ g⋅ b ⋅ y2
2
+ y 2 = E1 − h
This is a nonlinear implicit equation for y 2 and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We select y 2
so the left side of the equation equals E1 − h = 2.22⋅ ft
2
For
y 2 = 1 ⋅ ft
Q
2
2⋅ g⋅ b ⋅ y2
2
+ y 2 = 2.55⋅ ft
2
For
+ y 2 = 2.19⋅ ft
2
For
y 2 = 1.5⋅ ft
2
For
y 2 = 1.4⋅ ft
Q
2
2⋅ g⋅ b ⋅ y2
Q
V2 =
b⋅ y2
so we have
Fr 1 =
V1
g⋅ y1
2
2⋅ g⋅ b ⋅ y2
2
+ y 2 = 2.19⋅ ft
2
+ y 2 = 2.22⋅ ft
2
y 2 = 1.3⋅ ft
y 2 = 1.30⋅ ft
Hence
Note that
Q
ft
V2 = 7.69⋅
s
Fr 1 = 1.76
and
Fr 2 =
V2
g⋅ y2
Fr 2 = 1.19
Q
2
2⋅ g⋅ b ⋅ y2
Problem 11.22
Given:
Data on venturi flume
Find:
Flow rate
[2]
Solution:
Basic equation:
p1
ρ⋅ g
+
V1
2
V2
p2
2
+ y1 =
+
+ y2
2⋅ g
ρ⋅ g
2⋅ g
At each section
Q = V⋅ A = V⋅ b ⋅ y
The given data is
b 1 = 2 ⋅ ft
The Bernoulli equation applies because we have steady,
incompressible, frictionless flow
V=
or
y 1 = 1 ⋅ ft
Q
b⋅ y
b 2 = 1 ⋅ ft
y 2 = 0.75⋅ ft
2
Hence the Bernoulli equation becomes (with p 1 = p 2 = p atm)
Solving for Q
Q =
(
2⋅ g⋅ y1 − y2
2
)
⎛
⎞ −⎛
⎞
⎜ b ⋅y
⎜ b ⋅y
⎝ 2 2⎠ ⎝ 1 1⎠
1
1
2
⎛ Q ⎞
⎜ b ⋅y
⎝ 1 1⎠ + y =
1
2⋅ g
Q = 3.24⋅
ft
3
s
2
⎛ Q ⎞
⎜ b ⋅y
⎝ 2 2⎠ + y
2
2⋅ g
Problem 11.21
Given:
Data on trapezoidal channel
Find:
Critical depth
[Difficulty: 3]
Solution:
2
E=y+
Basic equation:
V
2⋅ g
In Section 11-2 we prove that the minimum specific energy is when we have critical flow; here we rederive the minimum energy point
For a trapezoidal channel (Table 11.1) A = ( b + cot(α)⋅ y ) ⋅ y
Q
V=
Hence for V
A
=
E=y+
Using this in the basic equation
( b + cot(α)⋅ y ) ⋅ y
b = 10⋅ ft
and
Q = 400 ⋅
=1−
dy
2
2
⎝1⎠
ft
α = 71.6 deg
3
s
g ⋅ y ⋅ ( b + y ⋅ cot(α))
3
Q
−
3
g ⋅ y ⋅ ( b + y ⋅ cot(α))
2
=0
2
Q ⋅ cot(α)
2
3⎞
2
Q ⋅ cot(α)
g ⋅ y ⋅ ( b + y ⋅ cot(α))
Hence we obtain for y
α = atan⎛⎜
2
Q
⎤ ⋅ 1
⎡
⎢
⎥
⎣ ( b + cot(α)⋅ y ) ⋅ y⎦ 2 ⋅ g
2
dE
E is a minimum when
Q
and
3
+
2
Q
3
g ⋅ y ⋅ ( b + y ⋅ cot(α))
2
=1
Q ⋅ ( b + 2 ⋅ y ⋅ cot(α))
or
3
g ⋅ y ⋅ ( b + y ⋅ cot(α))
3
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below, to make
the left side equal unity
2
y = 5 ⋅ ft
Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) )
3
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
3
2
= 0.3
y = 4 ⋅ ft
Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) )
3
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
3
3
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
2
y = 3.5⋅ ft
Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) )
Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) )
3
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
3
= 0.7
2
= 1.03
y = 3.55⋅ ft
Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) )
3
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
2
y = 3.53⋅ ft
3
= 1.00
The critical depth is
y = 3.53⋅ ft
3
= 0.98
=1
Problem 11.20
Given:
Trapezoidal channel
Find:
Critcal depth
[Difficulty: 2]
Solution:
2
Basic equation:
E=y+
V
2⋅ g
The critical depth occurs when the specific energy is minimized
For a trapezoidal channel (Table 11.1) A = y ⋅ ( b + cot(α)⋅ y )
Hence for V
Using this in Eq. 11.14
E is a minimum when
Q
V=
A
=
E=y+
dE
dy
Q
y ⋅ ( b + cot(α)⋅ y )
2
Q
⎤ ⋅ 1
⎡
⎢
⎥
⎣ y ⋅ ( b + cot(α)⋅ y) ⎦ 2 ⋅ g
2
2
Q ⋅ cot(α)
=1−
2
g ⋅ y ⋅ ( b + y ⋅ cot(α))
2
2
g ⋅ y ⋅ ( b + y ⋅ cot(α))
This can be simplified to
3
Q
+
g ⋅ y ⋅ ( b + y ⋅ cot(α))
This expression is the simplest one for y; it is implicit
3
3
g ⋅ y ⋅ ( b + y ⋅ cot(α))
2
Q ⋅ ( b + 2 ⋅ y ⋅ cot(α))
3
3
g ⋅ y ⋅ ( b + y ⋅ cot(α))
2
Q ⋅ cot(α)
Hence we obtain for y
3
Q
−
=1
2
=1
2
=0
Problem 11.19
Given:
Data on rectangular channel
Find:
Depths for twice the minimum energy
[Difficulty: 3]
Solution:
2
E=y+
Basic
equation:
V
2⋅ g
3
ft
Q = V⋅ b ⋅ y
For a rectangular channel
or
2
Hence, using this in the basic eqn.
E=y+
We have a nonlinear implicit equation for y
y+
V=
⎛ Q ⎞ ⋅ 1 =y+
⎜
⎝ b⋅ y ⎠ 2⋅ g
Q
b⋅ y
Q
with
⎛ Q2 ⎞ 1
⎜
⋅
⎜ 2 ⋅ b 2⋅ g y 2
⎝
⎠
b
and
= 10⋅
s
ft
= constant
E = 2 × 2.19⋅ ft
E = 4.38⋅ ft
⎛ Q2 ⎞ 1
⎜
⋅
=E
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
This is a nonlinear implicit equation for y and must be solved numerically. We can use one of a number of numerical root finding
techniques, such as Newton's method, or we can use Excel's Solver or Goal Seek, or we can manually iterate, as below. We start
with a y larger than the critical, and evaluate the left side of the equation so that it is equal to E = 4.38⋅ ft
For
y = 2 ⋅ ft
y+
⎛ Q2 ⎞ 1
⎜
⋅
= 2.39⋅ ft
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
For
y = 4 ⋅ ft
y+
⎛ Q2 ⎞ 1
⎜
⋅
= 4.10⋅ ft
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
For
y = 4.5⋅ ft
y+
⎛ Q2 ⎞ 1
⎜
⋅
= 4.58⋅ ft
⎜ 2
2
⎝ 2⋅ b ⋅ g ⎠ y
For
y = 4.30⋅ ft
y+
⎛ Q2 ⎞ 1
⎜
⋅
= 4.38⋅ ft
⎜ 2
2
⎝ 2⋅ b ⋅ g ⎠ y
Hence
y = 4.30⋅ ft
y = 0.5⋅ ft
y+
⎛ Q2 ⎞ 1
⎜
⋅
= 6.72⋅ ft
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
⎛ Q2 ⎞ 1
y+⎜
⋅
= 4.33⋅ ft
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
For the shallow depth
y = 1 ⋅ ft
y+
⎛ Q2 ⎞ 1
⎜
⋅
= 2.55⋅ ft
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
For
For
y = 0.6⋅ ft
⎛ Q2 ⎞ 1
y+⎜
⋅
= 4.92⋅ ft
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
For
y = 0.65⋅ ft
For
y = 0.645 ⋅ ft
y+
Hence
y = 0.645 ⋅ ft
For
⎛ Q2 ⎞ 1
⎜
⋅
= 4.38⋅ ft
⎜ 2 ⋅ b2⋅ g y 2
⎝
⎠
Problem 11.18
Given:
Data on rectangular channel
Find:
Minimum specific energy; Flow depth; Speed
[Difficulty: 2]
Solution:
2
Basic equation:
E=y+
V
2⋅ g
In Section 11-2 we prove that the minimum specific energy is when we have critical flow; here we rederive the minimum energy point
3
ft
For a rectangular channel
Q = V⋅ b ⋅ y
or
2
Hence, using this in the basic equation
E is a minimum when
The speed is then given by
E=y+
⎛ Q ⎞ ⋅ 1 =y+
⎜
⎝ b⋅ y ⎠ 2⋅ g
⎛ Q2 ⎞ 1
=1−⎜
⋅
=0
⎜ b2⋅ g y 3
dy
⎝
⎠
dE
V =
V=
Q
Q
Q
b
= 10⋅
s
ft
⎛ Q2 ⎞ 1
⎜
⋅
⎜ 2 ⋅ b 2⋅ g y 2
⎝
⎠
= constant
1
⎛ Q2 ⎞
y = ⎜
⎜ b2⋅ g
⎝
⎠
or
V = 6.85⋅
b⋅ y
with
b⋅ y
3
ft
s
1
⎛ g⋅ Q ⎞
⎜
⎝ b ⎠
Note that from Eq. 11.22 we also have
Vc =
The minimum energy is then
Emin = y +
3
ft
Vc = 6.85⋅
s
2
V
2⋅ g
Emin = 2.19⋅ ft
which agrees with the above
y = 1.46⋅ ft
Problem 11.17
Given:
Data on trapezoidal channel
Find:
Critical depth and velocity
[Difficulty: 3]
Solution:
2
V
Basic equation:
E=y+
The given data is:
b = 20⋅ ft
2⋅ g
α = atan ( 2)
S0 = 0.0016
α = 63.4deg
n = 0.025
ft
Q = 400⋅
3
s
2
In terms of flow rate
E=y+
Q
A = y ⋅ ( b + y ⋅ cot ( α ) )
where (Table 11.1)
2
2⋅ A ⋅ g
2
Hence in terms of y
Q
E=y+
2 2
2⋅ ( b + y ⋅ cot ( α ) ) ⋅ y ⋅ g
For critical conditions
dE
dy
2
2
Q
=0=1−
3
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
3
3
2
−
2
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
=1−
3
Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) )
3
g ⋅ y ⋅ ( b + y ⋅ cot( α) )
3
2
Hence
g ⋅ y ⋅ ( b + y ⋅ cot( α) ) − Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) ) = 0
Let
f ( y ) = g ⋅ y ⋅ ( b + y ⋅ cot( α) ) − Q ⋅ ( b + 2 ⋅ y ⋅ cot( α) )
3
2
Q ⋅ cot( α)
3
2
We can iterate or use Excel's Goal Seek or Solver to find y when f(y) = 0
Guess
y = 2 ⋅ ft
f ( y ) = −1.14 × 10
6 ft
7
2
y = 2.25⋅ ft
f ( y ) = −1.05 × 10
5 ft
7
2
s
y = 2.35⋅ ft
5 ft
f ( y ) = 3.88 × 10
s
y = 2.3⋅ ft
5 ft
f ( y ) = 1.36 × 10
7
2
y = 2.275 ⋅ ft
s
Hence critical depth is y = 2.27⋅ ft
and critical speed is
V =
Q
A
and
4 ft
f ( y ) = 1.38 × 10
7
2
y = 2.272 ⋅ ft
s
A = y ⋅ ( b + y ⋅ cot( α) )
V = 8.34⋅
ft
s
2
s
The solution is somewhere between y = 2.25 ft and y = 2.35 ft, as the sign of f(y) changes here.
f ( y ) = −657
ft
2
s
A = 48.0 ft
2
7
7
Problem 11.16
Given:
Rectangular channel flow
Find:
Critical depth
[Difficulty: 1]
1
Solution:
Basic equations:
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
Given data:
b = 2.5⋅ m
3
3
Q = 3⋅
m
s
1
Hence
⎛ Q2 ⎞
yc = ⎜
⎜ g⋅ b2
⎝
⎠
3
y c = 0.528 m
Problem 11.15
[Difficulty: 3]
Given:
Rectangular channel
Find:
Plot of specific energy curves; Critical depths; Critical specific energy
Solution:
Given data:
b=
20
ft
Specific energy:
⎛ Q2
E = y + ⎜⎜
2
⎝ 2 gb
⎞ 1
⎟⎟ 2
⎠ y
Critical depth:
yc
Specific Energy, E (ft·lb/lb)
y (ft)
0.5
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.5
4.0
4.5
5.0
⎛ Q
= ⎜⎜
⎝ gb
2
2
⎞
⎟⎟
⎠
1
3
5
Q =
0
0.50
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.50
4.00
4.50
5.00
Q =
25
0.60
0.67
0.84
1.02
1.22
1.41
1.61
1.81
2.01
2.21
2.40
2.60
2.80
3.00
3.50
4.00
4.50
5.00
Q =
75
1.37
1.21
1.14
1.22
1.35
1.51
1.69
1.87
2.05
2.25
2.44
2.63
2.83
3.02
3.52
4.01
4.51
5.01
Q =
125
2.93
2.28
1.75
1.61
1.62
1.71
1.84
1.99
2.15
2.33
2.51
2.69
2.88
3.07
3.55
4.04
4.53
5.02
Q =
200
6.71
4.91
3.23
2.55
2.28
2.19
2.21
2.28
2.39
2.52
2.67
2.83
3.00
3.17
3.63
4.10
4.58
5.06
y c (ft)
E c (ft)
0.365
0.547
0.759
1.14
1.067
1.60
1.46
2.19
4
3
y (ft)
Q=0
2
Q = 25 cfs
Q = 75 cfs
Q = 125 cfs
Q = 200 cfs
1
0
0
2
4
E (ft)
6
Problem 11.14
Given:
Data on sluice gate
Find:
Downstream depth; Froude number
[Difficulty: 2]
Solution:
Basic equation:
p1
ρ⋅ g
+
V1
2
2⋅ g
V2
p2
2
+ y1 =
+
+ y2 + h
2⋅ g
ρ⋅ g
The Bernoulli equation applies because we have steady,
incompressible, frictionless flow.
Noting that p 1 = p 2 = p atm, (1 = upstream, 2 = downstream) the Bernoulli equation becomes
V1
2
2⋅ g
V2
2
+ y1 =
+ y2
2⋅ g
3
The given data is
For mass flow
m
b = 5⋅ m
y 1 = 2.5⋅ m
Q = 10⋅
Q = V⋅ A
so
Q
and
V1 =
b⋅ y1
2
Using these in the Bernoulli equation
⎛ Q ⎞
⎜ b⋅ y
⎝ 1⎠ + y =
1
2⋅ g
s
Q
V2 =
b⋅ y2
2
⎛ Q ⎞
⎜ b⋅ y
⎝ 2⎠ + y
2
2⋅ g
(1)
2
The only unknown on the right is y2. The left side evaluates to
⎛ Q ⎞
⎜ b⋅ y
⎝ 1 ⎠ + y = 2.53 m
1
2⋅ g
To find y 2 we need to solve the non-linear equation. We must do this numerically; we may use the Newton method or similar, or
Excel's Solver or Goal Seek. Here we interate manually, starting with an arbitrary value less than y 1.
2
⎛ Q ⎞
⎜ b⋅ y
⎝ 2 ⎠ + y = 2.57 m
2
2⋅ g
2
⎛ Q ⎞
⎜ b⋅ y
⎝ 2 ⎠ + y = 2.54 m
2
2⋅ g
y 2 = 0.25⋅ m
⎛ Q ⎞
⎜ b⋅ y
⎝ 2 ⎠ + y = 3.51 m For y = 0.3⋅ m
2
2
2⋅ g
For
y 2 = 0.305 ⋅ m
⎛ Q ⎞
⎜ b⋅ y
⎝ 2 ⎠ + y = 2.50 m For y = 0.302 ⋅ m
2
2
2⋅ g
Hence
y 2 = 0.302 m
is the closest to three figs.
Then
Q
V2 =
b⋅ y2
m
V2 = 6.62
s
For
Fr 2 =
V2
g⋅ y2
2
2
Fr 2 = 3.85
Problem 11.12
Given:
Flow in a rectangular channel with wavy surface
Find:
Froude numbers
[Difficulty: 2]
Solution:
V
Basic equation
Fr =
Available data
b = 10⋅ ft
g⋅ y
y = 6⋅ ft
A "wavy" surface indicates an unstable flow, which suggests critical flow
Hence
Then
V = Fr ⋅ g ⋅ y
V = 13.9
ft
Q = V⋅ b ⋅ y
Q = 834
ft
Fr = 1
s
3
s
5
Q = 3.74 × 10 gpm
Problem 11.12
Given:
Flow in a rectangular channel
Find:
Froude numbers
[Difficulty: 1]
Solution:
V
Basic equation
Fr =
Available data
y = 750 ⋅ mm
Hence
Fr 1 =
Fr 2 =
g⋅ y
V1
g⋅ y
V2
g⋅ y
m
V1 = 1 ⋅
s
m
V2 = 4 ⋅
s
Fr 1 = 0.369
Subcritical flow
Fr 2 = 1.47
Supercritical flow
Problem 11.11
Given:
Motion of sumerged body
Find:
Speed versus ship length
[Difficulty: 2]
Solution:
c=
Basic equation
g⋅ y
We assume a shallow water wave (long wave compared to water depth)
In this case we want the Froude number to be 0.5, with
Fr = 0.5 =
V
and
c
c=
g⋅ x
where x is the ship length
V = 0.5⋅ c = 0.5⋅ g ⋅ x
Hence
Ship Speed (m/s)
100
10
1
1
10
100
Ship Length (m)
3
1× 10
Problem 11.10
Given:
Shallow water waves
Find:
Speed versus depth
[Difficulty: 2]
Solution:
c( y ) =
Basic equation
g⋅ y
We assume a shallow water wave (long wave compared to water depth)
10
Wave Speed (m/s)
Rapid Flow: Fr > 1
1
Tranquil Flow: Fr < 1
0.1
−3
1× 10
0.01
0.1
Depth (m)
1
10
Problem 11.9
Given:
Sharp object causing waves
Find:
Flwo speed and Froude number
[Difficulty: 1]
Solution:
Basic equation
c=
g⋅ y
Available data
y = 150 ⋅ mm
θ = 30⋅ deg
We assume a shallow water wave (long wave compared to water depth)
c =
g⋅ y
so
c = 1.21
m
s
From geometry
Hence
Also
sin( θ) =
Fr =
c
V
V
c
so
Fr = 2
V =
c
sin( θ)
or
V = 2.43
Fr =
m
s
1
sin( θ)
Fr = 2
Problem 11.8
[Difficulty: 2]
Given:
Expression for surface wave speed
Find:
Plot speed versus wavelength for water and mercury waves
Solution:
⎛ g ⋅ λ + 2 ⋅ π⋅ σ ⎞ ⋅ tanh⎛ 2 ⋅ π⋅ y ⎞
⎜
⎜
ρ⋅ λ ⎠
⎝ 2⋅ π
⎝ λ ⎠
Basic equation
c=
Available data
Table A.2 (20oC)
SG Hg = 13.55
SG w = 0.998
ρ = 1000⋅
kg
3
m
Table A.4 (20oC)
Hence
cw( λ) =
σHg = 484 × 10
−3 N
⋅
m
2 ⋅ π⋅ σw ⎞
⎛ g⋅ λ
2 ⋅ π⋅ y ⎞
+
⋅ tanh⎛⎜
⎜
⎝ λ ⎠
⎝ 2 ⋅ π SGw⋅ ρ⋅ λ ⎠
σw = 72.8 × 10
cHg( λ) =
−3 N
⋅
m
y = 7 ⋅ mm
2 ⋅ π⋅ σHg ⎞
⎛ g⋅ λ
2 ⋅ π⋅ y ⎞
+
⋅ tanh⎛⎜
⎜
⎝ λ ⎠
⎝ 2⋅ π SGHg⋅ ρ⋅ λ ⎠
0.7
Water
Mercury
Wave speed (m/s)
0.6
0.5
0.4
0.3
0.2
0.1
20
40
60
Wavelength (mm)
80
100
Problem 11.7
Given:
Expression for capillary wave length
Find:
Length of water and mercury waves
[Difficulty: 1]
Solution:
σ
Basic equation
λ = 2 ⋅ π⋅
Available data
Table A.2 (20oC)
ρ⋅ g
SG Hg = 13.55
SG w = 0.998
ρ = 1000⋅
kg
3
m
σHg = 484 × 10
Table A.4 (20oC)
Hence
λHg = 2 ⋅ π⋅
λ w = 2 ⋅ π⋅
σHg
SG Hg⋅ ρ⋅ g
σw
SGw⋅ ρ⋅ g
−3 N
⋅
m
σw = 72.8 × 10
λHg = 12 mm
λHg = 0.472 in
λw = 17.1 mm
λw = 0.675 in
−3 N
⋅
m
Problem 11.6
[Difficulty: 3]
Given:
Speed of surface waves with no surface tension
Find:
Speed when λ/y approaches zero or infinity; Value of λ/y for which speed is 99% of this latter value
Solution:
g⋅ λ
Basic equation
c=
For λ/y << 1
tanh⎛⎜
(1)
2 ⋅ π⋅ y ⎞
2 ⋅ π⋅ tanh⎛⎜
⎝ λ ⎠
2 ⋅ π⋅ y ⎞
approaches 1
⎝ λ ⎠
Hence c is proportional to
so as λ/y approaches ∞
λ
We wish to find λ/y when
c = 0.99⋅ g ⋅ y
Combining this with Eq 1
0.99⋅ g ⋅ y =
g⋅ λ
2 ⋅ π⋅ tanh⎛⎜
⎝
0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2
Hence
2 ⋅ π⋅ y ⎞
⎝ λ ⎠
=
λ
y
c=
c=
so
g⋅ λ
2⋅ π
g⋅ y
g⋅ λ
2
0.99 ⋅ g ⋅ y =
or
2 ⋅ π⋅ y ⎞
λ
tanh( ∞) → 1
⎠
2 ⋅ π⋅ tanh⎛⎜
2 ⋅ π⋅ y ⎞
⎝ λ ⎠
Letting λ/y = x
we find
0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2
2⋅ π ⎞
⎝ x ⎠
=x
This is a nonlinear equation in x that can be solved by iteration or using Excel's Goal Seek or Solver
Hence
x = 1
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 6.16
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 4.74
x = 4.74
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.35
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.09
x = 5.09
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.2
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.15
x = 5.15
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.17
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.16
x = 5.16
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.17
x = 0.99 ⋅ 2 ⋅ π⋅ tanh⎛⎜
2⋅ π ⎞
x = 5.16
λ
y
= 5.16
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
2
⎝ x ⎠
Problem 11.5
Given:
Pebble dropped into flowing stream
Find:
Estimate of water depth and speed
[Difficulty: 2]
Solution:
Basic equation
c=
g⋅ y
Available data
−5 ⋅ ft
Vwaveupstream =
1⋅ s
ft
Vwaveupstream = −5
s
13⋅ ft
Vwavedownstream =
1⋅ s
ft
Vwavedownstream = 13
s
and relative speeds will be
But we have
Vwavedownstream = Vstream + c
Adding
Vstream =
Subtracting
c =
and
Vwavedownstream + Vwaveupstream
2
Vwavedownstream − Vwaveupstream
2
Vwave = Vstream + c
Vwaveupstream = Vstream − c
ft
Vstream = 4
s
c=9
ft
s
We assume a shallow water wave (long wave compared to water depth)
Hence
c=
g⋅ y
so
y =
c
2
g
y = 2.52⋅ ft
Problem 11.4
Given:
Pebble dropped into flowing stream
Find:
Estimate of water speed
[Difficulty: 1]
Solution:
Basic equation
c=
g⋅ y
Available data
y = 2⋅ m
and relative speeds will be
and
7⋅ m
Vwave =
1⋅ s
We assume a shallow water wave (long wave compared to water depth)
c =
Hence
g⋅ y
so
Vstream = Vwave − c
c = 4.43
m
s
m
Vstream = 2.57
s
Vwave = Vstream + c
m
Vwave = 7
s
Problem 11.3
Given:
Wave from a passing boat
Find:
Estimate of water depth
[Difficulty: 1]
Solution:
Basic equation
c=
g⋅ y
Available data
c = 10⋅ mph
or
c = 14.7
ft
s
We assume a shallow water wave (long wave compared to water depth)
c=
g⋅ y
so
y =
c
2
g
y = 6.69 ft
Problem 11.2
[Difficulty: 2]
Given:
Circular channel
Find:
Derive expression for hydraulic radius; Plot R/D versus D for a range of depths
Solution:
The area is (from simple geometry - a segment of a circle plus two triangular sections)
2
A=
2
2
A=
2
1 D
α D
α
D
D
α
α
⋅ α + 2 ⋅ ⋅ ⋅ sin⎛⎜ π − ⎞ ⋅ ⋅ cos⎛⎜ π − ⎞ =
⋅α +
⋅ sin⎛⎜ π − ⎞ ⋅ cos⎛⎜ π − ⎞
8
8
4
2 2
2⎠ 2
2⎠
2⎠
2⎠
⎝
⎝
⎝
⎝
D
D
8
2
⋅α +
D
2
⋅ sin( 2 ⋅ π − α) =
8
D
8
2
⋅α −
D
8
2
D
⋅ sin( α) =
P=
The wetted perimeter is (from simple geometry)
D
2
8
⋅ ( α − sin( α) )
⋅α
2
D
Hence the hydraulic radius is
R=
A
P
=
8
⋅ ( α − sin( α) )
D
2
R
We are to plot
D
=
1
4
⋅ ⎛⎜ 1 −
⎝
y=
We will need y as a function of α:
=
⋅α
1
4
⋅ ⎛⎜ 1 −
⎝
sin( α) ⎞
α
⎠
⋅D
which is the same as that listed in Table 11.1
sin( α) ⎞
⎠
α
D
2
+
D
2
⋅ cos⎛⎜ π −
⎝
α⎞
2⎠
=
D
2
⋅ ⎛⎜ 1 − cos⎛⎜
⎝
α ⎞⎞
⎝ 2 ⎠⎠
or
y
D
=
1
2
⋅ ⎛⎜ 1 − cos⎛⎜
⎝
⎝ 2 ⎠⎠
The graph can be plotted in Excel.
0.4
R/D
0.3
0.2
0.1
0
0.2
0.4
0.6
y/D
0.8
α ⎞⎞
1
Problem 11.1
[Difficulty: 2]
Given:
Trapezoidal channel
Find:
Derive expression for hydraulic radius; Plot R/y versus y for two different side slopes
Solution:
b = 2⋅ m
Available data
α1 = 30⋅ deg
α2 = 60⋅ deg
The area is (from simple geometry of a rectangle and triangles)
1
A = b ⋅ y + 2 ⋅ ⋅ y ⋅ y ⋅ cot( α) = y ⋅ ( b + y ⋅ cot( α) )
2
The wetted perimeter is (from simple geometry)
P = b + 2⋅
Hence the hydraulic radius is
R=
R
We are to plot
y
=
A
P
=
y ⋅ ( b + y ⋅ cot( α) )
b + 2⋅
y
sin( α)
which is the same as that listed in Table 11.1
sin( α)
b + y ⋅ cot( α)
b + 2⋅
y
with
y
b = 2⋅ m
for α = 30o and 60o, and 0.5 < y < 3 m.
sin( α)
The graph is shown below; it can be plotted in Excel.
0.75
30 Degrees
60 Degrees
R/y
0.5
0.25
0
0.5
1
1.5
2
2.5
3
y (m)
As the depth increases, the hydraulic radius becomes smaller relative to depth y - wetted perimeter becomes dominant over area
Problem 10.120
[Difficulty: 4]
Discussion: When we change the working fluid, we need to be sure that we use the correct similitude
relationships. Specifically, we would need to keep fluid-specific parameters (gas constant and specific heat
ratio) in the relationships. The functional relationships are:
h0 s
ND 2
, ,
m
01 ND 2 ND
P
,
,
,k
f
1
3
c01
01 N 3 D 5
01 ND
So these dimensionless groups need to be considered. When we replace air with helium, both the gas constant R
and the specific heat ratio k will increase. Given a fixed inflow pressure and temperature and a fixed geometry,
the effect would be to decrease density and increase sound speed. Therefore, replacing air with helium should
result in decreased mass flow rate and power, and an increased operating speed.
When considering dimensional parameters, the important thing to remember is that the operability maps for
compressors and/or turbines were constructed for a single working fluid. Therefore, to be safe, an engineer
should reconstruct an operability map for a new working fluid.
Problem 10.119
Given:
Find:
Solution:
Design conditions for jet turbine, off-design actual conditions
New operating speed, mass flow rate, and exit conditions for similar operation
Basic equations:
M T01
η f1
p 01
Given data:
[Difficulty: 3]
p 01d 160 psi
ωd
M T01
T01
M d T01d
p 02
T01
p 01
ω
lbm
s
p 02
M T01
f3
p 02d 80 psi
p 01
T01
ω
T02d 1350 °F
ΔT0d
Solving for the actual mass flow rate:
Solving for the temperature drop:
T01d
p 01d
p 02d
T01
ω ωd
T01d
p 01d
M d 500
Solving for the required speed:
T01
T01 T02 T01d T02d
T01d
p 01
p 01
T01d
T01
ΔT0
T01
M T01
f2
T01 1600 °F
ω
p 01
ΔT0
T01d 1700 °F ωd 500 rpm
p 01 140 psi
At similar conditions:
T01
ω
M Md
T01
ΔT0 ΔT0d
T01d
T01
T02 T01 T01d T02d
T01d
Solving for the exit pressure:
p 02d
p 02 p 01
p 01d
ω 488 rpm
T01d p 01
T01 p 01d
M 448
lbm
s
Substituting in temperatures:
T02 1266 °F
p 02 70 psi
Problem 10.118
Given:
Find:
Solution:
Prototype air compressor equipped with throttle to control entry pressure
Speed and mass flow rate of compressor at off-design entrance conditions
Basic equations:
M T01
η f1
Given data:
[Difficulty: 3]
p 01
p 01d 14.7 psi
T01
ΔT01
ω
T01
T01d 70 °F
M T01
f2
p 01
T01
ω
ωd 3200 rpm T01 58 °F
Since the normalized speed is equal to that of the design point, it follows that:
ω
T01
Solving for the required speed:
At similar conditions:
M T01
p 01
M d 125
lbm
s
p 01 8.0 psi
ωd
T01d
T01
ω ωd
T01d
M d T01d
p 01d
Solving for the actual mass flow rate:
ω 3164 rpm
M Md
T01d p 01
T01 p 01d
M 68.8
lbm
s
Problem 10.117
[Difficulty: 2]
Given:
Prototype air compressor, 1/5 scale model to be built
Find:
Mass flow rate and power requirements for operation at equivalent efficiency
Solution:
M R T01 ω D
p D2 c01
01
M p 8.9
Given data:
M R T01 ω D
f2
3 5
2
c01
ρ01 ω D
p 01 D
Wc
η f1
Basic equations:
kg
s
ωp 600 rpm
Dm
Dp
1
5
Wcp 5.6 MW
Since the efficiencies are the same for the prototype and the model, it follows that:
M m Rm T01m
2
M p Rp T01p
p 01m Dm
p 01p Dp
2
ωm Dm
c01m
ωp Dp
Wcm
c01p
ρ01m ωm Dm
3
5
Wcp
3
ρ01p ωp Dp
5
Given identical entrance conditions for model and prototype and since the working fluid for both is air:
Mm
2
Dm
Mp
Dp
Solving for the mass flow rate of the model:
2
ωm Dm ωp Dp
Wcm
3
5
ωm Dm
Solving for the speed of the model:
3
2
M m 0.356
Solving for the power requirement for the model:
5
kg
s
Dp
ωm ωp
3000 rpm
Dm
3
Wcp
ωp Dp
Dm
Mm Mp
Dp
ωm Dm
Wcm Wcp
ωp Dp
5
Wcm 0.224 MW
Problem 10.116
[Difficulty: 5] Part 1/3
Problem 10.116
[Difficulty: 5] Part 2/3
Problem 10.116
[Difficulty: 5] Part 3/3
Problem 10.115
[Difficulty: 5] Part 1/3
Problem 10.115
[Difficulty: 5] Part 2/3
Problem 10.115
[Difficulty: 5] Part 3/3
Problem 10.114
[Difficulty: 2]
Given:
NASA-DOE wind turbine generator
Find:
Estimate rotor tip speed and power coefficient at maximum power condition
Solution:
Pm
CP
Basic equations:
1
2
and we have ρ 1.23
kg
X
3
ρ V π R
ω 70 rpm
3
2
R 5 m
ω R
V
U ω R
H 18 m
η
Pm
Pideal
2
A 110 m
U ω R 36.652
m
2
m
s
From Fig. 10.45: CP 0.34 when X 5.3 (maximum power condition) If we replace the π R term in the power coefficient
1
3
with the swept area we will get: P CP ρ V A
2
Here are the results, calculated using Excel:
A = 110.00 m
ρ =
U=
2
Power coefficient data were taken from Fig. 10.45
1.23 kg/m
36.65 m/s
V (kt) V (m/s)
10.0
12.5
15.0
17.5
20.0
22.5
25.0
30.0
5.14
6.43
7.72
9.00
10.29
11.57
12.86
15.43
3
X
7.125
5.700
4.750
4.071
3.562
3.167
2.850
2.375
CP
0.00
0.30
0.32
0.20
0.10
0.05
0.02
0.00
P (kW)
0.00
5.40
9.95
9.87
7.37
4.72
2.88
0.00
Power Versus Wind Speed
12
10
P (kW)
8
6
4
2
0
5
10
15
20
V (knots)
25
30
35
Problem 10.113
Given:
Model of farm windmill
Find:
Angular speed for optimum power; Power output
[Difficulty: 2]
Solution:
Basic equations:
CP
P
1
2
From Fig. 10.45
Hence, for
Also
3
ρ V π R
X
2
CPmax 0.3
V 10
m
s
1
3
2
P CPmax ρ V π R
2
ω R
V
ρ 1.225
kg
3
m
X 0.8
at
ω
and we have
X V
R
P 144 W
and
D 1 m
ω 16
R
rad
s
D
2
R 0.5 m
ω 153 rpm
Problem 10.112
[Difficulty: 3]
Problem 10.111
[Difficulty: 2]
Given:
NASA-DOE wind turbine generator
Find:
Estimate rotor tip speed and power coefficient at maximum power condition
Solution:
CP
Basic equations:
Pm
1
2
and we have ρ 0.00237
slug
ft
3
3
ρ V π R
X
2
ω 45 rpm 4.712
rad
s
ω R
V
U ω R
η
Pm
Pideal
R 63 ft V 16 knot 27.005
ft
s
P 135 hp
U ω R 297
The blade tip speed is:
The tip speed ratio is:
X
ω R
V
10.994
η 74%
ft
s
(X will decrease at the wind speed increases.)
P
The mechanical work out is: Pm
182.4 hp
η
From this we can calculate the power coefficient:
CP
Pm
1
2
3
ρ V π R
0.345
2
Problem 10.
[
5]
Problem 10.109
[Difficulty: 4]
9.174
Problem 10.108
[Difficulty: 4]
V2 = V3 = V
y
2h
x
V1
CS
V4
Given:
Definition of propulsion efficiency η
Find:
η for moving and stationary boat
Solution:
Assumption: 1) Atmospheric pressure on CS 2) Horizontal 3) Steady w.r.t. the CV 4) Use velocities relative to CV
The x-momentum (Example 10.3): T u 1 mrate u 4 mrate mrate V4 V1
Applying the energy equation to steady, incompressible, uniform flow through the moving CV gives the minimum power input
requirement
V 2 V 2
4
1
Pmin mrate
2
2
On the other hand, useful work is done at the rate of
Puseful V1 T V1 mrate V4 V1
Combining these expressions
or
η
η
V1 mrate V4 V1
V 2 V 2
1
4
mrate
2
2
V1 V4 V1
1
2
V4 V1 V4 V1
2 V1
V1 V4
When in motion
V1 30 mph
For the stationary case
V1 0 mph
and
V4 90 mph
η
η
2 V1
V1 V4
2 V1
V1 V4
η 50 %
η 0 %
Problem 10.107
9.89
[Difficulty: 4]
Problem 10.106
Given:
Data on jet-propelled aircraft
Find:
Propulsive efficiency
[Difficulty: 3]
y
x
U
V
FD
CS
Y
Solution:
X
Basic equation:
(4.26)
(4.56)
Assumption: 1) Atmospheric pressure on CS 2) Horizontal 3) Steady w.r.t. the CV 4) Use velocities relative to CV
The x-momentum is then
kg
where mrate 50
is the mass flow rate
s
FD U mrate ( V U) U
The useful work is then
The energy equation simplifies to W
η
Hence
U2
mrate
2
mrate ( V U) U
mrate
2
U 225
FD mrate ( V U)
or
With
FD u 1 mrate u 2 mrate ( U) mrate ( V) mrate
m
s
and
2
2
V U
η 45%
V2
mrate
2
2
mrate
V U
2
2
2 ( V U) U
V2 U2
2
1
V
U
V U
2
η
1
V 775
m
s
Problem 10.105
Given:
Data on fanboat and propeller
Find:
Thrust at rest; Thrust at 12.5 m/s
[Difficulty: 3]
Solution:
Assume the aircraft propeller coefficients in Fi.g 10.40 are applicable to this propeller.
At V = 0, J = 0. Extrapolating from Fig. 10.40b
We also have
D 1.5 m
n 1800 rpm n 30
The thrust at standstill (J = 0) is found from
At a speed V 12.5
m
s
CF 0.16
J
V
n D
The thrust and power at this speed can be found
rev
and
s
ρ 1.225
kg
3
m
2
4
FT CF ρ n D
J 0.278
and so from Fig. 10.40b
2
4
FT CF ρ n D
FT 893 N
(Note: n is in rev/s)
FT 809 N
CP 0.44
3
and
CF 0.145
5
P 111 kW
P CP ρ n D
Problem 10.104
[Difficulty: 3]
V1
Given:
Data on boat and propeller
Find:
Propeller diameter; Thrust at rest; Thrust at 15 m/s
V2 = V3 = V
y
2h
x
Solution:
CS
V4
Basic equation:
(4.26)
Assumption: 1) Atmospheric pressure on CS 2) Horizontal 3) Steady w.r.t. the CV 4) Use velocities relative to CV
The x-momentum is then
V
It can be shown (see Example 10.13) that
For the static case
T u 1 mrate u 4 mrate V4 V1 mrate
m
V1 0
s
1
2
V4 V1
kg
where mrate 50
is the mass flow rate
s
m
V4 45
s
so
V
1
V4 V1
2
2
From continuity
π D
mrate ρ V A ρ V
4
4 mrate
Hence
D
For V1 = 0
T mrate V4 V1
When in motion
m
V1 15
s
Hence for V1 = 15 m/s
T mrate V4 V1
V 22.5
m
s
kg
3
m
D 1.52 m
ρ π V
ρ 1.23
with
T 2250 N
and
V
1
V4 V1
2
T 750 N
so
V4 2 V V1
m
V4 30
s
Problem 10.103
[Difficulty: 4]
Problem 10.102
Given:
Hydraulic turbine site
Find:
Minimum pipe size; Fow rate; Discuss
[Difficulty: 4]
Solution:
2
hl
L V
Hl
f
g
D 2 g
Basic equations:
Δz
and also, from Example 10.15 the optimum is when Hl
3
As in Fig. 10.41 we assume L 2 Δz
and
Then, for a given pipe diameter D
V
2
Q V
Also
f =
f 0.02
2 g D Hl
f L
g D
3 f
2
π D
V
Ph ρ Q
2
4
Pm η Ph
Here are the results in Excel:
0.02
ρ = 998.00 kg/m3
η = 83%
3
25
30
35
40
45
50
6.39
7.00
7.56
8.09
8.58
9.04
0.314
0.495
0.728
1.016
1.364
1.775
6.40
12.12
20.78
33.16
50.09
72.42
Pm (kW)
5.31
10.06
17.25
27.53
41.57
60.11
41.0
8.19
1.081
36.14
30.00
D (cm) V (m/s) Q (m /s) P h (kW)
Turbine efficiency varies with specific speed
Pipe roughness appears to the 1/2 power, so has a secondary effect.
A 20% error in f leads to a 10% change in water speed
and 30% change in power.
A Pelton wheel is an impulse turbine that does not flow full of water;
it directs the stream with open buckets.
A diffuser could not be used with this system.
Use Goal Seek or Solver to vary D to make Pm 30 kW!
Power Versus Pipe Diameter
70
60
Pm (kW)
50
40
30
20
10
0
20
25
30
35
D (cm)
40
45
50
55
Problem 10.101
Given:
Find:
[Difficulty: 3]
Published data for the Tiger Creek Power Plant
(a) Estimate net head at the site, turbine specific speed, and turbine efficiency
(b) Comment on consistency of the published data
Solution:
NScu
Basic Equations:
N P
5
H
N P
NS
5
4
ρ ( g H)
η
P
ρ Q g Hnet
4
The given or available data is
kg
ρ 999
3
3
P 58 MW
Q 21
m
m
Hgross 373 m
s
ν 1.14 10
2
6 m
s
Using data from Fig. 10.37, we will assume η 87% We can take this to estimate the net head:
Therefore:
Hnet
Hgross
86.875 %
P
Hnet
324 m
ρ Q g η
This is close to 87%, so the assumption for the efficiency was a good one.
From the same figure, we will assume NScu 5
Therefore the dimensionless specific speed is
NScu
NS
0.115
43.46
5
We may then calculate the rotational speed for the turbine:
N
NS ρ g Hnet
4
108.8 rpm
P
The power output seems low for a turbine used for electricity generation; several turbines are probably used in this one plant.
To check the claims:
58 MW
58 MW
24 hr
1 day
s
3
21 m
365 day
yr
hr
3600 s
8 kW hr
5.081 10
0.767
kW hr
2
m m
This number is 50% higher than the claim.
yr
This is in excellent agreement with the claim.
Problem 10.100
[Difficulty: 4]
Problem 10.99
Given:
Data on impulse turbine
Find:
Plot of power and efficiency curves
[Difficulty: 2]
Solution:
T F R
Basic equations:
H =
33
ρ =
R =
1.94
0.50
P ω T
η
P
Here are the results calculated in Excel:
ρ Q g H
ω (rpm) Q (cfm) F (lbf) T (ft-lbf) P (hp)
ft
slug/ft3
ft
0
1000
1500
1900
2200
2350
2600
2700
7.74
7.74
7.74
7.44
7.02
5.64
4.62
4.08
2.63
2.40
2.22
1.91
1.45
0.87
0.34
0.09
1.32
1.20
1.11
0.96
0.73
0.44
0.17
0.05
0.000
0.228
0.317
0.345
0.304
0.195
0.084
0.023
η (%)
0.0%
47.3%
65.6%
74.4%
69.3%
55.3%
29.2%
9.1%
Turbine Performance Curves
100%
0.40
90%
0.35
80%
0.30
70%
60%
50%
0.20
40%
0.15
30%
0.10
20%
0.05
10%
0.00
0%
0
500
1000
1500
ω (rpm)
2000
2500
3000
η (%)
P (hp)
0.25
Problem 10.98
[Difficulty: 3]
Given:
Impulse turbine requirements
Find:
1) Operating speed 2) Wheel diameter 4) Jet diameter 5) Compare to multiple-jet and double-overhung
1
Solution:
Vj
Basic
equations:
2 g H
NS
ω P
2
1
5
2
4
ρ h
Model as optimum. This means. from Fig. 10.10
Given or available data
H 350 m
η
P
U 0.47 Vj
and from Fig. 10.17 NScu 5
P 15 MW
ρ 1.94
Vj
m
2 g H
Vj 82.9
s
U 0.47 Vj
U 38.9
D
Q 4.91
η ρ g H
1
5
2
4
ρ ( g H)
2 U
s
NScu
NS
43.46
NS 0.115
m
s
(1)
1
P
The wheel radius is
m
3
P
ω NS
For a single jet
η 89 %
3
We need to convert from N Scu (from Fig. 10.17) to NS (see discussion after Eq. 10.18b).
The water consumption is Q
with
slug
ft
Then
Q Vj Aj
ρ Q g H
ω 236 rpm
Dj
4 Q
(2)
π Vj
Dj 0.275 m
2
D 3.16 m
(3)
ω
For multiple (n) jets, we use the power and flow per jet
ωn ω n
From Eq 1
Results:
From Eq. 2
ωn ( n )
n
rpm
Djn
Dj
an
d
n
Djn( n )
0.275
Dn
n
Dn ( n )
1
236
2
333
0.194
2.23
3
408
0.159
1.82
4
471
0.137
1.58
5
527
0.123
1.41
A double-hung wheel is equivalent to having a single wheel with two jets
D
m
3.16
m
from Eq.
3
Problem 10.97
Given:
Find:
Solution:
[Difficulty: 3]
Data on Pelton wheel
Rotor radius, jet diameter, water flow rate.
The given or available data is
ρ 999
kg
3
2
6 m
Wmech 26.8 MW ω 225 rpm H 360 m
ν 1.14 10
m
From Bernoulli, the jet velocity is:
m
Vj Cv 2 g H 82.35
s
Vi
2 g H Assuming a velocity coefficient of
s
Cv 0.98
From Fig. 10.36, at maximum efficiency: U R ω 0.47 Vj
(4% loss in the nozzle):
So the radius can be calculated:
R 0.47
From Fig. 10.37 the efficiency at full load is η 86% Thus: η
Wmech
Q ρ g H
ω
1.643m
Solving for the flow rate:
Q
π 2
Q
We can now calculate the jet velocity: Aj Dj
Therefore,
Vj
4
Vj
Q
Dj 2
0.37 m
π Vj
3
Wmech
η ρ g H
8.836
m
s
Dj 37.0 cm
mrate ρ Q 8.83 10
3 kg
s
Problem 10.96
[Difficulty: 3]
10.39
10.39
Problem 10.95
[Difficulty: 2]
Problem 10.94
Given:
Find:
Solution:
Basic
equations:
[Difficulty: 2]
Data on Francis turbines at Niagra Falls
Specific speed, volume flow rate to each turbine, penstock size
1
Wh ρ Q g H
η
Wmech
NS
Wh
ω P
2
2
1
5
2
4
ρ h
h g H
h lT f
L V
D 2
The given or available data is
ρ 998
kg
3
Wmech 54 MW
ω 107 rpm
η 93.8%
H 65 m
Lpenstock 400 m Hnet H 83%
m
1
2
h g H 637.4
The specific energy of the turbine is:
m
NS
The specific speed is:
2
s
ω Wmech
1
5
2
4
ρ h
Solving for the flow rate of the turbine:
Wmech
Q
ρ h η
NS 0.814
3
90.495
3
m
Q 90.5
s
2
m
h lT g H Hnet 108.363
2
s
Based on the head loss:
2
Since
V
Q
A
4 Q
m
s
into the head loss equation:
2
π D
1
2
2
L 1 4 Q
8 f L Q
h lT f
2
2
5
D 2
π
D
π
D
Assuming concrete-lined penstocks:
D (m)
2.000
3.510
3.414
3.418
V (m/s)
28.807
9.354
9.888
9.862
8 f L Q2
Solving for the diameter: D
π2 h
lT
e 3 mm
Re
5.70E+07
3.25E+07
3.34E+07
3.34E+07
5
This will require an iterative solution.
If we assume a diameter of 2 m, we can iterate to find the actual diameter:
e /D
0.001500
0.000855
0.000879
0.000878
f
0.02173
0.01892
0.01904
0.01904
D (m)
3.510
3.414
3.418
3.418
D 3.42 m
Problem 10.93
[Difficulty: 2]
V1
U = R
Vj
D
Given:
Pelton turbine
Find:
1) Power 2) Operating speed 3) Runaway speed 4) Torque 5) Torque at zero speed
Solution:
2
2
p
h
p
V1
Vj
1
j
lT
α
z
α
z
ρ g
1
j
2 g
2 g
g
ρ g
Basic equations
2
V
h lT h l h lm K
2
and from Example
Tideal ρ Q R Vj U ( 1 cos( θ) )
θ 165 deg
10.5
Assumptions: 1) p j = pamt 2) Incompressible flow 3) α at 1 and j is approximately 1 4) Only minor loss at nozzle 5) z 1 = z j
Given data
Then
and
Hence
p 1g 700 psi
V1 15 mph
d 7.5 in
D 8 ft
p 1g
V1
2
2
K Vj
ρ g
2 g
2 g
g 2
π d
Vj
2
Vj
o
r
ft
3
Q Vj
4
Q 97.2
P η ρ Q g H
P 15392 hp
s
Urun Vj
T η Tideal
Stall occurs when
U 0
K 0.04
p 1g
ρ g
U 149
ρ 1.94
slug
ft
1K
ωrun
From Example 10.5 Tideal ρ Q R Vj U ( 1 cos( θ) )
Hence
H
η 86 %
2
p
V1
1g
2
2
ρ
2
From Fig. 10.10, normal operating speed is around U 0.47 Vj
At runaway
ft
V1 22
s
D
R
2
V1
H 1622 ft
ω
s
ft
Vj 317
s
2
2 g
ft
3
U
R
Urun
ω 37.2
rad
s
rad
ωrun 79.2
s
D
2
ω 356 rpm
ωrun 756 rpm
5
Tideal 2.49 10 ft lbf
5
T 2.14 10 ft lbf
Tstall η ρ Q R Vj ( 1 cos ( θ) )
5
Tstall 4.04 10 ft lbf
Problem 10.92
[Difficulty: 2]
Problem 10.91
Given:
Data on turbine system
Find:
Model test speed; Scale; Volume flow rate
[Difficulty: 3]
1
Solution:
Wh ρ Q g H
Basic equations:
η
Wmech
NS
Wh
slug
ft
3
2
1
5
2
4
ρ h
The given or available data is
ρ 1.94
ω P
Wp 36000 hp
Hp 50 ft
ωp 95 rpm
Hm 15 ft
Wm 50 hp
where sub p stands for prototype and sub m stands for model
Note that we need h (energy/mass), not H (energy/weight) h p Hp g
h p 1609
Hence for the prototype
NS
5
NS
2
4
ωm Wm
For dynamically similar conditions
5
2
4
Hp
2
Also
Qp
ωp Dp
2
ωp Dp
3
2
1
ρ hm
ωm NS
1
5
2
4
ρ hm
Hm
2
so
Dp
2
Qm
Dm
so
3
ωm Dm
rad
ωm 59.3
s
2
ωp
ωm
Hm
Hp
0.092
Dm
Qm Qp
ωp Dp
ωm
3
To find Q p we need efficiency. At Wp 36000 hp and Hp 50 ft from F ig. 10.17 we find (see below), for
1
NScu
ft
2
2
s
1
Wm
ωm Dm
2
h m 482.6
NS 3.12
1
Then for the model
h m Hm g
2
1
ρ hp
2
s
1
ωp Wp
ft
N( rpm) P( hp)
5
H( ft)
4
2
135.57
η 93 %
ωm 566 rpm
Hence from
and also
η
Wmech
Wh
Wmech
ρ Q g H
Wm
Qm
ρ g Hm η
Wp
Qp
ρ g Hp η
6
Qp 3.06 10 gpm
4
Qm 1.418 10 gpm
Problem 10.90
[Difficulty: 3]
Problem 10.89
[Difficulty: 4]
10.88
10.88
Problem 10.88
[Difficulty: 4]
Problem 10.87
[Difficulty: 3]
Given:
Data on centrifugal fan
Find:
Fan outlet area; Plot total pressure rise and power; Best effiiciency point
Solution:
ηp
Basic equations:
Wh
p dyn
At Q 200
ft
Wh Q Δpt
Wm
1
2
Δp ρw g Δht
(Note: Software cannot render a dot!)
2
ρair V
3
we have h dyn 0.25in
s
ρw
V
Hence
ρair
Q V A
2 g h dyn
and
h dyn
and
A
Δht Δh h dyn
ρw g
ρair V2
ρw 2
Q
V
The velocity V is directly proportional to Q, so the dynamic pressure at any flow rate Q is
The total pressure Δh t will then be
p dyn
3
ft
200 s
h dyn 0.25 in
Q
2
Δh is the tabulated static pressure rise
Here are the results, generated in Excel:
At Q =
h dyn =
200
0.25
3
ft /s
in
Hence
V = 33.13
ft/s
A = 6.03749 ft2
ρw =
1.94
slug/ft3
ρ air = 0.00237 slug/ft
Fitting a 2nd order polynomial to each set of data we find
3
-5
2
-3
h t =-3.56x10 Q + 6.57x10 Q + 1.883
3
Q (ft /s) Δp (psi) Pm (hp)
h dyn (in)
h t (in) Ph (hp) η (%)
106
141
176
211
0.075
0.073
0.064
0.050
2.75
3.18
3.50
3.51
0.07
0.12
0.19
0.28
2.15
2.15
1.97
1.66
2.15
2.86
3.27
3.32
246
282
0.033
0.016
3.50
3.22
0.38
0.50
1.29
0.94
3.01
2.51
-4
2
P h = -1.285x10 Q + 0.0517Q - 1.871
-5
2
78.2%
η =-3.37x10 Q + 0.0109Q -0.0151
90.0%
93.5% Finally, we use Solver to maximize η by varying Q :
94.5%
3
Q (ft /s)
η (%)
P h (hp)
h t (in)
85.9%
77.9%
A plot of the performance curves is shown on the next page.
161.72
2.01
3.13
86.6%
Fan Performance Curve
3.5
100%
3.0
h t (cm), Ph (kW)
2.0
50%
1.5
1.0
25%
0.5
0.0
100
120
140
160
180
200
3
Q (ft /s)
220
240
260
280
0%
300
η (%)
75%
2.5
Problem 10.86
Given:
Data on centrifugal fan and various sizes
Find:
Suitable fan; Fan speed and input power
[Difficulty: 3]
Solution:
Q'
Basic
equations:
Q
ω' D'
ω D
3
h'
h
ω'
ω
2
D'
D
2
P'
P
ω'
ω
3
5
D'
D
We choose data from the middle of the table above as being in the region of the best efficiency
Q 176
ft
3
s
Δp 0.064 psi
P 3.50 hp and
ω 750 rpm
D 3 ft
ρw 1.94
slug
ft
The flow and head are
Q' 600
ft
3
h' 1 in At best efficiency point: h
s
Δp
ρw g
3
1.772 in
These equations are the scaling laws for scaling from the table data to the new fan. Solving for scaled fan
speed, and diameter using the first two equations
1
ω' ω
2
Q
3
h'
1
4
Q' h
ω' 265 rpm
D' D
Q'
2
1
h
4
Q h'
D' 76.69 in
This size is too large; choose (by trial and error)
Q 246
ω' ω
ft
3
h
s
3
2
4
Q
1
h'
Q' h
0.033 psi
ρw g
ω' 514 rpm
0.914 in
P 3.50 hp
D' D
Q'
1
1
2
4
h
Q h'
D' 54.967 in
Hence it looks like the 54-inch fan will work; it must run at about 500 rpm. Note that it will NOT be running at best
efficiency. The power will be
P' P
ω'
3
D'
ω D
5
P' 9.34 hp
Problem 10.85
[Difficulty: 3]
Given:
Data on centrifugal fan and square metal duct
Find:
Minimum duct geometry for flow required; Increase if fan speed is increased
Solution:
Wh
Basic
equations:
ηp
Wh Q Δp
and for the duct
L V
Δp ρair f
Dh 2
and fan scaling
Q 200
Wm
2
ft
Δp ρw g Δh
(Note: Software cannot render a dot!)
2
4 A
4 H
Dh
H
P
4 H
3
s
ω 750 rpm
ω' 1000 rpm
Q'
ω'
ω
Q
Q' 266.67
Here are the results, calculated using Excel:
ρw =
1.94
slug/ft3
ρ air = 0.00237 slug/ft
Fitting a 2nd order polynomial to each set of data we find
3
-6
ν air = 1.58E-04 ft /s
L=
50
ft
Assume smooth ducting
Note: Efficiency curve not needed for this problem.
We use the data to get a relationship for pressure increase.
Q (ft3 /s) Δp (psi) Pm (hp) P h (hp) η (%)
106
141
176
211
246
282
0.075
0.073
0.064
0.050
0.033
0.016
2.75
3.18
3.50
3.51
3.50
3.22
2
-4
Δp =-1.51x10 Q + 2.37x10 Q + 0.0680
2
2.08
2.69
2.95
2.76
2.13
1.18
75.7%
84.7%
84.3%
78.7%
60.7%
36.7%
Now we need to match the pressure loss in the duct with
the pressure rise across the fan. To do this, we use
Solver to vary H so the error in p is zero
Fan
Q (ft3 /s) Δp (psi)
266.67
H (ft)
V (ft/s)
1.703
91.94
Answers:
Q (ft3/s)
200.00
A plot of the performance curve is shown on the next page.
Re
9.91.E+05
f
0.0117
0.0238
Duct
Δp (psi)
0.0238
Error in Δp
0.00%
H (ft)
Q (ft3 /s)
H (ft)
1.284
266.67
1.703
ft
3
s
Fan Performance Curve
0.08
100%
0.07
0.06
75%
0.04
50%
0.03
0.02
25%
0.01
0.00
100
120
140
160
180
200
3
Q (m /s)
220
240
260
280
0%
300
η (%)
Δp (mm)
0.05
Problem 10.84
Given:
Data on centrifugal fan
Find:
Plot of performance curves; Best effiiciency point
[Difficulty: 3]
Solution:
ηp
Basic
equations:
Wh
Wh Q Δp
Wm
Δp ρw g Δh (Note: Software cannot render a dot!)
Here are the results, calculated using Excel:
ρw =
slug/ft3
1.94
Fitting a 2nd order polynomial to each set of data we find
-6
2
-4
Δp =-1.51x10 Q + 2.37x10 Q + 0.0680
-5
3
Q (ft /s) Δp (psi) Pm (hp) P h (hp) η (%)
106
141
176
0.075
0.073
0.064
2.75
3.18
3.50
2.08
2.69
2.95
75.7%
84.7%
84.3%
211
246
282
0.050
0.033
0.016
3.51
3.50
3.22
2.76
2.13
1.18
78.7%
60.7%
36.7%
2
η =-3.37x10 Q + 0.0109Q -0.0151
Finally, we use Solver to maximize η by varying Q :
3
Q (ft /s)
Δp (psi)
η (%)
161.72
0.0668
86.6%
Fan Performance Curve
BEP
0.08
100%
0.07
0.06
75%
η
Δp
0.04
50%
0.03
0.02
25%
0.01
0.00
100
120
140
160
180
200
3
Q (ft /s)
220
240
260
280
0%
300
η (%)
Δp (psi)
0.05
Problem 10.83
[Difficulty: 4] Part 1/2
Problem 10.83
[Difficulty: 4] Part 2/2
Problem 10.82
[Difficulty: 4]
Given:
Swimming pool filtration system, filter pressure drop is Δp=0.6Q2, with Δp in psi and Q in gpm
Find:
Speed and impeller diameter of suitable pump; estimate efficiency
Solution:
We will apply the energy equation for steady, incompressible pipe flow.
Basic equations:
2
2
p
p
2
2
V1
V2
Le V2
1
2
L V
V
ρ α1 2 g z1 ρ α2 2 g z2 h lT hp h lT f D 2 Σ f D 2 Σ K 2
Q 30 gpm Q 1.893 10
The given or available data are:
ρ 1.93
slug
ft
3
ρ 995
3
3m
s
kg
ν 1.06 10
D 20 mm
3
2
5 ft
s
ν 9.848 10
H
h
g
2
7m
s
e 0 mm
m
Setting state 1 at the pump discharge, state 2 at the tee, state 3a downstream of the filter, and state 3b after the 40 ft pipe, we can look
at the pressure drop between 1 and 2:
V1 V2
e
D
e
D
0
0
V
Q
A
4 Q
2
V 6.025
π D
m
Re
s
V D
ν
1.224 10
5
e 1.11 6.9
f 1.8 log
Re
3.7 D
Therefore we can calculate the friction factor:
2
0.017
2
Le 0 K 0 and therefore the pressure drop is:
Since this is a straight run of pipe:
Δp12 ρ f
V
L
2 D
Δp12 47.04 kPa
Since both legs exhaust to the same pressure, the pressure drops between the two must be equal, and the flow rates must equal the
total flow rate of the system. This requires an iterative solution, using Solver in Excel. The result is:
3
3 m
Qa 1.094 10
s
3
4 m
Qb 7.99 10
s
The resulting pressure drop is
Δp23 42.96 kPa
Neglecting any pressure at the pump inlet, the pump must supply:
Δppump Δp12 Δp23 90.0 kPa
Δppump
The resulting head is: Hpump
9.226 m in U.S. units: Hpump 30.269 ft
ρ g
This head is too low for any of the pumps in Fig. D.1. Therefore, assuming a speed of 3500 rpm:
In customary units:
The pump power is:
Ncu 2733 N 1485 So from Figure 10.9 we can estimate the efficiency:
Wp
ρ Q g Hpump
η
262.056 W
N
ω Q
g Hpump
0.75
0.544
η 65 %
Wp 262.1 W
Problem 10.81
[Difficulty: 4]
Given:
Manufacturer data for a pump
Find:
(a) Plot performance and develop curve-fit equation.
(b) Calculate pump delivery vs discharge height for length of garden hose
Solution:
h lT f
Basic equations:
2
2
Le V2
L V
V
f
K
2
D 2
D 2
H
h
2
Hp H0 A Q
g
2
h lT f
For this case, Le = K = 0, therefore:
Given data:
L =
e =
D =
ρ =
Here are the results calculated in Excel:
Here are the data for the head generated by the pump, as well as the head losses for the hose and the pipe:
15
0
20
m
ft
mm
998
kg/m3
D =
e =
2
ν = 1.01E-06 m /s
H 0= 7.48727 m
A=
L V
D 2
0.0012 m/(L/min)
2
2
20
0
D =
e =
mm
mm
25
0.15
z (m)
Q (L/min)
Q
z fit (m)
V (m/s)
Re a
fa
H L (m)
V (m/s)
0.3
0.7
1.5
77.2
75.0
71.0
5959.840
5625.000
5041.000
0.320
0.722
1.425
4.096
3.979
3.767
8.11E+04
7.88E+04
7.46E+04
0.0188
0.0189
0.0191
12.1
11.4
10.4
2.621
2.546
2.411
6.49E+04 0.0334
6.30E+04 0.0334
5.97E+04 0.0335
7.0
6.6
6.0
3.0
4.5
6.0
8.0
61.0
51.0
26.0
0.0
3721.000
2601.000
676.000
0.000
3.012
4.359
6.674
7.487
3.236
2.706
1.379
0.000
6.41E+04
5.36E+04
2.73E+04
0.00E+00
0.0198
0.0206
0.0240
0.0000
7.9
5.8
1.7
0.0
2.071
1.732
0.883
0.000
5.13E+04
4.29E+04
2.19E+04
0.00E+00
4.4
3.1
0.8
0.0
Head (m)
Head Versus Flow Rate for Pump
10
9
8
7
6
5
4
3
2
1
0
Data
Fit
Hose
Pipe
0
10
mm
mm
20
30
40
Q (L/min)
50
60
70
80
Re a
fa
0.0337
0.0340
0.0356
0.0000
H L (m)
To determine the discharge heights for the hose and the pipe,
For the hose:
Re a
Q (L/min) V (m/s)
0.0
0.000
0.00E+00
10.0
0.531
1.05E+04
20.0
1.061
2.10E+04
30.0
1.592
3.15E+04
40.0
2.122
4.20E+04
50.0
2.653
5.25E+04
60.0
3.183
6.30E+04
we subtract the head loss from the head generated by the pump.
For the pipe:
Re a
fa
H L (m) Disch (m) V (m/s)
fa
0.0000
0.000
7.487
0.000 0.00E+00 0.0000
0.0305
0.328
7.039
0.340 8.40E+03 0.0398
0.0256
1.101
5.906
0.679 1.68E+04 0.0364
0.0232
2.248
4.157
1.019 2.52E+04 0.0351
0.0217
3.740
1.823
1.358 3.36E+04 0.0345
0.0207
5.558
-1.077
1.698 4.20E+04 0.0340
0.0199
7.689
-4.531
2.037 5.04E+04 0.0337
H L (m) Disch (m) % Diff
0.000
7.487
0%
0.140
7.227
-3%
0.514
6.492
-9%
1.115
5.290
-21%
1.943
3.620
-50%
2.998
1.483
4.279
-1.122
Flow Rate Versus Discharge Height
Flow Rate (L/min)
60
50
Hose
Pipe
40
30
20
10
0
0
1
2
3
4
5
6
7
Discharge Height (m)
The results show that the 15% performance loss is an okay "ball park" guess at the lower flow rates, but not very good
at flow rates above 30 L/min.
8
Problem 10.80
[Difficulty: 4]
Given:
Fire nozzle/pump system
Find:
Appropriate pump; Impeller diameter; Pump power input needed
Solution:
Basic equations
2
2
2
p
V2
V3
p3
2
L V2
ρ α 2 g z2 ρ α 2 g z3 h l h l f D 2
for the hose
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 2 and 3 is approximately 1 4) No minor loss
2
2
p
V2
V1
p1
2
ρ α 2 g z2 ρ α 2 g z1 h pump
for the pump
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) No minor loss
The first thing we need is the flow rate. Below we repeat Problem 8.179 calculations
Hence for the hose
Δp
ρ
p2 p3
ρ
2
f
L V
D 2
or
2 Δp D
V
ρ f L
We need to iterate to solve this for V because f is unknown until Re is known. This can be done using Excel's Solver, but here:
Δp 750 kPa
L 100 m
Make a guess for f f 0.01
Given
2 Δp D
V 7.25
ρ f L
2 Δp D
ρ f L
V 5.92
2 Δp D
ρ f L
V 5.81
m
Re
s
2 Δp D
ρ f L
s
Re
V D
kg
ν 1.01 10
3
2
6 m
m
ν
Re 2.51 10
5
V D
ν
Re 2.05 10
5
Re 2.01 10
5
Re 2.01 10
5
f 0.0156
m
Re
s
e
D
1
2.51
2.0 log
f
3.7 Re f
V
m
ρ 1000
f 0.0150
e
D
1
2.51
2.0 log
f
3.7 Re f
V
Given
V
D 3.5 cm
e
D
1
2.51
2.0 log
f
3.7 Re f
V
Given
e 0
V 5.80
V D
ν
f 0.0156
m
s
Re
V D
ν
s
2
Q
π D
V
4
Q 5.578 10
3
3m
3
Q 0.335
s
We have
p 1 350 kPa
For the pump
2
2
p
V2
V1
p1
2
ρ α 2 g z2 ρ α 2 g z1 h pump
so
h pump
p 2 700 kPa 750 kPa
p2 p1
Hpump
or
ρ
m
min
p 2 1450 kPa
p2 p1
Hpump 112 m
ρ g
3
We need a pump that can provide a flow of Q 0.335
m
min
or Q 88.4 gpm, with a head of Hpump 112 m or Hpump 368 ft
From Appendix D, Fig. D.1 we see that a Peerless 2AE11 can provide this kind of flow/head combination; it could also handle four
such hoses (the flow rate would be 4 Q 354 gpm). An impeller diameter could be chosen from proprietary curves.
The required power input is
Wh
Wm
ηp
Wm
Prequired
Ppump
η
where we choose ηp 75 % from Fig. 10.15
ρ Q g Hpump
Prequired
ηp
6.14 kW
70 %
Wm 8.18 kW
for one hose or
Prequired 8.77 kW
or
4 Wm 32.7 kW
for four
4 Prequired 35.1 kW for four
Problem 10.79
Given:
Find:
[Difficulty: 4]
Sprinkler system for lakeside home
(a) Head loss on suction side of pump
(b) Gage pressure at pump inlet
(c) Hydraulic power requirement for the pump
(d) Change in power requirement if pipe diameter is changed
(e) Change in power requirement if the pump were moved
Solution:
L34 = 45 m
30 m
L12 = 20 m
3m
We will apply the energy equation for steady, incompressible pipe flow.
Basic equations:
2
2
p
p
V1
V2
1
2
α
g
z
α
g
z
ρ
1 2
1 ρ
2 2
2 h lT h p
2
h lT f
2
2
Le V
L V
V
Σ f
Σ K
2
D 2
D 2
H
h
g
Assumptions: 1) p 1 = patm 2) V 1 = 0
The given or available data is
Q 40
L
min
D 2 cm
e 0.15 mm
z1 0 m z2 3 m z3 z2
From Table A.8 at 20 oC ν 1.01 10
2
6 m
s
At the specified flow rate, the speed of the water is:
e
D
7.5 10
3
Between 1 and 2:
p atm 101.3 kPa
z4 33 m
p v 2.34 kPa
ρ 998
p 4 300 kPa (gage)
L12 20 m L34 45 m
kg
3
m
V
Q
A
4 Q
2
π D
Therefore we can calculate the friction factor:
V 2.122
m
s
Re
V D
ν
e 1.11 6.9
f 1.8 log
Re
3.7 D
2
2
p2
L12 Le V2
V
V
α2
g z2 f
K
2
D 2
2
ρ
D
4.202 10
4
2
0.036
In this case: Le ( 30 16) D
K 0.78
L12 Le V
V
HlT12 f
K
D 2 g
2 g
D
2
The head loss before the pump is:
V2
p 2 ρ
Solving for pressure at 2:
2
HlT12 8.844 m
2
L12 Le V
V
g z2 f
K
2
D 2
2
D
2
p 2 54.946 kPa (gage)
To find the pump power, we need to analyze between 3 and 4:
2
p3
p4
L34 Le V2
V
g z3 g z4 f
K
D 2
2
ρ
ρ
D
L34 Le V
p 3 p 4 ρ g z4 z3 f
D 2
D
2
Now we can calculate the power:
In this case: Le ( 16 16) D
p 3 778.617 kPa
Hp
Thus the pump head is:
V
A
4 Q
2
V 0.531
π D
m
s
V D
Re
ν
2.101 10
4
e
D
2
p 2 ρ
Le ( 16 16) D
K 0
L34 Le V2
p 3 p 4 ρ g z4 z3 f
D 2
D
Hp
p3 p2
ρ g
58.44 m
2
L12 Le V
V
g z2 f
K
2
D 2
2
D
K 0.78
D 4 cm
e 1.11 6.9
f 1.8 log
Re
3.7 D
3
3.75 10
V2
85.17 m
Wp 556 W
Le ( 30 16) D
ρ g
Wp ρ g Q Hp
Changing to 4 centimeter pipe would reduce the mean velocity and hence the head loss and minor loss:
Q
p3 p2
K 0
2
0.032
p 2 26.922 kPa (gage)
p 3 778.617 kPa (gage)
Wpnew ρ g Q Hp 381.283 W
ΔWp
Wpnew Wp
Wp
31 %
2
The pump should not be moved up the hill. The NPSHA is:
NPSHA
V
p 2 p atm ρ
pv
2
If anything, the pump should be moved down the hill to increase the NPSHA.
ρ g
4.512 m
for 2-cm pipe.
Problem 10.78
[Difficulty: 4]
8.158
Problem 10.77
8.124
[Difficulty: 4] Part 1/2
Problem 10.77
[Difficulty: 4] Part 2/2
Problem 10.76
Given:
Data on flow from reservoir/pump
Find:
Appropriate pump; Reduction in flow after 10 years
Solution:
[Difficulty: 4]
2
2
p
p
V1
V4
1
4
ρ g α 2 g z1 ρ g α 2 g z4 HlT Hp
Basic equation:
for flow from 1 to 4
2
2
Le V2
L V
V
HlT f
f
K
2 g
D 2 g
D 2 g
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V 2 = V3 = V4 (constant area pipe)
ρ 1000
Given or available data
kg
ν 1.01 10
3
2
6 m
m
p 2 150 kPa
For minor losses we have
p v 2.34 kPa
(Table A.8)
Q 0.075
3
p 3 450 kPa
D 15 cm
e 0.046 mm
z1 20 m
z4 35 m
V
Le
Four elbows:
At the pump inlet
s
NPSHA
The head rise through the pump is Hp
p2
p3 p2
ρ g
D
1
2
4 Q
V 4.24
2
π D
4 12 48
(Fig. 8.16)
Square inlet:
m
s
m
s
Kent 0.5
2
ρ V p v
NPSHA 16.0 m
ρ g
Hp 30.6 m
3
Hence for a flow rate of Q 0.075
m
or Q 1189 gpm and Hp 30.6 m or Hp 100 ft, from
s
Appendix D. Fig. D3 a Peerless4AE11 would suffice
2
2
Le V2
L V
V
We do not know the pipe length L! Solving the energy equation for it:z1 z4 HlT Hp f
f
Kent
Hp
2 g
D 2 g
D 2 g
For f
Given
Re
V D
ν
5
Re 6.303 10
e
D
1
2.51
2.0 log
f
3.7 Re f
and
f 0.0161
e
D
4
3.07 10
L
Hence, substituting values
Le Kent D
z1 z4 Hp D
2
f
D
f V
2 g D
From Problem 10.63, for a pipe D 0.15 m or D 5.91 in, the aging over 10 years leads to
L 146 m
fworn 2.2 f
We need to solve the energy equation for a new V
Vworn
2
Hence
Qworn
π D
4
2 g z1 z4 Hp
L Le
fworn
Kent
D D
3
Vworn
m
Qworn 0.0510
s
3
ΔQ Qworn Q
Check f
Reworn
m
Vworn 2.88
s
Vworn D
ν
ΔQ 0.0240
Given
Hence using 2.2 x 0.0161 is close enough to using 2.2 x 0.0165
m
ΔQ
s
Q
e
1
2.51
D
2.0 log
3.7
f
Reworn f
32.0 %
f 0.0165
Problem 10.75
Given:
Pump and supply pipe system
Find:
Head versus flow curve; Flow for a head of 85 ft
[Difficulty: 4]
Solution:
2
2
p
p
V1
V2
1
2
Basic equations:
α1
g z1
α2
g z2 h lT h pump
2
2
ρ
ρ
2
Applying to the 70 ft branch (branch a)
2
h lT f
2
2
Le V2
L V
V
f
K
2
D 2
D 2
2
Le Va
Va
L Va
g Ha f
f
K
g Hpump
2
D 2
D 2
Lea
where Ha 70 ft and
is due to a standard T branch (= 60) and a standard elbow (= 30) from Table 8.4, and
D
K Kent Kexit 1.5 from Fig. 8.14
L Lea Va
(1)
Hpump Ha f
K
D
D
2 g
Applying to the 50 ft branch (branch b)
L Leb Vb
Hpump Hb f
K
D
D
2 g
(2)
Leb
where Hb 50 ft and
is due to a standard T run (= 20) and two standard elbows (= 60), and K Kent Kexit 1.5
D
Here are the calculations, performed in Excel:
Given data:
L
e
D
K
L e a /D
L e b /D
Ha
Hb
= 1000 ft
= 0.00085 ft
= 6.065 in
=
1.5
=
90
=
80
=
70
ft
=
50
ft
ρ =
3
1.94 slug/ft
ν = 1.06E-05 ft2/s
Computed results: Set up Solver so that it varies all flow rates to make the total head error zero
H p ump (ft) Q (ft3 /s) Q a (ft3 /s) V a (ft/s)
72.0
1.389
0.313
1.561
74.0
1.574
0.449
2.237
76.0
1.724
0.553
2.756
78.0
1.857
0.641
3.195
80.0
1.978
0.718
3.581
82.0
2.090
0.789
3.931
84.0
2.195
0.853
4.252
Re a
7.44E+04
1.07E+05
1.31E+05
1.52E+05
1.71E+05
1.87E+05
2.03E+05
fa
H pump (Eq. 1) Q b (ft3 /s) V b (ft/s)
0.0248
72.0
1.076
5.364
0.0241
74.0
1.125
5.607
0.0238
76.0
1.171
5.839
0.0237
78.0
1.216
6.063
0.0235
80.0
1.260
6.279
0.0234
82.0
1.302
6.487
0.0234
84.0
1.342
6.690
Re b
2.56E+05
2.67E+05
2.78E+05
2.89E+05
2.99E+05
3.09E+05
3.19E+05
fb
H pu mp (Eq. 2) H (Errors)
0.0232
72.0
0.00
0.0231
74.0
0.00
0.0231
76.0
0.00
0.0231
78.0
0.00
0.0231
80.0
0.00
0.0230
82.0
0.00
0.0230
84.0
0.00
85.0
2.246
0.884
4.404
2.10E+05 0.0233
85.0
1.362
6.789
3.24E+05 0.0230
85.0
0.00
86.0
88.0
90.0
92.0
94.0
2.295
2.389
2.480
2.567
2.651
0.913
0.970
1.023
1.074
1.122
4.551
4.833
5.099
5.352
5.593
2.17E+05
2.30E+05
2.43E+05
2.55E+05
2.67E+05
86.0
88.0
90.0
92.0
94.0
1.382
1.420
1.457
1.494
1.529
6.886
7.077
7.263
7.445
7.622
3.28E+05
3.37E+05
3.46E+05
3.55E+05
3.63E+05
86.0
88.0
90.0
92.0
94.0
0.00
0.00
0.00
0.00
0.00
0.0233
0.0233
0.0232
0.0232
0.0231
For the pump head less than the upper reservoir head flow will be out of the reservoir (into the lower one)
0.0230
0.0230
0.0230
0.0230
0.0229
Total Error:
0.00
Head Versus Flow Rate
100
Head (ft)
95
90
85
80
75
70
1.0
1.5
2.0
3
Q (ft /s)
2.5
3.0
Problem 10.74
[Difficulty: 3]
Problem 10.73
Given:
Water pipe system
Find:
Pump suitable for 300 gpm
[Difficulty: 3]
Solution:
2
2
p
p
V1
V2
1
2
α
g
z
α
g
z
ρ
1 2
1 ρ
2 2
2 h l
f
64
(Laminar)
Re
2
h lT f
L V
D 2
e
1
2.51
D
2.0 log
(Turbulent)
f
3.7 Re f
2
The energy equation can be simplified to Δp ρ f
L V
D 2
This can be written for each pipe section
2
Pipe A (first section)
LA VA
ΔpA ρ fA
DA 2
Pipe B (1.5 in branch)
LB VB
ΔpB ρ fB
DB 2
Pipe C (1 in branch)
LC VC
ΔpC ρ fC
DC 2
(1)
2
(2)
2
(3)
2
LD VD
ΔpD ρ fD
DD 2
In addition we have the following contraints
Pipe D (last section)
(4)
QA QD Q
Q QB QC
(5)
Δp ΔpA ΔpB ΔpD
ΔpB ΔpC
(7)
(6)
(8)
We have 2 unknown flow rates (or, equivalently, velocities); We solve the above eight equations simultaneously
Once we compute the flow rates and pressure drops, we can compute data for the pump
Δppump Δp
and
The calculations, performed in Excel, are shown on the next page.
Qpump QA
Wpump Δppump Qpump
Pipe Data:
Pipe
A
B
C
D
L (ft)
D (in)
e (ft)
150
150
150
150
1.5
1.5
1
1.5
0.00085
0.00085
0.00085
0.00085
Fluid Properties:
3
ρ=
1.94
slug/ft
μ =
2.10E-05
lbf-s/ft
Q=
300
gpm
0.668
ft /s
2
Flow Rate:
=
Flows:
Heads:
Constraints:
3
3
3
3
3
Q A (ft /s)
0.668
Q B (ft /s)
0.499
Q C (ft /s)
0.169
Q D (ft /s)
0.668
V A (ft/s)
54.47
V B (ft/s)
40.67
V C (ft/s)
31.04
V D (ft/s)
54.47
Re A
6.29E+05
Re B
4.70E+05
Re C
2.39E+05
Re D
6.29E+05
fA
0.0335
fB
0.0336
fC
0.0384
fD
0.0335
Δp A (psi)
804.0
Δp B (psi)
448.8
Δp C (psi)
448.8
Δp D (psi)
804.0
(6) Q = Q B + Q C
0.00%
Error:
(8) Δp B = Δp C
0.00%
0.00%
Vary Q B and Q C
using Solver to minimize total error
Q (gpm)
Δp (psi)
P (hp)
2057
300
360
This is a very high pressure; a sequence of pumps would be needed
For the pump:
Problem 10.72
Given:
Flow from pump to reservoir
Find:
Select a pump to satisfy NPSHR
Solution:
Basic equations
[Difficulty: 3]
2
2
p
V1
V2
p2
1
α
g
z
α
g
z
ρ
1 ρ
2 h lT h p
2
2
2
V1
L V1
h lT h l h lm f
Kexit
2
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 is approximately 1 4) V 2 << V 1
2
2
p1 V2
L V
V
z2 f
Kexit
Hp
2 g
D 2 g
ρ g 2 g
Note that we compute head per unit weight, H, not head per unit mass, h, so
the energy equation between Point 1 and the free surface (Point 2) becomes
Solving for H p
2
2
2
p1
V
L V
V
Hp z2
f
Kexit
ρ g
2 g
2 g
D 2 g
From Table A.7 (68oF)
ρ 1.94
slug
ft
For commercial steel pipe
3
e 0.00015 ft
ν 1.08 10
(Table
8.1)
Given
For the exit
Kexit 1.0
Note that for an NPSHR of 15 ft this means
2
Re
s
e
so
D
V D
Re 6.94 10
ν
p1
ρ g
π D
4
V
0.0002
f 0.0150
2
p1
L V
Hp z2
f
ρ g
D 2 g
15 ft
Q 4.42
5
2
p1
L V
Hp z2
f
ρ g
D 2 g
so we
find
2
Q
e
1
2.51
D
2.0 log
f
3.7 Re f
Flow is turbulent:
Note that
5 ft
ft
Hp 691 ft
3
Q 1983 gpm
s
For this combination of Q and Hp, from Fig. D.11 the best pump appears to be a Peerless two-stage 10TU22C operating at 1750 rpm
After 10 years, from Problem 10.63, the friction factor will have increased by a factor of 2.2 f 2.2 0.150
We now need to solve
2
p1
L V
Hp z2
f
ρ g
D 2 g
V
for the new velocity
V
p1
Hp z2
f L
ρ g
2 D g
2
Q
π D
4
V
f 0.330
Q 0.94
V 2.13
ft
ft
s
and f will still be 2.2 0.150
3
s
Q 423 gpm
Much less!
Problem 10.71
[Difficulty: 3]
Problem 10.70
[Difficulty: 3]
Given:
Flow system and data of Problem 10.68; data for pipe aging from Problem 10.63
Find:
Pumps to maintain system flow rates; compare delivery
to that with pump sized for new pipes only
Solution:
We will apply the energy equation for steady, incompressible pipe flow.
Basic equations:
2
2
p
p
V1
V2
1
2
α
g
z
α
g
z
ρ
1 2
1 ρ
2 2
2 h lT h p
2
h lT f
2
2
Le V
L V
V
Σ f
Σ K
2
D 2
D 2
H
h
g
Assumptions: 1) p ent = pexit = patm 2) V ent = Vexit = 0
The given or available data is
Q 800
L
min
2
6 m
From Table A.8 at 20 oC ν 1.01 10
s
At the specified flow rate, the speed of the water is:
e
D
4
4.6 10
e 0.046 mm
p v 2.34 kPa
ρ 998
p atm 101.3 kPa
kg
3
m
V
Q
A
4 Q
V 1.698
2
π D
m
s
Re
V D
ν
e 1.11 6.9
f 1.8 log
Re
3.7 D
Therefore we can calculate the friction factor:
For the entire system:
In this case:
D 10 cm
5
1.681 10
2
0.019
2
2
Le V2
L V
V
g z1 z2 f
f
K
hp
2
D 2
D 2
z1 7.2 m z2 87 m
L 2 m 400 m 402 m
Solving for the required head at the pump:
Hp z2 z1
Le ( 75 55 8 2 30) D
L Le V2
f K
D D 2 g
For old pipes, we apply the multipliers from Problem 10.63: f20 5.00 fnew
The results of the analysis, computed in Excel, are shown on the next page.
f40 8.75 fnew
K 0.78 1
The required pump head is computed and plotted below.
New
Q (L/min) V (m/s)
H p (m) Q (gpm)
Re
f
0
200
400
600
800
922
1000
1136
1200
1400
1600
1800
2000
0.000
0.424
0.849
1.273
1.698
1.957
2.122
2.410
2.546
2.971
3.395
3.820
4.244
0.00E+00
4.20E+04
8.40E+04
1.26E+05
1.68E+05
1.94E+05
2.10E+05
2.39E+05
2.52E+05
2.94E+05
3.36E+05
3.78E+05
4.20E+05
0.0000
0.0231
0.0207
0.0196
0.0189
0.0187
0.0185
0.0183
0.0182
0.0180
0.0178
0.0177
0.0176
79.80
80.71
83.07
86.77
91.80
95.52
98.14
103.20
105.80
114.77
125.04
136.62
149.51
0.00
52.84
105.68
158.52
211.36
243.64
264.20
300.08
317.04
369.88
422.72
475.56
528.40
New
H p (ft)
20 yo
H p (ft)
40 yo
H p (ft)
Pump (ft)
261.81
264.81
272.52
284.67
301.17
313.38
321.99
338.59
347.12
376.54
410.25
448.24
490.51
261.81
276.57
314.52
374.19
455.19
515.10
557.37
638.8
680.64
824.95
990.27
1176.6
1383.9
261.81
287.59
353.89
458.10
599.58
704.21
778.03
920.2
993.31
1245.3
1534.0
1859.4
2221.4
856.54
840.48
792.30
712.00
599.58
515.10
455.04
338.59
278.38
69.59
-171.31
-444.33
-749.47
If we assume that the head at 800 L/min for 40 year old pipe is 70% of the maximum head for the pump,
2
and that the pump curve has the form H = H 0 - AQ :
H 800 =
599.58 ft
We plot the pump curve along with the head loss on the graph below:
856.54 ft
H0=
A = 0.005752 ft/gpm2
Required Pump Head
800
New Pipe
20 Years Old
40 Years Old
Pump Curve
700
H (ft)
600
500
400
300
200
100
150
200
Q (gpm)
250
300
Sizing the pump for 800 L/min for at 40 years would (assuming no change in the pump characteristics) produce
922 L/min at 20 years and 1136 L/min for new pipe.
Since the head increases by a factor of two, the extra head could be obtained by placing a second identical pump
in series with the pump of Problem 10.68.
350
Problem 10.69
8.
8
8.155
.15
155
[Difficulty: 3]
Problem 10.68
[Difficulty: 3]
Given:
System shown, design flow rate
Find:
Head losses for suction and discharge lines, NPSHA,
select a suitable pump
Solution:
We will apply the energy equation for steady, incompressible pipe flow.
Basic equations:
2
2
p
p
V1
V2
1
2
ρ α1 2 g z1 ρ α2 2 g z2 h lT hp
2
h lT f
2
2
Le V
L V
V
Σ f
Σ K
2
D 2
D 2
H
h
g
Assumptions: 1) p ent = pexit = patm 2) V ent = Vexit = 0
The given or available data is
Q 800
From Table A.8 at 20 oC ν 1.01 10
L
min
2
6 m
s
D 10 cm
p v 2.34 kPa
D
4.6 10
At the inlet:
4
ρ 998
V
Q
A
4 Q
3
2
V 1.698
p 2tabs p v
ρ g
Re
s
V D
ν
NPSHA
1.681 10
5
2
0.019
In this case: Le 75 D K 0.78 L 2 m
z2 8.7 m z1 7.2 m
2
L Le V2
V
p 2t ρ g z2 z1 f
K
2
D D 2
NPSHA
m
e 1.11 6.9
f 1.8 log
Re
3.7 D
2
p
2
V2
L Le V2
2
V
g z1
α2
g z2 f
Σ K
2
2
ρ
D D 2
The NPSHA can be calculated:
kg
π D
Therefore we can calculate the friction factor:
Solving for total pressure at 2:
p atm 101.3 kPa
m
At the specified flow rate, the speed of the water is:
e
e 0.046 mm
p 2t 18.362 kPa (gage)
p 2t p atm p v
ρ g
NPSHA 8.24 m
For the entire system:
In this case:
2
2
Le V2
L V
V
g z1 z2 f
f
K
hp
2
D 2
D 2
z1 7.2 m z2 88 m
L 2 m 400 m 402 m
Solving for the required head at the pump:
In U.S. Customary units:
Q 211 gpm
Hp z2 z1
Le ( 75 55 8 2 30) D
L Le V2
f K
D D 2 g
K 0.78 1
Hp 92.7 m
Hp 304 ft
A pump would be selected by finding one for which the NPSHR is less than the NPSHA. Based on these data and the information
in Appendix D, a 2AE11 or a 4AE12 pump would be capable of supplying the required head at the given flow rate. The pump
should be operated at a speed between 1750 and 3500 rpm, but the efficiency may not be acceptable. One should consult a
complete catalog to make a better selection.
Problem 10.67
[Difficulty: 3]
Given:
Water supply for Englewood, CO
Find:
(a) system resistance curve
(b) specify appropriate pumping system
(c) estimate power required for steady-state operation at two specified flow rates
Solution:
2
2
p
p
V1
V2
1
2
α
g
z
α
g
z
ρ
1 2
1 ρ
2 2
2 h lT h p
Basic equations:
2
h lT f
Assumptions: 1) p 1 = p2 = patm
Hence
2
2) V 1 = V2 = 0
3) Kent = 0
2
L
V
g z1 z2 f 1
h p or
D
2
2
Le V
L V
V
Σ f
Σ K
2
D 2
D 2
H
4) Kexit = 1
h
Wp
g
5) L e/D = 0
2
L
V
Hp z2 z1 f 1
D
2 g
The results calculated using Excel are shown below:
Given or available data (Note: final results will vary depending on fluid data selected):
L =
e =
1770
0.046
m
mm (Table 8.1)
z 1 = 1610 m
z 2 = 1620 m
D =
68.5
cm
ρ =
2
ν = 1.01E-06 m /s (Table A.8)
The required pump head is computed and plotted below.
3
Q (m /hr)
V (m/s)
Re
f
H p (m)
0
500
1000
1500
2000
2500
3000
3200
3500
3900
4000
0.00
0.38
0.75
1.13
1.51
1.88
2.26
2.41
2.64
2.94
3.01
0.00E+00
2.56E+05
5.11E+05
7.67E+05
1.02E+06
1.28E+06
1.53E+06
1.64E+06
1.79E+06
1.99E+06
2.04E+06
0.0000
0.0155
0.0140
0.0133
0.0129
0.0126
0.0124
0.0124
0.0123
0.0122
0.0122
10.0
10.3
11.1
12.3
14.0
16.1
18.6
19.8
21.6
24.3
25.0
998
kg/m
3
ρ Q g Hp
ηp
Required Pump Head
30
H (m)
25
Pump Head
20
Flow Rates of Interest
15
10
5
0
0
500
1000
1500
2000
2500
3000
3500
3
Q (m /hr)
The maximum flow rate is:
17172 gpm
The associated head is:
80 ft
Based on these data and the data of Figures D.1 and D.2, we could choose two 16A 18B pumps in parallel,
or three 10AE14 (G) pumps in parallel. The efficiency will be approximately
90%
Therefore, the required power would be:
3
191.21 kW at Q =
3200 m /hr
286.47 kW at Q =
3900 m /hr
3
4000
Problem 10.66
Given:
Data on pump and pipe system
Find:
Delivery through series pump system; reduction after 20 and 40 years
[Difficulty: 4]
Solution:
Given or available data (Note: final results will vary depending on fluid data selected) :
L =
D =
e =
1200
12
0.00015
ft
in
ft (Table 8.1)
K ent =
K exp =
L e/D elbow =
0.5
1
30
(Fig. 8.14)
=
z =
1.23E-05
-50
ft2/s (Table A.7)
ft
L e/D valve =
8
(Table 8.4)
The pump data is curve-fitted to H pump = H 0 - AQ 2.
The system and pump heads are computed and plotted below.
To find the operating condition, Solver is used to vary Q
so that the error between the two heads is minimized.
Q (gpm)
Q 2 (gpm)
H pump (ft)
H pump (fit)
V (ft/s)
0
500
1000
1500
2000
2500
3000
3250
0
250000
1000000
2250000
4000000
6250000
9000000
179
176
165
145
119
84
43
180
176
164
145
119
85
43
0.00
1.42
2.84
4.26
5.67
7.09
8.51
9.22
Re
0
115325
230649
345974
461299
576623
691948
749610
f
0.0000
0.0183
0.0164
0.0156
0.0151
0.0147
0.0145
0.0144
H pumps (par)
H lT + z (ft)
359
351
329
291
237
169
85
38
50.0
50.8
52.8
56.0
60.3
65.8
72.4
76.1
H0 =
180
A =
1.52E-05
Q (gpm)
V (ft/s)
3066
8.70
ft
ft/(gpm)2
Re
707124
f
0.0145
H pumps (par)
H lT + z (ft)
73.3
73.3
Error)
0%
Pump and System Heads
Pump Curve Fit
Pump Data
Total Head Loss
Pumps in Series
400
350
300
H (ft) 250
200
150
100
50
0
0
1000
2000
3000
Q (gal/min)
20-Year Old System:
f = 2.00 f new
Q (gpm)
V (ft/s)
2964
8.41
Re
683540
f
0.0291
H pumps (par)
H lT + z (ft)
92.1
92.1
Re
674713
f
0.0349
H pump (fit)
H lT + z (ft)
98.9
98.9
f
0.0291
H pump (fit)
H lT + z (ft)
90.8
90.8
f
0.0351
H pump (fit)
H lT + z (ft)
94.1
94.1
Error)
0%
Flow reduction:
102 gpm
3.3% Loss
Error)
0%
Flow reduction:
141 gpm
4.6% Loss
Error)
0%
Flow reduction:
151 gpm
4.9% Loss
Error)
0%
Flow reduction:
294 gpm
9.6% Loss
40-Year Old System:
f = 2.40 f new
Q (gpm)
V (ft/s)
2925
8.30
20-Year Old System and Pumps:
f = 2.00 f new
H pump = 0.90 H new
Q (gpm)
V (ft/s)
2915
8.27
Re
672235
40-Year Old System and Pumps:
f = 2.40 f new
H pump = 0.75 H new
Q (gpm)
V (ft/s)
2772
7.86
Re
639318
4000
Problem 10.65
[Difficulty: 4]
Given:
Data on pump and pipe system
Find:
Delivery through parallel pump system; reduction in delivery after 20 and 40 years
Solution:
Given or available data (Note: final results will vary depending on fluid data selected) :
L =
D =
e =
1200
12
0.00015
=
z =
1.23E-05
-50
ft
in
ft (Table 8.1)
2
ft /s (Table A.7)
ft
The pump data is curve-fitted to H pump = H 0 - AQ 2.
The system and pump heads are computed and plotted below.
To find the operating condition, Solver is used to vary Q
so that the error between the two heads is minimized.
K ent =
K exp =
L e/D elbow =
0.5
1
30
(Fig. 8.14)
L e/D valve =
8
(Table 8.4)
2
Q (gpm)
Q (gpm)
H pump (ft)
H pump (fit)
V (ft/s)
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
0
250000
1000000
2250000
4000000
6250000
9000000
179
176
165
145
119
84
43
180
176
164
145
119
85
43
0.00
1.42
2.84
4.26
5.67
7.09
8.51
9.93
11.35
12.77
14.18
Re
0
115325
230649
345974
461299
576623
691948
807273
922597
1037922
1153247
f
0.0000
0.0183
0.0164
0.0156
0.0151
0.0147
0.0145
0.0143
0.0142
0.0141
0.0140
f
0.0141
H pumps (par)
H lT + z (ft)
100.3
100.3
Error)
0%
H0 =
180
A =
1.52E-05
Q (gpm)
V (ft/s)
4565
12.95
H pumps (par)
H lT + z (ft)
180
179
176
171
164
156
145
133
119
103
85
50.0
50.8
52.8
56.0
60.3
65.8
72.4
80.1
89.0
98.9
110.1
ft
ft/(gpm)2
Re
1053006
Pump and System Heads
Pump Curve Fit
Pump Data
Total Head Loss
Pumps in Parallel
200
150
H (ft)
100
50
0
0
1000
2000
Q (gal/min)
3000
4000
20-Year Old System:
f = 2.00 f new
Q (gpm)
V (ft/s)
3906
11.08
Re
900891
f
0.0284
H pumps (par)
H lT + z (ft)
121.6
121.6
Re
855662
f
0.0342
H pump (fit)
H lT + z (ft)
127.2
127.2
f
0.0285
H pump (fit)
H lT + z (ft)
114.6
114.6
f
0.0347
H pump (fit)
H lT + z (ft)
106.4
106.4
Error)
0%
Flow reduction:
660 gpm
14.4% Loss
Error)
0%
Flow reduction:
856
18.7%
Error)
0%
Flow reduction:
860 gpm
18.8% Loss
Error)
0%
Flow reduction:
1416
31.0%
40-Year Old System:
f = 2.40 f new
Q (gpm)
V (ft/s)
3710
10.52
20-Year Old System and Pumps:
f = 2.00 f new
H pump = 0.90 H new
Q (gpm)
V (ft/s)
3705
10.51
Re
854566
40-Year Old System and Pumps:
f = 2.40 f new
H pump = 0.75 H new
Q (gpm)
V (ft/s)
3150
8.94
Re
726482
5000
Problem 10.64
[Difficulty: 3]
Given:
Data on pump and pipe system
Find:
Delivery through parallel pump system; valve position to reduce delivery by half
Solution:
Governing Equations:
For the pumps and system
where the total head loss is comprised of major and minor losses
Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 - z2 = z) we have
g Δz h lT Δhpump
h lT g Δz g Hsystem g Δz Δhpump g Hpump
or
HlT Δz Hpump
where
Le
Le
L
V
HlT f 2
Kent Kexit 2 g
Delbow
Dvalve
D
2
For pumps in parallel
1
2
Hpump H0 A Q
4
where for a single pump
Hpump H0 A Q
2
The calculations performed using Excel are shown on the next page.
Given or available data (Note: final results will vary depending on fluid data selected):
L =
D =
e =
1200
12
0.00015
ft
in
ft (Table 8.1)
ν =
Δz =
1.23E-05
-50
ft /s (Table A.7)
ft
2
K ent =
K exp =
L e /D elbow =
0.5
1
30
(Fig. 8.14)
L e/D valve =
8
(Table 8.4)
2
The pump data is curve-fitted to H pump = H 0 - AQ .
The system and pump heads are computed and plotted below.
To find the operating condition, Solver is used to vary Q
so that the error between the two heads is minimized.
Q (gpm)
Q 2 (gpm)
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
0
250000
1000000
2250000
4000000
6250000
9000000
H0=
180
A=
1.52E-05
Q (gpm)
V (ft/s)
4565
12.95
H pum p (ft)
179
176
165
145
119
84
43
H pump (fit)
180
176
164
145
119
85
43
V (ft/s)
0.00
1.42
2.84
4.26
5.67
7.09
8.51
9.93
11.35
12.77
14.18
Re
0
115325
230649
345974
461299
576623
691948
807273
922597
1037922
1153247
f
0.0000
0.0183
0.0164
0.0156
0.0151
0.0147
0.0145
0.0143
0.0142
0.0141
0.0140
f
0.0141
H pumps (par)
100.3
H lT +Δz (ft)
100.3
Error)
0%
H pum ps (par)
180
179
176
171
164
156
145
133
119
103
85
H lT + Δz (ft)
50.0
50.8
52.8
56.0
60.3
65.8
72.4
80.1
89.0
98.9
110.1
ft
2
ft/(gpm)
Re
1053006
Pump and System Heads
Pump Curve Fit
Pump Data
Total Head Loss
Pumps in Parallel
200
150
H (ft)
100
50
0
0
1000
2000
Q (gal/min)
3000
For the valve setting to reduce the flow by half, use Solver to vary the value below to minimize the error.
L e /D valve =
Q (gpm)
V (ft/s)
2283
6.48
9965
Re
526503
f
0.0149
H pumps (par)
159.7
H lT + z (ft)
159.7
Error)
0%
4000
5000
Problem 10.63
[Difficulty: 4]
Given:
Data on pump and pipe system, and their aging
Find:
Reduction in delivery through system after 20 and 40 years (aging and non-aging pumps)
Solution:
Given or available data (Note: final results will vary depending on fluid data selected) :
L =
D =
e =
1200
12
0.00015
ft
in
ft (Table 8.1)
K ent =
K exp =
L e/D elbow =
0.5
1
30
(Fig. 8.14)
=
z =
1.23E-05
-50
ft2/s (Table A.7)
ft
L e/D valve =
8
(Table 8.4)
The pump data is curve-fitted to H pump = H 0 - AQ 2.
The system and pump heads are computed and plotted below.
To find the operating condition, Solver is used to vary Q
so that the error between the two heads is minimized.
New System:
Q (gpm)
Q 2 (gpm)
H pump (ft)
V (ft/s)
0
500
1000
1500
2000
2500
3000
0
250000
1000000
2250000
4000000
6250000
9000000
179
176
165
145
119
84
43
0.00
1.42
2.84
4.26
5.67
7.09
8.51
Re
0
115325
230649
345974
461299
576623
691948
f
0.0000
0.0183
0.0164
0.0156
0.0151
0.0147
0.0145
H pump (fit)
H lT + z (ft)
68.3
68.3
Error)
0%
H0 =
180
A =
1.52E-05
H pump (fit)
H lT + z (ft)
180
176
164
145
119
84.5
42.7
50.0
50.8
52.8
56.0
60.3
65.8
72.4
ft
Q (gpm)
V (ft/s)
2705
7.67
ft/(gpm)2
Re
623829
f
0.0146
Pump and System Heads -When New
200
180
160
140
H (ft) 120
100
80
60
40
20
0
Pump Curve Fit
Pump Data
Total Head Loss
0
500
1000
1500
Q (gal/min)
2000
2500
3000
20-Year Old System:
f = 2.00 f new
Q (gpm)
V (ft/s)
2541
7.21
Re
586192
f
0.0295
H pump (fit)
H lT + z (ft)
81.4
81.4
Re
572843
f
0.0354
H pump (fit)
H lT + z (ft)
85.8
85.8
H pump (fit)
H lT + z (ft)
79.3
79.3
H pump (fit)
H lT + z (ft)
78.8
78.8
Error)
0%
Flow reduction:
163 gpm
6.0% Loss
Error)
0%
Flow reduction:
221 gpm
8.2% Loss
Error)
0%
Flow reduction:
252 gpm
9.3% Loss
Error)
0%
Flow reduction:
490 gpm
18.1% Loss
40-Year Old System:
f = 2.40 f new
Q (gpm)
V (ft/s)
2484
7.05
20-Year Old System and Pump:
f = 2.00 f new
H pump = 0.90 H new
Q (gpm)
V (ft/s)
2453
6.96
Re
565685
f
0.0296
40-Year Old System and Pump:
f = 2.40 f new
H pump = 0.75 H new
Q (gpm)
V (ft/s)
2214
6.28
Re
510754
f
0.0358
3500
Problem 10.62
[Difficulty: 3]
Given:
Data on pump and pipe system
Find:
Delivery through series pump system; valve position to reduce delivery by half
Solution:
Governing Equations:
For the pumps and system
where the total head loss is comprised of major and minor losses
Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 - z2 = z) we have
g Δz h lT Δhpump
h lT g Δz g Hsystem g Δz Δhpump g Hpump
or
HlT Δz Hpump
where
Le
Le
L
V
HlT f 2
Kent Kexit 2 g
Delbow
Dvalve
D
For pumps in series
Hpump 2 H0 2 A Q
where for a single pump
Hpump H0 A Q
2
2
The calculations in Excel are shown on the next page.
2
Given or available data (Note: final results will vary depending on fluid data selected):
L =
D =
e =
1200
12
0.00015
ft
in
ft (Table 8.1)
ν =
Δz =
1.23E-05
-50
ft /s (Table A.7)
ft
2
K ent =
K exp =
L e /D elbow =
0.5
1
30
(Fig. 8.14)
L e/D valve =
8
(Table 8.4)
2
The pump data is curve-fitted to H pump = H 0 - AQ .
The system and pump heads are computed and plotted below.
To find the operating condition, Solver is used to vary Q
so that the error between the two heads is minimized.
Q (gpm)
Q 2 (gpm)
H pum p (ft)
H pump (fit)
V (ft/s)
0
500
1000
1500
2000
2500
3000
3250
0
250000
1000000
2250000
4000000
6250000
9000000
179
176
165
145
119
84
43
180
176
164
145
119
85
43
0.00
1.42
2.84
4.26
5.67
7.09
8.51
9.22
Re
0
115325
230649
345974
461299
576623
691948
749610
f
0.0000
0.0183
0.0164
0.0156
0.0151
0.0147
0.0145
0.0144
f
0.0145
H pumps (par)
H lT + Δz (ft)
73.3
73.3
Error)
0%
H0=
180
A=
1.52E-05
Q (gpm)
V (ft/s)
3066
8.70
H pum ps (par)
H lT + Δz (ft)
359
351
329
291
237
169
85
38
50.0
50.8
52.8
56.0
60.3
65.8
72.4
76.1
ft
2
ft/(gpm)
Re
707124
Pump and System Heads
Pump Curve Fit
Pump Data
Total Head Loss
Pumps in Series
400
350
300
H (ft) 250
200
150
100
50
0
0
1000
2000
Q (gal/min)
3000
For the valve setting to reduce the flow by half, use Solver to vary the value below to minimize the error.
L e /D valve =
50723
Q (gpm)
V (ft/s)
1533
4.35
Re
353562
f
0.0155
H pumps (par)
287.7
H lT + Δz (ft)
287.7
Error)
0%
4000
Problem 10.61
Given:
Data on pump and pipe system
Find:
Delivery through system, valve position to reduce delivery by half
[Difficulty: 3]
Solution:
Governing Equations:
For the pump and system
where the total head loss is comprised of major and minor losses
Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 - z2 = z) we have
g Δz h lT Δhpump
h lT g Δz g Hsystem g Δz Δhpump g Hpump
or
HlT Δz Hpump
2
where
Le
Le
L
V
HlT f 2
Kent Kexit 2 g
Delbow
Dvalve
D
The calculations performed using Excel are shown on the next page:
Given or available data (Note: final results will vary depending on fluid data selected):
L = 1200 ft
D =
12
in
e = 0.00015 ft (Table 8.1)
2
ν = 1.23E-05 ft /s (Table A.7)
z =
-50
ft
K e nt =
K e xp =
L e/D elbow =
0.5
1
30
(Fig. 8.14)
L e/D valve =
8
(Table 8.4)
2
The pump data is curve-fitted to H pump = H 0 - AQ .
The system and pump heads are computed and plotted below.
To find the operating condition, Solver is used to vary Q
so that the error between the two heads is minimized.
Q (gpm) Q 2 (gpm) H pump (ft)
0
500
1000
1500
2000
2500
3000
0
250000
1000000
2250000
4000000
6250000
9000000
H0=
180
179
176
165
145
119
84
43
0.00
1.42
2.84
4.26
5.67
7.09
8.51
Re
0
115325
230649
345974
461299
576623
691948
H pum p (fit) H lT + z (ft)
f
0.0000
0.0183
0.0164
0.0156
0.0151
0.0147
0.0145
180
176
164
145
119
84.5
42.7
50.0
50.8
52.8
56.0
60.3
65.8
72.4
ft
A = 1.52E-05 ft/(gpm)
Q (gpm) V (ft/s)
2705
V (ft/s)
7.67
2
Re
f
H pum p (fit) H lT + z (ft) Error)
623829 0.0146
68.3
68.3
0%
Pump and System Heads
200
180
160
140
H (ft) 120
100
80
60
40
20
0
Pump Curve Fit
Pump Data
Total Head Loss
0
500
1000
1500
2000
Q (gal/min)
2500
3000
3500
For the valve setting to reduce the flow by half, use Solver to vary the value below to minimize the error.
L e/D valve =
Q (gpm) V (ft/s)
1352
3.84
26858
H pum p (fit) H lT + z (ft) Error)
Re
f
311914 0.0158
151.7
151.7
0%
Problem 10.60
Given:
Pump and reservoir/pipe system
Find:
Flow rate using different pipe sizes
[Difficulty: 3]
Solution:
2
2
p
p
V1
V2
1
2
α
g
z
α
g
z
ρ
1 2
1 ρ
2 2
2 h lT h p
Basic equations:
2
h lT f
2
2
Le V
L V
V
Σ f
Σ K
2
D 2
D 2
H
and also
Le for the elbows, and K for the square entrance and exit
h
g
Le
2) V = V = 0 3) α = 1 4) z1 0, z2 24 ft 4) K Kent Kexp 5)
is for two elbows
atm
1
2
D
Assumptions: 1) p = p = p
1
Hence
2
h lT f
2
2
Le V2
L V
V
f
K
2
D 2
D 2
z2 HlT Hp or
and also
We want to find a flow that satisfies these equations, rewritten as energy/weight rather than energy/mass
2
L Le V
HlT f
K
D D 2 g
H1T z2 Hp
Here are the results calculated in Excel:
Given or available data (Note: final results will vary depending on fluid data selected):
L = 1750 ft
e = 0.00015 ft (Table 8.1)
D = 7.981 in
2
ν = 1.06E-05 ft /s (Table A.8)
z2 =
24
ft
K ent =
K exp =
L e/D elbow =
0.5
1
60
(Fig. 8.14)
L e/D va lve =
8
(Table 8.4)
(Two)
H1T Hp z2
2
The pump data is curve-fitted to H pump = H 0 - AQ .
The system and pump heads are computed and plotted below.
To find the operating condition, Solver is used to vary Q
so that the error between the two heads is minimized.
A plot of the pump and system heads is shown for the 8 in case - the others will look similar.
Q (cfm)
Q2
V (ft/s)
H p (ft)
H p (fit) H lT + z 2 (ft)
Re
f
0.000
0
90.0
0.00
0
0.0000
89
24.0
50.000
2500
87.0
2.40
150504
0.0180
87
28.5
100.000 10000
81.0
4.80
301007
0.0164
81
40.4
150.000 22500
70.0
7.20
451511
0.0158
72
59.5
200.000 40000
59.0
9.59
602014
0.0154
59
85.8
250.000 62500
43.0
11.99
752518
0.0152
42.3
119.1
300.000 90000
22.0
14.39
903022
0.0150
21.9
159.5
H0=
89
ft
A = 7.41E-04 ft/(cfm)
Q (cfm)
V (ft/s)
167.5
8.03
Repeating for:
2
Re
H p (fit) H lT + z 2 (ft) Error)
f
504063 0.0157
67.9
67.9
0.00%
D = 10.02 in
Q (cfm)
V (ft/s)
179.8
8.63
H p (fit) H lT + z 2 (ft) Error)
Re
f
541345 0.0156 64.7
64.7
0.00%
D =
Repeating for:
Q (cfm)
V (ft/s)
189.4
9.09
12
in
H p (fit) H lT + z 2 (ft) Error)
Re
f
570077 0.0155 62.1
62.1
0.00%
Pump and System Heads (8 in pipe)
180
160
140
120
H (ft)
100
80
60
40
20
0
Pump Curve Fit
Pump Data
Total Head Loss
0
50
100
150
Q (cfm)
200
250
300
350
Problem 10.59
Given:
Data on pump and pipe system
Find:
Delivery through system
[Difficulty: 3]
Solution:
Governing Equations:
For the pump and system
where the total head loss is comprised of major and minor losses
and the pump head (in energy/mass) is given by (from Example 10.6)
Hpump( ft) 55.9 3.44 10
5
Q( gpm)
2
Hence, applied between the two reservoir free surfaces (p 1 = p 2 = 0, V1 = V2 = 0, z1 = z2 ) we have
0 h lT Δhpump
h lT g Hsystem Δhpump g Hpump
or
HlT Hpump
where
L1
V1
L2
V2
HlT f1
Kent
f2
Kexit
D1
2 g D2
2
(1)
2
Results generated in Excel are shown on the next page.
2
Given or available data:
L1 =
D1 =
L2 =
D2 =
e =
3000
9
1000
6
0.00085
ft
ν=
in
K ent =
ft
K exp =
in
Q loss =
ft (Table 8.1)
2
1.23E-05 ft /s (Table A.7)
0.5
(Fig. 8.14)
1
75
gpm
The system and pump heads are computed and plotted below.
To find the operating condition, Goal Seek is used to vary Q 1
so that the error between the two heads is zero.
Q 1 (gpm)
Q 2 (gpm)
V 1 (ft/s)
V 2 (ft/s)
Re 1
Re 2
f1
f2
H lT (ft)
H pump (ft)
100
200
300
400
500
600
700
25
125
225
325
425
525
625
0.504
1.01
1.51
2.02
2.52
3.03
3.53
0.284
1.42
2.55
3.69
4.82
5.96
7.09
30753
61506
92260
123013
153766
184519
215273
11532
57662
103792
149922
196052
242182
288312
0.0262
0.0238
0.0228
0.0222
0.0219
0.0216
0.0215
0.0324
0.0254
0.0242
0.0237
0.0234
0.0233
0.0231
0.498
3.13
8.27
15.9
26.0
38.6
53.6
55.6
54.5
52.8
50.4
47.3
43.5
39.0
Q 1 (gpm)
Q 2 (gpm)
V 1 (ft/s)
V 2 (ft/s)
Re 1
Re 2
f1
f2
H lT (ft)
H pump (ft)
627
552
3.162
6.263
192785
254580
0.0216
0.0232
42.4
42.4
Error)
0%
700
800
Pump and System Heads
60
50
H (ft)
40
30
Pump
System
20
10
0
0
100
200
300
400
Q (gal/min)
500
600
Problem 10.58
[Difficulty: 3] Part 1/2
Problem 10.58
[Difficulty: 3] Part 2/2
Problem 10.57
[Difficulty: 2]
Problem 10.56
[Difficulty: 3]
Given:
Pump and reservoir system
Find:
System head curve; Flow rate when pump off; Loss, Power required and cost for 1 m 3/s flow rate
Solution:
Basic equations:
2
2
p
p
2
2
V1
V2
1
2
L V
V
ρ α1 2 g z1 ρ α2 2 g z2 h lT hp h lT f D 2 Σ K 2 (K for the exit)
where points 1 and 2 are the reservoir free surfaces, and h p is the pump head
H
Note also
h
g
ηp
Pump efficiency:
Wh
Wm
Assumptions: 1) p 1 = p2 = patm 2) V1 = V2 = 0 3) α 2 = 0 4) z1 0, z2 15 m 4) K Kent Kent 1.5
2
From the energy equation g z2 f
2
L V
V
h p K
2
D 2
Given or available data L 300 m
ρ 1000
2
D 40 cm
kg
ν 1.01 10
3
2
2
L V
V
h p g z2 f
K
2
D 2
e 0.26 mm
2
6 m
m
s
(Table 8.1)
(Table A.8)
The set of equations to solve for each flow rate Q are
4 Q
V
2
Re
V D
π D
ν
e
D
2.51
2.0 log
f
3.7 Re f
3
For example, for
Q 1
m
s
2
V 7.96
m
Re 3.15 10
s
2
L V
V
Hp z2 f
K
2 g
D 2 g
1
6
f 0.0179
Hp 33.1 m
40
Head (m)
30
20
10
10
0
0.2
0.4
0.6
20
Q (cubic meter/s)
0.8
2
L V
V
Hp z2 f
K
2 g
D 2 g
1
3
The above graph can be plotted in Excel. In Excel, Solver can be used to find Q for H p = 0 Q 0.557
3
At
Q 1
m
s
we saw that
Hp 33.1 m
4
Assuming optimum efficiency at Q 1.59 10 gpm from Fig.
10.15
ηp 92 %
Then the hydraulic power is
Wh ρ g Hp Q
Wh 325 kW
The pump power is then
Wh
Wm
ηp
Wm 2 706 kW
If electricity is 10 cents per kW-hr then the hourly cost is about $35
If electricity is 15 cents per kW-hr then the hourly cost is about $53
If electricity is 20 cents per kW-hr then the hourly cost is about $71
m
s
(Zero power rate)
Problem 10.55
[Difficulty: 5]
H
Given:
Pump and supply pipe system
Find:
Maximum operational flow rate as a function of temperature
Solution:
Basic equations:
2
2
p
p
2
2
V1
V2
Le V2
1
2
L V
V
ρ α1 2 g z1 ρ α2 2 g z2 h lTh lT f D 2 f D 2 K 2
NPSHA
Le for the elbow, and K for the square
entrance
2
Hr H0 A Q
pt pv
ρ g
Assumptions: 1) p 1 = 0 2) V1 = 0 3) α 2 = 0 4) z 2 = 0
We must match the NPSHR (=Hr) and NPSHA
From the energy equation
NPSHA
g H
pt pv
ρ g
2
2
p2 V2
Le V2
L V
V
f f K
2
2
D 2
D 2
ρ
p2
ρ g
p atm
ρ g
V2
2
2 g
The results generated using Excel are shown on the next page.
Given data:
Computed results:
pv
ρ g
p2
ρ g
2
H
L Le
K
D D
1 f
L Le patm pv
K
2 g D
ρ g
D
2
NPSHA H
2 g
V
V
f
Given data:
Computed results:
o
T ( C) p v (kPa) ρ
0
0.661
5
0.872
10
1.23
15
1.71
20
2.34
L=
e =
D=
K ent =
L e /D =
H0 =
6
0.26
15
0.5
30
3
m
mm
cm
A=
H=
p atm =
3000
6
101
m/(m /s)
m
kPa
=
1000
kg/m
m
3
2
3
2
3
(kg/m
1000
1000
1000
999
998
3
3
) ν (m /s) Q (m /s) V (m/s) Re
1.76E-06 0.06290 3.56 3.03E+05
1.51E-06 0.06286 3.56 3.53E+05
1.30E-06 0.06278 3.55 4.10E+05
1.14E-06 0.06269 3.55 4.67E+05
1.01E-06 0.06257 3.54 5.26E+05
f
NPSHA (m)NPSHR (m) Error
0.0232
0.0231
0.0230
0.0230
0.0229
14.87
14.85
14.82
14.79
14.75
14.87
14.85
14.82
14.79
14.75
0.00
0.00
0.00
0.00
0.00
25
30
35
3.17
4.25
5.63
997
996
994
8.96E-07 0.06240
8.03E-07 0.06216
7.25E-07 0.06187
3.53
3.52
3.50
5.91E+05 0.0229
6.57E+05 0.0229
7.24E+05 0.0228
14.68
14.59
14.48
14.68
14.59
14.48
0.00
0.00
0.00
40
7.38
992
6.59E-07 0.06148
3.48
7.92E+05 0.0228
14.34
14.34
0.00
= 1.01E-06 m /s
45
9.59
990
6.02E-07 0.06097 3.45 8.60E+05 0.0228
14.15
14.15
50
12.4
988
5.52E-07 0.06031 3.41 9.27E+05 0.0228
13.91
13.91
55
15.8
986
5.09E-07 0.05948 3.37 9.92E+05 0.0228
13.61
13.61
60
19.9
983
4.72E-07 0.05846 3.31 1.05E+06 0.0228
13.25
13.25
65
25.0
980
4.40E-07 0.05716 3.23 1.10E+06 0.0227
12.80
12.80
70
31.2
978
4.10E-07 0.05548 3.14 1.15E+06 0.0227
12.24
12.24
75
38.6
975
3.85E-07 0.05342 3.02 1.18E+06 0.0227
11.56
11.56
80
47.4
972
3.62E-07 0.05082 2.88 1.19E+06 0.0227
10.75
10.75
85
57.8
969
3.41E-07 0.04754 2.69 1.18E+06 0.0227
9.78
9.78
90
70.1
965
3.23E-07 0.04332 2.45 1.14E+06 0.0227
8.63
8.63
95
84.6
962
3.06E-07 0.03767 2.13 1.05E+06 0.0228
7.26
7.26
100
101
958
2.92E-07 0.02998 1.70 8.71E+05 0.0228
5.70
5.70
Use Solver to make the sum of absolute errors between NPSHA and NPSHR zero by varying the Q 's
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
NPSHR increases with temperature because the p v increases; NPHSA decreases because ρ decreases and p v increases
Maximum Flow Rate Versus Water temperature
0.07
0.06
0.04
3
Q (m /s)
0.05
0.03
0.02
0.01
0.00
0
10
20
30
40
50
o
T ( C)
60
70
80
90
100
Problem 10.54
[Difficulty: 2]
Problem 10.53
Given:
Pump and supply pipe system
Find:
Maximum operational flow rate
[Difficulty: 3]
Solution:
H
2
2
p
p
V1
V2
1
2
α
g
z
α
g
z
ρ
1 2
1 ρ
2 2
2 h lT
Basic equations:
h lT f
NPSHA
2
2
Le V2
L V
V
f
K
2
D 2
D 2
Le for the elbow, and K for the square entrance
pt pv
2
Hr H0 A Q
ρ g
Assumptions: 1) p 1 = 0 2) V1 = 0 3) α 2 = 1 4) z 2 = 0
We must match the NPSHR (=Hr) and NPSHA
g H
From the energy equation
2
2
p2 V2
Le V2
L V
V
f f K
2
2
D 2
D 2
ρ
NPSHA
pt pv
ρ g
p2
ρ g
p atm
ρ g
V2
2
2 g
pv
ρ g
2
L
ρ g
D
2
V L
NPSHA H
f
2 g D
p2
H
2 g
V
1 f
Le
K
D
Le
patm pv
K
D
ρ g
Calculated results and plot were generated using Excel:
Given data:
Computed results:
L =
20
ft
e = 0.00085 ft
D = 6.065 in
0.5
K e nt =
L e /D =
30
H0=
10
ft
Q (cfs) V (ft/s)
0.2
1.00
0.4
1.99
0.6
2.99
0.8
3.99
1.0
4.98
A=
H =
p atm =
pv=
7.9
22
14.7
0.363
ft/(cfs)2
ft
psia
psia
=
1.93
slug/ft
3
2
= 1.06E-05 ft /s
Crossover point:
1.2
1.4
1.6
1.8
Re
4.75E+04
9.51E+04
1.43E+05
1.90E+05
2.38E+05
f
NPSHA (ft) NPSHR (ft)
0.0259
55.21
10.32
0.0243
55.11
11.26
0.0237
54.95
12.84
0.0234
54.72
15.06
0.0232
54.43
17.90
5.98
6.98
7.98
8.97
2.85E+05
3.33E+05
3.80E+05
4.28E+05
0.0231
0.0230
0.0229
0.0229
54.08
53.66
53.18
52.63
21.38
25.48
30.22
35.60
2.0
9.97
4.75E+05 0.0228
52.02
41.60
2.2
2.4
2.6
10.97
11.96
12.96
5.23E+05 0.0228
5.70E+05 0.0227
6.18E+05 0.0227
51.35
50.62
49.82
48.24
55.50
63.40
2.28
11.36
5.42E+05 0.0228
51.07
51.07
Error
0.00
NPSHA and NPSHR
70
60
Head (ft)
50
40
NPSHA
NPSHR
30
20
10
0
0.0
0.5
1.0
1.5
Q (cfs)
2.0
2.5
3.0
Problem 10.52
[Difficulty: 3]
Given:
Data on a boiler feed pump
Find:
NPSHA at inlet for field temperature water; Suction head to duplicate field conditions
Solution:
Basic equation:
1
2
NPSHA p t p v p g p atm ρ V p v
2
Given or available data is
Ds 10 cm
Dd 7.5 cm
H 125 m
Q 0.025
p inlet 150 kPa
p atm 101 kPa
zinlet 50 cm
ρ 1000
3
m
s
kg
3
ω 3500 rpm
m
For field conditions
p g p inlet ρ g zinlet
From continuity
Vs
4 Q
2
π Ds
From steam tables (try Googling!) at 115oC
Hence
p g 145 kPa
m
Vs 3.18
s
p v 169 kPa
1
2
NPSHA p g p atm ρ Vs p v
2
Expressed in meters or feet of water
NPSHA 82.2 kPa
NPSHA
ρ g
8.38m
In the laboratory we must have the same NPSHA. From Table A.8 (or steam tables - try Googling!) at 27 oC
Hence
1
2
p g NPSHA p atm ρ Vs p v
2
The absolute pressure is
p g p atm 80.7 kPa
p g 20.3 kPa
NPSHA
ρ g
27.5 ft
p v 3.57 kPa
Problem 10.51
Given:
Data on a NPSHR for a pump
Find:
Curve fit; Maximum allowable flow rate
Solution:
[Difficulty: 2]
The results were generated in Excel:
2
Q (cfm)
20
40
60
80
100
120
140
Q
4.00E+02
1.60E+03
3.60E+03
6.40E+03
1.00E+04
1.44E+04
1.96E+04
NPSHR (ft)
7.1
8.0
8.9
10.3
11.8
14.3
16.9
NPSHR (fit)
7.2
7.8
8.8
10.2
12.0
14.2
16.9
The fit to data is obtained from a least squares fit to NPSHR = a + bQ
a =
b =
7.04
ft
2
5.01E-04 ft/(cfm)
2
Q (cfm)
NPSHR (ft)
160.9
20.00
Use Goal Seek to find Q !
NPSHR Curve for a Pump
NPSHR (m)
18
16
Data at 1450 rpm
14
12
10
Curve Fit
8
6
4
2
0
0
20
40
60
80
3
100
3
Q (m /s x 10 )
120
140
160
Problem 10.50
[Difficulty: 4]
Problem 10.49
[Difficulty: 2]
Given:
Data on a model pump
Find:
Temperature for dynamically similar operation at 1800 rpm; Flow rate and head; Comment on NPSH
Solution:
Basic equation:
Re1 Re2
Q1
and similarity
rules
ω1 D1
3
Q2
ω2 D2
H1
3
2
ω1 D1
2
H2
2
ω2 D2
2
3
The given or available data is
ω1 3600 rpm
From Table A.8 at 15 oC
ν1 1.14 10
For D = constant
V1 D
Re1
ν1
Q1
ω1 D
Re2
2
ω1 D
or
ν2
ω2
ν2 ν1
ω1
2
7m
ν2 5.7 10
( 5.52 6.02)
Q2
( 5.70 6.02)
H2
T2 48
or
ω2
Q2 Q1
ω1
or
ω2
H2 H1
ω1
3
2
ω2 D
degrees C
3
m
Q2 0.0500
s
2
H2 6.75 m
The water at 48 oC is closer to boiling. The inlet pressure would have to be changed to avoid cavitation. The increase between
runs 1 and 2 would have to be Δp p v2 p v1 where p v2 and pv1 are the vapor pressures at T 2 and T1. From the steam tables:
p v1 1.71 kPa
s
, we find, by linear interpolation
( 50 45)
2
s
ω2 D D
ω2 D
H1
2
ν1
s
3
ω1 D D
2
7m
T2 45
and also
H1 27 m
2
6 m
From Table A.8, at ν2 5.7 10
From similar operation
m
Q1 0.1
s
ω2 1800 rpm
p v2 11.276 kPa
Δp p v2 p v1
Δp 9.57 kPa
Problem 10.48
Given:
Data on a model pump
Find:
Prototype flow rate, head, and power at 125 rpm
[Difficulty: 3]
Solution:
Wh ρ Q g H
Basic equation:
Q1
ω1 D1
3
Q2
ω2 D2
and similarity rules
h1
(10.19a)
3
The given or available data is
2
ω1 D1
2
h2
2
ω2 D2
Nm 100 rpm
P1
(10.19b)
2
3
ρ1 ω1 D1
Np 125 rpm
ρ 1000
P2
5
3
(10.19a)
ρ2 ω2 D2
5
kg
3
m
3
From Eq. 10.8a
From Eq. 10.19a (with Dm/Dp = 1/3)
m
Qm 1
s
Hm 4.5 m
Whm ρ Qm g Hm
Whm 44.1 kW
Qp
ωp Dp
Qm
3
3
ωm Dm
3
Np
Qp 27 Qm
Nm
From Eq. 10.19b (with Dm/Dp = 1/3)
hp
2
ωp Dp
2
m
Qp 33.8
s
hm
g Hp
or
2
2
2
ωm Dm
2
ωp Dp
2
ωp Dp
ωp
2
Hp Hm
3 Hm
ωm Dm
ωm
From Eq. 10.19c (with Dm/Dp = 1/3)
Pp
3
ρ ωp Dp
5
3
ωp
Dp
3
Qp Qm
3 Qm
ωm Dm
ωm
ωp
or
or
5
ρ ωm Dm
Np
Whp 243 Whm
Nm
2
2
ωm Dpm
Np
Hp 9 Hm
Nm
3
Pm
3
2
2
g Hm
5
2
Hp 63.3 m
ωp Dp
ωp
5
Whp Whm
3 Whm
ωm Dm
ωm
3
Whp 20.9 MW
3
Problem 10.47
[Difficulty: 3]
Given:
Data on a model fan, smaller scale similar fan
Find:
Scale factor and volumetric flow rate of similar fan
Solution:
Basic equations:
Q1
ω1 D1
3
The given or available data is
Q2
ω2 D2
H1
3
2
ω1 D1
2
H2
2
ω2 D2
ω1 1440 rpm
2
ω2 1800 rpm
3
m
Q1 6.3
s
Solving the head equation for the scale D 2/D1:
We can use this to find the new flowrate:
D2
D1
H1 0.15 m
ω1
ω2
H2
H1
0.8
D2
Q2 Q1
ω1 D1
ω2
3
H2 H1 0.15 m
D2
D1
0.8
3
m
Q2 4.03
s
Problem 10.46
[Difficulty: 5]
Problem 10.45
10.20
10
0.2
20
10-4
1
10
0-4
10.20
[Difficulty: 3] Part 1/2
Problem 10.45
[Difficulty: 3] Part 2/2
Problem 10.44
[Difficulty: 4]
Problem 10.43
[Difficulty: 3]
10.6:
Problem 10.42
[Difficulty: 3]
10.6
Problem 10.41
[Difficulty: 3]
Problem 10.40
[Difficulty: 3]
Problem 10.39
[Difficulty: 3]
Given:
Data on Peerless Type 10AE12 pump at 1720 rpm
Find:
Data at speeds of 1000, 1200, 1400, and 1600 rpm
Solution:
Q1
The governing equations are the similarity rules:
ω1 D1
For scaling from speed ω1 to speed ω2:
Speed (rpm) = 1760
Q (gal/min)
0
500
1000
1500
2000
2500
3000
3500
4000
3
Q2 Q1
ω1
Q (gal/min)
0
284
568
852
1136
1420
1705
1989
2273
ω2 D2
h1
3
2
ω1 D1
ω2
H2 H1
ω1
ω2
Speed (rpm) = 1000
2
H (ft) H (fit)
Q
0
170
161
250000
160
160
1000000 155
157
2250000 148
152
4000000 140
144
6250000 135
135
9000000 123
123
12250000 110
109
16000000 95
93
Q2
Q (gal/min)
0
341
682
1023
1364
1705
2045
2386
2727
2
ω2 D2
2
where
h g H
2
Here are the results generated in Excel:
Speed (rpm) = 1200
H (ft)
52.0
51.7
50.7
49.0
46.6
43.5
39.7
35.3
30.2
2
h2
H (ft)
74.9
74.5
73.0
70.5
67.1
62.6
57.2
50.8
43.5
Speed (rpm) = 1400
Q (gal/min)
0
398
795
1193
1591
1989
2386
2784
3182
H (ft)
102.0
101.3
99.3
96.0
91.3
85.3
77.9
69.2
59.1
Speed (rpm) = 1600
Q (gal/min)
0
455
909
1364
1818
2273
2727
3182
3636
H (ft)
133.2
132.4
129.7
125.4
119.2
111.4
101.7
90.4
77.2
Data from Fig. D.8 is "eyeballed"
The fit to data is obtained from a least squares fit to H = H 0 - AQ
2
H0=
161
ft
A = 4.23E-06 ft/(gal/min)
Performance Curves for Pump at various Speeds
Fig. D.8 Data
180
1000 rpm
160
1200 rpm
1400 rpm
140
H (ft)
1600 rpm
120
100
80
60
40
20
0
0
500
1000
1500
2000
2500
Q (gal/min)
3000
3500
4000
4500
Problem 10.38
[Difficulty: 2]
Problem 10.37
Given:
Data on pumping system
Find:
Total delivery; Operating speed
[Difficulty: 3]
Solution:
Basic equations:
Wh
Wh ρ Q g H
ηp
Wm 30 kW
H 30 m
Wm
The given or available data is
ρ 1000
kg
3
H 98.425 ft η 65 %
m
Then for the system
WmTotal 8 Wm
WmTotal 240 kW
The hydraulic total power is WhTotal WmTotal η
The total flow rate will then be QTotal
The flow rate per pump is
Q
WhTotal 156 kW
3
WhTotal
m
QTotal 0.53
s
ρ g H
QTotal
8
3
Q 0.066
From Fig. 10.15 the peak effiiciency is at a specific speed of
about
NScu 2500
3
Hence
N NScu
H
4
1
Q
7 L
QTotal 4.58 10
day
N 2410
2
The nearest standard speed to N 2410 rpm should be used
m
s
Q 1051 gpm
Problem 10.36
Given:
Data on centrifugal pump
Find:
Head at 1150 rpm
[Difficulty: 2]
Solution:
Basic equation:
(Eq. 10.2c)
The given or available data is
ρ 1000
kg
3
3
Q 0.025
m
ω 1750 rpm
m
Q
Vn2
2 π r2 b 2
Hence
Q
r2
2 π b 2 Vn2
Then
V'n2
From the outlet geometry
Finally
ω'
ω
r2 0.0909 m
Vn2
V'n2 2.30
U'2 ω' r2
V't2 U'2 V'n2 cos β2
H'
U'2 V't2
g
b 2 1.25 cm
m
Vn2 3.5
s
ω' 1150 rpm
From continuity
Also
β2 60 deg
s
m
s
U'2 11.0
m
V't2 9.80
m
H' 10.9 m
s
s
r2 9.09 cm
Problem 10.35
Given:
Data on pumping system
Find:
Number of pumps needed; Operating speed
[Difficulty: 3]
Solution:
Wh ρ Q g H
Basic equations:
ηp
Wh
Wm
The given or available data is
kg
3
6 L
Qtotal 110 10
day
m
Qtotal 1.273
s
Then for the system
Wh ρ Qtotal g H
Wh 125 kW
The required total power is
Wh
Wm
η
Wm 192 kW
ρ 1000
3
m
Hence the total number of pumps must be
The flow rate per pump will then be Q
192
37.5
5.12 , or at least six pumps
Qtotal
6
From Fig. 10.15 the peak effiiciency is at
a specific speed of about
NScu 2000
We also need
H 32.8 ft
Q 3363 gpm
3
Hence
N NScu
H
4
1
Q
H 10 m
N 473
2
The nearest standard speed to N 473 rpm should be used
3
Q 0.212
m
s
Q 212
L
s
η 65 %
Problem 10.34
[Difficulty: 3]
Given:
Data on a pump at BEP
Find:
(a) Specific Speed
(b) Required power input
(c) Curve fit parameters for the pump performance curve.
(d) Performance of pump at 820 rpm
Solution:
The given or available data is
ρ 1.94
slug
ft
3
η 87%
D 16 in
The governing equations are
Ns
Q 2500 cfm H 140 ft
ω Q
( g H)
Wh ρ Q g H
0.75
ω 1350 rpm
W
Wh
η
ω' 820 rpm
2
H0
U2
g
Ns 1.66
The specific speed is:
W 761 hp
The power is:
At shutoff
Since
D
U2 ω
2
2
H H0 A Q
ft
U2 94.248
s
A
it follows that
H0
Therefore:
U2
2
H0 276.1 ft
g
H0 H
2
5 min
A 2.18 10
2
ft
Q
Another way to write this is:
ω'
H'0 H0
ω
ω' 820 rpm
At BEP: Q' Q
5
H( ft) 276.1 2.18 10
ω'
ω
Q( cfm)
2
2
and
Q' 1519 cfm
A' A
H'0 101.9 ft
Thus:
A' 2.18 10
2
5 min
ft
H' H
ω'
ω
5
2
H' 51.7 ft
η' η 87 %
ω'
Wm W
ω
At
5
3
Wm 170.5 hp
Problem 10.33
[Difficulty: 3]
Given:
Data on a pump
Find:
Shutoff head; best efficiency; type of pump; flow rate, head, shutoff head and power at 900 rpm
Solution:
The given or available data is
ρ 999
3
kg
Ns 1.74
3
D 500 mm
Q 0.725
m
H 10 m
s
m
Wm 90 kW
ω' 900 rpm
1
Wh ρ Q g H
The governing equations are
ω Q
Ns
Q1
ω1 D1
3
Q2
h1
ω2 D2
3
2
ω1 D1
2
H0 C1
3
h
Similarity rules:
2
2
P1
2
ω2 D2
g
4
h2
U2
2
3
ρ1 ω1 D1
5
P2
3
ρ2 ω2 D2
5
3
h g H 98.1
J
ω
Hence
kg
ω 63.7
1
Q
H0
The shutoff head is given by
4
Ns h
rad
Wh ρ Q g H 71.0 kW
s
Wh
ηp
Wm
78.9 %
2
U2
2
D
m
U2 ω
2
g
U2 15.9
s
H0
Hence
U2
g
2
H0 25.8 m
with D1 = D2:
Q1
ω1
Q2
Q
or
ω2
ω
Q'
Q' Q
ω'
H0
Also
2
ω
P1
ρ ω1
3
P2
ρ ω2
ω
3
1.073
m
s
H'0
h1
ω1
2
h2
ω2
2
Wm
3
ω
W' m
3
ω'
or
2
ω'
H'0 H0
ω
ω'
or
3
ω'
H
2
ω
H'
2
ω'
H' H
2
21.9 m
ω
2
H'0 56.6 m
ω'
W' m Wm
ω
ω'
3
W' m 292 kW
Problem 10.32
[Difficulty: 2]
Problem 10.31
[Difficulty: 2]
Given:
Data on small centrifugal pump
Find:
Specific speed; Sketch impeller shape; Required power input
Solution:
Basic equation:
(Eq. 7.22a)
(Eq. 10.3c)
The given or available data is
ρ 1000
kg
3
3
ω 2875 rpm
ηp 70 %
m
Q 0.016
m
s
2
Hence
h g H
h 392
1
Then
NS
ω Q
(H is energy/weight. h is energy/mass)
2
s
2
3
h
m
NS 0.432
4
From the figure we see the impeller will be centrifugal
The power input is (from Eq. 10.3c)
Wh
Wm
ηp
Wm
ρ Q g H
ηp
Wm 8.97 kW
H 40 m
Problem 10.30
[Difficulty: 2]
Problem 10.29
[Difficulty: 2]
Problem 10.28
Given:
Data on centrifugal pump
Find:
Electric power required; gage pressure at exit
[Difficulty: 3]
Solution:
Basic equations:
(Eq. 10.8a)
(Eq. 10.8b)
(Eq. 10.8c)
The given or available data is
ρ 1.94
slug
ft
3
T 4.75 lbf ft
ηp 75 %
ηe 85 %
Q 65 gpm
Q 0.145
p 1 12.5 psi
z1 6.5 ft
ft
V1 6.5
s
z2 32.5 ft
ft
V2 15
s
From Eq. 10.8c
ω T ηp
Hp
ρ Q g
Hence, from Eq. 10.8b
ρ
2
2
p 2 p 1 V1 V2 ρ g z1 z2 ρ g Hp
2
p 2 53.7 psi
Also
Wh ρ g Q Hp
Wh 1119
The shaft work is then
Hence, electrical input is
ft
ω 3000 rpm
s
Hp 124 ft
Wh
Wm
ηp
Wm
We
ηe
3
ft lbf
s
Wm 1492
We 1756
ft lbf
s
ft lbf
s
Wh 2.03 hp
Wm 2.71 hp
We 2.38 kW
Problem 10.27
[Difficulty: 2]
Problem 10.26
[Difficulty: 3]
Given:
Data on axial flow fan
Find:
Volumetric flow rate, horsepower, flow exit angle
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 0.002377
slug
ft
3
ω 1350 rpm
d tip 3 ft
The mean radius would be half the mean diameter:
Therefore, the blade speed is:
U r ω
d root 2.5 ft
r
U 194.39
1 d tip d root
2
2
U
Vn1 V1 cos α1
So the entrance velocity components are:
The volumetric flow rate would then be:
Since axial velocity does not change:
The exit speed relative to the blade is:
Vt2 U w2 cos β2
The flow exit angle is:
β2 60 deg
r 1.375 ft
s
V1 sin α1 w1 cos β1 U
V1 cos α1 w1 sin β1
ft
V1 107.241
s
cos α1
sin α1
tan β1
β1 30 deg
ft
From velocity triangles we can generate the following two equations:
Combining the two equations: V1
α1 55 deg
cos α1
w1 V1
sin β1
ft
Vn1 61.511
s
(axial component)
(tangential component)
ft
w1 123.021
s
Vt1 V1 sin α1
π
2
2
Q Vn1 d tip d root
4
ft
Vt1 87.846
s
Q 132.9
ft
3
s
Vn2 Vn1
Vn2
w2
sin β2
ft
Vt2 158.873
s
Vt2
α2 atan
Vn2
ft
so the tangential component of absolute velocity is:
w2 71.026
s
Into the expression for power:
Wm U Vt2 Vt1 ρ Q
Wm 7.93 hp
α2 68.8 deg
Problem 10.25
[Difficulty: 3]
Given:
Data on suction pump
Find:
Plot of performance curves; Best effiiciency point
Solution:
ηp
Basic equations:
ρ = 1.94 slug/ft
Ph
Ph ρ Q g H
Pm
3
Ns
N Q
( g H)
(Note: Software cannot render a dot!)
0.75
Fitting a 2nd order polynomial to each set of data we find
-5
2
-4
H =-1.062x10 Q + 6.39x10 Q + 22.8
-6
Q (cfm) H (ft) P m (hp) P h (hp) η (%)
0
200
400
600
800
1000
23.0
22.3
21.0
19.5
17.0
12.5
15.2
17.2
24.4
27.0
32.2
36.4
0.0
8.4
15.9
22.1
25.7
23.6
2
η =-1.752x10 Q + 0.00237Q + 0.0246
0.0%
49.0%
65.1%
82.0%
79.9%
65.0%
Finally, we use Solver to maximize η by varying Q :
Q (cfm)
H (ft)
η (%)
676
18.4
82.6%
Pump Performance Curve
100%
25
H
BEP
20
η
75%
50%
10
25%
5
0
0%
0
200
400
600
800
1000
Q (cfm)
The Specific Speed for this pump is:
2.639
1200
η (%)
H (ft)
15
Problem 10.24
[Difficulty: 3]
Given:
Data on suction pump
Find:
Plot of performance curves; Best effiiciency point
Solution:
ηp
Basic equations:
ρ = 1.94 slug/ft
Ph
Ph ρ Q g H
Pm
3
(Note: Software cannot render a dot!)
Fitting a 2nd order polynomial to each set of data we find
2
H =-0.00759Q + 0.390Q + 189.1
-5
Q (cfm) H (ft) P m (hp) P h (hp) η (%)
36
50
74
88
125
190
195
176
162
120
25
30
35
40
46
12.9
18.4
24.6
27.0
28.4
2
η =-6.31x10 Q + 0.01113Q + 0.207
51.7%
61.5%
70.4%
67.4%
61.7%
Finally, we use Solver to maximize η by varying Q :
Q (cfm)
H (ft)
η (%)
88.2
164.5
69.8%
Pump Performance Curve
250
100%
H
BEP
200
η
75%
50%
100
25%
50
0
0%
0
20
40
60
80
Q (cfm)
100
120
140
η (%)
H (ft)
150
Problem 10.23
[Difficulty: 2]
Problem 10.22
[Difficulty: 2]
Problem 10.21
[Difficulty: 4]
Problem 10.20
[Difficulty: 4]
Given:
Geometry of centrifugal pump with diffuser casing
Find:
Flow rate; Theoretical head; Power; Pump efficiency at maximum efficiency point
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 1000
kg
3
m
ω 1750 rpm
r2 7.5 cm
ω 183
b 2 2 cm
β2 65 deg
rad
s
m
U2 13.7
s
Using given data
U2 ω r2
Illustrate the procedure with
Q 0.065
From continuity
Q
Vn2
2 π r2 b 2
From geometry
Vn2
Vt2 U2 Vrb2 cos β2 U2
cos β2
sin β2
Hence
Q
Vt2 U2
cot β2
2 π r2 b 2
3
m
s
m
Vn2 6.9
s
m
Vt2 10.5
s
V2
2
m
V2 12.6
s
2
Vn2 Vt2
Hideal
U2 Vt2
Hideal 14.8 m
g
Tfriction 10 %
Vt1 0
Wmideal
Tfriction 10 %
ω
10 %
Q ρ g Hideal
ω
ρ Q Hideal
ω
Tfriction 5.13 N m
(axial inlet)
V2
2
2
Vn2
Hactual 60 %
0.75
2 g
2 g
η
Q ρ g Hactual
Q ρ g Hideal ω Tfriction
Hactual 3.03 m
η 18.7 %
25
Efficiency (%)
20
15
10
5
0
0.02
0.04
0.06
0.08
0.1
Q (cubic meter/s)
The above graph can be plotted in Excel. In addition, Solver can be used to vary Q to maximize η. The results are
3
Q 0.0282
m
s
Wm Q ρ g Hideal ω Tfriction
η 22.2 %
Hideal 17.3 m
Wm 5.72 kW
Hactual 4.60 m
Problem 10.19
[Difficulty: 3]
Given:
Geometry of centrifugal pump
Find:
Draw inlet velocity diagram; Design speed for no inlet tangential velocity; Outlet angle; Head; Power
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
r1 4 in
ρ 1.94
slug
ft
3
Velocity diagrams:
r2 7 in
b 1 0.4 in
Q 70 cfm
Q 1.167
w1
b 2 0.3 in
ft
β1 20 deg
3
s
Vn1 = V 1 (Vt1 = 0)
V t2
w2
2
1
V2
2
Vn2
U2
U1
Vn
Vn1
sin( β)
Vn2
From continuity
Q
Vn
w sin( β)
2 π r b
From geometry
Vn
Q
Vt U Vrb cos( β) U
cos( β) U
cot( β)
sin( β)
2 π r b
For Vt1 0 we obtain
Q
U1
cot β1 0
2 π r1 b 1
Solving for ω
ω
Q
2
2 π r1 b 1
Hence
β2 45 deg
U1 ω r1
w
or
ω r1
Q
2 π r1 b 1
rad
ω 138
ft
U1 45.9
s
U2 ω r2
cot β1
s
A2
A1
cot β1 0
ω 1315 rpm
ft
U2 80.3
s
r2 b 2
r1 b 1
Q
Vn2
2 π r2 b 2
From the sketch
ft
Vn2 12.73
s
Vt2
α2 atan
Vn2
Q
Vt2 U2
cot β2
2 π r2 b 2
ft
Vt2 67.6
s
α2 79.3 deg
Hence
Wm U2 Vt2 ρ Q
The head is
H
Wm
ρ Q g
4 ft lbf
Wm 1.230 10
s
Wm 22.4 hp
H 169 ft
Problem 10.18
Given:
Data on centrifugal pump
Find:
Pressure rise; Express as ft of water and kerosene
[Difficulty: 1]
Solution:
Basic equations:
The given or available data is
η
ρ Q g H
Wm
ρw 1.94
slug
ft
Wm 18 hp
3
Q 350 gpm
H
For kerosene, from Table A.2
SG 0.82
ft
3
s
η 82 %
η Wm
Solving for H
Q 0.780
H 166.8 ft
ρw Q g
η Wm
Hk
SG ρw Q g
Hk 203 ft
Problem 10.17
Given:
Impulse turbibe
Find:
Optimum speed using the Euler turbomachine equation
[Difficulty: 1]
Solution:
The governing equation is the Euler turbomachine equation
In terms of the notation of Example 10.13, for a stationary CV
r1 r2 R
U1 U2 U
Vt1 V U
Vt2 ( V U) cos( θ)
and
mflow ρ Q
Hence
Tshaft [ R ( V U) cos( θ) R ( V U) ] ρ Q
Tout Tshaft ρ Q R ( V U) ( 1 cos( θ) )
The power is
Wout ω Tout ρ Q R ω ( V U) ( 1 cos( θ) )
Wout ρ Q U ( V U) ( 1 cos( θ) )
These results are identical to those of Example 10.13. The proof that maximum power is when U = V/2 is hence also the same
and will not be repeated here.
Problem 10.16
[Difficulty: 3]
Given:
Data on a centrifugal pump
Find:
Flow rate for zero inlet tangential velocity; outlet flow angle; power; head
developed
Solution:
The given or available data is
ρ 999
kg
3
ω 1200 rpm
η 70 %
β1 25 deg
r2 150 mm
m
r1 90 mm
b 1 10 mm
b 2 7.5 mm
β2 45 deg
The governing equations (derived directly from the Euler turbomachine equation) are
We also have from geometry
Vt2
α2 atan
Vn2
From geometry
Vn1
Vt1 0 U1 Vrb1 cos β1 r1 ω
cos β1
sin β1
and from continuity
Q
Vn1
2 π r1 b 1
Hence
(1)
Q
r1 ω
0
2 π r1 b 1 tan β1
2
Q 2 π r1 b 1 ω tan β1
Q 29.8
L
s
3
Q 0.0298
The power, head and absolute angle α at the exit are obtained from direct computation using Eqs. 10.2b, 10.2c, and 1 above
U1 r1 ω
m
U1 11.3
s
m
U2 18.8
s
U2 r2 ω
From geometry
Vn2
Vt2 U2 Vrb2 cos β2 r2 ω
cos β2
sin β2
and from continuity
Q
Vn2
2 π r2 b 2
m
Vn2 4.22
s
m
Vt1 0
s
m
s
Hence
Vn2
Vt2 r2 ω
tan β2
m
Vt2 14.6
s
Using these results in Eq. 1
Vt2
α2 atan
Vn2
α2 73.9 deg
Using them in Eq. 10.2b
Wm U2 Vt2 U1 Vt1 ρ Q
Using them in Eq. 10.2c
H
1
g
Wm 8.22 kW
H 28.1 m
U2 Vt2 U1 Vt1
This is the power and head assuming no inefficiency; with η = 70%, we have (from Eq. 10.4c)
Wh η Wm
Wh 5.75 kW
Hp η H
Hp 19.7 m
(This last result can also be obtained from Eq. 10.4a Wh ρ Q g Hp)
Problem 10.15
Given:
Data on a centrifugal pump
Find:
Estimate exit angle of impeller blades
[Difficulty: 3]
Solution:
The given or available data is
ρ 999
kg
3
Q 50
m
ω 1750 rpm
L
Win 45 kW
s
b 2 10 mm
η 75 %
D 300 mm
The governing equation (derived directly from the Euler turbomachine
equation) is
Wm
Vt2
U2 ρ Q
For an axial inlet
Vt1 0
hence
We have
D
U2 ω
2
m
U2 27.5
s
Hence
Wm
Vt2
U2 ρ Q
m
Vt2 24.6
s
From continuity
Q
Vn2
π D b 2
m
Vn2 5.31
s
an
d
Wm η Win
Wm 33.8 kW
With the exit velocities determined, β can be determined from exit geometry
tan( β)
Vn2
U2 Vt2
or
U2 Vt2
β atan
Vn2
β 61.3 deg
Problem 10.14
[Difficulty: 3]
Problem 10.13
[Difficulty: 2]
Given:
Geometry of centrifugal pump
Find:
Inlet blade angle for no tangential inlet velocity at 125,000 gpm; Head; Power
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 1.94
slug
ft
3
ω 575 rpm
r1 15 in
r2 45 in
b 1 4.75 in
β2 60 deg
Q 125000 gpm
Q 279
ft
3
s
Vn
Vrb
sin( β)
From continuity
Q
Vn
Vrb sin( β)
2 π r b
From geometry
Vn
Q
Vt U Vrb cos( β) U
cos( β) U
cot( β)
2 π r b
sin( β)
For Vt1 0 we obtain
Q
U1
cot β1 0
2 π r1 b 1
Using given data
U1 ω r1
Hence
β1 acot
Also
U2 ω r2
b 2 3.25 in
or
cot β1
2 π r1 b 1 U1
Q
ft
U1 75.3
s
2 π r1 b 1 U1
Q
β1 50 deg
ft
U2 226
s
Q
Vt2 U2
cot β2
2 π r2 b 2
ft
Vt2 201
s
The mass flow rate is
mrate ρ Q
slug
mrate 540
s
Hence
Wm U2 Vt2 U1 Vt1 mrate
The head is
H
Wm
mrate g
7 ft lbf
Wm 2.45 10
s
Wm 44497 hp
H 1408 ft
Problem 10.12
[Difficulty: 3]
Problem 10.11
[Difficulty: 3]
Given:
Geometry of centrifugal pump
Find:
Shutoff head; Absolute and relative exit velocitiesTheoretical head; Power input
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
kg
ρ 999
3
m
ω 1800 rpm
R1 2.5 cm
R2 18 cm
β2 75 deg
Q 30
b 2 1 cm
3
m
min
3
Q 0.500
m
s
m
At the exit
U2 ω R2
U2 33.9
s
At shutoff
Vt2 U2
m
Vt2 33.9
s
At design. from continuity
Q
Vn2
2 π R 2 b 2
m
Vn2 44.2
s
From the velocity diagram
Vn2 w2 sin β2
Vn2
w2
sin β2
V2
with
Vt2
α2 atan
Vn2
For Vt1 0 we get
Wm U2 Vt2 ρ Q 374 kW
H0 117 m
m
Vt2 22.1
s
m
V2 49.4
s
2
Hence we obtain
m
w2 45.8
s
Vt2 U2 Vn2 cot β2
2
1
H0 U2 Vt2
g
Vn2 Vt2
α2 26.5 deg
H
Wm
ρ Q g
76.4 m
Problem 10.10
[Difficulty: 2]
Given:
Geometry of centrifugal pump
Find:
Draw inlet and exit velocity diagrams; Inlet blade angle; Power
Solution:
Q
Vn
2 π r b
Basic equations:
The given or available data is
R1 1 in
R2 7.5 in
b 2 0.375 in
Q 800 gpm
Q 1.8
U1 ω R1
ft
U1 17.5
s
U2 ω R2
ft
U2 131
s
Q
Vn2
2 π R 2 b 2
ft
Vn2 14.5
s
R2
Vn1
V
R1 n2
ft
Vn1 109
s
ρ 1.94
slug
ft
Velocity diagrams:
3
ft
ω 2000 rpm
3
β2 75 deg
s
Vt2
Vrb1
V n1 = V1 (Vt1 = 0)
Vrb2
2
1
V2
2
Vn2
U2
U1
Then
Vn1
β1 atan
U1
From geometry
Vt1 U1 Vn1 cos β1
Then
Wm U2 Vt2 U1 Vt1 ρ Q
β1 80.9 deg
(Essentially radial entry)
ft
Vt1 0.2198
s
Vt2 U2 Vn2 cos β2
4 ft lbf
Wm 5.75 10
s
ft
Vt2 127.1
s
Wm 105 hp
Problem 10.9
[Difficulty: 2]
Problem 10.8
[Difficulty: 2]
Given:
Geometry of centrifugal pump
Find:
Theoretical head; Power input for given flow rate
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 1000
kg
3
r2 7.5 cm
m
ω 1750 rpm
From continuity
b 2 2 cm
β2 65 deg
3
Q 225
3
m
Q 0.0625
hr
Q
m
s
m
Vn2
2 π r2 b 2
Vn2 6.63
s
From geometry
Vn2
Vt2 U2 Vrb2 cos β2 U2
cos β2
sin β2
Using given data
U2 ω r2
Hence
Q
Vt2 U2
cot β2
2 π r2 b 2
m
Vt2 10.7
s
The mass flow rate is
mrate ρ Q
kg
mrate 62.5
s
Hence
Wm U2 Vt2 mrate
The head is
H
m
U2 13.7
s
Wm
mrate g
Vt1 0
(axial inlet)
Wm 9.15 kW
H 14.9 m
Problem 10.7
[Difficulty: 2]
Given:
Geometry of centrifugal pump
Find:
Rotational speed for zero inlet velocity; Theoretical head; Power input
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 1.94
slug
ft
3
r1 3 in
r2 9.75 in
b 1 1.5 in
b 2 1.125 in
β1 60 deg
β2 70 deg
Q 4000 gpm
Q 8.91
From continuity
Vn
Vrb sin( β)
2 π r b
From geometry
Vn
Q
Vt U Vrb cos( β) U
cos( β) U
cot( β)
2 π r b
sin( β)
For Vt1 0 we get
Q
U1
cot β1 0
2 π r1 b 1
Hence, solving for ω
ω
Q
2
2 π r1 b 1
We can now find U2
or
ω r1
ω 105
ft
U2 85.2
s
Q
2 π r1 b 1
ω 1001 rpm
s
ft
Vt2 78.4
s
The mass flow rate is
mrate ρ Q
slug
mrate 17.3
s
Hence Eq 10.2b becomes
Wm U2 Vt2 mrate
Wm 1.15 10
The head is
H
Wm
mrate g
cot β1 0
rad
Q
Vt2 U2
cot β2
2 π r2 b 2
s
Vrb
sin ( β)
cot β1
U2 ω r2
3
Vn
Q
ft
5 ft lbf
s
Wm 210 hp
H 208 ft
Problem 10.6
[Difficulty: 2]
Given:
Geometry of centrifugal pump
Find:
Theoretical head; Power input for given flow rate
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 1.94
slug
ft
3
ω 1250 rpm
r1 3 in
r2 9.75 in
b 1 1.5 in
b 2 1.125 in
β1 60 deg
β2 70 deg
Q 1500 gpm
Q 3.34
Q
Vn
Vrb sin( β)
2 π r b
From geometry
Vn
Q
Vt U Vrb cos( β) U
cos( β) U
cot( β)
2 π r b
sin( β)
Using given data
U1 ω r1
U2 ω r2
ft
Vt1 22.9
s
Q
Vt2 U2
cot β2
2 π r2 b 2
ft
Vt2 104
s
The mass flow rate is
mrate ρ Q
slug
mrate 6.48
s
Hence
Wm U2 Vt2 U1 Vt1 mrate
The head is
H
Wm
mrate g
s
Vrb
sin( β)
ft
U1 32.7
s
3
Vn
From continuity
Q
Vt1 U1
cot β1
2 π r1 b 1
ft
Wm 66728
ft
U2 106.4
s
ft lbf
s
Wm 121 hp
H 320 ft
Problem 10.5
[Difficulty: 2]
Given:
Geometry of centrifugal pump
Find:
Theoretical head; Power input for given flow rate
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 1.94
slug
ft
3
ω 575 rpm
r1 15 in
r2 45 in
b 1 4.75 in
b 2 3.25 in
β1 40 deg
β2 60 deg
Q 80000 gpm
Q 178
Q
Vn
Vrb sin( β)
2 π r b
From geometry
Vn
Q
Vt U Vrb cos( β) U
cos( β) U
cot( β)
2 π r b
sin( β)
Using given data
U1 ω r1
U2 ω r2
ft
Vt1 6.94
s
Q
Vt2 U2
cot β2
2 π r2 b 2
ft
Vt2 210
s
The mass flow rate is
mrate ρ Q
slug
mrate 346
s
Hence
Wm U2 Vt2 U1 Vt1 mrate
The head is
H
Wm
mrate g
s
Vrb
sin( β)
ft
U1 75.3
s
3
Vn
From continuity
Q
Vt1 U1
cot β1
2 π r1 b 1
ft
ft
U2 226
s
7 ft lbf
Wm 1.62 10
s
4
Wm 2.94 10 hp
H 1455 ft
Problem 10.4
[Difficulty: 2]
Problem 10.3
Given:
Data on centrifugal pump
Find:
Estimate basic dimensions
[Difficulty: 2]
Solution:
Basic equations:
(Eq. 10.2b, directly derived from the Euler turbomachine equation)
The given or available data is
ρ 999
kg
3
3
Q 0.6
m
ω 3000 rpm
3
m
Q 0.0100
min
m
w2 5.4
s
Vt1 0
From the outlet geometry
Vt2 U2 Vrb2 cos β2 U2
Hence, in Eq. 10.2b
Wm U2 mrate r2 ω mrate
with
Wm η Win
and
mrate ρ Q
Hence
r2
From continuity
Hence
s
Win 5 kW
η 72 %
and
U2 r2 ω
β2 90 deg
For an axial inlet
Also
m
2
2
2
Wm 3.6 kW
kg
mrate 9.99
s
Wm
2
mrate ω
Vn2 w2 sin β2
Q
Vn2
2 π r2 b 2
Q
b2
2 π r2 Vn2
r2 0.06043 m
r2 6.04 cm
m
Vn2 5.40
s
3
b 2 4.8776 10
m
b 2 0.488 cm
Problem 10.2
Given:
Geometry of centrifugal pump
Find:
Estimate discharge for axial entry; Head
[Difficulty: 2]
Solution:
Basic equations:
(Eq. 10.2b)
(Eq. 10.2c)
The given or available data is
ρ 999
kg
3
r1 10 cm
r2 20 cm
b 1 4 cm
β1 30 deg
β2 15 deg
b 2 4 cm
m
ω 1600 rpm
From continuity
Q
Vn
w sin( β)
2 π r b
w
Vn
sin( β)
From geometry
Vn
Q
Vt U w cos( β) U
cos( β) U
cot( β)
2 π r b
sin( β)
For an axial entry
Vt1 0
so
Using given data
U1 ω r1
m
U1 16.755
s
Hence
Q 2 π r1 b 1 U1 tan β1
Q
U1
cot β1 0
2 π r1 b 1
3
Q 0.2431
m
s
To find the power we need U 2, Vt2, and m rate
The mass flow rate is
kg
mrate 242.9
s
mrate ρ Q
m
U2 33.5
s
U2 ω r2
Q
Vt2 U2
cot β2
2 π r2 b 2
Hence
Wm U2 Vt2 U1 Vt1 mrate
The head is
H
Wm
mrate g
m
Vt2 15.5
s
5 J
Wm 1.258 10
s
Wm 126 kW
H 52.8 m
Problem 10.1
[Difficulty: 2]
Problem 9.185
[Difficulty: 4]
x
R
L
Given:
Soccer free kick
Find:
Spin on the ball
Solution:
1
2
2
ρ A V
ρ 1.21
The given or available data is
Σ F M a
FL
CL
Basic equations:
2
5 m
kg
ν 1.50 10
3
m
M 420 gm
C 70 cm
Compute the Reynolds number
D
2
C
D 22.3 cm
π
V D
Re
L 10 m
s
A
Re 4.46 10
ν
π D
4
5
The ball follows a trajectory defined by Newton's second law. In the horizontal plane ( x coordinate)
2
and
1
2
FL ρ A V CL
2
where R is the instantaneous radius of curvature of the trajectory
2 M
Hence, solving for R
R
From the trajectory geometry
x R cos( θ) R
Hence
x R 1
Solving for R
R
Hence, from Eq 1
CL
For this lift coefficient, from Fig. 9.27
Hence
ω D
2 V
L
where
sin( θ)
2
R
R
L2 x2
2 x
2 M
R A ρ
R 50.5 m
CL 0.353
1.2
ω 1.2
(And of course, Beckham still kind of rules!)
(1)
CL A ρ
2 V
D
ω 3086 rpm
L
R
2
A 0.0390 m
This Reynolds number is beyond the range of Fig. 9.27; however, we use Fig. 9.27 as a rough estimate
V
FL M aR M ax M
R
x 1 m
V 30
m
s
Problem 9.184
[Difficulty: 3]
Problem 9.183
[Difficulty: 4]
x
R
L
Given:
Baseball pitch
Find:
Spin on the ball
Solution:
Basic equations:
1
2
The given or available data is
Σ F M a
FL
CL
2
ρ A V
ρ 0.00234
slug
ft
M 5 oz
Compute the Reynolds number
C 9 in
ν 1.62 10
3
D
2
L 60 ft
s
2
C
D 2.86 in
π
V D
Re
4 ft
Re 1.73 10
ν
A
π D
2
A 6.45 in
4
V 80 mph
5
This Reynolds number is slightly beyond the range of Fig. 9.27; we use Fig. 9.27 as a rough estimate
The ball follows a trajectory defined by Newton's second law. In the horizontal plane ( x coordinate)
2
V
FL M aR M ax M
R
1
2
FL ρ A V CL
2
and
where R is the instantaneous radius of curvature of the trajectory
From Eq 1 we see the ball trajectory has the smallest radius (i.e. it curves the most) when C L is as large as
possible. From Fig. 9.27 we see this is when CL 0.4
Solving for R
Also, from Fig. 9.27
Hence
From the trajectory geometry
R
ω D
2 V
2 M
(1)
CL A ρ
1.5
ω 1.5
to
2 V
D
x R cos( θ) R
Solving for x
x R R 1
L
R
2 V
1.8
ω 1.8
where
sin( θ)
L R
R
x R 1
ω D
ω 14080 rpm
2
Hence
R 463.6 ft
2
x 3.90 ft
2 V
D
L
R
defines the best range
ω 16896 rpm
Problem 9.182
[Difficulty: 2]
Data on original Flettner rotor ship
Given:
Find:
Maximum lift and drag forces, optimal force at same wind speed, power requirement
Solution:
CL
Basic equations:
FL
1
2
The given or available data is
2
ρ A V
ρ 0.00234
slug
ft
ν 1.62 10
The spin ratio is:
The area is
ω D
2 V
9.52
A D L 500 ft
L 50 ft
3
4 ft
D 10 ft
ω 800 rpm
V 30 mph 44
ft
s
2
s
From Fig. 9.29, we can estimate the lift and drag coefficients: CL 9.5 CD 3.5
2
Therefore, the lift force is:
1
2
FL CL ρ A V
2
FL 1.076 10 lbf
The drag force is:
1
2
FD CD ρ A V
2
FD 3.964 10 lbf
This appears to be close to the optimum L/D ratio. The total force is:
F
2
4
3
2
4
FL FD
To determine the power requirement, we need to estimate the torque on the cylinder.
F 1.147 10 lbf
T τ A R τ π L D
D
2
2
π τ D L
2
In this expression τ is the average wall shear stress. We can estimate this stress using the flat plate approximation:
V ω D D
2
7
Re
2.857 10
ν
τ
FD
A
τ
1
2
For a cylinder at this Reynolds number: CD 0.003 Therefore, the shear stress is:
ρ V CD 6.795 10
2
3 lbf
ft
2
2
So the torque is:
T
π τ D L
2
The power is: P T ω 4471
53.371 ft lbf
ft lbf
s
P 8.13 hp
Problem 9.181
[Difficulty: 2]
Problem 9.180
Given:
Data on rotating cylinder
Find:
Lift force on cylinder
[Difficulty: 2]
Solution:
CL
Basic equations:
FL
1
2
The given or available data is
2
ρ A V
ρ 1.21
kg
3
2
5 m
ν 1.50 10
m
The spin ratio is:
The area is
ω D
2 V
0.419
s
L 30 cm
D 5 cm
ω 240 rpm
V 1.5
m
s
From Fig. 9.29, we can estimate the maximum lift coefficient: CL 1.0
2
A D L 0.015 m
Therefore, the lift force is:
1
2
FL CL ρ A V
2
FL 0.0204 N
Problem 9.179
[Difficulty: 5]
Problem 9.178
[Difficulty: 2]
Problem 9.177
[Difficulty: 5]
Problem 9.176
[Difficulty: 5]
Problem 9.175
Given:
Car spoiler
Find:
Whether they are effective
[Difficulty: 4]
Solution:
To perform the investigation, consider some typical data
For the spoiler, assume
b 4 ft
c 6 in
ρ 1.23
kg
3
A b c
m
From Fig. 9.17 a reasonable lift coefficient for a conventional airfoil section is
Assume the car speed is
V 55 mph
Hence the "negative lift" is
1
2
FL ρ A V CL
2
CL 1.4
FL 21.7 lbf
This is a relatively minor negative lift force (about four bags of sugar); it is not likely to produce a noticeable
difference in car traction
The picture gets worse at 30 mph:
FL 6.5 lbf
For a race car, such as that shown on the cover of the text, typical data might be
b 5 ft
In this case:
c 18 in
A b c
FL 1078 lbf
Hence, for a race car, a spoiler can generate very significant negative lift!
A 7.5 ft
2
V 200 mph
A 2 ft
2
Problem 9.174
[Difficulty: 5] Part 1/2
Problem 9.174
[Difficulty: 5] Part 2/2
Problem 9.173
[Difficulty: 4]
Problem 9.172
[Difficulty: 3]
Problem 9.171
[Difficulty: 4]
Given:
Aircraft in circular flight
Find:
Maximum and minimum speeds; Drag and power at these extremes
Solution:
Basic equations:
FD
CD
1
CL
2
ρ A V
2
The given data or available data are
ρ 0.002377
A 225 ft
1
2
slug
ft
2
FL
2
ρ A V
R 3250 ft
3
P FD V
Σ F M a
M 10000 lbm
M 311 slug
ar 7
The minimum velocity will be when the wing is at its maximum lift condition. From Fig . 9. 17 or Fig. 9.19
CL 1.72
CDinf 0.02
where CDinf is the section drag coefficient
CL
2
CD CDinf
π ar
The wing drag coefficient is then
CD 0.155
Assuming the aircraft is flying banked at angle β, the vertical force balance is
FL cos( β) M g 0
or
1
2
1
2
ρ A V CL cos( β) M g
2
(1)
The horizontal force balance is
2
M V
FL sin( β) M ar
R
or
ρ A V CL sin( β)
2
2
M V
R
(2)
Equations 1 and 2 enable the bank angle β and the velocity V to be determined
2
2
M V
2
M g
R
2
2
1
sin( β) cos( β)
1
1
2
2
ρ
A
V
C
ρ
A
V
C
2
L
L
2
2
or
R
2
4
M V
2
M g
4
2
4
ρ A V CL
2 2
2
4
2 2
M g
V
2
2
ρ A CL
4
2
tan( β)
V
R g
V 149
2
M
R
2
ft
s
V 102 mph
2
V2
R g
β atan
β 12.0 deg
The drag is then
1
2
FD ρ A V CD
2
FD 918 lbf
5 ft lbf
P FD V
The power required to overcome drag is
P 1.37 10
P 249 hp
s
The analysis is repeated for the maximum speed case, when the lift/drag coefficient is at its
minimum value. From Fig. 9.19, reasonable values are
CL 0.3
corresponding to α = 2 o (Fig. 9.17)
47.6
CL
2
CD CDinf
π ar
The wing drag coefficient is then
4
From Eqs. 1 and 2
CL
CDinf
2 2
M g
V
2
2
ρ A CL
V ( 309.9 309.9i)
2
4
M
2
CD 0.0104
ft
s
Obviously unrealistic (lift is just too
low, and angle of attack is too low
to generate sufficient lift)
2
R
We try instead a larger, more reasonable, angle of attack
CL 0.55
CDinf 0.0065
CL
4
2 2
M g
V
2
2
ρ A CL
4
V 91.2
2
2
tan( β)
The drag is then
2
CD CDinf
π ar
The wing drag coefficient is then
From Eqs. 1 and 2
corresponding to α = 4 o (Fig. 9.17)
V
R g
1
2
FD ρ A V CD
2
The power required to overcome drag is
M
R
2
m
s
CD 0.0203
V 204 mph
2
V2
R g
β atan
β 40.6 deg
FD 485 lbf
P FD V
5 ft lbf
P 1.45 10
s
P 264 hp
Problem 9.170
Given:
Aircraft in circular flight
Find:
Drag and power
[Difficulty: 3]
Solution:
Basic equations:
CD
FD
1
CL
2
ρ A V
2
The given data or available data are
ρ 0.002377
FL
1
2
slug
ft
2
ρ A V
R 3250 ft
3
V 150 mph
V 220
ft
P FD V
Σ F M a
M 10000 lbm
M 311 slug
A 225 ft
s
2
ar 7
Assuming the aircraft is flying banked at angle β, the vertical force balance is
FL cos( β) M g 0
or
1
2
2
ρ A V CL cos( β) M g
(1)
The horizontal force balance is
2
M V
FL sin( β) M ar
R
or
1
2
2
ρ A V CL sin( β)
2
Then from Eq 1
M g
FL
cos( β)
Hence
CL
2
(2)
V
β atan
R g
β 24.8 deg
4
FL 1.10 10 lbf
FL
1
tan( β)
R
V2
R g
2
Equations 1 and 2 enable the bank angle β to be found
M V
2
ρ A V
CL 0.851
CL
For the section, CDinf 0.0075 at CL 0.851 (from Fig. 9.19), so
2
CD CDinf
π ar
Hence
CD
FD FL
CL
FD 524 lbf
The power is
P FD V
P 1.15 10
5 ft lbf
s
P 209 hp
CD 0.040
Problem 9.169
[Difficulty: 3]
Problem 9.168
Given:
Data on an airfoil
Find:
Solution:
Maximum payload; power required
The given data or available data is
Vold 150 mph ρ 0.00234
[Difficulty: 3]
slug
ft
3
A 192.5 ft
2
35
arold
5.5
Assuming the old airfoil operates at close to design lift, from Fig. 9.19 CL 0.3 CDi 0.0062
CL
2
Then
CDold CDi
π arold
The new wing aspect ratio is
arnew 8
Hence
The power required is
CL
CDold 0.0107
2
CDnew CDi
π arnew
CDnew 0.00978
1
2
P T V FD V ρ A V CD V
2
If the old and new designs have the same available power, then
1
2
2
ρ A Vnew CDnew Vnew ρ A Vold CDold Vold
2
2
1
3
or
CDold
Vnew Vold
CDnew
ft
Vnew 227
s
arold 6.36
(CDi is the old airfoil's
section drag coefficient)
Problem 9.167
[Difficulty: 3]
Given:
Data on a light airplane
Find:
Cruising speed achieved using a new airfoil design
Solution:
V 150 mph 220.00
The given data or available data is
Then the area is
A b c
and the aspect ratio is
ar
ft
ρ 0.00234
s
ft
A 192.50 ft
b
slug
3
c 5.5 ft
b 35 ft
2
ar 6.36
c
The governing equations for steady flight are
W FL
and
T FD
where W is the total weight and T is the thrust
CL 0.3
For the NACA 23015 airfoil:
CDi 0.0062
where CDi is the section drag coefficient
The wing drag coefficient is given by Eq. 9.42
The drag is given by
1
2
FD ρ A V CD
2
Engine thrust required
T FD
The power required is
P T V
The wing drag coefficient is given by Eq. 9.42
P
1
3
ρ A V CD
2
2
CD CDi
π ar
T 116.7 lbf
P 46.66 hp
CDi 0.0031
CL
2
CD CDi
π ar
3
so the new speed is:
CD 0.011
FD 116.7 lbf
CL 0.2
For the NACA 66 2-215 airfoil:
The power is:
CL
Vnew
2 P
ρ A CD
3
CD 5.101 10
ft
Vnew 282
s
Vnew 192.0 mph
Problem 9.166
[Difficulty: 3]
Given:
Data on a light airplane
Find:
Angle of attack of wing; power required; maximum "g" force
Solution:
The given data or available data is
ρ 1.23
kg
3
2
M 1000 kg
A 10 m
CL 0.72
CD 0.17
W M g FL
T FD
m
V 63
m
s
The governing equations for steady flight are
where W is the weight T is the engine thrust
The lift coeffcient is given by
1
2
FL ρ A V Cd
2
Hence the required lift coefficient is
CL
M g
1
2
2
ρ A V
From Fig 9.17, for at this lift coefficient
α 3 deg
and the drag coefficient at this angle of attack is
CD 0.0065
CL 0.402
(Note that this does NOT allow for aspect ratio effects on lift and drag!)
Hence the drag is
1
2
FD ρ A V CD
2
FD 159 N
and
T FD
T 159 N
The power required is then
P T V
P 10 kW
The maximum "g"'s occur when the angle of attack is suddenly increased to produce the maximum lift
From Fig. 9.17
CL.max 1.72
1
2
FLmax ρ A V CL.max
2
The maximum "g"s are given by application of Newton's second law
M aperp FLmax
where a perp is the acceleration perpendicular to the flight direction
FLmax 42 kN
Hence
In terms of "g"s
aperp
aperp
g
FLmax
aperp 42
M
m
2
s
4.28
Note that this result occurs when the airplane is banking at 90 o, i.e, when the airplane is flying momentarily in a
circular flight path in the horizontal plane. For a straight horizontal flight path Newton's second law is
M aperp FLmax M g
Hence
In terms of "g"s
aperp
aperp
g
FLmax
M
3.28
g
aperp 32.2
m
2
s
Problem 9.165
Given:
Data on an airfoil
Find:
Maximum payload; power required
[Difficulty: 3]
Solution:
V 40
The given data or available data is
Then the area is
A b c
and the aspect ratio is
ar
ft
s
ρ 0.00234
slug
ft
A 35.00 ft
b
3
c 5 ft
b 7 ft
2
ar 1.4
c
The governing equations for steady flight are
W FL
and
T FD
where W is the model total weight and T is the thrust
CL 1.2
At a 10o angle of attack, from Fig. 9.17
CDi 0.010
where CDi is the section drag coefficient
The wing drag coefficient is given by Eq. 9.42
The lift is given by
2
CD CDi
π ar
1
2
FL ρ A V CL
2
CD 0.337
FL 78.6 lbf
W M g FL
The payload is then given by
M
or
CL
FL
M 78.6 lb
g
The drag is given by
1
2
FD ρ A V CD
2
Engine thrust required
T FD
The power required is
P T V
FD 22.1 lbf
T 22.1 lbf
P 1.61 hp
NOTE: Strictly speaking we have TWO extremely stubby wings, so a recalculation of drag effects (lift is unaffected) gives
b 3.5 ft
and
A b c
2.00
A 1.63 m
ar
b
c
c 5.00 ft
ar 0.70
CL
2
CD CDi
π ar
so the wing drag coefficient is
The drag is
1
2
FD 2 ρ A V CD
2
Engine thrust is
T FD
The power required is
P T V
CD 0.665
FD 43.6 lbf
T 43.6 lbf
P 3.17 hp
In this particular case it would seem that the ultralight model makes more sense - we need a smaller engine and smaller lift
requirements. However, on a per unit weight basis, the motor required for this aircraft is actually smaller. In other words, it should
probably be easier to find a 3.5 hp engine that weighs less than 80 lb (22.9 lb/hp) than a 1 hp engine that weighs less than 50 lb (50
lb/hp).
Problem 9.164
[Difficulty: 3]
Given:
Data on F-16 fighter
Find:
Minimum speed at which pilot can produce 5g acceleration; flight radius, effect of altitude on results
Solution:
The given data or available data is
ρ 0.00234
slug
ft
3
A 300 ft
2
CL 1.6
W 26000 lbf
At 5g acceleration, the corresponding force is:
FL 5 W 130000 lbf
The minimum velocity corresponds to the maximum lift coefficient:
Vmin
2 FL
ρ A CL
481
ft
ft
Vmin 481
s
s
To find the flight radius, we perform a vertical force balance:
β 90 deg asin
FL sin ( 90 deg β) W 0
Now set the horizontal force equal to the centripetal acceleration:
W
78.5 deg
FL
W
FL cos ( 90 deg β)
a
g c
ac g
FL
W
cos ( 90 deg β)
ac 157.6
ft
s
2
The flight radius corresponding to this acceleration is:
R
As altitude increases, the density decreases, and both the velocity and radius will increase.
Vmin
ac
R 1469 ft
2
Problem 9.163
Given:
Data on an airfoil
Find:
Maximum payload; power required
[Difficulty: 2]
Solution:
The given data or available data is
ρ 0.00234
slug
ft
3
L 5 ft
w 7 ft
V 40
Then
A w L
A 35 ft
The governing equations for steady flight are
W FL
and
ft
s
CL 0.75
CD 0.19
2
T FD
where W is the model total weight and T is the thrust
The lift is given by
1
2
FL ρ A V CL
2
W M g FL
The payload is then given by
or
FL 49.1 lbf
M
FL
g
M 49.1 lb
The drag is given by
1
2
FD ρ A V CD
2
FD 12.4 lbf
Engine thrust required
T FD
T 12.4 lbf
The power required is
P T V
P 498
ft lbf
s
P 0.905 hp
The ultralight model is just feasible: it is possible to find an engine that can produce about 1 hp that weighs less than about 50 lb.
Problem 9.162
Given:
[Difficulty: 2]
Data on a hydrofoil
FL
V
Find:
Minimum speed, power required, top speed
Solution:
Assumption:
y
FD
W
x
The drag on the hydrofoil is much greater than any other drag force on the craft once the foil supports the craft.
The given data or available data is
ρ 1.94
slug
ft
A 7.5 ft
3
To support the hydrofoil, the lift force must equal the weight:
Based on the required lift force, the speed must be:
Vmin
2
CL 1.5
CD 0.63
W 4000 lbf
Pmax 150 hp
FL W 4000 lbf
2 FL
ft
Vmin 19.1
s
ρ A CL
The drag force at this speed is
1
2
FD ρ A Vmin CD
2
FD 1680 lbf
Engine thrust required
T FD
T 1680 lbf
The power required is
P T Vmin
P 58.5 hp
As the speed increases, the lift will increase such that the lift and weight are still balanced. Therefore:
CD
Pmax
W Vmax
CL
Solving for the maximum speed:
Vmax
Pmax CL
W CD
ft
Vmax 49.1
s
Problem 9.161
Given:
Aircraft in level flight
Find:
Effective lift area; Engine thrust and power
[Difficulty: 1]
Solution:
Basic equation:
CD
FD
1
2
ρ A V
For level, constant speed
2
FD T
Given or available data is
V 225
km
ρ 1.21
kg
hr
CL
FL
1
2
FL W
P T V
2
ρ A V
V 62.5
m
s
CL 0.45
CD 0.065
M 900 kg
(Table A.10, 20 oC)
3
m
Hence
Also
1
2
FL CL ρ A V M g
2
FL
FD
The power required is then
CL
CD
FL M g
T FD
T 1275 N
P T V
P 79.7 kW
A
2 M g
2
2
CL ρ V
FL 8826 N
A 8.30 m
CD
FD FL
CL
FD 1275 N
Problem 9.160
[Difficulty: 1]
Problem 9.159
[Difficulty: 5]
Problem 9.158
[Difficulty: 4]
Problem 9.157
[Difficulty: 2]
Given:
Antique airplane guy wires
Find:
Maximum power saving using optimum streamlining
Solution:
Basic equation:
Given or available data is
CD
FD
1
2
ρ A V
2
L 50 m
The Reynolds number is
Hence
Re
V 175
km
hr
V 48.6
m
s
A 0.25 m
kg
3
m
V D
ν
D 5 mm
2
A L D
ρ 1.21
P FD V
ν 1.50 10
2
5 m
Re 1.62 10
(Table A.10, 20 oC)
s
4
1
2
P CD ρ A V V
2
so from Fig. 9.13
CD 1.0
P 17.4 kW
with standard wires
Figure 9.19 suggests we could reduce the drag coefficient to CD 0.06
Hence
1
2
Pfaired CD ρ A V V
2
Pfaired 1.04 kW
The maximum power saving is then
ΔP P Pfaired
ΔP 16.3 kW
Thus
ΔP
P
94 %
which is a HUGE savings! It's amazing the antique planes flew!
Problem 9.156
[Difficulty: 3]
Given:
Data on airfoil and support in wind tunnel, lift and drag measurements
Find:
Lift and drag coefficients of airfoil
FL
Solution:
V
Basic equations: CD
1
2
The given or available data is
FD
2
ρ A V
L 6 in
CL
FL
1
2
2
ρ A V
W 30 in
FL 10 lbf
FD
V 100
y
ft
s
Dcyl 1 in
FD 1.5 lbf
ρ 0.00233
slug
ft
Re
V Dcyl
4
Re 5.112 10
ν
3
ν 1.63 10
4 ft
2
s
FD FDcyl FDairfoil
We need to determine the cylindrical support's contribution to the total drag force:
Compute the Reynolds number
x
Lcyl 10 in
Therefore: CDcyl 1
1
2
So the drag force on the support is: FDcyl CDcyl ρ V Lcyl Dcyl 0.809 lbf
2
So the airfoil drag is: FDairfoil FD FDcyl 0.691 lbf The reference area for the airfoil is: A L W 1.25 ft
The lift and drag coefficients are:
CL
FL
1
2
CD
2
ρ V A
2
CL 0.687
FDairfoil
1
2
2
ρ V A
CD 0.0474
Problem 9.155
[Difficulty: 5] Part 1/2
Problem 9.155
[Difficulty: 5] Part 2/2
Problem 9.154
[Difficulty: 5] Part 1/2
Problem 9.154
[Difficulty: 5] Part 2/2
Problem 9.153
[Difficulty: 5]
Problem 9.152
[Difficulty: 5]
Problem 9.151
Given:
Find:
Solution:
[Difficulty: 4]
Baseball popped up, drag estimates based on Reynolds number
Time of flight and maximum height
CD
Basic equation:
FD
1
2
Given or available data is
2
ρ A V
M 0.143 kg
ΣFy M ay
Here are the calculations performed in Excel:
ρ =
1.21
dVy
dt
m
V0y 25
D 0.073 m
s
We solve this problem by discretizing the flight of the ball:
Given or available data:
M = 0.143
25
V 0y =
D = 0.073
ay
ΣFy
ΔVy ay Δt
Δt
M
Δy Vy Δt
kg
m/s
m
kg/m
3
2
ν = 1.50E-05 m /s
Computed results:
A = 0.00419 m
Δt =
0.25 s
2
CD
a y (m/s2 )
V ynew (m/s)
25.0
22.3
19.6
17.0
14.4
11.9
9.3
6.8
4.4
1.9
0.0
-2.5
Re
1.22E+05
1.08E+05
9.54E+04
8.26E+04
7.01E+04
5.77E+04
4.54E+04
3.33E+04
2.13E+04
9.30E+03
0.00E+00
1.19E+04
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.47
0.47
0.10
-10.917
-10.688
-10.490
-10.320
-10.177
-10.059
-9.964
-9.893
-9.844
-9.840
-9.810
-9.799
22.3
19.6
17.0
14.4
11.9
9.3
6.8
4.4
1.9
0.0
-2.5
-4.9
-4.9
-7.3
-9.8
-12.2
-14.6
-16.9
-19.3
-21.5
-23.7
2.39E+04
3.57E+04
4.76E+04
5.93E+04
7.09E+04
8.24E+04
9.37E+04
1.05E+05
1.15E+05
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
-9.767
-9.714
-9.641
-9.547
-9.434
-9.303
-9.154
-8.988
-8.816
-7.3
-9.8
-12.2
-14.6
-16.9
-19.3
-21.5
-23.7
t n (s)
y (m)
V y (m/s)
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.44
2.69
0.0
5.9
11.1
15.7
19.6
22.9
25.6
27.6
29.0
29.8
30.0
29.7
2.94
3.19
3.44
3.69
3.94
4.19
4.44
4.69
4.93
28.7
27.2
25.1
22.3
19.0
15.0
10.5
5.4
0.0
The results are plotted below.
The answers are:
height =
30.0 m
time =
4.93 s
Trajectory of Baseball
35
30
25
y (m)
20
15
10
5
0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
t (s)
3.5
4.0
4.5
5.0
Problem 9.150
[Difficulty: 4]
Given: Data on a rocket
Find: Plot of rocket speed with and without drag
Solution:
From Example 4.12, with the addition of drag the momentum equation becomes
FB y FS y
CV
a rf y dV
t
CV
v xyz dV
CV
v xyz V xyz dA
where the surface force is
FS y
1
AV 2 C D
2
Following the analysis of the example problem, we end up with
2
dVCV Ve m e 12 AVCV C D
g
dt
M 0 m e t
This can be written (dropping the subscript for convenience)
dV
f V , t
dt
(1)
where
f V , t
Ve m e 12 AV 2 C D
M 0 m e t
g
(2)
Equation 1 is a differential equation for speed V.
It can be solved using Euler’s numerical method
Vn 1 Vn t f n
where Vn+1 and Vn are the n + 1th and nth values of V, fn is the function given by Eq. 2 evaluated at the nth
step, and t is the time step.
The initial condition is
V0 0 at t 0
Given or available data:
M 0 = 400 kg
m e = 5 kg/s
V e = 3500 m/s
= 1.23 kg/m
D = 700 mm
C D = 0.3
3
Computed results:
2
A = 0.38 m
N = 20
t = 0.50 s
Without drag:
V n (m/s) f n V n+1 (m/s)
With drag:
n t n (s) V n (m/s)
0 0.0
0.0
1 0.5
17.0
2 1.0
34.1
3 1.5
51.2
4 2.0
68.3
5 2.5
85.5
6 3.0
102
7 3.5
119
8 4.0
136
9 4.5
152
10 5.0
168
11 5.5
184
33.9
34.2
34.3
34.3
34.2
34.0
33.7
33.3
32.8
32.2
31.5
30.7
17.0
34.1
51.2
68.3
85.5
102
119
136
152
168
184
200
0.0
17.0
34.1
51.3
68.7
86.2
104
122
140
158
176
195
33.9
34.2
34.5
34.8
35.1
35.4
35.6
35.9
36.2
36.5
36.9
37.2
17.0
34.1
51.3
68.7
86.2
104
122
140
158
176
195
213
12 6.0
13 6.5
14 7.0
15 7.5
16 8.0
17 8.5
18 9.0
19 9.5
20 10.0
29.8
28.9
27.9
26.9
25.8
24.7
23.6
22.5
21.4
214
229
243
256
269
282
293
305
315
213
232
251
270
289
308
328
348
368
37.5
37.8
38.1
38.5
38.8
39.1
39.5
39.8
40.2
232
251
270
289
308
328
348
368
388
200
214
229
243
256
269
282
293
305
f n V n+1 (m/s)
Trajectory of a Rocket
400
300
V (m/s)
200
Without Drag
100
With Drag
0
0
2
4
6
8
t (s)
10
12
Problem 9.149
[Difficulty: 4]
Problem 9.148
[Difficulty: 4]
Given:
Data on sonar transducer
Find:
Drag force at required towing speed; minimum depth necessary to avoid cavitation
Solution:
CD
Basic equation:
FD
1
2
Given or available data is
2
ρ A V
D 15 in
A
π
4
2
D 1.227 ft
2
p p inf
1
2
V 55
The Reynolds number of the flow is: Re
The area is:
CP
V D
ν
ft
s
p min 5 psi ρ 1.93
6.486 10
h
6
ρ g
3
ν 1.06 10
5 ft
2
(Table A.7, 70oF)
s
From Fig. 9.11, we estimate the drag coefficient:
Therefore the drag force is:
p inf p atm
slug
ft
From Fig. 9.12 the minimum pressure occurs where CP 1.2
Solving for the required depth:
p p atm ρ g h
2
ρ V
1
2
FD CD ρ V A
2
Therefore:
CD 0.18
FD 645 lbf
1
2
p inf p min CP ρ V 29.326 psi
2
h 33.9 ft
Problem 9.147
[Difficulty: 4]
Problem 9.146
Given:
Data on barge and river current
Find:
Speed and direction of barge
Solution:
Basic
equation:
CD
FD
1
2
Given or available data is
2
ρ A V
W 8820 kN w 10 m
kg
CDa 1.3 ρw 998
3
m
[Difficulty: 4]
2
6 m
νw 1.01 10
s
L 30 m
h 7 m
kg
ρa 1.21
3
m
m
Vriver 1
s
νa 1.50 10
h sub
W
ρw g w L
3.00 m
Vsub w L h sub
CDw 1.3
2
5 m (Water data from Table A.8, air
s data from Table A.10, 20 oC)
First we need to calculate the amount of the barge submerged in the water. From Archimedes' Principle:
The submerged volume can be expressed as:
m
Vwind 10
s
W ρw g Vsub
Combining these expressions and solving for the depth:
h air h h sub 4.00 m
Therefore the height of barge exposed to the wind is:
Assuming the barge is floating downstream, the velocities of the water and air relative to the barge is:
Vw Vriver Vbarge
Assuming that the barge is rectangular, the areas exposed to the air and water are:
Va Vwind Vbarge
2
Aa L w 2 ( L w) h air 620 m
2
Aw L w 2 ( L w) h sub 540 m
In order for the barge to be traveling at a constant speed, the drag forces due to the air and water must match:
1
2
2
CDw ρw Vw Aw CDa ρa Va Aa
2
2
1
Solving for the speed relative to the water:
2
2
2 ρa Aa
Vw Va
ρw Aw
ρa Aa
In terms of the barge speed:
Vw Va
ρw Aw
So solving for the barge speed:
2
Since the drag coefficients are equal, we can simplify: ρw Vw Aw ρa Va Aa
Since the speeds must be in opposite directions:
ρa Aa
Vriver Vbarge Vwind Vbarge
ρw Aw
ρa Aa
Vriver Vwind
ρw Aw
m
Vbarge
Vbarge 1.426
s
ρa Aa
1
ρw Aw
downstream
Problem 9.145
[Difficulty: 3]
Given:
Data on rooftop carrier
Find:
Drag on carrier; Additional fuel used; Effect on economy; Effect of "cheaper" carrier
Solution:
Basic equation:
Given or available data is
CD
FD
1
2
ρ A V
2
w 1 m
V 100
h 50 cm
km
V 27.8
hr
kg
ρH2O 1000
3
m
ρ 1.225
kg
r 10 cm
m
FE 12.75
s
r
h
Additional power is
Additional fuel is
0.2
ΔP
FD V
s
o
FE 30.0
L
A 0.5 m
mi
gal
BSFC 0.3
2
5 m
m
From the diagram
km
2
A w h
ν 1.50 10
3
ηd 85 %
kg
kW hr
(Table A.10,
20oF)
s
1
2
FD CD ρ A V
2
CD 0.25
FD 59.1 N
ΔP 1.93 kW
ηd
ΔFC BSFC ΔP
4 kg
ΔFC 1.61 10
ΔFC 0.00965
s
kg
min
Fuel consumption of the car only is (with SGgas 0.72 from Table A.2)
FC
FE
Fuel economy with the carrier is
r
h
Additional power is
0
ΔP
FE
SG gas ρH2O
FCT FC ΔFC
The total fuel consumption is then
For the square-edged:
V
s
o
FD V
ηd
V
FCT
SG gas ρH2O
CD 0.9
ΔP 6.95 kW
FC 1.57 10
3 kg
s
3 kg
FCT 1.73 10
s
FE 11.6
km
L
1
2
FD CD ρ A V
2
FC 0.0941
kg
min
FCT 0.104
FE 27.2
mi
gal
FD 213 N
kg
min
Additional fuel is
ΔFC BSFC ΔP
The total fuel consumption is then
Fuel economy withy the carrier is now
4 kg
ΔFC 5.79 10
s
3 kg
FCT FC ΔFC
FE
V
FCT
FCT 2.148 10
SG gas ρH2O
The cost of the trip of distance d 750 km for fuel costing p
$ 3.50
gal
ΔFC 0.0348
FE 9.3
km
L
s
FCT 0.129
FE 21.9
The cost of the trip of with the rounded carrier ( FE 11.6
Cost
d
FE
p discount
Cost 69.47
plus the rental fee
Cost 59.78
plus the rental fee
km
) is then
L
d
FE
p
min
kg
min
mi
gal
with a rental discount $ 5 less than the rounded carrier is
then
Cost
kg
Hence the "cheaper" carrier is more expensive (AND the environment is significantly more damaged!)
Problem 9.144
[Difficulty: 4]
Problem 9.143
Given:
Data on a tennis ball
Find:
Maximum height
[Difficulty: 4]
Solution:
The given data or available data is M 57 gm
2
5 m
m
Vi 50
s
D 64 mm
ν 1.45 10
2
Then
A
From Problem 9.132
CD
CD
π D
24
m
Re 1
Re
24
1 Re 400
0.646
Re
400 Re 3 10
0.4275
5
5
6
CD 0.000366 Re
3 10 Re 2 10
CD 0.18
Re 2 10
6
1
2
FD ρ A V CD
2
dV
1
2
ρ V A CD M g
dt
2
For motion before terminal speed, Newton's second law (x upwards) is
M a M
For the maximum height Newton's second law is written in the form
M a M V
0
Hence the maximum height is
s
3 2
A 3.22 10
4
CD 0.5
The drag at speed V is given by
V
dV
x max
ρ A CD 2
V g
2 M
V
i
dV
1
2
ρ V A CD M g
dx
2
V
i
V
dV
ρ A CD 2
V g
2 M
0
This integral is quite difficult because the drag coefficient varies with Reynolds number, which
varies with speed. It is best evaluated numerically. A form of Simpson's Rule is
ΔV
f ( V) dV
f V0 4 f V1 2 f V2 4 f V3 f VN
3
where ΔV is the step size, and V0, V1 etc., are the velocities at points 0, 1, ... N.
ρ 1.23
kg
3
m
Here
V0 0
From the associated Excel workbook
(shown here)
If we assume
the integral
VN Vi
x max 48.7 m
ΔV
V (m/s)
0.0
2.5
5.0
7.5
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
42.5
45.0
47.5
50.0
Re
0
11034
22069
33103
44138
55172
66207
77241
88276
99310
110345
121379
132414
143448
154483
165517
176552
187586
198621
209655
220690
CD
0.000
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
W
1
4
2
4
2
4
2
4
2
4
2
4
2
4
2
4
2
4
2
4
1
Vi
N
f (V ) W xf (V )
0.000
0.252
0.488
0.695
0.866
1.00
1.09
1.16
1.19
1.21
1.21
1.20
1.18
1.15
1.13
1.10
1.06
1.03
1.00
0.970
0.940
0.000
1.01
0.976
2.78
1.73
3.99
2.19
4.63
2.39
4.84
2.42
4.80
2.36
4.62
2.25
4.38
2.13
4.13
2.00
3.88
0.940
CD 0.5
V
i
V
dV
x max
ρ A CD 2
V g
2 M
0
becomes
x max
ρ A CD 2
V 1
ρ A CD 2 M g i
M
ln
x max 48.7 m
The two results agree very closely! This is because the integrand does not vary much after the first few steps so
the numerical integral is accurate, and the analytic solution assumes CD = 0.5, which it essentially does!
Problem 9.142
Given:
Data on an air bubble
Find:
Time to reach surface
[Difficulty: 4]
Solution:
The given data or available data is
h 100 ft 30.48 m
ρ 1025
kg
CD 0.5 (Fig. 9.11)
(Table A.2)
3
p atm 101 kPa
m
1
dx V dt where
To find the location we have to integrate the velocity over time:
V
patm ρ g h
3 CD p atm ρ g ( h
4 g d 0
The results (generated in Excel) for each bubble diameter are shown below:
d 0 = 0.3 in
d 0 = 7.62 mm
d0=
t (s) x (m) V (m/s)
t (s) x (m) V (m/s)
0
5
10
15
20
25
30
35
40
45
50
63.4
0
2.23
4.49
6.76
9.1
11.4
13.8
16.1
18.6
21.0
23.6
30.5
0.446
0.451
0.455
0.460
0.466
0.472
0.478
0.486
0.494
0.504
0.516
0.563
5
mm
0
5
10
15
20
25
30
35
40
45
50
55
0
1.81
3.63
5.47
7.32
9.19
11.1
13.0
14.9
16.9
18.8
20.8
0.362
0.364
0.367
0.371
0.374
0.377
0.381
0.386
0.390
0.396
0.401
0.408
60
65
70
75
77.8
22.9
25.0
27.1
29.3
30.5
0.415
0.424
0.435
0.448
0.456
d0 =
15
mm
t (s) x (m) V (m/s)
0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.1
0
3.13
6.31
9.53
12.8
16.1
19.5
23.0
26.6
30.5
0.626
0.635
0.644
0.655
0.667
0.682
0.699
0.721
0.749
0.790
Use Goal Seek for the last time step to make x = h !
Depth of Air Bubbles versus Time
30
25
20
x (m)
15
10
Initial Diameter = 5 mm
Initial Diameter = 0.3 in
5
Initial Diameter = 15 mm
0
0
10
20
30
40
50
t (s)
60
70
80
x)
6
Problem 9.141
[Difficulty: 4]
Given:
Data on a tennis ball
Find:
Terminal speed time and distance to reach 95% of terminal speed
Solution:
The given data or available data is
M 57 gm
2
Then
A
From Problem 9.132
CD
CD
2
5 m
D 64 mm
π D
ν 1.45 10
3 2
A 3.22 10
4
24
m
Re 1
Re
24
1 Re 400
0.646
Re
CD 0.5
400 Re 3 10
0.4275
5
5
6
CD 0.000366 Re
3 10 Re 2 10
CD 0.18
Re 2 10
6
At terminal speed drag equals weight FD M g
The drag at speed V is given by
1
2
FD ρ A V CD
2
Assume
CD 0.5
Hence the terminal speed is
Vt
Check the Reynolds number
Re
2 M g
ρ A CD
Vt D
ν
This is consistent with the tabulated CD values!
m
Vt 23.8
s
Re 1.05 10
5
s
ρ 1.23
kg
3
m
For motion before terminal speed, Newton's second law is M a M
dV
1
2
M g ρ V A CD
dt
2
Hence the time to reach 95% of terminal speed is obtained by separating variables and integrating
t
0.95 Vt
1
g
ρ A CD
2 M
dV
2
V
0
For the distance to reach terminal speed Newton's second law is written in the form
M a M V
dV
1
2
M g ρ V A CD
dx
2
Hence the distance to reach 95% of terminal speed is obtained by separating variables and
integrating
x
0.95 V t
V
g
ρ A CD
2 M
dV
2
V
0
These integrals are quite difficult because the drag coefficient varies with Reynolds number, which
varies with speed. They are best evaluated numerically. A form of Simpson's Rule is
ΔV
f ( V) dV
f V0 4 f V1 2 f V2 4 f V3 f VN
3
where ΔV is the step size, and V0, V1 etc., are the velocities at points 0, 1, ... N.
Here
From the associated Excel
calculations (shown below):
V0 0
0.95 Vt
VN 0.95 Vt
ΔV
t 4.69 s
x 70.9 m
N
These results compare to 4.44 s and 67.1 m from Problem 9.132, which assumed the drag coefficient was constant and
analytically integrated. Note that the drag coefficient IS essentially constant, so numerical integration was not really necessary!
For the time:
V (m/s)
Re
0
1.13
2.26
3.39
4.52
5.65
6.78
7.91
9.03
10.2
11.3
12.4
0
4985
9969
14954
19938
24923
29908
34892
39877
44861
49846
54831
5438
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
W f (V ) W xf (V )
1 0.102 0.102
4 0.102 0.409
2 0.103 0.206
4 0.104 0.416
2 0.106 0.212
4 0.108 0.432
2 0.111 0.222
4 0.115 0.458
2 0.119 0.238
4 0.125 0.499
2 0.132 0.263
4 0.140 0.561
13.6
14.7
15.8
16.9
18.1
19.2
20.3
21.5
22.6
59815
64800
69784
74769
79754
84738
89723
94707
99692
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
2
4
2
4
2
4
2
4
1
CD
0.151
0.165
0.183
0.207
0.241
0.293
0.379
0.550
1.05
0.302
0.659
0.366
0.828
0.483
1.17
0.758
2.20
1.05
Total time:
4.69
For the distance:
f (V ) W xf (V )
s
0.00
0.115
0.232
0.353
0.478
0.610
0.752
0.906
1.08
1.27
1.49
1.74
0.000
0.462
0.465
1.41
0.955
2.44
1.50
3.62
2.15
5.07
2.97
6.97
2.05
2.42
2.89
3.51
4.36
5.62
7.70
11.8
23.6
4.09
9.68
5.78
14.03
8.72
22.5
15.4
47.2
23.6
Total distance:
70.9
m
Problem 9.140
[Difficulty: 4]
Problem 9.139
[Difficulty: 4]
Problem 9.138
[Difficulty: 3]
Problem 9.137
Given:
Data on model airfoil
Find:
Lift and drag coefficients
[Difficulty: 3]
Solution:
Basic equation:
CD
FD
1
2
ρ A V
2
Given or available data is D 2 cm
V 30
m
CL
FL
where A is plan area for airfoil, frontal area for rod
1
2
ρ A V
2
(Rod)
L 25 cm
FL 50 N
s
b 60 cm
c 15 cm (Airfoil)
FH 6 N
Note that the horizontal force F H is due to drag on the airfoil AND on the rod
ρ 1.225
kg
ν 1.50 10
3
2
5 m
m
For the rod
Rerod
V D
4
Rerod 4 10
ν
Arod L D
Arod 5 10
1
2
FDrod CDrod ρ Arod V
2
Hence for the airfoil
A b c
CD
2
so from Fig. 9.13
3
2
ρ A V
CD 0.0654
CDrod 1.0
2
m
FDrod 2.76 N
FD FH FDrod
FD
1
(Table A.10, 20 oC)
s
CL
FD 3.24 N
FL
1
2
2
ρ A V
CL 1.01
CL
CD
15.4
Problem 9.136
[Difficulty: 3]
Problem 9.135
[Difficulty: 4]
Problem 9.134
[Difficulty: 3]
Given:
Data on a tennis ball
Find:
Terminal speed time and distance to reach 95% of terminal speed
Solution:
The given data or available data is
M 57 gm
2
A
Then
π D
4
2
5 m
D 64 mm
ν 1.45 10
3
A 3.22 10
At terminal speed drag equals weight
FD M g
The drag at speed V is given by
1
2
FD ρ A V CD
2
Hence the terminal speed is
Vt
Check the Reynolds number
Re
s
kg
3
m
2
CD 0.5
(from Fig. 9.11)
M g
2
ρ 1.23
m
Assuming high Reynolds number
1
ρ A CD
Vt D
m
Vt 23.8
s
Re 1.05 10
ν
5
Check!
For motion before terminal speed Newton's second law applies
M a M
dV
1
2
M g ρ V A CD
dt
2
Separating variables
2
d
V g k V
dt
or
V
1
2
dV t
g k V
0
Hence
Evaluating at V = 0.95Vt
For distance x versus time, integrate
g
V( t)
k
0.95 Vt
dx
dt
g
k
k
where
1
2
ρ A CD
k 0.0174
2 M
dV
g k V
1
g k
k V
g
atanh
tanh g k t
g
k
tanh g k t
tanh g k t
1
t
g k
x
atanh 0.95 Vt
t
0
g
k
tanh g k t dt
k
g
t 4.44 s
1
m
Note that
1
tanh( a t) dt ln( cosh( a t) )
a
Hence
x ( t)
Evaluating at V = 0.95Vt
1
k
t 4.44 s
ln cosh g k t
so
x ( t) 67.1 m
Problem 9.133
[Difficulty: 3]
F n2
Fn1
W
Given:
Circular disk in wind
Find:
Mass of disk; Plot α versus V
Solution:
Basic equations:
CD
ΣM 0
FD
1
2
2
ρ V A
1
D
Summing moments at the pivot W L sin( α) Fn1 L L Fn2 0 (1) and for each normal drag
2
2
1
2
Fn ρ Vn A CD
2
Assume 1) No pivot friction 2) CD is valid for Vn = Vcos(α)
The data is
ρ 1.225
kg
μ 1.8 10
3
5 N s
m
2
m
D 25 mm
d 3 mm
CD1 1.17 (Table
9.3)
Red
ρ V d
μ
V 15
m
s
L 40 mm
α 10 deg
Red 3063
so from Fig. 9.13
CD2 0.9
2
Hence
1
2 π D
Fn1 ρ ( V cos( α) )
CD1
2
4
Fn1 0.077 N
1
D
2
Fn2 ρ ( V cos( α) ) L d CD2
2
2
Fn2 0.00992 N
The drag on the support is much less than on the disk (and moment even less), so results will not be much different from those of
Problem 9.105
2
Hence Eq. 1 becomes
1
1
D
1
D
2 π D
2
M L g sin( α) L ρ ( V cos( α) )
CD1 L ρ ( V cos( α) ) L d CD2
2
2
2 2
2
4
2
M
ρ V cos( α)
2
D
D
1
2
π D CD1 1
L d CD2
2 L
2
4 g sin( α) 2
M 0.0471 kg
V
Rearranging
4 M g
ρ
tan( α)
cos( α)
We can plot this by choosing α and computing V
1
1 π D2 C
D1
2
1 D L D d C
D2
2 L
2
V 35.5
80
V (m/s)
60
40
20
0
10
20
30
40
Angle (deg)
This graph can be easily plotted in Excel
50
60
70
m
s
tan( α)
cos( α)
Problem 9.132
[Difficulty: 3]
Problem 9.131
[Difficulty: 3]
Problem 9.130
Given:
3 mm raindrop
Find:
Terminal speed
[Difficulty: 2]
Solution:
Basic equation:
1
2
FD ρ A V CD
2
Given or available data is
D 3 mm
Summing vertical forces
ΣF 0
kg
ρH2O 1000
3
m
kg
ρair 1.225
3
m
1
2
M g FD M g ρair A V CD 0
2
M 1.41 10
Assume the drag coefficient is in the flat region of Fig. 9.11 and verify Re later
V
Re
2 M g
CD ρair A
V D
ν
V 8.95
2
5 m
(Table A.10, 20 oC)
s
Buoyancy is negligible
3
π D
M ρH2O
6
Check Re
ν 1.50 10
5
2
kg
A
π D
4
CD 0.4
m
s
3
Re 1.79 10 which does place us in the flat region of the curve
Actual raindrops are not quite spherical, so their speed will only be approximated by this result
6
A 7.07 10
2
m
Problem 9.129
[Difficulty:
[
2]
Problem 9.128
Given:
Find:
Solution:
[Difficulty: 2]
Data on helium-filled balloon, angle balloon string makes when subjected to wind
FB
V
Drag coefficient for the balloon
FD
1
2
Basic equations:
FD ρ A V CD
2
The given or available data is
D 20 in
FB 0.3 lbf
ρ 0.00233
slug
ft
Based on a free body diagram of the balloon,
The reference area for the balloon is: A
π
4
ΣFy 0
3
y
V 10
ν 1.63 10
ft
s
4 ft
x
θ 55 deg
2
T
(Table A.9, 70oF)
s
FD FB tan( 90 deg θ) 0.2101 lbf
2
D 2.182 ft
2
so the drag coefficient is:
CD
FD
1
2
2
ρ V A
0.826
Problem 9.127
[Difficulty: 2]
Problem 9.126
[Difficulty: 2]
Problem 9.125
Given:
A runner running during different wind conditions.
Find:
Calories burned for the two different cases
Solution:
Governing equation:
CD
FD
FD
1
V 2 A
2
1
C D V 2 A
2
Assumption: 1) CDA = 9 ft2 2) Runner maintains speed of 7.5 mph regardless of wind conditions
No wind:
0.00238 slug/ft 3
V 7.5 mph 11 ft/s
The drag force on the runner is:
FD
2
1
slug
2 ft
9 ft 2 0.00238 3 11 2 1.296 lbf
2
ft
s
Energy burned:
E Power time FD Vrunner time
Where
time 4 mi
Hence
E 1.296 lbf
hr
3600 s
1920 s
hr
7.5 mi
11f
0.0003238 kcal
1920 s
ft lbf
s
E 8.86 kcal
With 5 mph wind:
Going upwind:
Vrel 12.5 mph 18.33
The drag force on the runner is:
time 2 mi
FD
ft
s
2
1
slug
2 ft
9 ft 2 0.00238 3 18.33 2 3.598 lbf
2
ft
s
hr
3600 s
960 s
hr
7.5 mi
[Difficulty: 2]
Eupwind 3.598 lbf
11f
0.0003238 kcal
960 s
12.30 kcal
ft lbf
s
Vrel 2.5 mph 3.67
Going downwind:
The drag force on the runner is:
FD
ft
s
2
1
slug
2 ft
9 ft 2 0.00238 3 3.67 2 0.144 lbf
2
ft
s
hr
3600 s
960 s
hr
7.5 mi
11f
0.0003238 kcal
0.144 lbf
960 s
0.49 kcal
ft lbf
s
time 2 mi
E downwind
Hence the total energy burned to overcome drag when the wind is 5 mph is:
E 12.30 kcal 0.49 kcal 12.79 kcal
this is 44% higher
Problem 9.124
Given:
Data on wind turbine blade
Find:
Power required to maintain operating speed
[Difficulty: 4]
Solution:
Basic equation:
1
2
FD ρ A V CD
2
The given or available data is
ω 25 rpm
ρ 0.00233
L 1.5 ft
slug
ft
W 115 ft
ν 1.63 10
3
4 ft
2
(Table A.9, 70oF)
s
The velocity is a function of radial position, V( r) r ω, so Re varies from 0 to Remax
V( W) L
Remax 2.77 10
ν
6
The transition Reynolds number is 500,000 which therefore occurs at about 1/4 of the maximum radial distance; the
boundary layer is laminar for the first quarter of the blade. We approximate the entire blade as turbulent - the first 1/4 of
the blade will not exert much moment in any event
Re( r)
Hence
L
ν
V( r)
L ω
ν
r
1
CD
Using Eq. 9.37a
0.0742
1
ReL
The drag on a differential area is
Hence
M 1 dM
dFD
2
W
ReL
0.0742
1
L ω r
ν
5
1740
L ω
ν
1
2
2
ρ dA V CD ρ L V CD dr
2
2
W
1
1740
1
2
ρ L V CD r dr
0
1
2
M ρ L ω
2
5
M
W
r
L ω
0.0742
r
ν
1
5
1
5
1
ν
2 3
ρ L ω r 0.0742
r
2
L ω
1
5
1
r
L ω
1740
The bending moment is then
ν
dM dFD r
ν 1
1740
r dr
L ω
0
1
1
14
19
5
5
1
1740 ν 3
2 5 0.0742 ν
5
5
ν
ν 2
0.0742 L ω r 1740 L ω r dr M 2 ρ L ω 19 L ω W 3 L ω W
0
M 1666 ft lbf
5
Hence the power is
P M ω
P 7.93 hp
Problem 9.123
Given:
Data on wind turbine blade
Find:
Bending moment
[Difficulty: 1]
FD
V
Solution:
Basic equation:
1
2
FD ρ A V CD
2
The given or available data is
V 85 knot 143.464
A L W
slug
ft
ReL
CD
The bending moment is then
s
L 1.5 ft
W 115 ft
3
2
4 ft
ReL 1.32 10
ν
0.0742
1
1740
ReL
x
M
ν 1.63 10
V L
ReL
ft
A 172.5 ft
ρ 0.00233
For a flat plate, check Re
y
CD 0.00311
5
1
2
FD 2 ρ A V CD
2
FD 25.7 lbf
W
M FD
2
M 1480 ft lbf
6
2
(Table A.9, 70oF)
s
so use Eq. 9.37a
Problem 9.122
Given:
Data on car antenna
Find:
Bending moment
[Difficulty: 1]
FD
Solution:
V
1
2
Basic equation:
FD ρ A V CD
2
The given or available data is
V 120
km
33.333
hr
s
L 1.8 m
D 10 mm
x
A 0.018 m
kg
3
M
2
ν 1.50 10
5 m
m
For a cylinder, check Re
Re
V D
ν
y
2
A L D
ρ 1.225
m
Re 2.22 10
(Table A.10, 20 oC)
s
4
From Fig. 9.13
CD 1.0
1
2
FD ρ A V CD
2
The bending moment is then
L
M FD
2
M 11.0 N m
FD 12.3 N
Problem 9.121
[Difficulty: 5]
Problem 9.120
[Difficulty: 2]
Given:
Data on advertising banner
Find:
Power to tow banner; Compare to flat plate; Explain discrepancy
Solution:
Basic equation:
1
2
FD ρ A V CD
2
P FD V
V 55 mph
The given data or available data is
1
2
FD ρ A V CD
2
FD 771 lbf
P FD V
7
0.455
2.58
log ReL
1
2
FD ρ A V CD
2
s
A 180 ft
4 ft
V L
ReL
ν
ft
A L h
ν 1.62 10
For a flate plate, check Re
CD
V 80.7
2
L 45 ft
h 4 ft
slug
ft
CD 0.05
L
CD 0.563
h
4 ft lbf
P 6.22 10
s
P 113 hp
2
(Table A.9, 69oF)
s
ReL 2.241 10
so flow is fully turbulent. Hence use Eq 9.37b
1610
CD 0.00258
ReL
ρ 0.00234
FD 3.53 lbf
This is the drag on one side. The total drag is then 2 FD 7.06 lbf . This is VERY much less than the banner drag.
The banner drag allows for banner flutter and other secondary motion which induces significant form drag.
3
Problem 9.119
[Difficulty: 4]
Problem 9.118
[Difficulty: 4]
Problem 9.117
[Difficulty: 4]
Problem 9.116
Given:
Data on dimensions of anemometer
Find:
Calibration constant
[Difficulty: 5]
Solution:
The given data or available data is
D 2 in
R 3 in
ρ 0.00234
slug
ft
3
The drag coefficients for a cup with open end facing the airflow and a cup with open end facing downstream are, respectively,
from Table 9.3
CDopen 1.42
CDnotopen 0.38
Assume the anemometer achieves steady speed ω due to steady wind speed V
k
The goal is to find the calibration constant k, defined by
V
ω
We will analyze each cup separately, with the following assumptions
1) Drag is based on the instantaneous normal component of velocity (we ignore possible effects on drag
coefficient of velocity component parallel to the cup)
2) Each cup is assumed unaffected by the others - as if it were the only object present
3) Swirl is neglected
4) Effects of struts is neglected
R
Relative velocity
= Vcos - R
V
Vcos
In this more sophisticated analysis we need to compute the instantaneous normal relative velocity. From the
sketch, when a cup is at angle θ, the normal component of relative velocity is
Vn V cos( θ) ω R
(1)
The relative velocity is sometimes positive, and sometimes negatiive. From Eq. 1, this is determined by
ω R
θc acos
V
For
V n( θ)
0 θ θc
Vn 0
θc θ 2 π θc
Vn 0
θc θ 2 π
Vn 0
0
90
(2)
180
270
360
θ
The equation for computing drag is
1
2
FD ρ A Vn CD
2
where
A
(3)
2
π D
2
A 3.14 in
4
In Eq. 3, the drag coefficient, and whether the drag is postive or negative, depend on the sign of the relative velocity
For
0 θ θc
CD CDopen
FD 0
θc θ 2 π θc
CD CDnotopen
FD 0
θc θ 2 π
CD CDopen
FD 0
The torque is
1
2
T FD R ρ A Vn CD R
2
The average torque is
Tav
2 π
1
2 π
0
1
T dθ
π
π
T dθ
0
where we have taken advantage of symmetry
Evaluating this, allowing for changes when θ = θ c
θc
π
1
1
1
1
2
2
Tav
ρ A Vn CDopen R dθ
ρ A Vn CDnotopen R dθ
π 2
π 2
θ
0
c
Using Eq. 1
and note that
θ
π
c
2
2
Tav
C
( V cos( θ) ω R) dθ CDnotopen ( V cos( θ) ω R) dθ
2 π Dopen 0
θ
c
θ
π
c
2
2
2
ρ A R ω
V
V
Tav
CDopen
cos( θ) R dθ CDnotopen cos( θ) R dθ
ω
2 π
ω
θ
0
c
ρ A R
V
ω
k
The integral is
1
2
2 1
2
( k cos( θ) R) dθ k 2 cos( θ) sin( θ) 2 θ 2 k R sin( θ) R θ
For convenience define
1
2 1
2
f ( θ) k cos( θ) sin( θ) θ 2 k R sin( θ) R θ
2
2
Hence
Tav
ρ A R
2 π
CDopen f θc CDnotopen f ( π) f θc
For steady state conditions the torque (of each cup, and of all the cups) is zero. Hence
0
CDopen f θc CDnotopen f ( π) f θc
or
CDnotopen
f θc
f ( π)
CDopen CDnotopen
Hence
CDnotopen
1
2 1
2
2 π
2
k cos θc sin θc θc 2 k R sin θc R θc
k R π
2
CDopen CDnotopen
2
2
Recall from Eq 2 that
ω R
θc acos
V
Hence
or
R
θc acos
k
CDnotopen
R
1
R
R
R
2 1 R
2
2 π
2
k sin acos acos 2 k R sin acos R acos
k R π
k
2
k
k
k
C
C
2
k
2
Dopen
Dnotopen
This equation is to be solved for the coefficient k. The equation is highly nonlinear; it can be solved by
iteration or using Excel's Goal Seek or Solver
From the associated Excel workbook
k 0.990 ft
The result from Problem 9.106 was k 0.0561
k 0.0707
mph
rpm
mph
rpm
This represents a difference of 20.6%. The difference can be attributed
to the fact that we had originally averaged the flow velocity, rather than integrated over a complete revolution.
Problem 9.115
[Difficulty: 3]
Given: zero net force acting on the particle; drag force and electrostatic force
Find:
Solution:
(a) Under steady-state, the net force acting on the particle is zero. The forces
acting on the particle contain the electrostatic force FE and the drag force FD
(Page 418, the first equation right after Fig.8.11).
FE Fd 0 Qs E 6Ua 0
(1)
where U is the particle velocity relative to the stationary liquid.
Qs E
Then one obtains U 6a
(2)
V
FE
y
Qs
FD
x
(b) From the solution, we can know that the particle velocity depends on its size. Smaller particles run faster
than larger ones, and thus they can be separated.
(c) Substituting the values of a, Qs, E, and into equation (2), we obtains the velocity for a=1 m
U
N
10 12
C
1000 V/m
0.053
0.053m/s
3
6
m
Pa s m
6 10 Pa s 1 10
and U = 0.0053 m/s for a = 10 m.
The negatively charged particle moves in the direction opposite to that of the electric field applied.
Problem 9.114
[Difficulty: 4]
Given:
Data on a sports car
Find:
Speed for aerodynamic drag to exceed rolling resistance; maximum speed & acceleration at 100
km/h; Redesign change that has greatest effect
Solution:
1
2
Basic equation: FD ρ A V CD
2
P FD V
The given data or available data is
M 1250 kg
2
A 1.72 m
CD 0.31
Pengine 180 hp 134.23 kW FR 0.012 M g
To find the speed at which aerodynamic drag first equals rolling resistance, set the two forces equal
2 FR
ρ A CD
V 21.2
m
V 76.2
s
V 100
1
2
FD ρ V A CD
2
η
Pused
1
2
ρ V A CD FR
2
hr
The power consumed by drag and rolling resistance at this speed is
Hence the drive train efficiency is
km
V 27.8
hr
m
s
Pengine 17 hp 12.677 kW
FD 253 N
Pused FD FR V
Pused 11.1 kW
η 87.7 %
Pengine
The acceleration is obtained from Newton's second law
where T is the thrust produced by the engine, given by
M a ΣF T FR FD
P
T
V
The maximum acceleration at 100 km/h is when full engine power is used. Pengine 180 hp 134.2 kW
Because of drive train inefficiencies the maximum power at the wheels is Pmax η Pengine
Hence the maximum thrust is Tmax
The maximum acceleration is then
Pmax
V
3
km
To find the drive train efficiency we use the data at a speed of
The aerodynamic drag at this speed is
kg
m
FR 147.1 N
The rolling resistance is then
Hence V
ρ 1.23
Pmax 118 kW
Tmax 4237 N
amax
Tmax FD FR
M
amax 3.07
m
2
s
The maximum speed is obtained when the maximum engine power is just balanced by power consumed by drag and rolling resistance
Pmax
For maximum speed:
1 ρ V 2 A C F V
max
D
R max
2
This is a cubic equation that can be solved by iteration or by using Excel's Goal Seek or Solver
km
Vmax 248
hr
We are to evaluate several possible improvements:
For improved drive train
η η 6 %
η 93.7 %
Pmax
Pmax η Pengine
Pmax 126 kW
1 ρ V 2 A C F V
max
D
R max
2
km
Vmax 254
hr
Solving the cubic (using Solver)
Improved drag coefficient:
CDnew 0.29
Pmax
Pmax 118 kW
1 ρ V 2 A C
max
Dnew FR Vmax
2
km
This is the
Vmax 254
hr best option!
Solving the cubic (using Solver)
Reduced rolling resistance:
FRnew 0.91 % M g
FRnew 111.6 N
1
2
Pmax ρ Vmax A CD FRnew Vmax
2
Solving the cubic (using Solver)
km
Vmax 250
hr
The improved drag coefficient is the best option.
Problem 9.113
[Difficulty: 2]
Given:
Data on 1970's and current sedans
Find:
Plot of power versus speed; Speeds at which aerodynamic drag exceeds rolling drag
Solution:
CD
Basic equation:
FD
1
2
2
ρ V A
The aerodynamic drag is
1
2
FD CD ρ V A
2
Total resistance
FT FD FR
ρ =
The rolling resistance is
FR 0.015 W
The results generated in Excel are shown below:
0.00234
3
slug/ft
(Table A.9)
Computed results:
V (mph)
F D (lbf)
1970's Sedan
F T (lbf)
P (hp)
F D (lbf)
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
12.1
18.9
27.2
37.0
48.3
61.2
75.5
91.4
109
128
148
170
193
218
245
273
302
79.6
86.4
94.7
104
116
129
143
159
176
195
215
237
261
286
312
340
370
4.24
5.76
7.57
9.75
12.4
15.4
19.1
23.3
28.2
33.8
40.2
47.5
55.6
64.8
74.9
86.2
98.5
6.04
9.44
13.6
18.5
24.2
30.6
37.8
45.7
54.4
63.8
74.0
84.9
96.6
109
122
136
151
58.5
61.9
66.1
71.0
76.7
83.1
90.3
98.2
107
116
126
137
149
162
175
189
204
3.12
4.13
5.29
6.63
8.18
10.0
12.0
14.4
17.1
20.2
23.6
27.5
31.8
36.6
42.0
47.8
54.3
V (mph)
F D (lbf)
67.5
F R (lbf)
67.5
V (mph)
F D (lbf)
52.5
F R (lbf)
52.5
47.3
The two speeds above were obtained using Solver
59.0
Current Sedan
F T (lbf)
P (hp)
Power Consumed by Old and New Sedans
120
1970's Sedan
Current Sedan
100
80
P (hp)
60
40
20
0
20
30
40
50
60
V (mph)
70
80
90
100
Problem 9.112
[Difficulty: 3]
Given:
Data on a bus
Find:
Power to overcome drag; Maximum speed; Recompute with new fairing; Time for fairing to pay for itself
Solution:
Basic
equation:
1
2
FD ρ A V CD
2
The given data or available data is
1
2
FD ρ A V CD
2
P FD V
V 80
km
V 22.2
hr
FD 2096 N P FD V
m
s
2
A 7.5 m
CD 0.92
ρ 1.23
kg
3
m
P 46.57 kW The power available is
Pmax 465 hp 346.75 kW
The maximum speed corresponding to this maximum power is obtained from
1
1
2
Pmax ρ A Vmax CD Vmax
2
or
Pmax
Vmax
1 ρ A CD
2
We repeat these calculations with the new fairing, for which
1
2
FD ρ A V CD
2
FD 1959 N
3
m
km
Vmax 43.4
Vmax 156.2
s
hr
CD 0.86
Pnew FD V
Pnew 43.53 kW
1
The maximum speed is now
The initial cost of the fairing is
Pmax
Vmax
1 ρ A CD
2
3
Cost 4500 dollars
The cost per day is reduced by improvement in the bus performance at 80 km/h
The new cost per day is then
Hence the savings per day is
The initial cost will be paid for in
Costdaynew Gain Costday
Saving Costday Costdaynew
τ
Cost
Saving
m
km
Vmax 44.4
Vmax 159.8
s
hr
Costday 300
The fuel cost is
Gain
Pnew
P
Gain 93.5 %
Costdaynew 280
Saving 19.6
dollars
day
dollars
τ 7.56 month
day
dollars
day
Problem 9.111
[Difficulty: 3]
Problem 9.110
[Difficulty: 3]
Problem 9.109
[Difficulty: 3]
Problem 9.108
[Difficulty: 2]
Given:
Mass, maximum and minimum drag areas for a skydiver
Find:
(a) Terminal speeds for skydiver in each position
(b) Time and distance needed to reach percentage of terminal speed from given altitude
Solution:
CD
Basic equation
FD
1
2
2
ρ U A
W 170 lbf ACDmin 1.2 ft
The given or available data are:
ρ
From Table A.3 we can find the density:
ρSL
0.7433
2
ρ 0.002377
Ut
2 W
ρ A CD
slug
ft
To find terminal speed, we take FBD of the skydiver: ΣFy 0
Solving for the speed:
ACDmax 9.1 ft
3
2
H 9800 ft 2987 m
3 slug
0.7433 1.767 10
ft
M g FD 0
For the minimum drag area:
Utmax
For the maximum drag area:
Utmin
3
1
2
FD ρ U A CD M g W
2
2 W
ρ ACDmin
2 W
ρ ACDmax
ft
Utmax 400
s
ft
Utmin 145.4
s
To find the time needed to reach a fraction of the terminal velocity, we re-write the force balance:
ΣFy M ay
M g FD M ay
W
In terms of the weight:
W Ut
W dU
d U
g dt Ut
g dt
1
2
and
M g
2
ρ U A CD
W
g
U
dU
dy
1
2
2
ρ U A CD M
W dU
W
dU
U
g dt
g
dy
W Ut
g
2
U d U
Ut dy Ut
dU
dt
2
Simplifying this expression:
U
2
dU
dy
To normalize the derivatives by the terminal speed:
We may now re-write the above equation as:
2
2
W Ut
U
2 W W Ut d U
U
U d
W ρ A CD
g
2
g dt Ut
Ut
ρ A CD
Ut dy Ut
1
M U
where we have substituted for the terminal
speed.
Ut
t U d U
U
d U
Now we can integrate with respect to time and
1
g Ut dy Ut distance:
Ut
g dt Ut
U
Un
Ut
If we let
Ut dUn
1 Un
g dt
2
we can rewrite the equations:
t
g
Separating variables:
dt
Ut
0
0.90
1
1 Un
2
g t
dUn Integrating we get:
Ut
atanh( 0.9) atanh( 0 )
0
Evaluating the inverse hyperbolic tangents:
Now to find the distance:
2
Ut
2
t
1.472 Ut
dUn
1 Un
Un
g
dy
g
so: tmin
1.472 Utmin
g
Separating variables:
y
0
Integrating we get:
1.472 Utmax
6.65 s tmax
dy
2
Ut
g
18.32 s
0.9
Un
g
1 Un
dUn
2
0
2
1 0.92
0.8304 Ut
0.8304 Solving for the distance: y
ln
2
g
2 10
Ut
g y
1
so: y min
0.8304 Utmin
g
2
166.4 m y max
0.8304 Utmax
g
2
1262 m
Problem 9.107
Given:
Circular disk in wind
Find:
Mass of disk; Plot α versus V
[Difficulty: 2]
Solution:
CD
Basic
equations:
ΣM 0
FD
1
2
2
ρ V A
Summing moments at the pivotW L sin( α) Fn L 0
Hence
M g sin( α)
The data is
ρ 1.225
1
2
2
2 π D
ρ ( V cos( α) )
kg
3
m
M
V
Rearranging
4
V 15
2
2
1
2
Fn ρ Vn A CD
2
and
CD
m
D 25 mm
s
M 0.0451 kg
8 g sin( α)
2
π ρ D CD
CD 1.17
2
π ρ V cos( α) D CD
8 M g
α 10 deg
tan( α)
V 35.5
cos( α)
m
s
tan( α)
cos( α)
We can plot this by choosing α and computing V
80
V (m/s)
60
40
20
0
10
20
30
40
Angle (deg)
This graph can be easily plotted in Excel
50
60
70
(Table 9.3)
Problem 9.106
Given:
Data on dimensions of anemometer
Find:
Calibration constant; compare to actual with friction
[Difficulty: 3]
Solution:
The given data or available data is
D 2 in
R 3 in
ρ 0.00234
slug
ft
3
The drag coefficients for a cup with open end facing the airflow and a cup with open end facing downstream are, respectively,
from Table 9.3
CDopen 1.42
CDnotopen 0.38
1
2
The equation for computing drag is FD ρ A V CD
2
(1)
2
A
where
π D
A 0.0218 ft
4
2
Assuming steady speed ω at steady wind speed V the sum of moments will be zero. The two cups that are momentarily parallel
to the flow will exert no moment; the two cups with open end facing and not facing the flow will exert a moment beacuse of their
drag forces. For each, the drag is based on Eq. 1 (with the relative velocity used!). In addition, friction of the anemometer is
neglected
1
1
2
2
ΣM 0 ρ A ( V R ω) CDopen R ρ A ( V R ω) CDnotopen R
2
2
or
2
2
( V R ω) CDopen ( V R ω) CDnotopen
This indicates that the anemometer reaches a steady speed even in the abscence of friction because it is the
relative velocity on each cup that matters: the cup that has a higher drag coefficient has a lower relative
velocity
Rearranging for
k
V
ω
2
V R C
Dopen
ω
2
V R C
Dnotopen
ω
Hence
CDnotopen
1
CDopen
R
k
CDnotopen
1
CDopen
k 9.43 in
k 0.0561
mph
rpm
For the actual anemometer (with friction), we first need to determine the torque produced when the anemometer is
stationary but about to rotate
Minimum wind for rotation is
Vmin 0.5 mph
The torque produced at this wind speed is
Tf
1 ρ A V 2 C
min Dopen R
2
Tf 3.57 10
1 ρ A V 2 C
min Dnotopen R
2
6
ft lbf
A moment balance at wind speed V, including this friction, is
ΣM 0
or
1 ρ A ( V R ω) 2 C
Dopen R
2
1 ρ A ( V R ω) 2 C
Dnotopen R Tf
2
2 Tf
2
2
( V R ω) CDopen ( V R ω) CDnotopen
R ρ A
This quadratic equation is to be solved for ω when
V 20 mph
After considerable calculations
ω 356.20 rpm
This must be compared to the rotation for a frictionless model, given by
V
ωfrictionless
k
The error in neglecting friction is
ωfrictionless 356.44 rpm
ω ωfrictionless
ω
0.07 %
Problem 9.105
[Difficulty: 2]
Problem 9.104
[Difficulty: 2]
FB
V
FD
y
T
W
x
Given:
Sphere dragged through river
Find:
Relative velocity of sphere
Solution:
CD
Basic
equations:
FD
1
FB ρ g Vol
2
ρ V A
ΣF 0
2
The above figure applies to the sphere
For the horizontal forces FD T sin( θ) 0
For the vertical forces
Here
V 5
m
s
(1)
T cos( θ) FB W 0
D 0.5 m
The Reynolds number is Red
(2)
SG 0.30
V D
6
Red 1.92 10
ν
and from Table A.8 ν 1.30 10
2
6 m
s
ρ 1000
kg
3
m
Therefore we estimate the drag coefficient: CD 0.15 (Fig 9.11)
FB W
FD T sin( θ)
sin( θ) ρ g Vol ( 1 SG ) tan( θ)
cos( θ)
3
π D
Hence
FD ρ g
6
( 1 SG ) tan( θ)
Therefore
1
π D
2 π D
CD ρ V
ρ g
( 1 SG) tan( θ)
2
4
6
2
Solving for θ:
tan( θ)
3
But we have
1
2
1
2
2 π D
FD CD ρ V A CD ρ V
2
2
4
3
2
CD V
4 g D ( 1 SG )
2
CD V
3
θ atan
4 g D ( 1 SG)
The angle with the horizontal is:
α 90 deg θ
α 50.7 deg
Problem 9.103
[Difficulty: 3]
FBnet
V
FD
y
x
T
Given:
Series of party balloons
Find:
Wind velocity profile; Plot
Wlatex
Solution:
Basic equations:
CD
FD
1
FB ρair g Vol
2
ρ V A
2
The above figure applies to each balloon
For the horizontal forces FD T sin( θ) 0
T cos( θ) FBnet Wlatex 0
Here
π D
FBnet FB W ρair ρHe g
6
(2)
3
D 20 cm
M latex 3 gm
RHe 2077
Rair 287
N m
p He 111 kPa
kg K
N m
p air 101 kPa
kg K
Applying Eqs 1 and 2 to the top balloon, for which
Wlatex 0.02942 N
p He
kg
THe 293 K ρHe
ρHe 0.1824
RHe THe
3
m
p air
kg
Tair 293 K ρair
ρair 1.201
Rair Tair
3
m
FBnet 0.0418 N
θ 65 deg
FBnet Wlatex
cos( θ)
Wlatex M latex g
3
π D
FBnet ρair ρHe g
6
FD T sin( θ)
This problem is ideal for computing and
plotting in Excel, but we will go through
the details here.
(1)
For the vertical forces
We have (Table A.6)
ΣF 0
sin( θ)
Hence
FD FBnet Wlatex tan( θ)
FD 0.0266 N
But we have
1
1
2
2 π D
FD CD ρair V A CD ρair V
2
2
4
2
V
8 FD
2
CD ρair π D
From Table A.9
ν 1.50 10
2
5 m
s
V 1.88
CD 0.4
with
from Fig. 9.11 (we will
check Re later)
m
s
The Reynolds number is Red
V D
ν
4
Red 2.51 10
We are okay!
For the next balloon
θ 60 deg
8 FD
V
2
V 1.69
CD ρair π D
The Reynolds number is Red
For the next balloon
V D
2
V 1.40
CD ρair π D
The Reynolds number is Red
For the next balloon
V D
2
V 1.28
CD ρair π D
The Reynolds number is Red
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red
For the next balloon
V D
ν
θ 10 deg
CD 0.4
We are okay!
FD 0.00452 N
with
CD 0.4
with
CD 0.4
m
s
4
Red 1.03 10
with
s
V 0.77
FD 0.00717 N
m
FD FBnet Wlatex tan( θ)
8 FD
V
4
θ 20 deg
CD 0.4
We are okay!
Red 1.30 10
ν
with
s
V 0.97
FD 0.00870 N
m
FD FBnet Wlatex tan( θ)
8 FD
V
4
θ 30 deg
CD 0.4
We are okay!
Red 1.43 10
ν
with
s
V 1.07
FD 0.01043 N
m
FD FBnet Wlatex tan( θ)
8 FD
V
4
θ 35 deg
CD 0.4
We are okay!
Red 1.57 10
ν
with
s
V 1.18
FD 0.01243 N
m
FD FBnet Wlatex tan( θ)
8 FD
V
4
θ 40 deg
CD 0.4
We are okay!
Red 1.71 10
ν
with
s
FD FBnet Wlatex tan( θ)
8 FD
FD 0.01481 N
m
4
θ 45 deg
V
Red 1.87 10
ν
CD 0.4
We are okay!
FD FBnet Wlatex tan( θ)
8 FD
with
s
4
θ 50 deg
FD 0.0215 N
m
Red 2.25 10
ν
V
FD FBnet Wlatex tan( θ)
We are okay!
FD FBnet Wlatex tan( θ)
FD 0.002191 N
8 FD
V
2
V 0.54
CD ρair π D
The Reynolds number is Red
V D
m
s
Red 7184.21
ν
We are okay!
V ( 0.54 0.77 0.97 1.07 1.18 1.28 1.40 1.69 1.88 )
In summary we have
m
s
h ( 1 2 3 4 5 6 7 8 9 )m
10
h (m)
8
6
4
2
0
0.5
1
1.5
2
V (m/s)
This does not seem like an unreasonable profile for the lowest portion of an atmospheric boundary layer - over cities or rough terrain
the atmospheric boundary layer is typically 300-400 meters, so a near-linear profile over a small fraction of that distance is not out of
the question.
Problem 9.102
[Difficulty: 3]
Given:
Data on cyclist performance on a calm day
Find:
Performance hindered and aided by wind; repeat with high-tech tires; with fairing
Solution:
The given data or available data is
FR 7.5 N
M 65 kg
CD 1.2
ρ 1.23
2
A 0.25 m
kg
V 30
3
m
The governing equation is
1
2
FD ρ A V CD
2
Power steady power generated by the cyclist is
P FD FR V
Now, with a headwind we have
km
Vw 10
hr
km
hr
FD 12.8 N
P 169 W
The aerodynamic drag is greater because of the greater effective wind speed
1
2
FD ρ A V Vw CD
2
(1)
The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed is
P V FD FR
(2)
Combining Eqs 1 and 2 we obtain an expression for the cyclist's maximum speed V cycling into a
headwind (where P = 169 W is the cyclist's power)
Cycling into the wind:
1
2
P FR ρ A V Vw CD V
2
(3)
This is a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use Excel's
Goal Seek (or Solver). From the associated Excel workbook
V 24.7
From Solver
km
hr
By a similar reasoning:
Cycling with the wind:
1
2
P FR ρ A V Vw CD V
2
(4)
P 0.227 hp
V 35.8
From Solver
km
hr
With improved tires
FR 3.5 N
Maximum speed on a calm day is obtained from
1
2
P FR ρ A V CD V
2
This is a again a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use
Excel's Goal Seek (or Solver). From the associated Excel workbook
V 32.6
From Solver
km
hr
Equations 3 and 4 are repeated for the case of improved tires
From Solver
Against the wind
V 26.8
km
V 29.8
km
With the wind
hr
V 39.1
km
V 42.1
km
hr
For improved tires and fairing, from Solver
V 35.7
km
hr
Against the wind
hr
With the wind
hr
Problem 9.101
Given:
Data on cyclist performance on a calm day
Find:
Performance on a hill with and without wind
[Difficulty: 3]
Solution:
The given data or available
data is
FR 7.5 N
CD 1.2
2
M 65 kg
A 0.25 m
kg
ρ 1.23
V 30
3
m
The governing equation is
1
2
FD ρ A V CD
2
Power steady power generated by the cyclist is
P FD FR V
Riding up the hill (no wind)
θ 5 deg
km
hr
FD 12.8 N
P 169 W
P 0.227 hp
For steady speed the cyclist's power is consumed by working against the net force (rolling resistance, drag, and gravity)
Cycling up the
hill:
1
2
P FR ρ A V CD M g sin( θ) V
2
This is a cubic equation for the speed which can be solved analytically, or by iteration, or using Excel's
Goal Seek or Solver. The solution is obtained from the associated Excel workbook
V 9.47
From Solver
Now, with a headwind we have
km
hr
km
Vw 10
hr
The aerodynamic drag is greater because of the greater effective wind speed
1
2
FD ρ A V Vw CD
2
The power required is that needed to overcome the total force (rolling resistance, drag, and gravity) moving at the cyclist's speed is
Uphill against the wind:
1
2
P FR ρ A V Vw CD M g sin( θ) V
2
This is again a cubic equation for V
From Solver
V 8.94
km
hr
Pedalling downhill (no wind) gravity helps increase the speed; the maximum speed is obtained from
Cycling down the hill:
1
2
P FR ρ A V CD M g sin( θ) V
2
This cubic equation for V is solved in the associated Excel workbook
V 63.6
From Solver
km
hr
Pedalling downhill (wind assisted) gravity helps increase the speed; the maximum speed is obtained from
Wind-assisted downhill:
1
2
P FR ρ A V Vw CD M g sin( θ) V
2
This cubic equation for V is solved in the associated Excel workbook
V 73.0
From Solver
km
hr
Freewheeling downhill, the maximum speed is obtained from the fact that the net force is zero
Freewheeling downhill:
1
2
FR ρ A V CD M g sin( θ) 0
2
V
M g sin( θ) FR
1
2
Wind assisted:
V 58.1
km
V 68.1
km
ρ A CD
hr
1
2
FR ρ A V Vw CD M g sin( θ) 0
2
V Vw
M g sin( θ) FR
1
2
ρ A CD
hr
Problem 9.100
Given:
[Difficulty: 2]
Ballistic data for .44 magnum revolver bullet
m
m
Vi 250
Vf 210
Δx 150 m M 15.6 gm D 11.2 mm
s
s
Average drag coefficient
Find:
Solution:
Basic
1
2
FD CD ρ V A
equations:
2
(Drag)
Assumptions: (1) Standard air
(2) Neglect all losses other than aerodynamic drag
Newton's 2nd law:
Separating variables:
1
2
d
d
d
ΣFx FD M V M V V Therefore: M V V CD ρ V A
2
d
d
d
t
s
x
dx
2 M
dV
CD ρ A V
Solving this expression for the drag coefficient:
Integrating both sides yields:
CD
Vf
Δx ρ A
Vi
2 M
ln
Δx
Vf
CD ρ A
Vi
2 M
The area is:
ln
A
π
4
2
2
D 98.52 mm
Therefore the drag coefficent is:
CD 0.299
Problem 9.99
Given:
Data on cyclist performance on a calm day
Find:
Performance hindered and aided by wind
[Difficulty: 2]
Solution:
The given data or available data is
FR 7.5 N
M 65 kg
CD 1.2
ρ 1.23
2
A 0.25 m
kg
V 30
3
m
The governing equation is
1
2
FD ρ A V CD
2
km
hr
FD 12.8 N
The power steady power generated by the cyclist is
Now, with a headwind we have
P FD FR V
P 169 W
km
Vw 10
hr
V 24
km
hr
The aerodynamic drag is greater because of the greater effective wind speed
1
2
FD ρ A V Vw CD
2
FD 16.5 N
The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed
P V FD FR
P 160 W
This is less than the power she can generate
She wins the bet!
With the wind supporting her the effective wind speed is substantially lower
km
VW 10
hr
V 40
1
2
FD ρ A V VW CD
2
km
hr
FD 12.8 N
The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed
P V FD FR
This is more than the power she can generate
P 226 W
She loses the bet
P 0.227 hp
Problem 9.98
Given:
Bike and rider at terminal speed on hill with 8% grade.
W 210 lbf A 5 ft
Find:
[Difficulty: 2]
ft
Vt 50
s
2
CD 1.25
(a) Verify drag coefficient
(b) Estimate distance needed for bike and rider to decelerate to 10 m/s after reaching level road
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
(Drag)
Assumptions: (1) Standard air
(2) Neglect all losses other than aerodynamic drag
θ atan( 9 %) 5.143 deg Summing forces in the x-direction: ΣFx FG FD 0
1
2
Expanding out both force terms: M g sin( θ) CD ρ Vt A Solving this expression for the drag coefficient:
2
The angle of incline is:
CD
2 W sin( θ)
CD 1.26
2
ρ Vt A
The original estimate for the
drag coefficient was good.
W d W d
Once on the flat surface: ΣFx FD
V
V V Therefore:
g dt
g
ds
Separating variables:
ds
2 W
dV
CD ρ g A V
W
g
Integrating both sides yields:
d V C 1 ρ V2 A
D 2
ds
V
Δs
V2
CD ρ g A
V1
2 W
ln
Δs 447 ft
Problem 9.97
Given:
Windmills are to be made from surplus 55 gallon oil drums
D 24 in
Find:
[Difficulty: 2]
H 29 in
Which configuration would be better, why, and by how much
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
(Drag)
Assumptions: (1) Standard air
(2) Neglect friction in the pivot
(3) Neglect interference between the flow over the two halves
For the first configuration:
ΣM
D
ΣM
D
2
Fu
D
2
Fd
D
2
Fu Fd
Where Fu is the force on the half "catching" the wind and F d is the
force on the half "spilling" the wind.
1
1
D
1
2
2
2
CDu ρ V A CDd ρ V A CDu CDd ρ V A
2
2
2
2
2
For the second configuration:
ΣM
H
ΣM
H
2
Fu
H
2
Fd
H
2
Fu Fd
1
1
H
1
2
2
2
CDu ρ V A CDd ρ V A CDu CDd ρ V A
2
2
2
2
2
Since H > D, the second
configuration will be superior.
The improvement will be:
H D
D
20.8 %
Problem 9.96
[Difficulty: 3]
Given:
Data on airplane and parachute
Find:
Time and distance to slow down; plot speed against distance and time; maximum "g"'s
Solution:
The given data or available data is
M 8500 kg
km
Vi 400
hr
km
Vf 100
hr
π
2
2
Asingle Dsingle 28.274 m
4
Newton's second law for the aircraft is
CD 1.42
ρ 1.23
kg
3
Dsingle 6 m
Dtriple 3.75 m
m
π
2
2
Atriple Dtriple 11.045 m
4
M
dV
1
2
CD ρ A V
dt
2
where A and C D are the single parachute area and drag coefficient
Separating variables
dV
2
V
Integrating, with IC V = Vi
CD ρ A
2 M
Vi
V( t)
1
Integrating again with respect to t
x ( t)
Eliminating t from Eqs. 1 and 2
x
dt
CD ρ A
2 M
(1)
Vi t
CD ρ A
2 M
ln 1
CD ρ A
2 M
Vi t
Vi
CD ρ A V
2 M
ln
(2)
(3)
To find the time and distance to slow down to 100 km/hr, Eqs. 1 and 3 are solved with V = 100
km/hr (or use Goal Seek)
dV
The "g"'s are given by
dt
g
2
CD ρ A V
2 M g
which has a maximum at the initial instant (V = Vi)
For three parachutes, the analysis is the same except A is replaced with 3A. leading to
Vi
V( t)
1
x ( t)
3 CD ρ A
2 M
Vi t
3 CD ρ A
2 M
ln 1
3 CD ρ A
2 M
Vi t
The results generated in Excel are shown here:
t (s) x (m) V (km/hr)
t (s) x (m) V (km/hr)
0.0 0.0
1.0 96.3
2.0 171
3.0 233
4.0 285
5.0 331
6.0 371
7.0 407
8.0 439
9.0 469
9.29 477
0.0 0.0
1.0 94.2
2.0 165
3.0 223
4.0 271
5.0 312
6.0 348
7.0 380
7.93 407
9.0 436
9.3 443
400
302
243
203
175
153
136
123
112
102
100
400
290
228
187
159
138
122
110
100
91
89
"g "'s = -3.66 Max
Aircraft Velocity versus Time
400
350
One Parachute
Three Parachutes
300
V (km/hr)
250
200
150
100
50
0
0
1
2
3
4
5
6
t (s)
7
8
9
10
450
500
Aircraft Velocity versus Distance
400
350
V (km/hr)
300
One Parachute
250
Three Parachutes
200
150
100
50
0
0
50
100
150
200
250
300
350
x (m)
400
Problem 9.95
Given:
Data on airplane landing
M 9500 kg
Find:
[Difficulty: 3]
km
Vi 350
hr
km
Vf 100
hr
x f 1200 m CD 1.43 (Table 9.3)
Solution:
1
2
FD CD ρ V A
2
(Drag)
Assumptions: (1) Standard air
(2) Parachute behaves as open hemisphere
(3) Vertical speed is constant
Newton's second law for the aircraft is
M
dV
1
2
CD ρ A V
dt
2
where A and CD are the single parachute area and drag coefficient
Separating variables
dV
2
V
Integrating, with IC V = Vi
CD ρ A
2 M
1
Integrating again with respect to t
x ( t)
Eliminating t from Eqs. 1 and 2
x
dt
Vi
V( t)
CD ρ A
2 M
(1)
Vi t
CD ρ A
2 M
ln 1
CD ρ A
2 M
Vi t
Vi
CD ρ A V
2 M
ln
(2)
(3)
To find the minimum parachute area we must solve Eq 3 for A with x = xf when V = Vf
A
2 M
CD ρ x f
Vi
Vf
ln
(4)
For three parachutes, the analysis is the same except A is replaced with 3A, leading to
A
2 M
3 CD ρ x f
Vi
Vf
ln
kg
3
m
Single and three-parachute sizes; plot speed against distance and time; maximum "g''s
Basic
equations:
ρ 1.23
(5)
dV
The "g"'s are given by
2
dt
CD ρ A V
g
which has a maximum at the initial instant (V = Vi)
2 M g
The results generated in Excel are shown below:
Single:
A =
D =
Triple:
11.4 m
3.80 m
2
A = 3.8 m2
D = 2.20 m
"g "'s = -1.01 Max
t (s) x (m) V (km/hr)
0.00
2.50
5.00
7.50
10.0
12.5
15.0
17.5
20.0
22.5
24.6
0.0
216.6
393.2
542.2
671.1
784.7
886.3
978.1
1061.9
1138.9
1200.0
350
279
232
199
174
154
139
126
116
107
100
Aircraft Velocity versus Time
350
300
250
V (km/hr)
200
150
100
50
0
0
5
10
15
t (s)
20
25
Aircraft Velocity versus Distance
350
300
250
V (km/hr) 200
150
100
50
0
0
200
400
600
800
x (m)
1000
1200
Problem 9.94
Given:
Man with parachute
W 250 lbf V 20
Find:
[Difficulty: 2]
ft
ρ 0.00234
s
slug
ft
3
Minimum diameter of parachute
FD
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
(Drag)
Assumptions: (1) Standard air
x
(2) Parachute behaves as open hemisphere
(3) Vertical speed is constant
For constant speed:
ΣFy M g FD 0
In terms of the drag coefficient:
Solving for the area:
A
Setting both areas equal:
1
Therefore:
FD W
W
2
CD ρ V A W
2
2 W
From Table 9.2: CD 1.42 for an open hemisphere.
2
CD ρ V
π
4
V
y
2
D
2 W
Solving for the diameter of the parachute:
2
CD ρ V
The area is:
D
8
π
A
π
4
2
D
W
2
CD ρ V
Therefore the diameter is:
D 21.9 ft
Problem 9.93
Given:
Data on a rotary mixer
Find:
New design dimensions
[Difficulty: 3]
Solution:
The given data or available data is
R 0.6 m
P 350 W
ω 60 rpm
ρ 1099
kg
3
m
For a ring, from Table 9.3
CD 1.2
The torque at the specified power and speed is
T
P
T 55.7 N m
ω
The drag on each ring is then
1 T
FD
2 R
FD 46.4 N
The linear velocity of each ring is
V R ω
V 3.77
m
s
The drag and velocity of each ring are related using the definition of drag coefficient
FD
CD
1
2
Solving for the ring area
A
FD
1
2
But
A
2
ρ A V
π
4
3
A 4.95 10
2
ρ V CD
do di
2
The outer diameter is
d o 125 mm
Hence the inner diameter is
di
2
do
2
4 A
π
d i 96.5 mm
2
m
Problem 9.92
Given:
[Difficulty: 2]
Rotary mixer rotated in a brine solution
R 0.6 m
ω 60 rpm
d 100 mm SG 1.1
ρ ρw SG
ρ 1100
kg
3
m
ν 1.05 1.55 10
Find:
2
6 m
s
2
6 m
1.63 10
s
(a) Torque on mixer
(b) Horsepower required to drive mixer
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
(Drag)
T 2 R FD
(Torque)
P T ω
(Power)
Assumptions: Drag on rods and motion induced in the brine can be neglected.
The speed of the disks through the brine is:
The area of one disk is:
A
π
4
2
V R ω 3.77
m
s
From Table 9.2: CD 1.17 for a disk.
2
d 0.00785 m
So the drag force is:
1
2
FD CD ρ V A 71.8 N
2
and the torque is: T 2 R FD
The power consumed to run the mixer is:
P T ω 542 W
T 86.2 N m
P 0.726 hp
Problem 9.91
Given:
Fishing net
Find:
Drag; Power to maintain motion
[Difficulty: 3]
3
8
in 9.525 mm
Solution:
Basic equations:
CD
FD
1
2
ρ V A
2
We convert the net into an equivalent cylinder (we assume each segment does not interfere with its neighbors)
L 12 m
W 2 m
d 0.75 mm Spacing: D 1 cm
Total number of threads of length L is
Total number of threads of length W is
Total length of thread
n1
W
n2
L
LT L1 L2
ρ 999
D
3
Red
V d
ν
s
L1 n 1 L
L1 2400 m
n 2 1200
Total length
L2 n 2 W
L2 2400 m
LT 4800 m
LT 2.98 mile A lot!
2
Note that L W 24.00 m
ν 1.01 10
2
6 m
m
The Reynolds number is
m
Total length
A 3.60 m
kg
V 3.09
n 1 200
2
The frontal area is then A LT d
From Table A.8
D
V 6 knot
s
Red 2292
For a cylinder in a crossflow at this Reynolds number, from Fig. 9.13, approximately
Hence
1
2
FD CD ρ V A
2
FD 13.71 kN
The power required is
P FD V
P 42.3 kW
CD 0.8
Problem 9.90
Given:
Flag mounted vertically
H 194 ft W 367 ft
Find:
[Difficulty: 2]
V 10 mph 14.67
ft
s
ρ 0.00234
slug
ft
3
ν 1.62 10
4 ft
2
s
Force acting on the flag. Was failure a surprise?
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
(Drag)
We should check the Reynolds number to be sure that the data of Fig. 9.10 are applicable:
ReW
V W
ν
3.32 10
7
(We used W as our length scale here since it is the lesser of the two dimensions of the flag.) Since the Reynolds number is less than
1000, we may use Figure 9.10 to find the drag coefficient.
4
The area of the flag is: A H W 7.12 10 ft
So the drag force is:
2
AR
W
H
1.89
1
2
FD CD ρ V A
2
From Fig. 9.10: CD 1.15
4
FD 2.06 10 lbf
This is a large force.
Failure should have
been expected.
Problem 9.89
Given:
"Resistance" data on a ship
Lp 130 m
Lm
Find:
[Difficulty: 4]
Lp
80
1.625 m
ρ 1023
2
Ap 1800 m
Am
Ap
80
2
kg
3
3 N s
μ 1.08 10
m
2
m
2
0.281 m
Plot of wave, viscous and total drag (prototype and model); power required by prototype
Solution:
Basic
equations:
CD
FD
1
2
From Eq. 9.32
(9.32)
2
Fr
U
gL
ρ U A
1
2
FD CD A ρ U
2
This applies to each component of the drag (wave and viscous) as well as to the total
The power consumed is
P FD U
From the Froude number
U Fr gL
1
3
P CD A ρ U
2
The solution technique is: For each speed Fr value from the graph, compute U; compute the drag from
the corresponding "resistance" value from the graph. The results were generated in Excel and are shown
below:
Model
Fr
Wave
"Resistance"
Viscous
"Resistance"
0.10
0.20
0.30
0.35
0.40
0.45
0.50
0.60
0.00050
0.00075
0.00120
0.00150
0.00200
0.00300
0.00350
0.00320
0.0052
0.0045
0.0040
0.0038
0.0038
0.0036
0.0035
0.0035
Wave
Total
U (m/s)
Drag (N)
"Resistance"
0.0057
0.0053
0.0052
0.0053
0.0058
0.0066
0.0070
0.0067
0.40
0.80
1.20
1.40
1.60
1.80
2.00
2.40
0.0057
0.0344
0.1238
0.2107
0.3669
0.6966
1.0033
1.3209
Viscous
Drag (N)
0.0596
0.2064
0.4128
0.5337
0.6971
0.8359
1.0033
1.4447
Total
Power (W)
Drag (N)
0.0654
0.2408
0.5366
0.7444
1.0640
1.5324
2.0065
2.7656
Drag on a Model Ship
3.0
2.5
Total
Wave
Viscous
2.0
F (N)
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
U (m/s)
2.5
3.0
2.5
3.0
Power Requirements for a Model Ship
7.0
6.0
5.0
P (W)
4.0
3.0
2.0
1.0
0.0
0.0
0.5
1.0
1.5
2.0
U (m/s)
0.0261
0.1923
0.6427
1.0403
1.6993
2.7533
4.0057
6.6252
Prototype
Fr
Wave
"Resistance"
Viscous
"Resistance"
0.10
0.20
0.30
0.35
0.40
0.45
0.50
0.60
0.00050
0.00075
0.00120
0.00150
0.00200
0.00300
0.00350
0.00320
0.0017
0.0016
0.0015
0.0015
0.0013
0.0013
0.0013
0.0013
Total
U (m/s)
"Resistance"
0.0022
0.0024
0.0027
0.0030
0.0033
0.0043
0.0048
0.0045
3.6
7.1
10.7
12.5
14.3
16.1
17.9
21.4
Wave
Drag
(MN)
0.0029
0.0176
0.0634
0.1079
0.1879
0.3566
0.5137
0.6763
Viscous
Drag (MN)
0.0100
0.0376
0.0793
0.1079
0.1221
0.1545
0.1908
0.2747
Total
Drag
(MN)
0.0129
0.0552
0.1427
0.2157
0.3100
0.5112
0.7045
0.9510
Drag on a Prototype Ship
1.0
F (MN)
0.8
Total
0.6
Wave
Viscous
0.4
0.2
0.0
0
5
10
15
U (m/s)
20
25
Power Required by a Prototype Ship
25000
20000
P (kW)
15000
10000
5000
0
0
5
10
15
U (m/s)
20
For the prototype wave resistance is a much more significant factor at high speeds! However, note that for
both scales, the primary source of drag changes with speed. At low speeds, viscous effects dominate, and so
the primary source of drag is viscous drag. At higher speeds, inertial effects dominate, and so the wave drag
is the primary source of drag.
25
Power
(kW)
Power
(hp)
46.1
394.1
1528.3
2696.6
4427.7
8214.7
12578.7
20377.5
61.8
528.5
2049.5
3616.1
5937.6
11015.9
16868.1
27326.3
Problem 9.88
Given:
[Difficulty: 4]
Supertanker in seawater at 40oF
L 1000 ft B 270 ft D 80 ft
ν 1.05 1.65 10
Find:
5 ft
V 15 knot 25.32
2
s
5 ft
1.73 10
2
ρ 1.9888
s
(a) Thickness of the boundary layer at the stern of the ship
(b) Skin friction drag on the ship
(b) Power required to overcome the drag force
ft
SG 1.025
s
slug
ft
3
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
The Reynolds number is
δ
x
0.382
Rex
ReL
V L
At the stern of the ship:
0.20
ν
(Drag)
9
1.4613 10 So the BL is turbulent. The BL thickness is calculated using:
δ L
0.382
ReL
The wetted area of the hull is:
δ 5.61 ft
0.20
5
2
A L ( B 2 D) 4.30 10 ft For this Reynolds number: CD
So the drag force is:
1
2
FD CD ρ V A
2
The power consumed to overcome the skin friction drag is:
3
0.455
logReL
2.58
1.50 10
5
FD 4.11 10 lbf
P FD V
4
P 1.891 10 hp
7 ft lbf
P 1.040 10
s
Problem 9.87
Given:
[Difficulty: 4]
600-seat jet transport to operate 14 hr/day, 6 day/wk
L 240 ft D 25 ft
Find:
V 575 mph z 12 km
TSFC 0.6
lbm
hr lbf
(a) Skin friction drag on fuselage at cruise
(b) Annual fuel savings if drag is reduced by 1%
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
(Drag)
T 216.7 K 390.1 R ρ 0.2546 0.002377
From the atmosphere model:
slug
ft
From the Sutherland model for viscosity: μ
b T
1
ReL
ρ V L
μ
4.1247 10
8
S
5 kg
1.422 10
ft
3
So the Reynolds number is
CD
4
A π D L 1.885 10 ft
3
0.455
logReL
2.58
1.76 10
2
So the drag force is:
If there were a 1% savings in drag, the drop in drag force would be:
The savings in fuel would be:
T
For this Reynolds number:
The wetted area of the fuselage is:
m s
3
4 slug
6.0518 10
hr
Δmfuel TSFC ΔFD 14
day
1
2
3
FD CD ρ V A FD 7.13 10 lbf
2
ΔFD 1 % FD 71.31 lbf
6 52 day
7
yr
4 lbm
Δmfuel 2.670 10
yr
If jet fuel costs $1 per gallon, this would mean
a savings of over $4,400 per aircraft per year.
Problem 9.86
Given:
Plastic sheet falling in water
Find:
Terminal speed both ways
[Difficulty: 3]
Solution:
Basic
equations:
h 0.5 in
ΣFy 0
FD
CD
for terminal
speed
1
2
W 4 ft
L 2 ft
SG 1.7
2
ρ V A
CD
0.0742
(9.34) (assuming 5 x 105 < ReL < 107)
1
ReL
5
From Table A.8 at 70 oF ν 1.06 10
5 ft
A W L
2
ρ 1.94
s
slug
ft
3
A free body diagram of the sheet is shown here. Summing the forces in the vertical (y) direction:
FD Fb Wsheet 0
FD Wsheet Fb ρ g h A ( SG 1 )
FD
Fb
Also, we can generate an expression for the drag coefficient in terms of the geometry of the sheet
and the water properties:
y
V
x
4
1
9
W sheet
1
0.0742
1
0.0742
2
2
2
5 5
5
FD 2 CD A ρ V 2
A ρ V
W L ρ V 0.0742 W L ν ρ V
1
1
2
2
ReL
5
V L
ν
(Note that we double FD because drag
acts on both sides of the sheet.)
5
5
9
Hence
ρH2O g h W ( SG 1 ) 0.0742 W L
Check the Reynolds number
Repeating for
ReL
ReL
5
1
9
5
5
ν ρ V
Solving for V
V L
5
ν
1
g h ( SG 1) L 5
V
0.0742
ν
L 4 ft
Check the Reynolds number
1
V L
ν
1
g h ( SG 1) L 5
ft
V
V 15.79
s
0.0742
ν
ReL 2.98 10
6
Hence Eq. 9.34 is reasonable
6
Eq. 9.34 is still reasonable
9
V 17.06
ft
s
ReL 6.44 10
The short side vertical orientation falls more slowly because the largest friction is at the region of the leading edge (τ tails
off as the boundary layer progresses); its leading edge area is larger. Note that neither orientation is likely - the plate will
flip around in a chaotic manner.
Problem 9.85
Given:
Racing shell for crew approximated by half-cylinder:
L 7.32 m
Find:
[Difficulty: 3]
D 457 mm
V 6.71
m
s
(a) Location of transition on hull
(b) Thickness of turbulent BL at the rear of the hull
(c) Skin friction drag on hull
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
5
Transition occurs at Ret 5 10
so the location of transition would be:
δ
For the turbulent boundary layer
x
The Reynolds number at x = L is:
The wetted area of the hull is:
A
(Drag)
0.382
Rex
ReL
π D
2
0.2
V L
ν
Therefore δ
x t 0.0745 m
V
0.2
so the BL thickness is:
δ
0.382 L
ReL
2
L 5.2547 m
So the drag force is:
7
Ret ν
0.382 L
ReL
4.91 10
xt
For this Reynolds number:
1
2
FD CD ρ V A
2
Note that the rowers must produce an average power of
CD
δ 0.0810 m
0.2
3
0.455
logReL
2.58
2.36 10
FD 278 N
P FD V 1.868 kW to move the shell at this speed.
Problem 9.84
Given:
Nuclear submarine cruising submerged. Hull approximated by circular cylinder
L 107 m
Find:
[Difficulty: 4]
D 11.0 m
V 27 knot
(a) Percentage of hull length for which BL is laminar
(b) Skin friction drag on hull
(c) Power consumed
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
5
Transition occurs at Ret 5 10
(Drag)
so the location of transition would be:
xt
L
Ret ν
xt
V L
L
0.0353%
We will therefore assume that the BL is completely turbulent.
The Reynolds number at x = L is:
The wetted area of the hull is:
ReL
V L
ν
1.42 10
9
For this Reynolds number:
CD
3
0.455
logReL
2.58
1.50 10
2
A π D L 3698 m
So the drag force is:
The power consumed is:
1
2
FD CD ρ V A
2
P FD V
5
FD 5.36 10 N
P 7.45 MW
Problem 9.83
Given:
Stabilizing fin on Bonneville land speed record auto
z 1340 m
Find:
[Difficulty: 2]
V 560
km
hr
H 0.785 m
L 1.65 m
(a) Evaluate Reynolds number of fin
(b) Estimate of location for transition in the boundary layer
(c) Power required to overcome skin friction drag
Solution:
Basic
equations:
Assumptions:
At this elevation:
1
2
FD CD ρ V A
2
(Drag)
(1) Standard atmosphere (use table A.3)
T 279 K ρ 0.877 1.23
kg
1.079
3
m
The Reynolds number on the fin is:
Assume transition occurs at:
ReL
μ 1.79 10
3
m
m s
ρ V L
5
Ret 5 10
7
ReL 1.547 10
μ
The location for transition would then be:
From Figure 9.8, the drag coefficient is: CD 0.0029
The drag force would then be:
5 kg
kg
The area is:
xt
x t 53.3 mm
2
FD 98.0 N
P FD V
If we check the drag coefficient using Eq. 9.37b:
ρ V
A 2 L H 2.591 m (both sides of the fin)
1
2
FD CD ρ V A
2
The power required would then be:
Ret μ
P 15.3 kW
CD
0.455
log ReL
2.58
1610
ReL
0.0027
This is slightly less than the graph, but still reasonable agreement.
Problem 9.82
Given:
[Difficulty: 3]
Towboat model at 1:13.5 scale to be tested in towing tank.
Lm 7.00 m
Find:
Bm 1.4 m
d m 0.2 m
Vp 10 knot
(a) Model speed in order to exhibit similar wave drag behavior
(b) Type of boundary layer on the prototype
(c) Where to place BL trips on the model
(d) Estimate skin friction drag on prototype
Solution:
Basic
equations:
1
2
FD CD ρ V A
2
(Drag)
Vm
The test should be conducted to match Froude numbers:
Rep
The Reynolds number is:
g Lm
A L ( B 2 d )
0.0594
ReL
Therefore
CDm
0.0743
0.2
0.2
0.2
2.97 10
3
Rem
For the prototype:
CDp
Vm 2.72 knot
8
Rep 4.85 10
ν
5
Ret 5 10
so
xt
L
Ret
Rep
0.00155
x t 0.0109 m
We calculate the drag coefficient from turbulent BL theory:
0.0743
ReL
g Lp
x t 0.00155 Lm
Thus the location of transition would be:
CD 1.25 Cf 1.25
Lm
Vm Vp
Lp
Vp
Vp Lp
Therefore the boundary layer is turbulent. Transition occurs at
The wetted area is:
0.455
logRep
2.56
For the model: Lm 7 m
Rem
Vm Lm
ν
6
2
9.77 10 Am 12.6 m
1
2
and the drag force is: FDm CDm ρ Vm Am
2
1610
Rep
CDp 1.7944 10
3
FDm 36.70 N
3
2
Ap 2.30 10 m
1
2
FDp CDp ρ Vp Ap
2
FDp 54.5 kN
Problem 9.81
Given:
Aircraft cruising at 12 km
Find:
Skin friction drag force; Power required
[Difficulty: 3]
Solution:
Basic
equations:
CD
FD
1
2
ρ V A
2
We "unwrap" the cylinder to obtain an equivalent flat plate
L 38 m
From Table A.3, with
D 4 m
ρ
z 12000 m
ρSL
ρ 0.2546 ρSL
2
A L π D
A 478 m
kg
ρSL 1.225
3
m
0.2546
kg
ρ 0.3119
and also
T 216.7 K
kg
S 110.4 K
3
m
1
From Appendix A-3
μ
b T
2
1
S
with
6
b 1.458 10
T
1
m s K
2
1
Hence
b T
2
1
S
μ
μ 1.42 10
5 N s
2
m
T
ReL
Next we need the Reynolds number
0.455
CD
log ReL
2.58
ρ V L
μ
CD 0.00196
The drag is then
1
2
FD CD ρ V A
2
FD 7189 N
The power consumed is
P FD V
P 1.598 MW
ReL 1.85 10
8
so use Eq. 9.35
V 800
km
hr
Problem 9.80
Given:
[Difficulty: 3]
Towboat model at 1:13.5 scale to be tested in towing tank.
Lm 3.5 m
Find:
Bm 1 m
d m 0.2 m
m
Up 7 knot 3.601
s
Disp m 5500 N
(a) Estimate average length of wetted surface on the hull
(b) Calculate skin friction drag force on the prototype
Solution:
Basic
equations:
1
2
FD CD ρ U A
2
CD
(Drag)
0.455
logReL
2.58
1610
(Drag Coefficient)
ReL
We will represent the towboat as a rectangular solid of length L av, with the displacement of the boat. From buoyancy:
W ρ g V ρ g Lav Bm d m thus:
For the prototype:
Lav
W
ρ g Bm d m
Lp 13.5 Lav
Lp 37.9 m
ReL
The Reynolds number is:
Lav 2.80 m
Up Lp
ReL 1.36 10
ν
8
This flow is predominantly turbulent, so we will use a turbulent analysis.
The drag coefficient is: CD
The area is:
2
0.455
logReL
2.58
1610
ReL
0.00203
2
A 13.5 Lav Bm 2 d m 716 m
The drag force would then be:
1
2
FD CD ρ Up A
2
FD 9.41 kN
This is skin friction only.
Problem 9.79
Given:
Pattern of flat plates
Find:
Drag on separate and composite plates
[Difficulty: 3]
Solution:
Basic
equations:
CD
FD
1
2
For separate plates
From Table A.8 at 70 oF
2
ρ V A
L 3 in
W 3 in
ν 1.06 10
5 ft
First determine the Reynolds number ReL
CD
2
s
V L
ν
0.0742
1
ReL
ρ 1.93
A W L
ft
3
5
ReL 7.08 10 so use Eq. 9.34
CD 0.00502
5
FD 0.272 lbf
This is the drag on one plate. The total drag is then
FTotal 4 FD
For the composite plate
L 4 3 in
L 12.000 in
ReL
First determine the Reylolds number
CD
V 30
slug
1
2
FD CD ρ V A
2
The drag (one side) is then
2
A 9.000 in
0.0742
1
V L
ν
FTotal 1.09 lbf
For both sides:
2 FTotal 2.18 lbf
A W L
A 36 in
2
6
ReL 2.83 10 so use Eq. 9.34
CD 0.00380
5
The drag (one side) is
then
ReL
1
2
FD CD ρ V A
2
FD 0.826 lbf
For both
sides:
2 FD 1.651 lbf
The drag is much lower on the composite compared to the separate plates. This is because τ w is largest
near the leading edges and falls off rapidly; in this problem the separate plates experience leading edges
four times!
ft
s
Problem 9.78
Given:
[Difficulty: 3]
Barge pushed upriver
L 80 ft B 35 ft
Find:
2
5 ft
D 5 ft
From Table A.7: ν 1.321 10
s
ρ 1.94
slug
ft
3
Power required to overcome friction; Plot power versus speed
Solution:
CD
Basic
equations:
FD
1
2
CD
(9.32)
2
ρ U A
From Eq. 9.32
1
2
FD CD A ρ U
2
The power consumed is
P FD U
0.455
logReL
2.58
A L ( B 2 D)
and
1
3
P CD A ρ U
2
1610
(9.37b)
ReL
A 3600 ft
Re L
CD
P (hp)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
9.70E+06
1.94E+07
2.91E+07
3.88E+07
4.85E+07
5.82E+07
6.79E+07
7.76E+07
8.73E+07
9.70E+07
1.07E+08
1.16E+08
1.26E+08
1.36E+08
1.45E+08
0.00285
0.00262
0.00249
0.00240
0.00233
0.00227
0.00222
0.00219
0.00215
0.00212
0.00209
0.00207
0.00205
0.00203
0.00201
0.0571
0.421
1.35
3.1
5.8
9.8
15
22
31
42
56
72
90
111
136
150
120
P (hp) 90
60
30
0
6
ν
The calculated results and the plot were generated in Excel:
U (mph)
3
U L
2
Power Consumed by Friction on a Barge
0
ReL
9
U (mph)
12
15
Problem 9.77
Given:
[Difficulty: 5]
Laboratory wind tunnel of Problem 9.76 with a movable top wall:
The given or available data (Table A.9) is
ft
U1 80
s
Find:
H1 1 ft
W 1 ft
δ 0.4 in
L 10 in
ν 1.57 10
4 ft
2
ρ 0.00234
s
slug
ft
3
(a) Velocity distribution needed for constant boundary layer thickness
(b) Tunnel height distribution h(x) from 0 to L
Solution:
Basic
equations:
ρ dV ρ V dA 0
t
τw
ρ
Assumptions:
(Continuity)
2
d
d
U θ δdisp U U
dx
dx
(Momentum integral equation)
(1) Steady flow
(2) Incompressible flow
(3) Turbulent, 1/7-power velocity profile in boundary layer
(4) δ = constant
δ
From the 1/7-power profile: δdisp
8
θ
7
72
δ
H
72
56
τw
After applying assumptions, the momentum integral equation is:
2
( H 2)
ρ U
To integrate, we need to make an assumption about the wall shear stress:
τw
Case 1: assume constant τw:
Integrating:
U dU
dx
ρ θ ( H 2 )
U
U1
1
2 τw
ρ U1
x
2 θ ( H 2 )
2
d U
U dx
θ
U U1
2
2
τw
ρ θ ( H 2 )
which may be rewritten as:
x
U
U1
1
Cf
θ ( H 2 )
1
ν
2
Case 2: assume τ w has the form: τw 0.0233 ρ U
U δ
4
Substituting and rearranging yields the following expression:
x
1
τw
2
ρ U
4
( H 2) θ d U
U dx
U δ
0.0233
ν
dU
or
0.75
0.0233
ν
0.25
δ
U
dx
Integrating this yields:
( H 2) θ
or:
0.25
ν
x
1 0.00583
U1
( H 2 ) θ
U1 δ
From continuity: U1 A1 U A which may be rewritten as:
U1 W 2 δdisp H1 2 δdisp U W 2.δdisp h 2 δdisp
0.25
0.25
ν
4 U
U1
0.0233 δ
Thus:
A
A1
U
U
1
0.25
1
Evaluating using the given data:
Cf 0.0466 Reδ1
0.25
x
( H 2) θ
h 2 δdisp
and
H1 2 δdisp
U1
δ
δdisp
0.0500 in
8
U
θ
7
72
W
δ 0.0389 in
h
solving for h:
1 2
Reδ1
U1 δ
ν
δdisp U1
δdisp
2
H1
H1
U
4
1.699 10
3
4.082 10
The results for both wall profiles are shown in the plot here:
Top Surface Height (in)
14
13.5
13
12.5
Case 1
Case 2
12
0
2
4
U
4
6
Distance along tunnel (in)
8
10
Problem 9.76
Given:
[Difficulty: 5]
Laboratory wind tunnel has fixed walls. BL's
are well represented by 1/7-power profile.
Information at two stations are known:
The given or available data (Table A.9) is
ft
U1 80
s
dp
dx
0.035
Find:
H1 1 ft
in H2 O
W1 1 ft
L 10 in
in
δ1 0.4 in
ν 1.62 10
4 ft
2
ρ 0.00234
s
slug
ft
3
(a) Reduction in effective flow area at section 1
(b) dθ/dx at section 1
(c) θ at section 2
Solution:
Basic
equations:
ρ dV ρ V dA 0
t
τw
ρ
Assumptions:
(Continuity)
2
d
d
U θ δdisp U U
dx
dx
(Momentum integral equation)
(1) Steady flow
(2) Turbulent, 1/7-power velocity profile in boundary layer
(3) z = constant
The percent reduction in flow area at 1 is given as:
The displacement thickness is determined from:
Aeff A
A
δdisp δ
1
0
Substituting the velocity profile and valuating the integral:
W1 2 δdisp H1 2 δdisp W1 H1
W1 H1
1
1 u dη
U
δdisp δ
1
0
u
where
U
1
δ
7
1 η dη
8
η
7
η
Therefore:
Thus:
y
δ
δdisp1 0.0500 in
Aeff A
A
1.66 %
Solving the momentum integral equation for the momentum thickness gradient:
1
At station 1:
τw1
ρ U1
2
0.0233
U1 δ1
L
1
u
θ u
1 dη
U
δ U
0
ν
4
0.0233
8
Solving for the velocity gradient:
1
2
τw
2
( H 2)
ρ U
θ dU
U dx
4
2.057 10 3
U1 δ1
ν
δdisp1
7
Thus: θ1
δ1 0.0389 in H
1.286
72
θ1
72
9
0
p
dx
1
2
1
7
7
7
7
7
η η dη
Now outside the boundary layer
dθ
2
ρ U constant
1 dU
U dx
1
dp
2 dx
ρ U
from the Bernoulli equation. Then:
0.1458
1
ft
dp
dx
ρ U
dx
Substituting all of this information into the above expression:
dθ
dx
We approximate the momentum thickness at 2 from:
dU
dθ
θ2 θ1
L
dx
4.89 10
4
0.00587
in
ft
θ2 0.0438 in
Problem 9.75
[Difficulty: 4]
Discussion: Shear stress decreases along the plate because the freestream flow speed remains constant while the
boundary-layer thickness increases.
The momentum flux decreases as the flow proceeds along the plate. Momentum thickness θ (actually proportional
to the defect in momentum within the boundary layer) increases, showing that momentum flux decreases. The
forct that must be applied to hold the plate stationary reduces the momentum flux of the stream and boundary
layer.
The laminar boundary layer has less shear stress than the turbulent boundary layer. Therefore laminar boundary
layer flow from the leading edge produces a thinner boundary layer and less shear stress everywhere along the
plate than a turbulent boundary layer from the leading edge.
Since both boundary layers continue to grow with increasing distance from the leading edge, and the turbulent
boundary layer continues to grow more rapidly because of its higher shear stress, this comparison will be the
same no matter the distance from the leading edge.
Problem 9.74
Given:
[Difficulty: 3]
u
Laminar boundary layer with velocity profile
U
2
a b λ c λ d λ
3
λ
y
δ
Separation occurs when shear stress at the surface becomes zero.
Find:
(a) Boundary conditions on the velocity profile at separation
(b) Appropriate constants a, b, c, d for the profile
(c) Shape factor H at separation
(d) Plot the profile and compare with the parabolic approximate profile
Solution:
Basic
equations:
u
U
y
δ δ
2
y
2
(Parabolic profile)
The boundary conditions for the separation profile are:
The velocity gradient is defined as:
du
dy
Applying the boundary conditions:
at y 0
u0
τ μ
du
at y δ
uU
τ μ
du
dy
dy
0
Four boundary
conditions for four
coefficients a, b, c, d
0
U d u U
2
b 2 c λ 3 d λ
δ
δ dλ U
y0 λ0
u
U
du
dy
2
3
a b 0 c 0 d 0 0
The velocity profile and gradient may now be written as:
δ
U
b 2 c 0 3 d 0
u
c λ d λ
U
2
0
2
du
3
dy
U
δ
Therefore:
a0
Therefore:
b0
2 c λ 3 d λ
2
Applying the other boundary conditions:
yδ λ1
u
U
du
dy
The velocity profile is:
u
U
δdisp
δ
1
0
1
2
3 λ 2 λ
3
2
3
c 1 d 1 1
U
δ
2 c 1 3 d 1
0
2
1
δdisp
θ
c3
d 2
δdisp δ
δ θ
2
3
2
3
3 λ 2 λ 1 3 λ 2 λ dλ Expanding out the
δ 0
integrand yields:
θ
9
4
9
1
2
3
4
5
6
2
3 λ 2 λ 9 λ 12 λ 4 λ dλ 1
5
7
70
δ 0
2
θ
2 c 3 d 0
H
The shape parameter is defined as:
1 3λ2 2λ3 dλ 1 1 12 12
Solving this system
of equations yields:
cd1
Thus
H
1
2
70
9
H 3.89
The two velocity profiles are plotted here:
Height y/δ
1
0.5
Separated
Parabolic
0
0
0.5
Velocity Distribution u/U
1
Problem 9.73
[Difficulty: 5]
Given:
Channel flow with laminar boundary layers
Find:
Maximum inlet speed for laminar exit; Pressure drop for parabolic velocity in boundary layers
Solution:
Basic
equations:
Retrans 5 10
δ
5
x
5.48
p
Rex
ρ
2
V
g z const
2
Assumptions: 1) Steady flow 2) Incompressible 3) z = constant
From Table A.10 at 20oC
Then
For
For a parabolic profile
ν 1.50 10
2
5 m
ρ 1.21
s
Umax L
Retrans
Umax
ν
5
Retrans 5 10
δdisp
δ
1
0
δ2 L
3
m
Retrans ν
L
5.48
Retrans
L 3 m
h 15 cm
m
Umax 2.50
s
U1 Umax
δdisp2 δ2
3
m
U1 2.50
s
δ2 0.0232 m
1
1
2
1 u dλ
1 2 λ λ dλ
U
3
0
1
From continuity
kg
where δtrans is the displacement
thickness
δdisp2 0.00775 m
U1 w h U2 w h 2 δdisp2
h
U2 U1
h 2 δdisp2
m
U2 2.79
s
Since the boundary layers do not meet Bernoulli applies in the core
p1
ρ
Δp
From hydrostatics
U1
2
ρ
2
2
p2
ρ
U2 U1
2
Δp ρH2O g Δh
Δh
U2
Δp
ρH2O g
2
2
2
ρ
2
2
Δp p 1 p 2 U2 U1
2
Δp 0.922 Pa
with
kg
ρH2O 1000
3
m
Δh 0.0940 mm
Δh 0.00370 in
Problem 9.72
[Difficulty: 4]
Given:
Laminar (Blasius) and turbulent (1/7-power) velocity distributions
Find:
Plot of distributions; momentum fluxes
Solution:
δ
The momentum flux is given by
Using the substitutions
the momentum flux becomes
2
mf ρ u dy
u
U
per unit width of the boundary layer
0
y
f ( η)
δ
η
1
2
mf ρ U δ f ( η ) dη
2
0
For the Blasius solution a numerical evaluation (a Simpson's rule) of the integral is needed
2 Δη
2
2
2
2
mflam ρ U δ
f η0 4 f η1 2 f η2 f ηN
3
where Δη is the step size and N the number of steps
The result for the Blasius profile is
2
mflam 0.525 ρ U δ
1
For a 1/7 power velocity profile
2
2
7
mfturb ρ U δ η dη
0
7
2
mfturb ρ U δ
9
The laminar boundary has less momentum, so will separate first when encountering an adverse pressure gradient. The
computed results were generated in Excel and are shown below:
(Table 9.1) (Simpsons Rule)
η
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
1.00
Laminar Weight Weight x
u/U
0.000
0.166
0.330
0.487
0.630
0.751
0.846
0.913
0.956
0.980
0.992
w
1
4
2
4
2
4
2
4
2
4
1
Simpsons':
y /δ = η
2
(u/U )
0.00
0.11
0.22
0.95
0.79
2.26
1.43
3.33
1.83
3.84
0.98
0.525
0.0
0.0125
0.025
0.050
0.10
0.15
0.2
0.4
0.6
0.8
1.0
t
u/U
0.00
0.53
0.59
0.65
0.72
0.76
0.79
0.88
0.93
0.97
1.00
Laminar and Turbulent Boundary Layer
Velocity Profiles
0.75
y /δ
0.50
Laminar
Turbulent
0.25
0.00
0.00
0.25
0.50
0.75
u/U
1.00
Problem 9.71
Given:
Plane-wall diffuser
Find:
(a) For inviscid flow, describe flow pattern and pressure distribution as φ is increased from zero
(b) Redo part (a) for a viscous fluid
(c) Which fluid will have the higher exit pressure?
[Difficulty: 2]
Solution:
For the inviscid fluid:
With φ = 0 (straight channel) there will be no change in the velocity, and hence no pressure gradient.
With φ > 0 (diverging channel) the velocity will decrease, and hence the pressure will increase.
For the viscous fluid:
With φ = 0 (straight channel) the boundary layer will grow, decreasing the effective flow area. As a result, velocity
will increase, and the pressure will drop.
With φ > 0 (diverging channel) the pressure increase due to the flow divergence will cause in increase in the rate of
boundary layer growth. If φ is too large, the flow will separate from one or both walls.
The inviscid fluid will have the higher exit pressure. (The pressure gradient with the real fluid is reduced by the
boundary layer development for all values of φ.)
Problem 9.70
[Difficulty: 3]
Given:
Data on a large tanker
Find:
Cost effectiveness of tanker; compare to Alaska pipeline
Solution:
The given data is
L 360 m
B 70 m
D 25 m
kg
ρ 1020
U 6.69
3
m
s
m
4
P 1.30 10 hp
(Power consumed by drag)
P 9.7 MW
The power to the propeller is
P
Pprop
70 %
Pprop 1.86 10 hp
The shaft power is
Ps 120% Pprop
Ps 2.23 10 hp
The efficiency of the engines is
η 40 %
Hence the heat supplied to the engines is
Q
t
The journey time is
Ps
4
4
8 BTU
Q 1.42 10
η
x
hr
t 134 hr
U
10
Qtotal Q t
The total energy consumed is
x 2000 mi
Qtotal 1.9 10 BTU
From buoyancy the total ship weight equals the displaced seawater volume
M ship g ρ g L B D
9
M ship ρ L B D
M ship 1.42 10 lb
Hence the mass of oil is
M oil 75% M ship
M oil 1.06 10 lb
The chemical energy stored in the petroleum is
q 20000
E q M oil
The total chemical energy is
The equivalent percentage of petroleum cargo used is then
9
BTU
lb
13
E 2.13 10 BTU
Qtotal
E
The Alaska pipeline uses
epipeline 120
BTU
but for the
ton mi ship
The ship uses only about 15% of the energy of the pipeline!
0.089 %
eship
Qtotal
M oil x
eship 17.8
BTU
ton mi
Problem 9.69
[Difficulty: 3]
Given:
Linear, sinusoidal and parabolic velocity profiles
Find:
Momentum fluxes
Solution:
δ
The momentum flux is given by
2
mf ρ u w dy
0
where w is the width of the boundary layer
For a linear velocity profile
u
U
For a sinusoidal velocity profile
u
U
For a parabolic velocity profile
u
U
y
δ
η
(1)
π y
π
sin η
2 δ
2
(2)
sin
2
y
δ
2
y 2 η ( η) 2
δ
For each of these
u U f ( η)
Using these in the momentum flux equation
2
mf ρ U δ w f ( η) dη
(3)
y δ η
1
2
(4)
0
1
For the linear profile Eqs. 1 and 4 give
2
mf ρ U δ w η dη
mf
1
2
π
mf ρ U δ w
sin η dη
2
mf
1
mf ρ U δ w
mf
8
2
0
1
For the sinusoidal profile Eqs. 2 and 4 give
2
3
2
2
ρ U δ w
2
ρ U δ w
0
1
For the parabolic profile Eqs. 3 and 4 give
2
2
2 η ( η) 2 dη
0
The linear profile has the smallest momentum, so would be most likely to separate
15
2
ρ U δ w
Problem 9.68
[Difficulty: 3]
Given:
Data on flow in a duct
Find:
Velocity at location 2; pressure drop; length of duct; position at which boundary layer is 20 mm
Solution:
The given data is
D 6 in
δ1 0.4 in
Table A.9
ρ 0.00234
slug
ft
Governing
equations
Mass
In the boundary layer
δ
x
ν 1.56 10
3
ft
U1 80
s
δ2 1.2 in
0.382
4 ft
2
s
(9.26)
1
5
Rex
In the the inviscid core, the Bernoulli equation holds
p
ρ
2
V
2
g z constant
(4.24)
Assumptions: (1) Steady flow
(2) No body force (gravity) in x direction
For a 1/7-power law profile, from Example 9.4 the displacement thickness is
Hence
δ
δdisp
8
δ1
δdisp1
8
δdisp1 0.0500 in
δ2
δdisp2
8
δdisp2 0.1500 in
From the definition of the displacement thickness, to compute the flow rate, the uniform flow at locations 1
and 2 is assumed to take place in the entire duct, minus the displacement thicknesses
π
2
A1 D 2 δdisp1
4
A1 0.1899 ft
2
π
2
A2 D 2 δdisp2
4
2
A2 0.1772 ft
Mass conservation (Eq. 4.12) leads to U2
ρ U1 A1 ρ U2 A2 0
or
A1
U2 U1
A2
ft
U2 85.7
s
The Bernoulli equation applied between locations 1 and 2 is
p1
ρ
or the pressure drop is
U1
2
2
p2
ρ
U2
2
2
ρ
2
2
p 1 p 2 Δp U2 U1
2
Δp 7.69 10
3
psi (Depends on ρ value selected)
The static pressure falls continuously in the entrance region as the fluid in the central core accelerates into a decreasing core.
If we assume the stagnation pressure is atmospheric, a change in pressure of about 0.008 psi is not significant; in addition, the
velocity changes by about 5%, again not a large change to within engineering accuracy
To compute distances corresponding to boundary layer thicknesses, rearrange Eq.9.26
1
δ
x
0.382
1
Rex
U
x
0.382
ν
5
5
so
x
δ
0.382
4
1
U
ν
5
Applying this equation to locations 1 and 2 (using U = U1 or U2 as approximations)
5
1
4
4
δ1 U1
x1
0.382 ν
For location 3
x 1 1.269 ft
5
1
δ2 U2
x2
0.382 ν
4
4
x 2 x 1 3.83 ft
(Depends on ν value selected)
δ3 0.6 in
δ3
δdisp3
8
x 2 5.098 ft
δdisp3 0.075 in
π
2
A3 D 2 δdisp3
4
A3 0.187 ft
A1
U3 U1
A3
ft
U3 81.4
s
5
1
δ3 U2
x3
0.382 ν
4
4
x 3 x 1 0.874 ft
(Depends on ν value selected)
x 3 2.143 ft
2
4
Problem 9.67
Given:
[Difficulty: 3]
Laboratory wind tunnel has fixed walls. BL's are well represented by 1/7-power
profile. Information at two stations are known:
The given or available data (Table A.9) is
m
U1 26.1
s
Find:
H 305 mm W 305 mm δ1 12.2 mm δ2 16.6 mm
ν 1.46 10
2
5 m
ρ 1.23
s
kg
3
m
(a) Change in static pressure between 1 and 2
(b) Estimate length of tunnel between stations 1 and 2.
Solution:
Basic
equations:
ρ dV ρ V dA 0
t
Assumptions:
(Continuity)
(1) Steady flow
(2) Turbulent, 1/7-power velocity profile in boundary layer
(3) z = constant
A1 U1 A2 U2
Applying continuity between 1 and 6:
where A is the effective flow area. In terms of the duct dimensions:
W 2 δdisp1H 2 δdisp1 U1 W 2 δdisp2 H 2 δdisp2 U2
solving for the speed at 2:
W 2 δdisp1 H 2 δdisp1
U2 U1
W 2 δdisp2 H 2 δdisp2
δdisp δ
The displacement thickness is determined from:
1
1
u
1
dη
U
u
where
U
η
7
η
y
δ
0
Substituting the velocity profile and valuating the integral:
δdisp δ
1
0
1
δ
7
1 η dη
δdisp1 1.525 mm
Therefore:
8
We may now determine the speed at 2:
Applying Bernoulli between 1 and 2:
p1
ρ
U1
2
2
p2
ρ
U2
2
2
Solving for the pressure change:
Δp
Substituting given values:
δdisp2 2.075 mm
m
U2 26.3
s
1
2
ρ U1 U2
2
Δp 6.16Pa
2
1
For a flat plate turbulent boundary layer with 1/7-power law profile: δ x
5
1
4
4
δ1 U1
x1
0.494 m
0.382 ν
0.382
1
Rex
ν
0.382
4
5
5
Solving for location at 1:
x
U
5
To estimate the length between 1 and 6, we determine length necessary to build the BL at section 2:
5
1
4
4
δ2 U2
x2
0.727 m
0.382 ν
Therefore, the distance between 1 and 2 is:
L x2 x1
L 0.233 m
Problem 9.66
Given:
[Difficulty: 3]
Laboratory wind tunnel has flexible wall to accomodate BL growth. BL's are well
represented by 1/7-power profile. Information at two stations are known:
The given or available data (Table A.9) is
U 90
ft
s
Find:
H1 1 ft
W1 1 ft
δ1 0.5 in
δ6 0.65 in
ν 1.57 10
4 ft
2
ρ 0.00238
s
slug
ft
3
(a) Height of tunnel walls at section 6.
(b) Equivalent length of flat plate that would produce the inlet BL
(c) Estimate length of tunnel between stations 1 and 6.
Solution:
Basic
equations:
Assumptions:
ρ dV ρ V dA 0
t
(Continuity)
(1) Steady flow
(2) Turbulent, 1/7-power velocity profile in boundary layer
(3) z = constant
(4) p = constant
Applying continuity between 1 and 6:
A1 U1 A6 U6
where A is the effective flow area. The velocities at 1 and 6 must be
equal since pressure is constant. In terms of the duct dimensions:
W1 2 δdisp1H1 2 δdisp1 W1 2 δdisp6 H6 2 δdisp6
solving for the height at 6:
H6
W1 2 δdisp1H1 2 δdisp1
W1 2 δdisp6
The displacement thickness is determined from:
δdisp δ
1
2 δdisp6
1
1 u dη
U
u
where
U
η
7
η
y
δ
0
Substituting the velocity profile and valuating the integral:
δdisp δ
1
0
1
δ
7
1 η dη
δdisp1 0.0625 in
Therefore:
8
We may now determine the height at 6:
δdisp6 0.0813 in
H6 1.006 ft
1
For a flat plate turbulent boundary layer with 1/7-power law profile: δ1 L1
5
1
4
δ1 U 4
L1
0.382 ν
0.382
1
Re1
ν
0.382
4
5
5
L1 Solving for L1:
U
5
L1 1.725 ft
To estimate the length between 1 and 6, we determine length necessary to build the BL at section 6:
5
1
4
L6
δ6 U 4
2.394 ft
0.382 ν
Therefore, the distance between 1 and 6 is:
L L6 L1
L 0.669 ft
Problem 9.65
Given:
[Difficulty: 3]
Air at standard conditions flowing through a plane-wall diffuser with negligible
BL thickness. Walls diverge slightly to accomodate BL growth, so p = constant.
The given or available data (Table A.9) is
U 60
m
s
Find:
L 1.2 m
W1 75 mm
2
5 m
ν 1.46 10
s
ρ 1.23
kg
3
m
(a) why Bernoulli is applicable to this flow.
(b) diffuser width W2 at x = L
Solution:
p1
Basic
equations:
ρ
V1
2
2
p2
g z1
ρ
V2
2
g z2
2
ρ dV ρ V dA 0
t
Assumptions:
(Bernoulli Equation)
(Continuity)
(1) Steady flow
(2) Turbulent, 1/7-power velocity profile in boundary layer
(3) z = constant
(4) p = constant
The Bernoulli equation may be applied along a streamline in any steady, incompressible flow in the absence of
friction. The given flow is steady and incompressible. Frictional effects are confined to the thin wall boundary
layers. Therefore, the Bernoulli equation may be applied along any streamline in the core flow outside the boundary
layers. (In addition, since there is no streamline curvature, the pressure is uniform across sections 1 and 2.
From the assumptions, Bernoulli reduces to: V1 V2 and from continuity: ρ V1 A1 ρ V2 A2eff 0
or A2eff W2 2 δdisp2 b W1 b
The Reynolds number is:
ReL
U L
ν
Therefore: W2 W1 2 δdisp2
6
4.932 10
From turbulent BL theory:
δ2 L
0.382
1
ReL
5
21.02 mm
The displacement thickness is determined from:
δdisp2 δ2
1
1
u
1
dη
U
where
u
U
η
7
η
y
δ
0
Substituting the velocity profile and valuating the integral:
δdisp2 δ2
1
1
δ2
7
1 η dη
0
Therefore:
W2 W1 2 δdisp2
8
δdisp2 2.628 mm
W2 80.3 mm
Problem 9.64
[Difficulty: 3]
Air at standard conditions flowing over a flat plate
Given:
The given or available data (Table A.10) is
U 30
ft
s
Find:
x 3 ft
ν 1.57 10
4 ft
2
ρ 0.00238
s
slug
ft
3
δ and τw at x assuming:
(a) completely laminar flow (parabolic velocity profile)
(b) completely turbulent flow (1/7-power velocity profile)
Solution:
(Laminar Flow)
Basic
equations:
δ
x
(Turbulent Flow)
5.48
δ
Rex
x
0.382
Rex
Cf
τw
1
2
2
ρ U
0.730
Rex
The Reynolds number is:
ν
δlam x
For laminar flow:
1
2
U x
5.73 10
δturb x
5.48
0.382
Comparing results:
δlam
1
ρ U
Rex
(Skin friction factor)
5
τwlam 7.17 10
5
The turbulent boundary layer has a much larger skin friction, which causes it to
grow more rapidly than the laminar boundary layer.
4.34
5
τwturb 3.12 10
3.72
τwlam
psi
5
Rex
τwturb
6
δturb 0.970 in
1
1
2 0.0594
τwturb ρ U
1
2
δturb
0.0594
δlam 0.261 in
Rex
Rex
2
5
1
2 0.730
τwlam ρ U
2
Rex
For turbulent flow:
5
τw
Cf
Rex
(Boundary Layer Thickness)
1
psi
Problem 9.63
[Difficulty: 3]
Turbulent boundary layer flow of water, 1/7-power profile
Given:
The given or available data (Table A.9) is
U 20
m
s
L 1.5 m
b 0.8 m ν 1.46 10
2
5 m
s
ρ 1.23
kg
3
m
x 1 0.5 m
(a) δ at x = L
(b) τw at x = L
(c) Drag force on the portion 0.5 m < x < L
Find:
Solution:
Basic
equations:
δ
x
0.382
Rex
Cf
(Boundary Layer Thickness)
1
5
τw
1
2
2
0.0594
(Skin friction factor)
1
ρ U
Rex
5
Assumptions: 1) Steady flow
2) No pressure force
3) No body force in x direction
At the trailing edge of the plate:
ReL
U L
ν
2.05 10
6
δL L
Therefore
δL 31.3 mm
1
ReL
1
2 0.0594
Similarly, the wall shear stress is: τwL ρ U
1
2
ReL
0.382
5
τwL 0.798 Pa
5
L
To find the drag:
L
1
1
1
L
5
1
U
2
5
5
dx where c is defined:
FD τw b dx 0.0594 ρ U
x
b dx c b x
2
ν
x
0
x
1
1
1
U
2
c 0.0594 ρ U
2
ν
1
4
5
Therefore the drag is:
5
FD c b L
4
5
5 1
2
ρ U b L CfL x 1 Cfx1
4 2
At x = x1:
Rex1
U x 1
ν
6.849 10
5
Cfx1
0.0594
1
Rex1
3
4.043 10
and at x = L CfL
5
0.0594
1
ReL
3
3.245 10
5
FD 0.700 N
Therefore the drag is:
Alternately, we could solve for the drag using the momentum thickness:
At x = L
δL 31.304 mm
7
θL
δ 3.043 mm At x = x1:
72 L
2
FD ρ U b θL θx1
δx1 x 1
0.382
1
Rex1
where θ
12.999 mm θx1
7
72
δ
7
δ 1.264 mm
72 x1
5
Therefore the drag is:
FD 0.700 N
Problem 9.62
[Difficulty: 3]
1
u
8
1
y η8
δ
Given:
Turbulent boundary layer flow with 1/8 power velocity profile:
Find:
Expressions for δ/x and Cf using the momentum integral equation; compare to 1/7-power rule results.
Solution:
We will apply the momentum integral equation
τw
Governing
Equations:
ρ
2
d
d
U θ δdisp U U
dx
dx
τw
Cf
1
2
U
(Momentum integral equation)
(Skin friction coefficient)
2
ρ U
(1) Zero pressure gradient, so U is constant and dp/dx = 0
(2) δ is a function of x only, and δ = 0 at x = 0
0.25
(3) Incompressible flow
2 ν
τ
0.0233
ρ
U
(4) Wall shear stress is:
w
U δ
Assumptions:
1
u
u
2 d
τw ρ U θ ρ U
δ
1 dη
U
U
dx
dx
0
1
2
1
2
2 8
8
8
Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ
Setting our two τ w's equal:
dx
90
d
x
0
Applying the assumptions to the momentum integral equation yields:
0.0233 ρ U
2
U
δ
ν
2 d
1
1
0.25
d δ
56 dx
2 6
ρ U
Simplifying and separating variables:
δ dδ 0.262
4
ν
4
dx
U
4
1
5
Integrating both sides:
4
5
δ
4
0.262
ν
4
x C
U
1
4
5
ν
but C = 0 since δ = 0 at x = 0. Therefore: δ 0.262 x
4
U
In terms of the Reynolds number:
δ
x
5
0.410
1
Rex
5
For the skin friction factor:
1
0.0233 ρ U
2
Cf
τw
1
2
2
ρ U
1
2
U δ
ν
2
ρ U
4
1
4
1
1
Re 5
4
4
ν x
x
4
0.0466
0.0466 Rex 0.410 Upon simplification:
U x δ
1
1
Cf
0.0582
1
Rex
These results compare to
δ
x
0.353
1
Rex
5
and
Cf
0.0605
1
Rex
5
for the 1/7-power profile.
5
Problem 9.61
[Difficulty: 3]
1
u
Turbulent boundary layer flow with 1/6 power velocity profile:
Given:
The given or available data (Table A.9) is
U 1
m
s
U
L 1 m
6
1
y η 6
δ
2
6 m
ν 1.00 10
s
ρ 999
Expressions for δ/x and Cf using the momentum integral equation; evaluate drag for the conditions given
Solution:
We will apply the momentum integral equation
τw
ρ
2
d
d
U θ δdisp U U
dx
dx
τw
Cf
1
2
3
m
Find:
Governing
Equations:
kg
(Momentum integral equation)
(Skin friction coefficient)
2
ρ U
(1) Zero pressure gradient, so U is constant and dp/dx = 0
(2) δ is a function of x only, and δ = 0 at x = 0
0.25
(3) Incompressible flow
ν
2
τw 0.0233 ρ U
(4) Wall shear stress is:
U δ
Assumptions:
1
u
u
2 d
Applying the assumptions to the momentum integral equation yields:
τw ρ U θ ρ U
δ
1 dη
U
U
dx
dx
0
1
2
1
2
2 6
6
6
Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ
Setting our two τ w's equal:
dx
56 dx
0
2 d
0.0233 ρ U
2
U δ
ν
0.25
1
1
d δ
56 dx
2 6
ρ U
Simplifying and separating variables:
4
δ dδ 0.0233
56
ν
4
dx
6 U
4
1
4
4
5
56 ν
56 ν
4 4
δ 0.0233 x C but C = 0 since δ = 0 at x = 0. Therefore: δ 0.0233 x
6 U
6 U
5
4
5
Integrating both sides:
1
5
δ
In terms of the Reynolds number:
x
0.353
1
Rex
5
For the skin friction factor:
1
0.0233 ρ U
2
Cf
τw
1
2
2
1
ρ U
2
U δ
ν
4
2
ρ U
1
4
1
1
Re 5
4
4
ν
x
x
0.0466 Re 4
0.0466
Upon simplification:
x
U
x
δ
0.353
1
1
Cf
0.0605
1
Rex
5
L
The drag force is:
1
1 L
1
1
L
5
5
0.0605
1
2 ν
2 ν
5
5
dx
FD τw b dx 0.0605 ρ U x
b dx
ρ U b x
2
2
U
U
0
0
0
1
Evaluating the integral:
FD
0.0605
2
5
4
5 5
2 ν
ρ U b L
4
U
2
In terms of the Reynolds number: FD
0.0378 ρ U b L
1
ReL
For the given conditions and assuming that b = 1 m:
6
ReL 1.0 10
5
and therefore:
FD 2.38 N
Problem 9.60
[Difficulty: 3]
1
u
6
1
y η 6
δ
Given:
Turbulent boundary layer flow with 1/6 power velocity profile:
Find:
Expressions for δ/x and Cf using the momentum integral equation; compare to 1/7-power rule results.
Solution:
We will apply the momentum integral equation
τw
Governing
Equations:
ρ
2
d
d
U θ δdisp U U
dx
dx
τw
Cf
1
2
U
(Momentum integral equation)
(Skin friction coefficient)
2
ρ U
(1) Zero pressure gradient, so U is constant and dp/dx = 0
(2) δ is a function of x only, and δ = 0 at x = 0
0.25
(3) Incompressible flow
2 ν
τ
0.0233
ρ
U
(4) Wall shear stress is:
w
U δ
Assumptions:
1
u
u
2 d
τw ρ U θ ρ U
δ
1 dη
U
U
dx
dx
0
1
2
1
2
2 6
6
6
Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ
Setting our two τ w's equal:
dx
56
d
x
0
Applying the assumptions to the momentum integral equation yields:
0.0233 ρ U
2
U
δ
ν
0.25
2 d
1
1
d δ
56 dx
2 6
ρ U
Simplifying and separating variables:
4
δ dδ 0.0233
56
6
ν
4
dx
U
4
1
4
4
5
56 ν
56 ν
4 4
δ 0.0233 x C but C = 0 since δ = 0 at x = 0. Therefore: δ 0.0233 x
6 U
6 U
5
4
5
Integrating both sides:
1
5
In terms of the Reynolds number:
δ
x
0.353
1
Rex
5
For the skin friction factor:
1
0.0233 ρ U
2
Cf
τw
1
2
2
ρ U
1
2
U δ
ν
2
ρ U
4
1
4
1
1
Re 5
4
4
ν x
x
4
0.0466
0.0466 Rex 0.353 Upon simplification:
U x δ
1
1
Cf
0.0605
1
Rex
These results compare to
δ
x
0.353
1
Rex
5
and
Cf
0.0605
1
Rex
5
for the 1/7-power profile.
5
Problem 9.59
Given:
Pattern of flat plates
Find:
Drag on separate and composite plates
[Difficulty: 3]
Solution:
Basic
equations:
cf
τw
1
2
cf
2
ρ U
0.0594
1
Rex
For separate plates
L 3 in
From Table A.7 at 70 oF
5 ft
ν 1.06 10
s
W 3 in
2
ρ 1.93
We also have
3
ReL 1.89 10
ν
so turbulent
0
1
2
2 0.0594
ρ U
1
2
FD ρ U W
2
1
9
L
0
5
0.0594
1
U x
ν
dx
0.0594
2
4
1
5
5
5
dx L
x
4
0
This is the drag on one plate. The total drag is then
L
1
5
5
5
dx
ρ U W ν x
0
5
L
The integral is
6
L
Rex
Hence
s
FD τw W dx
τw cf ρ U
2
2
1
ft
slug
U L
ReL
FD τw dA
1
U 80
ft
First determine the nature of the boundary layer
The drag (one side) is
5
1
so
4
9
FD 0.0371 ρ W ν L U
5
FTotal 4 FD
FD 1.59 lbf
FTotal 6.37 lbf
For both sides:
2 FTotal 12.73 lbf
For the composite plate
L 4 3 in
L 12.00 in and since the Reynolds number for the single plate was turbulent, we
know that the flow around the composite plate will be turbulent as well.
1
4
9
FComposite 0.0371 ρ W ν L U
5
FComposite 4.82 lbf
For both sides:
2 FComposite 9.65 lbf
The drag is much lower on the composite compared to the separate plates. This is because τ w is largest
near the leading edges and falls off rapidly; in this problem the separate plates experience leading edges
four times!
Problem 9.58
Given:
Parabolic plate
Find:
Drag
[Difficulty: 3]
Solution:
Basic
equations:
cf
τw
1
2
cf
2
ρ U
0.0594
1
Rex
5
W
2
L
W 1 ft
2
L 3 in
1 ft
U 80
ft
s
Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L
From Table A.9 at 70 oF
ν 1.63 10
4 ft
2
ρ 0.00233
s
ft
ReL
First determine the nature of the boundary layer
The drag (one side) is
We also have
3
U L
ReL 1.23 10
ν
5
so still laminar, but we are
told to assume turbulent!
L
FD τw dA
FD τw w( x ) dx
w( x ) W
0
x
L
1
2 1
2 0.0594
τw cf ρ U ρ U
1
2
2
Rex
Hence
slug
1
2
FD ρ U W
2
5
L
x
0.0594
L
1
U x
ν
9
dx
0.0594
2
1
1
L
3
5
2 5
10
ρ U W L
ν x dx
0
5
0
1
4
9
FD 0.0228 ρ W ν L U
5
FD 0.00816 lbf
Note: For two-sided solution
2 FD 0.01632 lbf
Problem 9.57
Given:
Triangular plate
Find:
Drag
[Difficulty: 3]
Solution:
Basic
equations:
cf
τw
1
2
cf
2
ρ U
L 2 ft
0.0594
1
Rex
3
5
L 1.732 ft
2
W 2 ft
2
From Table A.10 at 70oF
4 ft
ν 1.62 10
s
ρ 0.00234
ReL
The drag (one side) is
We also have
3
U L
ReL 9 10
ν
5
so definitely still laminar over a
significant portion of the plate,
but we are told to assume
turbulent!
FD τw w( x ) dx
w( x ) W
0
x
L
1
2 1
2 0.0594
τw cf ρ U ρ U
1
2
2
1
2 W
FD ρ U
2
L
L
The integral is
s
L
FD τw dA
Rex
Hence
ft
slug
ft
First determine the nature of the boundary layer
U 80
4
9
5
5
5
x dx L
9
0
L
5
1
9
0.0594 x
1
U x
ν
dx
0.0594
2
L
4
5
5
ν x dx
L
0
5 W
ρ U
5
0
1
so
4
9
FD 0.0165 ρ W L ν U
Note: For two-sided solution
5
FD 0.0557 lbf
2 FD 0.1114 lbf
Problem 9.56
Turbulent boundary layer flow of water
L 1 m
Find:
Solution:
Governing
Equations:
Plot δ,
δ*,
U 1
1
2
6 m
m
ν 1.00 10
s
u
s
U
y
δ
7
and τw versus x/L for the plate
We will determine the drag force from the shear stress at the wall
δ
x
0.382
(Boundary layer thickness)
1
Rex
δdisp
δ
Cf
5
1
(Displacement thickness)
8
τw
1
2
2
0.0594
1
ρ U
Rex
(Skin friction factor)
5
Assumption: Boundary layer is turbulent from x = 0
For the conditions given:
ReL
U L
ν
6
1.0 10
q
1
2
2
ρ U 500 Pa
τw
0.0594
1
Rex
30
Boundary Layer and Displacement Thicknesses (mm)
Here is the plot of boundary layer thickness
and wall shear stress:
q 29.7 Pa Rex
1
5
5
3
BL Thickness
Disp. Thickness
Wall Shear
20
2
10
1
0
0
0.5
x (m)
0
1
Wall Shear Stress (Pa)
Given:
[Difficulty: 2]
Problem 9.55
U 10
[Difficulty: 3]
m
L 5 m
ν 1.45 10
2
5 m
Given:
Data on flow over a flat plate
Find:
Plot of laminar and turbulent boundary layer; Speeds for transition at trailing edge
s
(from Table A.10)
s
Solution:
Governing For laminar flow δ 5.48
Equations:
x
Rex
The critical Reynolds number is
(9.21)
and
U x
Rex
ν
δ 5.48
so
Recrit 500000 Hence, for velocity U the critical length xcrit is
δ
x
0.382
(9.26)
δ 0.382
so
1
Rex
ν
5
For (a) completely laminar flow Eq. 1 holds; for (b) completely turbulent flow Eq. 3 holds; for (c)
transitional flow Eq.1 or 3 holds depending on xcrit in Eq. 2. Results are shown below from Excel.
x (m)
Re x
0.00
0.125
0.250
0.375
0.500
0.700
0.75
1.00
1.50
2.00
3.00
4.00
0.00E+00
8.62E+04
1.72E+05
2.59E+05
3.45E+05
4.83E+05
5.17E+05
6.90E+05
1.03E+06
1.38E+06
2.07E+06
2.76E+06
5.00
3.45E+06
(a) Laminar (b) Turbulent (c) Transition
δ (mm)
δ (mm)
δ (mm)
0.00
0.00
0.00
2.33
4.92
2.33
3.30
8.56
3.30
4.04
11.8
4.04
4.67
14.9
4.67
5.52
19.5
5.5
5.71
20.6
20.6
6.60
26.0
26.0
8.08
35.9
35.9
9.3
45.2
45.2
11.4
62.5
62.5
13.2
78.7
78.7
14.8
94.1
94.1
4
5
5
x
U
(1)
U
x crit 500000
1
For turbulent flow
ν x
(3)
ν
U
(2)
Boundary Layer Profiles on a Flat Plate
100
75
δ (mm)
Laminar
Turbulent
Transitional
50
25
0
0
0.5
1
1.5
2
2.5
x (m)
3
The speeds U at which transition occurs at specific points are shown below
x trans
(m)
5
4
3
2
1
U (m/s)
1.45
1.81
2.42
3.63
7.25
3.5
4
4.5
5
Problem 9.54
[Difficulty: 3]
Note: Figure data applies to problem 9.18 only
Given:
Data on fluid and turbulent boundary layer
Find:
Solution:
Mass flow rate across ab; Momentum flux across bc; Distance at which turbulence occurs
CV
Mass
Basic
equations:
d
Momentum
c
Rx
Assumptions: 1) Steady flow 2) No pressure force 3) No body force in x direction 4) Uniform flow at ab
The given or available data (Table A.10) is
U 50
m
s
δ 19 mm
Consider CV abcd
b 3 m
ρ 1.23
kg
ν 1.50 10
3
2
5 m
m
kg
mad 3.51
s
mad ρ U b δ
(Note: Software cannot render a dot)
1
δ
mad ρ u b dy mab 0
Mass
s
and in the boundary layer
u
U
0
y
1
7
7
η
δ
dy dη δ
1
Hence
1
7
7
mab ρ U b δ ρ U η δ dη ρ U b δ ρ U b δ
8
0
1
mab ρ U b δ
8
kg
mab 0.438
s
1
δ
The momentum flux
across bc is
δ
mfbc u ρ V dA
0
0
2
7
2
2
7
u ρ u b dy ρ U b δ η dη ρ U b δ
9
0
7
2
mfbc ρ U b δ
9
mfbc 136.3
2
s
From momentum
Rx U ( ρ U δ) mab u ab mfbc
Transition occurs at
Rex 5 10
5
kg m
and
2
Rx ρ U b δ mab U mfbc
U x
Rex
ν
x trans
Rx 17.04 N
Rex ν
U
x trans 0.1500 m
Problem 9.53
Given:
[Difficulty: 3]
Turbulent boundary layer flow of water, 1/7-power profile
The given or available data (Table A.9) is
U 1
m
s
Find:
L 1 m
ν 1.00 10
2
6 m
ρ 999
s
kg
3
m
(a) Expression for wall shear stress
(b) Integrate to obtain expression for skin friction drag
(c) Evaluate for conditions shown
Solution:
Basic
Equation:
Cf
τw
1
2
0.0594
(Skin friction factor)
1
2
ρ U
Rex
5
Assumptions: 1) Steady flow
2) No pressure force
3) No body force in x direction
4) Uniform flow at ab
τw 0.0594 ρ U Rex
2
1
Solving the above expression for the wall shear stress:
2
1
5
U
2
τw 0.0594 ρ U
2
ν
1
1
5
x
1
5
L
L
1
1
1
L
5
1
U
2
5
5
dx where c is defined:
FD τw b dx 0.0594 ρ U
x
b dx c b x
2
ν
0
0
Integrating to find the drag:
0
1
U
2
c 0.0594 ρ U
2
ν
1
4
5
Therefore the drag is:
5
5
1
U L
2
5
FD c b L 0.0594 ρ U b L
4
4
2
ν
Upon simplification:
1
5
1
0.0721
2
FD ρ U b L
1
2
ReL
Evaluating, with b 1 m
ReL
U L
ν
1 10
6
1
0.0721
2
FD ρ U b L
1
2
ReL
5
FD 2.27 N
5
Problem 9.52
Given:
Data on flow in a channel
Find:
Static pressures; plot of stagnation pressure
[Difficulty: 3]
Solution:
The given data is
h 1.2 in
Appendix A
ρ 0.00239
δ2 0.4 in
w 6 in
slug
ft
Governing
equations:
ft
U2 75
s
3
Mass
Before entering the duct, and in the the inviscid core, the Bernoulli
equation holds
2
p
V
g z constant
2
ρ
(4.24)
Assumptions: (1) Steady flow
(2) No body force in x direction
For a linear velocity profile, from Table 9.2 the displacement thickness at location 2 is
δ2
δdisp2
2
δdisp2 0.2 in
From the definition of the displacement thickness, to compute the flow rate, the uniform flow at location 2 is
assumed to take place in the entire duct, minus the displacement thicknesses at top and bottom
Then
2
A2 w h 2 δdisp2
A2 4.80 in
Q A2 U2
Q 2.50
ft
3
s
Mass conservation (Eq. 4.12) leads to U2
U1 A1 U2 A2
A2
U1
U
A1 2
where
A1 w h
2
A1 7.2 in
ft
U1 50
s
The Bernoull equation applied between atmosphere and location 1 is
p atm
ρ
p1
ρ
U1
2
2
or, working in gage pressures
1
2
p 1 ρ U1
2
p 1 0.0207 psi
(Static pressure)
Similarly, between atmosphere and location 2 (gage pressures)
1
2
p 2 ρ U2
2
p 2 0.0467 psi
(Static pressure)
The static pressure falls continuously in the entrance region as the fluid in the central core accelerates into a decreasing core.
The stagnation pressure at location 2 (measured, e.g., with a Pitot tube as in Eq. 6.12), is indicated by an application of the
Bernoulli equation at a point
pt
ρ
p
ρ
u
2
2
where p t is the total or stagnation pressure, p = p 2 is the static pressure, and u is the local velocity, given by
u
U2
y
y δ2
δ2
h
δ2 y
2
u U2
(Flow and pressure distibutions are symmetric about centerline)
Hence
y (in)
0.00
0.04
0.08
0.12
0.16
0.20
0.24
0.28
0.32
0.36
0.40
0.44
0.48
0.52
0.56
0.60
1
2
p t p 2 ρ u
2
The plot of stagnation pressure is shown in the Excel sheet below
Stagnation Pressure Distibution in a Duct
u (ft/s) p t (psi)
0.00
7.50
15.00
22.50
30.00
37.50
45.00
52.50
60.00
67.50
75.00
75.00
75.00
75.00
75.00
75.00
0.000
0.000
0.002
0.004
0.007
0.012
0.017
0.023
0.030
0.038
0.047
0.047
0.047
0.047
0.047
0.047
0.6
0.5
0.4
y (in) 0.3
0.2
0.1
0.0
0.00
0.01
0.02
0.03
p t (psi gage)
The stagnation pressure indicates total mechanical energy - the curve indicates significant loss close to the walls
and no loss of energy in the central core.
0.04
0.05
Problem 9.51
Given:
Horizontal surface immersed in a stream of standard air. Laminar BL with linear profile forms.
L 0.8 m b 1.9 m
Find:
Solution:
[Difficulty: 2]
U 5.3
m
s
ν 1.46 10
2
5 m
u
s
U
y
δ
Algebraic expressions for wall shear stress and drag; evaluate at given conditions
We will determine the drag force from the shear stress at the wall
Governing τw ρ U2 d θ μ u at y = 0
dx
y
Equations:
δ
1
θ
u
U
0
1
u
dη
(Wall shear stress)
(Momentum thickness)
U
1
1
2
η ( 1 η ) dη η η dη
δ 0
0
θ
For the linear velocity profile:
δ
1
Therefore it follows that d θ d θ δ δ To determine the wall shear stress:
dx
dδ x 6 x
Separating variables yields:
6 μ
ρ U
2
dx δ dδ
Substituting this back into the expression for wall shear stress:
The drag force is given by:
For the given conditions:
FD τw dA
ReL
U L
ν
δ
Integrating yields:
2
τw
μ U
δ
6 μ
ρ U
x
x
μ U
δ
1
6
2
ρ U
6
12
1
12
μ U
Rex
x
0.1667
δ
x
Solving this expression for δ/x:
μ U
τw
δ
x
12
Rex
μ U
τw 0.289
Rex
x
Rex
θ
L
L
L
2 dθ
2
ρ U b dx b ρ U dθ
τw b dx
dx
0
0
2.90 10
θ
Evaluating this integral:
2
FD ρ U b θL
0
5
12
δL L
5.14 mm
ReL
δL
θL
0.857 mm
6
2
FD ρ U b θL 0.0563 N
FD 0.0563 N
Problem 9.50
[Difficulty: 2]
Horizontal surface immersed in a stream of standard air. Laminar BL with linear profile forms.
Given:
L 0.8 m b 1.9 m
Find:
Solution:
Plot δ,
δ*,
U 5.3
m
s
ν 1.46 10
2
5 m
u
s
U
y
δ
and τw versus x/L for the plate
We will determine the drag force from the shear stress at the wall
Governing τw ρ U2 d θ μ u at y = 0
dx
y
Equations:
δdisp
δ
1
0
δ
1
θ
u
U
0
1 u dη
U
1
For the linear velocity profile:
(Wall shear stress)
(Displacement thickness)
u
dη
(Momentum thickness)
U
1
1
2
η ( 1 η ) dη η η dη
δ 0
0
θ
δ
1
Therefore it follows that d θ d θ δ δ To determine the wall shear stress:
dx
dδ x 6 x
Separating variables yields:
Also,
δdisp
δ
6 μ
ρ U
2
dx δ dδ
1
( 1 η ) dη
Evaluating this integral:
0
The Reynolds number is related to x through:
δ
Integrating yields:
2
δdisp
δ
5
Rex 3.63 10 x
Plots of δ, δdisp and τ w as functions of x are shown on the next page.
6 μ
ρ U
θ
Evaluating this integral:
x
τw
μ U
δ
1
6
2
ρ U
6
0.1667
δ
x
Solving this expression for δ/x:
δ
x
1
δdisp
2
δ
where x is measured in meters.
3.46
Rex
1
2
BL Thickness
Disp. Thickness
Wall Shear
0.04
4
0.03
2
0.02
0
0
0.2
0.4
x (m)
0.6
0.01
0.8
Wall Shear Stress (Pa)
Boundary Layer and Displacement Thicknesses (mm)
6
Problem 9.49
Given:
Water flow over flat plate
Find:
Drag on plate for linear boundary layer
[Difficulty: 3]
Solution:
Basic
equations:
FD 2 τw dA
du
τw μ
dy
L 0.35 m
From Table A.8 at 10 oC ν 1.30 10
6 m
ρ 1000
s
ReL
First determine the nature of the boundary
layer
y
The velocity profile is
u U U η
δ
du
U
Hence
τw μ
μ
dy
δ
We also have
The integral is
2 dδ
τw ρ U
1
0
u
dη
U
s
ReL 2.15 10
5
so laminar
but we need δ(x)
1
so
1
2 dδ
2 dδ
τw ρ U
ρ U
dx
6
dx
(2)
U
1
2 dδ
τw μ ρ U
δ
6
dx
δ dδ
Hence
δ
6 μ
dx
δ
x
or
δ
ρ U
12 μ
ρ U
FD 2
2
or
2
x
L
U
τw dA 2 W
μ dx 2 W
δ
0
L
0
6 μ
ρ U
x c
12
Rex
but δ(0) = 0 so c = 0
3.46
Rex
1
2
dx 2 L
x
so
0
FD
2
3
3
ρ W ν L U
L
1
ρ U
μWU U
2
2
x
dx
μ U
x
dx
12 μ
ν 0
3
1
L
The integral is
U
1
3
m
U L
ν
m
u
kg
u
η η2 dx 16
Separating variables
Then
1
(1)
U 0.8
dx
1
u
2 dδ
1 dη ρ U η ( 1 η) dη
dx
U
U
dx 0
0
0
Comparing Eqs 1 and 2
at y = 0, and also
W 1 m
2
2 dδ
τw ρ U
FD
2 μ W U
3
FD 0.557N
U L
ν
Problem 9.48
[Difficulty: 2]
Horizontal surface immersed in a stream of standard air. Laminar BL with sinusoidal profile forms.
Given:
L 1.8 m b 0.9 m
Find:
Solution:
Plot δ,
δ*,
U 3.2
m
s
ν 1.46 10
2
5 m
s
δ
1
0
δ
1
θ
u
U
0
1 u dη
U
dη
(Wall shear stress)
(Displacement thickness)
u
1
(Momentum thickness)
U
1
π
π
θ
sin η 1 sin η dη
δ
2
2
For the sinusoidal velocity profile:
0
Evaluating this integral:
θ
δ
4π
Separating variables yields:
2 π
2 4 π
dx δ dδ or
δ dδ
2 ρ U
1
0
π π 2
sin η sin η dη
2 2
4 π
d
d
θ θ δ
δ
2 π x
dx
dδ x
π 0 π
π μ U
2 4 π
τw μ U cos
ρ U
δ
2 δ
2 π x
2 δ 2 δ
π μ U
Solving this expression for δ/x:
1
0
0.1366 Therefore it follows that
2 π
To determine the wall shear stress:
δ
π y
2 δ
We will determine the drag force from the shear stress at the wall
δdisp
Also,
U
sin
and τw versus x/L for the plate
Governing τw ρ U2 d θ μ u at y = 0
dx
y
Equations:
δdisp
u
δ
x
π
2
4π
1 sin π η dη
2
The Reynolds number is related to x through:
π
2
μ
4 π ρ U
2
dx Integrating yields:
δ
2
π
2
μ
x
4 π ρ U
μ
δ
ρ U x
x
Evaluating this integral:
δdisp
δ
5
Rex 2.19 10 x
Plots of δ, δdisp and τ w as functions of x are shown on the next page.
1
2
π
0.363
where x is measured in meters.
δdisp
δ
4.80
Rex
0.363
BL Thickness
Disp. Thickness
Wall Shear
0.04
10
0.03
0.02
5
0.01
0
0
0.5
1
x (m)
1.5
0
Wall Shear Stress (Pa)
Boundary Layer and Displacement Thicknesses (mm)
15
Problem 9.47
[Difficulty: 3]
u
y
Given:
Laminar boundary layer flow with linear velocity profile:
Find:
Expressions for δ/x and Cf using the momentum integral equation
Solution:
We will apply the momentum integral equation
Governing
Equations:
τw
ρ
Cf
1
δ
η
(Momentum integral equation)
τw
2
Assumptions:
2
d
d
U θ δdisp U U
dx
dx
U
(Skin friction coefficient)
2
ρ U
(1) Zero pressure gradient, so U is constant and dp/dx = 0
(2) δ is a function of x only, and δ = 0 at x = 0
(3) Incompressible flow
1
u
u
2 d
Applying the assumptions to the momentum integral equation yields:
τw ρ U θ ρ U
δ
1 dη
U
U
dx
dx
0
1
2
2
2 1
Substituting for the velocity profile: τw ρ U d δ η η dη ρ U d δ
dx
6 dx
0
2 d
at y = 0
u
y
τw μ
Now the wall shear stress is also:
2
δ
2
6 μ
ρ U
Solving for the boundary layer thickness:
Substituting the velocity profile:
μ U
2 1 d
ρ U δ
6 dx
δ
Setting both expressions for the wall shear stress equal:
Integrating this expression:
τw
μ U
δ
6 μ
Separating variables: δ dδ
ρ U
dx
2
x C However, we know that C = 0 since δ = 0 when x = 0. Therefore:
δ
From the definition for skin friction coefficient:
12 μ
ρ U
Cf
x
δ
or
x
τw
1
2
2
ρ U
μ U
δ
δ
2
12 μ
δ
ρ U x
x
2
2
ρ U
2 μ
ρ U δ
2
μ
x
ρ U x δ
2
6 μ
ρ U
x
3.46
Rex
Rex
Rex 3.46
Upon simplification:
Cf
0.577
Rex
Problem 9.46
Given:
Pattern of flat plates
Find:
Drag on separate and composite plates
Solution:
Basic
equations:
cf
1
0.730
cf
2
ρ U
Rex
2
Parabolic boundary layer profile
Assumption:
For separate plates
τw
[Difficulty: 3]
L 3 in
W 3 in
U 3
We also have
Hence
From Table A.7 at 70 oF ν 1.06 10
s
ReL
First determine the nature of the boundary layer
The drag (one side) is
ft
U L
ReL 7.08 10
ν
4
5 ft
2
ρ 1.93
slug
s
so definitely laminar
L
FD τw dA
FD τw W dx
0
1
2 1
2 0.730
τw cf ρ U ρ U
2
2
Rex
1
2
FD ρ U W
2
L
L
1
0.730
0.730
2
2
dx
dx
ρ U W ν x
2
U x
0
3
ν
0
L
The integral is
1
1
2
2
dx 2 L
x
so
0
This is the drag on one plate. The total drag is then
3
FD 0.730 ρ W ν L U
FD 0.0030 lbf
FTotal 4 FD
FTotal 0.0119 lbf
For both sides:
For the composite plate
L 4 3 in
L 1.00 ft
ReL
U L
ν
2.83 10
5
2 FTotal 0.0238 lbf
so still laminar
3
FComposite 0.730 ρ W ν L U
FComposite 0.0060 lbf
For both sides:
2 FComposite 0.0119 lbf
The drag is much lower on the composite compared to the separate plates. This is because τ w is largest near the
leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times!
ft
3
Problem 9.45
[Difficulty: 4]
Note: Plate is now reversed!
Given:
Parabolic plate
Find:
Drag
Solution:
Basic
equations:
cf
τw
1
2
0.730
cf
2
ρ U
Rex
W
2
L
W 1 ft
2
L 0.25 ft
1 ft
U 15
ft
s
Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = 0
From Table A.10 at 70oF
ν 1.63 10
4 ft
2
ρ 0.00234
s
ft
U L
ReL
First determine the nature of the boundary layer
The drag (one side) is
slug
ν
3
4
ReL 2.3 10
so just laminar
L
FD τw dA
FD τw w( x ) dx
w( x ) W 1
0
We also have
1
2 1
2 0.730
τw cf ρ U ρ U
2
2
Rex
Hence
1
2
FD ρ U W
2
L
x
0.730 1
L
U x
3
dx
0.730
2
ρ U W ν
L
2
1
x
1
L
x
L
dx
0
ν
0
The tricky integral is (this might
be easier to do numerically!)
0.730
2
x
1
3
FD
2
L x x so
dx x
L ln
L
x
L
2
L x x
1
2
ρ U W ν
L
0
1
x
i
1
L
dx
Note: For two-sided solution
The drag is much higher compared to Problem 9.44. This is because τ w is largest near the leading
edge and falls off rapidly; in this problem the widest area is also at the front
L
1
x
1
L
dx 0.434 m
0
FD 4.98 10
4
lbf
4
2 FD 9.95 10
lbf
Problem 9.44
Given:
Parabolic plate
Find:
Drag
[Difficulty: 3]
Solution:
Basic equations:
cf
τw
1
2
0.730
cf
2
ρ U
Rex
W
2
L
W 1 ft
2
L 0.25 ft
1 ft
U 15
ft
s
Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L
From Table A.9 at 70 oF
ν 1.63 10
4 ft
2
ρ 0.00233
s
ft
ReL
First determine the nature of the boundary layer
The drag (one side) is
slug
3
U L
4
ReL 2.3 10
ν
so just laminar
L
FD τw dA
FD τw w( x ) dx
w( x ) W
0
We also have
1
2 1
2 0.730
τw cf ρ U ρ U
2
2
Rex
Hence
1
2
FD ρ U W
2
x
L
L
0.730
U x
x
L
3
dx
0.730
2
L
ν
ρ U W
1 dx
L
2
0
ν
0
3
FD 0.365 ρ W ν L U
FD 3.15 10
Note: For two-sided solution
4
lbf
4
2 FD 6.31 10
lbf
Problem 9.43
[Difficulty: 3]
Plate is reversed from
this!
Given:
Triangular plate
Find:
Drag
Solution:
Basic equations:
cf
τw
1
2
2
ρ U
L 2 ft
From Table A.9 at 70 oF
cf
3
4 ft
2
ρ 0.00233
s
W 2 ft
ReL
U 15
U L
ν
ReL 2 10
5
so definitely laminar
FD τw w( x ) dx
w( x ) W 1
0
We also have
1
2 1
2 0.730
τw cf ρ U ρ U
2
2
Rex
Hence
1
2
FD ρ U W
2
L
0.730 1
U x
x
L
s
3
L
FD τw dA
ft
slug
ft
First determine the nature of the boundary layer
The drag (one side) is
Rex
L 1.732 ft
2
ν 1.63 10
0.730
0.730
2
dx
ρ U W ν
2
L
3
0
ν
x
1
2
x
L
2
x
dx
L
1
0
The integral is
L
0
x
1
2
3
1
2
2
2 L
4
x
2
d
2
L
x
L
L
3 L
3
1
3
FD 0.487 ρ W ν L U
FD 2.22 10
Note: For two-sided solution
The drag is much higher (twice as much) compared to Problem 9.42. This is because τ w is largest
near the leading edge and falls off rapidly; in this problem the widest area is also at the front
3
lbf
3
2 FD 4.43 10
lbf
Problem 9.42
Given:
Triangular plate
Find:
Drag
[Difficulty: 3]
Solution:
Basic
equations:
cf
τw
1
2
cf
2
ρ U
L 2 ft
3
0.730
Rex
L 1.732 ft
2
W 2 ft
U 15
ft
s
Assumptions:
(1) Parabolic boundary layer profile
(2) Boundary layer thickness is based on distance from leading edge (the "point" of the triangle).
From Table A.9 at 70 oF
ν 1.63 10
4 ft
2
ρ 0.00233
s
ft
ReL
First determine the nature of the boundary layer
The drag (one side) is
We also have
slug
U L
ν
3
ReL 2 10
5
so definitely laminar
L
FD τw dA
FD τw w( x ) dx
w( x ) W
0
x
L
1
2 1
2 0.730
τw cf ρ U ρ U
2
2
Rex
L
Hence
1
2 W
FD ρ U
2
L
L
1
0.730
0.730 x
W
2
2
dx
ρ U ν x dx
2
L
U x
0
3
ν
0
L
The integral is
1
3
2
2
2
x dx L
3
0
so
3
3
FD 0.243 ρ W ν L U
FD 1.11 10
lbf
Note: For two-sided solution
2 FD 2.21 10
3
lbf
Problem 9.41
[Difficulty: 2]
Given:
Data on fluid and plate geometry
Find:
Drag at both orientations using boundary layer equation
Solution:
The given data is
ρ 1.5
slug
ft
μ 0.0004
3
lbf s
ft
ReL
First determine the nature of the boundary layer
2
ρ U L
μ
U 10
ft
L 10 ft
s
ReL 3.75 10
b 3 ft
5
The maximum Reynolds number is less than the critical value of 5 x 105
Hence:
Governing equations:
cf
τw
1
2
cf
(9.22)
2
ρ U
0.730
(9.23)
Rex
L
The drag (one side) is
FD τw b dx
0
Using Eqs. 9.22 and 9.23
FD ρ U b
2
1
L
2
0.73
ρ U x
dx
μ
0
3
Repeating for
FD 0.73 b μ L ρ U
FD 5.36 lbf
L 3 ft
b 10 ft
3
FD 0.73 b μ L ρ U
FD 9.79 lbf
(Compare to 6.25 lbf for Problem 9.18)
(Compare to 12.5 lbf for Problem 9.19)
Problem 9.40
[Difficulty: 3]
Thin flat plate installed in a water tunnel. Laminar BL's with parabolic profiles form on both sides of the plate.
Given:
L 0.3 m
b 1 m
U 1.6
m
s
ν 1 10
2
6 m
u
s
U
2
y
y
δ δ
2
Total viscous drag force acting on the plate.
Find:
Solution:
We will determine the drag force from the shear stress at the wall
U L
First we will check the Reynolds number of the flow: ReL
ν
5
4.8 10
Therefore the flow is laminar throughout.
L
The viscous drag for the two sides of the plate is:
FD 2 τw b dx
The wall shear stress τw is:
0
2
at y = 0, which for the parabolic profile yields:
2 0 2 μ U
u
τw μ U
2
δ
δ
y
δ
τw μ
1
δ 5.48
The BL thickness δ is:
L
2
x Therefore: FD 2 b
U
L
ν
1
4
2 μ U
U
2
x
dx
dx
b μ U
1
5.48
ν 0
ν 2
5.48
x
U
0
Evaluating this integral:
FD
8 b μ U
5.48
U L
ν
FD 1.617 N
Problem 9.39
Given:
Parabolic solution for laminar boundary layer
Find:
Derivation of FD; Evaluate FD and θ L
[Difficulty: 2]
Solution:
Basic
equations:
u
U
2
y
δ
y
δ
L 9 in
Assumptions:
2
δ
x
b 3 ft
5.48
Rex
U 5
ft
ρ 1.94
s
slug
ft
p 0, and U = const
x
2) δ is a function of x only
3) Incompressible
4) Steady flow
3
1) Flat plate so
The momentum integral equation then simplifies to
τw
ρ
U θ
dx
d
2
where
The drag force is then
FD τw dA
For the given profile
θ
δ
0
θ
2
15
u
U
1
0
τw ρ U
dx
1
δ
θ
2 dθ
For U = const
From Table A.7 at 68 oF
u
dy
U
θL
L
L
2 dθ
2
ρ U b dx ρ U b 1 dθ
τw b dx
dx
0
0
1
0
1
2
FD ρ U b θL
u
u
2
2
2
2
3
4
1 dη 2 η η 1 2 η η dη 2 η 5 η 4 η η dη
U
U
15
0
0
δ
ν 1.08 10
δL L
5 ft
5.48
ReL
2
θL
δ
15 L
2
FD ρ U b θL
2
s
ReL
U L
ν
δL 0.0837 in
θL 0.01116 in
FD 0.1353 lbf
ReL 3.47 10
5
Problem 9.38
Given:
Parabolic solution for laminar boundary layer
Find:
Plot of δ, δ*, and τ w versus x/L
[Difficulty: 2]
Solution:
Given or available data:
Basic
equations:
u
U
2
y
ν 1.08 10
y
δ δ
2
5 ft
δ
x
2
s
(From Table A.8 at 68 oF) L 9 in
5.48
cf
Rex
τw
1
2
2
ρ U
U 5
ft
s
0.730
Rex
1
1
1
1 3
u y
u
2
2
Hence: * 1 dy 1 d 1 2 d
U
U
3 0 3
0
0
0
The computed results are from Excel, shown below:
Laminar Boundary Layer Profiles
δ (in)
0.000
0.019
0.026
0.032
0.037
0.042
0.046
0.050
0.053
0.056
0.059
0.062
0.065
0.067
0.070
0.072
0.075
0.077
0.079
0.082
0.084
δ * (in) τ w (psi)
0.000
0.006 0.1344
0.009 0.0950
0.011 0.0776
0.012 0.0672
0.014 0.0601
0.015 0.0548
0.017 0.0508
0.018 0.0475
0.019 0.0448
0.020 0.0425
0.021 0.0405
0.022 0.0388
0.022 0.0373
0.023 0.0359
0.024 0.0347
0.025 0.0336
0.026 0.0326
0.026 0.0317
0.027 0.0308
0.028 0.0300
0.09
0.16
0.08
0.14
0.07
0.12
δ
0.06
0.10
0.05
τw (psi)
0.00
0.45
0.90
1.35
1.80
2.25
2.70
3.15
3.60
4.05
4.50
4.95
5.40
5.85
6.30
6.75
7.20
7.65
8.10
8.55
9.00
Re x
0.00.E+00
1.74.E+04
3.47.E+04
5.21.E+04
6.94.E+04
8.68.E+04
1.04.E+05
1.22.E+05
1.39.E+05
1.56.E+05
1.74.E+05
1.91.E+05
2.08.E+05
2.26.E+05
2.43.E+05
2.60.E+05
2.78.E+05
2.95.E+05
3.13.E+05
3.30.E+05
3.47.E+05
δ and δ * (in)
x (in)
0.08
0.04
τw
0.03
0.02
0.06
0.04
δ*
0.02
0.01
0.00
0.00
0
3
6
x (in)
9
Problem 9.37
Given:
Blasius nonlinear equation
Find:
Blasius solution using Excel
[Difficulty: 5]
Solution:
The equation to be solved is
2
d3 f
d 3
f
d2 f
d 2
0
(9.11)
The boundary conditions are
f 0 and
df
0 at 0
d
df
1 at
d
Recall that these somewhat abstract variables are related to physically meaningful variables:
f
(9.12)
u
f
U
and
y
U
x
y
Using Euler’s numerical method
f n1 f n f n
(1)
f n1 f n f n
(2)
f n1 f n f n
h
In these equations, the subscripts refer to the nth discrete value of the variables, and = 10/N is the step
size for (N is the total number of steps).
But from Eq. 9.11
f
1
f f
2
so the last of the three equations is
1
f n1 f n f n f n
2
(3)
Equations 1 through 3 form a complete set for computing f , f , f . All we need is the starting condition
for each. From Eqs. 9.12
f 0 0 and f 0 0
We do NOT have a starting condition for f ! Instead we must choose (using Solver) f 0 so that the last
condition of Eqs. 9.12 is met:
f N 1
Computations (only the first few lines of 1000 are shown):
=
0.01
Make a guess for the first f ''; use Solver to vary it until f 'N = 1
Count
0
1
2
3
4
5
6
7
10
8
9
10
8
11
12
6
13
14
4
15
16
2
17
18
0
19
0.0 20
21
22
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.20
0.21
0.22
f
f'
f''
0.0000
0.0000
0.3303
0.0000
0.0033
0.3303
0.0000
0.0066
0.3303
0.0001
0.0099
0.3303
0.0002
0.0132
0.3303
0.0003
0.0165
0.3303
0.0005
0.0198
0.3303
Blasius
0.0007 Velocity
0.0231 Profile
0.3303
0.0009
0.0264
0.3303
0.0012
0.0297
0.3303
0.0015
0.0330
0.3303
0.0018
0.0363
0.3303
0.0022
0.0396
0.3303
0.0026
0.0429
0.3303
0.0030
0.0462
0.3303
0.0035
0.0495
0.3303
0.0040
0.0528
0.3303
0.0045
0.0562
0.3303
0.0051
0.0595
0.3303
0.0056
0.0628
0.3303
0.6
0.0063 0.4 0.0661
0.3302
0.0069
0.0694
0.3302
u/U = f '
0.0076
0.0727
0.3302
0.8
1.0
Problem 9.36
Given:
Data on flow over flat plate
Find:
Plot of laminar thickness at various speeds
Solution:
Given or available data:
Governing
Equations:
δ
x
5.48
2
5 m
ν 1.5 10
(9.21)
and
Rex
The critical Reynolds number is
[Difficulty: 2]
(from Table A.10 at 20oC)
s
U x
Rex
ν
δ 5.48
so
ν x
U
Recrit 500000
Hence, for velocity U the critical length xcrit is
x crit 500000
ν
U
The calculations and plot were generated in Excel and are shown below:
U (m/s)
x c rit (m)
1
7.5
2
3.8
3
2.5
4
1.9
5
1.5
10
0.75
x (m)
δ (mm)
δ (mm)
δ (mm)
δ (mm)
δ (mm)
δ (mm)
0.000
0.025
0.050
0.075
0.100
0.2
0.5
1.5
1.9
2.5
3.8
5.0
0.00
3.36
4.75
5.81
6.71
9.49
15.01
25.99
29.26
33.56
41.37
47.46
0.00
2.37
3.36
4.11
4.75
6.71
10.61
18.38
20.69
23.73
29.26
0.00
1.94
2.74
3.36
3.87
5.48
8.66
15.01
16.89
19.37
0.00
1.68
2.37
2.91
3.36
4.75
7.50
13.00
14.63
0.00
1.50
2.12
2.60
3.00
4.24
6.71
11.62
0.00
1.06
1.50
1.84
6.0
7.5
51.99
58.12
Laminar Boundary Layer Profiles
δ (mm)
70
60
U = 1 m/s
50
U = 2 m/s
U = 3 m/s
40
U = 4 m/s
30
U = 5 m/s
20
U = 10 m/s
10
0
0
2
4
x (m)
6
8
Problem 9.35
[Difficulty: 4]
Given:
Blasius solution for laminar boundary layer
Find:
Point at which u = 0.95U; Slope of streamline; expression for skin friction coefficient and total drag; Momentum
thickness
Solution:
Basic equation: Use results of Blasius solution (Table 9.1 on the web), and
f'
f'
u
U
u
U
at
η 3.5
0.9555
at
η 4.0
f' 0.95
From Table A.10 at 20oC
ν 1.50 10
Hence
y η
dy
dx
dy
dx
We have
Rex
U
0.9130
Hence by linear interpolation,
when
The streamline slope is given by
ν x
η y
η 3.5
2
5 m
U 5
and
s
ν x
v
u
2
m
( 0.95 0.9130)η 3.94
x 20 cm
s
u U f'
where
ν U
( 0.9555 0.9130)
y 0.305 cm
U
1
( 4 3.5)
x
( η f' f )
1
U f'
1
2
U x
ν
U x
v
and
( η f' f )
f'
1
2 Rex
1
2
ν U
x
( η f' f )
( η f' f )
f'
4
Rex 6.67 10
ν
From the Blasius solution (Table 9.1 on the web)
Hence by linear
interpolation
f 1.8377
at
η 3.5
f 2.3057
at
η 4.0
f 1.8377
dy
dx
The shear stress is
1
2 Rex
( 2.3057 1.8377)
( 4.0 3.5)
( η f' f )
f'
( 3.94 3.5)
f 2.25
0.00326
u v μ u at y = 0 (v = 0 at the wall for all x, so the derivative is zero there)
x
y
y
τw μ
2
U d f
τw μ U
2
ν x
dη
2
and at η = 0
d f
dη
2
0.3321
(from Table 9.1)
τw 0.3321 U
The friction drag is
2
ρ U μ
μ
ρ U
2
τw 0.3321 ρ U
0.3321
ρ U x
Rex
x
FD τw dA
L
τw b dx
where b is the plate width
0
L
L
2
ρ U
1
2 ν
dx
FD 0.3321
b dx 0.3321 ρ U
1
U
Rex
2
0
x
1
2 ν
2
FD 0.3321 ρ U
b 2 L
U
For the momentum integral
τw
2
ρ U
θL
We have
dθ
or
dx
2
FD ρ U b L
dθ
τw
2
0.6642
ReL
dx
ρ U
L
1 FD
0.6642 L
τw dx
2
2 b
ReL
ρ U 0
ρ U
1
L 1 m
θL
0
ReL
0.6642 L
ReL
U L
ν
ReL 3.33 10
θL 0.115 cm
5
Problem 9.34
[Difficulty: 3]
Given:
Find:
Blasius exact solution for laminar boundary layer flow
Solution:
We will apply the stream function definition to the Blasius solution.
(a) Prove that the y component of velocity in the solution is given by Eq. 9.10.
(b) Algebraic expression for the x component of a fluid particle in the BL
(c) Plot ax vs η to determine the maximum x component of acceleration at a given x
For Blasius: u U f'( η) and η y
U
1 ν U
From the stream function: v ψ
f ( η)
x
2 x
Thus
1 ν U f ( η)
v
2 x
ψ
The stream function is:
ν x
U ν x f ( η)
d f η But η 1 y U 1 η
2 x
2 x ν x
x
dη x
ν U x
d f 1 η 1 ν U ( η f'( η) f ( η) )
dη 2 x 2 x
ν U x
which is Eq. 9.10.
v
The acceleration in the x-direction is given by:
x
ax u
x
u v
η
1 η U f''( η)
d
d
f'( η) η U f''( η)
2
x
dη
2 x
dx
u U
y
u
y
where u U f'( η)
u U
1
2
ν U
x
( η f'( η) f ( η) )
Evaluating the partial derivatives:
U
d
d
f'( η) η U f''( η)
ν x
dη
dy
2
Therefore:
2
1 η U f'' 1 ν U
U
1 U
1 U
ax U f'( η)
( η f' f ) U f''
η f' f'' ( η f' f'' f f'')
ν x
2 x
2 x
2 x 2 x
Simplifying yields:
2
ax
U
2x
f ( η) f''( η)
If we plot f(η)f''(η) as a function of η:
f(η)f''(η)
The maximum value of this function is
0.23 at η of approximately 3.
0.2
2
axmax 0.115
0.1
0
0
1
2
3
η
4
5
U
x
Problem 9.33
[Difficulty: 3]
Given:
Blasius exact solution for laminar boundary layer flow
Find:
Solution:
Plot v/U versus y/δ for Rex 10
We will apply the stream function definition to the Blasius solution.
5
For Blasius: u U f'( η) and η y
U
ν x
The stream function is:
1 ν U
From the stream function: v ψ
f ( η)
x
2 x
Thus
1 ν U f ( η)
v
2 x
ψ
U ν x f ( η)
d f η But η 1 y U 1 η
2 x
2 x ν x
x
dη x
ν U x
d f 1 η 1 ν U ( η f'( η) f ( η) ) and
dη 2 x 2 x
ν U x
v
U
1
2
ν
U x
( η f'( η) f ( η) )
v
U
Since y δ at η 5 it follows that
y
δ
η
5
η f'( η) f ( η)
2 Rex
Plotting v/U as a function of y/δ:
Dimensionless Height y/δ
0.8
0.6
0.4
0.2
0
0
3
1 10
3
2 10
Dimensionless flow Velocity v/U
3
3 10
Problem 9.32
Blasius exact solution for laminar boundary layer flow
Given:
Find:
(a) Evaluate shear stress distribution
(b) Plot τ/τw versus y/δ
(c) Compare with results from sinusoidal velocity profile: u 2 y
δ
U
y
δ
2
We will apply the shear stress definition to both velocity profiles.
Solution:
Governing
Equation:
τ μ
y
τ
2
ρ U
μ
ρ U
From the above equation:
f''( η)
τ
τw
For the parabolic profile: τ
(Shear stress in Newtonian fluid)
u
U
The shear stress is: τ μ ( U f'( η) ) U μ ( f''( η) ) η U μ f''( η)
ν x
ν x
y
y
U
For Blasius: u U f'( η) and η y
Therefore:
[Difficulty: 3]
U
ν x
f''( η)
f''( 0 )
f''( η)
τ is proportional to f''(η)
Rex
f''( η)
0.33206
Since y δ at η 5 it follows that
μ U d
u μ U 2 2 y
δ
δ
δ
U
y
d
δ
τw
μ U
δ
2
y
δ
Thus:
η
5
τ
τw
1
y
δ
Both profiles are plotted here:
Dimensionless Height y/δ
0.8
0.6
0.4
0.2
Parabolic
Blasius
0
0
0.2
0.4
0.6
Dimensionless Shear Stress τ/τw
0.8
Problem 9.31
Blasius exact solution for laminar boundary layer flow
Given:
Find:
(a) Evaluate shear stress distribution
(b) Plot τ/τw versus y/δ
π y
u
(c) Compare with results from sinusoidal velocity profile:
sin
U
2 δ
We will apply the shear stress definition to both velocity profiles.
Solution:
Governing
Equation:
τ μ
y
τ
2
ρ U
μ
ρ U
From the above equation:
U
f''( η)
τ
τw
For the sinusoidal profile: τ
ν x
(Shear stress in Newtonian fluid)
u
For Blasius: u U f'( η) and η y
Therefore:
[Difficulty: 3]
U
The shear stress is: τ μ ( U f'( η ) ) U μ ( f''( η ) ) η U μ f''( η )
ν x
νx
y
y
U
f''( η)
f''( 0 )
f''( η)
τ is proportional to f''(η)
Rex
f''( η)
0.33206
Since y δ at η 5 it follows that
μ U d
u μ U π cos π y
δ
δ 2
U
y
2 δ
d
δ
τw
μ U π
δ 2
y
δ
η
5
Thus:
τ
τw
π y
2 δ
cos
Both profiles are plotted here:
Dimensionless Height y/δ
0.8
0.6
0.4
0.2
Sinusoidal
Blasius
0
0
0.2
0.4
0.6
Dimensionless Shear Stress τ/τw
0.8
Problem 9.30
[Difficulty: 2]
Given:
Find:
Blasius exact solution for laminar boundary layer flow
Solution:
The Blasius solution is given in Table 9.1; it is plotted below.
u
Plot and compare to parabolic velocity profile:
U
2
y
δ
y
δ
2
Parabolic
Blasius
Dimensionless Height y/δ
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
Dimensionless Velocity u/U
0.8
Problem 9.29
[Difficulty: 5]
Given:
Air flow in laboratory wind tunnel test section.
Find:
(a) Displacement thickness at station 2
(b) Pressure drop between 1 and 2
(c) Total drag force caused by friction on each wall
We will apply the continuity, x-momentum, and Bernoulli equations to this problem.
Solution:
Governing
Equations:
δdisp
infinity
0
Assumptions:
1 u dy
U
δ
0
1 u dy
U
(Definition of displacement thickness)
dV V dA 0
CS
t CV
2
p V
gz const
2
(Continuity)
udV uV dA Fsx Fbx
CS
t CV
(x- Momentum)
(1) Steady, incompressible flow
(2) No body forces in the x-direction
(3) No viscous forces outside boundary layer
(4) Streamline exists between stations 1 and 2
If we divide both sides of the displacement thickness definition by δ, we get:
(Bernoulli)
(5) Uniform flow outside boundary layer
(6) Boundary layer is the same on all walls
(7) Neglect corner effects
(8) Constant elevation between 1 and 2
δdisp
δ
1
δ
δ
0
However, we can change the variable of integration to η = y/δ, resulting in: dη
1
δ
1 u dy
U
dy Therefore:
δdisp
δ
1
0
1 u dη
U
1
1
u
If we assume the power law profile (turbulent BL):
Evaluating this integral:
δdisp
δ
1
7
8
1
7
η
U
Into the displacement thickness:
2
H
U2 U1
H 2 δdisp2
1
7
1 η dη
δdisp2 2.54 mm
2
U1 A1 U2 A2 or U1 H U2 H 2 δdisp2
After applying the assumptions from above, continuity reduces to:
Solving for the speed at 2:
1
δdisp2
20.3 mm
8
So the displacement thickness is:
8
δdisp
δ
0
2
m
U2 50.2
s
Substituting known values:
305
305
2
2.54
2
m
U2 51.9
s
From Bernoulli equation, since z = constant:
p1
ρ
Δp12
U U2
2 1
ρ
2
2
Δp12
1
2
1.23
U1
2
kg
3
2
p2
ρ
U2
2
2
2
50.2 51.9
along a streamline. Therefore:
2
2 m
m
2
2
s
N s
Δp12 106.7 Pa
kg m
To determine the drag on the walls, we choose the control volume shown above and apply the x-momentum equation.
From the assumptions, the equation reduces to:
CS
uV dA Fsx
Applying this to the control volume:
δ
2
p 1 H δ2 FD p 2 H δ2 U1 ρ U1 H δ2 Uavg mtop u ρ u H dy
0
The mass flow rate through the top of the CV
δ
2
mtop m1 m2 ρ U1 H δ2 ρ u H dy
can be determined using the continuity equation across the control volume:
0
1
1
7
7
This integral can be evaluated using the power law profile: ρ u H dy ρ U2 H δ2 η dη ρ U2 H δ2 Therefore:
8
0
0
7
U1 U2
mtop ρ H δ2 U1 U2
The
average
speed
can
be
approximated
as
the
mean
of
the
speeds
at
1
and
2:
U
8
avg
2
δ2
Finally the integral in the momentum equation may also be evaluated using the power law profile:
1
2
7
2
2
7
u ρ u H dy ρ U2 H δ2 η dη ρ U2 H δ2
9
0
0
δ2
p 1 H δ2 FD p 2 H δ2 U1 ρ U1 H δ2
U1 U2
2
Thus, the momentum equation may be rewritten as:
ρ H δ2 U1
7
8
U2
7
9
2
ρ U2 H δ2Solving for the drag force:
FD
p 1 p 2 ρ U1 2
U1 U2
7
7
U1 U2 U2 2 H δ2 Substituting in all known values yields:
8
2
9
N
kg
FD 106.7
1.23
2
3
m
m
2
2
2
2
50.2 m ( 50.2 51.9) 50.2 7 51.9 m 7 51.9 m N s 0.305 m 0.020
s
8
s kg m
2
s2 9
FD 2.04 N
The viscous drag force acts on the CV in the direction shown. The viscous
drag force on the wall of the test section is equal and opposite:
Problem 9.28
Given:
Data on fluid and boundary layer geometry
Find:
Gage pressure at location 2; average wall stress
[Difficulty: 3]
Solution:
The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it
reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the
boundary layer; an easier approach is to simply use the displacement thickness!
The average wall stress can be estimated using the momentum equation for a CV
The given and available (from Appendix A) data is
ρ 0.00234
slug
ft
3
ft
U1 50
s
L 20 ft
D 15 in
δ2 4 in
Governing equations:
Mass
Momentum
Bernoulli
p
ρ
2
V
g z constant
2
(4.24)
Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction
The displacement thickness at location 2 can be computed from boundary layer thickness using Eq. 9.1
1
δdisp2
δ2
0
Hence
u
1 dy δ
2
U
0
δ2
δdisp2
8
Applying mass conservation at locations 1 and 2
1
δ2
7
1 η dη
8
δdisp2 0.500 in
ρ U1 A1 ρ U2 A2 0
π 2
A1 D
4
A1
U2 U1
A2
or
A1 1.227 ft
2
The area at location 2 is given by the duct cross section area minus the displacement boundary layer
π
2
A2 D 2 δdisp2
4
A2 1.069 ft
2
Hence
A1
U2 U1
A2
ft
U2 57.4
s
For the pressure change we can apply Bernoulli to locations 1 and 2 to find
Hence
ρ
2
2
p 1 p 2 Δp U2 U1
2
Δp 6.46 10
p 2 ( gage ) p 1 ( gage ) Δp
p 2 6.46 10
3
psi
3
p 2 Δp
psi
For the average wall shear stress we use the momentum equation, simplified for this problem
D
2
2
2 π
2
Δp A1 τ π D L ρ U1 A1 ρ U2 D 2 δ2 ρ
4
D
2
2
2 π r u dr
δ2
1
where
y
u ( r) U2
δ2
7
r
and
D
2
y
dr dy
0
2
2
2
ρ
2 π r u dr 2 π ρ U2
D δ
δ
2
D
The integral is
2
2
7
D y y dy
2
δ2
2
D
2
ρ
D
2
Hence
τ
δ2
D δ2
2
2
2 π r u dr 7 π ρ U2 δ2
8
9
D δ2
2
2 π
2
2
Δp A1 ρ U1 A1 ρ U2 D 2 δ2 7 π ρ U2 δ2
8
4
9
τ 6.767 10
π D L
5
psi
Problem 9.27
[Difficulty: 3]
Given:
Find:
Air flow in laboratory wind tunnel test section.
Solution:
We will apply the continuity and Bernoulli equations to this problem.
(a) Freestream speed at exit
(b) Pressure at exit
δdisp
Governing
Equations:
infinity
0
1 u dy
U
δ
0
1 u dy
U
(Definition of displacement thickness)
dV V dA 0
CS
t CV
2
p V
gz const
2
(Continuity)
(Bernoulli)
(1) Steady, incompressible flow
(2) No body forces in the x-direction
(3) No viscous forces outside boundary layer
(4) Streamline exists between stations 1 and 2
Assumptions:
(5) Uniform flow outside boundary layer
(6) Boundary layer is the same on all walls
(7) Neglect corner effects
(8) Constant elevation between 1 and 2
If we divide both sides of the displacement thickness
definition by δ, we get:
δdisp
δ
1
δ
δ
0
1
dy
U
dη
Therefore:
δ
1
0
1
δdisp
δ
0
1
7
1 η dη
1
δdisp1
0.8 in
8
W = 1 ft
L = 2 ft
However, we can change the variable of integration
to η = y/δ, resulting in:
δdisp
U1 = 80 ft/s
u
1
δ
dy
1
1 u dη
u
7
For the power law profile:
η
U
U
Evaluating this integral:
δdisp
δ
δdisp1 0.100 in
1
7
8
1
8
1
δdisp2
1 in
8
Into the displacement thickness:
So the displacement thicknesses are:
δdisp2 0.125 in
2
W 2 δdisp1
U2 U1
Substituting known values:
W 2 δdisp2
Solving for the speed at 2:
2 U2 W 2 δdisp22
U1 A1 U2 A2 or U1 W 2 δdisp1
After applying the assumptions from above, continuity reduces to:
ft
U2 80
s
2
1 2 0.100
1 2 0.125
ft
U2 91.0
s
From Bernoulli equation, since z = constant:
p1
ρ
Δp12
U U2
2 1
2
ρ
2
Δp12
1
2
U1
2
0.00239
2
slug
ft
3
p2
U2
ρ
2
2
2
along a streamline. Therefore:
80 91
2 ft
2
2
s
2
lbf s
slug ft
ft
12 in
2
Δp12 0.01561 psi
From ambient to station 1 we see a loss at the tunnel entrance:
2
2
p
U0 p 1
U1
0
ρ 2 ρ 2 h lT Since U0 0 and p 0 p atm 0 we can solve for the pressure at 1:
p 1 ρ h lT
1
2
ρ U1
2
where
ρhlT
0.3
12
ft 1.94
slug
ft
3
32.2
ft
2
2
s
2
2
lbf
1
slug
ft
lbf s
Therefore: p 1 0.01085
0.00239
80
2
3
2
slug ft
s
in
ft
p 2 p 1 Δp12
it follows that:
lbf s
slug ft
ft
12 in
2
ρhlT 0.01085 psi
2
ft 0.0640 psi So the pressure at 2 is:
12 in
p 2 0.0640 psi 0.01561 psi 0.0796 psi Since the pressure drop can be expressed as
h2
p2
ρ g
lbf
So in terms of water height: h 2 0.0796
2
in
2
3
p 2 ρ g h 2
2
s
slug ft 12 in
12 in ft
1.94 slug 32.2 ft
2
ft
ft
lbf s
p 2 0.0796 psi
h 2 2.20 in
Problem 9.26
[Difficulty: 3]
Given:
Find:
Developing flow of air in flat horizontal duct. Assume 1/7-power law velocity profile in boundary layer.
Solution:
We will apply the continuity and x-momentum equations to this problem.
(a) Displacement thickness is 1/8 times boundary layer thickness
(b) Static gage pressure at section 1.
(c) Average wall shear stress between entrance and section 2.
Governing
Equations:
δdisp
infinity
δ
1 u dy
U
0
0
dV V dA 0
1 u dy
U
(Definition of displacement thickness)
CS
t CV
ud
V
u
V
CS dA Fsx Fbx
t CV
Assumptions:
(Continuity)
(x- Momentum)
(1) Steady, incompressible flow
(2) No body forces in the x-direction
(3) No viscous forces outside boundary layer
(4) Boundary layers only grow on horizontal walls
L = 20 ft
H = 1 ft
V1 = 40 ft/s
δ 2 = 4 in
δdisp
If we divide both sides of the displacement thickness definition by δ, we get:
δ
1
δ
δ
0
However, we can change the variable of integration to η = y/δ, resulting in:
dη
1
δ
dy
1 u dy
U
Therefore:
δdisp
δ
1
0
1
1
For the power law profile:
u
U
η
7
Into the displacement thickness:
δdisp
δ
0
Evaluating this integral:
δdisp
δ
1 u dη
U
1
7
1 η dη
1
7
8
1
δdisp
8
δ
1
8
V1 A1 V2 A2 or
After applying the assumptions from above, continuity reduces to:
Solving for the velocity at 2:
H
V2 V1
V1
H 2 δdisp2
H
H
p0
ρ
1
2
p 1g p 1 p 0 ρ V1
2
p 1g
1
2
p 2g p 2 p 0 ρ V2
2
p 2g
1
2
1
2
Substituting known values:
4
2
p
V
ρ
0.00234
2
slug
ft
0.00234
3
slug
ft
3
δ2
ft
V2 40 1 ft
s
From Bernoulli equation, since z = constant:
V1 w H V2 w H 2 δdisp2
1
1
1
4 ft
1
4 12
ft
V2 43.6
s
along a streamline. Therefore:
40
43.6
2
2
lbf s
ft
s
slug ft 12 in
ft
2
2
2
lbf s
ft
s
slug ft 12 in
ft
p 1g 0.01300 psi
2
Now if we apply the momentum equation to the control volume (considering the assumptions shown):
p 2g 0.01545 psi
Fsx uV dA
CS
H
p1 p2 w 2
H
τ w L V1 ρ V1 w
2
δ2
0
H
u ρ u w dy V2 ρ V2 δ2 w
2
1
2
2
2
2 7
7
The integral is equal to: ρ w u dy ρ V2 δ2 w η dη ρ V2 δ2 w
9
0
0
δ2
H
p1 p2 w 2
τ
1
L
Therefore the momentum equation becomes:
2
2 H
w ρ V2 δ2 w Simplifying and solving for the shear stress we get:
2
2
9
2 H
τ w L ρ V1
H
2
2 H
2 H
p 1 p 2 ρ V1 V2 δ2 Substituting in known values we get:
2
2
2
9
2
2
2
2
lbf 1 ft
slug
ft 1 ft
ft 1
2 4 lbf s ft
τ
[ ( 0.01328 ) ( 0.01578 ) ]
0.00234
40
43.6 ft
2 2
3
20 ft
s 2
s 2
9 12 slug ft 12 in
in
ft
1
5
τ 5.46 10
psi
Problem 9.25
Given:
Data on wind tunnel and boundary layers
Find:
Pressure change between points 1 and 2
[Difficulty: 2]
Solution:
Basic
equations
(4.12)
p
ρ
2
V
2
g z const
Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal
For this flow
ρ U A const
The given data is
ft
U0 100
s
We also have
δdisp2 0.035 in
Hence at the Point 2
A2 h 2 δdisp2
U1 U0
h 3 in
2
p1
ρ
Hence
Δp
U U2
2 1
ρ
2
2
A1 9 in
2
The pressure change is found from Bernoulli
2
A2 8.58 in
Applying mass conservation between Points 1
and 2
ρ U1 A1 ρ U2 A2 0
A1 h
o
r
U1
2
2
2
The pressure drops by a small amount as the air accelerates
p2
ρ
U2
2
A1
U2 U1
A2
ft
U2 105
s
wit
h
ρ 0.00234
2
slug
ft
3
Δp 8.05 10
psi
Δp 1.16
lbf
ft
2
3
Problem 9.24
[Difficulty: 2]
Given:
Data on wind tunnel and boundary layers
Find:
Uniform velocity at Point 2; Change in static pressure through the test section
Solution:
Basic
equations
(4.12)
δdisp
δ
1 u dy
U
p
ρ
2
V
2
g z const
0
Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal
1
u
ρ U A const
and
The given data is
m
U1 20
s
W 40 cm
We also have
δ1 1 cm
δ2 1.3 cm
For this flow
Hence
δdisp
δ
0
U
1 u dy
U
δ
1
1
7
y
dy
δ
1
y
δ
7
2
2
A W
1
7
1 η dη
A 0.1600 m
η
where
0
0
Hence at the inlet and exit
δ1
δdisp1
δdisp1 0.125 cm
8
Hence the areas are
δ
δ2
δdisp2
8
y
δ
δ
δdisp
8
δdisp2 0.1625 cm
2
2
A2 W 2 δdisp2
2
A1 W 2 δdisp1
A1 0.1580 m
2
A2 0.1574 m
Applying mass conservation between Points 1 and 2
ρ U1 A1 ρ U2 A2 0
The pressure change is found from Bernoulli
p1
ρ
Hence
Δp
ρ
2
U1 U2
2
2
or
U1
2
2
p2
ρ
U2
2
A1
U2 U1
A2
m
U2 20.1
s
with
ρ 1.21
2
kg
3
m
4
Δp 2.66 10
psi
Δp 1.835 Pa
Problem 9.23
[Difficulty: 2]
Given:
Data on wind tunnel and boundary layers
Find:
Uniform velocity at exit; Change in static pressure through the test section
Solution:
Basic
equations
δdisp
(4.12)
δ
1 u dy
U
p
ρ
2
V
2
g z const
0
Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal
1
For this flow
The given data is
We also have
Hence
δdisp
δ
0
u
an
d
U
m
U1 25
s
δ1 20 mm
h 25 cm
A h
1 u dy
U
δ
y
δ
ρ U A const
7
2
A 625 cm
2
δ2 30 mm
1
1
7
y
dy
δ
δ
1
1
7
1 η dη
0
0
η
wher
e
y
δ
δ
δdisp
8
Hence at the inlet and exit
δ1
δdisp1
8
Hence the areas are
δdisp1 2.5 mm
2
2
A2 h 2 δdisp2
δ2
δdisp2
8
δdisp2 3.75 mm
A1 h 2 δdisp1
A1 600 cm
2
A2 588 cm
2
Applying mass conservation between Points 1 and 2
ρ U1 A1 ρ U2 A2 0
p1
The pressure change is found from Bernoulli
ρ
Hence
Δp
U U2
2 1
ρ
2
2
A1
U2 U1
A2
or
U1
2
2
p2
Δp 15.8 Pa
ρ
U2
2
m
U2 25.52
s
2
with
ρ 1.21
kg
3
m
The pressure drops slightly through the test section
Problem 9.22
[Difficulty: 2]
Given:
Data on boundary layer in a cylindrical duct
Find:
Velocity U2 in the inviscid core at location 2; Pressure drop
Solution:
The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it
reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the
boundary layer; an easier approach is to simply use the displacement thickness!
The given or available data (from Appendix A) is
ρ 1.23
kg
m
U1 12.5
s
3
m
D 100 mm
δ1 5.25 mm
δ2 24 mm
Governing
equations: Mass
p
Bernoulli
ρ
2
V
2
g z constant
(4.24)
The displacement thicknesses can be computed from boundary layer thicknesses using Eq. 9.1
1
u
δdisp
1 dy δ
U
0
1
δ
7
1 η dη
δ
8
0
Hence at locations 1 and 2
δ1
δdisp1
8
Applying mass conservation at locations 1 and 2
δdisp1 0.656 mm
δ2
δdisp2
8
δdisp2 3 mm
ρ U1 A1 ρ U2 A2 0
A1
U2 U1
A2
or
The two areas are given by the duct cross section area minus the displacement boundary layer
π
2
A1 D 2 δdisp1
4
Hence
A1 7.65 10
3
2
m
A1
U2 U1
A2
For the pressure drop we can apply Bernoulli to locations 1 and 2 to find
π
2
A2 D 2 δdisp2
4
A2 6.94 10
m
U2 13.8
s
ρ
2
2
p 1 p 2 Δp U2 U1 Δp 20.6 Pa
2
3 2
m
Problem 9.21
[Difficulty: 2]
Given:
Data on wind tunnel and boundary layers
Find:
Displacement thickness at exit; Percent change in uniform velocity through test section
Solution
:The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary
layer grows it reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to
compute the mass flow in the boundary layer; an easier approach is to simply use the displacement thickness!
Basic
equations
δdisp
(4.12)
δ
1 u dy
U
0
Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal
1
u
7
and
The design data is
ft
Udesign 160
s
w 1 ft
The volume flow rate is
Q Udesign Adesign
Q 160
δin 0.4 in
δexit 1 in
We also have
Hence
δdisp
δ
0
U
1 u dy
U
δ
1
y
δ
ρ U A const
For this flow
h 1 ft
1
7
y
dy
δ
Adesign w h
ft
Adesign 1 ft
3
s
δ
0
1
1
7
1 η dη where
0
η
y
δ
Hence at the inlet and exit
δin
δdispin
8
δdispin 0.05 in
δdispexit
δexit
8
δdispexit 0.125 in
δ
δdisp
8
2
Hence the areas are
Ain w 2 δdispin h 2 δdispin
Ain 0.9834 ft
Aexit w 2 δdispexit h 2 δdispexit
2
Aexit 0.9588 ft
2
Applying mass conservation between "design" conditions and the inlet
ρ Udesign Adesign ρ Uin Ain 0
or
Also
Uin Udesign
Adesign
Ain
Adesign
Uexit Udesign
Aexit
ft
Uin 162.7
s
ft
Uexit 166.9
s
The percent change in uniform velocity is then
Uexit Uin
Uin
2.57 % The exit displacement thickness is
δdispexit 0.125 in
Problem 9.20
[Difficulty: 3]
δ = 1 in
Flow over a flat plate with parabolic laminar boundary layer profile
Given:
Find:
(a) Mass flow rate across ab
(b) x component (and direction) of force needed to hold the plate in place
We will apply the continuity and x-momentum equations to this system.
Solution:
Governing
(Continuity)
d
V
V
dA 0
Equations:
CV
CS
t
(x- Momentum)
udV uV dA Fsx Fbx
CS
t CV
(1) Steady flow
Assumptions:
(2) No net pressure forces
(3) No body forces in the x-direction
(4) Uniform flow at da
CV
d
ρ U b δ ρ u b dy mab 0
From the assumptions, the continuity equation becomes:
c
Rx
δ
The integral can be written as:
0
δ
δ
ρ u b dy ρ b u dy ρ U b δ
0
0
1
2η η2 dη
where η
0
y
δ
This integral is equal to: ρ U b δ 1
1
2
ρ U b δ
3 3
2
1
mab ρ U b δ ρ U b δ ρ U b δ Substituting known values:
3
3
Solving continuity for the mass flux through ab we get:
1
slug
ft
ft
mab
1.5
10 3.0 ft 1 in
3
s
12 in
3
ft
slug
mab 1.250
s
δ
From the assumptions, the momentum equation becomes:
Rx u da ( ρ U b δ) u ab mab u ρ u b dy where u da u ab U
0
1
2
2
Thus: Rx ρ U b δ ρ U b δ
3
δ
2
2
u ρ u b dy ρ U b δ
3
0
δ
δ
2
2
u ρ u b dy ρ b u dy ρ U b δ
0
0
ρ U b δ
2
4
3
1
1
2 η η
2
0
2
δ
u ρ u b dy The integral can be written as:
0
dη ρ U b δ
2
1
4η2 4η3 η4 dη
This integral is equal to:
0
1
8
2
2
2
8 2
ρ U b δ Therefore the force on the plate is: Rx ρ U b δ ρ U b δ
5 15
15
15 3
Substituting known values:
Rx
2
15
1.5
slug
ft
3
10
This force must be applied to the control volume by the plate.
ft
s
2
3.0 ft 1 in
ft
12 in
2
lbf s
slug ft
Rx 5.00 lbf
(to the left)
Problem 9.19
[Difficulty: 3]
Given:
Data on fluid and boundary layer geometry
Find:
Mass flow rate across ab; Drag; Compare to Problem 9.18
Solution:
The given data is
Governing
equations:
ρ 1.5
slug
ft
U 10
3
ft
L 3 ft
s
δ 0.6 in
b 10 ft
Mass
Momentum
Assumptions:
(1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a
Applying these to the CV abcd
δ
Mass
For the boundary layer
( ρ U b δ) ρ u b dy mab 0
u
U
y
δ
0
dy
η
δ
dη
1
Hence
1
mab ρ U b δ ρ U η δ dy ρ U b δ ρ U b δ
2
0
1
mab ρ U b δ
2
slug
mab 3.75
s
δ
Momentum
Rx U ( ρ U δ) mab u ab u ρ u b dy
0
u ab U
Note that
and
1
δ
2
2
u ρ u b dy ρ U b δ η dη
0
1
0
1
2
2
2
Rx ρ U b δ ρ U b δ U ρ U b δ η dy
2
0
2
Rx ρ U b δ
1
2
1
2
Rx ρ U b δ
6
2
ρ U δ
1
3
2
ρ U δ
Rx 12.50 lbf
We should expect the drag to be larger than for Problem 9.18 because the viscous friction is mostly concentrated near the leading
edge (which is only 3 ft wide in Problem 9.18 but 10 ft here). The reason viscous stress is highest at the front region is that the
boundary layer is very small (δ <<) so τ = μdu/dy ~ μU/δ >>
Problem 9.18
Given:
Data on fluid and boundary layer geometry
Find:
Mass flow rate across ab; Drag
[Difficulty: 3]
CV
Solution:
The given data is
ρ 1.5
slug
ft
Governing
equations:
U 10
3
ft
s
d
L 10 ft δ 1 in b 3 ft
c
Rx
Mass
Momentum
Assumptions:
(1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a
Applying these to the CV abcd
δ
Mass
( ρ U b δ) ρ u b dy mab 0
0
For the boundary layer
u
U
y
δ
dy
η
δ
dη
1
Hence
1
mab ρ U b δ ρ U η δ dy ρ U b δ ρ U b δ
2
0
1
mab ρ U b δ
2
slug
mab 1.875
s
δ
Momentum
Rx U ( ρ U δ) mab u ab u ρ u b dy
0
u ab U
Note that
and
1
δ
2
2
u ρ u b dy ρ U b δ η dη
0
2
Rx ρ U b δ
0
1
2
2
ρ U b δ U ρ U b δ η dy
2
1
0
2
Rx ρ U b δ
1
2
2
ρ U δ
1
3
2
ρ U δ
1
2
Rx ρ U b δ
6
Rx 6.25 lbf
We are able to compute the boundary layer drag even though we do not know the viscosity because it is the viscosity
that creates the boundary layer in the first place
Problem 9.17
Given:
Find:
Solution:
[Difficulty: 2]
Power law and parabolic velocity profiles
The displacement and momentum thicknesses expressed as δ*/δ and θ/δ for each
profile
We will apply the definition of the displacement and momentum thickness to each profile.
δdisp
Governing
Equations:
infinity
0
θ
δ
u
U
1
0
1 u dy
U
δ
0
1 u dy
U
u
dy
(Definition of momentum thickness)
U
If we divide both sides of the equations by δ, we get:
δdisp
δ
1
δ
δ
0
δ
1
0
1 u dη
U
δ
1
u
θ
U
0
For the power law profile:
U
η
7
dη
δdisp
δ
0
δdisp
δ
Into the momentum thickness:
δ
1
dy
1
u
dy
U
Therefore:
U
Evaluating this integral:
1
U
0
1
1
1
1
θ
7
7
η 1 η dη
δ 0
0
u
dη
Into the displacement thickness:
δ
θ
u
1
1
u
1
δ
δ
1 u dy
U
However, we can change the variable of integration to η = y/δ, resulting in:
δdisp
(Definition of displacement thickness)
1
7
1 η dη
1
2
1
7
7
η η dη
7
8
1
δdisp
8
δ
Evaluating this integral:
θ
δ
θ
δ
For the parabolic profile:
u
U
2 η η
2
Into the displacement thickness:
δdisp
δ
1
0
1 2 η η2 dη
1
0
1 2η η2 dη
7
8
0.1250
7
9
0.0972
7
72
Evaluating this integral:
δdisp
δ
Into the momentum thickness:
11
1
1
3
1
δdisp
3
δ
1
0.3333
2
2
2
3
4
2 η η 1 2 η η dη 2 η 5 η 4 η η dη
δ 0
0
θ
Evaluating this integral:
θ
δ
1
5
3
1
Profile
1
5
2
θ
15
δ
Power Law
δdisp
0.1250 δ
θ
0.0972 δ
Parabolic
0.3333 δ
0.1333 δ
0.1333
Problem 9.16
Given:
Find:
Solution:
[Difficulty: 2]
Linear, sinusoidal, and parabolic velocity profiles
The displacement thickness expressed as δ*/δ for each profile
We will apply the definition of the displacement thickness to each profile.
δdisp
Governing
Equation:
infinity
0
1 u dy
U
δ
0
1 u dy
U
If we divide both sides of the equation by δ, we get:
(Definition of displacement thickness)
1
δ
δdisp
δ
δ
0
dη
the variable of integration to η = y/δ, resulting in:
For the linear profile:
δdisp
δ
u
U
Evaluating this integral:
δ
1
0
u
δ
1
0
U
sin
π
2
η
Therefore:
δ
u
U
2 η η
2
1
0
1
1
2
1 u dη
U
1
δdisp
2
δ
0.5000
Into the displacement thickness:
1 sin π η dη Evaluating this integral:
2
For the parabolic profile:
δdisp
δdisp
δ
0
For the sinusoidal profile:
δdisp
δ
dy
However, we can change
δdisp
η Into the displacement thickness:
1
( 1 η ) dη
1
1 u dy
U
δdisp
δ
1
2
δdisp
π
δ
0.3634
Into the displacement thickness:
1
2
1 2 η η2 dη
1 2 η η dη
Evaluating this integral:
0
δdisp
δ
11
1
3
1
δdisp
3
δ
0.3333
Problem 9.15
Given:
Find:
Solution:
[Difficulty: 2]
Linear, sinusoidal, and parabolic velocity profiles
The momentum thickness expressed as θ/δ for each profile
We will apply the definition of the momentum thickness to each profile.
θ
Governing
Equation:
δ
u
U
1
0
u
dy
(Definition of momentum thickness)
U
1
δ
δ
δ
θ
If we divide both sides of the equation by δ, we get:
u
U
0
the variable of integration to η = y/δ, resulting in:
For the linear profile:
u
U
dη
1
δ
dy
1
u
For the sinusoidal profile:
U
sin
π
2
η
0
1
0
θ
δ
δ
u
U
2
π
0
θ
δ
1
2
2
1
3
u
U
1
u
dη
U
1
θ
6
δ
0.1667
2
π
π
sin η sin η dη
2 2
2 π
π 4
2 η η
1
θ
Into the momentum thickness:
1
π
π
θ
sin η 1 sin η dη
δ
2
2
1
dy However, we can change
U
Therefore:
1
2
η ( 1 η) dη η η dη Evaluating this integral:
δ 0
0
For the parabolic profile:
u
η Into the momentum thickness:
θ
Evaluating this integral:
1
θ
δ
0.1366
Into the momentum thickness:
1
2
2
2
3
4
2 η η 1 2 η η dη 2 η 5 η 4 η η dη
δ 0
0
θ
Evaluating this integral:
θ
δ
1
5
3
1
1
5
2
θ
15
δ
0.1333
Problem 9.14
Given:
Find:
Solution:
[Difficulty: 2]
Power law velocity profiles
Plots of y/δ vs u/U for this profile and the parabolic profile of Problem 9.10
Here are the profiles:
Boundary Layer Velocity Profiles
Power
Parabolic
Dimensionless Distance y/δ
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
Dimensionless Velocity u/U
Note that the power law profile gives and infinite value of du/dy as y approaches zero:
du U d u U U y
dy d y 7
6
7
as
y0
Problem 9.13
Given:
Laminar boundary layer profile
Find:
If it satisfies BC’s; Evaluate */ and /
[Difficulty: 3]
Solution:
The boundary layer equation is
u
y
2
U
u
y
2 2
U
u 0 0
The BC’s are
At y = 0
At y =
du
dy
0 y
2 1
2
2
y for which u = U at y =
0
y
u
2 0 0
U
du
1
U 2 2
0 so it fails the outer BC.
dy
y
This simplistic distribution is a piecewise linear profile: The first half of the layer has velocity gradient
second half has velocity gradient
U
1.414
U
, and the
2 2 U 0.586 U . At y = , we make another transition to zero velocity gradient.
For *:
u
u
* 1 dy 1 dy
U
U
0
0
Then
1
1
* 1 u
u y
u
1 dy 1 d 1 d
0 U
U 0 U
0
u
1
2
0
U
2
u
2 2 2 1
U
with
2
1
1
2
Hence
*
u
1 d
U
0
1
1
12
0
2 d
1 2 2
1
12
1
2 1 d
2 2
12
1
2
2 1 1
2
0
2
1
2 2
1 2
2 1
2 3
2
* 1
0.396
2
8
4
8
4
4
For :
u
u
u
u
1 dy 1 dy
U U
U U
0
0
Then
1 u
0 U
u
u
1 dy
U
U
0
1
u y
u
1 d
U 0 U
1
u
1 d
U
Hence, after a LOT of work
u
u
1 d
0U U
1
12
2 1 2 d
0
12
2 1
1
2 2
2
3
3
0
2 2
1
2 1 1 2 2
2 1 d
12
1
1
2 1
2
2 1
2
2 2 1 2 2 1
0.152
2
8 12 24
6 12
1 2
Problem 9.12
Given:
Laminar boundary layer profile
Find:
If it satisfies BC’s; Evaluate */ and /
[Difficulty: 2]
Solution:
3
4
The boundary layer equation is
u
y
y y
2 2 for which u = U at y =
U
The BC’s are
u 0 0
0
y
u
3
4
20 20 0 0
U
1
1
2
3
du
y2
y3
U 2 6 3 4 4
U 2 6 3 4 4 0
y
dy
At y = 0
At y =
du
dy
0
u
u
dy 1 dy
U
U
0
For *:
* 1
Then
1
1
* 1 u
u y
u
1 dy 1 d 1 d
0 U
U 0 U
0
with
u
2 2 3 4
U
*
u
1
1
3
1 d 1 2 2 3 4 d 2 4 5
0 .3
U
2
5
10
0
0
0
1
Hence
1
1
For :
u
u
u
u
1 dy 1 dy
U U
U U
0
0
Then
1
1
1 u u
u
u y
u
u
1 dy 1 d 1 d
0U U
U U 0 U U
0
Hence
u
u
1 d 2 3 4 1 2 3 4 d 2 4 2 2 3 9 4 4 5 4 6 4 7 8 d
0U U
0
0
1
1
1
2 4 3 1 4 9 5 4 7 1 8 1 9
37
0.117
3
2
5
7
2
9 0 315
1
Problem 9.11
Given:
Laminar boundary layer profile
Find:
If it satisfies BC’s; Evaluate */ and /
[Difficulty: 2]
Solution:
3
The boundary layer equation is
u 3 y 1 y
for which u = U at y =
U 2 2
The BC’s are
u 0 0
At y = 0
At y =
du
dy
0
y
u 3
1 3
0 0 0
U 2
2
3 1 3 y2
3 1 32
du
0
U
U
3
3
dy
2 2 y
2 2
For *:
u
u
* 1 dy 1 dy
U
U
0
0
Then
1
1
* 1 u
u
u y
1 dy 1 d 1 d
0 U
U 0 U
0
with
Hence
u 3
1
3
U 2
2
1
1
1
*
u
1 3
3 2 1 4 3
3
1 d 1 d 0.375
2
2
4
8 0 8
0 U
0
u
u
u
u
1 dy 1 dy
U U
U U
0
0
For :
Then
1
1
1 u u
u
u
u
u y
1 dy 1 d 1 d
0U U
U U 0 U U
0
Hence
1
1
1
9
1
3
1
1 u u
1 3
3
3
1 d 3 1 3 d 2 3 4 6 d
0U U
2
2
4
2
2
4
2
2 2
0
0
3 2 3 3 1 4 3 5 1 7
39
0.139
4
4
8
10
28 0 280
1
Problem 9.10
Given:
Find:
Linear, sinusoidal, and parabolic velocity profiles
Solution:
Here are the profiles:
[Difficulty: 2]
Plots of y/δ vs u/U for all three profiles
Laminar Boundary Layer Velocity Profiles
Linear
Sinusoidal
Parabolic
Dimensionless Distance y/δ
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
Dimensionless Velocity u/U
0.8
Problem 9.9
Given:
[Difficulty: 2]
Sinusoidal velocity profile for laminar boundary layer:
u A sin( B y ) C
Find:
(a) Three boundary conditions applicable to this profile
(b) Constants A, B, and C.
Solution:
For the boundary layer, the following conditions apply:
u 0 at
y 0 (no slip condition)
u U at
y δ (continuity with freestream)
y
u 0 at
y δ (no shear stress at freestream)
Applying these boundary conditions:
( 1 ) u ( 0 ) A sin( 0 ) C 0
C0
( 2 ) u ( δ) A sin( B δ) U
( 3)
y
u A B cos( B y )
Thus: u ( δ) A B cos( B δ) 0
y
Therefore: B δ
δ U Therefore: A U
2δ
Back into (2): A sin
π
So the expression for the velocity profile is:
π y
2 δ
u U sin
π
2
or
B
π
2 δ
Problem 9.8
Given:
Aircraft or missile at various altitudes
Find:
Plot of boundary layer length as function of altitude
Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
Re crit =
UL crit/ = 500000
The critical length is then
L crit = 500000/U
Let L 0 be the length at sea level (density 0 and viscosity 0). Then
L crit/L 0 = (/0)/(/0)
The viscosity of air increases with temperature so generally decreases with elevation;
the density also decreases with elevation, but much more rapidly.
Hence we expect that the length ratio increases with elevation
For the density , we use data from Table A.3.
For the viscosity , we use the Sutherland correlation (Eq. A.1)
= bT 1/2/(1+S /T )
b =
S =
1.46E-06
110.4
kg/m.s.K1/2
K
[Difficulty: 2]
Computed results:
z (km)
T (K)
/0
/0
L crit/L 0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
6.0
288.2
284.9
281.7
278.4
275.2
271.9
268.7
265.4
262.2
258.9
255.7
249.2
1.0000
0.9529
0.9075
0.8638
0.8217
0.7812
0.7423
0.7048
0.6689
0.6343
0.6012
0.5389
1.000
0.991
0.982
0.973
0.965
0.955
0.947
0.937
0.928
0.919
0.910
0.891
1.000
1.04
1.08
1.13
1.17
1.22
1.28
1.33
1.39
1.45
1.51
1.65
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0
16.0
17.0
18.0
19.0
20.0
22.0
24.0
26.0
28.0
30.0
242.7
236.2
229.7
223.3
216.8
216.7
216.7
216.7
216.7
216.7
216.7
216.7
216.7
216.7
218.6
220.6
222.5
224.5
226.5
0.4817
0.4292
0.3813
0.3376
0.2978
0.2546
0.2176
0.1860
0.1590
0.1359
0.1162
0.0993
0.0849
0.0726
0.0527
0.0383
0.0280
0.0205
0.0150
0.872
0.853
0.834
0.815
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.800
0.806
0.812
0.818
0.824
1.81
1.99
2.19
2.41
2.67
3.12
3.65
4.27
5.00
5.85
6.84
8.00
9.36
10.9
15.2
21.0
29.0
40.0
54.8
Length of Laminar Boundary Layer
versus Elevation
60
50
40
L/L 0
30
20
10
0
0
10
20
z (m)
30
Problem 9.7
[Difficulty: 2]
Given:
Laminar boundary layer (air & water)
Find:
Plot of boundary layer length as function of speed (at various altitudes for air)
Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
Re crit = UL crit/ = 500000
The critical length is then
L crit = 500000/U
For air at sea level and 10 km, we can use tabulated data for density from Table A.3.
For the viscosity , use the Sutherland correlation (Eq. A.1)
= bT 1/2/(1+S /T )
b = 1.46E-06 kg/m.s.K1/2
S = 110.4 K
Air (sea level, T = 288.2 K):
=
1.225
kg/m3
(Table A.3)
= 1.79E-05 N.s/m2
(Sutherland)
Water (20 oC):
Air (10 km, T = 223.3 K):
=
0.414
kg/m3 =
(Table A.3)
= 1.46E-05 N.s/m2
(Sutherland)
998
slug/ft3
= 1.01E-03 N.s/m2
(Table A.8)
Computed results:
Water Air (Sea level) Air (10 km)
U (m/s)
L crit (m)
L crit (m)
L crit (m)
0.05
0.10
0.5
1.0
5.0
15
20
25
30
50
100
200
10.12
5.06
1.01
0.506
0.101
0.0337
0.0253
0.0202
0.0169
0.0101
0.00506
0.00253
146.09
73.05
14.61
7.30
1.46
0.487
0.365
0.292
0.243
0.146
0.0730
0.0365
352.53
176.26
35.25
17.63
3.53
1.18
0.881
0.705
0.588
0.353
0.176
0.0881
1000
0.00051
0.0073
0.0176
Length of Laminar Boundary Layer
for Water and Air
100.0
1.0
L crit (m)
0.0
0.0
1.E-02
Water
Air (Sea level)
Air (10 km)
1.E+00
1.E+02
U (m/s)
1.E+04
Problem 9.6
[Difficulty: 2]
Given:
Sheet of plywood attached to the roof of a car
Find:
Speed at which boundary layer becomes turbulent; Speed at which 90% is turbulent
Solution:
Rex
Basic equation
ρ U x
μ
ν 1.50 10
For air
U x
Rex 5 10
and transition occurs at about
ν
2
5 m
5
(Table A.10)
s
Now if we assume that we orient the plywood such that the longer dimension is parallel to the motion of the car, we can say:
Hence
U
ν Rex
x
U 3.8
m
s
When 90% of the boundary layer is turbulent
x 0.1 2 m
Hence
U
ν Rex
x
U 37.5
m
s
x 2 m
U 13.50
km
U 135.0
km
hr
hr
Problem 9.5
[Difficulty: 2]
Given:
Flow around American and British golf balls, and soccer ball
Find:
Speed at which boundary layer becomes turbulent
Solution:
Basic equation
For air
ReD
ρ U D
μ
ν 1.62 10
For the American golf ball D 1.68 in
For the British golf ball
For soccer ball
U D
5
ReD 2.5 10
and transition occurs at about
ν
4 ft
2
(Table A.9)
s
Hence
D 41.1 mm
Hence
D 8.75 in
Hence
U
U
U
ν ReD
D
ν ReD
D
ν ReD
D
U 289
ft
U 300
ft
s
U 55.5
s
ft
s
U 197 mph
U 88.2
m
U 205 mph
U 91.5
m
U 37.9 mph
U 16.9
m
s
s
s
Problem 9.4
[Difficulty: 2]
Given:
Experiment with 1 cm diameter sphere in SAE 10 oil
Find:
Reasonableness of two flow extremes
Solution:
Basic equation
ReD
ρ U D
μ
U D
ν 1.1 10
For
ReD 1
For
ReD 2.5 10
Note that for
ReD 2.5 10
For water
ν 1.01 10
For
ReD 2.5 10
ReD 2.5 10
(Fig. A.3 at 20 oC)
and
ν
2
4 m
For SAE 10
s
we find
5
ν ReD
D
ν ReD
D
D 1 cm
U 0.011
U 2750
m
s
m
s
U 1.10
cm
s
which is reasonable
which is much too high!
we need to increase the sphere diameter D by a factor of about 1000, or reduce the
viscosity ν by the same factor, or some combination of these. One possible solution is
2
6 m
5
U
U
5
5
and transition occurs at about
s
(Table A.8 at 20 oC)
we find
U
D 10 cm
and
ν ReD
D
Hence one solution is to use a 10 cm diameter sphere in a water tank.
U 2.52
m
s
which is reasonable
Problem 9.3
[Difficulty: 3]
Given:
Boeing 757
Find:
Point at which BL transition occurs during takeoff and at cruise
Solution:
Basic equation
For air at 68oF
ρ U x
Rex
μ
ν 1.62 10
xp
At 33,000 ft
T 401.9 R
U
5
and transition occurs at about
Rex 5 10
(Table A.9)
and we are given
x p 0.345 ft
x p 4.14 in
(Intepolating from Table A.3)
T 57.8 °F
ν
4 ft
ν Rex
Hence
U x
2
s
U 160
mi
hr
234.7
ft
s
We need to estimate ν or μ at this temperature. From Appendix A-3
b T
μ
S
1
Hence
μ
kg
6
b 1.458 10
1
T
b T
1
S
m s K
5 N s
μ 1.458 10
2
S 110.4 K
2
7 lbf s
μ 3.045 10
m
T
ft
2
For air at 10,000 m (Table A.3)
ρ
ρSL
ν
Hence
xp
0.3376
ρSL 0.002377
slug
ft
4 ft
μ
ν 3.79 10
ρ
ν Rex
U
x p 0.244 ft
3
ρ 0.3376 ρSL
ρ 8.025 10
s
x p 2.93 in
ft
2
and we are given
4 slug
U 530
mi
hr
3
Problem 9.2
Given:
Model of riverboat
Find:
Distance at which transition occurs
[Difficulty: 2]
Solution:
Basic equation
For water at 10oC
Hence
For the model
Rex
ρ U x
μ
ν 1.30 10
xp
xm
ν Rex
U
xp
18
U x
2
6 m
5
and transition occurs at about
Rex 5 10
(Table A.8)
and we are given
ν
s
x p 0.186 m
x p 18.6 cm
x m 0.0103 m
x m 10.3 mm
U 3.5
m
s
Problem 9.1
Given:
Minivan traveling at various speeds
Find:
Plot of boundary layer length as function of speed
[Difficulty: 2]
Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
VL crit/ =500000
Re crit =
The critical length is then
L crit = 500000/V
Tabulated or graphical data:
=
3.79E-07
=
0.00234
lbf.s/ft
2
3
slug/ft
(Table A.9, 68oF)
Computed results:
V (mph)
L crit (ft)
10
13
15
18
20
30
40
50
60
70
80
90
5.52
4.42
3.68
3.16
2.76
1.84
1.38
1.10
0.920
0.789
0.690
0.614
Length of Laminar Boundary Layer
on the Roof of a Minivan
6
5
4
L crit (ft)
3
2
1
0
0
10
20
30
40
50
V (mph)
60
70
80
90
100
Problem 8.205
[Difficulty: 5] Part 1/2
Problem 8.205
[Difficulty: 5] Part 2/2
Problem 8.204
[Difficulty: 5] Part 1/2
Problem 8.204
[Difficulty: 5] Part 2/2
Problem 8.203
[Difficulty: 3]
Problem 8.202
[Difficulty: 4] Part 1/2
Problem 8.202
[Difficulty: 4] Part 2/2
Problem 8.201
[Difficulty: 1]
V 1, A 1
Given:
Flow through a diffuser
Find:
Derivation of Eq. 8.42
Solution:
Basic equations
Cp =
p2 − p1
p1
1
ρ
2
⋅ ρ⋅ V1
2
+
2
V1
2
p2
+ g ⋅ z1 =
+
ρ
V2
2
V 2, A 2
2
+ g ⋅ z2
Q = V⋅ A
Assumptions: 1) All the assumptions of the Bernoulli equation 2) Horizontal flow 3) No flow separation
From Bernoulli
p2 − p1
ρ
2
2
2
2
⎛ A1 ⎞
=
−
=
−
⋅⎜
2
2
2
2
⎝ A2 ⎠
V1
V2
V1
V1
2
using continuity
Hence
⎡ V 2 V 2 ⎛ A ⎞ 2⎤
1
1 ⎥
⎢ 1
Cp =
=
⋅
−
⋅⎜
⎢
⎥=1−
1
2
2
1
2
2
⎝ A2 ⎠ ⎦
⋅ ρ⋅ V1
⋅ V1 ⎣
2
2
Finally
Cp = 1 −
p2 − p1
1
1
⎛ A1 ⎞
⎜
⎝ A2 ⎠
which is Eq. 8.42.
2
AR
This result is not realistic as a real diffuser is very likely to have flow separation
2
Problem 8.200
Given:
Flow through venturi
Find:
Maximum flow rate before cavitation
[Difficulty: 3]
Solution:
Basic equation
C⋅ At
mactual =
4
(
C⋅ At
)
⋅ 2 ⋅ ρ⋅ p 1 − p 2 =
4
1−β
⋅ 2 ⋅ ρ⋅ ∆p
1−β
Note that mactual is mass flow rate (the
software cannot render a dot!)
For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re
Available data
D1 = 100 ⋅ mm
Dt = 50⋅ mm
p 1g = 200 ⋅ kPa C = 0.99
p atm = 101 ⋅ kPa
p v = 1.23⋅ kPa
Steam tables - saturation pressure at 10oC
ρ = 1000⋅
kg
ν = 1.3⋅ 10
3
2
−6 m
⋅
s
m
Then
At =
β =
π⋅ Dt
At = 1963⋅ mm
4
Dt
A1 =
Hence the largest ∆p is
∆p = p 1 − p t
Then
mrate =
4
pt = pv
2
A1 = 7854⋅ mm
p t = 1.23 kPa
∆p = 300 ⋅ kPa
C⋅ At
4
kg
mrate = 49.2
s
⋅ 2 ⋅ ρ⋅ ∆p
1−β
mrate
ρ
Q
V1 =
A1
Re1 =
2
p 1 = 301 ⋅ kPa
The smallest allowable throat pressure is the saturation pressure
Check the Re
π⋅ D1
β = 0.5
D1
p 1 = p atm + p 1g
Q =
(Table A.8)
2
2
3
Q = 0.0492
m
s
m
V1 = 6.26
s
V1 ⋅ D1
ν
5
Re1 = 4.81 × 10
3
Thus ReD1 > 2 x 105. The volume flow rate is
Q = 0.0492
m
s
(Asumption - verify later)
Q = 49.2
L
s
Problem 8.199
[Difficulty: 3]
Problem 8.198
[Difficulty: 3]
Given:
Flow through a venturi meter
Find:
Maximum flow rate for incompressible flow; Pressure reading
Solution:
Basic equation
C⋅ At
mactual =
4
(
C⋅ At
)
⋅ 2 ⋅ ρ⋅ p 1 − p 2 =
⋅ 2 ⋅ ρ⋅ ∆p
Note that mactual is mass flow rate (the
software cannot render a dot!)
4
1−β
1−β
Assumptions: 1) Neglect density change 2) Use ideal gas equation for density
ρ=
Then
p
ρ = 60⋅
Rair⋅ T
lbf
2
2
×
in
1
− 3 slug
⎛ 12⋅ in ⎞ × lbm⋅ R × 1 ⋅ slug ⋅
ρ = 9.53 × 10 ⋅
⎜
3
53.33 ⋅ ft⋅ lbf
32.2⋅ lbm ( 68 + 460 ) ⋅ R
⎝ 1⋅ ft ⎠
ft
For incompressible flow V must be less than about 100 m/s or 330 ft/s at the throat. Hence
mactual = ρ⋅ V2 ⋅ A2
mactual = 9.53 × 10
− 3 slug
ft
β=
Dt
β =
D1
3
3
∆p = ρHg⋅ g ⋅ ∆h
∆h =
and in addition
⎛ mactual ⎞
4
∆p =
⋅⎜
⋅ 1−β
2⋅ ρ
⎝ C⋅ At ⎠
2
(
s
×
π
4
×
⎛ 1 ⋅ ft⎞
⎜
⎝4 ⎠
2
slug
mactual = 0.154 ⋅
s
β = 0.5
6
Also
1
ft
× 330 ⋅
)
∆h =
so
∆p
ρHg⋅ g
(1 − β4) ⋅⎛ mactual ⎞ 2
2 ⋅ ρ⋅ ρHg⋅ g
⎜
⎝ C⋅ At ⎠
For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re
∆h =
(1 − 0.54) ×
2
Hence
At 68oF,(Table A.7)
2
⎡
slug
4 ⎛ 4 ⎞⎤
1
⎥
×
×
× ⎢0.154
×
×
×⎜
−3
13.6⋅ 1.94⋅ slug 32.2⋅ ft ⎣
s
0.99 π ⎝ 1 ⋅ ft ⎠ ⎦
9.53 × 10
slug
ft
V=
3
Q
A
ft
=
4 ⋅ mactual
π⋅ ρ⋅ D1
ν = 1.08 × 10
ReD1 =
2
− 5 ft
⋅
3
2
2
s
V =
4
π
ft
×
3
9.53 × 10
−3
1
×
⎛ 1 ⋅ ft⎞
⎜2
⎝ ⎠
slug
2
× 0.154
slug
s
∆h = 6.98⋅ in
V = 82.3⋅
2
s
V⋅ D1
ν
Thus ReD1 > 2 x 105. The mass flow rate is
ft 1
ReD1 = 82.3⋅ × ⋅ ft ×
s
2
slug
mactual = 0.154 ⋅
s
s
−5
1.08 × 10
⋅ ft
2
and pressure
ReD1 = 3.81 × 10
∆h = 6.98⋅ in
6
Hg
ft
s
Problem 8.197
[Difficulty: 2]
Problem 8.196
Given:
Flow through an venturi meter
Find:
Flow rate
[Difficulty: 2]
Solution:
Basic equation
C⋅ At
mactual =
4
(
)
⋅ 2⋅ ρ⋅ p 1 − p 2 =
1− β
C⋅ At
⋅ 2⋅ ρ⋅ ∆p
4
1− β
Note that mactual is mass flow rate (the
software cannot render a dot!)
For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re
Available data
D1 = 2⋅ in
Dt = 1⋅ in
∆p = 25⋅ psi
ρ = 1.94⋅
slug
ft
β =
Then
Q=
Q =
Dt
β = 0.5
D1
mactual
ρ
C⋅ At
=
At 68oF(Table A.7)
V=
C = 0.99
and assume
⋅ 2⋅ ρ⋅ ∆p
4
ρ⋅ 1 − β
π⋅ C⋅ Dt
2
2 ⋅ ∆p
⋅
4⋅ 1 − β
ρ
Q
V =
4
Hence
3
A
Q = 0.340
4⋅ Q
π⋅ D1
− 5 ft
ν = 1.08⋅ 10
⋅
2
V = 15.6⋅
2
s
Thus ReD1 > 2 x 105. The volume flow rate is
ReD1 =
ft
3
s
Q = 152 ⋅ gpm
ft
s
V⋅ D1
ν
Q = 152 ⋅ gpm
5
ReD1 = 2.403 × 10
Problem 8.195
[Difficulty: 2]
Given:
Flow through a venturi meter (NOTE: Throat is obviously 3 in not 30 in!)
Find:
Flow rate
Solution:
Basic equation
C⋅ At
mactual =
4
(
C⋅ At
)
⋅ 2 ⋅ ρ⋅ p 1 − p 2 =
⋅ 2 ⋅ ρ⋅ ∆p
Note that mactual is mass flow rate (the
software cannot render a dot!)
4
1−β
1−β
For ReD1 > 2 x 105, 0.980 < C < 0.995. Assume C = 0.99, then check Re
β=
Also
Then
Dt
β =
D1
β = 0.5
6
∆p = ρHg⋅ g ⋅ ∆h = SG Hg⋅ ρ⋅ g ⋅ ∆h
Q=
mactual
ρ
V=
C⋅ At
=
Q
4⋅ Q
ReD1 =
⋅ 2 ⋅ ρ⋅ SGHg⋅ ρ⋅ g ⋅ ∆h =
4 ⋅ ρ⋅ 1 − β
π⋅ C⋅ Dt
π⋅ D1
⎛ 1 ⋅ ft⎞ × 2 × 13.6 × 32.2⋅ ft × 1⋅ ft
⎜
2
⎝4 ⎠
s
V =
2
− 6 ft
⋅
4
π
1
×
⎛ 1 ⋅ ft⎞
⎜
⎝2 ⎠
2
2
4
⋅ 2 ⋅ SGHg⋅ g ⋅ ∆h
4⋅ 1 − β
2
4
At 75oF,(Table A.7) ν = 9.96 × 10
2
4
× 0.99 ×
1 − 0.5
A
⋅ 2 ⋅ ρ⋅ ∆p =
ρ⋅ 1 − β
4×
=
π⋅ C⋅ Dt
4
π
Q =
Hence
3
× 1.49⋅
ft
Q = 1.49⋅
ft
V = 7.59⋅
ft
3
s
3
s
s
2
s
V⋅ D1
ν
Thus ReD1 > 2 x 105. The volume flow rate is
ft 1
ReD1 = 7.59⋅ × ⋅ ft ×
s
2
Q = 1.49⋅
ft
3
s
s
−6
9.96 × 10
⋅ ft
2
ReD1 = 3.81 × 10
5
Problem 8.194
Given:
Reservoir-pipe system
Find:
Orifice plate pressure difference; Flow rate
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT = hl + Σh lm(8.29)
⎝
⎠ ⎝
⎠
2
hl = f ⋅
f =
64
Re
L V
⋅
D 2
2
(8.34)
(Laminar)
h lm = K⋅
(8.36)
V
(8.40a)
2
⎞
⎛⎜ e
1
2.51
D
⎟
= −2.0⋅ log ⎜
+
0.5
3.7
0.5
⎜
f
Re⋅ f ⎠
⎝
(Turbulent)
(8.37)
2
There are three minor losses: at the entrance; at the orifice plate; at the exit. For each
2
The energy equation (Eq. 8.29) becomes (α = 1)
g ⋅ ∆H =
V
2
⋅ ⎛⎜ f ⋅
L
⎝ D
h lm = K⋅
+ Kent + Korifice + Kexit⎞
⎠
V
2
(1)
(∆H is the difference in reservoir heights)
This cannot be solved for V (and hence Q) because f depends on V; we can solve by manually iterating, or by using Solver
The tricky part to this problem is that the orifice loss coefficient Korifice is given in Fig. 8.23 as a percentage of pressure differential
∆p across the orifice, which is unknown until V is known!
The mass flow rate is given by
mrate = K⋅ At⋅ 2 ⋅ ρ⋅ ∆p
(2)
where K is the orifice flow coefficient, At is the orifice area, and ∆p is the pressure drop across the orifice
Equations 1 and 2 form a set for solving for TWO unknowns: the pressure drop ∆p across the orifice (leading to a value for Korifice)
and the velocity V. The easiest way to do this is by using Solver
In Excel:
Problem 8.193
Given:
Flow through an orifice
Find:
Pressure drop
[Difficulty: 2]
Solution:
(
)
Basic equation
mactual = K⋅ At⋅ 2 ⋅ ρ⋅ p 1 − p 2 = K⋅ At⋅ 2 ⋅ ρ⋅ ∆p
For the flow coefficient
K = K⎜ ReD1 ,
At 65oC,(Table A.8)
⎛
Dt
⎝
D1
⎠
ρ = 980 ⋅
⎞
kg
ν = 4.40 × 10
3
2
−7 m
⋅
m
V=
Q
V =
A
ReD1 =
β=
Note that mactual is mass flow rate (the
software cannot render a dot!)
V⋅ D
4
π
×
s
1
( 0.15⋅ m)
2
× 20⋅
L
s
m
ReD1 = 1.13⋅ × 0.15⋅ m ×
s
ν
Dt
β =
D1
3
×
0.001 ⋅ m
V = 1.13
1⋅ L
s
−7
4.40 × 10
2
⋅m
β = 0.5
150
K = 0.624
Then
2
2
⎛ mactual ⎞ 1
ρ⋅ Q ⎞ 1
ρ
Q ⎞
∆p = ⎜
⋅
= ⎛⎜
⋅
= ⋅ ⎛⎜
⎝ K⋅ At ⎠ 2⋅ ρ ⎝ K⋅ At ⎠ 2⋅ ρ 2 ⎝ K⋅ At ⎠
s
ReD1 = 3.85 × 10
75
From Fig. 8.20
m
2
⎡ L 0.001 ⋅ m3
⎤
1
4
1
⎥
∆p =
× 980 ⋅
× ⎢20⋅ ×
×
×
×
3 ⎢
2⎥
2
s
0.624
π
1⋅ L
m
( 0.075 ⋅ m) ⎦
⎣
1
kg
2
∆p = 25.8⋅ kPa
5
Problem 8.192
[Difficulty: 2]
Problem 8.191
[Difficulty: 5] Part 1/2
Problem 8.191
[Difficulty: 5] Part 2/2
Problem 8.190
[Difficulty: 4]
Problem 8.189
Given:
Water pipe system
Find:
Flow rates
[Difficulty: 5]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = hl
⎝
⎠ ⎝ρ
⎠
f =
64
2
h lT = f ⋅
L V
⋅
D 2
⎞
⎛ e
⎜
1
2.51
D
(Turbulent)
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
(Laminar)
Re
2
The energy equation can be simplified to
∆p = ρ⋅ f ⋅
L V
⋅
D 2
This can be written for each pipe section
2
Pipe A (first section)
LA VA
∆pA = ρ⋅ fA ⋅
⋅
DA 2
Pipe B (1.5 in branch)
LB VB
∆pB = ρ⋅ fB⋅
⋅
DB 2
Pipe C (1 in branch)
LC VC
∆pC = ρ⋅ fC⋅
⋅
DC 2
Pipe D (last section)
LD VD
∆pD = ρ⋅ fD⋅
⋅
DD 2
(4)
QA = QD
(5)
QA = QB + QC
(6)
∆p = ∆pA + ∆pB + ∆pD
(7)
∆pB = ∆pC
(8)
(1)
2
(2)
2
(3)
2
In addition we have the following contraints
We have 4 unknown flow rates (or velocities) and four equations (5 - 8); Eqs 1 - 4 relate pressure drops to flow rates (velocities)
In Excel:
Problem 8.188
Given:
Pipe system
Find:
Flow in each branch if pipe 3 is blocked
Solution:
Governing equations:
[Difficulty: 5]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
f =
64
(Laminar)
Re
(8.36)
2
(8.29)
⎞
⎛⎜ e
1
2.51
D
⎟
= −2.0⋅ log ⎜
+
0.5
3.7
0.5
⎜
f
Re⋅ f ⎠
⎝
h lT = f ⋅
L V
⋅
D 2
(Turbulent)
(8.37)
2
The energy equation (Eq. 8.29) can be simplified to
∆p = ρ⋅ f ⋅
L V
⋅
D 2
This can be written for each pipe section
In addition we have the following contraints
Q0 = Q1 + Q4
(1)
∆p = ∆p0 + ∆p1
(3)
Q4 = Q2
∆p = ∆p0 + ∆p4 + ∆p2
We have 4 unknown flow rates (or, equivalently, velocities) and four equations
In Excel:
(8.34)
(2)
(4)
Problem 8.187
Given:
Pipe system
Find:
Flow in each branch
Solution:
Governing equations:
[Difficulty: 5]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
f =
64
(Laminar)
Re
(8.36)
2
(8.29)
⎞
⎛⎜ e
1
2.51
D
⎟
= −2.0⋅ log ⎜
+
0.5
3.7
0.5
⎜
f
Re⋅ f ⎠
⎝
h lT = f ⋅
L V
⋅
D 2
(Turbulent)
(8.37)
2
The energy equation (Eq. 8.29) can be simplified to
∆p = ρ⋅ f ⋅
L V
⋅
D 2
This can be written for each pipe section
In addition we have the following contraints
Q0 = Q1 + Q4
(1)
∆p = ∆p0 + ∆p1
(3)
∆p2 = ∆p3
(5)
Q4 = Q2 + Q3
∆p = ∆p0 + ∆p4 + ∆p2
We have 5 unknown flow rates (or, equivalently, velocities) and five equations
In Excel:
(8.34)
(2)
(4)
Problem 8.186
[Difficulty: 4] Part 1/2
Problem 8.186
[Difficulty: 4] Part 2/2
Problem 8.185
Given:
Fan/duct system
Find:
Flow rate
[Difficulty: 3]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT − ∆hfan
⎝
⎠ ⎝
⎠
2
h lT = f ⋅
2
L V
V
⋅
+ K⋅
2
Dh 2
The energy equation becomes for the system (1 = duct inlet, 2 = duct outlet)
2
∆hfan = f ⋅
2
or
∆ppump =
2
L V
V
⋅
+ K⋅
2
Dh 2
ρ⋅ V
2
⋅ ⎛⎜ f ⋅
L
⎝ Dh
2
+ K⎞
(1)
⎠
where
4⋅ A
4⋅ h
Dh =
=
=h
Pw
4⋅ h
This must be matched to the fan characteristic equation; at steady state, the pressure generated by the fan just equals that lost to
friction in the circuit
2
∆pfan = 1020 − 25⋅ Q − 30⋅ Q
In Excel:
(2)
Fan and Duct Pressure Heads
2500
Dp (Pa)
2000
1500
Duct
1000
Fan
500
0
0.0
0.5
1.0
1.5
3
Q (m /s)
2.0
2.5
3.0
Problem 8.184 Equations
Given:
Pump/pipe system
Find:
Flow rate, pressure drop, and power supplied; Effect of roughness
[Difficulty: 4]
Solution:
Re =
f =
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = h lT − ∆hpump
⎝
⎠ ⎝ρ
⎠
⎞
⎛ e
⎜ D
1
2.51
(Laminar)
(Turbulent)
= −2.0⋅ log ⎜
+
3.7
f
Re
f
⋅
⎝
⎠
ρ⋅ V⋅ D
μ
64
Re
2
h lT = f ⋅
L V
⋅
D 2
The energy equation becomes for the system (1 = pipe inlet, 2 = pipe outlet)
2
∆hpump = f ⋅
2
L V
⋅
D 2
∆ppump = ρ⋅ f ⋅
or
L V
⋅
D 2
(1)
This must be matched to the pump characteristic equation; at steady state, the pressure generated by the pump just equals that
lost to friction in the circuit
2
∆ppump = 145 − 0.1⋅ Q
(2)
Finally, the power supplied to the pump, efficiency η,
Power =
In Excel:
Q⋅ ∆p
η
(3)
is
Pum p and Pipe Pressure Heads
Pipe (e = 0.5 in)
160
Pipe (e = 0.25 in)
Pump
Dp (psi)
120
80
40
0
10
15
20
3
Q (ft /s)
25
30
Problem 8.183
Given:
Flow in a pump testing system
Find:
Flow rate; Pressure difference; Power
[Difficulty: 4]
Solution:
Governing equations:
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Re =
f =
2
ρ⋅ V⋅ D
hl = f ⋅
μ
64
L V
⋅
D 2
(8.36)
Re
(8.34)
(Laminar)
h lm = f ⋅
∑
major
Le V2
⋅
D 2
hl +
∑
h lm (8.29)
minor
(8.40b)
⎞
⎛ e
⎜ D
1
2.51
(8.37)
= −2.0⋅ log ⎜
+
3.7
f
Re
f
⋅
⎝
⎠
(Turbulent)
The energy equation (Eq. 8.29) becomes for the circuit (1 = pump outlet, 2 = pump inlet)
p1 − p2
ρ
2
= f⋅
2
or
∆p = ρ⋅ f ⋅
2
Lelbow
Lvalve ⎞
⎛L
+ 4⋅
+
2 ⎝D
D
D ⎠
V
2
L V
V
V
⋅
+ 4 ⋅ f ⋅ Lelbow⋅
+ f ⋅ Lvalve⋅
2
2
D 2
⋅⎜
(1)
This must be matched to the pump characteristic equation; at steady state, the pressure generated by the pump just equals that
lost to friction in the circuit
4
2
∆p = 750 − 15 × 10 ⋅ Q
(2)
Finally, the power supplied to the pump, efficiency η, is
Power =
In Excel:
Q⋅ ∆p
η
(3)
1200
Circuit and Pump Pressure Heads
Dp (kPa)
1000
800
600
Circuit
400
Pump
200
0
0.00
0.01
0.02
0.03
0.04
3
Q (m /s)
0.05
0.06
0.07
Problem 8.182
[Difficulty: 3]
Problem 8.181
Given:
Flow in water fountain
Find:
Daily cost
[Difficulty: 2]
Solution:
Basic equations
Wpump = Q⋅ ∆p
∆p = ρ⋅ g ⋅ H
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
3
Available data
m
Q = 0.075 ⋅
ρ = 999 ⋅
s
kg
3
m
Hence
H = 10⋅ m
Cost =
ηp = 85⋅ %
0.14
kW⋅ hr
(dollars)
∆p = ρ⋅ g ⋅ H
∆p = 98⋅ kPa
Wpump = Q⋅ ∆p
Wpump = 7.35⋅ kW
Power =
Wpump
ηp ⋅ ηm
C = Cost⋅ Power⋅ day
Power = 10.2⋅ kW
C = 34.17
(dollars)
ηm = 85⋅ %
Problem 8.180
[Difficulty: 4] Part 1/2
Problem 8.180
[Difficulty: 4] Part 2/2
Problem 8.179
[Difficulty: 4]
d
e
f
c
Given:
Fire nozzle/pump system
Find:
Design flow rate; nozzle exit velocity; pump power needed
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V2
V3
2
3
⎜ ρ + α⋅ 2 + g⋅ z2 − ⎜ ρ + α⋅ 2 + g⋅ z3 = h l
⎝
⎠ ⎝
⎠
L V2
hl = f ⋅ ⋅
D 2
2
for the hose
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 2 and 3 is approximately 1 4) No minor loss
p3
ρ
+
V3
2
p4
+ g ⋅ z3 =
+
ρ
2
V4
2
2
+ g ⋅ z4
for the nozzle (assuming Bernoulli applies)
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V2
V1
2
1
⎜ ρ + α⋅ 2 + g⋅ z2 − ⎜ ρ + α⋅ 2 + g⋅ z1 = h pump
⎝
⎠ ⎝
⎠
for the pump
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) No minor loss
Hence for the hose
∆p
ρ
=
p2 − p3
ρ
2
= f⋅
L V
⋅
D 2
or
V=
2 ⋅ ∆p⋅ D
ρ⋅ f ⋅ L
We need to iterate to solve this for V because f is unknown until Re is known. This can be done using Excel's Solver, but here:
∆p = 750 ⋅ kPa
L = 100 ⋅ m
e = 0
D = 3.5⋅ cm
ρ = 1000⋅
kg
ν = 1.01 × 10
3
2
−6 m
⋅
m
Make a guess for f
Given
f = 0.01
2 ⋅ ∆p⋅ D
ρ⋅ f ⋅ L
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
V =
Given
V =
2 ⋅ ∆p⋅ D
ρ⋅ f ⋅ L
V = 5.92
m
s
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
V = 7.25
m
s
Re =
V⋅ D
Re = 2.51 × 10
ν
f = 0.0150
Re =
V⋅ D
ν
f = 0.0156
Re = 2.05 × 10
5
5
s
2 ⋅ ∆p⋅ D
V =
V = 5.81
ρ⋅ f ⋅ L
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
V = 5.80
ρ⋅ f ⋅ L
2
Q = V⋅ A =
p3
For the nozzle
ρ
+
V3
V4 =
2
π⋅ D
4
⋅V
p4
+ g ⋅ z3 =
+
ρ
3 N
2 × 700 × 10 ⋅
2
m
Re =
s
× ( 0.035 ⋅ m) × 5.80⋅
p 1 = 350 ⋅ kPa
4
Ppump = ρ⋅ Q⋅
(p2 − p1)
ρ
Ppump
η
(
= Q⋅ p 2 − p 1
)
ν
Re = 2.01 × 10
5
V⋅ D
Re = 2.01 × 10
5
ν
Q = 5.58 × 10
s
+ g ⋅ z4
3
m
1000⋅ kg
×
so
kg⋅ m
2
s ⋅N
+ ⎛⎜ 5.80⋅
⎝
m⎞
s
2
3
Q = 0.335 ⋅
s
(
2
p 2 = 700 ⋅ kPa + 750 ⋅ kPa
The pump power is Ppump = mpump⋅ h pump
V⋅ D
3
−3m
m
2⋅ p3 − p4
V4 =
ρ
)
m
min
+ V3
2
m
V4 = 37.9
s
⎠
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V2
V1
2
1
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
2
1 = h pump
2
2
⎝
⎠ ⎝ρ
⎠
For the pump
Prequired =
2
m
V4
×
s
2
π
Q =
2
Re =
f = 0.0156
2 ⋅ ∆p⋅ D
V =
m
so
h pump =
p2 − p1
ρ
p 2 = 1450⋅ kPa
P pump and mpump are pump power and mass flow rate (software can't do a dot!)
Ppump = 5.58 × 10
Prequired =
3
−3 m
6.14⋅ kW
70⋅ %
⋅
s
3 N
× ( 1450 − 350 ) × 10 ⋅
2
m
Ppump = 6.14⋅ kW
Prequired = 8.77⋅ kW
Problem 8.178
Given:
Flow in air conditioning system
Find:
Pressure drop; cost
[Difficulty: 3]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
3
Available data
L = 5 ⋅ km
ρ = 1000
D = 0.75⋅ m
kg
3
− 3 N⋅ s
μ = 1.3⋅ 10
m
V =
Then
so
Q
2⎞
⎛ π⋅ D
⎜
⎝ 4 ⎠
Given
e = 0.046 ⋅ mm
⋅
2
Q = 0.65⋅
m
s
(Table A.8)
m
V = 1.47
m
Re =
s
⎞
⎛ e
⎜ D
1
2.51
= −2 ⋅ log⎜
+
f
⎝ 3.7 Re⋅ f ⎠
L
μ
f = 0.0131
2
⋅ ρ⋅
V
The energy equation becomes
∆p = f ⋅
and
Wpump = Q⋅ ∆p
The required power is
Power =
The daily cost is then
C = cost⋅ Power⋅ day
D
ρ⋅ V⋅ D
2
Wpump
ηp ⋅ ηm
∆p = 94.4⋅ kPa
Wpump = 61.3⋅ kW
Power = 84.9⋅ kW
C = 285 dollars
ηp = 85⋅ %
cost =
ηm = 85⋅ %
0.14
(dollars)
kW⋅ hr
Re = 8.49 × 10
5
Problem 8.177
[Difficulty: 3]
Problem 8.176
[Difficulty: 3]
Problem 8.175
Given:
Flow in pipeline with pump
Find:
Pump pressure ∆p
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞ ∆p
V1
V2
1
2
pump
= h lT
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 +
ρ
⎝
⎠ ⎝
⎠
2
hl = f ⋅
L V
⋅
D 2
h lm = f ⋅
Le V2
⋅
D 2
2
h lm = K⋅
V
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
From Section 8.8
Hence
and
l
L = 50⋅ m
D = 125 ⋅ mm
Q = 50⋅
p 1 = 150 ⋅ kPa
p 2 = 0 ⋅ kPa
z1 = 15⋅ m
z2 = 30⋅ m
Kent = 0.5
Lelbow90 = 30⋅ D
Lelbow90 = 3.75 m
LGV = 8 ⋅ D
LAV = 150 ⋅ D
LAV = 18.75 m
ρ = 1000
V =
Q
V = 4.07
⎛ π⋅ D
⎜
⎝ 4 ⎠
2⎞
e = 0.15⋅ mm
s
− 3 N⋅ s
kg
μ = 1.3⋅ 10
3
m
m
Re =
s
⋅
2
(Table A.8)
m
ρ⋅ V⋅ D
5
Re = 3.918 × 10
μ
⎛ e
⎞
⎜
1
2.51
D
= −2 ⋅ log⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
LGV = 1 m
f = 0.0212
The loss is then
2
h lT =
Lelbow90
LGV
LAV
⎛ L
⎞
+ 7⋅ f ⋅
+ 5⋅ f ⋅
+ f⋅
+ Kent
2 ⎝ D
D
D
D
⎠
V
The energy equation becomes
⋅⎜f ⋅
p1 − p2
ρ
2
h lT = 145
m
2
s
2
∆ppump
V
+ g ⋅ z1 − z2 −
+
= h lT
2
ρ
(
(
)
)
2
V
(
∆ppump = ρ⋅ h lT + ρ⋅ g ⋅ z2 − z1 + ρ⋅
+ p2 − p1
2
)
∆ppump = 150 ⋅ kPa
Problem 8.174
Given:
Flow through water pump
Find:
Power required
Solution:
Basic equations
[Difficulty: 1]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
Vd
Vs
d
s
h pump = ⎜
+
+ g ⋅ zd − ⎜
+
+ g ⋅ zs
2
2
⎝ρ
⎠ ⎝ρ
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
In this case we assume
Ds = Dd
The available data is
∆p = 35⋅ psi
Then
h pump =
Vs = Vd
so
Q = 500 ⋅ gpm
pd − ps
ρ
=
∆p
η = 80⋅ %
Wpump = mpump⋅ h pump
and
ρ
∆p
∆p
Wpump = mpump⋅
= ρ⋅ Q⋅
ρ
ρ
Wpump = Q⋅ ∆p
Wpump = 5615
ft⋅ lbf
s
Wpump = 10.2⋅ hp
Note that the software cannot render a dot, so the power is Wpump and mass flow rate is mpump!
For an efficiency of
η = 80 %
Wrequired =
Wpump
η
Wrequired = 12.8⋅ hp
Problem 8.173
Given:
Flow through water pump
Find:
Power required
Solution:
Basic equations
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
Vd
Vs
d
s
h pump = ⎜
+
+ g ⋅ zd − ⎜
+
+ g ⋅ zs
2
2
⎝ρ
⎠ ⎝ρ
⎠
V=
Q
A
=
4⋅ Q
2
π⋅ D
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
3
Hence for the inlet
4
lbm
1 ⋅ slug
ft
Vs =
× 25⋅
×
×
×
s
32.2⋅ lbm 1.94⋅ slug
π
For the outlet
4
lbm
1 ⋅ slug
ft
Vd =
× 25⋅
×
×
×
s
32.2⋅ lbm 1.94⋅ slug
π
3
Then
h pump =
pd − ps
ρ
2
+
⎛ 12 ⋅ 1 ⎞
⎜
⎝ 3 ft ⎠
2
⎛ 12 ⋅ 1 ⎞
⎜
⎝ 2 ft ⎠
2
ft
Vs = 8.15⋅
s
p s = −2.5⋅ psi
ft
Vd = 18.3⋅
s
p d = 50⋅ psi
2
Vd − Vs
Wpump = mpump⋅ h pump
and
2
2
2
⎛⎜ p − p
Vd − Vs ⎞
d
s
Wpump = mpump⋅ ⎜
+
2
⎝ ρ
⎠
Note that the software cannot render a dot, so the power is Wpump and mass flow rate is mpump!
⎡
lbm
1 ⋅ slug
lbf
Wpump = 25⋅
×
× ⎢( 50 − −2.5) ⋅
×
2
s
32.2⋅ lbm ⎢
in
⎣
Wpump = 5.69⋅ hp
For an efficiency
of
η = 70⋅ %
2
3
2
2
1
lbf ⋅ s ⎥⎤
1 ⋅ hp
2
2 ⎛ ft ⎞
⎛ 12⋅ in ⎞ × ft
+
×
18.3
−
8.15
⋅
×
×
⎜
⎜
ft⋅ lbf
1.94⋅ slug
2
slug⋅ ft⎥
⎝ 1 ⋅ ft ⎠
⎝s⎠
⎦ 550 ⋅
(
)
s
Wrequired =
Wpump
η
Wrequired = 8.13⋅ hp
Problem 8.172
Problem 8.151
[Difficulty: 5] Part 1/2
Problem 8.172
[Difficulty: 5] Part 2/2
Problem 8.171
[Difficulty: 4]
Problem 8.170
[Difficulty: 3]
Problem 8.169
[Difficulty: 3] Part 1/2
Problem 8.169
[Difficulty: 3] Part 2/2
Problem 8.167
Given:
Flow in a tube
Find:
Effect of diameter; Plot flow rate versus diameter
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = hl
⎝
⎠ ⎝ρ
⎠
Re =
f =
ρ⋅ V⋅ D
64
μ
(8.29)
2
hl = f ⋅
L V
⋅
D 2
(8.36)
(8.34)
(Laminar)
Re
⎞
⎛ e
⎜
1
2.51
D
(8.37)
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
(Turbulent)
The energy equation (Eq. 8.29) becomes for flow in a tube
2
L V
p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅
D 2
This cannot be solved explicitly for velocity V (and hence flow rate Q), because f depends on V; solution for a given diameter D
requires iteration (or use of Solver)
Flow Rate versus Tube Diameter for Fixed Dp
0.8
0.6
3
Q (m /s)
x 10
4
Laminar
Turbulent
0.4
0.2
0.0
0.0
2.5
5.0
D (mm)
7.5
10.0
Problem 8.166
Given:
Flow of air in square duct
Find:
Minimum required size
Solution:
Basic equations
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
2
V1
V2
1
2
L V
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
=
h
h
=
f
⋅
⋅
⎜ρ
⎜
1
2
l l
2
2
Dh 2
⎝
⎠ ⎝ρ
⎠
4⋅ A
Dh =
Pw
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses
Available data
Q = 1500⋅ cfm
ρH2O = 1.94⋅
L = 1000⋅ ft
slug
ft
e = 0.00015 ⋅ ft
− 4 ft
ν = 1.47⋅ 10
3
⋅
2
s
ρ = 0.00247 ⋅
(Table 8.1)
slug
ft
∆h = 0.75⋅ in
(Table A.9)
3
Hence for flow between the inlet (Point 1) and the exit (2) the energy equation becomes
p1
ρ
−
p2
ρ
∆p
=
ρ
2
= f⋅
For a square duct
4⋅ h⋅ h
Dh =
=h
2⋅ ( h + h)
Hence
∆p = ρ⋅ f ⋅ L⋅
L V
⋅
Dh 2
2
V
2⋅ h
A = h⋅ h = h
and also
2
= ρ⋅ f ⋅ L⋅
∆p = ρH2O⋅ g ⋅ ∆h
and
∆p = 3.90
lbf
ft
2
∆p = 0.0271⋅ psi
2
2
Q
2
2⋅ h⋅ A
=
ρ⋅ f ⋅ L⋅ Q
2⋅ h
5
1
Solving for h
⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞
h=⎜
⎝ 2⋅ ∆p ⎠
5
(1)
Equation 1 is tricky because h is unknown, so Dh is unknown, hence V is unknown (even though Q is known), and Re and hence f
are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simmultaneously by varying h, but here we try
guesses:
1
f = 0.01
Dh = h
⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞
h = ⎜
⎝ 2 ⋅ ∆p ⎠
Dh = 1.15⋅ ft
5
h = 1.15⋅ ft
V =
Q
h
Re =
V⋅ Dh
ν
V = 19.0⋅
2
Re = 1.48 × 10
5
ft
s
⎛ e
⎞
⎜
Dh
1
2.51
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
f = 0.0174
1
⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞
h = ⎜
⎝ 2 ⋅ ∆p ⎠
5
h = 1.28⋅ ft
V =
Q
h
Dh = h
Dh = 1.28⋅ ft
Re =
⎛ e
⎞
⎜
Dh
1
2.51
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
V = 15.2⋅
2
V⋅ Dh
ν
ft
s
Re = 1.33 × 10
5
f = 0.0177
1
⎛ ρ⋅ f ⋅ L⋅ Q2 ⎞
h = ⎜
⎝ 2 ⋅ ∆p ⎠
Dh = h
V =
V⋅ Dh
ν
Q
h
Re =
⎛ e
⎞
⎜
D
1
h
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
h = 1.28⋅ ft
Re =
h = 1.28⋅ ft
Dh = 1.28⋅ ft
Given
Hence
5
Dh = h
Re = 1.32 × 10
Dh = 1.28⋅ ft
V = 15.1⋅
2
V⋅ Dh
ν
ft
s
Re = 1.32 × 10
f = 0.0177
V =
Q
h
2
V = 15.1⋅
ft
s
5
In this process h and f have converged to a solution. The minimum dimensions are 1.28 ft square (15.4 in square)
5
Problem 8.165
[Difficulty: 4]
Problem 8.164
Given:
Flow of air in rectangular duct
Find:
Minimum required size
Solution:
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
hl = f ⋅
4⋅ A
Dh =
Pw
L V
⋅
Dh 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses
3
Q = 1⋅
Available data
m
L = 100 ⋅ m
s
kg
ρH2O = 999 ⋅
3
m
ρ = 1.25⋅
∆h = 25⋅ mm
ar = 3
e = 0⋅ m
2
−5 m
kg
ν = 1.41⋅ 10
3
⋅
(Table A.10)
s
m
Hence for flow between the inlet (Point 1) and the exit (2) the energy equation becomes
p1
ρ
−
p2
ρ
=
2
∆p
= f⋅
ρ
L V
⋅
Dh 2
∆p = ρH2O⋅ g ⋅ ∆h
and
∆p = 245 Pa
2
For a rectangular duct
4⋅ b⋅ h
2 ⋅ h ⋅ ar
2 ⋅ h ⋅ ar
Dh =
=
=
1 + ar
2⋅ ( b + h)
h ⋅ ( 1 + ar)
Hence
∆p = ρ⋅ f ⋅ L⋅
2
V
2
⋅
( 1 + ar)
2 ⋅ h ⋅ ar
2
= ρ⋅ f ⋅ L⋅
1
Solving for h
⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar) ⎤
⎥
h=⎢
⋅
⎢ 4⋅ ∆p
3 ⎥
ar
⎣
⎦
Q
2
2 b
A = b⋅ h = h ⋅
and also
⋅
( 1 + ar)
2⋅ A
2 ⋅ h ⋅ ar
2
=
ρ⋅ f ⋅ L⋅ Q
4
⋅
h
2
= h ⋅ ar
( 1 + ar) 1
⋅
3
5
h
ar
5
(1)
Equation 1 is tricky because h is unknown, so Dh is unknown, hence V is unknown (even though Q is known), and Re and hence f
are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simmultaneously by varying h, but here we try
guesses:
1
f = 0.01
⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar)⎤
⎥
h = ⎢
⋅
⎢ 4 ⋅ ∆p
3 ⎥
ar
⎣
⎦
5
h = 0.180 m
V =
Q
2
h ⋅ ar
V = 10.3
m
s
2 ⋅ h ⋅ ar
Dh =
1 + ar
Given
Dh = 0.270 m
Re =
⎛ e
⎞
⎜
D
1
h
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V⋅ Dh
ν
Re = 1.97 × 10
5
f = 0.0157
1
⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar)⎤
⎥
h = ⎢
⋅
⎢ 4 ⋅ ∆p
3 ⎥
ar
⎣
⎦
2 ⋅ h ⋅ ar
Dh =
1 + ar
Given
5
h = 0.197 m
V =
Q
V = 8.59
2
h ⋅ ar
Dh = 0.295 m
Re =
⎛ e
⎞
⎜
D
1
h
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V⋅ Dh
ν
m
s
5
Re = 1.8 × 10
f = 0.0160
1
⎡ ρ⋅ f ⋅ L⋅ Q2 ( 1 + ar)⎤
⎥
h = ⎢
⋅
⎢ 4 ⋅ ∆p
3 ⎥
ar
⎣
⎦
2 ⋅ h ⋅ ar
Dh =
1 + ar
Given
Hence
5
h = 0.198 m
Dh = 0.297 m
h = 198 mm
Q
2
Re =
V⋅ Dh
ν
Re = 1.79 × 10
Dh = 0.297 m
m
s
5
f = 0.0160
b = 2⋅ h
b = 395 ⋅ mm
V =
Q
2
h ⋅ ar
2 ⋅ h ⋅ ar
Dh =
1 + ar
V = 8.53
h ⋅ ar
⎛ e
⎞
⎜
D
1
h
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
h = 0.198 m
V =
Re =
V⋅ Dh
In this process h and f have converged to a solution. The minimum dimensions are 198 mm by 395 mm
ν
V = 8.53
m
s
Re = 1.79 × 10
5
Problem 8.163
Given:
Flow out of reservoir by pump
Find:
Smallest pipe needed
Solution:
[Difficulty: 4]
2
2
2
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V2
V1
V2
Le V2
1
2
L V2
⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g⋅ z2 = h lT h lT = h l + hlm = f ⋅ D ⋅ 2 + Kent ⋅ 2 + f ⋅ D ⋅ 2
⎝
⎠ ⎝
⎠
Basic equations
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl <<
Hence for flow between the free surface (Point 1) and the pump inlet (2) the energy equation becomes
−
Solving for h 2 = p 2/ρg
2
p2
− g ⋅ z2 −
ρ
V2
2
=−
2
2
2
Le V2
V
L V
V
− g ⋅ z2 −
= f⋅ ⋅
+ Kent ⋅
+ f ⋅ ⋅ and
2
2
D 2
ρ
D 2
p2
2
Le ⎞
⎤
V ⎡ ⎛L
h 2 = −z2 −
⋅ ⎢f ⋅ ⎜ +
+ Kent⎥
2⋅ g ⎣ ⎝ D
D⎠
⎦
From Table 8.2
Kent = 0.78
We also have
e = 0.046 ⋅ mm (Table 8.1) ν = 1.51 × 10
and we are given
Q = 6⋅
(1)
for rentrant, and from Table 8.4 two standard elbows lead to
2
−6 m
⋅
3
−3m
L
Q = 6 × 10
s
p = ρ⋅ g ⋅ h
s
Le
D
= 2 × 30 = 60
(Table A.8)
z2 = 3.5⋅ m L = ( 3.5 + 4.5) ⋅ m L = 8 m
s
h 2 = −6 ⋅ m
Equation 1 is tricky because D is unknown, so V is unknown (even though Q is known), L/D and Le/D are unknown, and Re and
hence f are unknown! We COULD set up Excel to solve Eq 1, the Reynolds number, and f, simultaneously by varying D, but here
we try guesses:
D = 2.5⋅ cm
V =
4⋅ Q
2
π⋅ D
Given
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V = 12.2
m
s
Re =
V⋅ D
ν
f = 0.0238
2
Le ⎞
⎤
V ⎡ ⎛L
h 2 = −z2 −
⋅ ⎢f ⋅ ⎜ +
+ Kent⎥ h 2 = −78.45 m
2⋅ g ⎣ ⎝ D
D⎠
⎦
but we need -6 m!
Re = 2.02 × 10
5
D = 5 ⋅ cm V =
4⋅ Q
V = 3.06
2
π⋅ D
Given
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
m
s
Re =
V⋅ D
ν
Re = 1.01 × 10
5
Re = 9.92 × 10
4
f = 0.0219
2
⎡ ⎛ L Le ⎞
⎤
h 2 = −z2 −
⋅ ⎢f ⋅ ⎜ +
+ Kent⎥ h 2 = −6.16 m
2⋅ g ⎣ ⎝ D
D⎠
⎦
V
D = 5.1⋅ cm
V =
4⋅ Q
2
π⋅ D
Given
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V = 2.94
m
s
but we need -6 m!
Re =
f = 0.0219
2
Le ⎞
⎤
V ⎡ ⎛L
h 2 = −z2 −
⋅ ⎢f ⋅ ⎜ +
+ Kent⎥ h 2 = −5.93 m
2⋅ g ⎣ ⎝ D
D⎠
⎦
To within 1%, we can use 5-5.1 cm tubing; this corresponds to standard 2 in pipe.
V⋅ D
ν
Problem 8.162
Given:
Hydraulic press system
Find:
Minimum required diameter of tubing
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
L V2
hl = f ⋅ ⋅
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses
The flow rate is low and it's oil, so try assuming laminar flow. Then, from Eq. 8.13c
1
∆p =
128⋅ μ ⋅ Q⋅ L
D=
or
4
π⋅ D
0.0209⋅
− 2 N⋅s
μ = 3.5 × 10
For SAE 10W oil at 100 oF (Fig. A.2, 38oC)
⋅
2
1⋅
m
4
lbf ⋅ s
ft
×
⎛ 128⋅ μ ⋅ Q⋅ L ⎞
⎜
⎝ π⋅ ∆p ⎠
2
− 4 lbf ⋅ s
μ = 7.32 × 10
N⋅s
⋅
ft
2
2
m
1
Hence
3
2
2
⎡ 128
ft
in
1⋅ ft ⎞ ⎤⎥
− 4 lbf ⋅ s
D = ⎢
× 7.32 × 10
× 0.02⋅
× 165⋅ ft ×
× ⎛⎜
⎢π
2
s
( 3000 − 2750) ⋅ lbf ⎝ 12⋅ in ⎠ ⎥
ft
⎣
⎦
Check Re to assure flow is laminar
From Table A.2
SG oil = 0.92
Re = 0.92 × 1.94⋅
V=
Q
A
4⋅ Q
=
2
π⋅ D
Re =
so
slug
ft
3
× 15.4⋅
ft
s
×
3
12 1 ⎞
V=
× 0.02⋅
× ⎛⎜
⋅
s
π
⎝ 0.488 ft ⎠
4
0.488
⋅ ft ×
12
ft
4
D = 0.0407⋅ ft
D = 0.488⋅ in
2
V = 15.4⋅
ft
s
SG oil⋅ ρH2O⋅ V⋅ D
μ
ft
2
−4
7.32 × 10
×
lbf ⋅ s
lbf ⋅ s
2
slug ⋅ ft
Hence the flow is laminar, Re < 2300. The minimum diameter is 0.488 in, so 0.5 in ID tube should be chosen
Re = 1527
Problem 8.161
Given:
Flow from large reservoir
Find:
Diameter for flow rates in two pipes to be same
Solution:
Basic equations
[Difficulty: 5]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
2
hl = f ⋅
2
L V
⋅
D 2
V
h lm = Kent⋅
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 50⋅ mm
H = 10⋅ m
L = 10⋅ m
e = 0.15⋅ mm
(Table 8.1)
ν = 1 ⋅ 10
2
−6 m
⋅
Kent = 0.5
(Table A.8)
s
For the pipe of length L the energy equation becomes
2
V2
L V2
g ⋅ z1 − z2 − ⋅ V2 = f ⋅ ⋅
+ Kent⋅
2
2
D 2
(
Solving for V
)
2
2
and
V2 = V
z1 − z2 = H
2⋅ g⋅ H
V=
f⋅
We also have
1
L
+ Kent + 1
D
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
(1)
Re =
(2)
V⋅ D
(3)
ν
We must solve Eqs. 1, 2 and 3 iteratively.
Make a guess for V
and
Then
V = 1⋅
m
s
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
2⋅ g⋅ H
V =
f⋅
Repeating
Then
Re =
L
+ Kent + 1
D
V⋅ D
ν
Re =
V⋅ D
Re = 5.00 × 10
ν
f = 0.0286
V = 5.21
m
s
Re = 2.61 × 10
5
4
and
Then
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
Repeating
and
Then
Re =
L
D
V = 5.36
+ Kent + 1
V⋅ D
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
π
Q =
4
s
5
f = 0.0267
2⋅ g⋅ H
V =
m
Re = 2.68 × 10
ν
f⋅
Hence
f = 0.0267
V = 5.36
L
+ Kent + 1
D
m
s
3
2
⋅D ⋅V
Q = 0.0105
m
Q = 10.5⋅
s
l
s
This is the flow rate we require in the second pipe (of length 2L)
For the pipe of length 2L the energy equation becomes
2
V2
2 ⋅ L V2
g ⋅ z1 − z2 − ⋅ V2 = f ⋅
⋅
+ Kent⋅
2
2
2
D
(
)
2
Hence
H=
1
V
2⋅ g
⋅ ⎛⎜ f ⋅
⎝
2⋅ L
D
+ Kent + 1⎞
and
Re =
V⋅ D
Re = 2.23 × 10
ν
3
and
V2 = V
z1 − z2 = H
Q = 0.0105
m
D = 0.06⋅ m
Then we have
Q
π
4
5
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
V =
2
V = 3.72
⋅D
f = 0.0256
2
Using Eq 4 to find H
V ⎛ 2⋅ L
Hiterate =
⋅⎜f ⋅
+ Kent + 1⎞
2⋅ g ⎝ D
⎠
Hence the diameter is too large: Only a head of
s
(4)
⎠
We must make a guess for D (larger than the other pipe)
Then
2
2
Hiterate = 7.07 m
Hiterate = 7.07 m
But
H = 10 m
would be needed to generate the flow. We make D smaller
m
s
Try
Then
and
D = 0.055 ⋅ m
Re =
V⋅ D
Re = 2.43 × 10
ν
Q
V =
Then we have
π
4
5
⎛ e
⎞
⎜ D
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
1
2
V = 4.43
⋅D
m
s
f = 0.0261
2
Using Eq 4 to find H
2⋅ L
Hiterate =
⋅ ⎛⎜ f ⋅
+ Kent + 1⎞
2⋅ g ⎝ D
⎠
V
Hence the diameter is too small: A head of
Try
Then
and
D = 0.056 ⋅ m
Re =
Q
V =
Re = 2.39 × 10
ν
But
H = 10 m
Hiterate = 10.97 m would be needed. We make D slightly larger
Then we have
V⋅ D
Hiterate = 10.97 m
π
4
5
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V = 4.27
2
⋅D
m
s
f = 0.0260
2
Using Eq 4 to find H
V ⎛ 2⋅ L
Hiterate =
⋅⎜f ⋅
+ Kent + 1⎞
2⋅ g ⎝ D
⎠
Hence the diameter is too large A head of
Try
Then
and
Hiterate = 10.02 m
D = 0.05602 ⋅ m Then we have
Re =
V⋅ D
Q
π
4
5
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
But
H = 10 m
would be needed. We can make D smaller
V =
Re = 2.39 × 10
ν
Hiterate = 10.02 m
2
V = 4.27
⋅D
m
s
f = 0.0260
2
Using Eq 4 to find H
V ⎛ 2⋅ L
Hiterate =
⋅⎜f ⋅
+ Kent + 1⎞
2⋅ g ⎝ D
⎠
Hiterate = 10 m
Hence we have
D = 0.05602 m
V = 4.27
Check
Q = 0.0105
3
D = 56.02 ⋅ mm
m
π
s
4
2
⋅ D ⋅ V = 0.0105
3
m
s
m
s
But
H = 10 m
Problem 8.160
[Difficulty: 5]
Applying the energy equation between inlet and exit:
∆p
ρ
= f
L V 2
D 2
"Old school":
or
∆p ρf V 2
=
L
D 2
∆p ⎛ ∆p ⎞ ⎛ Q0 ⎞
=⎜ ⎟ ⎜ ⎟
L ⎝ L ⎠0 ⎜⎝ Q ⎟⎠
Q (gpm)
20
18
16
Flow (gpm)
14
12
10
8
6
4
2
0
0.00
0.01
Your boss was wrong!
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
0.02
8.00
8.25
8.50
8.75
9.00
Q (ft3/s)
D=
e=
2
ν =
ρ =
V (ft/s)
Re
f
1 in
0.00015 ft
2
1.08E-05 ft /s
3
1.94 slug/ft
∆p (old
∆p (psi/ft)
school) (psi)
0.00279
0.511
3940
0.0401
0.00085
0.00085
0.00334
0.613
4728
0.0380
0.00122
0.00115
0.00390
0.715
5516
0.0364
0.00166
0.00150
0.00446
0.817
6304
0.0350
0.00216
0.00189
0.00501
0.919
7092
0.0339
0.00274
0.00232
1.021Pressure
7881 Drop 0.0329
0.00338
0.00278
Flow0.00557
Rate versus
0.00613
1.123
8669
0.0321
0.00409
0.00328
0.00668
1.226
9457
0.0314
0.00487
0.00381
0.00724
1.328
10245
0.0307
0.00571
0.00438
0.00780
1.430
11033
0.0301
0.00663
0.00498
0.00836
1.532
11821
0.0296
0.00761
0.00561
0.00891
1.634
12609
0.0291
0.00865
0.00628
0.00947
1.736
13397
0.0286
0.00977
0.00698
0.01003
1.838
14185
0.0282
0.01095
0.00771
0.01058
1.940
14973
0.0278
0.01220
0.00847
0.01114
2.043
15761
0.0275
0.01352
0.00927
0.01170
2.145
16549
0.0272
0.01491
0.01010
0.01225
2.247
17337
0.0268
0.01636
0.01095
0.01281
2.349
18125
0.0265
0.01788
0.01184
0.01337
2.451
18913
0.0263
0.01947
0.01276
0.01393
2.553
19701
0.0260
0.02113
0.01370
0.01448
2.655
20489
0.0258
0.02285
0.01468
School"
0.01504
2.758
21277
0.0255 "Old 0.02465
0.01569
0.01560
2.860
22065
0.0253 Exact0.02651
0.01672
0.01615
2.962
22854
0.0251
0.02843
0.01779
0.01671
3.064
23642
0.0249
0.03043
0.01888
0.01727
3.166
24430
0.0247
0.03249
0.02000
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.01783
3.268
25218
0.0245
0.03462
0.02115
Pressure Drop (psi/ft)
0.01838
3.370
26006
0.0243
0.03682
0.02233
0.01894
3.472
26794
0.0242
0.03908
0.02354
0.01950
3.575
27582
0.0240
0.04142
0.02477
0.02005
3.677
28370
0.0238
0.04382
0.02604
Problem 8.159
[Difficulty: 4] Part 1/2
Problem 8.159
[Difficulty: 4] Part 2/2
Problem 8.158
[Difficulty: 4] Part 1/2
Problem 8.158
[Difficulty: 4] Part 2/2
Problem 8.157
[Difficulty: 4]
Problem 8.156
[Difficulty: 5]
Given:
Tank with drain hose
Find:
Flow rate at different instants; Estimate of drain time
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
Basic equations
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor loss at entrance (L >>; verify later)
Available data
L = 1⋅ m
D = 15⋅ mm
e = 0.2⋅ mm
3
Vol = 30⋅ m
Hence the energy equation applied between the tank free surface (Point 1) and the hose exit (Point 2, z = 0) becomes
g ⋅ z1 −
Solving for V
V=
We also have
Re =
From Fig. A.2 (20oC)
In addition
V2
2
2
2
2
V
L V
= g ⋅ z1 −
= f⋅ ⋅
2
D 2
2⋅ g⋅ h
(1)
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
V⋅ D
Re = c⋅ V (2)
or
ν
2
−6 m
ν = 1.8 × 10
⋅
c =
s
D
c = 8333
ν
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
h = 10⋅ m
initially
where
c=
and
D
ν
s
m
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
f = 0.01
V =
then
2⋅ g⋅ h
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
V = 10.8
m
s
Re = c⋅ V
Re = 9.04 × 10
4
Given
Given
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f = 0.0427
3.7
f
Re⋅ f ⎠
⎝
V =
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f = 0.0427
3.7
f
Re⋅ f ⎠
⎝
V =
2⋅ g⋅ h
L⎞
⎛1 + f ⋅
⎜
D⎠
⎝
2⋅ g⋅ h
L⎞
⎛1 + f ⋅
⎜
D⎠
⎝
Note that we could use Excel's Solver for this problem
π⋅ D
4
Given
3
−3m
Q = 1.26 × 10
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f = 0.0430
f
⎝ 3.7 Re⋅ f ⎠
V =
2
Q = V⋅
π⋅ D
4
Given
Q = 8.9 × 10
⎞
⎛ e
⎜ D
2.51
= −2.0⋅ log ⎜
+
f = 0.0444
f
⎝ 3.7 Re⋅ f ⎠
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
2⋅ g⋅ h
L⎞
⎛1 + f ⋅
⎜
D⎠
⎝
m
s
L
D
Re = c⋅ V
Re = 5.95 × 10
4
Re = c⋅ V
Re = 5.95 × 10
4
= 2.8
Ke = 0.5
h lm < h l
l
s
V = 5.04
V = 5.04
Q = 0.890 ⋅
s
V =
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f = 0.0444
f
⎝ 3.7 Re⋅ f ⎠
2
Q = V⋅
2⋅ g⋅ h
3
−4m
1
The flow rate is then
s
m
s
m
s
Re = c⋅ V
Re = 4.20 × 10
4
Re = c⋅ V
Re = 4.20 × 10
4
Re = c⋅ V
Re = 1.85 × 10
4
Re = c⋅ V
Re = 1.85 × 10
4
l
s
h = 1⋅ m
Next we recompute everything for
Given
Q = 1.26⋅
s
V =
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f = 0.0430
3.7
f
Re⋅ f ⎠
⎝
The flow rate is then
f⋅
m
h = 5⋅ m
Next we recompute everything for
Given
V = 7.14
Note:
2
Q = V⋅
The flow rate is then
V = 7.14
π⋅ D
4
V =
3
−4m
Q = 3.93 × 10
s
2⋅ g⋅ h
L⎞
⎛1 + f ⋅
⎜
D⎠
⎝
2⋅ g⋅ h
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
V = 2.23
V = 2.23
Q = 0.393 ⋅
m
s
m
s
l
s
Initially we have dQ/dt = -1.26L/s, then -.890 L/s, then -0.393 L/s. These occur at h = 10 m, 5 m and 1 m. The corresponding
volumes in the tank are then Q = 30,000 L, 15,000 L, and 3,000 L3. Using Excel we can fit a power trendline to the dQ/dt versus Q
data to find, approximately
1
dQ
dt
= −0.00683 ⋅ Q
t = 293 ⋅ ( 30 −
2
Q)
where dQ/dt is in L/s and t is s. Solving this with initial condition Q = -1.26 L/s when t = 0 gives
Hence, when Q = 3000 L (h = 1 m)
t = 293 ⋅ ( 30000 −
3000) ⋅ s
4
t = 3.47 × 10 s
t = 9.64⋅ hr
Problem 8.155
[Difficulty: 4]
Given:
Tank with drainpipe
Find:
Flow rate for rentrant, square-edged, and rounded entrances
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT
⎝
⎠ ⎝
⎠
Basic equations
2
h lT = f ⋅
2
L V
V
⋅
+ Kent⋅
2
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 1 ⋅ in
L = 2 ⋅ ft
e = 0.00085 ⋅ ft
h = 3 ⋅ ft
(Table 8.1)
r = 0.2⋅ in
Hence the energy equation applied between the tank free surface (Point 1) and the pipe exit (Point 2, z = 0) becomes
V2
g ⋅ z1 −
Solving for V
V=
2
2
2
2
2⋅ g⋅ H
(1)
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
H = h+L
where now we have
We also have
From Table A.7 (20oC)
In addition
Re =
2
V
L V
V
= g ⋅ z1 −
= f⋅ ⋅
+ Kent⋅
2
2
D 2
V⋅ D
Re = c⋅ V (2)
or
ν
ν = 1.01 × 10
H = 5 ft
2
−6 m
⋅
− 5 ft
ν = 1.09 × 10
s
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
c=
where
2
s
c =
D
ν
Make a guess for f
f = 0.01
(3)
Re = c⋅ V
Given
Kent = 0.78
V =
then
Re = 9.67 × 10
2⋅ g⋅ H
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
4
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
f = 0.0388
ν
c = 7665⋅
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
For a reentrant entrance, from Table 8.2
D
V = 3.85
m
s
s
ft
2⋅ g⋅ H
V =
Given
V = 3.32
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V =
m
s
Re = c⋅ V
Re = 8.35 × 10
4
Re = c⋅ V
Re = 8.35 × 10
4
f = 0.0389
2⋅ g⋅ H
V = 3.32
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
m
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
Q = V⋅
π⋅ D
Q = 0.0594
4
f = 0.01
Given
Given
then
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V =
V =
2⋅ g⋅ H
L⎞
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
V = 3.51
L⎞
V = 3.51
L⎞
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
π⋅ D
m
s
m
Make a guess for f
f = 0.01
V = 4.83
Q = 0.0627
4
For a rounded entrance, from Table 8.2
r
D
s
s
ft
5
Re = c⋅ V
Re = 8.82 × 10
4
Re = c⋅ V
Re = 8.82 × 10
4
Re = c⋅ V
s
3
Q = 28.2⋅ gpm
s
= 0.2
then
m
m
Re = 1.04 × 10
V = 4.14
f = 0.0388
2⋅ g⋅ H
Q = V⋅
Q = 26.7⋅ gpm
2⋅ g⋅ H
2
The flow rate is then
s
f = 0.0387
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V =
3
Kent = 0.5
For a square-edged entrance, from Table 8.2
Make a guess for f
ft
Kent = 0.04
V =
2⋅ g⋅ H
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
Re = c⋅ V
Re = 1.22 × 10
5
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
V =
f = 0.0386
2⋅ g⋅ H
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
V =
V =
s
Re = c⋅ V
Re = 9.80 × 10
4
Re = c⋅ V
Re = 9.80 × 10
4
Re = c⋅ V
Re = 9.80 × 10
4
f = 0.0388
2⋅ g⋅ H
m
V = 3.89
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
m
V = 3.90
s
f = 0.0388
2⋅ g⋅ H
m
V = 3.89
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
In summary:
Q = V⋅
π⋅ D
4
Renentrant: Q = 26.7⋅ gpm
Q = 0.0697
ft
3
s
Square-edged:
Q = 31.3⋅ gpm
Q = 28.2⋅ gpm
Rounded:
Q = 31.3⋅ gpm
Problem 8.154
[Difficulty: 4]
Given:
Tank with drainpipe
Find:
Flow rate for rentrant, square-edged, and rounded entrances
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
h lT = f ⋅
2
L V
V
⋅
+ Kent⋅
2
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 1 ⋅ in
L = 2 ⋅ ft
e = 0.00085 ⋅ ft
h = 3 ⋅ ft
(Table 8.1)
r = 0.2⋅ in
Hence the energy equation applied between the tank free surface (Point 1) and the pipe exit (Point 2, z = 0) becomes
V2
g ⋅ z1 −
Solving for V
V=
We also have
Re =
From Table A.7 (20oC)
In addition
2
2
2
2
2
V
L V
V
= g ⋅ z1 −
= f⋅ ⋅
+ Kent⋅
2
2
D 2
2⋅ g⋅ h
(1)
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
V⋅ D
Re = c⋅ V (2)
or
ν
ν = 1.01 × 10
2
−6 m
⋅
− 5 ft
ν = 1.09 × 10
s
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
c=
where
2
s
c =
D
ν
Make a guess for f
f = 0.01
(3)
Re = c⋅ V
Given
Kent = 0.78
V =
then
Re = 7.49 × 10
2⋅ g⋅ h
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
4
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
f = 0.0389
ν
c = 7665⋅
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
For a reentrant entrance, from Table 8.2
D
V = 2.98
m
s
s
ft
2⋅ g⋅ h
V =
Given
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V = 2.57
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V =
m
s
Re = c⋅ V
Re = 6.46 × 10
4
Re = c⋅ V
Re = 6.46 × 10
4
Re = c⋅ V
Re = 6.46 × 10
4
f = 0.0391
2⋅ g⋅ h
V =
Given
V = 2.57
m
s
f = 0.0391
2⋅ g⋅ h
V = 2.57
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
m
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
Q = V⋅
π⋅ D
Q = 0.0460
4
Given
then
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V =
Given
f = 0.01
V =
2⋅ g⋅ h
L⎞
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
2⋅ g⋅ h
V = 2.71
L⎞
m
s
2⋅ g⋅ h
V = 2.71
L⎞
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
Q = V⋅
π⋅ D
m
Q = 0.0485
4
For a rounded entrance, from Table 8.2
r
D
m
Re = 8.07 × 10
4
Re = c⋅ V
Re = 6.83 × 10
4
Re = c⋅ V
Re = 6.82 × 10
4
V = 3.21
s
Re = c⋅ V
f = 0.0390
2
The flow rate is then
Q = 20.6⋅ gpm
s
f = 0.0389
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
V =
3
Kent = 0.5
For a square-edged entrance, from Table 8.2
Make a guess for f
ft
= 0.2
s
ft
3
s
Q = 21.8⋅ gpm
Kent = 0.04
Make a guess for f f = 0.01
V =
then
V = 3.74
m
V =
V =
2⋅ g⋅ h
V =
m
V = 3.02
L⎞
4
s
Re = c⋅ V
Re = 7.59 × 10
4
Re = c⋅ V
Re = 7.58 × 10
4
Re = c⋅ V
Re = 7.58 × 10
4
f = 0.0389
2⋅ g⋅ h
m
V = 3.01
L⎞
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
Re = 9.41 × 10
f = 0.0388
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
Re = c⋅ V
s
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
2⋅ g⋅ h
s
f = 0.0389
2⋅ g⋅ h
m
V = 3.01
L⎞
⎛1 + K + f ⋅
⎜
ent
D⎠
⎝
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
In summary:
Q = V⋅
π⋅ D
4
Renentrant: Q = 20.6⋅ gpm
Q = 0.0539
ft
3
s
Square-edged:
Q = 24.2⋅ gpm
Q = 21.8⋅ gpm
Rounded:
Q = 24.2⋅ gpm
Problem 8.153
Given:
Syphon system
Find:
Flow rate; Minimum pressure
Solution:
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
h lT = f ⋅
L V
⋅
+ h lm
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Hence the energy equation applied between the tank free surface (Point 1) and the tube exit (Point 2, z = 0) becomes
g ⋅ z1 −
V2
2
2
2
2
2
Le V2
V
L V
V
= g ⋅ z1 −
= f⋅ ⋅
+ Kent⋅
+ f⋅ ⋅
2
2
D 2
D 2
Kent = 0.78
From Table 8.2 for reentrant entrance
For the bend
R
D
=9
V=
The two lengths are
Le = 56⋅ D
We also have
Re =
In addition
= 28
D
(1)
⎡
⎛ L Le ⎞⎤
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
Le = 2.8 m
V⋅ D
or
ν
ν = 1.14 × 10
2
−6 m
⋅
s
then
= 56
h = 2.5⋅ m
and
L = 4.51 m
Re = c⋅ V (2)
where
c = 0.05⋅ m ×
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
f = 0.01
D
L = ( 0.6 + π⋅ 0.45 + 2.5) ⋅ m
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
Le
for a 90o bend so for a 180 o bend
2⋅ g⋅ h
Solving for V
From Table A.7 (15oC)
Le
so from Fig. 8.16
s
D
ν
4 s
−6
1.14 × 10
(3)
c=
2
c = 4.39 × 10 ⋅
⋅m
e = 0.0015⋅ mm (Table 8.1)
m
V =
2⋅ g⋅ h
⎡
⎛ L Le ⎞⎤
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
2⋅ g⋅ h
V =
⎡
⎛ L Le ⎞⎤
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
2⋅ g⋅ h
V =
⎡
⎛L
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
V =
Le ⎞⎤
2⋅ g⋅ h
Le ⎞⎤
⎡
⎛L
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
V = 3.89
m
Re = c⋅ V
s
Re = 1.71 × 10
5
f = 0.0164
V = 3.43
m
s
Re = c⋅ V
Re = 1.50 × 10
5
Re = c⋅ V
Re = 1.49 × 10
5
Re = c⋅ V
Re = 1.49 × 10
5
f = 0.0168
V = 3.40
m
s
f = 0.0168
V = 3.40
m
s
Note that we could use Excel's Solver for this problem.
2
Q =
The flow rate is then
π⋅ D
4
3
−3m
⋅V
Q = 6.68 × 10
s
The minimum pressure occurs at the top of the curve (Point 3). Applying the energy equation between Points 1 and 3
2
⎛⎜ p
⎞
2
2
V3
⎛ p3 V2
⎞
Le V2
3
L V
V
g ⋅ z1 − ⎜
+
+ g ⋅ z3 = g ⋅ z1 − ⎜
+
+ g ⋅ z3 = f ⋅ ⋅
+ Kent⋅
+ f⋅ ⋅
2
2
2
D 2
D 2
⎝ρ
⎠
⎝ρ
⎠
where we have
Le
D
= 28
for the first 90o of the bend, and
L = ⎛⎜ 0.6 +
⎝
π × 0.45 ⎞
2
⎠
⋅m
L = 1.31 m
⎡
⎛ L Le ⎞⎤⎤
V ⎡
p 3 = ρ⋅ ⎢g ⋅ z1 − z3 −
⋅ ⎢1 + Kent + f ⋅ ⎜ +
⎥⎥
2 ⎣
⎣
⎝ D D ⎠⎦⎦
(
)
2
⎡
⎢
kg ⎢
m
p 3 = 1000⋅
× 9.81⋅ × ( −0.45⋅ m) −
3 ⎢
2
m
s
⎣
2
⎤
⎛ 3.4⋅ m ⎞
⎥
⎜
2
s⎠ ⎡
1.31
⎝
⎥ N⋅ s p = −20.0⋅ kPa
⎞
⎤
⎛
⋅ ⎢1 + 0.78 + 0.0168⋅ ⎜
+ 28 ⎥ ×
3
2
⎣
⎝ 0.05
⎠⎦⎥⎦ kg⋅ m
Problem 8.152
[Difficulty: 4]
Given:
System for fire protection
Find:
Height of water tower; Maximum flow rate; Pressure gage reading
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
∑
hl +
major
∑
h lm(8.29)
minor
2
ρ⋅ V⋅ D
hl = f ⋅
μ
64
(8.36)
Re
L V
⋅
D 2
(8.34)
h lm = 0.1⋅ h l
h lm = f ⋅
⎞
⎛ e
⎜ D
1
2.51
(8.37)
= −2.0⋅ log ⎜
+
3.7
f
Re
f
⋅
⎝
⎠
(Laminar)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
For no flow the energy equation (Eq. 8.29) applied between the water tower free surface (state 1; height H) and pressure gage is
g⋅ H =
p2
or
ρ
H=
p2
ρ⋅ g
(1)
The energy equation (Eq. 8.29) becomes, for maximum flow (and α = 1)
2
g⋅ H −
2
V
= h lT = ( 1 + 0.1) ⋅ h l
2
or
g⋅ H =
V
2
⋅ ⎛⎜ 1 + 1.1⋅ f ⋅
⎝
L⎞
(2)
D⎠
This can be solved for V (and hence Q) by iterating, or by using Solver
The energy equation (Eq. 8.29) becomes, for restricted flow
g⋅ H −
p2
ρ
2
+
The results in Excel are shown below:
V
= h lT = ( 1 + 0.1) ⋅ h l
2
2
p 2 = ρ⋅ g ⋅ H − ρ⋅
V
2
⋅ ⎛⎜ 1 + 1.1⋅ ρ⋅ f ⋅
⎝
L⎞
D⎠
(3)
Problem 8.151
[Difficulty: 5]
Given:
Flow from a reservoir
Find:
Effect of pipe roughness and pipe length on flow rate; Plot
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
ρ⋅ V⋅ D
64
Re
μ
2
hl = f ⋅
hl +
major
∑
h lm (8.29)
minor
2
L V
⋅
D 2
(8.36)
∑
h lm = K⋅
(8.34)
V
2
(8.40a)
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
(Laminar)
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
The energy equation (Eq. 8.29) becomes for this flow (see Example 8.5)
⎛
p pump = ∆p = ρ⋅ ⎜ g ⋅ d + f ⋅
⎝
2⎞
L V
⋅
D 2
⎠
We need to solve this for velocity V, (and hence flow rate Q) as a function of roughness e, then length L. This cannot be solved
explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given relative roughness e/D or length L
requires iteration (or use of Solver)
It is not possible to roughen the tube sufficiently to slow
the flow down to a laminar flow for this ∆p.
Flow Rate versus Tube Relative Roughness for fixed Dp
0.020
0.015
3
Q (m /s)
0.010
0.005
0.000
0.00
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
0.05
e/D
Flow Rate versus Tube Length for fixed Dp
0.010
Q (m3/s)
0.005
0.000
0
200
400
600
L (m)
800
1000
1200
Problem 8.150
Given:
Flow in a tube
Find:
Effect of tube length on flow rate; Plot
[Difficulty: 3]
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Re =
f =
ρ⋅ V⋅ D
64
Re
μ
2
hl = f ⋅
L V
⋅
D 2
(8.36)
∑
hl +
major
∑
h lm (8.29)
minor
2
(8.34)
(Laminar)
h lm = K⋅
V
2
(8.40a)
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
The energy equation (Eq. 8.29) becomes for flow in a tube
2
L V
p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅
D 2
This cannot be solved explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given L requires
iteration (or use of Solver)
Flow Rate vs Tube Length for Fixed Dp
10.0
Laminar
Turbulent
Q (m3/s)
4
x 10 1.0
0.1
0
5
10
15
L (km)
20
25
30
35
The "critical" length of tube is between 15 and 20 km. For this range, the fluid is making a transition between laminar and
turbulent flow, and is quite unstable. In this range the flow oscillates between laminar and turbulent; no consistent solution is
found (i.e., an Re corresponding to turbulent flow needs an f assuming laminar to produce the ∆p required, and vice versa!) More
realistic numbers (e.g., tube length) are obtained for a fluid such as SAE 10W oil (The graph will remain the same except for scale)
Problem 8.149
Given:
Flow in a tube
Find:
Effect of tube roughness on flow rate; Plot
[Difficulty: 3]
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
ρ⋅ V⋅ D
64
Re
μ
2
hl = f ⋅
L V
⋅
D 2
(8.36)
∑
hl +
major
∑
h lm (8.29)
minor
2
(8.34)
(Laminar)
h lm = K⋅
V
2
(8.40a)
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
The energy equation (Eq. 8.29) becomes for flow in a tube
2
L V
p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅
D 2
This cannot be solved explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given relative
roughness e/D requires iteration (or use of Solver)
Flow Rate versus Tube Relative Roughness
for fixed Dp
8
6
3
Q (m /s)
x 10
4
4
2
0
0.00
0.01
0.02
e/D
0.03
0.04
0.05
It is not possible to roughen the tube sufficiently to slow the flow down to a laminar flow for this ∆p. Even a relative roughness of
0.5 (a physical impossibility!) would not work.
Problem 8.148
[Difficulty: 3] Part 1/2
Problem 8.148
[Difficulty: 3] Part 2/2
Problem 8.147
Given:
Galvanized drainpipe
Find:
Maximum downpour it can handle
Solution:
[Difficulty: 3]
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Available data
D = 50⋅ mm
e = 0.15⋅ mm
h=L
(Table 8.1)
From Table A.7 (20oC)
2
The energy equation becomes
L V
g ⋅ z1 − g ⋅ z2 = g ⋅ z1 − z2 = g ⋅ h = f ⋅ ⋅
D 2
Solving for V
V=
(
k=
2 ⋅ D⋅ g ⋅ h
L⋅ f
2 ⋅ D⋅ g
=
)
2 ⋅ D⋅ g
V=
f
k =
k
(1)
f
2 × 0.05⋅ m × 9.81⋅
m
k = 0.99
2
s
Re =
We also have
V⋅ D
Re = c⋅ V
or
ν
s
c = 0.05⋅ m ×
(2)
m
s
c=
where
D
ν
4 s
−6
1.01 × 10
c = 4.95 × 10 ⋅
2
⋅m
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
In addition
ν = 1.01 × 10
m
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
Given
f = 0.01
V =
then
k
V = 9.90
f
m
s
Re = c⋅ V
5
Re = 4.9 × 10
⎛ e
⎞
⎜ D
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
1
f = 0.0264
V =
k
f
V = 6.09
m
s
Re = c⋅ V
Re = 3.01 × 10
5
2
−6 m
⋅
s
Given
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
f = 0.0266
Given
k
V =
V = 6.07
f
m
s
Re = c⋅ V
Re = 3.00 × 10
5
Re = c⋅ V
Re = 3.00 × 10
5
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
f = 0.0266
k
V =
V = 6.07
f
2
The flow rate is then
Q = V⋅
m
s
3
π⋅ D
Q = 0.0119⋅
4
m
s
3
The downpour rate is then
Q
Aroof
0.0119⋅
=
m
s
2
×
500 ⋅ m
Note that we could use Excel's Solver for this problem
100 ⋅ cm
1⋅ m
×
60⋅ s
1 ⋅ min
= 0.143 ⋅
cm
min
The drain can handle 0.143 cm/min
Problem 8.146
Given:
Flow from large reservoir
Find:
Flow rates in two pipes
Solution:
Basic equations
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g ⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
hl = f ⋅
2
L V
⋅
D 2
V
h lm = Kent ⋅
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 50⋅ mm
H = 10⋅ m
L = 10⋅ m
e = 0.15⋅ mm
(Table 8.1)
ν = 1⋅ 10
2
−6 m
⋅
Kent = 0.5
(Table A.8)
s
The energy equation becomes
2
V2
L V2
g ⋅ z1 − z2 − ⋅ V2 = f ⋅ ⋅
+ Kent⋅
2
2
D 2
(
Solving for V
)
2
2
and
V2 = V
z1 − z2 = H
2⋅ g⋅ H
V=
f⋅
We also have
1
L
+ Kent + 1
D
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
(1)
Re =
(2)
V⋅ D
(3)
ν
We must solve Eqs. 1, 2 and 3 iteratively.
Make a guess for V
and
Then
V = 1⋅
m
s
Then
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
L
D
+ Kent + 1
Re =
V⋅ D
ν
f = 0.0286
V = 5.21
m
s
Re = 5.00 × 10
4
Repeating
and
Then
Re =
V⋅ D
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
Repeating
and
Then
Re =
2⋅ g⋅ H
π
4
V = 5.36
m
s
5
Re = 2.68 × 10
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Q =
f = 0.0267
+ Kent + 1
D
ν
f⋅
Hence
L
V⋅ D
V =
5
Re = 2.61 × 10
ν
L
f = 0.0267
V = 5.36
+ Kent + 1
D
m
s
3
2
⋅D ⋅V
Q = 0.0105
m
Q = 10.5⋅
s
l
s
We repeat the analysis for the second pipe, using 2L instead of L:
Make a guess for V
and
Then
V = 1⋅
m
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
2⋅ g⋅ H
V =
f⋅
Repeating
and
Then
Then
s
Re =
2⋅ L
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
2⋅ L
+ Kent + 1
D
Re = 5.00 × 10
ν
f = 0.0286
V = 3.89
m
s
Re = 1.95 × 10
ν
f⋅
V⋅ D
+ Kent + 1
D
V⋅ D
V =
Re =
f = 0.0269
V = 4.00
m
s
5
4
Repeating
and
Then
Re =
V⋅ D
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
Hence
Re = 2.00 × 10
ν
Q =
π
4
2⋅ L
+ Kent + 1
D
2
⋅D ⋅V
As expected, the flow is considerably less in the longer pipe.
5
f = 0.0268
V = 4.00
m
s
Q = 7.861 × 10
3
−3m
s
Q = 7.86⋅
l
s
Problem 8.145
[Difficulty: 3]
Problem 8.144
[Difficulty: 3]
Problem 8.143
[Difficulty: 3]
Problem 8.142
[Difficulty: 3]
Problem 8.141
Given:
Kiddy pool on a porch.
Find:
Time to fill pool with a hose.
[Difficulty: 3]
Solution:
⎛ p1
⎞ ⎛ p2
⎞
V1 2
V 22
⎜
⎟
⎜
Basic equations:
⎜ ρ + α 1 2 + gz1 ⎟ − ⎜ ρ + α 2 2 + gz 2 ⎟⎟ = hlT
⎝
⎠ ⎝
⎠
2
2
⎛e/ D
1
2.51
LV
V
hlT = hl + hlm = f
+K
= −2.0 log⎜
+
⎜
D 2
2
f
⎝ 3.7 Re f
Assumptions: 1) Steady flow 2) Incompressible 3) Neglect minor losses 4)
Given data
p1 = 60
z1 = 0 ft
D = 0.625 in
z 2 = 20.5 ft
e
=0
D
L = 50 ft
k=
V =
Q = VA
V1 2
V2
= α2 2
2
2
lbf
gage
in 2
The energy equation becomes:
Solving for V:
α1
⎞
⎟
⎟
⎠
p2 = 0
lbf
gage
in 2
LV2
− gz 2 = f
ρ
D 2
p1
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
⎝ρ
⎠
f ⋅L
V =
k
(1)
f
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
2
3
⎞
slug ⋅ ft
ft
1
⎝ρ
⎠ = 2 × 0.625 in × ft × ⎛⎜ 60 lbf × 144 in × ft
×
− 32.2 2 × 20.5 ft ⎟⎟ ×
2
2
2
⎜
L
12 in ⎝ in
1.94 slug lbf ⋅ s
ft
s
⎠ 50 ft
k = 2.81
ft
s
We also have
Re =
ρ ⋅V ⋅ D
µ
Re = c ⋅ V (2)
or
where
c=
ρ⋅D
µ
Assuming water at 68oF (ρ = 1.94 slug/ft3, µ = 2.1 x 10-5 lbf·s/ft2):
c = 1.94
s
ft
ft 2
lbf ⋅ s 2
slug
= 4811.5
×
×
×
×
0
.
625
in
3
-5
ft
12 in 2.1 × 10 lbf ⋅ s slug ⋅ ft
ft
⎛ 2.51
= −2.0 log⎜
⎜ Re f
f
⎝
1
In addition:
⎞
⎟ (3)
⎟
⎠
Equations 1, 2 and 3 form a set of simultaneous equations for V , Re and f
Given
Given
V =
k
f = 0.0177
V =
k
f = 0.0179
V =
k
f = 0.015
Make a guess for f
⎛ 2.51
= −2.0 log⎜
⎜ Re f
f
⎝
⎛ 2.51
1
= −2.0 log⎜
⎜ Re f
f
⎝
1
then
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
f
f
f
= 22.94
ft
s
Re = c ⋅ V = 1.1 × 10 5
= 21.12
ft
s
Re = c ⋅ V = 1.02 × 10 5
= 21.0
ft
s
Re = c ⋅ V = 1.01 × 10 5
The flowrate is then:
Q =
π
× (0.625) in 2 ×
2
4
ft 2
ft
ft 3
21
.
0
0
.
0447
×
=
s
s
144 in 2
Volume pool = 2.5 ft ×
time =
Volume pool
Q
=
π
4
× (5) ft 2 = 49.1 ft 3
2
49.1 ft 3
= 1097 s =
ft 3
0.0447
s
This problem can also be solved explicitly in the following manner:
p1
The energy equation becomes:
ρ
f =
or:
1
V
− gz 2 = f
LV2
D 2
⎛p
⎞
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
L
Plugging this into the Colebrook equation:
1
=V
f
L
⎛p
⎞
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
⎛
⎛
⎜
⎜
⎜ 2.51µ ⎜
= −2.0 log⎜
⎜V
⎜ ρV D ⎜
⎜
⎜
⎝
⎝
⎛
2 D⎜⎜
⎝
⎞⎞
⎟⎟
⎟⎟
L
⎟⎟
p1
⎞
− gz 2 ⎟⎟ ⎟ ⎟
ρ
⎠ ⎟⎠ ⎟⎠
18.3 min.
Noting that the
V s on the right hand side cancel provides:
⎞
⎞⎛
⎛
⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟
⎜
⎜
⎟
⎟⎜
⎜ 2.51µ
L
⎝ρ
⎠⎟
V = −2.0 log⎜
⎟
⎜
⎟
ρ
D
L
p
⎛
⎞
1
⎟
⎜
⎜
2 D⎜⎜ − gz 2 ⎟⎟ ⎟
⎟
⎜
⎟
⎜
⎝ρ
⎠ ⎠⎝
⎝
⎠
Assuming water at 68oF (ρ = 1.94 slug/ft3, µ = 2 x 10-5 lbf·s/ft2) and g = 32.2 ft/s2 gives the remaining information needed to perform
the calculation finding:
⎞
⎛
slug ⋅ ft
1
12 in
lbf ⋅ s
ft 3
⎟
⎜ 2.51 × 2.1 × 10 −5
×
×
×
×
2
2
0.625 in
ft
1.94 slug lbf ⋅ s
ft
⎟
⎜
⎟
⎜
V = −2.0 log⎜
50 ft
12 in
1
⎟
×
×
×
3
⎟
⎜
2 × 0.625 in
ft
⎛ lbf 144 in 2
⎞
slug ⋅ ft
ft
ft
⎜⎜ 60 2 ×
×
×
− 32.2 2 × 20.5 ft ⎟⎟ ⎟
⎜⎜
2
2
1.94 slug lbf ⋅ s
ft
s
⎝ in
⎠ ⎟⎠
⎝
⎛ 2 × 0.625 in
⎛
⎞ ⎞⎟
ft
lbf 144 in 2
ft 3
slug ⋅ ft
ft
⎟⎟
32
.
2
20
.
5
ft
×⎜
×
× ⎜⎜ 60 2 ×
×
×
−
×
⎜
50 ft
12 in ⎝
1.94 slug lbf ⋅ s 2
s2
in
ft 2
⎠ ⎟⎠
⎝
V = 21.0
ft
s
and:
Q =
π
4
× (0.052) ft 2 × 21.0
2
Volume pool = 2.5 ft ×
time =
Volume pool
Q
=
π
4
ft
ft 3
= 0.0447
s
s
× (5) ft 2 = 49.1 ft 3
2
49.1 ft 3
= 1097 s =
ft 3
0.0447
s
18.3 min.
Problem 8.140
[Difficulty: 4]
.
Given:
Two potential solutions to improve flowrate.
Find:
Which solution provides higher flowrate
Solution:
Basic equations:
⎛ p1
⎞ ⎛p
⎞
V2
V2
⎜⎜ + α 1 1 + gz1 ⎟⎟ − ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ = hlT
2
2
⎝ρ
⎠ ⎝ ρ
⎠
⎛e/ D
1
2.51
LV2
V2
hlT = hl + hlm = f
+K
= −2.0 log⎜
+
⎜ 3.7 Re f
D 2
2
f
⎝
Assumptions: 1) Steady flow 2) Incompressible 3) Neglect minor losses 4)
Option 1: let z1 = 0
Given data
Q = VA
V1 2
V2
= α2 2
2
2
p 2 = p atm = 0 kPa gage
p1 = 200 kPa gage
The energy equation becomes:
Solving for V:
α1
⎞
⎟
⎟
⎠
V =
D = 0.019 m
p1
ρ
− gz 2 = f
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
⎝ρ
⎠
f ⋅L
e
=0
D
z 2 = 15 m
LV2
D 2
V =
k
f
(1)
L = 23 m
k=
⎞
⎛p
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
3
⎠ = 2 × 0.019 m × ⎛⎜ 200,000 N × m × kg ⋅ m − 9.81 m × 15 m ⎞⎟ × 1
⎝ρ
⎜
⎟ 23 m
L
m 2 999 kg N ⋅ s 2
s2
⎝
⎠
k = 0.296
m
s
Re =
We also have
ρ ⋅V ⋅ D
µ
Re = c ⋅ V (2)
or
where
c = 999
Assuming water at 20oC (ρ = 999 kg/m3, µ = 1 x 10-3 kg/(m·s)):
⎛ 2.51
= −2.0 log⎜
⎜ Re f
f
⎝
1
In addition:
Given
1
f
Given
1
f
Given
1
f
f
⎛ 2.51
= −2.0 log⎜
⎜ Re f
⎝
⎛ 2.51
= −2.0 log⎜
⎜ Re f
⎝
⎛ 2.51
= −2.0 log⎜
⎜ Re f
⎝
The flowrate is then:
Option 2: let
Given data
f = 0.015
z1 = 0
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
π
kg
m⋅s
s
× 0.019 m ×
= 18981
3
-3
m
m
1 × 10 kg
V , Re and f
V =
k
f = 0.0213
V =
k
f = 0.0222
V =
k
f = 0.0223
V =
k
then
m
Q1 = × (0.019) m × 1.98 =
4
s
2
ρ⋅D
µ
⎞
⎟ (3)
⎟
⎠
Equations 1, 2 and 3 form a set of simultaneous equations for
Make a guess for
c=
2
f
f
f
f
= 2.42
m
s
Re = c ⋅ V = 4.59 × 10 4
= 2.03
m
s
Re = c ⋅ V = 3.85 × 10 4
= 1.99
m
s
Re = c ⋅ V = 3.77 × 10 4
= 1.98
m
s
Re = c ⋅ V = 3.76 × 10 4
5.61 × 10
−4
m3
s
p 2 = p atm = 0 kPa gage
p1 = 300 kPa gage
D = 0.0127 m
The analysis for Option 2 is identical to Option 1:
The energy equation becomes:
p1
ρ
− gz 2 = f
LV2
D 2
z 2 = 15 m
e
= 0.05
D
L = 16 m
V =
Solving for V:
k=
⎞
⎛p
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
⎠
⎝ρ
f ⋅L
V =
k
(4)
f
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
3
⎝ρ
⎠ = 2 × 0.0127 m × ⎛⎜ 300,000 N × m × kg ⋅ m − 9.81 m × 15 m ⎞⎟ × 1
⎜
⎟ 16 m
L
m 2 999 kg N ⋅ s 2
s2
⎝
⎠
k = 0.493
We also have
m
s
Re =
ρ ⋅V ⋅ D
µ
c = 999
Re = c ⋅ V (5)
or
ρ⋅D
µ
s
kg
m⋅s
× 0.0127 m ×
= 12687.3
3
-3
m
1 × 10 kg
m
⎛e/ D
2.51
= −2.0 log⎜
+
⎜
f
⎝ 3.7 Re f
1
In addition:
c=
where
⎞
⎛
⎟ = −2.0 log⎜ 0.05 + 2.51
⎟
⎜ 3.7 Re f
⎠
⎝
⎞
⎟
⎟
⎠
(6)
Equations 4, 5 and 6 form a set of simultaneous equations for V , Re and f
Make a guess for
Given
Given
f
f = 0.07
⎛ 0.05
2.51
= −2.0 log⎜
+
⎜ 3.7 Re f
f
⎝
⎛ 0.05
1
2.51
= −2.0 log⎜
+
⎜
f
⎝ 3.7 Re f
1
The flowrate is then:
V =
k
f = 0.0725
V =
k
f = 0.0725
V =
k
then
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
π
f
f
m
s
Re = c ⋅ V = 2.36 × 10 4
= 1.83
m
s
Re = c ⋅ V = 2.32 × 10 4
= 1.83
m
s
Re = c ⋅ V = 2.32 × 10 4
3
m
−4 m
Q2 = × (0.0127 ) m × 1.83 = 2.32 × 10
s
4
s
2
2
This problem can also be solved explicitly:
LV2
− gz 2 = f
D 2
ρ
p1
The energy equation becomes:
or:
f
= 1.86
f =
Plugging this into the Colebrook equation:
1
V
⎛p
⎞
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
L
Option 1 is 2.42 times more effective!
1
f
Noting that the
=V
L
⎛p
⎞
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
⎛
⎛
⎜
⎜
⎜ e / D 2.51µ ⎜
= −2.0 log⎜
+
V
ρV D ⎜⎜
⎜ 3.7
⎜
⎜
⎝
⎝
V s on the right hand side cancel provides:
⎛
⎜
⎜ e / D 2.51µ
V = −2.0 log⎜
+
ρD
3
.
7
⎜
⎜
⎝
⎛
2 D⎜⎜
⎝
⎛
2 D⎜⎜
⎝
⎞⎞
⎟⎟
⎟⎟
L
⎟⎟
p1
⎞
− gz 2 ⎟⎟ ⎟ ⎟
ρ
⎠ ⎟⎠ ⎟⎠
⎞
⎞⎛
⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟
⎜ρ
⎟⎟
⎟⎜
L
⎝
⎠
⎟
⎟⎜
L
p1
⎞ ⎟⎜
⎟
− gz 2 ⎟⎟
⎟
⎜
⎟
ρ
⎠ ⎠⎝
⎠
Assuming water at 20oC (ρ = 999 kg/m3, µ = 1 x 10-3 kg/(m·s)) gives the remaining information needed to perform the calculation.
For Option 1:
⎛
⎞
kg
m3
1
⎜ 2.51 × 1 × 10 −3
⎟
×
×
m ⋅ s 999 kg 0.019 m
⎜
⎟
⎜
⎟
V = −2.0 log⎜
23 m
1
⎟
×
×
⎜
2 × 0.019 m ⎛
⎞⎟
m
N
m3
kg ⋅ m
⎜⎜ 200,000 2 ×
×
− 9.81 2 × 15 m ⎟⎟ ⎟
⎜⎜
2
999
kg
⋅
m
N
s
s
⎝
⎠ ⎟⎠
⎝
⎛ 2 × 0.019 m ⎛
⎞ ⎞⎟
N
m3
kg ⋅ m
m
⎟⎟
9
.
81
15
m
×⎜
× ⎜⎜ 200,000 2 ×
×
−
×
⎜
23 m
999 kg N ⋅ s 2
m
s2
⎝
⎠ ⎟⎠
⎝
m
V = 1.98
s
and:
Q1 =
Option 2: let
Given data
z1 = 0
π
4
× (0.019) m 2 × 1.98
2
m3
m
= 5.61 × 10 − 4
s
s
p 2 = p atm = 0 kPa gage
p1 = 300 kPa gage
z 2 = 15 m
D = 0.0127 m
e
= 0.05
D
The analysis for Option 2 results in the same equations as used in Option 1 once again giving:
⎛
⎜
⎜ e / D 2.51µ
V = −2.0 log⎜
+
ρD
⎜ 3.7
⎜
⎝
⎛
2 D⎜⎜
⎝
⎞
⎞⎛
⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟
⎟⎟
⎜ρ
⎟⎜
L
⎠
⎝
⎟
⎟⎜
L
p1
⎞ ⎟⎜
⎟
− gz 2 ⎟⎟
⎟
⎜
⎟
ρ
⎠ ⎠⎝
⎠
L = 16 m
⎞
⎛ 0.05
kg
m3
1
⎟
⎜
+ 2.51 × 1 × 10 −3
×
×
m ⋅ s 999 kg 0.0127 m
⎟
⎜ 3.7
⎟
⎜
V = −2.0 log⎜
16 m
1
⎟
×
×
3
⎜
2 × 0.0127 m ⎛
⎞⎟
kg ⋅ m
N
m
m
⎜⎜ 300,000 2 ×
×
− 9.81 2 × 15 m ⎟⎟ ⎟
⎜⎜
2
999 kg N ⋅ s
m
s
⎝
⎠ ⎟⎠
⎝
⎛ 2 × 0.0127 m ⎛
⎞ ⎞⎟
m
N
m3
kg ⋅ m
⎟⎟
9
.
81
15
m
×⎜
× ⎜⎜ 300,000 2 ×
×
−
×
⎜
16 m
999 kg N ⋅ s 2
m
s2
⎝
⎠ ⎟⎠
⎝
V = 1.83
The flowrate is then:
Q2 =
π
4
× (0.0127 ) m 2 × 1.83
2
m
=
s
m
s
2.32 × 10 − 4
m3
s
Problem 8.139
[Difficulty: 2]
Problem 8.138
Given:
Flow in horizontal pipe
Find:
Flow rate
Solution:
Basic equations
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Available data
L = 200 ⋅ m
D = 75⋅ mm
e = 2.5⋅ mm
∆p = 425 ⋅ kPa
ρ = 1000⋅
kg
3
μ = 1.76⋅ 10
− 3 N⋅ s
⋅
m
2
m
Hence the energy equation becomes
p1
ρ
p2
−
Solving for V
V=
We also have
Re =
ρ
=
∆p
ρ
2
= f⋅
L V
⋅
D 2
2 ⋅ D⋅ ∆p
V=
L⋅ ρ⋅ f
ρ⋅ V⋅ D
k
f
Re = c⋅ V
or
μ
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
In addition
k =
(1)
(2)
2 ⋅ D⋅ ∆p
k = 0.565
L⋅ ρ
c =
where
ρ⋅ D
μ
m
s
c = 4.26 × 10
4 s
m
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
Given
Given
f = 0.1
V =
then
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
f = 0.0573
V =
f = 0.0573
V =
2
The flow rate is then
Q = V⋅
π⋅ D
4
k
f
k
Note that we could use Excel's Solver for this problem
m
s
V = 7.74⋅
f
k
V= ⋅
f
3
Q = 0.0104
V = 5.86⋅
Q = 10.42
l
s
ft
s
ft
s
ft
s
Q = 165 ⋅ gpm
Re = c⋅ V
Re = 7.61 × 10
4
Re = c⋅ V
Re = 1.01 × 10
5
Re = c⋅ V
Re = 1.01 × 10
5
Problem 8.137
Given:
Draining of swimming pool with larger hose
Find:
Flow rate and average velocity
Solution:
Basic equations
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
2
L V
⋅
D 2
hl = f ⋅
V
h lm = Kent⋅
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Available data
D = 25⋅ mm
L = 30⋅ m
e = 0.2⋅ mm
h = 3⋅ m
2
g ⋅ ( ∆z + h ) = f ⋅
Hence the energy equation becomes
Solving for V
V=
2 ⋅ g ⋅ ( ∆z + h )
f⋅
We also have
Re =
L
D
∆z = 1.5⋅ m
2
2
−6 m
Kent = 0.5
ν = 1 ⋅ 10
⋅
s
2
L V
V
V
⋅
+ Kent⋅
+
2
2
D 2
(1)
+ Kent + 1
V⋅ D
(2)
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
In addition
ν
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f, which we can solve iteratively
Make a guess for f
f = 0.1
then
V =
2 ⋅ g ⋅ ( ∆z + h )
f⋅
using Eq 3, at this Re
V =
Re =
V⋅ D
ν
Re = 2.13 × 10
4
L
D
V = 1.37
+ Kent + 1
m
s
Re =
V⋅ D
ν
Re = 3.41 × 10
4
Re = 3.46 × 10
4
f = 0.0371
V =
Then, repeating
2 ⋅ g ⋅ ( ∆z + h )
f⋅
f = 0.0371
Q = V⋅
π⋅ D
4
L
D
+ Kent + 1
V = 1.38
m
s
which is the same as before, so we have convergence
2
The flow rate is then
s
+ Kent + 1
2 ⋅ g ⋅ ( ∆z + h )
f⋅
Using Eq 3, at this Re
D
m
f = 0.0382
Then, repeating
using Eq 3, at this Re
L
V = 0.852
3
−4m
Q = 6.79 × 10
Note that we could use Excel's Solver for this problem
s
Q = 0.679
l
s
Re =
V⋅ D
ν
Problem 8.136
[Difficulty: 3]
Given:
Drinking of a beverage
Find:
Fraction of effort of drinking of friction and gravity
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g ⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Hence the energy equation becomes, between the bottom of the straw (Point 1) and top (Point 2)
g ⋅ z1 −
2
⎛ p2
⎞
L V
⎜ + g ⋅ z2 = f ⋅ ⋅
D 2
⎝ρ
⎠
where p 2 is the gage pressure in the mouth
The negative gage pressure the mouth must create is therefore due to two parts
(
2
)
p grav = −ρ⋅ g ⋅ z2 − z1
p fric = −ρ⋅ f ⋅
12
128
⋅ gal
Assuming a person can drink 12 fluid ounces in 5 s
Q =
Assuming a straw is 6 in long diameter 0.2 in, with roughness
e = 5 × 10
V=
4⋅ Q
−5
4
Re =
Given
Then
and
− 5 ft
⋅
ν
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
p fric = −1.94⋅
Hence the fraction due to friction is
slug
ft
3
slug
ft
3
− 3 ft
Q = 2.51 × 10
7.48⋅ gal
⋅
3
s
in (from Googling!)
− 3 ft
3
s
2
×
⎛ 1 × 12⋅ in ⎞ V = 11.5⋅ ft
⎜
s
⎝ 0.2⋅ in 1 ⋅ ft ⎠
(for water, but close enough)
ft
Re = 11.5⋅
p grav = −1.94⋅
3
2
s
V⋅ D
1⋅ ft
× 2.51 × 10
π
π⋅ D
From Table A.7 (68oF) ν = 1.08 × 10
×
5⋅ s
V =
2
L V
⋅
D 2
ft
× 32.2⋅
2
p fric
p fric + p grav
0.2
12
⋅ ft ×
s
1.08 × 10
4
−5
ft
Re = 1.775 × 10
2
f = 0.0272
×
s
× 0.0272 ×
s
×
1
2
6
0.2
2
⋅ ft ×
×
= 77⋅ %
1
2
lbf ⋅ s
p grav = −31.2⋅
slug⋅ ft
× ⎛⎜ 11.5⋅
⎝
ft ⎞
s⎠
2
2
×
lbf ⋅ s
slug⋅ ft
and gravity is
These results will vary depending on assumptions, but it seems friction is significant!
p fric = −105 ⋅
p grav
p fric + p grav
lbf
ft
lbf
ft
2
2
p grav = −0.217 ⋅ psi
p fric = −0.727 ⋅ psi
= 23⋅ %
Problem 8.135
[Difficulty: 3]
Problem 8.134
Given:
Proposal for bench top experiment
Find:
Design it; Plot tank depth versus Re
[Difficulty: 4]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Re =
f =
ρ⋅ V⋅ D
μ
64
2
hl = f ⋅
(8.34)
(8.36)
(Laminar)
The energy equation (Eq. 8.29) becomes
2
2
2
L V
V
g ⋅ H − α⋅
= f⋅ ⋅
+ K⋅
2
2
D 2
This can be solved explicity for reservoir height H
2
H=
In Excel:
V
2⋅ g
⋅ ⎛⎜ α + f ⋅
⎝
L
D
+ K⎞
⎠
major
hl +
∑
h lm
(8.29)
minor
2
L V
⋅
D 2
Re
V
∑
h lm = K ⋅
V
(8.40a)
2
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Computed results:
Q (L/min) V (m/s)
0.200
0.225
0.250
0.275
0.300
0.325
0.350
0.375
0.400
0.425
0.450
0.472
0.531
0.589
0.648
0.707
0.766
0.825
0.884
0.943
1.002
1.061
Re
Regime
f
H (m)
1413
1590
1767
1943
2120
2297
2473
2650
2827
3003
3180
Laminar
Laminar
Laminar
Laminar
Laminar
Laminar
Turbulent
Turbulent
Turbulent
Turbulent
Turbulent
0.0453
0.0403
0.0362
0.0329
0.0302
0.0279
0.0462
0.0452
0.0443
0.0435
0.0428
0.199
0.228
0.258
0.289
0.320
0.353
0.587
0.660
0.738
0.819
0.904
The flow rates are realistic, and could easily be measured using a tank/timer system
The head required is also realistic for a small-scale laboratory experiment
Around Re = 2300 the flow may oscillate between laminar and turbulent:
Once turbulence is triggered (when H > 0.353 m), the resistanc e to flow increases
requiring H >0.587 m to maintain; hence the flow reverts to la minar, only to trip over
again to turbulent! This behavior will be visible: the exit flow will switch back and
forth between smooth (laminar) and chaotic (turbulent)
Required Reservoir Head
versus Reynolds Number
1.00
0.75
H (m)
0.50
Laminar
0.25
0.00
1000
Turbulent
1500
2000
Re
2500
3000
3500
Problem 8.133
Given:
Flow through fire hose and nozzle
Find:
Supply pressure
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
2
L V
h lT = h l + h lm = f ⋅ ⋅
+
D 2
∑
Minor
⎛ V2 ⎞
⎜ K⋅
⎝ 2⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) p 2 = p atm so p 2 = 0 gage
Hence the energy equation between Point 1 at the supply and the nozzle exit (Point n); let the velocity in the hose be V
p1
ρ
2
Vn
−
2
2
(
)
and
V=
2
From continuity
Vn =
⎛ D ⎞ ⋅V
⎜D
⎝ 2⎠
⎡⎢ L
+ Ke + 4⋅ Kc +
2 ⎢ D
⎣
2
ρ⋅ V
Solving for p 1
p1 =
From Table A.7 (68oF)
ρ = 1.94⋅
⋅ f⋅
slug
ft
Re =
For the hose
Flow is turbulent:
e
D
2
2
Vn
L V
V
= f⋅ ⋅
+ Ke + 4⋅ Kc ⋅
+ Kn ⋅
2
2
D 2
V⋅ D
Re = 15.3⋅
ν
A
=
4⋅ Q
4
V=
2
π
π⋅ D
× 0.75⋅
4
⎛ D ⎞ ⋅ 1 + K ⎤⎥
( n)⎥
⎜D
⎝ 2⎠
⎦
ν = 1.08 × 10
3
Q
ft
s
− 5 ft
×
⋅
3
12
ft
3
s
1
×
⎛ 1 ⋅ ft ⎞
⎜4
⎝ ⎠
2
V = 15.3⋅
ft
s
2
s
s
⋅ ft ×
−5
1.08 × 10
⋅ ft
2
Re = 3.54 × 10
5
Turbulent
= 0.004
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
p1 =
1
2
× 1.94⋅
slug
ft
3
4 lbf
p 1 = 2.58 × 10 ⋅
ft
2
× ⎛⎜ 15.3⋅
⎝
ft ⎞
s⎠
2
⎡⎢
⎢
⎢⎣
× 0.0287 ×
p 1 = 179 ⋅ psi
250
1
4
f = 0.0287
+ 0.5 + 4 × 0.5 +
4
2
⎛ 3 ⎞ × ( 1 + 0.02)⎥⎤ × lbf ⋅ s
⎜
⎥ slug⋅ ft
⎝1⎠
⎥⎦
Problem 8.132
Given:
Flow down corroded iron pipe
Find:
Pipe roughness; Power savings with new pipe
[Difficulty: 4]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
3
Available data
D = 50⋅ mm
∆z = 40⋅ m
L = ∆z
p 1 = 750⋅ kPa
p 2 = 250⋅ kPa
Q = 0.015⋅
m
s
ρ = 999⋅
kg
3
m
Hence the energy equation becomes
2
⎛ p1
⎞ ⎛ p2
⎞
L V
⎜ + g ⋅ z1 − ⎜ + g ⋅ z2 = f ⋅ ⋅
D 2
⎝ρ
⎠ ⎝ρ
⎠
Here
V=
Q
=
A
4⋅ Q
V=
2
π⋅ D
4
π
3
× 0.015⋅
m
s
×
1
( 0.05⋅ m)
V = 7.64
2
m
s
In this problem we can compute directly f and Re, and hence obtain e/D
Solving for f
f =
⎛ p1 − p2
2⋅ D
2
L⋅ V
f = 2×
⋅⎜
⎝
0.05
40
From Table A.8 (20oF) ν = 1.01 × 10
Flow is turbulent:
ρ
×
(
+ g z1 − z2
⎞
)⎠
2
3
⎤
kg⋅ m
m
⎛ s ⎞ × ⎡⎢( 750 − 250 ) × 103⋅ N × m
×
+ 9.81⋅ × 40⋅ m⎥ f = 0.0382
⎜
2
1000⋅ kg
2
2
⎥
⎝ 7.64⋅ m ⎠ ⎢⎣
m
s ⋅N
s
⎦
2
−6 m
⋅
s
Re =
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V⋅ D
ν
Re = 7.64⋅
m
s
× 0.05⋅ m ×
s
−6
1.01 × 10
2
⋅m
Re = 3.78 × 10
5
Solving for e
New pipe (Table 8.1)
⎛ −
⎜
e = 3.7⋅ D⋅ ⎜ 10
⎝
⎞
1
2⋅ f
−
e
e = 0.15⋅ mm
2.51
e
e = 0.507 mm
Re⋅ f ⎠
D
= 0.0101
= 0.003
D
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
f = 0.0326
In this problem
Hence
2⎤
⎡
L V
∆p = p 1 − p 2 = ρ⋅ ⎢g ⋅ z2 − z1 + f ⋅ ⋅
D 2
⎣
(
∆pnew = 1000⋅
kg
m
∆pold = p 1 − p 2
Compared to ∆pold = 500 ⋅ kPa we find
3
⎡
)
× ⎢9.81⋅
⎢
⎣
m
2
⎥
⎦
× ( −40⋅ m) +
s
0.0326
2
×
40
0.05
× ⎛⎜ 7.64⋅
⎝
∆pold = 500 kPa
∆pold − ∆pnew
∆pold
= 26.3⋅ %
As power is ∆pQ and Q is constant, the power reduction is the same as the above percentage!
2⎤
2
⎥ × N⋅ s
s ⎠ ⎥ kg⋅ m
⎦
m⎞
∆pnew = 369 ⋅ kPa
Problem 8.131
Given:
Same flow rate in various ducts
Find:
Pressure drops of each compared to round duct
Solution:
Basic equations
[Difficulty; 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
4⋅ A
Dh =
Pw
e = 0
(Smooth)
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses
The energy equation simplifies to
2
L V
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅
Dh 2
But we have
From Table A.9
Hence
For a round duct
Q
V=
V = 1250⋅
A
ν = 1.62 × 10
Re =
V⋅ Dh
− 4 ft
⋅
min
×
1 ⋅ min
60⋅ s
s
s
×
4
π
4⋅ A
4⋅ b⋅ h
2 ⋅ h ⋅ ar
Dh =
=
=
Pw
2⋅ ( b + h)
1 + ar
But
h=
ar
so
slug
2
h =
b⋅ h
ar
3
f
2
⋅
V
Dh 2
V = 20.8⋅
ft
s
at 68oF
5
× Dh = 1.284 × 10 ⋅ Dh
−4 2
1.62 × 10 ⋅ ft
Dh =
π
2
= ρ⋅
s
For a rectangular duct
b
1 ⋅ ft
ft
ft
L
1
×
ρ = 0.00234 ⋅
4⋅ A
Dh = D =
3
2
Re = 20.8⋅
ν
ft
∆p
or
=
A
ar
× 1 ⋅ ft
2
(Dh in ft)
Dh = 1.13⋅ ft
where
ar =
or
h=
b
h
A
and
ar
2 ⋅ ar
Dh =
⋅ A
1 + ar
The results are:
Round
Given
ar = 1
Dh = 1.13⋅ ft
5 1
Re = 1.284 × 10 ⋅ ⋅ Dh
ft
⎛ e
⎞
⎜
D
1
h
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2 ⋅ ar
Dh =
⋅ A
1 + ar
f = 0.0167
Dh = 1 ⋅ ft
Re = 1.45 × 10
∆p
L
= ρ⋅
f
5
2
⋅
V
Dh 2
∆p
L
= 7.51 × 10
5 1
5
Re = 1.284 × 10 ⋅ ⋅ Dh Re = 1.28 × 10
ft
− 3 lbf
⋅
ft
3
Given
⎛ e
⎞
⎜
D
1
h
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
f = 0.0171
∆p
L
= ρ⋅
f
2
⋅
V
∆p
Dh 2
L
8.68 × 10
Hence the square duct experiences a percentage increase in pressure drop of
= 8.68 × 10
−3
− 3 lbf
⋅
ft
3
−3
− 7.51 × 10
= 15.6⋅ %
−3
7.51 × 10
ar = 2
Given
2 ⋅ ar
Dh =
⋅ A
1 + ar
Dh = 0.943 ⋅ ft
⎛ e
⎞
⎜
Dh
1
2.51
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
5 1
Re = 1.284 × 10 ⋅ ⋅ Dh
ft
f = 0.0173
∆p
L
Re = 1.21 × 10
= ρ⋅
f
2
⋅
V
∆p
Dh 2
9.32 × 10
Hence the 2 x 1 duct experiences a percentage increase in pressure drop of
5
= 9.32 × 10
L
−3
− 3 lbf
ft
Given
2 ⋅ ar
Dh =
⋅ A
1 + ar
Dh = 0.866 ⋅ ft
= 24.1⋅ %
−3
⎛ e
⎞
⎜
Dh
1
2.51
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
5 1
Re = 1.284 × 10 ⋅ ⋅ Dh
ft
f = 0.0176
∆p
L
Re = 1.11 × 10
= ρ⋅
f
2
⋅
V
5
∆p
Dh 2
L
= 0.01⋅
lbf
ft
3
−3
Hence the 3 x 1 duct experiences a percentage increase in pressure drop of
0.01 − 7.51 × 10
−3
7.51 × 10
Note that f varies only about 7%; the large change in ∆p/L is primarily due to the 1/Dh factor
3
−3
− 7.51 × 10
7.51 × 10
ar = 3
⋅
= 33.2⋅ %
Problem 8.130
[Difficulty: 3] Part 1/2
Problem 8.130
[Difficulty: 3] Part 2/2
Problem 8.129
[Difficulty: 3]
c
h
LA
d
e
LB
Given:
Pipe friction experiment
Find:
Required average speed; Estimate feasibility of constant head tank; Pressure drop over 5 m
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT
⎝
⎠ ⎝
⎠
2
LA VA
LB VB
h lT = h A + h B = fA⋅
⋅
+ fB⋅
⋅
DA 2
DB 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses
We wish to have
ReB = 10
Hence, from
ReB =
5
VB⋅ DB
VB =
ν
2
−6 m
5
VB = 10 × 1.01 × 10
We will also need
⎛ DB ⎞
VA = VB⋅ ⎜
⎝ DA ⎠
ReA =
VA⋅ DA
ν
2
⋅
s
ReB⋅ ν
and for water at 20oC
DB
×
1
⎛ 2.5 ⎞
⎜
⎝ 5 ⎠
2
m
ReA = 1.01⋅ × 0.05⋅ m ×
s
m
VA = 1.01
s
s
1.01 × 10
4
−6
2
⋅m
ReA = 5 × 10
Both tubes have turbulent flow
For PVC pipe (from Googling!) e = 0.0015⋅ mm
For tube A
For tube B
Given
Given
2
−6 m
m
VB = 4.04
s
0.025 ⋅ m
m
VA = 4.04⋅ ×
s
ν = 1.01 × 10
⎛ e
⎞
⎜
D
1
A
2.51 ⎟
= −2.0⋅ log⎜
+
⎜ 3.7
fA
ReA⋅ fA
⎝
⎠
⎛ e
⎞
⎜
D
1
B
2.51 ⎟
= −2.0⋅ log⎜
+
⎜ 3.7
fB
ReB⋅ fB
⎝
⎠
fA = 0.0210
fB = 0.0183
⋅
s
2
Applying the energy equation between Points 1 and 3
(
VB
)
g ⋅ LA + h −
LA =
2
2
⎛
⎜
⎜g −
⎝
1
2
LA =
2
2
LA VA
LB VB
= fA⋅
⋅
+ fB⋅
⋅
DA 2
DB 2
LB ⎞
⎛
⋅ ⎜ 1 + fB⋅
− g⋅ h
2
DB
⎝
⎠
VB
Solving for LA
2
2⎞
fA VA
⋅
DA 2
× ⎛⎜ 4.04⋅
⎝
m⎞
s
2
⎠
9.81⋅
⎠
× ⎛⎜ 1 + 0.0183 ×
⎞ − 9.81⋅ m × 0.5⋅ m
2
0.025 ⎠
s
20
⎝
m
2
−
0.0210
2
s
×
1
0.05⋅ m
× ⎛⎜ 1.01⋅
⎝
m⎞
s
2
LA = 12.8 m
⎠
Most ceilings are about 3.5 m or 4 m, so this height is IMPRACTICAL
Applying the energy equation between Points 2 and 3
2
2
2
⎛⎜ p
VB ⎞ ⎛⎜ p 3
VB ⎞
2
L VB
⎜ ρ + 2 − ⎜ ρ + 2 = fB⋅ D ⋅ 2
⎝
⎠ ⎝
⎠
B
∆p = 1000⋅
kg
3
m
×
0.0183
2
×
5⋅ m
0.025 ⋅ m
× ⎛⎜ 4.04⋅
⎝
L
m⎞
s
⎠
2
2
×
VB
∆p = ρ⋅ fB⋅
⋅
DB 2
or
N⋅ s
kg⋅ m
∆p = 29.9⋅ kPa
2
Problem 8.128
Given:
Data on circuit
Find:
Plot pressure difference for a range of flow rates
[Difficulty: 3]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Basic equations:
Re =
f =
ρ⋅ V⋅ D
64
μ
2
hl = f ⋅
(8.36)
L V
⋅
D 2
∑
hl +
major
∑
h lm (8.29)
minor
2
h lm = K⋅
(8.34)
(Laminar)
Re
V
(8.40a)
2
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
The energy equation (Eq. 8.29) becomes for the circuit ( 1 = pump inlet, 2 = pump outlet)
p1 − p2
ρ
In Excel:
2
= f⋅
2
2
L V
V
V
⋅
+ 4 ⋅ f ⋅ Lelbow⋅
+ f ⋅ Lvalve⋅
2
2
D 2
Lelbow
Lvalve ⎞
⎛L
+ 4⋅
+
2 ⎝D
D
D ⎠
2
or
∆p = ρ⋅ f ⋅
V
⋅⎜
Required Pressure Head for a Circuit
1200
Dp (kPa)
1000
800
600
400
200
0
0.00
0.01
0.02
0.03
Q (m3/s)
0.04
0.05
0.06
0.07
Problem 8.127
Given:
Flow through rectangular duct
Find:
Pressure drop
Solution:
Basic equations
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
2
V1
V2
1
2
L V
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 +
⎝
⎠ ⎝
⎠
∑
Minor
⎛ Le V2 ⎞
⎜f ⋅ ⋅
⎝ D 2⎠
4 ⋅ a⋅ b
Dh =
2⋅ (a + b)
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
Q = 1750⋅ cfm
At 50oF, from Table A.9 ρ = 0.00242 ⋅
L = 1000⋅ ft
slug
ft
Hence
1
f
V = 15.6⋅
a⋅ b
ρ⋅ V⋅ Dh
⎞
⎝ Re⋅ f ⎠
Hence
∆p = f ⋅
or, in in water
h =
L
Dh
∆p
ρw⋅ g
⋅
ft
2.51
5
so
2
⋅ ρ⋅
V
2
∆p = 0.031 ⋅ psi
h = 0.848 ⋅ in
2
and
s
Re = 1.18 × 10
μ
= −2 ⋅ log⎛⎜
− 7 lbf ⋅ s
ft
Q
V =
Re =
For a smooth duct
3
μ = 3.69⋅ 10
b = 2.5⋅ ft
f = 0.017
a = 0.75⋅ ft
ρw = 1.94⋅
slug
ft
3
4 ⋅ a⋅ b
Dh =
2⋅ ( a + b)
Dh = 1.15⋅ ft
Problem 8.126
Given:
Flow through three different layouts
Find:
Which has minimum loss
Solution:
Basic equations
[Difficulty: 3]
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
2
V1
V2
1
2
L V
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
=
h
h
=
h
+
h
=
f
⋅
⋅
+
⎜ρ
⎜
1
2
lT lT
l
lm
2
2
D 2
⎝
⎠ ⎝ρ
⎠
∑
Minor
⎛ Le V2 ⎞
⎜f ⋅ ⋅
⎝ D 2⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore additional length of elbows
For a flow rate of
For water at 20oC
Q = 350 ⋅
L
min
ν = 1.01 × 10
For Case (a)
⋅
s
L =
4⋅ Q
2
p1
Re =
V⋅ D
ν
e
D
p2
ρ
m
s
min
3
×
0.001 ⋅ m
1⋅ L
× 0.05⋅ m ×
2
1 ⋅ min
×
×
60⋅ s
⎛ 1 ⎞ V = 2.97 m
⎜
s
⎝ 0.05⋅ m ⎠
s
−6
1.01 × 10
2
Re = 1.47 × 10
⋅m
−4
= 6.56 × 10
f = 0.0201
L = 5.81 m
2
= f⋅
L
× 350 ⋅
Re = 2.97⋅
2
−
π
π⋅ D
e = 0.15⋅ mm
2
4
V =
5.25 + 2.5 ⋅ m
ρ
Two 45o miter bends (Fig. 8.16), for each
Le
D
= 13
2
Le V
L V
⋅
+ 2⋅ f ⋅ ⋅
D 2
D 2
2
Le ⎞
V ⎛L
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅ ⎜ + 2⋅
2 ⎝D
D⎠
∆p = 1000⋅
kg
3
× .0201 × ⎛⎜ 2.97⋅
⎝
m
For Case (b)
=
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Hence the energy equation is
Solving for ∆p
A
2
−6 m
Flow is turbulent. From Table 8.1
Given
Q
V=
L = ( 5.25 + 2.5) ⋅ m
Hence the energy equation is
p1
ρ
−
p2
ρ
= f⋅
m⎞
s
⎠
2
×
2
⎛ 5.81 + 2⋅ 13⎞ × N⋅ s
⎜
⎝ 0.05
⎠ kg⋅ m
L = 7.75 m
2
Le V2
L V
⋅
+ f⋅ ⋅
D 2
D 2
One standard 90o elbow (Table 8.4)
∆p = 25.2⋅ kPa
Le
D
= 30
5
2
Solving for ∆p
Le ⎞
V ⎛L
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅⎜ +
2 ⎝D
D⎠
∆p = 1000⋅
kg
3
× .0201 × ⎛⎜ 2.97⋅
⎝
m
For Case (c)
Hence the energy equation is
L = ( 5.25 + 2.5) ⋅ m
p1
ρ
Solving for ∆p
−
p2
ρ
m⎞
s
2
×
⎠
L = 7.75 m
2
2
⎛ 7.75 + 30⎞ × N⋅ s
⎜
⎝ 0.05
⎠ kg⋅ m
Three standard 90o elbows, for each
∆p = 32.8⋅ kPa
Le
D
= 30
2
Le V
L V
⋅
+ 3⋅ f ⋅ ⋅
D 2
D 2
= f⋅
2
Le ⎞
V ⎛L
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅ ⎜ + 3⋅
2 ⎝D
D⎠
∆p = 1000⋅
kg
3
m
× .0201 × ⎛⎜ 2.97⋅
⎝
Hence we conclude Case (a) is the best and Case (c) is the worst
m⎞
s
⎠
2
×
2
⎛ 7.75 + 3 × 30⎞ × N⋅ s
⎜
⎝ 0.05
⎠ kg⋅ m
∆p = 43.4⋅ kPa
Problem 8.125
[Difficulty: 3]
Given: Data on reservoir/pipe system
Find: Plot elevation as a function of flow rate; fraction due to minor losses
Solution:
L =
D =
e/D =
K ent =
K exit =
250
50
0.003
0.5
1.0
Required Head versus Flow Rate
m
mm
200
150
∆z (m)
ν = 1.01E-06 m2/s
3
Q (m /s) V (m/s)
0.0000
0.0005
0.0010
0.0015
0.0020
0.0025
0.0030
0.0035
0.0040
0.0045
0.0050
0.0055
0.0060
0.0065
0.0070
0.0075
0.0080
0.0085
0.0090
0.0095
0.0100
0.000
0.255
0.509
0.764
1.02
1.27
1.53
1.78
2.04
2.29
2.55
2.80
3.06
3.31
3.57
3.82
4.07
4.33
4.58
4.84
5.09
Re
0.00E+00
1.26E+04
2.52E+04
3.78E+04
5.04E+04
6.30E+04
7.56E+04
8.82E+04
1.01E+05
1.13E+05
1.26E+05
1.39E+05
1.51E+05
1.64E+05
1.76E+05
1.89E+05
2.02E+05
2.14E+05
2.27E+05
2.40E+05
2.52E+05
100
f
∆z (m) h lm /h lT
0.000
0.0337 0.562
0.0306 2.04
0.0293 4.40
0.0286 7.64
0.0282 11.8
0.0279 16.7
0.0276 22.6
0.0275 29.4
0.0273 37.0
0.0272 45.5
0.0271 54.8
0.0270 65.1
0.0270 76.2
0.0269 88.2
0.0269 101
0.0268 115
0.0268 129
0.0268 145
0.0267 161
0.0267 179
50
0.882%
0.972%
1.01%
1.04%
1.05%
1.07%
1.07%
1.08%
1.09%
1.09%
1.09%
1.10%
1.10%
1.10%
1.10%
1.11%
1.11%
1.11%
1.11%
1.11%
0
0.0000
0.0025
0.0050
3
Q (m /s)
0.0075
0.0100
Minor Loss Percentage versus Flow Rate
1.2%
1.1%
h lm /h lT
1.0%
0.9%
0.8%
0.0000
0.0025
0.0050
3
Q (m /s)
0.0075
0.0100
Problem 8.124
Given:
Flow from pump to reservoir
Find:
Pressure at pump discharge
Solution:
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
V1
L V1
h lT = h l + h lm = f ⋅ ⋅
+ Kexit ⋅
2
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 <<
Hence the energy equation between Point 1 and the free surface (Point 2) becomes
2
2
⎛ p1 V2 ⎞
L V
V
⎜ +
− ( g ⋅ z2 ) = f ⋅ ⋅
+ Kexit ⋅
2 ⎠
2
D 2
⎝ρ
⎛
V
⎝
2
Solving for p 1
p 1 = ρ⋅ ⎜ g ⋅ z2 −
From Table A.7 (68oF)
ρ = 1.94⋅
slug
ft
Re =
2
3
V⋅ D
ν
For commercial steel pipe e = 0.00015 ⋅ ft
2⎞
2
+ f⋅
L V
V
⋅
+ Kexit ⋅
2
D 2
ν = 1.08 × 10
Re = 10⋅
ft
For the exit
Kexit = 1.0
so we find
ft
4 ⋅ mile
ft
3
⎡
× ⎢32.2⋅
⎢
⎣
2
s
× 50⋅ ft + .0150 ×
9
12
2
s
s
⋅ ft ×
−5
1.08 × 10
⋅ ft
Re = 6.94 × 10
2
e
so
D
⎛ e
⎞
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
slug
⋅
(Table 8.1)
Flow is turbulent:
p 1 = 1.94⋅
s
×
− 5 ft
⎠
0.75⋅ ft
×
⎛
⎝
1mile
×
1
2
× ⎛⎜ 10⋅
⎝
= 0.000200
2⎤
2⎞
L V
⋅
D 2
⎠
2
⎥ × lbf ⋅ s
s ⎠ ⎥ slug⋅ ft
⎦
ft ⎞
Turbulent
f = 0.0150
p 1 = ρ⋅ ⎜ g ⋅ z2 + f ⋅
5280⋅ ft
5
4 lbf
p 1 = 4.41 × 10 ⋅
ft
2
p 1 = 306 ⋅ psi
Problem 8.123
Given:
Data on water system
Find:
Minimum tank height; equivalent pressure
[Difficulty: 4]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Basic equations:
2
ρ⋅ V⋅ D
Re =
hl = f ⋅
μ
∑
hl +
major
∑
(8.29)
h lm
minor
2
L V
⋅
D 2
h lm = K⋅
(8.34)
⎞
⎛ e
⎜
1
2.51
D
(8.37)
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
V
(8.40a)
2
h lm = f ⋅
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Available data
D = 7.5⋅ mm
L = 1⋅ m
Re = 100000
From Section 8.7
Kent = 0.5
Lelbow45 = 16⋅ D
Lelbow90 = 30⋅ D
LGV = 8 ⋅ D
Lelbow45 = 0.12 m
Lelbow90 = 0.225 m
LGV = 0.06 m
From Table A.8 at 10oC
ρ = 1000
Q =
π⋅ μ⋅ D⋅ Re
4⋅ ρ
μ = 1.3⋅ 10
3
3
−4m
d =
⎛
2⋅ g ⎝
V
V
2⋅ g
⋅⎜1 + f ⋅
IF INSTEAD the reservoir was pressurized
Q = 0.766
s
2
d−
2
Hence
⋅
2
m
Q = 7.66 × 10
The energy equation becomes
f = 0.0180
− 3 N⋅ s
kg
m
Then
and so
L
D
V =
s
Q
V = 17.3
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
m
s
Lelbow90
Lelbow45
LGV ⎞
⎛ L
+ 2⋅ f ⋅
+ 2⋅ f ⋅
+ f⋅
2⋅ g ⎝ D
D
D
D ⎠
2
=
l
V
+ 2⋅ f ⋅
⋅⎜f ⋅
Lelbow90
D
+ 2⋅ f ⋅
∆p = ρ⋅ g ⋅ d
Lelbow45
D
+ f⋅
LGV ⎞
D
∆p = 781 ⋅ kPa
⎠
d = 79.6⋅ m
Unrealistic!
which is feasible
at this Re
Problem 8.122
Given:
Flow of oil in a pipe
Find:
Percentage change in loss if diameter is reduced
Solution:
Basic equations
Available data
2
L V
hl = f ⋅ ⋅
D 2
ν = 7.5⋅ 10
V=
Here
Re =
Then
Q
A
f =
− 4 ft
⋅
s
2
V =
π⋅ D
V⋅ D
4
π
D = 1 ⋅ in
× 0.1⋅
Re = 18.3⋅
ν
hl = f ⋅
Laminar
Re
L = 100 ⋅ ft
2
The flow is LAMINAR
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Turbulent
2
4⋅ Q
=
64
[Difficulty: 3]
ft
s
ft
3
×
s
1
×
12
⎛ 12 ⋅ 1 ⎞
⎜
⎝ 1 ft ⎠
hl =
2
V = 18.3⋅
7.5 × 10
−4
⋅ ft
ft
3
s
s
Re = 2033
2
2
⎛ 18.3 ft ⎞
⎜
64
s⎠
100
⎝
hl =
×
×
1
2
2033
64 L V
⋅ ⋅
Re D 2
Q = 0.100
ft
s
⋅ ft ×
2
L V
⋅
D 2
Q = 45⋅ gpm
h l = 6326⋅
ft
2
2
s
12
D = 0.75⋅ in
When the diameter is reduced to
V=
Re =
Q
A
4⋅ Q
=
2
π⋅ D
V⋅ D
V =
4
π
× 0.1⋅
Re = 32.6⋅
ν
The flow is TURBULENT For drawn tubing, from Table 8.1
Given
⎛ e
⎞
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
s
3
s
×
×
⎛ 12 ⋅ 1 ⎞
⎜
⎝ 0.75 ft ⎠
0.75
12
2
V = 32.6⋅
ft
s
s
⋅ ft ×
7.5 × 10
−4
⋅ ft
2
L V
⋅
D 2
Re = 2717
e = 0.000005⋅ ft
f = 0.0449
2
⎛ 32.6 ft ⎞
⎜
100
s⎠
⎝
h l = .0449 ×
×
0.75
2
2
hl = f ⋅
ft
ft
12
4
The increase in loss is
3.82 × 10 − 6326
6326
= 504 ⋅ %
4 ft
h l = 3.82 × 10 ⋅
This is a HUGH increase! The main increase is because the
diameter reduction causes the velocity to increase; the loss
goes as V2, and 1/D, so it increases very rapidly
2
2
s
Problem 8.121
Given:
Data on tube geometry
Find:
Plot of reservoir depth as a function of flow rate
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
ρ⋅ V⋅ D
μ
64
2
hl = f ⋅
(8.36)
L V
⋅
D 2
(Laminar)
g ⋅ d − α⋅
This can be solved expicitly for height d, or solved using Solver
2
In Excel:
V
2⋅ g
⋅ ⎛⎜ α + f ⋅
⎝
L
D
+ K⎞
⎠
major
∑
h lm (8.29)
minor
V
(8.40a)
2
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
2
d=
hl +
2
h lm = K⋅
(8.34)
Re
The energy equation (Eq. 8.29) becomes
∑
V
2
2
= f⋅
2
L V
V
⋅
+ K⋅
2
D 2
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Required Reservoir Head versus Flow Rate
75
50
d (m)
25
0
0
2
4
6
Q (L/min)
8
10
12
Problem 8.120
Given:
Data on a tube
Find:
"Resistance" of tube for flow of kerosine; plot
[Difficulty: 4]
Solution:
The basic equations for turbulent flow are
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = hl
⎝
⎠ ⎝ρ
⎠
⎞
⎛ e
2
⎜ D
L V
1
2.51
(8.34)
hl = f ⋅ ⋅
= −2.0⋅ log ⎜
+
D 2
f
⎝ 3.7 Re⋅ f ⎠
The given data is
L = 250 ⋅ mm
From Fig. A.2 and Table A.2
μ = 1.1 × 10
(8.37)
D = 7.5⋅ mm
− 3 N⋅ s
⋅
ρ = 0.82 × 990 ⋅
2
m
For an electrical resistor
(8.29)
kg
3
= 812 ⋅
m
kg
(Kerosene)
3
m
V = R⋅ I
(1)
Simplifying Eqs. 8.29 and 8.34 for a horizontal, constant-area pipe
⎛
⎜
⎜
2
p1 − p2
L V
L ⎝
= f⋅ ⋅
= f⋅ ⋅
D 2
ρ
D
⎞
Q
π
4
2
2
⋅D
⎠
2
∆p =
or
8 ⋅ ρ⋅ f ⋅ L
2
5
2
⋅Q
(2)
π ⋅D
By analogy, current I is represented by flow rate Q, and voltage V by pressure drop ∆p . Comparing Eqs. (1) and (2), the
"resistance" of the tube is
R=
∆p
Q
=
8 ⋅ ρ⋅ f ⋅ L⋅ Q
2
5
π ⋅D
The "resistance" of a tube is not constant, but is proportional to the "current" Q! Actually, the dependence is not quite linear,
because f decreases slightly (and nonlinearly) with Q. The analogy fails!
The analogy is hence invalid for
Re > 2300
or
ρ⋅ V⋅ D
μ
> 2300
ρ⋅
Writing this constraint in terms of flow rate
Q
π
4
2
⋅D
⋅D
μ
> 2300
or
Q>
2300⋅ μ⋅ π⋅ D
4⋅ ρ
3
−5m
Q = 1.84 × 10
Flow rate above which analogy fails
s
The plot of "resistance" versus flow rate cab be done in Excel.
"Resistance" of a Tube versus Flow Rate
9
"R"
3
(10 Pa/m /s)
1.E+01
1.0E-05
1.0E-04
1.0E-03
1.E-01
1.E-03
3
Q (m /s)
1.0E-02
Problem 8.119
Given:
Data on water flow from a tank/tubing system
Find:
Minimum tank level for turbulent flow
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g⋅ z1 − ⎜ ρ + α2⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
2
ρ⋅ V⋅ D
hl = f ⋅
μ
∑
hl +
major
∑
h lm (8.29)
minor
2
L V
⋅
D 2
h lm = K ⋅
(8.34)
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
(8.37)
V
2
h lm = f ⋅
(8.40a)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Velocity at free surface is <<
The available data is
D = 7.5⋅ mm
L = 500 ⋅ mm
From Table A.8 at 10oC
ρ = 1000
− 3 N⋅ s
kg
μ = 1.3⋅ 10
3
m
Re = 10000
From
Re =
Kent = 0.5
ρ⋅ V⋅ D
Re =
μ
π
4
Hence
V =
2
π⋅ μ⋅ D⋅ Re
4⋅ ρ
V = 1.73
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
f
Q =
⋅D
Q
1
Assuming a smooth tube
or
= −2 ⋅ log⎛⎜
⎞
⎝ Re⋅ f ⎠
2.51
so
m
s
2
2
V
⋅ ⎛⎜ f ⋅
L
+ Kent + Kexit⎞
Solving for d
d =
FOR r > 0.15D)
Kent = 0.04 (Table 8.2)
2⋅ g
⎝ D
⎠
3
−5m
Q = 7.66 × 10
f = 0.0309
g⋅ d = f ⋅
The energy equation (Eq. 8.29) becomes
2
m
Kexit = 1
(Table 8.2)
ρ⋅ Q⋅ D
⋅
2
2
L V
V
V
⋅
+ Kent⋅
+ Kexit ⋅
2
2
D 2
d = 545 ⋅ mm
d = 475 ⋅ mm
s
Q = 0.0766⋅
l
s
Problem 8.118
[Difficulty: 2]
Problem 8.117
Given:
Data on water flow from a tank/tubing system
Find:
Minimum tank level for turbulent flow
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Re =
f =
ρ⋅ V⋅ D
μ
64
2
hl = f ⋅
(8.36)
L V
⋅
D 2
hl +
major
∑
h lm (8.29)
minor
2
h lm = K⋅
(8.34)
(Laminar)
2
g ⋅ d − α⋅
This can be solved expicitly for height d, or solved using Solver
V
(8.40a)
2
⎞
⎛ e
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Re
The energy equation (Eq. 8.29) becomes
∑
V
2
2
= f⋅
2
L V
V
⋅
+ K⋅
2
D 2
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Problem 8.116
[Difficulty: 4]
Problem 8.115
[Difficulty: 2]
Problem 8.114
[Difficulty: 4]
e
d
Flow
Nozzle
Short pipe
Given:
Flow out of water tank through a nozzle
Find:
Change in flow rate when short pipe section is added; Minimum pressure; Effect of frictionless flow
Solution:
Basic equations
2
2
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
V2
1
2
L V2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 + K⋅ 2 Q = V⋅ A
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0
Available data
D2 = 25⋅ mm
r = 0.02⋅ D2
D3 = 50⋅ mm
r = 0.5⋅ mm
z1 = 2.5⋅ m
ρ = 999 ⋅
Hence for the nozzle case, between the free surface (Point 1) and the exit (2) the energy equation becomes
g ⋅ z1 −
V2
2
V2
2
= Knozzle⋅
2
2
2 ⋅ g ⋅ z1
Solving for V 2
V2 =
(1 + Knozzle)
Hence
V2 =
2 × 9.81⋅
m
2
× 2.5⋅ m ×
s
Q = V2 ⋅ A2
Q = 6.19⋅
m
s
1
m
V2 = 6.19
s
( 1 + 0.28)
×
π
4
× ( 0.025 ⋅ m)
2
3
−3m
Q = 3.04 × 10
s
Q = 3.04
L
When a small piece of pipe is added the energy equation between the free surface (Point 1) and the exit (3) becomes
g ⋅ z1 −
From continuity
V3
2
2
V2
= Knozzle⋅
2
A2
V3 = V2 ⋅
= V2 ⋅ AR
A3
2
V2
+ Ke⋅
2
2
3
m
Knozzle = 0.28
For a rounded edge, we choose the first value from Table 8.2
kg
s
V2 =
Solving for V 2
2 ⋅ g ⋅ z1
⎛ AR2 + K
⎞
nozzle + Ke⎠
⎝
2
⎛ D2 ⎞ ⎛ 25 ⎞ 2
AR =
=⎜
=⎜
= 0.25
A3
⎝ D3 ⎠ ⎝ 50 ⎠
A2
We need the AR for the sudden expansion
AR = 0.25
Ke = 0.6
From Fig. 8.15 for AR = 0.25
V2 =
V2 =
Hence
Q = V2 ⋅ A2
2 ⋅ g ⋅ z1
⎛ AR2 + K
⎞
nozzle + Ke⎠
⎝
2 × 9.81⋅
m
2
× 2.5⋅ m ×
s
Q = 7.21⋅
m
s
×
π
4
× ( 0.025 ⋅ m)
1
m
V2 = 7.21
s
(0.252 + 0.28 + 0.6)
3
−3m
2
Q = 3.54 × 10
∆Q
Comparing results we see the flow increases from 3.04 L/s to 3.54 L/s
=
Q
s
3.54 − 3.04
3.04
Q = 3.54
= 16.4⋅ %
The flow increases because the effect of the pipe is to allow an exit pressure at the nozzle LESS than atmospheric!
The minimum pressure point will now be at Point 2 (it was atmospheric before adding the small pipe). The energy equation
between 1 and 2 is
2
2
⎛⎜ p
V2
V2 ⎞
2
g ⋅ z1 − ⎜
+
= Knozzle⋅
2 ⎠
2
⎝ρ
Solving for p 2
2
⎡⎢
V2
p 2 = ρ⋅ ⎢g ⋅ z1 −
⋅ ( Knozzle +
2
⎣
Hence
p 2 = 999 ⋅
kg
3
m
⎡
m
⎢
⎣
s
× ⎢9.81⋅
2
⎤⎥
)⎦
1⎥
× 2.5⋅ m −
1
2
× ⎛⎜ 7.21⋅
⎝
m⎞
s
⎠
2
⎤
lbf ⋅ s
⎥
⎦
slug⋅ ft
× ( 0.28 + 1 )⎥ ×
2
p 2 = −8.736 ⋅ kPa
If the flow were frictionless the the two loss coeffcients would be zero. Instead of
Instead of
V2 =
2 ⋅ g ⋅ z1
⎛ AR2 + K
⎞
nozzle + Ke⎠
⎝
we'd have
If V2 is larger, then p2, through Bernoulli, would be lower (more negative)
V2 =
2 ⋅ g ⋅ z1
2
AR
which is larger
L
s
Problem 8.113
Given:
Sudden expansion
Find:
Expression for upstream average velocity
[Difficulty: 2]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g⋅ z1 − ⎜ ρ + α2⋅ 2 + g⋅ z2 = h lT
⎝
⎠ ⎝
⎠
The basic equation is
(8.29)
2
V
h lT = h l + K ⋅
2
Assume:
1) Steady flow 2) Incompressible flow 3) h l = 0 4) α1 = α2 = 1 5) Neglect gravity
The mass equation is
V1⋅ A1 = V2⋅ A2
so
V2 = AR⋅ V1
Equation 8.29 becomes
p1
ρ
or (using Eq. 1)
∆p
ρ
Solving for V1
If the flow were frictionless, K = 0, so
+
=
V1 =
V1
2
=
2
(1)
p1
+
ρ
p2 − p1
V1
2
+ K⋅
2
V1
=
ρ
2
2
2 ⋅ ∆p
(
compared to
(
(
2 ⋅ ∆p
2
< V1
)
2
ρ⋅ 1 − AR
∆pinvscid =
∆p =
2
)
2
V1
2
2
(
V1
2
2
(
)
2
⋅ 1 − AR
2
)
⋅ 1 − AR − K
Hence a given flow rate would generate a larger ∆p for inviscid flow
2
)
⋅ 1 − AR − K
Hence the flow rate indicated by a given ∆p would be lower
If the flow were frictionless, K = 0, so
V1
ρ⋅ 1 − AR − K
Vinviscid =
A1
V2 = V1⋅
A2
Problem 8.112
[Difficulty: 3]
Problem 8.111
[Difficulty: 4]
Given:
Sudden expansion
Find:
Expression for minor head loss; compare with Fig. 8.15; plot
Solution:
The governing CV equations (mass, momentum, and energy) are
Assume:
1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No
shaft work 6) Neglect viscous friction 7) Neglect gravity
The mass equation becomes
V1 ⋅ A1 = V2 ⋅ A2
The momentum equation becomes
p 1 ⋅ A2 − p 2 ⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2
or (using Eq. 1)
A1
p 1 − p 2 = ρ⋅ V1 ⋅
⋅ V2 − V1
A2
The energy equation becomes
p1
p2
⎛
⎛
2⎞
2⎞
Qrate = ⎜ u 1 +
+ V1 ⋅ −ρ⋅ V1 ⋅ A1 + ⎜ u 2 +
+ V2 ⋅ ρ⋅ V2 ⋅ A2
ρ
ρ
⎝
⎠
⎝
⎠
or (using Eq. 1)
(
)
(
h lm = u 2 − u 1 −
=
mrate
h lm =
V1 − V2
2
⎡
⎢
h lm =
⋅ 1−
2 ⎢
⎣
V1
2
2
)
V1 − V2
2
(
2⎤
(
2
A1
+ V1 ⋅
⋅ V2 − V1
A2
⎛ V2 ⎞
⎜
⎝ V1 ⎠
)
(2)
2
Qrate
(
)
(
2
Combining Eqs. 2 and 3
(1)
+
p1 − p2
ρ
)
⎤
⎥
2 A1 ⎡⎛ V2 ⎞
+ V1 ⋅
⋅ ⎢⎜
− 1⎥
⎥
A2
⎦
⎣⎝ V1 ⎠ ⎦
(3)
)
AR =
From Eq. 1
h lm =
Hence
h lm =
A1
A2
V1
2
V1
2
V1
(
2
(
2
⋅ 1 − AR
2
h lm = K⋅
V2
=
2
)
2
⋅ 1 − AR + 2 ⋅ AR − 2 ⋅ AR
V1
2
2
K = ( 1 − AR)
Finally
) + V12⋅AR⋅(AR − 1)
2 V1
= ( 1 − AR) ⋅
2
2
2
This result, and the curve of Fig. 8.15, are shown below as computed in Excel. The agreement is excellent.
AR
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
K CV
K Fig. 8.15
1.00
0.81
0.64
0.49
0.36
0.25
0.16
0.09
0.04
0.01
0.00
1.00
0.60
0.38
0.25
0.10
0.01
0.00
(Data from Fig. 8.15
is "eyeballed")
Loss Coefficient for a
Sudden Expansion
1.0
Theoretical Curve
0.8
Fig. 8.15
K
0.5
0.3
0.0
0.00
0.25
0.50
Area Ratio AR
0.75
1.00
Problem 8.110
[Difficulty: 4]
Problem 8.109
Given:
Data on geometry of conical diffuser; flow rate
Find:
Static pressure rise; loss coefficient
Solution:
Basic equations
Cp =
p2 − p1
1
2
V1
[Difficulty: 3]
2
(
⋅ ρ⋅ V1
(8.41) h lm = K⋅
2
2
Given data
D1 = 2 ⋅ in
From Eq. 8.41
1
2
∆p = p 2 − p 1 = ⋅ ρ⋅ V1 ⋅ Cp (1)
2
)
V1
2
= Cpi − Cp ⋅
2
D2 = 3.5⋅ in
1
Cpi = 1 −
(8.44)
(8.42)
2
AR
N = 6 ⋅ in
Q = 750 ⋅ gpm
(N = length)
K= 1−
Combining Eqs. 8.44 and 8.42 we obtain an expression for the loss coefficient K
1
2
− Cp
(2)
AR
The pressure recovery coefficient Cp for use in Eqs. 1 and 2 above is obtained from Fig. 8.15 once compute AR and the
dimensionless length N/R1 (where R1 is the inlet radius)
The aspect ratio AR is
⎛ D2 ⎞
AR = ⎜
⎝ D1 ⎠
R1 =
2
D1
R1 = 1 ⋅ in
2
2
⎛ 3.5 ⎞
⎜
⎝ 2 ⎠
AR =
AR = 3.06
N
Hence
R1
=6
From Fig. 8.15, with AR = 3.06 and the dimensionless length N/R1 = 6, we find Cp = 0.6
To complete the calculations we need V1
V1 =
π
4
3
4
gal
1 ⋅ ft
1 ⋅ min
V1 =
× 750 ⋅
×
×
×
π
min 7.48⋅ gal
60⋅ s
Q
⋅ D1
2
∆p =
We can now compute the pressure rise and loss coefficient from Eqs. 1 and 2
∆p =
1
2
K= 1−
× 1.94⋅
slug
ft
1
3
− Cp
2
AR
× ⎛⎜ 76.6⋅
⎝
ft ⎞
s⎠
K = 1−
2
2
× 0.6 ×
1
3.06
2
lbf ⋅ s
slug⋅ ft
− 0.6
×
1
2
⎛ 1 ⎞ V = 76.6⋅ ft
1
⎜ 2
s
⎜ ⋅ ft
⎝ 12 ⎠
2
⋅ ρ⋅ V1 ⋅ Cp
2
⎛ 1 ⋅ ft ⎞
⎜
⎝ 12⋅ in ⎠
2
∆p = 23.7⋅ psi
K = 0.293
Problem 8.108
Given:
Data on inlet and exit diameters of diffuser
Find:
Minimum lengths to satisfy requirements
[Difficulty: 2]
Solution:
Given data
D1 = 100 ⋅ mm
D2 = 150 ⋅ mm
The governing equations for the diffuser are
h lm = K⋅
V1
(
)
V1
= Cpi − Cp ⋅
2
2
1
Cpi = 1 −
and
2
2
(8.44)
(8.42)
2
AR
Combining these we obtain an expression for the loss coefficient K
1
K= 1−
− Cp
2
(1)
AR
The area ratio AR is
⎛ D2 ⎞
AR = ⎜
⎝ D1 ⎠
2
AR = 2.25
The pressure recovery coefficient Cp is obtained from Eq. 1 above once we select K; then, with Cp and AR specified, the minimum
value of N/R1 (where N is the length and R1 is the inlet radius) can be read from Fig. 8.15
(a)
K = 0.2
1
Cp = 1 −
2
−K
Cp = 0.602
AR
From Fig. 8.15
N
R1
= 5.5
R1 =
N = 5.5⋅ R1
(b)
K = 0.35
Cp = 1 −
D1
2
R1 = 50⋅ mm
N = 275 ⋅ mm
1
2
−K
Cp = 0.452
AR
From Fig. 8.15
N
R1
=3
N = 3 ⋅ R1
N = 150 ⋅ mm
Problem 8.107
[Difficulty: 3]
Given:
Flow out of water tank
Find:
Volume flow rate using hole; Using short pipe section; Using rounded edge
Solution:
Basic equations
2
2
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
V2
1
2
L V2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 + K⋅ 2 Q = V⋅ A
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0
Available data
D = 25⋅ mm
r = 5 ⋅ mm
h = 5⋅ m
Hence for all three cases, between the free surface (Point 1) and the exit (2) the energy equation becomes
2
g ⋅ z1 −
V2
2
2
= K⋅
V2
and solving for V 2
2
From Table 8.2 Khole = 0.5 for a hole (assumed to be square-edged)
Also, for a rounded edge
Hence for the hole
r
D
=
V2 =
5⋅ mm
25⋅ mm
= 0.2 > 0.15
2 × 9.81⋅
m
s
Q = V2 ⋅ A2
Hence for the pipe
V2 =
2 × 9.81⋅
Q = 8.09⋅
m
s
Q = V2 ⋅ A2
V2 =
Hence the change in flow rate is
2
× 5⋅ m ×
Q = 7.42⋅
1
s
×
π
4
× ( 0.025⋅ m)
2 × 9.81⋅
m
s
2
× 5⋅ m ×
2
1
s
×
π
4
× ( 0.025⋅ m)
L
s
1
( 1 + 0.04)
4.77 − 3.97 = 0.8⋅
L
s
Kround = 0.04
3
−3m
Q = 3.97 × 10
s
Q = 3.97⋅
L
s
m
V2 = 7.42
s
( 1 + 0.78)
m
(1 + K)
m
V2 = 8.09
s
( 1 + 0.5)
m
2⋅ g ⋅ h
Kpipe = 0.78 for a short pipe (rentrant)
so from Table 8.2
3.64 − 3.97 = −0.33⋅
Hence the change in flow rate is
Hence for the rounded
2
× 5⋅ m ×
V2 =
2
3
−3m
Q = 3.64 × 10
s
Q = 3.64⋅
L
s
The pipe leads to a LOWER flow rate
m
V2 = 9.71
s
Q = V2⋅ A2
Q = 4.77⋅
L
s
The rounded edge leads to a HIGHER flow rate
Problem 8.106
[Difficulty: 3]
Problem 8.105
[Difficulty: 3]
Problem 8.104
Given:
Flow through short pipe
Find:
Volume flow rate; How to improve flow rate
[Difficulty: 3]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT
⎝
⎠ ⎝
⎠
2
V2
L V2
h lT = h l + h lm = f ⋅ ⋅
+ K⋅
2
D 2
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) L << so ignore hl 5) Reentrant
Hence between the free surface (Point 1) and the exit (2) the energy equation becomes
V1
2
+ g ⋅ z1 −
2
From continuity
Hence
V2
2
= K⋅
2
V2
2
2
A2
V1 = V2 ⋅
A1
2
2
2
V2
V2
⎛ A2 ⎞
⋅⎜
+ g⋅ h −
= K⋅
2
2
2
⎝ A1 ⎠
V2
Solving for V 2
V2 =
Hence
V2 =
2⋅ g⋅ h
Q = V2 ⋅ A2
m
2
× 1⋅ m ×
s
Q = 3.33⋅
K = 0.78
and from Table 8.2
2
⎡
⎛ A2 ⎞ ⎤⎥
⎢
⎢1 + K − ⎜ A ⎥
⎣
⎝ 1⎠ ⎦
2 × 9.81⋅
2
m
s
1
m
V2 = 3.33
s
2
⎡
350 ⎞ ⎤
⎢1 + 0.78 − ⎛⎜
⎥
⎣
⎝ 3500 ⎠ ⎦
2
× 350 ⋅ mm ×
⎛ 1⋅ m ⎞
⎜
⎝ 1000⋅ mm ⎠
2
3
−3m
Q = 1.17 × 10
s
The flow rate could be increased by (1) rounding the entrance and/or (2) adding a diffuser (both somewhat expensive)
3
Q = 0.070 ⋅
m
min
Problem 8.103
[Difficulty: 4]
Given:
Contraction coefficient for sudden contraction
Find:
Expression for minor head loss; compare with Fig. 8.15; plot
Solution:
We analyse the loss at the "sudden expansion" at the vena contracta
The governing CV equations (mass, momentum, and energy) are
Assume:
1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No
shaft work 6) Neglect viscous friction 7) Neglect gravity
The mass equation becomes
Vc⋅ Ac = V2 ⋅ A2
The momentum equation becomes
p c⋅ A2 − p 2 ⋅ A2 = Vc⋅ −ρ⋅ Vc⋅ Ac + V2 ⋅ ρ⋅ V2 ⋅ A2
or (using Eq. 1)
Ac
p c − p 2 = ρ⋅ Vc⋅
⋅ V2 − Vc
A2
The energy equation becomes
pc
p2
⎛
⎛
2⎞
2⎞
Qrate = ⎜ u c +
+ Vc ⋅ −ρ⋅ Vc⋅ Ac + ⎜ u 2 +
+ V2 ⋅ ρ⋅ V2 ⋅ A2
ρ
ρ
⎝
⎠
⎝
⎠
or (using Eq. 1)
(1)
(
)
(
h lm = u 2 − u c −
=
mrate
)
)
(2)
(
Qrate
(
)
2
Vc − V2
2
(
2
+
pc − p2
ρ
(3)
)
2
h lm =
Combining Eqs. 2 and 3
Vc − V2
2
⎡
⎢
h lm =
⋅ 1−
2 ⎢
⎣
Vc
Ac
Cc =
From Eq. 1
h lm =
h lm =
Vc
Ac
+ Vc⋅
⋅ V2 − Vc
A2
(
2⎞
2
⎠ + Vc ⋅ Cc⋅ ( Cc − 1 )
⋅ ⎛ 1 − C c + 2 ⋅ C c − 2 ⋅ C c⎞
⎝
⎠
2
2
Vc
⎤
⎥
2 Ac ⎡⎛ V2 ⎞
+ Vc ⋅
⋅ ⎢⎜
− 1⎥
⎥
A2
⎦
⎣⎝ Vc ⎠ ⎦
Vc
2
2
2⎤
⎛ V2 ⎞
⎜
⎝ Vc ⎠
⋅ ⎛ 1 − Cc
2 ⎝
Vc
)
V2
=
A2
h lm =
Hence
2
2
2
2
(
⋅ 1 − Cc
2
2
)2
(4)
2
2
2
Vc
⎛ V2 ⎞
2
h lm = K⋅
= K⋅
⋅⎜
= K⋅
⋅ Cc
2
2
2
Vc
⎝ ⎠
V2
But we have
K=
Hence, comparing Eqs. 4 and 5
Vc
(5)
( 1 − Cc) 2
Cc
2
⎛ 1 − 1⎞
⎜C
⎝ c
⎠
2
So, finally
K=
where
⎛ A2 ⎞
Cc = 0.62 + 0.38⋅ ⎜
⎝ A1 ⎠
3
This result,can be plotted in Excel. The agreement with Fig. 8.15 is reasonable.
0.5
0.4
K
0.3
0.2
0.1
0
0.2
0.4
0.6
AR
0.8
1
Problem 8.102
Given:
Flow through a reentrant device
Find:
Head loss
[Difficulty: 3]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
2
V2
L V2
h lT = h l + h lm = f ⋅ ⋅
+ K⋅
2
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) L << so ignore hl 5) Reentrant
3
Available data
D1 = 100 ⋅ mm
D2 = 50⋅ mm
Q = 0.01⋅
π
2
A1 = ⋅ D1
4
A1 = 7.85 × 10 mm
3
2
m
K = 0.78
and from Table 8.2
s
π
2
A2 = ⋅ D2
4
3
2
A2 = 1.96 × 10 mm
Hence between the free surface (Point 1) and the exit (2) the energy equation becomes
p1
ρ
From continuity
Hence
Solving for h
+
V1
2
2
−
V2
2
2
−
Q = V1 ⋅ A1 = V2 ⋅ A2
2
p2
ρ
= K⋅
V2
2
2
ρ
2
1 Q ⎞
1 Q ⎞
Q ⎞
g ⋅ h + ⋅ ⎛⎜
− ⋅ ⎛⎜
= K⋅ ⋅ ⎛⎜
2 A1
2 A2
2 A2
⎝ ⎠
⎝ ⎠
⎝ ⎠
1
⎛Q⎞
⎜A
⎝ 2⎠
h =
p1 − p2
and also
=
2
2
2
⎡
⎛ A2 ⎞ ⎤⎥
⎢
⋅ 1+ K− ⎜
⎥
2⋅ g ⎢
⎣
⎝ A1 ⎠ ⎦
h = 2.27 m
ρ⋅ g ⋅ h
ρ
= g⋅ h
where h is the head loss
Q = V⋅ A
Problem 8.101
Given:
Data on a pipe sudden contraction
Find:
Theoretical calibration constant; plot
[Difficulty: 4]
Solution:
Given data
D1 = 45⋅ mm
D2 = 22.5⋅ mm
The governing equations between inlet (1) and exit (2) are
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
where
h l = K⋅
V2
(8.29)
2
(8.40a)
2
Hence the pressure drop is (assuming α = 1)
2
⎛⎜ V 2 V 2
V2 ⎞
1
2
∆p = p 1 − p 2 = ρ⋅ ⎜
−
+ K⋅
2
2 ⎠
⎝ 2
For the sudden contraction
so
π
π
2
2
V1 ⋅ ⋅ D1 = V2 ⋅ ⋅ D2 = Q
4
4
∆p =
⎡⎛ D ⎞ 4
⎢ 1
⋅ ⎜
⎢ D ( 1 + K) −
2
⎣⎝ 2 ⎠
ρ⋅ V1
2
⎛ D1 ⎞
V2 = V1 ⋅ ⎜
⎝ D2 ⎠
or
⎤
⎥
1
⎥
⎦
For the pressure drop we can use the manometer equation
∆p = ρ⋅ g ⋅ ∆h
Hence
In terms of flow rate Q
ρ⋅ g ⋅ ∆h =
⎡⎛ D ⎞ 4
⎢ 1
⋅ ⎜
⎢ D ( 1 + K) −
2
⎣⎝ 2 ⎠
ρ⋅ V1
2
⎤
⎥
1
⎥
⎦
⎡⎛ D 4
⎢ 1⎞
ρ⋅ g ⋅ ∆h = ⋅
⋅ ⎜
( 1 + K) −
2 ⎢ D2
2
⎣
⎝
⎠
π
2
⎛ ⋅D ⎞
⎜
1
⎝4
⎠
ρ
2
Q
⎤
⎥
1
⎥
⎦
2
or
⎡⎛ D ⎞ 4
⎢ 1
g ⋅ ∆h =
⋅ ⎜
( 1 + K) −
2
4 ⎢ D2
⎠
π ⋅ D1 ⎣⎝
Hence for flow rate Q we find
Q = k ⋅ ∆h
2
8⋅ Q
2
k=
where
g ⋅ π ⋅ D1
4
⎡⎛ D ⎞ 4
⎢ 1
8⋅ ⎜
⎢ D ( 1 + K) −
⎣⎝ 2 ⎠
For K, we need the aspect ratio AR
⎛ D2 ⎞
AR = ⎜
⎝ D1 ⎠
From Fig. 8.15
K = 0.4
⎤
⎥
⎥
⎦
1
⎤
⎥
⎥
⎦
1
2
AR = 0.25
5
2
Using this in the expression for k, with the other given values
k =
g ⋅ π ⋅ D1
4
⎡⎛ D ⎞ 4
⎢ 1
8⋅ ⎜
⎢ D ( 1 + K) −
⎣⎝ 2 ⎠
−3 m
k = 1.52 × 10
⎤
⎥
1
⎥
⎦
L
For ∆h in mm and Q in L/min
k = 2.89⋅
min
1
mm
2
The plot of theoretical Q versus flow rate ∆h can be done in Excel.
Calibration Curve for a
Sudden Contraction Flow Meter
60
Q (L/mm)
50
40
30
20
10
0
0
50
100
150
200
Dh (mm)
It is a practical device, but is a) Nonlinear and b) has a large energy loss
250
300
350
⋅
2
s
Problem 8.100
Given:
Flow through sudden expansion
Find:
Inlet speed; Volume flow rate
[Difficulty: 3]
Solution:
Basic equations
2
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
V1
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lm h lm = K⋅ 2
⎝
⎠ ⎝
⎠
Q = V⋅ A
∆p = ρH2O⋅ g ⋅ ∆h
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal
Hence the energy equation becomes
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V2 ⎞
V1
1
⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2
⎝
⎠ ⎝
⎠
From continuity
A1
V2 = V1 ⋅
= V1 ⋅ AR
A2
Hence
2
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V1 ⋅ AR ⎞
V1
1
= K⋅
⎜ρ + 2 −⎜ρ +
2
2
⎝
⎠ ⎝
⎠
Solving for V 1
V1 =
(
2⋅ p2 − p1
(
)
2
2
⎛ D1 ⎞ ⎛ 75 ⎞ 2
AR = ⎜
=⎜
= 0.111
⎝ D2 ⎠ ⎝ 225 ⎠
)
ρ⋅ 1 − AR − K
kg
m
so from Fig. 8.14
2
N⋅ s
5
Also
p 2 − p 1 = ρH2O⋅ g ⋅ ∆h = 1000⋅
× 9.81⋅ ×
⋅m ×
= 49.1⋅ Pa
3
2
1000
kg⋅ m
m
s
Hence
V1 =
2 × 49.1⋅
Q = V1 ⋅ A1 =
3
N
2
m
π⋅ D1
4
×
m
1.23⋅ kg
×
1
(1 − 0.1112 − 0.8)
2
⋅ V1
Q =
π
4
×
kg⋅ m
2
N⋅ s
m
V1 = 20.6
s
2
×
K = 0.8
⎛ 75 ⋅ m⎞ × 20.6⋅ m
⎜
s
⎝ 1000 ⎠
3
Q = 0.0910⋅
m
s
3
Q = 5.46⋅
m
min
Problem 8.99
Given:
Flow through sudden contraction
Find:
Volume flow rate
[Difficulty: 3]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = h lm
2
2
⎝
⎠ ⎝ρ
⎠
h lm = K⋅
V2
2
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal
Hence the energy equation becomes
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V2 ⎞
V2
1
⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2
⎝
⎠ ⎝
⎠
From continuity
A2
V1 = V2 ⋅
= V2 ⋅ AR
A1
Hence
2
2
2
2
⎛⎜ p
V2 ⋅ AR ⎞ ⎛⎜ p 2
V2 ⎞
V2
1
−⎜
+
= K⋅
⎜ρ +
2
2 ⎠
2
⎝
⎠ ⎝ρ
Solving for V 2
Hence
(
2⋅ p1 − p2
V2 =
(
2
)
ρ⋅ 1 − AR + K
V2 =
2 × 0.5⋅
lbf
2
π
4
π⋅ D2
4
2
×
⎛ D2 ⎞ ⎛ 1 ⎞ 2
AR = ⎜
=⎜
= 0.25
⎝ D1 ⎠ ⎝ 2 ⎠
2
×
in
Q = V2 ⋅ A2 =
Q =
2
)
so from Fig. 8.14
3
1
slug⋅ ft
⎛ 12⋅ in ⎞ × ft
×
×
⎜
1.94⋅ slug
2
2
⎝ 1 ⋅ ft ⎠
1 − 0.25 + 0.4
lbf ⋅ s
(
)
ft
V2 = 7.45⋅
s
2
⋅ V2
3
⎛ 1 ⋅ ft⎞ × 7.45⋅ ft Q = 0.0406⋅ ft
⎜
s
s
⎝ 12 ⎠
Q = 2.44⋅
ft
3
min
Q = 18.2⋅ gpm
K = 0.4
Problem 8.98
[Difficulty: 3]
Problem 8.97
[Difficulty: 3]
Given:
Flow through gradual contraction
Find:
Pressure after contraction; compare to sudden contraction
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1
2 = h lm
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
h lm = K⋅
V2
2
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal
Q = 25⋅
3
L
Q = 0.025
m
p 1 = 500⋅ kPa
ρ = 999⋅
⎛ D2 ⎞ ⎛ 37.5 ⎞ 2
For an included angle of 150 and an area ratio
=⎜
=⎜
= 0.25 we find from Table 8.3
A1
⎝ D1 ⎠ ⎝ 75 ⎠
K = 0.35
Available data
s
D1 = 75⋅ mm D2 = 37.5⋅ mm
s
Hence the energy equation becomes
3
m
2
A2
o
kg
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V2 ⎞
V2
1
⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2 with
⎝
⎠ ⎝
⎠
V1 =
4⋅ Q
π⋅ D1
2
V2 =
4⋅ Q
π⋅ D2
2
2
ρ
8 ⋅ ρ⋅ Q ⎡ ( 1 + K)
1 ⎤
2
2
p 2 = p 1 − ⋅ ⎡( 1 + K) ⋅ V2 − V1 ⎤ = p 2 −
⋅⎢
−
⎥
⎦
2 ⎢
4
4⎥
2 ⎣
π
D1
⎣ D2
⎦
8
kg ⎛
m
3 N
p 2 = 500 × 10 ⋅
−
× 999 ⋅
× ⎜ 0.025 ⋅
2
2
3 ⎝
s
m
π
m
3⎞
2
× ⎡( 1 + 0.35) ×
⎢
⎣
⎠
Repeating the above analysis for an included angle of 180 o (sudden contraction)
2
1
( 0.0375⋅ m)
4
−
2
⎤ × N⋅ s p = 170 ⋅ kPa
4⎥
kg⋅ m 2
( 0.075 ⋅ m) ⎦
1
K = 0.41
3
2
⎛
m ⎞
1
1
⎤ × N⋅ s p = 155 ⋅ kPa
p 2 = 500 × 10 ⋅
−
× 999 ⋅
× ⎜ 0.025 ⋅
× ⎡( 1 + 0.41) ×
−
⎢
2
2
3 ⎝
s ⎠
4
4⎥
kg⋅ m 2
m
π
m
( 0.0375⋅ m)
( 0.075 ⋅ m) ⎦
⎣
3 N
8
kg
Problem 8.96
[Difficulty: 3]
Using the above formula for f 0, and Eq. 8.37 for f 1
e/D =
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0.0309
0.0244
0.0207
0.0189
0.0178
0.0148
0.0131
0.0122
0.0116
0.0090
0.0081
0.0066
0.0060
0.0310
0.0245
0.0210
0.0193
0.0183
0.0156
0.0143
0.0137
0.0133
0.0123
0.0122
0.0120
0.0120
0.0311
0.0248
0.0213
0.0197
0.0187
0.0164
0.0153
0.0148
0.0146
0.0139
0.0139
0.0138
0.0138
0.0315
0.0254
0.0223
0.0209
0.0201
0.0183
0.0176
0.0173
0.0172
0.0168
0.0168
0.0167
0.0167
0.0322
0.0265
0.0237
0.0226
0.0220
0.0207
0.0202
0.0200
0.0199
0.0197
0.0197
0.0197
0.0197
0.0335
0.0285
0.0263
0.0254
0.0250
0.0241
0.0238
0.0237
0.0236
0.0235
0.0235
0.0235
0.0235
0.0374
0.0336
0.0321
0.0316
0.0313
0.0308
0.0306
0.0305
0.0305
0.0304
0.0304
0.0304
0.0304
0.0430
0.0401
0.0391
0.0387
0.0385
0.0382
0.0381
0.0381
0.0380
0.0380
0.0380
0.0380
0.0380
0.0524
0.0502
0.0495
0.0492
0.0491
0.0489
0.0488
0.0488
0.0488
0.0487
0.0487
0.0487
0.0487
0.0741
0.0727
0.0722
0.0720
0.0719
0.0718
0.0717
0.0717
0.0717
0.0717
0.0717
0.0717
0.0717
0.001
0.002
0.005
0.01
0.02
0.05
0.0338
0.0288
0.0265
0.0256
0.0251
0.0241
0.0238
0.0237
0.0236
0.0235
0.0234
0.0234
0.0234
0.0376
0.0337
0.0322
0.0316
0.0313
0.0308
0.0306
0.0305
0.0305
0.0304
0.0304
0.0304
0.0304
0.0431
0.0402
0.0391
0.0387
0.0385
0.0381
0.0380
0.0380
0.0380
0.0379
0.0379
0.0379
0.0379
0.0523
0.0502
0.0494
0.0492
0.0490
0.0488
0.0487
0.0487
0.0487
0.0486
0.0486
0.0486
0.0486
0.0738
0.0725
0.0720
0.0719
0.0718
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
f0
Using the add-in function Friction factor from the Web
e/D =
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0
0.0001
0.0002
0.0005
f
0.0309
0.0245
0.0209
0.0191
0.0180
0.0150
0.0132
0.0122
0.0116
0.0090
0.0081
0.0065
0.0059
0.0310
0.0248
0.0212
0.0196
0.0185
0.0158
0.0144
0.0138
0.0134
0.0123
0.0122
0.0120
0.0120
0.0312
0.0250
0.0216
0.0200
0.0190
0.0166
0.0154
0.0150
0.0147
0.0139
0.0138
0.0138
0.0137
0.0316
0.0257
0.0226
0.0212
0.0203
0.0185
0.0177
0.0174
0.0172
0.0168
0.0168
0.0167
0.0167
0.0324
0.0268
0.0240
0.0228
0.0222
0.0208
0.0202
0.0200
0.0199
0.0197
0.0197
0.0196
0.0196
The error can now be computed
e/D =
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0
0.01%
0.63%
0.85%
0.90%
0.92%
0.84%
0.70%
0.59%
0.50%
0.07%
0.35%
1.02%
1.31%
0.0001
0.15%
0.88%
1.19%
1.30%
1.34%
1.33%
1.16%
0.99%
0.86%
0.17%
0.00%
0.16%
0.18%
0.0002
0.26%
1.02%
1.32%
1.40%
1.42%
1.25%
0.93%
0.72%
0.57%
0.01%
0.09%
0.18%
0.19%
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
0.46%
1.20%
1.38%
1.35%
1.28%
0.85%
0.48%
0.30%
0.20%
0.11%
0.15%
0.19%
0.20%
Error (%)
0.64%
0.73%
1.22%
1.03%
1.21%
0.84%
1.07%
0.65%
0.94%
0.52%
0.47%
0.16%
0.19%
0.00%
0.07%
0.07%
0.01%
0.10%
0.15%
0.18%
0.18%
0.19%
0.20%
0.20%
0.20%
0.20%
0.55%
0.51%
0.28%
0.16%
0.09%
0.07%
0.13%
0.16%
0.17%
0.19%
0.20%
0.20%
0.20%
0.19%
0.11%
0.00%
0.06%
0.09%
0.15%
0.18%
0.18%
0.19%
0.20%
0.20%
0.20%
0.20%
0.17%
0.14%
0.16%
0.17%
0.18%
0.19%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
0.43%
0.29%
0.24%
0.23%
0.22%
0.21%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
The maximum discrepancy is 1.42% at Re = 100,000 and e/D = 0.0002
0.100
f
0.010
0.001
1E+04
e/D = 0
e/D = 0.0001
e/D = 0.0002
e/D = 0.0005
e/D = 0.001
e/D = 0.002
e/D = 0.005
e/D = 0.01
e/D = 0.02
e/D = 0.05
1E+05
1E+06
Re
1E+07
1E+08
Problem 8.95
[Difficulty: 2]
Problem 8.94
[Difficulty: 3]
Solution:
Using the add-in function Friction factor from the web site
e/D =
Re
500
1.00E+03
1.50E+03
2.30E+03
1.00E+04
1.50E+04
1.00E+05
1.50E+05
1.00E+06
1.50E+06
1.00E+07
1.50E+07
1.00E+08
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.04
0.1280
0.0640
0.0427
0.0489
0.0338
0.0313
0.0251
0.0246
0.0236
0.0235
0.0234
0.0234
0.0234
0.1280
0.0640
0.0427
0.0512
0.0376
0.0356
0.0313
0.0310
0.0305
0.0304
0.0304
0.0304
0.0304
0.1280
0.0640
0.0427
0.0549
0.0431
0.0415
0.0385
0.0383
0.0380
0.0379
0.0379
0.0379
0.0379
0.1280
0.0640
0.0427
0.0619
0.0523
0.0511
0.0490
0.0489
0.0487
0.0487
0.0486
0.0486
0.0486
0.1280
0.0640
0.0427
0.0747
0.0672
0.0664
0.0649
0.0648
0.0647
0.0647
0.0647
0.0647
0.0647
f
0.1280
0.0640
0.0427
0.0473
0.0309
0.0278
0.0180
0.0166
0.0116
0.0109
0.0081
0.0076
0.0059
0.1280
0.0640
0.0427
0.0474
0.0310
0.0280
0.0185
0.0172
0.0134
0.0130
0.0122
0.0121
0.0120
0.1280
0.0640
0.0427
0.0474
0.0312
0.0282
0.0190
0.0178
0.0147
0.0144
0.0138
0.0138
0.0137
0.1280
0.0640
0.0427
0.0477
0.0316
0.0287
0.0203
0.0194
0.0172
0.0170
0.0168
0.0167
0.0167
0.1280
0.0640
0.0427
0.0481
0.0324
0.0296
0.0222
0.0214
0.0199
0.0198
0.0197
0.0197
0.0196
Friction Factor vs Reynolds Number
1.000
0.100
f
e/D =
0.010
0.001
1.0E+02
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.04
1.0E+03
1.0E+04
Re
1.0E+05
1.0E+06
1.0E+07
1.0E+08
Problem 8.93
[Difficulty: 3]
Using the above formula for f 0, and Eq. 8.37 for f 1
e/D =
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0.0310
0.0244
0.0208
0.0190
0.0179
0.0149
0.0131
0.0122
0.0116
0.0090
0.0081
0.0066
0.0060
0.0311
0.0247
0.0212
0.0195
0.0185
0.0158
0.0145
0.0139
0.0135
0.0124
0.0122
0.0120
0.0120
0.0313
0.0250
0.0216
0.0200
0.0190
0.0167
0.0155
0.0150
0.0148
0.0140
0.0139
0.0138
0.0137
0.0318
0.0258
0.0226
0.0212
0.0204
0.0186
0.0178
0.0175
0.0173
0.0168
0.0168
0.0167
0.0167
0.0327
0.0270
0.0242
0.0230
0.0223
0.0209
0.0204
0.0201
0.0200
0.0197
0.0197
0.0196
0.0196
0.0342
0.0291
0.0268
0.0258
0.0253
0.0243
0.0239
0.0238
0.0237
0.0235
0.0235
0.0234
0.0234
0.0383
0.0342
0.0325
0.0319
0.0316
0.0309
0.0307
0.0306
0.0305
0.0304
0.0304
0.0304
0.0304
0.0440
0.0407
0.0395
0.0390
0.0388
0.0383
0.0381
0.0380
0.0380
0.0379
0.0379
0.0379
0.0379
0.0534
0.0508
0.0498
0.0494
0.0493
0.0489
0.0488
0.0487
0.0487
0.0487
0.0486
0.0486
0.0486
0.0750
0.0731
0.0724
0.0721
0.0720
0.0717
0.0717
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.001
0.002
0.005
0.01
0.02
0.05
0.0338
0.0288
0.0265
0.0256
0.0251
0.0241
0.0238
0.0237
0.0236
0.0235
0.0234
0.0234
0.0234
0.0376
0.0337
0.0322
0.0316
0.0313
0.0308
0.0306
0.0305
0.0305
0.0304
0.0304
0.0304
0.0304
0.0431
0.0402
0.0391
0.0387
0.0385
0.0381
0.0380
0.0380
0.0380
0.0379
0.0379
0.0379
0.0379
0.0523
0.0502
0.0494
0.0492
0.0490
0.0488
0.0487
0.0487
0.0487
0.0486
0.0486
0.0486
0.0486
0.0738
0.0725
0.0720
0.0719
0.0718
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
f0
Using the add-in function Friction factor from the Web
e/D =
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0
0.0001
0.0002
0.0005
f
0.0309
0.0245
0.0209
0.0191
0.0180
0.0150
0.0132
0.0122
0.0116
0.0090
0.0081
0.0065
0.0059
0.0310
0.0248
0.0212
0.0196
0.0185
0.0158
0.0144
0.0138
0.0134
0.0123
0.0122
0.0120
0.0120
The error can now be computed
0.0312
0.0250
0.0216
0.0200
0.0190
0.0166
0.0154
0.0150
0.0147
0.0139
0.0138
0.0138
0.0137
0.0316
0.0257
0.0226
0.0212
0.0203
0.0185
0.0177
0.0174
0.0172
0.0168
0.0168
0.0167
0.0167
0.0324
0.0268
0.0240
0.0228
0.0222
0.0208
0.0202
0.0200
0.0199
0.0197
0.0197
0.0196
0.0196
e/D =
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0.29%
0.39%
0.63%
0.69%
0.71%
0.65%
0.52%
0.41%
0.33%
0.22%
0.49%
1.15%
1.44%
0.36%
0.24%
0.39%
0.38%
0.33%
0.04%
0.26%
0.41%
0.49%
0.51%
0.39%
0.15%
0.09%
0.43%
0.11%
0.19%
0.13%
0.06%
0.28%
0.51%
0.58%
0.60%
0.39%
0.27%
0.09%
0.06%
0.61%
0.21%
0.25%
0.35%
0.43%
0.64%
0.64%
0.59%
0.54%
0.24%
0.15%
0.05%
0.03%
Error (%)
0.88%
1.27%
0.60%
1.04%
0.67%
1.00%
0.73%
0.95%
0.76%
0.90%
0.72%
0.66%
0.59%
0.47%
0.50%
0.37%
0.43%
0.31%
0.16%
0.10%
0.10%
0.06%
0.03%
0.02%
0.02%
0.01%
1.86%
1.42%
1.11%
0.93%
0.81%
0.48%
0.31%
0.23%
0.19%
0.06%
0.03%
0.01%
0.00%
2.12%
1.41%
0.98%
0.77%
0.64%
0.35%
0.21%
0.15%
0.12%
0.03%
0.02%
0.01%
0.00%
2.08%
1.21%
0.77%
0.58%
0.47%
0.24%
0.14%
0.10%
0.08%
0.02%
0.01%
0.00%
0.00%
1.68%
0.87%
0.52%
0.38%
0.30%
0.14%
0.08%
0.06%
0.05%
0.01%
0.01%
0.00%
0.00%
The maximum discrepancy is 2.12% at Re = 10,000 and e/D = 0.01
0.100
f0
0.010
0.001
1E+04
e/D = 0
e/D = 0.0001
e/D = 0.0002
e/D = 0.0005
e/D = 0.001
e/D = 0.002
e/D = 0.005
e/D = 0.01
e/D = 0.02
e/D = 0.05
1E+05
1E+06
Re
1E+07
1E+08
Problem 8.92
[Difficulty: 2]
Problem 8.91
[Difficulty: 2]
Given:
Data on flow in a pipe
Find:
Friction factor; Reynolds number; if flow is laminar or turbulent
Solution:
Given data
From Appendix A
∆p
D = 75⋅ mm
ρ = 1000⋅
L
kg
= 0.075 ⋅
μ = 4 ⋅ 10
3
Pa
m
kg
mrate = 0.075 ⋅
s
− 4 N⋅ s
⋅
m
2
m
The governing equations between inlet (1) and exit (2) are
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l (8.29)
⎝
⎠ ⎝
⎠
2
hl = f ⋅
L V
⋅
D 2
(8.34)
For a constant area pipe
V1 = V2 = V
Hence Eqs. 8.29 and 8.34 become
f =
2⋅ D
2
⋅
(p1 − p2)
ρ
L⋅ V
For the velocity
mrate
V =
ρ⋅
π
4
f =
The Reynolds number is
Re =
2 ⋅ D ∆p
⋅
2 L
ρ⋅ V
V = 0.017
2
⋅D
2 ⋅ D ∆p
⋅
2 L
ρ⋅ V
Hence
=
ρ⋅ V⋅ D
μ
m
s
f = 0.0390
Re = 3183
This Reynolds number indicates the flow is turbulent.
(From Eq. 8.37, at this Reynolds number the friction factor for a smooth pipe is f = 0.043; the friction factor computed above thus
indicates that, within experimental error, the flow corresponds to turbulent flow in a smooth pipe)
Problem 8.90
Given:
Data on flow from reservoir
Find:
Head from pump; head loss
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞ h
V3
V4
3
4
lT
+
α
⋅
+
z
−
+
α
⋅
+
z
=
= HlT
⎜ ρ⋅ g
⎜
3
4
2⋅ g
2⋅ g
g
⎝
⎠ ⎝ ρ⋅ g
⎠
for flow from 3 to 4
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞ ∆h
V3
V2
3
2
pump
= Hpump for flow from 2 to 3
⎜ ρ⋅ g + α⋅ 2⋅ g + z3 − ⎜ ρ⋅ g + α⋅ 2 ⋅ g + z2 =
g
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 = V3 = V4 (constant area pipe)
Then for the pump
Hpump =
p3 − p2
ρ⋅ g
3
2
m
kg⋅ m
s
3 N
Hpump = ( 450 − 150 ) × 10 ⋅
×
×
×
2
1000⋅ kg
2
9.81⋅ m
m
s ⋅N
In terms of energy/mass
h pump = g ⋅ Hpump
For the head loss from 3 to 4 HlT =
p3 − p4
ρ⋅ g
h pump = 9.81⋅
m
2
Hpump = 30.6 m
2
× 30.6⋅ m ×
s
N⋅ s
kg⋅ m
3
h lT = g ⋅ HlT
N⋅ m
kg
+ z3 − z4
2
m
kg⋅ m
s
3 N
HlT = ( 450 − 0 ) × 10 ⋅
×
×
×
+ ( 0 − 35) ⋅ m
2
1000⋅ kg
2
9.81⋅ m
m
s ⋅N
In terms of energy/mass
h pump = 300 ⋅
h lT = 9.81⋅
m
2
s
HlT = 10.9 m
2
× 10.9⋅ m ×
N⋅ s
kg⋅ m
h lT = 107 ⋅
N⋅ m
kg
Problem 8.89
[Difficulty: 2]
Problem 8.88
Given:
Data on flow through a tube
Find:
Head loss
Solution:
Basic equation
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞ h
V1
V2
1
2
lT
+
α
⋅
+
z
−
+
α
⋅
+
z
=
= HlT
⎜ ρ⋅ g
⎜
1
2
2⋅ g
2⋅ g
g
⎝
⎠ ⎝ ρ⋅ g
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1
Given or available data
The basic equation reduces to
Q = 10⋅
h lT =
L
min
∆p
ρ
D = 15⋅ mm
∆p = 85⋅ kPa
ρ = 999 ⋅
kg
3
m
2
h lT = 85.1
m
2
s
HlT =
h lT
g
HlT = 8.68 m
Problem 8.87
[Difficulty: 2]
Problem 8.86
Given:
Data on flow through Alaskan pipeline
Find:
Head loss
Solution:
Basic equation
[Difficulty: 2]
2
2
⎛ p
⎞ ⎛ p
⎞ h
V1
V2
lT
⎜ 1
⎜ 2
+ α⋅
+ z1 −
+ α⋅
+ z2 =
= HlT
⎜ ρ ⋅g
⎜ ρ ⋅g
2⋅ g
2⋅ g
g
oil
oil
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) SG = 0.9 (Table A.2)
Then
p1 − p2
HlT =
+ z1 − z2
SGoil⋅ ρH2O⋅ g
3
2
1
m
kg⋅ m
s
3 N
HlT = ( 8250 − 350 ) × 10 ⋅
×
×
×
×
+ ( 45 − 115 ) ⋅ m
2
2
9.81⋅ m
0.9 1000⋅ kg
m
s ⋅N
In terms of energy/mass h lT = g ⋅ HlT
h lT = 9.81⋅
m
2
s
HlT = 825 m
2
× 825 ⋅ m ×
N⋅ s
kg⋅ m
h lT = 8.09⋅
kN⋅ m
kg
Problem 8.85
[Difficulty: 2]
Problem 8.84
[Difficulty: 2]
Given:
Increased friction factor for water tower flow, and reduced length
Find:
How much flow is decreased
Solution:
Basic equation from Example 8.7
V2 =
(
2 ⋅ g ⋅ z1 − z2
f ⋅ ⎛⎜
L
⎝D
where now we have
L = 530 ⋅ ft
We need to recompute with f = 0.04
V2 =
)
+ 8⎞ + 1
⎠
D = 4 ⋅ in
2 × 32.2⋅
ft
2
× 80⋅ ft ×
s
z1 − z2 = 80⋅ ft
1
0.035 ⋅ ⎛⎜
⎜
⎝
530
4
12
+ 8⎞ + 1
ft
V2 = 9.51⋅
s
⎠
2
Hence
π⋅ D
Q = V2 ⋅ A = V2 ⋅
4
Q = 9.51⋅
ft
s
×
π
4
2
×
⎛ 4 ⋅ ft⎞ × 7.48⋅ gal × 60⋅ s
⎜
3
1 ⋅ min
⎝ 12 ⎠
1 ⋅ ft
Q = 372 ⋅ gpm
(From Table G.2 1 ft3 = 7.48 gal)
Problem 8.83
Given:
Increased friction factor for water tower flow
Find:
How much flow is decreased
[Difficulty: 2]
Solution:
Basic equation from Example 8.7
V2 =
(
2 ⋅ g ⋅ z1 − z2
f ⋅ ⎛⎜
L
⎝D
)
+ 8⎞ + 1
⎠
where
L = 680 ⋅ ft
D = 4 ⋅ in
With f = 0.0308, we obtain
ft
V2 = 8.97⋅
s
and Q = 351 gpm
We need to recompute with f = 0.035
V2 =
2 × 32.2⋅
ft
2
× 80⋅ ft ×
s
z1 − z2 = 80⋅ ft
1
0.035 ⋅ ⎛⎜
⎜
⎝
680
4
12
+ 8⎞ + 1
ft
V2 = 8.42⋅
s
⎠
2
Hence
π⋅ D
Q = V2 ⋅ A = V2 ⋅
4
Q = 8.42⋅
ft
s
×
π
4
2
×
⎛ 4 ⋅ ft⎞ × 7.48⋅ gal × 60⋅ s
⎜
3
1 ⋅ min
⎝ 12 ⎠
1 ⋅ ft
Q = 330 ⋅ gpm
(From Table G.2 1 ft3 = 7.48 gal)
Hence the flow is decreased by
( 330 − 309 ) ⋅ gpm = 21⋅ gpm
Problem 8.82
Given:
A given piping system and volume flow rate with two liquid choices.
Find:
Which liquid has greater pressure loss
[Difficulty: 2]
Solution:
Governing equation:
⎞
⎞ ⎛P
⎛ P1
V2
V2
⎜⎜ + α1 1 + gz1 ⎟⎟ − ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ = hlT
2
2
⎠
⎠ ⎝ρ
⎝ρ
2
2
LV
V
+K
hlT = hl + hlm = f
2
D 2
Assumption: 1) Steady flow 2) Incompressible 3) Neglect elevation effects 4)Neglect velocity effects
LV2
V2
∆P = ρf
+ ρK
2
D 2
From Table A.8 it is seen that hot water has a lower density and lower kinematic viscosity than cold water.
The lower density means that for a constant minor loss coefficient (K) and velocity the pressure loss due to minor losses will be less
for hot water.
The lower kinematic viscosity means that for a constant diameter and velocity the Reynolds number will increase. From Figure 8.13 it
is seen that increasing the Reynolds number will either result in a decreased friction factor (f) or no change in the friction factor. This
potential decrease in friction factor combined with a lower density for hot water means that the pressure loss due to major losses will
be less for hot water as well.
Cold water has a greater pressure drop
Problem 8.81
Given:
Data on flow through elbow
Find:
Inlet velocity
[Difficulty: 2]
Solution:
Basic equation
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞ h
V1
V2
1
2
lT
⎜ ρ⋅ g + α⋅ 2⋅ g + z1 − ⎜ ρ⋅ g + α⋅ 2 ⋅ g + z2 = g = HlT
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1
Then
2
2
(
V2 − V1 = 2 ⋅ V1
V1 =
V1 =
)2 − V12 = 3⋅ V12 =
(
2⋅ p1 − p2
)
(
)
+ 2 ⋅ g ⋅ z1 − z2 − 2 ⋅ g ⋅ HlT
ρ
⎡ (p1 − p2)
⎤
+ g ⋅ ( z1 − z2 ) − g ⋅ HlT⎥
ρ
3 ⎣
⎦
2
⋅⎢
2
3
⎡
3 N
⎢
⎣
m
× ⎢50 × 10 ⋅
2
3
×
m
1000⋅ kg
×
kg⋅ m
2
s ⋅N
+
9.81⋅ m
2
s
× ( −2 ) ⋅ m − 9.81⋅
m
2
s
⎤
× 1 ⋅ m⎥
⎥
⎦
m
V1 = 3.70
s
Problem 8.80
[Difficulty: 2]
Given:
Data on flow in a pipe
Find:
Head loss for horizontal pipe; inlet pressure for different alignments; slope for gravity feed
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
−
+
α
⋅
+
g
⋅
z
⎜ρ
⎜
1 2
1
2 2
2 = h lT
⎝
⎠ ⎝ρ
⎠
The basic equation between inlet (1) and exit (2) is
Given or available data
D = 75⋅ mm
Horizontal pipe data
p 1 = 275 ⋅ kPa
Equation 8.29 becomes
h lT =
V = 5⋅
m
s
p 2 = 0 ⋅ kPa
p1 − p2
ρ = 999 ⋅
kg
m
(Gage pressures)
h lT = 275 ⋅
ρ
3
μ = 0.001 ⋅
(8.29)
N⋅ s
2
m
z1 = z2
V1 = V2
J
kg
For an inclined pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data
z1 = 0 ⋅ m
Equation 8.29 becomes
z2 = 15⋅ m
(
)
p 1 = p 2 + ρ⋅ g ⋅ z2 − z1 + ρ⋅ h lT
p 1 = 422 ⋅ kPa
For a declining pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data
z1 = 0 ⋅ m
Equation 8.29 becomes
z2 = −15⋅ m
(
)
p 1 = p 2 + ρ⋅ g ⋅ z2 − z1 + ρ⋅ h lT
p 1 = 128 ⋅ kPa
For a gravity feed with the same flow rate, the head loss will be the same as above; in addition we have the following new data
p 1 = 0 ⋅ kPa
Equation 8.29 becomes
h lT
z2 = z1 −
g
(Gage)
z2 = −28.1 m
Problem 8.79
Given:
Data on flow through elbow
Find:
Head loss
Solution:
Basic equation
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞ h
V1
V2
1
2
lT
+
α
⋅
+
z
−
+
α
⋅
+
z
=
= HlT
⎜ ρ⋅ g
⎜
1
2
2⋅ g
2⋅ g
g
⎝
⎠ ⎝ ρ⋅ g
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1
Then
HlT =
p1 − p2
ρ⋅ g
2
+
V1 − V2
3
2⋅ g
2
+ z1 − z2
2
(
)
2
2
m
kg⋅ m
s
1
m
s
3 N
2
2
HlT = ( 70 − 45) × 10 ⋅
×
×
×
+ × 1.75 − 3.5 ⋅ ⎛⎜ ⎞ ×
+ ( 2.25 − 3 ) ⋅ mHlT = 1.33 m
2
1000⋅ kg
2
9.81⋅ m
9.81⋅ m
2
s⎠
⎝
m
s ⋅N
In terms of energy/mass
h lT = g ⋅ HlT
h lT = 9.81⋅
m
2
s
2
× 1.33⋅ m ×
N⋅ s
kg⋅ m
h lT = 13.0⋅
N⋅ m
kg
Problem 8.78
[Difficulty: 3]
Given:
Definition of kinetic energy correction coefficient α
Find:
α for the power-law velocity profile; plot
Solution:
Equation 8.26b is
α=
⌠
⎮
3
⎮ ρ⋅ V dA
⌡
2
mrate⋅ Vav
where V is the velocity, mrate is the mass flow rate and Vav is the average velocity
1
⎞
R⎠
n
For the power-law profile (Eq. 8.22)
V = U⋅ ⎛⎜ 1 −
For the mass flow rate
mrate = ρ⋅ π⋅ R ⋅ Vav
Hence the denominator of Eq. 8.26b is
mrate⋅ Vav = ρ⋅ π⋅ R ⋅ Vav
We next must evaluate the numerator of Eq. 8.26b
⎝
r
2.
2
R
3
2 2 3
n
2 ⋅ π⋅ ρ⋅ R ⋅ n ⋅ U
r⎞
⎛
dr =
ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎜ 1 −
R⎠
( 3 + n) ⋅ ( 3 + 2⋅ n)
⎝
3
0
r
To integrate substitute
m=1−
Then
r = R⋅ ( 1 − m)
⌠
⎮
⎮
⎮
⎮
⌡
3
⌠
3
⎮
⎮
n
r
3
3
ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr
⎮
R⎠
⎝
⌡
⌠
⎮
⎮
⌡
⌠
⎮
⎮
⎮
⎮
⌡
2
R
0
R
dm = −
dr
R
dr = −R⋅ dm
3
0
⌠
3
⎮
n
⎮
r
3
n
ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr = −⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ( 1 − m) ⋅ m ⋅ R dm
⌡
R⎠
⎝
1
1
⌠
3 ⎞
⎛ 3
⎮
+1
⎜
⎮
3
n
n
ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ⎝ m − m
⎠ ⋅ R dm
⌡
⌠
⎮
⎮
⌡
Hence
0
2 2
3
⌠
2 ⋅ R ⋅ n ⋅ ρ⋅ π⋅ U
⎮
3
ρ
⋅
V
d
A
=
⎮
( 3 + n) ⋅ ( 3 + 2⋅ n)
⌡
α=
Putting all these results together
⌠
⎮
3
⎮ ρ⋅ V dA
⌡
2
2
=
mrate⋅ Vav
α=
2
3
2⋅ R ⋅ n ⋅ ρ⋅ π⋅ U
( 3+ n) ⋅ ( 3+ 2⋅ n)
2
3
ρ⋅ π⋅ R ⋅ Vav
3
2
2⋅ n
⎛ U ⎞ ⋅
⎜V
⎝ av ⎠ ( 3 + n) ⋅ ( 3 + 2⋅ n )
To plot α versus ReVav we use the following parametric relations
( )
n = −1.7 + 1.8⋅ log Reu
Vav
U
=
2⋅ n
(Eq. 8.23)
2
(Eq. 8.24)
( n + 1) ⋅ ( 2⋅ n + 1)
Vav
ReVav =
⋅ ReU
U
α=
3
2
2⋅ n
⎛ U ⎞ ⋅
⎜V
⎝ av ⎠ ( 3 + n) ⋅ ( 3 + 2⋅ n )
(Eq. 8.27)
A value of ReU leads to a value for n; this leads to a value for Vav/U; these lead to a value for ReVav and α
The plots of α, and the error in assuming α = 1, versus ReVav can be done in Excel.
Re U
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
2.50E+06
5.00E+06
7.50E+06
1.00E+07
n
5.50
6.22
6.76
7.08
7.30
8.02
8.56
8.88
9.10
9.82
10.4
10.7
10.9
V av/U
0.776
0.797
0.811
0.818
0.823
0.837
0.846
0.851
0.854
0.864
0.870
0.873
0.876
Re Vav
Alpha
7.76E+03 1.09
1.99E+04 1.07
4.06E+04 1.06
6.14E+04 1.06
8.23E+04 1.05
2.09E+05 1.05
4.23E+05 1.04
6.38E+05 1.04
8.54E+05 1.04
2.16E+06 1.03
4.35E+06 1.03
6.55E+06 1.03
8.76E+06 1.03
Error
8.2%
6.7%
5.9%
5.4%
5.1%
4.4%
3.9%
3.7%
3.5%
3.1%
2.8%
2.6%
2.5%
Kinetic Energy Coefficient
vs Reynolds Number
Alpha
1.10
1.08
1.05
1.03
1.00
1E+03
1E+04
1E+05
1E+06
1E+07
1E+06
1E+07
Re Vav
Error in assuming Alpha = 1
vs Reynolds Number
10.0%
Error
7.5%
5.0%
2.5%
0.0%
1E+03
1E+04
1E+05
Re Vav
Problem 8.77
[Difficulty: 3]
Problem 8.76
Given:
Laminar flow between parallel plates
Find:
Kinetic energy coefficient, α
[Difficulty: 3]
Solution:
Basic Equation: The kinetic energy coefficient, α is given by
∫
α=
A
From Section 8-2, for flow between parallel plates
ρ V 3dA
(8.26b)
m& V 2
2
2
⎡ ⎛
⎞ ⎤ 3 ⎡ ⎛ y ⎞ ⎤
y
⎟ ⎥ = V ⎢1 − ⎜
⎟ ⎥
u = umax ⎢1 − ⎜
⎢ ⎜a ⎟ ⎥ 2 ⎢ ⎜a ⎟ ⎥
⎢⎣ ⎝ 2 ⎠ ⎥⎦
⎢⎣ ⎝ 2 ⎠ ⎥⎦
since umax =
3
V .
2
Substituting
α=
∫
A
ρV 3dA
m& V 2
=
∫
A
ρu 3dA
ρV A V 2
3
=
1 ⎛u⎞
1
dA =
⎜
⎟
∫
A A⎝V ⎠
wa
a
2
a
2
3
3
2 ⎛u⎞
⎛u⎞
∫a ⎜⎝ V ⎟⎠ wdy = a ∫0 ⎜⎝ V ⎟⎠ dy
−
2
Then
3
31
3
1
3
2 a ⎛ u ⎞ ⎛ umax ⎞ ⎛⎜ y ⎞⎟ ⎛ 3 ⎞
⎜⎜
⎟⎟ ⎜
α=
= ⎜ ⎟ ∫ (1 − η 2 ) dη
⎟ d⎜
∫
a ⎟ ⎝2⎠ 0
a 2 0 ⎝ umax ⎠ ⎝ V ⎠
⎝ 2⎠
where η =
y
a
2
Evaluating,
(1 − η )
2 3
= 1 − 3η 2 + 3η 4 − η 6
The integral is then
⎛ 3⎞
α =⎜ ⎟
⎝2⎠
31
⎛ 3⎞
∫0 (1 − 3η + 3η − η )dη = ⎜⎝ 2 ⎟⎠
2
4
6
3
1
3 5 1 7 ⎤ 27 16
⎡
3
⎢⎣η − η + 5 η − 7 η ⎥⎦ = 8 35 = 1.54
0
Problem 8.75
[Difficulty: 3] Part 1/2
Problem 8.75
[Difficulty: 3] Part 2/2
Problem 8.74
[Difficulty: 3]
Problem 8.73
[Difficulty: 3]
Given: Data on mean velocity in fully developed turbulent flow
Find: Trendlines for each set; values of n for each set; plot
Solution:
y/R
0.898
0.794
0.691
0.588
0.486
0.383
0.280
0.216
0.154
0.093
0.062
0.041
0.024
u/U
0.996
0.981
0.963
0.937
0.907
0.866
0.831
0.792
0.742
0.700
0.650
0.619
0.551
y/R
0.898
0.794
0.691
0.588
0.486
0.383
0.280
0.216
0.154
0.093
0.062
0.037
u/U
0.997
0.998
0.975
0.959
0.934
0.908
0.874
0.847
0.818
0.771
0.736
0.690
Equation 8.22 is
Mean Velocity Distributions in a Pipe
u/U
1.0
0.1
0.01
0.10
1.00
y/R
Re = 50,000
Re = 500,000
Power (Re = 500,000)
Power (Re = 50,000)
Applying the Trendline analysis to each set of data:
At Re = 50,000
At Re = 500,000
u/U = 1.017(y/R )0.161
u/U = 1.017(y/R )0.117
2
with R = 0.998 (high confidence)
Hence
1/n = 0.161
n = 6.21
with R 2 = 0.999 (high confidence)
Hence
Both sets of data tend to confirm the validity of Eq. 8.22
1/n = 0.117
n = 8.55
Problem 8.72
[Difficulty: 3]
Problem 8.71
Given:
Data from a funnel viscometer filled with pitch.
Find:
Viscosity of pitch.
[Difficulty: 1]
Solution:
V πD 4 ρg ⎛ h ⎞
=
(Volume flow rate)
⎜1 + ⎟
t
128µ ⎝ L ⎠
where Q is the volumetric flow rate, V flow volume, t is the time of flow, D is the diameter of the funnel stem, ρ is the density of the
Basic equation: Q =
pitch, µ is the viscosity of the pitch, h is the depth of the pitch in the funnel body, and L is the length of the funnel stem.
Assumption:
Viscous effects above the stem are negligible and the stem has a uniform diameter.
The given or available data is:
Calculate the flow rate:
V = 4.7 ×10 −5 m 3
t = 17,708days
D = 9.4mm
h = 75mm
L = 29mm
ρ = 1.1×103
Q=
kg
m3
4.7 × 10 −5 m 3
m3
V
=
= 3.702 × 10 −14
s
t 17708day × 24hour × 3600s
day
hour
Solve the governing equation for viscosity:
µ=
µ=
πD 4 ρg ⎛
h⎞
⎜1 + ⎟
128Q ⎝ L ⎠
µ = 2.41×108
N ⋅s
m2
4
m
m
⎛
⎞
3 kg
⎟ ×1.1× 10 3 × 9.81 2
m
s ⎛ 75mm ⎞ N × s 2
⎝ 1000mm ⎠
⎟×
⎜1 +
3
⎝ 29mm ⎠ kg × m
−14 m
128 × 3.702 ×10
s
π × (9.4mm )4 × ⎜
Compare this to the viscosity of water, which is 10-3 N·s/m2!
Relate this equation to 8.13c for flow driven by a pressure gradient:
Q=
π∆pD 4 πD 4 ∆p
.
=
×
128µL 128µ L
For this problem, the pressure (∆p) is replaced by the hydrostatic force of the pitch.
Consider the pressure variation in a static fluid.
dp
∆p
∆p
= − ρg = − ρg =
=
.
dz
∆z L + h
Replacing the term in 8.13c
Q=
Hence
which is the same as the given equation.
πD 4 ∆p πD 4 ρg × (L + h ) πD 4
⎛ h⎞
×
=
×
=
× ρg × ⎜1 + ⎟
128µ L 128µ
128µ
L
⎝ L⎠
V πD 4 ρg ⎛ h ⎞
Q= =
⎜1 + ⎟
128µ ⎝ L ⎠
t
Problem 8.70
[Difficulty: 3]
Problem 8.69
Given:
Flow through channel
Find:
Average wall stress
[Difficulty: 2]
Solution:
Basic equation
(Eq. 4.18a)
Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow
With these assumptions the x momentum equation becomes
p 1 ⋅ W⋅ H + τw⋅ 2 ⋅ L⋅ ( W + H) − p 2 ⋅ W⋅ H = 0
or
H
W⋅ H
τw = p 2 − p 1 ⋅
2 ⋅ ( W + H) ⋅ L
(
)
τw = −∆p⋅
L
2 ⋅ ⎛⎜ 1 +
⎝
1
lbf
τw = − × 1 ⋅
×
2
2
in
1 ⋅ in ×
2
144 ⋅ in
ft
2
×
H⎞
W⎠
1 ⋅ ft
12⋅ in
30⋅ ft
×
1
⎞
⎛
⎜
1 ⋅ ft
9.5⋅ in ×
⎜
⎟
12⋅ in
⎜1 +
30⋅ ft
⎝
⎠
Since τw < 0, it acts to the left on the fluid, to the right on the channel wall
lbf
τw = −0.195 ⋅
2
ft
−3
τw = −1.35 × 10
⋅ psi
Problem 8.68
[Difficulty: 2]
Given:
Data on pressure drops in flow in a tube
Find:
Which pressure drop is laminar flow, which turbulent
Solution:
Given data
∂
∂x
p 1 = −4.5⋅
kPa
∂
m
∂x
p 2 = −11⋅
kPa
m
D = 30⋅ mm
From Section 8-4, a force balance on a section of fluid leads to
R ∂
D ∂
τw = − ⋅ p = − ⋅ p
2 ∂x
4 ∂x
Hence for the two cases
D ∂
τw1 = − ⋅ p 1
4 ∂x
τw1 = 33.8 Pa
D ∂
τw2 = − ⋅ p 2
4 ∂x
τw2 = 82.5 Pa
Because both flows are at the same nominal flow rate, the higher pressure drop must correspond to the turbulent flow, because, as
indicated in Section 8-4, turbulent flows experience additional stresses. Also indicated in Section 8-4 is that for both flows the
shear stress varies from zero at the centerline to the maximums computed above at the walls.
The stress distributions are linear in both cases: Maximum at the walls and zero at the centerline.
Problem 8.67
Given:
Pipe glued to tank
Find:
Force glue must hold when cap is on and off
[Difficulty: 2]
Solution:
Basic equation
(Eq. 4.18a)
First solve when the cap is on. In this static case
2
π⋅ D
Fglue =
4
⋅ p1
where p 1 is the tank pressure
Second, solve for when flow is occuring:
Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow
With these assumptions the x momentum equation becomes
2
p1⋅
π⋅ D
4
2
π⋅ D
+ τw⋅ π⋅ D⋅ L − p 2 ⋅
=0
4
Here p1 is again the tank pressure and p 2 is the pressure at the pipe exit; the pipe exit pressure is p atm = 0 kPa gage. Hence
2
Fpipe = Fglue = −τw⋅ π⋅ D⋅ L =
π⋅ D
4
⋅ p1
We conclude that in each case the force on the glue is the same! When the cap is on the glue has to withstand the tank pressure;
when the cap is off, the glue has to hold the pipe in place against the friction of the fluid on the pipe, which is equal in magnitude
to the pressure drop.
π
lbf
2
Fglue =
× ( 3 ⋅ in) × 30⋅
2
4
in
Fglue = 212 ⋅ lbf
Problem 8.66
Given:
Turbulent pipe flow
Find:
Wall shear stress
[Difficulty: 2]
Solution:
Basic equation
(Eq. 4.18a)
Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow
With these assumptions the x momentum equation becomes
2
p1⋅
π⋅ D
4
2
π⋅ D
+ τw⋅ π⋅ D⋅ L − p 2 ⋅
=0
4
or
τw =
( p2 − p1)⋅ D
4⋅ L
3
1
12
τw = − × 750 ⋅ psi ×
15
4
Since τw is negative it acts to the left on the fluid, to the right on the pipe wall
τw = −3.13⋅ psi
=−
∆p⋅ D
4⋅ L
Problem 8.65
Given:
Two-fluid flow in tube
Find:
Velocity distribution; Plot
[Difficulty: 3]
Solution:
D = 5 ⋅ mm
Given data
L = 5⋅ m
∆p = −5 ⋅ MPa
μ1 = 0.5⋅
N⋅ s
μ2 = 5 ⋅
2
m
N⋅ s
2
m
From Section 8-3 for flow in a pipe, Eq. 8.11 can be applied to either fluid
2
u=
⎛ ∂ ⎞ c1
⋅ ln( r) + c2
p +
4 ⋅ μ ⎝ ∂x ⎠
μ
r
⋅⎜
Applying this to fluid 1 (inner fluid) and fluid 2 (outer fluid)
2
u1 =
r
4 ⋅ μ1
⋅
∆p
L
+
2
c1
⋅ ln( r) + c2
μ1
r=
We need four BCs. Two are obvious
u2 =
D
r
4 ⋅ μ2
u2 = 0
2
⋅
∆p
L
+
(1)
c3
μ2
⋅ ln( r) + c4
r=
D
4
u1 = u2
(2)
The third BC comes from the fact that the axis is a line of symmetry
du1
r= 0
dr
=0
(3)
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
r=
du1
du2
μ1 ⋅
= μ2 ⋅
dr
dr
D
4
2
Using these four BCs
⎛ D⎞
⎜
⎝ 2 ⎠ ⋅ ∆p + c3 ⋅ ln⎛ D ⎞ + c = 0
⎜
4
4 ⋅ μ2 L
μ2 ⎝ 2 ⎠
lim
c1
r → 0 μ1 ⋅ r
(4)
2
⎛ D⎞
⎜
⎝ 4 ⎠ ⋅ ∆p + c1 ⋅ ln⎛ D ⎞ + c =
⎜
2
4 ⋅ μ1 L
μ1 ⎝ 4 ⎠
2
⎛ D⎞
⎜
⎝ 4 ⎠ ⋅ ∆p + c3 ⋅ ln⎛ D ⎞ + c
⎜
4
4 ⋅ μ2 L
μ2 ⎝ 4 ⎠
4 ⋅ c1
4 ⋅ c3
D ∆p
D ∆p
⋅
+
= ⋅
+
D
8 L
D
8 L
=0
Hence, after some algebra
c1 = 0
(To avoid singularity)
c2 = −
(
2
D ⋅ ∆p μ2 + 3 ⋅ μ1
64⋅ L
μ1 ⋅ μ2
)
2
c3 = 0
c4 = −
D ⋅ ∆p
16⋅ L⋅ μ2
u 1 ( r) =
The velocity distributions are then
⎡⎢ 2
4 ⋅ μ1 ⋅ L ⎢
⎣
∆p
⋅ r −
2
⎛ D ⎞ ⋅ ( μ2 + 3⋅ μ1 )⎤⎥
⎜
⎥
4 ⋅ μ2
⎝2⎠
⎦
u 2 ( r) =
∆p
4 ⋅ μ2 ⋅ L
⎡2
⋅ ⎢r −
⎣
2
⎛ D ⎞ ⎥⎤
⎜
⎝2⎠ ⎦
(Note that these result in the same expression if µ 1 = µ 2, i.e., if we have one fluid)
Evaluating either velocity at r = D/4 gives the velocity at the interface
2
u interface = −
3 ⋅ D ⋅ ∆p
u interface = −
64⋅ μ2 ⋅ L
3
64
× ( 0.005 ⋅ m) × ⎛ −5 × 10 ⋅
2
2
⎞× m × 1
2
5 ⋅ N⋅ s 5 ⋅ m
m ⎠
6 N
⎜
⎝
u interface = 0.234
Evaluating u 1 at r = 0 gives the maximum velocity
2
u max = −
(
D ⋅ ∆p⋅ μ2 + 3 ⋅ μ1
)
64⋅ μ1 ⋅ μ2 ⋅ L
u max = −
1
64
× ( 0.005 ⋅ m) × ⎛ −5 × 10 ⋅
2
2
m
⎞ × 5 + 3 × 0.5 ⋅ m × 1 u
= 1.02
max
2
N⋅ s 5 ⋅ m
5 × .5
s
m ⎠
6 N
⎜
⎝
2.5
Inner fluid
Outer fluid
r (mm)
2
1.5
1
0.5
0
0.2
0.4
0.6
Velocity (m/s)
The velocity distributions can be plotted in Excel
0.8
1
1.2
m
s
Problem 8.64
Given:
The expression of hydraulic resistance of straight channels with different cross sectional shapes
Find:
Hydraulic resistance
[Difficulty: 2]
Solution:
Based on the expressions of hydraulic resistance listed in the table, one obtains
Using the circle as the example,
Rhyd
1 8 1×10 −3 × 10 × 10 −3 Pa ⋅ s × m
= µL 4 =
= 0.254 ×1012 Pa ⋅ s/m 3
4
−4 4
π
π
a
(1×10 )
m
8
The results are
Shape
Rhyd (1012 Pa·s/m3)
Circle
0.25
Ellipse
3.93
Triangle
18.48
Two plates
0.40
Rectangle
0.51
Square
3.24
Comparing the values of the hydraulic resistances, a straight channel with a circular cross section is the most energy efficient to pump
fluid with a fixed volumetric flow rate; the triangle is the worst.
Problem 8.63
[Difficulty: 3]
Given:
Fully developed flow, velocity profile, and expression to calculate the flow rate
Find:
Velocity and flow rate
Solution:
For the fully developed flow, the N-S equations can be simplified to
Substituting the trial solution in equation (1), one obtains
⎛ ∂ 2 u ∂ 2 u ⎞ ∂p
µ ⎜⎜ 2 + 2 ⎟⎟ =
= constant
∂z ⎠ ∂x
⎝ ∂y
1 ⎞ ∂p
⎛ 1
− 2µu 0 ⎜ 2 + 2 ⎟ =
b ⎠ ∂x
⎝a
u0 = −
Rearrange it and one obtains
a 2b 2
∂p
2
2
2 µ (a + b ) ∂x
Q = ∫ u ( y, z )dydz = ab ∫
The flow rate is
2π
0
Substituting u ( ρ ,φ ) = u 0 (1 − ρ ) into Eq. (4) and integrating twice:
2
Substituting u0 into (5), one obtains
Q = ab ∫
2π
0
0
1
0
1
πa 3b 3
∂p
Q = πabu0 = −
2
2
2
4 µ (a + b ) ∂x
For a pipe radius R, a = b = R, from equation (6),
1 ⎛ πR 4 ∂p ⎞
⎟
Q pipe = ⎜⎜ −
8 ⎝ µ ∂x ⎟⎠
which is the same as equation (8.13b) in the book.
For a channel with an elliptic cross-section with a = R and b = 1.5R, from equation (6), one has
Q pipe =
29 ⎛ πR 4 ∂p ⎞
⎜−
⎟.
104 ⎜⎝ µ ∂x ⎟⎠
(2)
(3)
1
∫ ρu ( ρ , φ )dρdφ
∫ ρu (1 − ρ
0
(1)
2
(4)
1
)dρdφ = πabu0
2
(6)
(5)
Problem 8.62
[Difficulty: 2]
Given:
Fully developed flow in a pipe; slip boundary condition on the wall
Find:
Velocity profile and flow rate
Solution:
Similar to the example described in Section 8.3, one obtained
u=
r 2 ∂p
+ c2
4 µ ∂x
(1)
The constant c2 will be determined by the slip velocity boundary condition at r = R:
u =l
∂u
∂r
(2)
and one obtains
c2 =
R 2 ∂p ⎛ l
⎞
⎜ 2 − 1⎟
4 µ ∂x ⎝ R ⎠
(3)
Substituting c2 into Eq.(1), one obtains
u=−
1 ∂p 2
R − r 2 + 2lR
4 µ ∂x
(
)
(4)
The volume flow rate is
R
Q = ∫ u 2πrdr = −
0
πR 4 ∂p ⎡
l⎤
1+ 4 ⎥
⎢
R⎦
8µ ∂x ⎣
Substituting R = 10 µm, µ = 1.84 x 10-5 N⋅s/m2, mean free path l = 68 nm, and −
Q=−
π (10 ×10 −6 ) 4 m 4
8 1.84 × 10 −5 Pa.s
× (−1.0 × 10 6 )Pa/m × [1 + 4
(5)
∂p
= 1.0×106 Pa/m into eq. (5),
∂x
68 ×10 −9 m
] = 2.19 × 10 −10 m 3 /s .
10 × 10 −6 m
Problem 8.61
[Difficulty: 4]
Given:
Fully developed flow, Navier-Stokes equations; Non-Newtonian fluid
Find:
Velocity profile, flow rate and average velocity
Solution:
According to equation (8.10), we can write the governing equation for Non-Newtonian fluid velocity in a circular tube
r ∂p c1
⎛ du ⎞
+
τ rx = k ⎜ ⎟ =
2 ∂x r
⎝ dr ⎠
n
(1)
However, as for the Newtonian fluid case, we must set c1 = 0 as otherwise we’d have infinite stress at r = 0. Hence, equation (1)
becomes
r ∂p
⎛ du ⎞
k⎜ ⎟ =
2 ∂x
⎝ dr ⎠
n
(2)
The general solution for equation (2), obtained by integrating, is given by
1
1+
1
⎛ 1 ∂p ⎞ n
u =⎜
r n + c2
⎟
⎝ 2k ∂x ⎠ ⎛1 + 1 ⎞
⎜
⎟
⎝ n⎠
1
(3)
Apply the no slip boundary condition at r = R into equation (3), we get
1
1+
1
⎛ 1 ∂p ⎞ n
R n
c 2 = −⎜
⎟
⎝ 2k ∂x ⎠ ⎛1 + 1 ⎞
⎜
⎟
⎝ n⎠
1
(4)
The fluid velocity then is given as
1
n +1
n +1
⎞
n ⎛ 1 ∂p ⎞ n ⎛⎜ n
n ⎟
u (r ) =
r
R
−
⎟ ⎜
⎜
⎟
(n + 1) ⎝ 2k ∂x ⎠ ⎝
⎠
(5)
The volume flow rate is
1
Q = ∫ udA == ∫
A
Hence
R
0
n +1
n +1
⎞
n ⎛ 1 ∂p ⎞ n ⎛ n
2πr
⎜
⎟ ⎜⎜ r − R n ⎟⎟dr
(n + 1) ⎝ 2k ∂x ⎠ ⎝
⎠
(6)
1
1
R
3 n +1
3 n +1
n +1
r2 n ⎤
2nπ ⎛ 1 ∂p ⎞ n ⎡ n
2nπ ⎛ 1 ∂p ⎞ n n ⎛ n
1⎞
n
Q=
r
R
−
=
R
− ⎟
⎜
⎟ ⎢
⎜
⎟
⎜
⎥
(n + 1) ⎝ 2k ∂x ⎠ ⎣ 3n + 1
2
⎝ 3n + 1 2 ⎠
⎦ 0 (n + 1) ⎝ 2k ∂x ⎠
(7)
Simplifying
1
nπ ⎛ 1 ∂p ⎞ n
Q=−
⎜
⎟ R
(3n + 1) ⎝ 2k ∂x ⎠
When n = 1, then k = µ, and
3 n +1
n
(8)
πa 4 ∂p
Q=−
8µ ∂x , just like equation (8.13b) in the textbook.
The average velocity is given by
1
Q
Q
n ⎛ 1 ∂p ⎞ n
V = = 2 =−
⎟ R
⎜
(3n + 1) ⎝ 2k ∂x ⎠
A πR
n +1
n
(9)
Based on Eq. (7), the pressure gradient is
⎤
⎡
∂p
Q(3n + 1) ⎥
= −2k ⎢
3 n +1
⎥
⎢
∂x
⎣ nπR n ⎦
n
(10)
Substituting Q = 1µL/min= 1 × 10-9/60 m3/s, R = 1mm = 10-3 m, and n = 0.5 into eq.(10):
⎡ 10 −9
⎤
(
3(0.5) + 1) ⎥
⎢
∂p
= −2k ⎢ 60 3(0.5 )+1 ⎥
∂x
⎢ 0.5π 0.5
⎥
⎢⎣
⎥⎦
0.5
= −325k Pa/m for n = 0.5
(11)
Similarly, the required pressure gradients for n = 1 and n = 1.5 can be obtained:
∂p
= −42.4k Pa/m, for n = 1
∂x
(12)
∂p
≈ −5.42k Pa/m, for n = 1.5
∂x
(12)
Obviously, the magnitude of the required pressure gradient increases as n decreases. Among the three types of fluids (pseudoplastic
for n = 0.5, Newtonian for n = 1, and dilatant for n = 1.5), the dilatant fluid requires the lowest pressure pump for the same pipe
length.
Problem 8.60
[Difficulty: 4]
Given:
Relationship between shear stress and deformation rate; fully developed flow in a cylindrical blood vessel
Find:
Velocity profile; flow rate
Solution:
Similar to the Example Problem described in Section 8.3, based on the force balance, one obtains
τ rx =
r dp
2 dx
(1)
This result is valid for all types of fluids, since it is based on a simple force balance without any assumptions about fluid rheology.
Since the axial pressure gradient in a steady fully developed flow is a constant, Equation (1) shows that τ = 0 < τc at r = 0. Therefore,
there must be a small region near the center line of the blood vessel for which τ < τc. If we call Rc the radial location at which τ = τc,
the flow can then be divided into two regions:
r > Rc: The shear stress vs. shear rate is governed by
τ = τc + µ
du
dr
(2)
r < Rc: τ = 0 < τc.
We first determine the velocity profile in the region r > Rc. Substituting (1) into (2), one obtains:
r dp
du
= τc + µ
dr
2 dx
(3)
Using equation (3) and the fact that du/dr at r = Rc is zero, the critical shear stress can be written as
Rc dp
= τc .
2 dx
(4)
Rearranging eq. (4), Rc is
Rc = 2τ c /
dp
.
dx
Inserting (4) into (3), rearranging, and squaring both sides, one obtains
(5)
µ
du 1 dp
=
[r − 2 rRc + Rc ]
dr 2 dx
(6)
Integrating the above first-order differential equation using the non-slip boundary condition, u = 0 at r = R:
u=−
1 dp ⎡ 2
8
⎤
3/ 2
(R − r 2 ) −
Rc ( Rc − r 3 / 2 ) + 2 Rc ( R − r )⎥ for Rc ≤ r ≤ R
⎢
4µ dx ⎣
3
⎦
(7)
In the region r < Rc, since the shear stress is zero, fluid travels as a plug with a plug velocity. Since the plug velocity must match the
velocity at r = Rc, we set r = Rc in equation (7) to obtain the plug velocity:
u=−
[
]
1 dp 2
2
( R − Rc ) + 2 Rc ( R − Rc ) for r ≤ Rc
4µ dx
(8)
The flow rate is obtained by integrating u(r) across the vessel cross section:
R
Rc
R
Q = ∫ u (r )2πrdr = ∫ u (r )2πrdr + ∫ u (r )2πrdr
0
Rc
0
4
πR 4 dp ⎡ 16 Rc 4 Rc 1 ⎛ Rc ⎞ ⎤
=−
+
− ⎜ ⎟ ⎥
⎢1 −
8µ dx ⎣⎢ 7 R 3 R 21 ⎝ R ⎠ ⎦⎥
Given R = 1mm = 10-3 m, µ = 3.5 cP = 3.5×10-3 Pa⋅s, and τc = 0.05 dynes/cm2 = 0.05×10-1 Pa, and
From eq. (5), Rc = 2τ c /
(9)
dp
= −100 Pa / m .
dx
dp
dx
2 × 0.05 10 ×10 −6 N / m 2
Rc =
= 0.1mm
100
Pa / m
Substituting the values of R, µ, Rc, and
Q=−
π 1× 10−12 m 4
8 3.5 × 10
−3
Pa.s
dp
into eq. (9),
dx
× (−100) Pa / m × [1 −
16 0.1 mm 4 0.1 mm 1 0.1 mm 4
+
− (
) ] = 3.226 × 10 −9 m3 / s
7 1 mm 3 1 mm 21 1 mm
Problem 8.59
.
Given:
Definition of hydraulic resistance
Find:
Hydraulic resistance in a diffuser
Solution:
Basic equation:
Rhyd =
∆p 8µ z2 1
8µ z2
1
dz
dz
=
=
4
∫
∫
z
z
Q
π 1 r
π 1 (ri + αz ) 4
Rhyd =
=
8µ
π
=−
∆p
Q
∫
z
0
1
d (ri + αz )
(ri + αz ) 4
8µ 1
8µ
−3
(ri + αz ) −3 0z = −
[(ri + αz ) −3 − ri ]
3πα
πα 3
Rhyd = −
8µ ⎡
1
1⎤
−
⎢
⎥
3πα ⎣ (ri + αz ) 3 ri 3 ⎦
[Difficulty: 2]
Problem 8.58
Given:
Tube dimensions and volumetric flow rate
Find:
Pressure difference and hydraulic resistance
[Difficulty: 2]
Solution:
The flow rate of a fully developed pressure-driven flow in a pipe is Q =
flow rate
π∆pR 4
8µL
. Rearranging it, one obtains ∆p =
8µLQ
. For a
πR 4
Q = 10µl / min , L=1 cm, µ = 1.0 × 10 −3 Pa.s , and R = 1 mm,
∆p =
8 10 ×10 −9 m 3
0.01
m
×
×
×
× 4 ×1.0 ×10 −3 Pa.s = 0.00424 Pa
−12
π
60
s 1×10
m
Similarly, the required pressure drop for other values of R can be obtained.
The hydraulic resistance
Rhyd =
∆p 8µL
=
. Substituting the values of the viscosity, length and radius of the tube, one obtains the
Q πR 4
value of the hydraulic resistance.
R (mm)
1
10-1
10-2
10-3
10-4
∆p
0.00424 Pa
42.4 Pa
424 kPa
4.24 GPa
4.24 x 104 GPa
Rhyd (Pa.s/m3)
2.55 x 107
2.55 x 1011
2.55 x 1015
2.55 x 1019
2.55 x 1023
(3) To achieve a reasonable flow rate in microscale or nanoscale channel, a very high pressure difference is required since ∆p is
proportional to R−4. Therefore, the widely used pressure-driven flow in large scale systems is not appropriate in microscale or
nanoscale channel applications. Other means to manipulate fluids in microscale or nanoscale channel applications are required.
Problem 8.57
[Difficulty: 4]
Problem 8.52
8.56
Problem 8.56
[Difficulty: 4] Part 1/2
Problem 8.56
[Difficulty: 4] Part 2/2
Problem 8.54
[Difficulty: 3]
Problem 8.53
[Difficulty: 3]
Problem 8.52
[Difficulty: 3]
Given:
Data on a tube
Find:
"Resistance" of tube; maximum flow rate and pressure difference for which electrical analogy holds for
(a) kerosine and (b) castor oil
Solution:
L = 250 ⋅ mm
The given data is
D = 7.5⋅ mm
From Fig. A.2 and Table A.2
Kerosene:
μ = 1.1 × 10
− 3 N⋅ s
⋅
ρ = 0.82 × 990 ⋅
2
m
Castor oil:
μ = 0.25⋅
3
= 812 ⋅
m
N⋅ s
ρ = 2.11 × 990 ⋅
2
m
For an electrical resistor
kg
kg
3
3
m
= 2090⋅
m
V = R⋅ I
kg
kg
3
m
(1)
The governing equation for the flow rate for laminar flow in a tube is Eq. 8.13c
4
Q=
or
π⋅ ∆p⋅ D
128 ⋅ μ⋅ L
128 ⋅ μ⋅ L
∆p =
4
⋅Q
(2)
π⋅ D
By analogy, current I is represented by flow rate Q, and voltage V by pressure drop ∆p.
Comparing Eqs. (1) and (2), the "resistance" of the tube is
R=
128 ⋅ μ⋅ L
4
π⋅ D
The "resistance" of a tube is directly proportional to fluid viscosity and pipe length, and strongly dependent on the inverse
of diameter
The analogy is only valid for
Re < 2300
ρ⋅
Writing this constraint in terms of flow rate
Q
π
4
2
or
ρ⋅ V⋅ D
μ
< 2300
⋅D
⋅D
μ
< 2300
or
Qmax =
2300⋅ μ⋅ π⋅ D
4⋅ ρ
The corresponding maximum pressure gradient is then obtained from Eq. (2)
∆pmax =
128 ⋅ μ⋅ L
4
π⋅ D
2
⋅ Qmax =
32⋅ 2300⋅ μ ⋅ L
3
ρ⋅ D
Substituting values
3
−5m
(a) For kerosine
Qmax = 1.84 × 10
(b) For castor oil
Qmax = 1.62 × 10
s
3
−3m
s
l
Qmax = 1.10⋅
min
∆pmax = 65.0⋅ Pa
l
Qmax = 97.3⋅
min
∆pmax = 1.30⋅ MPa
The analogy fails when Re > 2300 because the flow becomes turbulent, and "resistance" to flow is then no longer linear with flow
rate
Problem 8.51
[Difficulty: 2]
d
p1 D
F
L
Given:
Hyperdermic needle
Find:
Volume flow rate of saline
Solution:
π⋅ ∆p⋅ d
4
Basic equation
Q=
For the system
F
4⋅ F
∆p = p 1 − p atm =
=
2
A
π⋅ D
4
∆p =
At 68oF, from Table A.7
(Eq. 8.13c; we assume laminar flow and verify this is correct after solving)
128⋅ μ ⋅ L
π
× 7.5⋅ lbf ×
⎛ 1 × 12⋅ in ⎞
⎜
⎝ 0.375⋅ in 1⋅ ft ⎠
∆p = 67.9⋅ psi
− 5 lbf ⋅ s
μH2O = 2.1 × 10
⋅
ft
Q =
π
128
× 67.9⋅
lbf
2
− 7 ft
V=
Q
A
⋅
Q
=
144 ⋅ in
1 ⋅ ft
2
2
4
2
ft
1
12⋅ in
⎞ ×
×
×
−4
12⋅ in ⎠
1 ⋅ in
1 ⋅ ft
1.05 × 10
lbf ⋅ s
⎝
3
− 3 in
Q = 1.43 × 10
s
π⋅ d
⋅
1 ⋅ ft
× ⎛⎜ 0.005 ⋅ in ×
3
V =
μ = 1.05 × 10
ft
2
×
− 4 lbf ⋅ s
μ = 5⋅ μH2O
2
in
Q = 8.27 × 10
Check Re:
2
4
π
× 8.27 × 10
− 7 ft
3
s
⋅
s
2
×
3
Q = 0.0857⋅
⎛ 1 ⎞ × ⎛ 12⋅ in ⎞
⎜
⎜
⎝ .005⋅ in ⎠ ⎝ 1⋅ ft ⎠
2
V = 6.07⋅
in
min
ft
s
4
Re =
ρ⋅ V⋅ d
ρ = 1.94⋅
μ
Re = 1.94⋅
slug
ft
slug
ft
3
× 6.07⋅
ft
s
(assuming saline is close to water)
3
× 0.005 ⋅ in ×
1 ⋅ ft
12⋅ in
×
ft
2
1.05 × 10
−4
⋅ lbf ⋅ s
×
slug⋅ ft
2
s ⋅ lbf
Re = 46.7
Flow is laminar
2
Problem 8.50
Given:
Data on water temperature and tube
Find:
Maximum laminar flow; plot
[Difficulty: 3]
B
Solution:
− 5 N⋅s
A = 2.414⋅ 10
From Appendix A
⋅
B = 247.8 ⋅ K
2
C = 140 ⋅ K
μ ( T) = A ⋅ 10
in
T −C
m
D = 7.5⋅ mm
ρ = 1000⋅
kg
Recrit = 2300
3
m
T1 = −20 °C
− 3 N⋅s
( )
T1 = 253 K
μ T1 = 3.74 × 10
T2 = 120 °C
2
T2 = 393 K
− 4 N⋅s
( )
μ T2 = 2.3 × 10
m
2
m
The plot of viscosity is
0.01
μ
N⋅ s
2
m
−3
1× 10
−4
1× 10
− 20
0
20
40
60
80
100
120
T (C)
For the flow rate
( )
ρ⋅ Vcrit⋅ D
Recrit =
Qmax T1 = 5.07 × 10
μ
3.00
−5m
s
=
ρ⋅ Qmax⋅ D
μ⋅
π
4
=
4 ⋅ ρ⋅ Qmax
2
⋅D
μ⋅ π⋅ D
( )
Qmax( T) =
π⋅ μ( T) ⋅ D⋅ Recrit
4⋅ ρ
( )
L
Qmax T1 = 182
hr
Qmax T2 = 3.12 × 10
3
−6m
s
( )
L
Qmax T2 = 11.2
hr
200
Q (L/hr)
150
100
50
− 20
0
20
40
60
T (C)
80
100
120
Problem 8.49
[Difficulty: 2]
Problem 8.48
[Difficulty: 2]
Problem 8.47
[Difficulty: 4]
Given:
Equation for fluid motion in the x-direction.
Find:
Expression for peak pressure
Solution:
Begin with the steady-state Navier-Stokes equation – x-direction
Governing equation:
The Navier-Stokes equations are
4
3
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
1
4
5
(5.1c)
3
4
3
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
⎛ ∂u
∂u
∂u
∂p
∂u ⎞
⎟
⎜
+
u
+
=
−
v
+
+
w
g
ρ⎜
ρ x
µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂x
∂z ⎟⎠
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
4
5
3
4
6
5
3
⎛ ∂ 2v ∂ 2v ∂ 2v ⎞
⎛ ∂v
∂v
∂v ⎞
∂v
∂p
ρ ⎜⎜ + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂y
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
3
3
3
3
3
3
(5.27a)
3
(5.27b)
3
⎛∂ w ∂ w ∂ w⎞
⎛ ∂w
∂w
∂w
∂w ⎞
∂p
+u
+v
+w
⎟⎟ = ρg z −
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂y
∂z ⎠
∂z
⎝ ∂t
⎝ ∂x
ρ ⎜⎜
2
2
2
(5.27c)
The following assumptions have been applied:
(1) Steady flow (given).
(2) Incompressible flow; ρ = constant.
(3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0.
(4) Fully developed flow, so no properties except possibly pressure p vary in the x direction; ∂/∂x = 0.
(5) See analysis below.
(6) No body force in the y direction; gy = 0
Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption
(3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the
Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4)
also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant (except of course in a more realistic model v ≠ 0 near the
transition. Since v is zero at the solid surface, then v must be zero everywhere. The fact that v = 0 reduces the Navier–Stokes
equations further, as indicated by (5). Hence for the y direction
∂p
=0
∂y
which indicates the pressure is a constant across the flow. Hence we conclude that p is a function at most of x.
In the x direction, we obtain
0=−
∂p
∂ 2u
+µ 2
∂x
∂y
(1)
Integrating this twice for the first region
u1 =
where
1 dp 2 c1
y + y + c2
2 µ dx 1
µ
dp
denotes the pressure gradient in region 1. Note that we change to regular derivative as p is a function of x only. Note that
dx 1
⎛ ∂p ⎞
⎟ and a function of y only
⎝ ∂x ⎠
Eq 1 implies that we have a function of x only ⎜
⎛ ∂ 2u ⎞
⎜⎜ 2 ⎟⎟ that must add up to be a constant (0); hence
⎝ ∂y ⎠
EACH is a constant! This means that
p
dp
= const = s
L1
dx 1
using the notation of the figure.
To evaluate the constants, c1 and c2, we must apply the boundary conditions. We do this separately for each region.
In the first region, at y = 0, u = U. Consequently, c2 = U. At y = h1, u = 0. Hence
0=
1 dp 2 c1
h1 + h1 + U
2 µ dx 1
µ
so
c1 = −
µU
1 dp
h1 −
2 dx 1
h1
Hence, combining results
u1 =
⎛ y ⎞
1 dp
y 2 − h1 y + U ⎜⎜ − 1⎟⎟
2 µ dx 1
⎝ h1 ⎠
(
)
Exactly the same reasoning applies to the second region, so
u2 =
where
1 dp
2 µ dx
(y
2
2
⎞
⎛ y
− h2 y + U ⎜⎜ − 1⎟⎟
⎝ h2 ⎠
)
p
dp
= const = − s
dx 2
L2
What connects these flow is the flow rate Q.
h1
h2
0
0
q = ∫ u1dy = ∫ u 2 dy = −
1 dp 3 Uh1
1 dp 3 Uh2
h1 −
h2 −
=−
12 µ dx 1
2
12µ dx 2
2
Hence
1 p s 3 Uh1
1 p s 3 Uh2
h1 +
h2 +
=−
12µ L1
2
12µ L2
2
Solving for ps,
p s ⎛ h13 h23 ⎞ Uh2 Uh1
⎜ + ⎟=
−
12 µ ⎜⎝ L1 L2 ⎟⎠
2
2
or
ps =
6 µU (h2 − h1 )
⎛ h13 h23 ⎞
⎜⎜ + ⎟⎟
⎝ L1 L2 ⎠
Problem 8.46
Given:
Paint flow (Bingham fluid)
Find:
Maximum thickness of paint film before flow occur
[Difficulty: 3]
Solution:
Basic equations:
du
τyx = τy + μp ⋅
dy
Bingham fluid:
Use the analysis of Example 8.3, where we obtain a force balance between gravity and shear stresses:
dτyx
dy
The given data is
τy = 40⋅ Pa
ρ = 1000⋅
= −ρ⋅ g
kg
3
m
From the force balance equation, itegrating
Hence
Motion occurs when
τyx = −ρ⋅ g ⋅ ( δ − y )
τmax ≥ τy
Hence the maximum thickness is
or
τyx = −ρ⋅ g ⋅ y + c
and we have boundary condition
τyx( y = δ) = 0
τmax = ρ⋅ g ⋅ δ
and this is a maximum at the wall
ρ⋅ g ⋅ δ ≥ τy
δ =
τy
ρ⋅ g
δ = 4.08 × 10
−3
m
δ = 4.08 mm
Problem 8.45
[Difficulty: 3]
Problem 8.44
Given:
Navier-Stokes Equations
Find:
Derivation of Example 8.3 result
[Difficulty: 2]
Solution:
The Navier-Stokes equations are (using the coordinates of Example 8.3, so that x is vertical, y is horizontal)
4
3
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
1
4
5
(5.1c)
3
4
3
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
⎛ ∂u
∂u
∂u
∂u ⎞
∂p
ρ ⎜⎜ + u + v + w ⎟⎟ = ρg x − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂x
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
4
5
3
6
4
5
3
⎛∂ v ∂ v ∂ v⎞
⎛ ∂v
∂v
∂v
∂v ⎞
∂p
+ u + v + w ⎟⎟ = ρg y −
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂y
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
1
3
3
3
3
2
3
3
2
(5.27a)
3
(5.27b)
3
⎛∂ w ∂ w ∂ w⎞
⎛ ∂w
∂w
∂w
∂w ⎞
∂p
+u
+v
+w
⎟⎟ = ρg z −
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂z
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
2
2
(5.27c)
The following assumptions have been applied:
(1) Steady flow (given).
(2) Incompressible flow; ρ = constant.
(3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0.
(4) Fully developed flow, so no properties except possibly pressure p vary in the x direction; ∂/∂x = 0.
(5) See analysis below.
(6) No body force in the y direction; gy = 0
Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption
(3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the
Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4)
also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero
everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction
∂p
=0
∂y
which indicates the pressure is a constant across the layer. However, at the free surface p = patm = constant. Hence we conclude that p
= constant throughout the fluid, and so
∂p
=0
∂x
In the x direction, we obtain
µ
∂ 2u
+ ρg = 0
∂y 2
Integrating twice
u=−
c
1
ρgy 2 + 1 y + c2
2µ
µ
To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, du/dy =
0 (we assume air friction is negligible). Hence
τ (y = δ ) = µ
which gives
du
dy
=−
y =δ
1
µ
ρgδ +
c1
µ
=0
c1 = ρgδ
and finally
2
1
ρg
ρg 2 ⎡ ⎛ y ⎞ 1 ⎛ y ⎞ ⎤
2
ρgy +
δ ⎢⎜ ⎟ − ⎜ ⎟ ⎥
u=−
y=
2µ
µ
µ ⎢⎣⎝ δ ⎠ 2 ⎝ δ ⎠ ⎥⎦
Problem 8.43
[Difficulty: 3]
Given:
Data on a journal bearing
Find:
Time for the bearing to slow to 100 rpm; visocity of new fluid
Solution:
The given data is
D = 35⋅ mm
L = 50⋅ mm
δ = 1 ⋅ mm
ωi = 500 ⋅ rpm
ωf = 100 ⋅ rpm
μ = 0.1⋅
2
I = 0.125 ⋅ kg⋅ m
N⋅ s
2
m
I⋅ α = Torque = −τ⋅ A⋅
The equation of motion for the slowing bearing is
D
2
where α is the angular acceleration and τ is the viscous stress, and A = π⋅ D⋅ L is the surface area of the bearing
τ = μ⋅
As in Example 8.2 the stress is given by
U
δ
=
μ⋅ D⋅ ω
2⋅ δ
where U and ω are the instantaneous linear and angular velocities.
I⋅ α = I⋅
Hence
dω
Separating variables
ω
dω
=−
dt
μ⋅ D⋅ ω
2⋅ δ
⋅ π⋅ D⋅ L⋅
D
2
3
=−
μ⋅ π⋅ D ⋅ L
4⋅ δ
⋅ω
3
=−
μ⋅ π⋅ D ⋅ L
4 ⋅ δ⋅ I
⋅ dt
3
−
Integrating and using IC ω = ω0
μ⋅ π⋅ D ⋅ L
ω( t) = ωi⋅ e
4⋅ δ⋅ I
⋅t
3
−
The time to slow down to ω f
= 10 rpm is obtained from solving
so
4 ⋅ δ⋅ I
t = −
3
μ⋅ π⋅ D ⋅ L
For the new fluid, the time to slow down is
t = 10⋅ min
Rearranging the equation
μ = −
4 ⋅ δ⋅ I
3
π ⋅ D ⋅ L⋅ t
μ⋅ π⋅ D ⋅ L
4⋅ δ⋅ I
ωf = ωi⋅ e
⎛ ωf ⎞
⋅ ln⎜
⎝ ωi ⎠
Hence
3
t = 1.19 × 10 s
⎛ ωf ⎞
⋅ ln⎜
⎝ ωi ⎠
μ = 0.199
kg
m⋅ s
⋅t
t = 19.9⋅ min
It is more viscous as it slows
down the rotation in a
shorter time
Problem 8.42
[Difficulty: 5] Part 1/2
Problem 8.42
[Difficulty: 5] Part 2/2
Problem 8.41
Problem 2.66
[Difficulty: 5]
Problem 8.40
[Difficulty: 2]
Given: Expression for efficiency
Find: Plot; find flow rate for maximum efficiency; explain curve
Solution:
η
0.0%
7.30%
14.1%
20.3%
25.7%
30.0%
32.7%
33.2%
30.0%
20.8%
0.0%
Efficiency of a Viscous Pump
35%
30%
25%
η
q
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
20%
15%
10%
5%
0%
0.00
0.10
0.20
0.30
0.40
q
For the maximum efficiency point we can use Solver (or alternatively differentiate)
q
0.333
η
33.3%
The efficiency is zero at zero flow rate because there is no output at all
The efficiency is zero at maximum flow rate ∆p = 0 so there is no output
The efficiency must therefore peak somewhere between these extremes
0.50
Problem 8.39
[Difficulty: 5]
Problem 8.38
[Difficulty: 5]
Problem 8.37
[Difficulty: 3]
Problem 8.36
Using the result for average velocity from Example 8.3
[Difficulty: 3]
Problem 8.35
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Shear stress on lower plate; pressure gradient for zero shear stress at y/a = 0.25; plot velocity and shear stress
Solution:
u(y) =
Basic equations
U⋅ y
a
q = 1.5⋅
Available data
+
⎡ y ⎞2
−
2 ⋅ μ dx ⎣⎝ a ⎠
a
2
⋅
dp
gpm
⋅ ⎢⎛⎜
y⎤
⎥ (1)
a⎦
a = 0.05⋅ in
ft
From Fig. A.2, Carbon tetrachloride at 20oC
μ = 0.001 ⋅
Q
=
l
U⋅ a
2
a
−
3
⋅
dp
12⋅ μ dx
τ = μ⋅
(2)
U
a
+ a⋅
U=
From Eq. 3, when y = 0, with
U = 1.60
ft
N⋅ s
s
y
⎝a
−
1⎞
2⎠
(3)
− 5 lbf ⋅ s
μ = 2.089 × 10
2
2⋅ Q
or
a⋅ l
τyx =
dx
⋅ ⎛⎜
68°F = 20°C
⋅
m
From Eq. 2, for zero pressure gradient
dp
ft
U =
2⋅ q
U = 1.60⋅
a
μ⋅ U
2
ft
s
−5
τyx = 5.58 × 10
a
⋅ psi
A mild adverse pressure gradient would reduce the flow rate.
For zero shear stress at y/a = 0.25, from Eq. 3
0 = μ⋅
U
a
+ a⋅
dp
dx
⋅ ⎛⎜
1
⎝4
−
1⎞
2⎠
dp
or
dx
dp
2
dx
a
= 0.0536⋅
0.75
y/a
0.75
y/a
4 ⋅ μ⋅ U
1
1
0.5
0.5
0.25
0.25
− 0.5
=
0
0.5
1
1.5
2
u (ft/s)
Note that the location of zero shear is also where u is maximum!
−4
− 1× 10
0
−4
1× 10
Shear Stress (psi)
−4
2× 10
psi
ft
Problem 8.34
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Pressure gradient for no flow; plot velocity and stress distributions; also plot for u = U at y = a/2
Solution:
Basic equations
Available data
From Eq 2 for Q = 0
u(y) =
U⋅ y
a
U = 1.5⋅
dp
dx
=
+
m
2
2
⋅
dp
⋅ ⎢⎛⎜
y⎤
Q
⎥ (1)
a⎦
a = 5 ⋅ mm
s
l
=
U⋅ a
2
−
a
3
⋅
dp
12⋅ μ dx
τ = μ⋅
(2)
U
a
μ = 1⋅
From Fig. A.2 for castor oil at 20oC
+ a⋅
dp
dx
N⋅ s
2
m
6 ⋅ μ⋅ U
a
⎡ y ⎞2
−
2 ⋅ μ dx ⎣⎝ a ⎠
a
= 6 × 1⋅
N⋅ s
2
× 1.5⋅
m
m
s
×
1
dp
( 0.005 ⋅ m)
dx
2
The graphs below, using Eqs. 1 and 3, can be plotted in Excel
1
y/a
0.75
0.5
0.25
− 0.5
0
0.5
1
1.5
u (m/s)
1
y/a
0.75
0.5
0.25
−1
− 0.5
0
0.5
1
Shear Stress (kPa)
The pressure gradient is adverse, to counteract the flow generated by the upper plate motion
1.5
= 360 ⋅
kPa
m
⋅ ⎛⎜
y
⎝a
−
1⎞
2⎠
(3)
u(y) =
For u = U at y = a/2 we need to adjust the pressure gradient. From Eq. 1
⎡⎢⎛ a ⎞ 2 a ⎤⎥
2
2
a dp ⎢⎜ 2
2⎥
U=
+
⋅ ⋅ ⎢⎜
−
a
2 ⋅ μ dx ⎣⎝ a ⎠
a ⎥⎦
U⋅
Hence
U⋅ y
a
⎡ y ⎞2
−
2 ⋅ μ dx ⎣⎝ a ⎠
a
+
2
⋅
dp
y⎤
⋅ ⎢⎛⎜
⎥
a⎦
a
dp
or
dx
dp
dx
=−
4 ⋅ U⋅ μ
a
2
= −240 ⋅
= −4 × 1 ⋅
N⋅ s
2
× 1.5⋅
m
m
s
kPa
m
1
y/a
0.75
0.5
0.25
0
0.5
1
1.5
2
u (m/s)
1
y/a
0.75
0.5
0.25
−1
− 0.5
0
0.5
1
Shear Stress (kPa)
The pressure gradient is positive to provide the "bulge" needed to satisfy the velocity requirement
1.5
×
1
( 0.005 ⋅ m)
2
Problem 8.33
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Location and magnitude of maximum velocity; Volume flow in 10 s; Plot velocity and shear stress
Solution:
From Section 8.2
For u max set du/dx = 0
Hence
From Fig. A.2 at
u(y) =
du
U⋅ y
b
=0=
dy
U
u = u max
2
dp
⋅
b
+
b
y⎤
⎡ y ⎞2
−
2 ⋅ μ dx ⎣⎝ b ⎠
b
+
2
⋅
⋅ ⎢⎛⎜
dp
2 ⋅ μ dx
⋅⎛
2⋅ y
b
59 °F = 15⋅ °C
2
μ = 4⋅
1⎞
−
⎜ 2
⎝b
y=
at
⎥
b⎦
a
=
⎠
U
b
1 dp
⋅ ⋅ (2⋅ y − b)
2 ⋅ μ dx
+
μ⋅ U
−
b⋅
dp
dx
N⋅ s
μ = 0.0835⋅
2
m
y =
Hence
0.1⋅ in
2
+ 0.0835⋅
lbf ⋅ s
ft
2
× 2⋅
ft
ft
s
⎡⎛ y ⎞ 2
u max =
+
⋅ ⋅ ⎢⎜
−
2 ⋅ μ dx ⎣⎝ b ⎠
b
U⋅ y
u max = 2 ⋅
b
ft
s
2
dp
lbf ⋅ s
2
1
×
2
0.1⋅ in
y⎤
⎥
×
in ⋅ ft
y = 0.0834⋅ in
50⋅ lbf
y = 0.0834⋅ in
with
b⎦
2
2
2
ft
50⋅ psi ⎛ 12⋅ in ⎞
⎛ .0834 ⎞ + 1 × ⎛ 0.1 ⋅ ft⎞ ×
×
−
×
×
⎜
⎜
⎜
.0835 ⋅ lbf ⋅ s
ft
⎝ 0.1 ⎠ 2 ⎝ 12 ⎠
⎝ 1 ⋅ ft ⎠
×
⎡⎛ .0834 ⎞ 2 ⎛ .0834 ⎞⎤
⎢⎜
⎥
−⎜
⎣⎝ 0.1 ⎠ ⎝ 0.1 ⎠⎦
u max = 2.083 ⋅
b
⌠
=⎮
w ⌡0
Q
Q
w
=
1
2
⌠
⎮
u ( y ) dy = w⋅ ⎮
⎮
⌡
b
⎡ U⋅ y b 2 dp ⎡⎛ y ⎞ 2
⎢
+
⋅ ⋅ ⎢⎜
−
2 ⋅ μ dx ⎣⎝ b ⎠
⎣ b
× 2⋅
ft
s
×
0.1
12
⋅ ft −
1
12
3
×
= 0.0125
s
Q
ft
w
2
ft
50⋅ psi ⎞
⎛ 0.1 ⋅ ft⎞ ×
× ⎛⎜ −
×
⎜
.0835 ⋅ lbf ⋅ s ⎝
ft ⎠
⎝ 12 ⎠
ft
w
U⋅ b
3
b
dp
⎥⎥ dy =
−
⋅
2
12⋅ μ dx
b⎦⎦
0
3
Q
y⎤⎤
= 5.61⋅
gpm
ft
⎛ 12⋅ in ⎞
⎜
⎝ 1⋅ ft ⎠
2
ft
s
Flow =
Q
w
⋅ ∆t = 5.61⋅
The velocity profile is
gpm
ft
1 ⋅ min
×
60⋅ s
u
U
=
y
b
× 10⋅ s
+
Flow = 0.935 ⋅
⎡ y ⎞2
−
2 ⋅ μ⋅ U dx ⎣⎝ b ⎠
b
2
⋅
dp
⋅ ⎢⎛⎜
y⎤
⎥
b⎦
gal
ft
τ = μ⋅
For the shear stress
du
dy
= μ⋅
U
b
+
b dp ⎡ ⎛ y ⎞
⋅ ⋅ ⎢2 ⋅ ⎜
− 1⎥⎤
2 dx ⎣ ⎝ b ⎠
⎦
The graphs below can be plotted in Excel
1
0.8
y/b
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
1.2
u/U
1
0.8
y/b
0.6
0.4
0.2
− 500
0
500
3
1× 10
Shear Stress (Pa)
3
1.5× 10
3
2× 10
2.5× 10
3
Problem 8.32
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Shear stress on lower plate; Plot shear stress; Flow rate for pressure gradient; Pressure gradient for zero shear; Plot
Solution:
u(y) =
From Section 8-2
U⋅ y
a
+
⎡ y ⎞2
−
2 ⋅ μ dx ⎣⎝ a ⎠
a
2
⋅
dp
⋅ ⎢⎛⎜
y⎤
⎥
a⎦
3
ft
a
u = U⋅
For dp/dx = 0
a
⌠
⌠
U⋅ a
y
= ⎮ u ( y ) dy = w⋅ ⎮ U⋅ dy =
⌡
⎮
2
a
l
0
⌡
y
Q
a
1
Q =
2
× 5⋅
ft
×
s
0.1
12
⋅ ft
Q = 0.0208⋅
s
ft
0
τ = μ⋅
For the shear stress
du
dy
=
μ⋅ U
− 7 lbf ⋅ s
μ = 3.79 × 10
when dp/dx =
0
a
⋅
ft
(Table A.9)
2
The shear stress is constant - no need to plot!
τ = 3.79 × 10
− 7 lbf ⋅ s
⋅
ft
2
× 5⋅
ft
s
×
12
0.1⋅ ft
×
⎛ 1⋅ ft ⎞
⎜
⎝ 12⋅ in ⎠
2
−6
τ = 1.58 × 10
Q will decrease if dp/dx > 0; it will increase if dp/dx < 0.
τ = μ⋅
For non- zero dp/dx:
du
dy
=
μ⋅ U
a
+ a⋅
τ( y = 0.25⋅ a) = μ⋅
At y = 0.25a, we get
U
a
dp
dx
⋅ ⎛⎜
+ a⋅
y
−
⎝a
dp
dx
⋅ ⎛⎜
1⎞
2⎠
1
⎝4
−
1⎞
2⎠
= μ⋅
U
a
−
a dp
⋅
4 dx
lbf
Hence this stress is zero when
dp
dx
=
4 ⋅ μ⋅ U
a
2
− 7 lbf ⋅ s
= 4 × 3.79 × 10
⋅
ft
2
× 5⋅
ft
s
2
×
2
⎛ 12 ⎞ = 0.109 ⋅ ft = 7.58 × 10− 4 psi
⎜
ft
ft
⎝ 0.1⋅ ft ⎠
0.1
y (in)
0.075
0.05
0.025
−4
− 1× 10
0
−4
1× 10
Shear Stress (lbf/ft3)
−4
2× 10
−4
3× 10
⋅ psi
Problem 8.31
[Difficulty: 2]
Given:
Velocity distribution on incline
Find:
Expression for shear stress; Maximum shear; volume flow rate/mm width; Reynolds number
Solution:
u(y) =
From Example 5.9
ρ⋅ g ⋅ sin( θ)
μ
du
⎛
y
⎝
2
⋅ ⎜ h⋅ y −
2⎞
⎠
For the shear stress
τ = μ⋅
τ is a maximum at y = 0
τmax = ρ⋅ g ⋅ sin( θ) ⋅ h = SG ⋅ ρH2O⋅ g ⋅ sin( θ) ⋅ h
dy
= ρ⋅ g ⋅ sin( θ) ⋅ ( h − y )
τmax = 1.2 × 1000
kg
3
× 9.81⋅
m
m
2
2
× sin( 15⋅ deg) × 0.007 ⋅ m ×
s
N⋅ s
kg⋅ m
τmax = 21.3 Pa
dy
Q=
This stress is in the x direction on the wall
⌠
h
⎮
⌠
⌠
Q = ⎮ u dA = w⋅ ⎮ u ( y ) dy = w⋅ ⎮
⌡
⎮
⌡
0
⌡
The flow rate is
h
ρ⋅ g ⋅ sin( θ)
μ
0
2⎞
⎛
y
⎝
2
⋅ ⎜ h⋅ y −
⎠
ρ⋅ g ⋅ sin( θ) ⋅ w⋅ h
3
Q
w
=
1
3
× 1.2 × 1000
kg
3
× 9.81⋅
m
m
2
2
m
3
× sin( 15⋅ deg) × ( 0.007 ⋅ m) ×
s
The average velocity is
V=
The gap Reynolds number is
Re =
Q
A
=
Q
V = 217 ⋅
w⋅ h
N⋅ s
−4
= 2.18 × 10
mm
s
Q
m
w
= 217
mm
3
s
mm
×
1
V = 31.0⋅
7 ⋅ mm
μ
kg
3
m
× 31⋅
mm
s
2
× 7 ⋅ mm ×
m
1.60⋅ N⋅ s
×
⎛ 1⋅ m ⎞
⎜
⎝ 1000⋅ mm ⎠
3
s
mm
ρ⋅ V⋅ h
Re = 1.2 × 1000
The flow is definitely laminar
2
⋅
1.60⋅ N⋅ s kg⋅ m
mm
m
3⋅ μ
2
Re = 0.163
s
3
Problem 8.30
[Difficulty: 3]
Given:
Data on flow of liquids down an incline
Find:
Velocity at interface; velocity at free surface; plot
Solution:
2
Given data
h = 10⋅ mm
θ = 60⋅ deg
ν1
ν2 =
5
m
ν1 = 0.01⋅
s
ν2 = 2 × 10
2
−3m
s
(The lower fluid is designated fluid 1, the upper fluid 2)
From Example 5.9 (or Exanple 8.3 with g replaced with gsinθ), a free body analysis leads to (for either fluid)
d
2
dy
2
u =−
ρ⋅ g ⋅ sin( θ)
μ
Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields
u1 = −
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ1
2
⋅ y + c1 ⋅ y + c2
We need four BCs. Two are
obvious
u2 = −
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ2
2
⋅ y + c3 ⋅ y + c4
y=0
u1 = 0
(1)
y=h
u1 = u2
(2)
The third BC comes from the fact that there is no shear stress at the free surface
y = 2⋅ h
du2
μ2 ⋅
=0
dy
(3)
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
du1
du2
μ1 ⋅
= μ2 ⋅
dy
dy
y=h
Using these four BCs
c2 = 0
−
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ1
2
⋅ h + c1 ⋅ h + c2 = −
−ρ⋅ g ⋅ sin( θ) ⋅ 2 ⋅ h + μ2 ⋅ c3 = 0
Hence, after some algebra
c1 =
2 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h
μ1
c2 = 0
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ2
(4)
2
⋅ h + c3 ⋅ h + c4
−ρ⋅ g ⋅ sin( θ) ⋅ h + μ1 ⋅ c1 = −ρ⋅ g ⋅ sin( θ) ⋅ h + μ2 ⋅ c3
c3 =
2 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h
μ2
(
2 μ2 − μ1
c4 = 3 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h ⋅
2 ⋅ μ1 ⋅ μ2
)
The velocity distributions are then
u1 =
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ1
(
⋅ 4⋅ y⋅ h − y
)
2
u2 =
⎡
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ2
(
2 μ2 − μ1
⋅ ⎢3 ⋅ h ⋅
μ1
⎣
)
+ 4⋅ y⋅ h − y
2⎤
⎥
⎦
Rewriting in terms of ν1 and ν2 (ρ is constant and equal for both fluids)
u1 =
g ⋅ sin( θ)
2 ⋅ ν1
(
⋅ 4⋅ y⋅ h − y
)
2
u2 =
g ⋅ sin( θ)
2 ⋅ ν2
⎡
(
2 ν2 − ν1
⋅ ⎢3 ⋅ h ⋅
⎣
ν1
)
+ 4⋅ y⋅ h − y
2⎤
⎥
⎦
(Note that these result in the same expression if ν1 = ν2, i.e., if we have one fluid)
2
u interface =
Evaluating either velocity at y = h, gives the velocity at the interface
Evaluating u 2 at y = 2h gives the velocity at the free surface
2
3 ⋅ g ⋅ h ⋅ sin( θ)
u freesurface = g ⋅ h ⋅ sin( θ) ⋅
u freesurface⋅ h
Note that a Reynolds number based on the free surface velocity is
ν2
2 ⋅ ν1
(3⋅ ν2 + ν1)
= 1.70
2 ⋅ ν1 ⋅ ν2
u interface = 0.127
0.000
0.0166
0.0323
0.0472
0.061
0.074
0.087
0.098
0.109
0.119
0.127
u 2 (m/s)
indicating laminar flow
Velocity Distributions down an Incline
24
Lower Velocity
20
0.127
0.168
0.204
0.236
0.263
0.287
0.306
0.321
0.331
0.338
0.340
y (mm)
0.000
1.000
2.000
3.000
4.000
5.000
6.000
7.000
8.000
9.000
10.000
11.000
12.000
13.000
14.000
15.000
16.000
17.000
18.000
19.000
20.000
Upper Velocity
16
12
8
4
0
0.0
0.1
0.2
u (m/s)
0.3
s
u freesurface = 0.340
The velocity distributions can be plotted in Excel.
y (mm) u 1 (m/s)
m
0.4
m
s
Problem 8.29
[Difficulty: 2]
Problem 8.28
[Difficulty: 2]
Problem 8.27
[Difficulty: 2]
Given:
Velocity profile between parallel plates
Find:
Pressure gradients for zero stress at upper/lower plates; plot
Solution:
U⋅ y
⎛ ∂ ⎞ ⎡⎢⎛ y ⎞ 2
−
p ⋅ ⎜
2 ⋅ μ ⎝ ∂x ⎠ ⎣⎝ a ⎠
2
⋅⎜
y⎤
⎥
u=
The shear stress is
y
1
⎛∂ ⎞
τyx = μ⋅
= μ⋅ +
⋅ ⎜ p ⋅ ⎛ 2⋅
− ⎞
⎜
2 ⎝ ∂x ⎠
2
dy
a
a
a
a
+
a
From Eq. 8.8, the velocity distribution is
du
U
a
a⎦
2
⎝
(a) For τyx = 0 at y = a
The velocity distribution is then
(b) For τyx = 0 at y = 0
The velocity distribution is then
0 = μ⋅
u=
The velocity distributions can be plotted in Excel.
a
U⋅ y
a
0 = μ⋅
u=
U
U
a
U⋅ y
a
+
−
−
+
a ∂
⋅ p
2 ∂x
a
2
2⋅ μ
⋅
∂
∂x
2 ⋅ U⋅ μ
a
2
⎡ y ⎞2
−
⎣⎝ a ⎠
⋅ ⎢⎛⎜
y⎤
⎥
a⎦
a ∂
⋅ p
2 ∂x
a
2
2⋅ μ
⋅
⎠
u
U
∂
∂x
2 ⋅ U⋅ μ
a
2
⎡ y ⎞2
−
⎣⎝ a ⎠
⋅ ⎢⎛⎜
y⎤
⎥
a⎦
u
U
p =−
2 ⋅ U⋅ μ
a
= 2⋅
p =
y
−
a
⎛y⎞
⎜
⎝a⎠
2 ⋅ U⋅ μ
a
=
2
⎛y⎞
⎜
⎝a⎠
2
2
2
y /a
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
(a) u /U
0.000
0.190
0.360
0.510
0.640
0.750
0.840
0.910
0.960
0.990
1.00
(b) u /U
0.000
0.010
0.040
0.090
0.160
0.250
0.360
0.490
0.640
0.810
1.000
Zero-Stress Velocity Distributions
1.00
Zero Stress Upper Plate
Zero Stress Lower Plate
y /a
0.75
0.50
0.25
0.00
0.00
0.25
0.50
u /U
0.75
1.00
Problem 8.26
[Difficulty: 2]
Given:
Computer disk drive
Find:
Flow Reynolds number; Shear stress; Power required
Solution:
For a distance R from the center of a disk spinning at speed ω
V = R⋅ ω
The gap Reynolds number is
Re =
V = 25⋅ mm ×
ρ⋅ V⋅ a
Re = 22.3⋅
m
s
1000⋅ mm
× 8500⋅ rpm ×
2⋅ π⋅ rad
rev
1⋅ min
×
ν = 1.45 × 10
ν
⋅m ×
−5
1.45 × 10
m
s
from Table A.10 at 15oC
s
s
−6
× 0.25 × 10
⋅
V = 22.3⋅
60⋅ s
2
−5 m
V⋅ a
=
μ
1⋅ m
Re = 0.384
2
⋅m
The flow is definitely laminar
The shear stress is then
τ = μ⋅
du
dy
= μ⋅
τ = 1.79 × 10
− 5 N⋅ s
V
μ = 1.79 × 10
a
− 5 N⋅ s
⋅
2
m
The power required is
P = T⋅ ω
T = τ⋅ A⋅ R
P = τ⋅ A⋅ R⋅ ω
P = 1600⋅
N
2
m
× 22.3⋅
m
s
2
from Table A.10 at 15oC
m
1
×
⋅
0.25 × 10
τ = 1.60⋅ kPa
−6
⋅m
where torque T is given by
A = ( 5 ⋅ mm)
with
× 2.5 × 10
−5
2
⋅ m × 25⋅ mm ×
2
A = 2.5 × 10
1⋅ m
1000⋅ mm
−5
× 8500⋅ rpm ×
2
m
2 ⋅ π⋅ rad
rev
×
1 ⋅ min
60⋅ s
P = 0.890 W
Problem 8.25
Given:
Laminar flow of two fluids between plates
Find:
Velocity at the interface
[Difficulty: 3]
Solution:
Using the analysis of Section 8.2, the sum of forces in the x direction is
⎡ ∂ dy ⎛ ∂ dy ⎞⎤
⎛ ∂ dx
dx ⎞
∂
⋅ b ⋅ dy = 0
⎢τ + τ ⋅ − ⎜ τ − τ ⋅ ⎥ ⋅ b ⋅ dx + ⎜ p − p ⋅ − p + p ⋅
∂x 2 ⎠
⎣ ∂y 2 ⎝ ∂y 2 ⎠⎦
⎝ ∂x 2
Simplifying
dτ
dy
=
dp
dx
2
=0
μ⋅
or
dy
y=0
u1 = 0
2
=0
u 1 = c1 ⋅ y + c2
Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields
We need four BCs. Three are obvious
d u
y = h u1 = u2
y = 2⋅ h
u 2 = c3 ⋅ y + c4
u2 = U
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
y=h
du1
du2
μ1⋅
= μ2⋅
dy
dy
Using these four BCs
0 = c2
c1⋅ h + c2 = c3⋅ h + c4
Hence
c2 = 0
From the 2nd and 3rd equations
c1⋅ h − U = −c3⋅ h
Hence
μ1
c1⋅ h − U = −c3⋅ h = − ⋅ h ⋅ c1
μ2
and
Hence for fluid 1 (we do not need to complete the analysis for fluid 2)
20⋅
Evaluating this at y = h, where u 1 = u interface
u interface =
ft
s
⎛1 + 1 ⎞
⎜
3⎠
⎝
U = c3⋅ 2⋅ h + c4
μ1 ⋅ c1 = μ2 ⋅ c3
c1 =
U
⎛
μ1 ⎞
⎝
μ2
⎠
h⋅ ⎜ 1 +
u1 =
U
⎛
h ⋅⎜1 +
⎝
μ1 ⎞
μ2
⎠
u interface = 15⋅
ft
s
⋅y
μ1⋅ c1 = μ2⋅ c3
Problem 8.24
[Difficulty: 3]
Given:
Properties of two fluids flowing between parallel plates; applied pressure gradient
Find:
Velocity at the interface; maximum velocity; plot velocity distribution
Solution:
dp
Given data
=k
dx
k = −50⋅
kPa
m
h = 5 ⋅ mm
μ1 = 0.1⋅
N⋅ s
μ2 = 4 ⋅ μ1
2
μ2 = 0.4⋅
m
N⋅ s
2
m
(Lower fluid is fluid 1; upper is fluid 2)
Following the analysis of Section 8.2, analyse the forces on a differential CV of either fluid
The net force is zero for steady flow, so
⎡τ + dτ ⋅ dy − ⎛ τ − dτ ⋅ dy ⎞⎤ ⋅ dx⋅ dz + ⎡p − dp ⋅ dx − ⎛ p + dp ⋅ dx ⎞⎤ ⋅ dy⋅ dz = 0
⎢
⎜
⎥
⎢
⎜
⎥
dy 2 ⎠⎦
dx 2 ⎠⎦
⎣ dy 2 ⎝
⎣ dx 2 ⎝
dτ
Simplifying
dy
=
dp
dx
=k
μ⋅
so for each fluid
d
2
dy
2
u =k
Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields
u1 =
k
2
⋅ y + c1 ⋅ y + c2
2 ⋅ μ1
u2 =
k
2
2 ⋅ μ2
⋅ y + c3 ⋅ y + c4
For convenience the origin of coordinates is placed at the centerline
y = −h
We need four BCs. Three are obvious
u1 = 0
y=0
(1)
u 1 = u 2 (2)
y=h
u2 = 0
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
du1
du2
μ1 ⋅
= μ2 ⋅
dy
dy
y=0
Using these four BCs
0=
k
2
⋅ h − c1 ⋅ h + c2
2 ⋅ μ1
c2 = c4
0=
k
2 ⋅ μ2
(4)
2
⋅ h + c3 ⋅ h + c4
μ1 ⋅ c1 = μ2 ⋅ c3
Hence, after some algebra
c1 =
k⋅ h
2 ⋅ μ1
⋅
(μ2 − μ1)
(μ2 + μ1)
c4 = −
k⋅ h
2
μ2 + μ1
c2 = c4
c3 =
k⋅ h
2 ⋅ μ2
⋅
(μ2 − μ1)
(μ2 + μ1)
(3)
c1 = −750
1
c2 = 2.5
s
m
c3 = −187.5
s
1
c4 = 2.5
s
m
s
The velocity distributions are then
u1( y) =
k
2 ⋅ μ1
⎡
2
⋅ ⎢y + y ⋅ h ⋅
⎣
(μ2 − μ1)⎤
⎥−
(μ2 + μ1)⎦
k⋅ h
2
u2( y) =
μ2 + μ1
k
2 ⋅ μ2
⎡
2
⋅ ⎢y + y ⋅ h ⋅
⎣
(μ2 − μ1)⎤
⎥−
(μ2 + μ1)⎦
k⋅ h
μ2 + μ1
Evaluating either velocity at y = 0, gives the velocity at the interface
u interface = −
k⋅ h
2
u interface = 2.5
μ2 + μ1
m
s
The plots of these velocity distributions can be done in Excel. Typical curves are shown below
5
y (mm)
2.5
0
0.5
1
1.5
2
2.5
3
3.5
− 2.5
−5
u (m/s)
Clearly, u 1 has the maximum velocity, when
(
(
h μ2 − μ1
y max = − ⋅
2 μ2 + μ1
)
)
du1
dy
=0
y max = −1.5 mm
(We could also have used Excel's Solver for this.)
2 ⋅ y max + h ⋅
or
(
)
u max = u 1 y max
u max = 3.06
(μ2 − μ1)
(μ2 + μ1)
m
s
=0
2
Problem 8.23
[Difficulty: 2]
Problem 8.22
[Difficulty: 2]
Given:
Laminar flow between moving plates
Find:
Expression for velocity; Volume flow rate per depth
Solution:
Given data
ft
U1 = 5⋅
s
d = 0.2⋅ in
ft
U2 = 2⋅
s
Using the analysis of Section 8.2, the sum of forces in the x direction is
⎡ ∂ dy ⎛
⎛
dy ⎞⎤
dx
dx ⎞
⎢τ + τ ⋅ − ⎜ τ − ∂ τ ⋅ ⎥ ⋅ b ⋅ dx + ⎜ p − ∂ p ⋅ − p + ∂ p ⋅ ⋅ b ⋅ dy = 0
2 ⎠⎦
∂y
∂x 2
∂x 2 ⎠
⎣ ∂y 2 ⎝
⎝
Simplifying
dτ
dy
=
dp
dx
2
=0
μ⋅
or
d u
dy
Integrating twice
u = c1 ⋅ y + c2
Boundary conditions:
u ( 0 ) = −U1
c2 = −U1
(
)
2
=0
ft
c2 = −5
s
u ( y = d ) = U2
u in ft/s, y in ft
Hence
y
u ( y ) = U1 + U2 ⋅ − U1
d
u ( y ) = 420 ⋅ y − 5
The volume flow rate is
⌠
⌠
Q = ⎮ u dA = b ⋅ ⎮ u dy
⌡
⌡
⌠
Q = b⋅ ⎮
⎮
⌡
d
Q = b⋅ d⋅
(U2 − U1)
Q
2
b
= d⋅
(U2 − U1)
2
3
ft
Q
b
= 0.2⋅ in ×
1 ⋅ ft
12⋅ in
×
1
2
× ( 2 − 5) ×
ft
Q
s
b
= −.025⋅
s
ft
3
ft
Q
b
= −.025⋅
s
ft
×
7.48⋅ gal
1 ⋅ ft
3
×
60⋅ s
Q
1 ⋅ min
b
U1 + U2
d
= −11.2⋅
gpm
ft
−1
c1 = 420 s
U + U2 d 2
⎞
⎡ U + U ⋅ y − U ⎤ dx = b⋅ ⎛⎜ 1
⋅
−
U
⋅
d
⎢( 1
)
⎥
2 d
1
1
2
⎣
⎦
⎝ d
⎠
0
Hence
c1 =
Problem 8.21
[Difficulty: 3]
Given:
Laminar velocity profile of power-law fluid flow between parallel plates
Find:
Expression for flow rate; from data determine the type of fluid
Solution:
⎡⎢
n
h ∆p ⎞
n⋅ h ⎢
u = ⎛⎜ ⋅
⋅
⋅ 1−
⎝ k L ⎠ n + 1 ⎢⎣
n+ 1⎤
1
The velocity profile is
⌠
Q = w⋅ ⎮
⌡
The flow rate is then
⎛y⎞
⎜
⎝h⎠
n
⎥
⎥
⎥⎦
h
h
or, because the flow is symmetric
u dy
−h
⌠
⎮
⎮
⎮ 1−
⎮
⌡
The integral is computed as
0
n+ 1
⎛y⎞
⎜
⎝h⎠
⌠
Q = 2 ⋅ w⋅ ⎮ u dy
⌡
n
2⋅ n+ 1⎤
⎡⎢
⎥
n ⎥
⎢
n
y
dy = y ⋅ ⎢1 −
⋅ ⎛⎜ ⎞
⎥⎦
⎣ 2⋅ n + 1 ⎝ h ⎠
1
2⋅ n+ 1⎤
⎡
n
⎢
⎥
h ∆p ⎞
n⋅ h
n
n
Q = 2 ⋅ w⋅ ⎛⎜ ⋅
⋅
⋅ h ⋅ ⎢1 −
⋅ ( 1)
⎥
⎝ k L ⎠ n + 1 ⎣ 2⋅ n + 1
⎦
Using this with the limits
An Excel spreadsheet can be used for computation of n.
The data is
dp (kPa)
Q (L/min)
10
0.451
20
0.759
30
1.01
40
1.15
50
1.41
60
1.57
70
1.66
80
1.85
90
2.05
1
n
This must be fitted to
Q=
1
2
⎛ h ⋅ ∆p ⎞ ⋅ 2 ⋅ n⋅ w⋅ h or
⎜
⎝ k L ⎠ 2⋅ n + 1
Q = k ⋅ ∆p
n
100
2.25
1
n
Q=
2
⎛ h ⋅ ∆p ⎞ ⋅ 2 ⋅ n⋅ w⋅ h
⎜
⎝ k L ⎠ 2⋅ n + 1
We can fit a power curve to the data
Flow Rate vs Applied Pressure for a
Non-Newtonian Fluid
10.0
Q (L/min)
Data
Power Curve Fit
1.0
Q = 0.0974dp0.677
2
R = 0.997
0.1
10
Hence
dp (kPa)
1/n =
It's a dilatant fluid
0.677
n =
1.48
100
Problem 8.20
[Difficulty: 2]
Problem 8.19
[Difficulty: 5]
Problem 8.18
[Difficulty: 4]
Problem 8.17
Given:
Navier-Stokes Equations
Find:
Derivation of Eq. 8.5
[Difficulty: 2]
Solution:
The Navier-Stokes equations are
4
3
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
1
4
5
3
(5.1c)
6
4
3
⎛∂ u ∂ u ∂ u⎞
⎛ ∂u
∂u
∂u
∂u ⎞
∂p
+u
+v
+ w ⎟⎟ = ρg x −
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂x
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
1
4
5
3
2
4
2
5
3
⎛ ∂ 2v ∂ 2v ∂ 2v ⎞
⎛ ∂v
∂v
∂v
∂v ⎞
∂p
ρ ⎜⎜ + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂y
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
3
3
3
3
3
3
(5.27a)
3
(5.27b)
3
⎛∂ w ∂ w ∂ w⎞
⎛ ∂w
∂w
∂w
∂w ⎞
∂p
+u
+v
+w
⎟⎟ = ρg z −
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂z
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
2
2
(5.27c)
The following assumptions have been applied:
(1) Steady flow (given).
(2) Incompressible flow; ρ = constant.
(3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0.
(4) Fully developed flow, so no properties except pressure p vary in the x direction; ∂/∂x = 0.
(5) See analysis below.
(6) No body force in the x direction; gx = 0
Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption
(3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the
Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4)
also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero
everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction
∂p
= ρg
∂y
which indicates a hydrostatic variation of pressure. In the x direction, after assumption (6) we obtain
∂ 2u ∂p
µ 2 − =0
∂y
∂x
Integrating twice
u=
1 ∂p 2 c1
y + y + c2
2 µ ∂x
µ
To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, u = 0.
Hence
0=
1 ∂p 2 c1
a + a
2 µ ∂x
µ
which gives
c1 = −
1 ∂p
a
2 µ ∂x
and finally
u=
2
a 2 ∂p ⎡⎛ y ⎞ ⎛ y ⎞⎤
⎢⎜ ⎟ − ⎜ ⎟ ⎥
2 µ ∂x ⎢⎣⎝ a ⎠ ⎝ a ⎠⎥⎦
Problem 8.16
[Difficulty: 3]
Problem 8.15
[Difficulty: 3]
Given:
Hydrostatic bearing
Find:
Required pad width; Pressure gradient; Gap height
Solution:
Basic equation
Q
l
Available data
=−
h
3
12⋅ μ
⋅ ⎛⎜
dp ⎞
⎝ dx ⎠
F = 1000⋅ lbf
l = 1 ⋅ ft
p i = 35⋅ psi
212 °F = 100 °C
At 100 oC from Fig. A.2, for SAE 10-30
(F is the load on width l)
μ = 0.01⋅
Q = 2.5⋅
N⋅ s
gal
per ft
hr
− 4 lbf ⋅ s
μ = 2.089 × 10
2
⋅
m
ft
2
For a laminar flow (we will verify this assumption later), the pressure gradient is constant
2⋅ x ⎞
p ( x ) = p i⋅ ⎛⎜ 1 −
W⎠
⎝
where p i = 35 psi is the inlet pressure (gage), and x = 0 to W/2
⌠
F = l ⋅ ⎮ p dx
⌡
Hence the total force in the y direction due to pressure is
where b is the pad width into the paper
W
⌠2
2⋅ x ⎞
⎮
dx
F = 2⋅ l⋅ ⎮
p i⋅ ⎛⎜ 1 −
W⎠
⎝
⌡
F=
1
2
⋅ p i⋅ l⋅ W
0
W =
This must be equal to the applied load F. Hence
dp
The pressure gradient is then
=−
dx
∆p
W
=−
2 ⋅ ∆p
W
= −2 ×
2 F
⋅
pi l
35⋅ lbf
2
W = 0.397 ⋅ ft
×
in
1
0.397 ⋅ ft
= −176 ⋅
psia
ft
1
2
Q
From the basic equation
l
Check Re:
From Fig. A.3
Re =
V⋅ D
ν
ν = 1.2⋅ 10
=
=−
h
3
12⋅ μ
⋅ ⎛⎜
dp ⎞
⎝ dx ⎠
⎛ 12⋅ μ⋅ Q ⎞
⎜
l
h = ⎜−
⎟
dp
⎜
dx
⎝
⎠
we can solve for
3
h = 2.51 × 10
−3
⋅ in
D Q
h Q
⋅ = ⋅
ν A
ν l⋅ h
2
−5 m
⋅
s
− 4 ft
ν = 1.29 × 10
⋅
2
s
Re =
Q
ν⋅ l
Re = 0.72
so flow is very laminar
Problem 8.14
[Difficulty: 3]
Problem 8.13
[Difficulty: 3]
Problem 8.12
Given:
Piston-cylinder assembly
Find:
Mass supported by piston
[Difficulty: 3]
Solution:
Basic equation
Available data
Q
l
3
=
a ⋅ ∆p
This is the equation for pressure-driven flow between parallel plates; for a small gap a,
the flow between the piston and cylinder can be modeled this way, with l = πD
12⋅ μ⋅ L
L = 4 ⋅ inD = 4 ⋅ in
a = 0.001 ⋅ in
From Fig. A.2, SAE10 oil at 20oF is
Q = 0.1⋅ gpm
μ = 0.1⋅
N⋅ s
2
68 °F = 20 °C
− 3 lbf ⋅ s
μ = 2.089 × 10
or
⋅
m
Hence, solving for ∆p
∆p =
12⋅ μ⋅ L⋅ Q
π⋅ D⋅ a
4
∆p = 2.133 × 10 ⋅ psi
3
F = ∆p⋅ A = ∆p⋅
A force balance for the piston involves the net pressure force
2
π⋅ D
M =
Note the following
Q
Vave =
a ⋅ π⋅ D
4
⋅
∆p
Hence
2
ft
π
4
ft
Vave = 2.55⋅
s
Hence an estimate of the Reynolds number in the gap is
W = M⋅ g
and the weight
5
M = 8331⋅ slug
g
2
⋅D
M = 2.68 × 10 ⋅ lb
ν = 10
Re =
2
−4 m
⋅
s
a⋅ Vave
ν
ν = 1.076 × 10
− 3 ft
⋅
2
s
Re = 0.198
This is a highly viscous flow; it can be shown that the force on the piston due to this motion is much less than that due to ∆p!
Note also that the piston speed is
Vpiston =
Vpiston
Vave
4⋅ Q
2
π⋅ D
= 0.1⋅ %
ft
Vpiston = 0.00255 ⋅
s
so the approximation of stationary walls is valid
Problem 8.11
[Difficulty: 2]
y
2h
Given:
Laminar flow between flat plates
Find:
Shear stress on upper plate; Volume flow rate per width
Solution:
du
τyx = μ⋅
dy
Then
τyx =
At the upper surface
y=h
The volume flow rate is
⌠
h
2
⌠
⌠
h ⋅ b dp ⎮
Q = ⎮ u dA = ⎮ u ⋅ b dy = −
⋅ ⋅⎮
⌡
2⋅ μ dx ⎮
⌡
−h
⌡
−h
2
2
⋅
u(y) = −
dp
dx
⋅⎛−
2⋅ y ⎞
⎜ 2
⎝ h ⎠
= −y ⋅
h
2
Basic equation
⋅
⎡
dp
2⋅ μ dx
⋅ ⎢1 −
⎣
b
=−
2
⎛ y ⎞ ⎤⎥
⎜
⎝h⎠ ⎦
(from Eq. 8.7)
dp
dx
1⋅ m
3 N
τyx = −1.5⋅ mm ×
× 1.25 × 10 ⋅
2
1000⋅ mm
m ⋅m
h
−h
Q
x
2
3
× ⎛⎜ 1.5⋅ mm ×
⎝
1⋅ m
3
⎡
⎢1 −
⎣
2⎤
⎛ y ⎞ ⎥ dy
⎜
⎝h⎠ ⎦
τyx = −1.88Pa
3
Q= −
2
⎞ × 1.25 × 103⋅ N × m
2
0.5⋅ N⋅ s
1000⋅ mm ⎠
m ⋅m
Q
b
2⋅ h ⋅ b dp
⋅
3⋅ μ dx
= −5.63 × 10
2
−6 m
s
Problem 8.10
[Difficulty: 2]
Problem 8.9
[Difficulty: 3]
D
p1
F
a
L
Given:
Piston cylinder assembly
Find:
Rate of oil leak
Solution:
Q
Basic equation
l
3
=
3
a ⋅ ∆p
Q=
12⋅ μ⋅ L
π⋅ D⋅ a ⋅ ∆p
(from Eq. 8.6c; we assume laminar flow and verify
this is correct after solving)
12⋅ μ⋅ L
F
4⋅ F
∆p = p 1 − p atm =
=
2
A
π⋅ D
For the system
4
∆p =
× 4500⋅ lbf ×
π
⎛ 1 × 12⋅ in ⎞
⎜
⎝ 4⋅ in 1⋅ ft ⎠
2
μ = 0.06 × 0.0209⋅
At 120 oF (about 50oC), from Fig. A.2
∆p = 358 ⋅ psi
lbf ⋅ s
ft
Q =
π
12
× 4 ⋅ in × ⎛⎜ 0.001 ⋅ in ×
Check Re:
⎝
V=
Re =
Q
A
μ = 1.25 × 10
2
− 3 lbf ⋅ s
⋅
ft
2
3
2
2
3
ft
1
− 5 ft
⎞ × 358 ⋅ lbf × 144 ⋅ in ×
× Q = 1.25 × 10 ⋅
2
2
−3
s
12⋅ in ⎠
2 ⋅ in
in
1 ⋅ ft
1.25 × 10
lbf ⋅ s
1 ⋅ ft
=
Q
a ⋅ π⋅ D
V⋅ a
1
π
× 1.25 × 10
ν = 6 × 10
ν
Re = 0.143 ⋅
V =
ft
s
× 0.001 ⋅ in ×
1 ⋅ ft
12⋅ in
−5
×
− 5 ft
3
s
× 10.8
ft
×
1
.001⋅ in
×
1
4 ⋅ in
2
×
⎛ 12⋅ in ⎞
⎜
⎝ 1⋅ ft ⎠
− 4 ft
ν = 6.48 × 10
s
s
−4
6.48 × 10
ft
2
⋅
Re = 0.0184
3
Q = 0.0216⋅
2
V = 0.143 ⋅
Q
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
3
4
− 5 ft
Vp =
× 1.25 × 10
×
s
π
⎛ 1 × 12⋅ in ⎞
⎜
⎝ 4 ⋅ in 1 ⋅ ft ⎠
s
ft
s
2
(at 120 oF, from Fig. A.3)
s
so flow is very much laminar
The speed of the piston is approximately
Vp =
in
2
The piston motion is negligible so our assumption of flow between parallel plates is reasonable
− 4 ft
Vp = 1.432 × 10
⋅
s
Problem 8.8
[Difficulty: 3]
Problem 8.7
[Difficulty: 2]
Problem 8.6
[Difficulty: 2]
Problem 8.5
[Difficulty: 4] Part 1/2
Problem 8.5
[Difficulty: 4] Part 2/2
Problem 8.4
[Difficulty: 2]
Given:
That transition to turbulence occurs at about Re = 2300
Find:
Plots of average velocity and volume and mass flow rates for turbulence for air and water
Solution:
The basic equations are
From Tables A.8 and A.10
For the average velocity
V⋅ D
Re =
Recrit = 2300
ν
kg
ρair = 1.23⋅
3
m
V=
νair = 1.45 × 10
⋅
s
Vair =
2
−5 m
⋅
2
s
Vair =
D
Hence for air
Vw =
⋅
π
4
2
⋅D ⋅V =
π
π
4
Vw =
2 Recrit⋅ ν
D
2
−5 m
Qair =
× 2300 × 1.45⋅ 10
4
⋅
s
=
π⋅ Recrit⋅ ν
4
m
s
D
⋅D
2
⋅D
2
For water
D
2
s
⋅D ⋅
s
0.00262 ⋅
D
Q = A⋅ V =
m
0.0334⋅
2
−6 m
For the volume flow rates
νw = 1.14 × 10
D
2300 × 1.14 × 10
For water
2
−6 m
kg
ρw = 999 ⋅
3
m
Recrit⋅ ν
2300 × 1.45 × 10
Hence for air
2
−5 m
π
−6 m
Qw =
× 2300 × 1.14⋅ 10 ⋅
⋅D
s
4
m
Qair = 0.0262⋅
×D
s
2
m
Qw = 0.00206 ⋅
×D
s
Finally, the mass flow rates are obtained from volume flow rates
mair = ρair⋅ Qair
kg
mair = 0.0322⋅
×D
m⋅ s
mw = ρw⋅ Qw
kg
mw = 2.06⋅
×D
m⋅ s
These results can be plotted in Excel as shown below in the next two pages
⋅
s
From Tables A.8 and A.10 the data required is
◊ air =
1.23
kg/m 3
2
◊ air = 1.45E-05 m /s
◊w =
999
kg/m
3
◊ w = 1.14E-06 m /s
2
0.0001
0.001
0.01
0.05
V air (m/s) 333.500
33.350
3.335
0.667
2.62
0.262
D (m)
V w (m/s)
26.2
1.0
2.5
5.0
7.5
10.0
3.34E-02 1.33E-02 6.67E-03 4.45E-03 3.34E-03
5.24E-02 2.62E-03 1.05E-03 5.24E-04 3.50E-04 2.62E-04
3
Q air (m /s) 2.62E-06 2.62E-05 2.62E-04 1.31E-03 2.62E-02 6.55E-02 1.31E-01 1.96E-01 2.62E-01
Q w (m 3/s) 2.06E-07 2.06E-06 2.06E-05 1.03E-04 2.06E-03 5.15E-03 1.03E-02 1.54E-02 2.06E-02
m air (kg/s) 3.22E-06 3.22E-05 3.22E-04 1.61E-03 3.22E-02 8.05E-02 1.61E-01 2.42E-01 3.22E-01
m w (kg/s) 2.06E-04 2.06E-03 2.06E-02 1.03E-01 2.06E+00 5.14E+00 1.03E+01 1.54E+01 2.06E+01
Average Velocity for Turbulence in a Pipe
1.E+04
V (m/s)
1.E+02
Velocity (Air)
Velocity (Water)
1.E+00
1.E-02
1.E-04
1.E-04
1.E-03
1.E-02
1.E-01
D (m )
1.E+00
1.E+01
Flow Rate for Turbulence in a Pipe
Q (m3/s)
1.E+01
1.E-01
Flow Rate (Air)
Flow Rate (Water)
1.E-03
1.E-05
1.E-07
1 .E-04
1.E-03
1.E-02
1.E-01
1.E+00
1.E+01
D (m)
Mass Flow Rate for Turbulence in a Pipe
m flow (kg/s)
1.E+02
1.E+00
Mas s Flow Rate (Air)
Mas s Flow Rate (Water)
1.E-02
1.E-04
1.E-06
1.E-04
1.E-03
1.E-02
1.E-01
D (m)
1.E+00
1.E+01
Problem 8.3
[Difficulty: 3]
Given:
Air entering pipe system
Find:
Flow rate for turbulence in each section; Which become fully developed
Solution:
From Table A.10
The given data is
ν = 1.69 × 10
2
−5 m
⋅
L = 2⋅ m
s
D1 = 25⋅ mm
D2 = 15⋅ mm
D3 = 10⋅ mm
or
Q=
Recrit = 2300
The critical Reynolds number is
Writing the Reynolds number as a function of flow rate
Re =
V⋅ D
ν
=
Q
π
4
⋅
D
2 ν
⋅D
Re⋅ π⋅ ν⋅ D
4
Then the flow rates for turbulence to begin in each section of pipe are
Q1 =
Q2 =
Q3 =
Recrit⋅ π⋅ ν⋅ D1
Q1 = 7.63 × 10
4
Recrit⋅ π⋅ ν⋅ D2
Q2 = 4.58 × 10
4
Recrit⋅ π⋅ ν⋅ D3
Q3 = 3.05 × 10
4
3
−4m
s
3
−4m
s
3
−4m
s
Hence, smallest pipe becomes turbulent first, then second, then the largest.
For the smallest pipe transitioning to turbulence (Q3)
For pipe 3
Re3 = 2300
Llaminar = 0.06⋅ Re3 ⋅ D3
or, for turbulent,
Lmin = 25⋅ D3
Lmin = 0.25 m
For pipes 1 and 2
Llaminar = 0.06⋅ ⎜
⎛ 4⋅ Q3 ⎞
⎝
π⋅ ν⋅ D1
⎠
⋅ D1
Llaminar = 1.38 m
Lmax = 40⋅ D3
Lmax = 0.4 m
Llaminar = 1.38 m
Llaminar < L: Fully developed
Lmax/min < L: Fully developed
Llaminar < L: Fully developed
⎛ 4⋅ Q3 ⎞
Llaminar = 0.06⋅ ⎜
⎝
π⋅ ν⋅ D2
⎠
⋅ D2
Llaminar = 1.38 m
Llaminar < L: Fully developed
For the middle pipe transitioning to turbulence (Q2)
For pipe 2
Re2 = 2300
Llaminar = 0.06⋅ Re2 ⋅ D2
Llaminar = 2.07 m
or, for turbulent,
Lmin = 25⋅ D2
Lmin = 1.23⋅ ft
Lmax = 40⋅ D2
Llaminar > L: Not fully developed
Lmax = 0.6 m
Lmax/min < L: Fully developed
For pipes 1 and 3
⎛ 4 ⋅ Q2 ⎞
L1 = 0.06⋅ ⎜
⎝
π⋅ ν⋅ D1
L3min = 25⋅ D3
⎠
⋅ D1
L3min = 0.25⋅ m
L1 = 2.07⋅ m
L3max = 40⋅ D3
Llaminar > L: Not fully developed
L3max = 0.4 m
Lmax/min < L: Fully developed
For the large pipe transitioning to turbulence (Q1)
For pipe 1
Re1 = 2300
Llaminar = 0.06⋅ Re1 ⋅ D1
Llaminar = 3.45 m
or, for turbulent,
Lmin = 25⋅ D1
Lmin = 2.05⋅ ft
Lmax = 40⋅ D1
Llaminar > L: Not fully developed
Lmax = 1.00 m
Lmax/min < L: Fully developed
For pipes 2 and 3
L2min = 25⋅ D2
L2min = 1.23⋅ ft
L2max = 40⋅ D2
L2max = 0.6 m
Lmax/min < L: Fully developed
L3min = 25⋅ D3
L3min = 0.82⋅ ft
L3max = 40⋅ D3
L3max = 0.4 m
Lmax/min < L: Fully developed
Problem 8.2
[Difficulty: 2]
Problem 8.1
Given:
Air entering duct
Find:
Flow rate for turbulence; Entrance length
Solution:
The basic equations are
The given data is
Re =
V⋅ D
π
2
Recrit = 2300
Q=
D = 125⋅ mm
From Table A.10
ν = 2.29 × 10
Llaminar = 0.06⋅ Recrit⋅ D
or, for turbulent, Lturb = 25D to 40D
ν
Q
π
Hence
[Difficulty: 1]
Recrit =
4
4
⋅D ⋅V
2
−5 m
⋅
s
⋅D
2
⋅D
or
ν
Q =
Recrit⋅ π⋅ ν ⋅ D
4
For laminar flow
Llaminar = 0.06⋅ Recrit⋅ D
Llaminar = 17.3 m
For turbulent flow
Lmin = 25⋅ D
Lmin = 3.13 m
3
−3m
Q = 5.171 × 10
Lmax = 40⋅ D
s
Lmax = 5.00 m
Problem 7.97
[Difficulty: 5]
Discussion: The equation given in Problem 7.2 contains three terms. The first term contains surface tension and gives a
speed inversely proportional to wavelength. These terms will be important when small wavelengths are
considered.
The second term contains gravity and gives a speed proportional to wavelength. This term will be important
when long wavelengths are considered.
The argument of the hyperbolic tangent is proportional to water depth and inversely proportional to
wavelength. For small wavelengths this term should approach unity since the hyperbolic tangent of a large
number approaches one.
The governing equation is:
2
c
σ 2 π g λ tanh 2 π h
2 π
ρ λ
λ
The relevant physical parameters are:
g 9.81
m
2
ρ 999
s
kg
3
σ 0.0728
m
A plot of the wave speed versus wavelength at different depths is shown here:
Wave Speed versus Wavelength
h = 1 mm
h = 5 mm
h = 10 mm
h > 50 mm
Wave Speed (m/s)
0.4
0.3
0.2
0.1
0
0
0.05
Wavelength (m)
0.1
N
m
Problem 7.96
[Difficulty: 3]
Given:
A 1:16 scale model of a bus (152 mm x 200 mm x 762 mm) is tested in a wind tunnel at 26.5 m/s. Drag force is 6.09
N. The axial pressure gradient is -11.8 N/m2/m.
Find:
(a) Horizontal buoyancy correction
(b) Drag coefficient for the model
(c) Aerodynamic drag on the prototype at 100 kph on a calm day.
Solution:
The horizontal buoyancy force is the difference in the pressure force between the front and back of the model due
to the pressure gradient in the tunnel:
dp
FB p f p b A
L A
dx m m
2
Am 152 mm 200 mm Am 30400 mm
where:
N
2
Thus: FB 11.8
762 mm 30400 mm
2
m m
So the corrected drag force is:
m
1000 mm
3
FB 0.273 N
FDc 6.09 N 0.273 N FDc 5.817 N
The corrected model drag coefficient would then be:
CDm
FDc
1
2
3
Substituting in values:
2
ρ V Am
2
2
s
1
1000 mm
kg m
CDm 2 5.82 N
1.23 kg 26.5 m
2
2
m
30400 mm
N s
m
CDm 0.443
If we assume that the test was conducted at high enough Reynolds number, then the drag coefficient should be the
same at both scales, i.e.: C
Dp CDm
1
2
FDp ρ V Ap CDp
2
where
m
2 2
Ap 30400 mm 16
1000
mm
2
2
2
Ap 7.782 m
2
1
kg
km 1000 m
hr
7.782 m2 0.443 N s
FDp
1.23
100
3
2
hr
km 3600 s
kg m
m
FDp 1.636 kN
(The rolling resistance must also be included to obtain the total tractive effort needed to propel the vehicle.)
Problem 7.95
Given:
Flapping flag on a flagpole
Find:
Explanation of the flapping
[Difficulty: 4]
Solution:
Discussion: The natural wind contains significant fluctuations in air speed and direction. These fluctuations tend to disturb the flag
from an initially plane position.
When the flag is bent or curved from the plane position, the flow nearby must follow its contour. Flow over a convex surface tends to
be faster, and have lower pressure, than flow over a concave curved surface. The resulting pressure forces tend to exaggerate the
curvature of the flag. The result is a seemingly random "flapping" motion of the flag.
The rope or chain used to raise the flag may also flap in the wind. It is much more likely to exhibit a periodic motion than the flag
itself. The rope is quite close to the flag pole, where it is influenced by any vortices shed from the pole. If the Reynolds number is
such that periodic vortices are shed from the pole, they will tend to make the rope move with the same frequency. This accounts for
the periodic thump of a rope or clank of a chain against the pole.
The vortex shedding phenomenon is characterized by the Strouhal number, St = fD/V∞, where f is the vortex shedding frequency, D is
the pole diameter, and D is the wind speed. The Strouhal number is constant at approximately 0.2 over a broad range of Reynolds
numbers.
Problem 7.94
[Difficulty: 3]
Given:
A scale model of a truck is tested in a wind tunnel. The axial pressure gradient and frontal area of the prototype
are known. Drag coefficient is 0.85.
Find:
(a) Horizontal buoyancy correction
(b) Express this correction as a fraction of the measured drag force.
Solution:
The horizontal buoyancy force is the difference in the pressure force between the front and back of the model due
to the pressure gradient in the tunnel:
Δp
FB p f p b A
L A
ΔL m m
where:
lbf
60 ft 110 ft
Thus: FB 0.07
2
2
16
ft ft
16
Lm
Lp
16
Am
Ap
2
16
2
FB 0.113 lbf
The horizontal buoyancy correction should be added to the measured drag force on the model. The measured drag
force on the model is given by:
1
1
2
2 Ap
FDm ρ V Am CD ρ V
CD
2
2
2
16
2
When we substitute in known values we get:
2
2
1
slug
ft
110 ft
lbf s
FDm
0.00238
250
0.85
3
2
s
2
slug ft
ft
16
Therefore the ratio of the forces is:
DragRatio
0.113
27.16
FDm 27.16 lbf
DragRatio 0.42 %
Problem 7.93
Given:
Find:
Solution:
[Difficulty: 2]
Kinetic energy ratio for a wind tunnel is the ratio of the kinetic energy flux in the test section to the drive power
Kinetic energy ratio for the 40 ft x 80 ft tunnel at NASA-Ames
nmi 6080 ft
hr
ft
From the text: P 125000 hp Vmax 300
Vmax 507
hr
nmi
3600 s
s
2
m
Therefore, the kinetic energy ratio is: KEratio
V
2
P
2
( ρ V A) V
2 P
3
3
ρ A V
2 P
Assuming standard conditions
and substituting values:
2
1
slug
ft
1
hp s
lbf s
KEratio
0.00238
( 40 ft 80 ft) 507
3
s
125000 hp 550 ft lbf
2
slug ft
ft
KEratio 7.22
Problem 7.92
Given:
Water drop mechanism
Find:
Difference between small and large scale drops
[Difficulty: 2]
Solution:
3
Given relation
5
d D ( We)
ρ V2 D
D
σ
For dynamic similarity
ρ V 2 D
m m
Dm
dm
σ
dp
ρ V 2 D
p
p
Dp
σ
2
Hence
dm
dp
5
1 5
20
1
3
5
3
5
3
2
Dm
Dp
5
Vm
Vp
6
5
where d p stands for dprototype not the original
d p!
5
6
5
The small scale droplets are 4.4% of the size of the large scale
dm
dp
0.044
Problem 7.91
[Difficulty: 3]
Given:
For a marine propeller (Problem 7.40) the thrust force is: FT FT( ρ D V g ω p μ)
For ship size propellers viscous and pressure effects can be neglected. Assume that power and torque depend on
the same parameters as thrust.
Find:
Solution:
Scaling laws for propellers that relate thrust, power and torque to other variables
We will use the Buckingham pi-theorem. Based on the simplifications given above:
1
FT
2
Select primary dimensions F, L, t:
3
FT
P
P
4
5
ρ
F L
t
ω
D
ρ
T
F L
F
ρ
T
F t
L
D
2
4
L
V
g
ω
V
g
ω
L
L
1
t
2
t
t
n = 8 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 5 dimensionless groups (3 dependent, 2 independent). Setting up dimensional equations:
a
a
b
c
Π1 FT ρ ω D
Thus:
Summing exponents:
F:
1a0
L:
4 a c 0
t:
2 a b 0
F t2 1 b c 0 0 0
F
L F L t
L4 t
The solution to this system is:
a 1
b 2
c 2
Π1
FT
2
4
ρ ω D
a
a
b
c
Π2 P ρ ω D
Summing exponents:
F:
1a0
L:
1 4 a c 0
t:
1 2 a b 0
Thus:
F t2 1 b c 0 0 0
L F L t
t 4 t
L
F L
The solution to this system is:
a 1
b 3
c 5
Π2
P
3
5
ρ ω D
a
a
b
c
Π3 T ρ ω D
F t2 1 b c 0 0 0
F L
L F L t
L4 t
Thus:
Summing exponents:
F:
1a0
L:
1 4 a c 0
t:
2 a b 0
The solution to this system is:
a 1
b 2
Π3
c 5
T
2
5
ρ ω D
a
a
b
c
Π4 V ρ ω D
F t2 1 b c 0 0 0
L F L t
t 4 t
L
L
Thus:
Summing exponents:
F:
a0
L:
1 4 a c 0
t:
1 2 a b 0
a
The solution to this system is:
a0
b
c
Π5 g ρ ω D
F t2
t 4
L
L
Thus:
Summing exponents:
F:
a0
L:
1 4 a c 0
t:
1 2 a b 0
Check using M, L, t dimensions:
a
1
t
c 1
b
c
0 0 0
L F L t
The solution to this system is:
a0
3
6
b 1
V
Π4
ω D
b 1
M L L 2 1
t
1
2 M
4
L
t
L
t
t
1
L
1
Π5
c 1
M L
t
2
3
L
3
M
3 1
t
L
5
1
M L
t
2
2
g
2
ω D
L
3
M
2 1
t
L
5
1
L 2 1
t
1
2
2
L
t
Based on the dependent and independent variables, the "scaling laws" are:
FT
V
g
f1
2 4
ω D 2
ρ ω D
ω D
V
g
f2
3 5
ω D 2
ρ ω D
ω D
P
V
g
f3
2 5
ω D 2
ρ ω D
ω D
T
Problem 7.90
[Difficulty: 3]
Given:
Data on model propeller
Find:
Speed, thrust and torque on prototype
Solution:
We will use the Buckingham Pi-theorem to find the functional relationships between these variables. Neglecting the
effects of viscosity:
1
F
2
Select primary dimensions M, L, t:
3
F
T
M L
M L
t
4
5
ρ
T
2
ρ
t
D
2
2
V
ρ
V
M
L
3
t
L
D
ω
D
ω
n = 6 parameters
1
L
r = 3 dimensions
t
m = r = 3 repeating parameters
ω
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 F ρ D ω
Thus:
M L
t
Summing exponents:
2
1 3 a b 0
t:
2 c 0
a
b
c
Π2 T ρ D ω
Thus:
M L
Summing exponents:
t:
2 c 0
a
b
c
Π3 V ρ D ω
L
3
L
b
1
c
0 0 0
M L t
t
2
2
b 4
M
a
3
L
L
b
c 2
1
t
a 1
Thus:
L
t
M
3
L
b 5
a
L
b
1
t
c
Π1
F
4
2
ρ D ω
c
0 0 0
M L t
The solution to this system is:
M: 1 a 0
2 3 a b 0
a
a 1
t
L:
M
The solution to this system is:
M: 1 a 0
L:
c 2
0 0 0
M L t
Π2
T
5
2
ρ D ω
Summing exponents:
The solution to this system is:
a0
M: 0 a 0
L:
1 3 a b 0
t:
1 c 0
L
F
6 Check using F, L, t dimensions:
4
F t
2
1
L
4
b 1
2
t 1
F L
V
Π3
D ω
c 1
L
4
F t
2
1
L
L 1
t 1
t L
2
5
t 1
For dynamically similar conditions:
Vm
Dm ωm
Vp
Dp ωp
Fm
4
2
ρm Dm ωm
Vp Dm
ωp ωm
Vm Dp
Thus:
Fp
4
ρp Dp ωp
Thus:
2
130 1
ωp 1800 rpm
50
8
Dp
Fp Fm
ρm Dm
ρp
4
ωp
ωm
ωp 585 rpm
2
1
Fp 100 N
1
4
8 585
1 1800
2
Fp 43.3 kN
Tm
5
2
ρm Dm ωm
Tp
5
ρp Dp ωp
Thus:
2
5
Dp ωp
Tp Tm
ρm Dm
ωm
ρp
2
Tp 10 N m
1
1
5
8 585
1 1800
2
Tp 34.6 kN m
Problem 7.89
Given:
Model of water pump
Find:
Model head, flow rate and diameter
Solution:
From Buckingham Π
h
2
2
Q ρ ω D2
ω D3
μ
f
ω D
Neglecting viscous effects
Qm
3
ωm Dm
Hence if
ωp Dp
P
and
3
hm
then
3
2
3
ωm
3
hp
2
ωp Dp
Pm
and
2
3
5
ωm Dm
Pp
3
ωp Dp
5
3
(1)
then
2
2
2
Dm
1000 Dm
2 4 2
hp
2
2
500
ωp Dp
Dp
Dp
and
Pm
We can find Pp from
kg
m
J
Pp ρ Q h 1000
0.75
15
11.25 kW
3
s
kg
m
hm
2
2
ωm Dm
Dm
Dm
1000 Dm
2
Qp
ωp Dp
500
Dp
Dp
Qm
5
Q ρ ω D2
ω D3
μ
f
ω D
Qp
[Difficulty: 3]
2
Dm
ωm
3
(2)
5
5
5
3
Dm
1000 Dm
5 8 5
Pp
3
5
500
ωp Dp
Dp
Dp
Dm
ωm
(3)
3
1
From Eq 3
From Eq 1
From Eq 2
Pm
Pp
5
8
Dm
Dp
5
Dm
2
Qp
Dp
Qm
Dm
4
hp
Dp
hm
so
1 Pm
Dm Dp
8 Pp
so
Dm
Qm Qp 2
Dp
so
Dm
hm hp 4
Dp
3
2
1
5
3
Dm 0.25 m
5
1 2.25
8 11.25
3
m
Qm 0.75
2
s
2
h m 15
J
kg
4
0.12
0.25
0.12
0.25
3
Dm 0.120 m
3
m
Qm 0.166
s
2
h m 13.8
J
kg
Problem 7.88 (In Excel)
[Difficulty: 4]
Given: Data on centrifugal water pump
Find: groups; plot pressure head vs flow rate for range of speeds
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of groups is:
=5
=3
=3
=2
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , g , d
M
1
D
L
-3
t
-1
1
GROUPS:
p
1:
M
1
L
-1
a =
b =
c =
-1
-2
-2
t
-2
Q
2:
M
0
L
3
a =
b =
c =
0
-1
-3
M
0
L
0
a =
b =
c =
0
0
0
t
-1
The following groups from Example 7.1 are not used:
3:
Hence
1
p
2 D 2
and
2
Q
M
0
L
0
a =
b =
c =
0
0
0
t
0
4:
with 1 = f(2).
D 3
Based on the plotted data, it looks like the relation between 1 and 2 may be parabolic
Hence
p
Q
Q
a b
c
D
D 3
D 3
2
2
2
The data is
Q (ft3/min)
p (psf)
0
50
75
100
120
140
150
165
7.54
7.29
6.85
6.12
4.80
3.03
2.38
1.23
t
0
D =
Q /(D 3)
p /( D )
2
2
1.94
800
1
(D is not given; use D = 1 ft as a scale)
0.00000
0.00995
0.01492
0.01989
0.02387
0.02785
0.02984
0.03283
0.000554
0.000535
0.000503
0.000449
0.000353
0.000223
0.000175
0.000090
Centifugal Pump Data and Trendline
0.0006
2 2
p /( D )
3
slug/ft
rpm
ft
0.0005
0.0004
Pump Data
Parabolic Fit
0.0003
0.0002
0.0001
0.0000
0.000
0.005
0.010
0.015
0.020
0.025
0.030
0.035
3
Q /(D )
2 2
3 2
3
p /( D ) = -0.6302 (Q /( D )) + 0.006476 (Q /( D )) + 0.0005490
The curve fit result is:
From the Trendline analysis
a = 0.000549
b = 0.006476
c = -0.6302
and
2
Q
Q
c
p 2 D 2 a b
D 3
D 3
Finally, data at 500 and 1000 rpm can be calculated and plotted
Q (ft3/min)
p (kPa)
Q (ft3/min)
p (kPa)
600
rpm
0
20
40
60
80
100
120
132
4.20
4.33
4.19
3.77
3.08
2.12
0.89
0.00
1200
rpm
0
50
75
100
120
140
150
165
16.82
17.29
16.88
16.05
15.09
13.85
13.12
11.91
p (psf)
Centifugal Pump Curves
Pump Data at 800 rpm
20
18
16
14
12
10
8
6
4
2
0
Pump Curve at 600 rpm
Pump Curve at 1200 rpm
0
20
40
60
80
100
3
Q (ft /min)
120
140
160
180
Problem 7.87
CD
For drag we can use
Model:
L=
=
For water
=
D
1
V
2
2
As a suitable scaling area for A we use L
2
3
[Difficulty: 3]
CD
A
D
1
V 2 L2
2
ft
1.94
slug/ft3
2.10E-05
lbf·s/ft2
Wave Drag
The data is:
V (ft/s)
D Wave (lbf)
D Friction (lbf)
Fr
Re
C D(Wave)
C D(Friction)
3.5E-05
10
0
0.022
20
0.028
0.079
30
0.112
0.169
40
0.337
0.281
50
0.674
0.45
60
0.899
0.618
70
1.237
0.731
1.017
2.77E+06
0.00E+00
2.52E-05
2.035
5.54E+06
8.02E-06
2.26E-05
3.052
8.31E+06
1.43E-05
2.15E-05
4.070
1.11E+07
2.41E-05
2.01E-05
5.087
1.39E+07
3.09E-05
2.06E-05
6.105
1.66E+07
2.86E-05
1.97E-05
7.122
1.94E+07
2.89E-05
1.71E-05
3.0E-05
2.5E-05
2.0E-05
CD
1.5E-05
1.0E-05
5.0E-06
0.0E+00
The friction drag coefficient becomes a constant, as expected, at high Re .
The wave drag coefficient appears to be linear with Fr , over most values
0
1
2
3
4
5
6
7
8
Fr
Ship:
V (knot)
V (ft/s)
Fr
Re
L=
15
25.32
0.364
3.51E+08
150
ft
20
33.76
0.486
4.68E+08
D
1
V 2 L2 C D
2
Friction Drag
3.5E-05
3.0E-05
Hence for the ship we have very high Re , and low Fr .
2.5E-05
From the graph we see the friction C D levels out at about 1.9 x 10-5
From the graph we see the wave C D is negligibly small
2.0E-05
CD
C D(Wave)
C D(Friction)
0
1.90E-05
0
1.90E-05
D Wave (lbf)
D Friction (lbf)
0
266
0
473
D Total (lbf)
266
473
1.5E-05
1.0E-05
5.0E-06
0.0E+00
0.0.E+00
5.0.E+06
1.0.E+07
1.5.E+07
Re
2.0.E+07
2.5.E+07
Problem 7.86
Given:
Find:
Solution:
Information relating to geometrically similar model test for a centrifugal pump.
The missing values in the table
We will use the Buckingham pi-theorem.
n = 5 parameters
1
Δp
2
Select primary dimensions M, L, t:
3
Δp
Q
M
L
L t
4
5
ρ
ρ
Q
3
t
2
ω
ω
ρ
ω
M
1
3
t
L
D
D
L
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 Δp ρ ω D
1 3 a c 0
t:
2 b 0
a
b
c
Π2 Q ρ ω D
a 1
t:
1 b 0
1
b
Check using F, L, t dimensions:
3
b 2
a
c 2
Π1
Δp
2
2
ρ ω D
b
1
M
c
0 0 0
L M L t
t
3
t
L
L
Thus:
The solution to this system is:
M: a 0
3 3 a c 0
a
The solution to this system is:
Summing exponents:
L:
M
c
0 0 0
L M L t
2
3
t
L t L
M: 1 a 0
L:
M
Thus:
Summing exponents:
6
[Difficulty: 2]
a0
F
L
2
L
4
F t
2
b 1
2 1
t
L
2
1
c 3
L
3
t
t
1
L
3
1
Π2
Q
3
ω D
Thus the relationship is:
Δp
2
2
ρ ω D
Q
3
ω D
f
The flows are geometrically similar, and we assume kinematic similarity. Thus, for dynamic similarity:
If
Qm
3
ωm Dm
Qp
ωp Dp
Dp
From the first relation: Qp Qm
ωm Dm
ωp
From the second relation:
then
3
2
2
3
Δpp
ρm ωm Dm
3
m
183
Qp 0.0928
min 367
ωm Dm
Δpm Δpp
ρp ωp Dp
ρm
Δpm
2
ρp ωp Dp
150
50
3
3
m
Qp 1.249
min
2
Δpm 52.5 kPa
2
999
800
367 50
183 150
2
Δpm 29.3 kPa
Problem 7.85 (In Excel)
[Difficulty: 3]
Given: Data on model of aircraft
Find: Plot of lift vs speed of model; also of prototype
Solution:
V m (m/s)
F m (N)
10
2.2
15
4.8
20
8.7
25
13.3
30
19.6
35
26.5
40
34.5
45
43.8
This data can be fit to
Fm
1
2
AmCDVm
2
2
Fm kmVm
or
From the trendline, we see that
N/(m/s)2
k m = 0.0219
(And note that the power is 1.9954 or 2.00 to three signifcant
figures, confirming the relation is quadratic)
Also, k p = 1110 k m
Hence,
kp =
2
24.3 N/(m/s)
F p = k p V m2
V p (m/s)
75
100
125
150
175
200
225
250
F p (kN)
(Trendline)
137
243
380
547
744
972
1231
1519
50
54.0
Lift vs Speed for an Airplane Model
60
y = 0.0219x1.9954
R2 = 0.9999
F m (N)
50
40
30
20
Model
10
Power Curve Fit
0
0
10
20
30
40
50
60
200
250
300
V m (m/s)
Lift vs Speed for an
Airplane Prototype
1600
F p (kN)
1400
1200
1000
800
600
400
200
0
0
50
100
150
V p (m/s)
Lift vs Speed for an Airplane Model
(Log-Log Plot)
F m (N)
100
y = 0.0219x1.9954
R2 = 0.9999
10
Model
Power Curve Fit
1
10
100
V m (m/s)
Lift vs Speed for an Airplane Prototype (Log-Log Plot)
F p (kN)
10000
1000
100
10
1
10
100
V p (m/s)
1000
Problem 7.84
[Difficulty: 4]
Power to drive a fan is a function of ρ, Q, D, and ω.
Given:
3
P
2
Select primary dimensions M, L, t:
3
P
4
5
Q
ρ
Q
M
L
3
t
2
L
D
D
ρ
3
ρ
D
3
L
n = 5 parameters
ω
ω
1
t
r = 3 dimensions
m = r = 3 repeating parameters
ω
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 P ρ D ω
Thus:
M L
M
3
L
a 1
L:
2 3 a b 0
t:
3 c 0
b
a
L
b
1
c
0 0 0
M L t
t
The solution to this system is:
M: 1 a 0
a
2
3
t
Summing exponents:
c
Π2 Q ρ D ω
Thus:
Summing exponents:
L
3
t
M
3
L
b 5
a
L
b
1
t
Π1
c 3
a0
L:
3 3 a b 0
t:
1 c 0
4
Check using F, L, t dimensions:
b 3
Π2
c 1
1
1
D2 8 in
5
Q
3
ω D
3
88 2500
15 1800
P
Thus the relationship is:
For dynamic similarity we must have geometric and kinematic similarity, and:
Q2 ω1
D2 D1
Q1 ω2
3
ρ ω D
0 0 0
M L t
1 3
1
F L L
L
t 1
t
1
2 5
t
3
t
L
F t L
3
P
c
The solution to this system is:
M: a 0
6
ω2 1800 rpm
We will use the Buckingham pi-theorem.
1
t
ft
Condition 2: Q2 88
s
Fan diameter for condition 2 to insure dynamic similarity
Find:
Solution:
M L
3
ft
ω1 2500 rpm Q1 15
s
Condition 1: D1 8 in
3
5
ρ ω D
Q1
ω1 D1
3
Q2
ω2 D2
3
Q
3
ω D
f
Solving for D2
3
D2 16.10 in
Problem 7.83
[Difficulty: 3]
Given:
Recommended procedures for wind tunnel tests of trucks and buses suggest:
-Model frontal area less than 5% of test section area
-Reynolds number based on model width greater than 2,000,000
-Model height less than 30% of test section height
-Model projected width at maximum yaw (20 deg) less than 30% of test section width
-Air speed less than 300 ft/s to avoid compressibility effects
Model of a tractor-trailer to be tested in a tunnel 1.5 ft high x 2 ft wide. Full scale rig is 13'6" high, 8' wide, and 65'
long.
Find:
(a) Max scale for tractor-trailer model in this tunnel
(b) If adequate Reynolds number can be achieved in this facility.
Solution:
Let s be the scale ratio. Then:
Area criterion:
h m s h p wm s wp lm s lp
Am 0.05 1.5 ft 2.0 ft Am 0.15 ft
Height criterion: h m 0.30 1.5 ft h m 0.45 ft
2
0.15
Therefore: s
Therefore: s
13.5 8
0.45
13.5
s 0.0373
s 0.0333
Width criterion: we need to account for the yaw in the model. We make a relationship for the maximum width as a
function of the model dimensions and the yaw angle and relate that to the full-scale dimensions.
wm20deg wm cos( 20 deg) lm sin( 20 deg) s wp cos( 20 deg) lp sin( 20 deg)
wm20deg 0.30 2.0 ft
Therefore: s
wm20deg 0.60 ft
0.60
8 cos( 20 deg) 65 sin( 20 deg)
To determine the acceptable scale for the model, we take the smallest of these scale factors:
1
s
49.58
We choose a round number to make the model scale easier to calculate:
For the current model conditions: Re
Re 300
ft
s
Vm wm
νm
s
1 8 ft
50
1.57 10 4 ft2
For standard air: νm 1.57 10
4 ft
s 0.0202
s 0.0202
Model
1
50
Prototype
2
Substituting known values:
s
5
Re 3.06 10 This is less than the minimum stipulated in the problem, thus:
An adequate Reynolds number can not be achieved.
Problem 7.82
[Difficulty: 3]
Given:
Model of tractor-trailer truck
Find:
Drag coefficient; Drag on prototype; Model speed for dynamic similarity
Solution
:For kinematic similarity we need to ensure the geometries of model and prototype are similar, as is the incoming flow field
The drag coefficient is
For air (Table A.10) at
20oC
CD
Fm
1
2
ρ V A
2 m m m
kg
ρm 1.21
3
m
μp 1.81 10
5 N s
2
m
3
2
m
CD 2 350 N
1.21 kg
2
s 1 N s
2
kg m
75 m
0.1 m
CD 1.028
This is the drag coefficient for model and prototype
For the rig
1
2
Fp ρp Vp Ap CD
2
2
Lp
100
Am
Lm
Ap
with
2
2
km 1000 m
1 hr
N s
2
Fp
1.21
90
10 m 1.028
3
2
hr
1 km
3600 s
kg m
m
1
For dynamic similarity
kg
ρm Vm Lm
μm
ρp Vp Lp
km 1000 m
1 hr
10
Vm 90
hr
1 km
3600 s
1
c
Hence we have
M
1.40 286.9
Vm
c
250
343
c
N m
kg K
0.729
Fp 3.89 kN
ρp Lp μm
Lp
Vm Vp
Vp
ρm Lm μp
Lm
μp
For air at standard conditions, the speed of sound is
2
Ap 10 m
m
Vm 250
s
k R T
( 20 273 ) K
kg m
2
s N
c 343
m
s
which indicates compressibility is significant - this model
speed is impractical (and unnecessary)
Problem 7.81
Given:
Find:
Solution:
[Difficulty: 4]
A circular container partiall filled with water is rotate about its axis at constant angular velocity ω. Velocity in the
θ direction is a function of r, τ, ω, ρ, and μ.
(a) Dimensionless parameters that characterize this problem
(b) If honey would attain steady motion as quickly as water if rotated at the same angular speed
(c) Why Reynolds number is not an important parameter in scaling the steady-state motion of liquid in the
container.
The functional relationship for drag force is: Vθ Vθ( ω r τ ρ μ) From the Buckingham Π-theorem, we have
6 variables and 3 repeating parameters. Therefore, we will have 3 dimensionless groups. The functional form of
these groups is:
Vθ
ω r
μ
From the above result Π2
Π2 Π3
μ
2
ρ ω r
νh τh νw τw
ω τ
2
containing the properties μ and ρ, and
ρ ω r
μ τ
2
ρ r
ν τ
2
r
νw
τh τw
νh
Now for steady flow:
νh τh
2
r
μ
Π3 ω τ containing the time τ
νw τw
2
ω τ
2
ρ ω r
g
and at the same radius:
r
Now since honey is more viscous than water, it follows that:
τh τw
At steady state, solid body rotation exists. There are no viscous forces, and therefore, the Reynolds number would
not be important.
Problem 7.80
Given:
Find:
Solution:
[Difficulty: 3]
The drag force on a circular cylinder immersed in a water flow can be expressed as a function of D, l, V, ρ, and μ.
Static pressure distribution can be expressed in terms of the pressure coefficient. At the minimum static
pressure, the pressure coefficient is equal to -2.4. Cavitation onset occurs at a cavitation number of 0.5.
(a) Drag force in dimensionless form as a function of all relevant variables
(b) Maximum speed at which a cylinder could be towed in water at atmospheric pressure without cavitation
The functional relationship for drag force is: FD FD( D l V ρ μ) From the Buckingham Π-theorem, we have
6 variables and 3 repeating parameters. Therefore, we will have 3 dimensionless groups. The functional form of
these groups is:
FD
2
2
ρ V D
The pressure coefficient is:
CP
p p inf
1
2
At the minimum pressure point
At the onset of cavitation
ρ Ca CPmin
2 p inf p v
ρ V
l
D
ρ V D
μ
p pv
1
2
2
ρ V
1
2
p min p inf ρ Vmax CPmin where CPmin 2.4
2
1
2
p min p v ρ Vmax Ca
2
p inf
Equating these two expressions:
Vmax
2
and the cavitation number is: Ca
g
where Ca 0.5
1
1
2
2
ρ Vmax CPmin p v ρ Vmax Ca and if we solve for Vmax:
2
2
At room temperature (68 deg F): p v 0.339 psi
ρ 1.94
slug
ft
3
Substituting values we get:
Vmax
2 ( 14.7 0.339 )
lbf
2
in
ft
3
1.94 slug
1
[ 0.5 ( 2.4) ]
slug ft
2
lbf s
2
144 in
ft
2
ft
Vmax 27.1
s
Problem 7.79
[Difficulty: 2]
Given:
Model size, model speed, and air temperatures.
Find:
Equivalent speed of the full scale vehicle corresponding to the different air temperatures.
Solution:
Re L
Governing
Equation:
VL
(Reynolds Number)
where V is the air velocity, L is the length of the rocket or model, and , ν is the kinematic viscosity of air.
Subscript m corresponds to the model and r is the rocket.
Assumption: Modeling follows the Reynolds equivalency.
The given or available data is:
Lm LR 12in
VT 100mph
VR 120mph
ft 2
ft 2
(Table A.9)
68 F 1.62 10 4
(Table A.9)
s
s
2
4 ft
150 F 2.09 10
(Table A.9)
s
m2
CO2 8.3 10 6
(Figure A.3 or other source)
s
40 F 1.47 10 4
Determine
Re L
VR LR
R
the
120
Reynolds
Number
for
expected
maximum
speed
mile 5280ft
hr
ft
12in
hr
mile 3600s
12in
2
ft
1.62 10-4
s
at
ambient
Re L 1.09 106
Re-arrange the Reynolds Number Equation for speed equivalents:
Re L
VL
VR VT
LM 68 F
LR T
In this problem, the only term that changes is νT
Solve for speed at the low temperature: VR VT
LM 68 F
LR 40 F
ft 2
12in
s 110mph
100mph
ft 2
12in
1.47 10 4
s
1.62 10 4
temperature:
Solve for speed at the high temperature: VR VT
Solve for CO2:
VR VT
VR @ 40 F 110mph
LM 68 F
LR CO2
LM 68 F
LR 150 F
ft 2
1.62 10
12in
s 77.5mph
100mph
2
ft
12in
2.09 10 4
s
4
ft 2
12in
s
100mph
181mph
2
2
12in
m
ft
8.3 10 6
s 0.305m
1.62 10 4
VR @150 F 77.5mph
VR @ CO2 181mph
Chilling the air to 40°F increases the model speed, but not enough to achieve the target. Heating the air works against the desired
outcome.
This shows that the equivalent speed can be increased by decreasing the kinematic viscosity. An inspection of figure A.3 shows that
cooling air decreases the kinematic viscosity. It also shows that CO2 has a lower kinematic viscosity than air resulting in much higher
model speeds.
Problem 7.78
[Difficulty: 3]
Given:
A scale model of a submarine is to be tested in fresh water under two conditions:
1 - on the surface
2 - far below the surface
Find:
(a) Speed for the model test on the surface
(b) Speed for the model test submerged
(c) Ratio of full-scale drag to model drag
Solution:
On the surface, we need to match Froude numbers:
Vm 24 knot
Thus for 1:50 scale:
Vm
g Lm
Lm
or: Vm Vp
Lp
g Lp
Vp
1
50
ρm Vm Lm
When submerged, we need to match Reynolds numbers:
μm
From Table A.2, SG seawater 1.025 and μseawater 1.08 10
1.025
50
Vm 0.35 knot
0.998
1
1.08 10
1.00 10
ρp Vp Lp
ρp Lp μm
or: Vm Vp
ρm Lm μp
μp
3 N s
at 20oC. Thus for 1:50 scale:
2
m
3
3
FDm
1
2
ρ V A
2 m m m
For surface travel:
FDp
FDm
For submerged travel:
2
FDp
1
2
ρ V Ap
2 p p
2
Vp Ap ρp Vp Lp
Substituting in known values:
FDm ρm Vm
Am
ρm Vm Lm
FDp
1.025
0.998
FDp
m
Vm 9.99
s
Vm 19.41 knot or
Under dynamically similar conditions, the drag coefficients will match:
Solving for the ratio of forces:
m
Vm 1.75
s
Vm 3.39 knot or
ρp
2
24 50 1.29 105
3.39 1
2
0.35
50
0.835
FDm 0.998 19.41
1
1.025
FDp
FDm
1.29 10
FDp
FDm
5
0.835
(on surface)
(submerged)
Problem 7.77
[Difficulty: 3]
Given:
Model of automobile
Find:
Factors for kinematic similarity; Model speed; ratio of protype and model drags; minimum pressure for no cavitation
Solution:
For dynamic similarity
ρm Vm Lm
μm
ρp Vp Lp
ρp Lp μm
Vm Vp
ρm Lm μp
μp
For air (Table A.9) and water (Table A.7) at 68 oF
ρp 0.00234
ρm 1.94
slug
ft
slug
ft
μp 3.79 10
3
3
ft
s
Vm 60 mph
60 mph
Fm
2
2
Hence
Fp
Fm
For Ca = 0.5
ρp Vp Lp
2
2
2
2
ρm Vm Lm
p min p v
0.5
1
2
ρ V
2
From steam tables, for water at 68oF
0.00234
1.94
ρp Vp Lp
2
2
5
1
2.10 10 5
3.79 10 7
ft
Vm 29.4
s
Fp
ρm Vm Lm
ft
5 lbf s
μm 2.10 10
2
ft
88
Then
7 lbf s
2
0.00234
1.94
1
Fp
Fm
slug
for the water tank
p tank 3.25 psi 0.7 1.94
slug
2
2
lbf s
1 ft
s
slug ft 12 in
29.4
ft
29.4
ft
2
ft
This is the minimum allowable pressure in the water tank; we can use it to find the required tank pressure
p min p tank
1.4
2
2
Cp 1.4
p tank p min
ρ V p min 0.7 ρ V
1
2
2
ρ V
2
4
0.270
so
1.94
p min 0.339 psi
2
1
2
p min p v ρ V
4
so we get
p v 0.339 psi
2
88 5
29.4 1
3
ft
3
2
2
lbf s
1 ft
s
slug ft 12 in
p min 3.25 psi
2
p tank 11.4 psi
Problem 7.76
[Difficulty: 4]
Given:
Model the motion of a glacier using glycerine. Assume ice as Newtonian fluid with density of glycerine but one
million times as viscous. In laboratory test the professor reappears in 9.6 hours.
Find:
(a) Dimensionless parameters to characterize the model test results
(b) Time needed for professor to reappear
Solution:
We will use the Buckingham pi-theorem.
1
V
2
Select primary dimensions F, L, t:
3
V
ρ
g
L
M
L
M
2
L t
t
4
5
ρ
g
ρ
L
g
3
t
μ
μ
D
H
L
D
H
L
L
L
L
n = 7 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a b
c
Π1 V ρ g D
Thus:
Summing exponents:
M: a 0
L:
1 3 a b c 0
t:
1 2 b 0
a b
c
Π2 μ ρ g D
Thus:
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 2 b 0
M
3
t
L
L
a
b
L
c
0 0 0
L M L t
2
t
The solution to this system is:
1
1
a0
b
c
2
2
M
M
a
L
Π1
V
g D
b
c
0 0 0
L M L t
3
2
L t
L t
The solution to this system is:
1
3
a 1
b
c
2
2
Π2
μ
3
ρ g D
(This is a gravity-driven
version of Reynolds #)
a b
c
Π3 H ρ g D
L
Summing exponents:
a0
L:
1 3 a b c 0
t:
2 b 0
L
Check using F, L, t dimensions:
t
t
1
L
For dynamic similarity:
2
b0
1
1
L
4
μm
2
L
μp
ρm g m Dm
Lm
Lp
2
L
L
1
L
1
2
Matching the last two terms insures geometric similarity.
SG ice 0.92
From Tables A.1 and A.2:
ρp g p Dp
2
So
t
1
F t L
1
2
2
1
3
L F t
1
3
Dp
H
Π3
D
c 1
Π1 f Π2 Π3 Π4
The functional relationship would be:
Dm
b
L
Π4
D
By inspection we can see that
Therefore:
L
The solution to this system is:
M: a 0
6
a
M
c
0 0 0
3 2 L M L t
L t
Thus:
3
SG glycerine 1.26
2
3
3
μm ρp
1 0.92 8.11 10 5 Since we have geometric similarity, the last two terms
μ ρ
6 1.26
must match for model and prototype:
p m
10
8.11 10
5
Lm 1850 m 8.11 10
5
Matching the first Π term:
Vm
Vp
Dm
Dp
0.00900
Lm 0.1500 m
The time needed to reappear would be:
τ
L
V
Thus:
Lm
Lm
τm
Vm
Vm
τm
Lp
Lp Vm
Lm Lp Vm
1
day
τp
τm
τp 9.6 hr
0.00900
Vp
Lm Vp
5
Vm Lm Vp
24 hr
8.11 10
Solving for the actual time:
τp 44.4 day
Your professor will be back before
the end of the semester!
Problem 7.75
[Difficulty: 2]
Given:
Model of boat
Find:
Model kinematic viscosity for dynamic similarity
Solution:
For dynamic similarity
Vm Lm
νm
Hence from Eq 2
Vm
Vp
Vp Lp
Vm
(1)
g Lm
νp
g Lm
Vp
(2)
g Lp
(from Buckingham Π; the first
is the Reynolds number, the
second the Froude number)
Lm
g Lp
Lp
3
Using this in Eq 1
Vm Lm
Lm Lm
Lm
νm νp
νp
νp
Vp Lp
Lp Lp
Lp
2
3
From Table A.8 at 50 oF
5 ft
νp 1.41 10
2
s
νm 1.41 10
5 ft
2
s
1
10
2
7 ft
νm 4.46 10
2
s
Note that there aren't any fluids in Figure A.3 with viscosities that low!
Problem 7.74
A model test of a 1:4 scale tractor-trailer rig is performed in standard air. The drag force is a function of A, V, ρ,
and μ.
(a) Dimensionless parameters to characterize the model test results
(b) Conditions for dynamic similarity
(c) Drag force on the prototype vehicle based on test results
(d) Power needed to overcome the drag force
We will use the Buckingham pi-theorem.
Given:
Find:
Solution:
1
FD
2
Select primary dimensions F, L, t:
3
FD
A
4
5
L
2
ρ
V
A
M L
t
V
2
ρ
n = 5 parameters
μ
V
ρ
L
M
M
3
L t
t
L
μ
r = 3 dimensions
m = r = 3 repeating parameters
A
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 F ρ V A
Thus:
M L
t
2
Summing exponents:
1 3 a b 2c 0
t:
2 b 0
a
b
c
M
a
L
b
c
2
0 0 0
3 t L M L t
L
a 1
L:
Π2 μ ρ V A
The solution to this system is:
M: 1 a 0
Thus:
M
L t
Summing exponents:
L:
1 3 a b 2 c 0
t:
1 b 0
Check using F, L, t dimensions: F
M
b 2
a
L
3 t
L
b
2
L
Π1
c 1
c
0
a 1
L
4
F t
2
t
2
L
2
b 1
1
L
2
c
4
1
FD
2
ρ V A
0 0
M L t
The solution to this system is:
M: 1 a 0
6
[Difficulty: 3]
1
2
t 1
F t L
1
2
2 L L
L F t
Π2
μ
ρ V A
For dynamic similarity:
We must have geometric and kinematic similarity, and
The Reynolds numbers must match.
FDm
Once dynamic similarity is insured, the drag coefficients must be equal:
1
2
ρ V A
2 m m m
FDp
1
2
2
Vp Ap
0.00237 75
2
So for the prototype: FDp FDm
F
550
lbf
4
Dp
ρm Vm
Am
0.00237 300
ρp
The power requirement would be:
2
ρ V Ap
2 p p
P FDp Vp
P 550 lbf 75
ft
s
hp s
550 ft lbf
FDp 550 lbf
P 75.0 hp
P 55.9 kW
Problem 7.73
[Difficulty: 2]
Given:
Model of flying insect
Find:
Wind tunnel speed and wing frequency; select a better model fluid
Solution:
For dynamic similarity the following dimensionless groups must be the same in the insect and model (these are
Reynolds number and Strouhal number, and can be obtained from a Buckingham Π analysis)
Vinsect Linsect
νair
Vm Lm
ωinsect Linsect
νm
Vinsect
From Table A.9 (68oF)
kg
ρair 1.21
3
m
νair 1.50 10
The given data is
ωinsect 60 Hz
m
Vinsect 1.5
s
ωm Lm
Vm
2
5 m
s
Linsect
Lm
1
8
Linsect νm
Linsect
m 1
m
Hence in the wind tunnel Vm Vinsect
Vinsect
Vm 1.5
Vm 0.1875
Lm νair
Lm
s
8
s
Vm Linsect
0.1875 1
Also
ωm ωinsect
ωm 60 Hz
ωm 0.9375 Hz
Vinsect Lm
1.5
8
It is unlikely measurable wing lift can be measured at such a low wing frequency (unless the measured lift was averaged, using an
integrator circuit, or perhaps a load cell and data acquisition system). Maybe try hot air (100 oC) for the model
2
5 m
νhot 2.29 10
For hot air try
Hence
Also
Vinsect Linsect
νair
Vm Lm
s
Linsect νhot
Vm Vinsect
Lm νair
νhot
Vm Linsect
ωm ωinsect
Vinsect Lm
ωm 60 Hz
Hot air does not improve things much. Try modeling in water
Hence
Vinsect Linsect
νair
Vm Lm
νw
νair 1.50 10
instead of
0.286
1.5
m
s
1
Vm 1.5
s
8
5
2.29 10
5
1.50 10
1
2
6 m
s
6
m 1 1.01 10
m
Vm 1.5
Vm 0.01262
5
s
8
s
1.50 10
Vm Linsect
Vm
0.01262
1
ωm ωinsect
ωinsect
Lratio
ωm 60 Hz
Vinsect Lm
Vinsect
1.5
8
This is even worse! It seems the best bet is hot (very hot) air for the wind tunnel. Alternatively, choose a much
smaller wind tunnel model, e.g., a 2.5 X model would lead to V m = 0.6 m/s and ωm = 9.6 Hz
Also
m
Vm 0.286
s
ωm 1.43 Hz
8
νw 1.01 10
Linsect νw
Vm Vinsect
Lm νair
2
5 m
ωm 0.0631 Hz
Problem 7.72
Given:
Flow around cruise ship smoke stack
Find:
Range of wind tunnel speeds
[Difficulty: 2]
Solution:
For dynamic similarity
Vm Dm
νm
Since
1 knot 1
Hence for
nmi
hr
and
Vp Dp
or
νp
Dp
15
Vm
Vp
V 15 Vp
Dm
1 p
1 nmi 6076.1 ft
nmi 6076.1 ft
hr
Vp 12
hr
nmi
3600 s
ft
Vp 20.254
s
ft
Vm 15 20.254
s
ft
Vm 304
s
nmi 6076.1 ft
hr
Vp 24
hr
nmi
3600 s
ft
Vp 40.507
s
ft
Vm 15 40.507
s
ft
Vm 608
s
Note that these speeds are very high - compressibility effects may become important, since the Mach number is
no longer much less than 1!
Problem 7.71
[Difficulty: 3]
Given:
Find:
1/8-scale model of a tractor-trailer rig was tested in a pressurized wind tunnel.
Solution:
We will use definitions of the drag coefficient and Reynolds number.
(a) Aerodynamic drag coefficient for the model
(b) Compare the Reynolds numbers for the model and the prototype vehicle at 55 mph
(c) Calculate aerodynamic drag on the prototype at a speed of 55 mph into a headwind of 10 mph
Governing
Equations:
CD
FD
1
2
Re
(Drag Coefficient)
2
ρ V A
ρ V L
(Reynolds Number)
μ
Am Wm Hm
Assume that the frontal area for the model is:
3
The drag coefficient would then be:
From the definition of Re:
Rem
Rep
Rem
Rep
3.23
1.23
75
m
s
hr
55 mi
CDm 2 128 N
ρm Vm Lm μp
ρp Vp Lp μm
mi
5280 ft
ft
0.3048 m
2
Am 0.305 m 0.476 m
m
3.23 kg
Am 0.1452 m
2
1
kg m
s
2
2
75.0 m
0.1452 m
N s
CDm 0.0970
Assuming standard conditions and equal viscosities:
3600 s
1
11
hr 8
Rem Rep
Since the Reynolds numbers match, assuming geometric and kinetic similarity we can say that the drag coefficients are equal:
1
2
FDp CD ρp Vp Ap
2
Susbstituting known values yields:
2
2
1
kg
mi 5280 ft 0.3048 m
hr
2
2 N s
FDp
0.0970 1.23
( 55 10)
0.1452 m 8
3
2
hr
ft
3600 s
mi
kg m
m
FDp 468 N
Problem 7.70
[Difficulty: 3]
Given:
The frequency of vortex shedding from the rear of a bluff cylinder is a function of ρ, d, V, and μ. Vortex shedding
occurs in standard air on two cylinders with a diameter ratio of 2.
Find:
(a) Functional relationship for f using dimensional analysis
(b) Velocity ratio for vortex shedding
(c) Frequency ratio for vortex shedding
Solution:
We will use the Buckingham pi-theorem.
1
f
2
Select primary dimensions F, L, t:
3
f
ρ
1
M
t
4
5
d
ρ
ρ
L
V
d
3
L
V
V
n = 5 parameters
μ
μ
L
M
t
L t
r = 3 dimensions
m = r = 3 repeating parameters
d
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b c
Π1 f ρ V d
Thus:
Summing exponents:
1
a
L
The solution to this system is:
a0
L:
3 a b c 0
t:
1 b 0
b c
Π2 μ ρ V d
b
c
0 0 0
L M L t
3
t
t
L
M: a 0
a
M
Thus:
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 b 0
M
M
b 1
a
L
b
c
0 0 0
L M L t
3
t
L t
L
The solution to this system is:
a 1
b 1
4
6
Check using F, L, t dimensions:
c1
f d
Π1
V
c 1
t 1
1 t
F t L
L 1
1
2
2 L L
t L
L F t
μ
Π2
ρ V d
The functional relationship is:
f d
Π1 f Π2
V
f
ρ V d
μ
To achieve dynamic similarity between geometrically similar flows, we must duplicate all but one of the dimensionless groups:
ρ1 V1 d 1
μ1
Now if
ρ2 V2 d 2
V1
μ2
V2
ρ1 V1 d 1
μ1
ρ2 V2 d 2
μ2
ρ2 d 2 μ1
1
1 1
ρ1 d 1 μ2
2
it follows that:
f1 d 1
V1
f2 d 2
V2
V1
V2
and
f1
f2
d 2 V1
1 1
d 1 V2
2 2
f1
f2
1
2
1
4
Problem 7.69
[Difficulty: 2]
Given:
Oil flow in pipe and dynamically similar water flow
Find:
Average water speed and pressure drop
Solution:
From Example 7.2
Δp
2
ρ V
f
l e
ρ V D D D
μ
μH2O
ρH2O VH2O DH2O
From Fig. A.3 at 77 oF
νOil 10.8 8 10
From Table A.8 at 60 oF
μOil
For dynamic similarity
5 ft
Hence
VH2O
Then
ΔpOil
ρOil VOil
From Table A.2
2
2
s
2
2
3
ft
s
ft
VH2O 0.0420
s
s
2
ΔpH2O
2
s
4 ft
8.64 10
4 ft
8.64 10
μH2O ρOil
νH2O
VH2O
Voil
V
ρH2O μOil
νOil Oil
s
5 ft
1.21 10
2
s
5 ft
νH2O 1.21 10
so
ρOil VOil DOil
ΔpH2O
2
ρH2O VH2O
ρH2O VH2O
ρOil VOil
2
ΔpOil
SG Oil 0.92
ΔpH2O
1
0.92
2
0.0420 7 psi
3
3
ΔpH2O 1.49 10
psi
Problem 7.68
[Difficulty: 3]
Given:
A 1:20 model of a hydrofoil is to be tested in water at 130 deg F. The prototype operates at a speed of 60 knots
in water at 45 deg F. To model the cavitation, the cavitation number must be duplicated.
Find:
Ambient pressure at which the test must be run
Solution:
To duplicate the Froude number between the model and the prototype requires:
Lm
Vm Vp
Lp
1
Vm 60 knot
20
p m p vm
1
2
Vm
p m p vm p p p vp
ρp Vp
2
g Lm
Vp
g Lp
Thus:
Vm 13.42 knot
To match the cavitation number between the model and the prototype:
ρm
Vm
2
ρm Vm
p p p vp
1
2
ρp Vp
Therefore:
2
Vm
Assuming that the densities are equal: p m p vm p p p vp
Vp
From table A.7: at 130 deg F p vm 2.23 psi
p m 2.23 psi ( 14.7 psi 0.15 psi)
13.42
60
at 45 deg F
p vp 0.15 psi
Thus the model pressure is:
2
p m 2.96 psi
2
Problem 7.67
Given:
Model of Frisbee
Find:
Dimensionless parameters; Prototype speed and angular speed
[Difficulty: 2]
Solution:
Assumption: Geometric, kinematic, and dynamic similarity between model and prototype.
F F( D V ω h ρ μ)
The functional dependence is
From Buckingham Π
F
2
2
ρ V D
For dynamic similarity
where F represents lift or drag
ρ V D ω D h
V D
μ
f
ρm Vm Dm
μm
ρp Vp Dp
μp
ρm Dm μp
Vp Vm
ρp Dp μm
ft
Vp 140 ( 1 )
s
1 ( 1)
7
ft
Vp 20
s
Also
ωm Dm
Vm
ωp Dp
Vp
Dm Vp
ωp ωm
Dp Vm
ωp 5000 rpm
1
7
20
140
ωp 102 rpm
Problem 7.66
[Difficulty: 3]
Given:
Model of water pump
Find:
Model flow rate for dynamic similarity (ignoring Re); Power of prototype
Solution:
Q
From Buckingham Π
Hence
3
Qp
ωm Dm
ωp Dp
3
3
Dm
Qm Qp
ωp Dp
ωm
3
Then
where Q is flow rate, ω is angular speed, d
is diameter, and ρ is density (these Π
groups will be discussed in Chapter 10)
5
ρ ω D
ft
Qm 15
s
From Table A.8 at 68 oF
3
ω D
Qm
For dynamic similarity
P
and
3
ρp 1.94
2400
750
slug
ft
3
5
ρm ωm Dm
3
ft
Qm 0.750
s
ρm 0.00234
slug
ft
3
Pp
3
ρp ωp Dp
3
5
ωp Dp
Pp Pm
ρm ωm
Dm
ρp
3
From Table A.9 at 68 oF
3
Pm
1
4
1.94
Pp 0.1 hp
0.00234
5
3
750 4
2400 1
5
3
Pp 2.59 10 hp
Note that if we had used water instead of air as the working fluid for the model pump, it would have drawn 83 hp. Water would have
been an acceptable working fluid for the model, and there would have been less discrepancy in the Reynolds number.
Problem 7.65
[Difficulty: 3]
The fluid dynamic charachteristics of a gold ball are the be tested using a model in a wind tunnel. The
dependent variables are the drag and lift forces. Independent variables include the angular speed and dimple
depth. A pro golfer can hit a ball at a speed of 75 m/s and 8100 rpm. Wind tunnel maximum speed is 25 m/s.
(a) Suitable dimensionless parameters and express the functional dependence between them.
(b) Required diameter of model
(c) Required rotational speed of model
Given:
Find:
Solution:
Assumption:
Wind tunnel is at standard conditions
FD FD( D V ω d ρ μ) FL FL( D V ω d ρ μ) n = 7 and m = r = 3, so
from the Buckingham pi theorem, we expect two sets of four Π terms. The application of the Buckingham pi theorem will not be
shown here, but the functional dependences would be:
FD
FL
ρ V D ω D d
ρ V D ω D d
f
g
2 2
2 2
μ
μ
V D
V D
ρ V D
ρ V D
The problem may be stated as:
To determine the required model diameter, we match Reynolds numbers between the model and prototype flows:
ρm Vm Dm
μm
ρp Vp Dp
μp
ρp Vp μm
Thus: Dm Dp
Substituting known values:
ρm Vm μp
75
Dm 4.27 cm 1
1
25
Dm 12.81 cm
To determine the required angular speed of the model, we match the dimensionless rotational speed between the flows:
ωm Dm
Vm
ωp Dp
Vp
Dp Vm
Thus: ωm ωp
Dm Vp
4.27
25
Substituting known values: ωm 8100 rpm
12.81
75
ωm 900 rpm
Problem 7.64
[Difficulty: 2]
Given:
Model of wing
Find:
Model test speed for dynamic similarity; ratio of model to prototype forces
Solution:
We would expect
From Buckingham Π
F F( l s V ρ μ)
F
2
ρ V l s
For dynamic similarity
where F is the force (lift or drag), l is the chord and s the span
ρ V l l
μ s
f
ρm Vm lm
μm
ρp Vp lp
μp
Hence
ρp lp μm
Vm Vp
ρm lm μp
From Table A.8 at 20 oC
μm 1.01 10
3 N s
From Table A.10 at 20oC
2
m
1.21 kg
3
m
m
Vm 7.5
kg
s
998 3
m
Then
Fm
2
ρm Vm lm sm
1.01 10 3 N s
2
m
10
5 N s
1
1.81 10 2
m
Fp
2
ρp Vp lp sp
2
μp 1.81 10
2
m
m
Vm 5.07
s
ρm Vm lm sm
998
Fp
ρp
2 l p sp
1.21
Vp
Fm
5 N s
2
5.07 1 1 3.77
10 10
7.5
Problem 7.63
[Difficulty: 2]
Given:
Model of weather balloon
Find:
Model test speed; drag force expected on full-scale balloon
Solution:
From Buckingham Π
F
2
2
ρ V D
f
ν
V
F( Re M)
V D c
For similarity
Rep Rem
Hence
Rep
Mp Mm
and
Vp Dp
νp
Rem
(Mach number criterion
satisified because M<<1)
Vm Dm
νm
νm Dp
Vm Vp
νp Dm
From Table A.7 at 68 oF
νm 1.08 10
5 ft
2
From Table A.9 at 68 oF
s
ft
1.08 10 5
ft
s
Vm 5
2
s
ft
1.62 10 4
s
4 ft
νp 1.62 10
2
s
2
Then
Fm
2
2
ρm Vm Dm
10 ft
1
ft
6
ft
Vm 20.0
s
2
Fp
2
ρp Vp Dp
2
0.00234 slug
3
ft
Fp 0.85 lbf
slug
1.94
3
ft
2
ρp Vp Dp
Fp Fm
ρm.
2
2
Vm Dm
2
5 ft
s 10 ft 2
ft
20 1 ft
s 6
Fp 0.231 lbf
Problem 7.62
[Difficulty: 3]
Given:
A 1/10 scale airfoil was tested in a wind tunnel at known test conditions. Prototype airfoil has a chord length of
6 ft and is to be flown at standard conditions.
Find:
(a) Reynolds number at which the model was tested
(b) Corresponding prototype speed
Solution:
Assumptions: (a) The viscosity of air does not vary appreciably between 1 and 5 atmospheres
(b) Geometric, kinematic, and dynamic similarity applies
F f ( ρ V L μ)
The problem may be stated as:
F
2
ρ V L
2
g ( Re)
where
Re
ρ V L
From the Buckingham pi theorem, we expect 2 Π terms:
Lm
The model chord length is
μ
At 59 deg F:
2116 lbf
atm ft
2
lbm R
53.33 ft lbf
μm 3.74 10
7 lbf s
ft
2
1
519 R
slug
32.2 lbm
ρm 0.0119
Rem 0.0119
Therefore:
1.20 ft
5
pm
Substituting known values:
ρm
R Tm
We can calculate the model flow density from the ideal gas equation of state:
ρm 5 atm
6 ft
slug
ft
3
slug
ft
130
3
ft
s
1.2 ft
ft
2
3.74 10
2
7
lbf s
lbf s
slug ft
6
Rem 5.0 10
Matching Reynolds numbers between the model and prototype flows:
ρm Vm Lm
μm
From the ideal gas equation of state:
ρm
ρp
ft
5
519
1
Vp 130
s
1 519
5
ρp Vp Lp
μp
ρm Lm μp
Thus: Vp Vm
ρp Lp μm
p m Tp Lm μp
p m Tp
Therefore: Vp Vm
So substituting in values yields:
p p Tm Lp μm
p p Tm
7
3.74 10
7
3.74 10
ft
Vp 130.0
s
Problem 7.61
[Difficulty: 3]
Given:
A torpedo with D = 533 mm and L = 6.7 m is to travel at 28 m/s in water. A 1/5 scale model of the torpedo is to be
tested in a wind tunnel. The maximum speed in the tunnel is fixed at 110 m/s, but the pressure can be varied at a
constant temperature of 20 deg C.
Find:
(a) Minimum pressure required in the wind tunnel for dynamically similar testing.
(b) The expected drag on the prototype if the model drag is 618 N.
Solution:
The problem may be stated as:
F
2
2
g ( Re)
F f ( ρ V D μ) From the Buckingham pi theorem, we expect 2 Π terms:
ρ V D
Re
where
μ
ρ V D
ρm Vm Dm
Matching Reynolds numbers between the model and prototype flows:
μm
At 20 deg C:
3 N s
μp 1.00 10
and
2
m
μm 1.81 10
5 N s
ρp Vp Dp
μp
Vp Dp μm
Thus: ρm ρp
Vm Dm μp
So substituting in values yields:
2
m
5
kg
28
5
1.81 10
kg
ρm 998
ρm 23.0
3
3
3
110 1
m
m
1.00 10
Substituting in values:
p m 23.0
kg
3
m
If the conditions are dynamically similar:
287
N m
kg K
p m ρm R Tm
2
293 K
Fm
2
2
ρm Vm Dm
Substituting in known values:
From the ideal gas equation of state:
998
Fp 618 N
23.0
Pa m
p m 1.934 MPa
N
Fp
2
ρp Vp Dp
2
28 5
110 1
2
Thus:
Vp
Fp Fm
ρm Vm
ρp
2
Dp
Dm
2
2
Fp 43.4 kN
Problem 7.60
[Difficulty: 2]
Given:
Flow around ship's propeller
Find:
Model propeller speed using Froude number and Reynolds number
Solution:
Basic equations:
Fr
V
Re
g L
V L
ν
Assumptions: (a) The model and the actual propeller are geometrically similar
(b) The flows about the propellers are kinematically and dynamically similar
Using the Froude number
But the angular velocity is given by
Comparing Eqs. 1 and 2
The model rotation speed is then
Using the Reynolds number
Vm
Fr m
g Lm
Fr p
Vp
g Lp
V L ω
or
Vp
so
Vm
Vp
Lm ωm
Lp ωp
Lm
ωm
Lp
ωp
Lp
ωm ωp
Lm
Rem
Vm
Vm Lm
νm
Lm
(1)
Lp
Lm ωm
Lp ωp
Lp
Lm
ωm 100 rpm
Rep
Vp Lp
or
νp
Vm
Vp
(2)
9
ωm 300 rpm
1
Lp νm
Lp
Lm νp
Lm
(3)
(We have assumed the viscosities of the sea water and model water are comparable)
Comparing Eqs. 2 and 3
Lm ωm
Lp
Lp ωp
Lm
The model rotation speed is then
Lp
ωm ωp
Lm
Lp
ωp
Lm
ωm
2
2
ωm 100 rpm
9
1
2
Of the two models, the Froude number appears most realistic; at 8100 rpm serious cavitation will occur, which would
invalidate the similarity assumptions. Both flows will likely have high Reynolds numbers so that the flow becomes
independent of Reynolds number; the Froude number is likely to be a good indicator of static pressure to dynamic
pressure for this (although cavitation number would be better).
ωm 8100 rpm
Problem 7.59
[Difficulty: 3]
Given:
Find:
Measurements of drag are made on a model car in a fresh water tank. The model is 1/5-scale.
Solution:
The flows must be geometrically and kinematically similar, and have equal Reynolds numbers to be dynamically
similar:
(a) Conditions requred to ensure dynamic similarity between the model and the prototype.
(b) Required fraction of speed in air at which the model needs to be tested in water to ensure dynamically similar
conditions.
(c) Drag force on the prototype model traveling at 90 kph in air if the model drag is 182 N traveling at 4 m/s in
water.
Geometric similarity requires a true model in all respects.
Kinematic similarity requires the same flow pattern, i.e., no free-surface or cavitation effects.
The problem may be stated as F = f(ρ,V,L,μ)
F
Dimensional analysis gives this relation:
2
ρ V L
2
g ( Re)
Vm Lm
Matching Reynolds numbers between the model and prototype
flows:
νw 1.00 10
From Tables A.8 and A.10 at 20 deg C:
Vm
Vp
6
1.00 10
5
1.51 10
5
1
νm
2
6 m
Vp Lp
μ
Thus:
νp
νa 1.51 10
s
Vm
Vp
V L
ν
νm Lp
νp Lm
Therefore:
Vm
0.331
If the conditions are dynamically similar:
Vp
Fm
2
2
ρm Vm Lm
Fp
2
ρp Vp Lp
2
Thus:
2
Vp Lp
Fp Fm
ρm Vm
Lm
ρp
2
Substituting in known values:
ρ V L
Re
2
5 m
and
s
where
1.20
km 1000 m
hr
s
Fp 182 N
90
999
km
3600 s 4 m
hr
5
1
0.331
2
2
Fp 213 N
Problem 7.58
Given:
[Difficulty: 5]
Vessel to be powered by a rotating circular cylinder. Model tests are planned to determine the required power
for the prototype.
(a) List of parameters that should be included in the analysis
(b) Perform dimensional analysis to identify the important dimensionless groups
Find:
From an inspection of the physical problem: P f ( ρ μ V ω D H)
Solution:
We will now use the Buckingham pi-theorem to find the dimensionless groups.
1
P
2
Select primary dimensions M, L, t:
3
P
ρ
ρ
M L
t
4
5
μ
2
3
ρ
V
ω
μ
V
ω
M
M
L
1
3
L t
t
t
L
ω
D
H
D
H
L
L
n = 7 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 P ρ ω D
Thus:
M L
t
Summing exponents:
2 3 a c 0
t:
3 b 0
a
b
c
Π2 μ ρ ω D
Thus:
M
L t
t:
1 b 0
a
b
c
Π3 V ρ ω D
Summing exponents:
M: a 0
L:
1 3 a c 0
t:
1 b 0
a
1
b
c
0 0 0
3 t L M L t
L
b 3
M
a
1
c 5
L
P
3
5
ρ ω D
c
0 0 0
3 t L M L t
L
a 1
Thus:
Π1
b
The solution to this system is:
M: 1 a 0
1 3 a c 0
M
a 1
Summing exponents:
L:
3
The solution to this system is:
M: 1 a 0
L:
2
M
b 1
a
1
c 2
Π2
μ
b
c
0 0 0
L M L t
3
t
t
L
The solution to this system is:
a0
b 1
c 1
2
ρ ω D
V
Π3
ω D
a
b
c
Π4 H ρ ω D
Thus:
L
t:
b 0
1
a0
b0
4
6
Check using F, L, t dimensions:
The functional relationship is:
b
The solution to this system is:
M: a 0
1 3 a c 0
a
c
0 0 0
3 t L M L t
L
Summing exponents:
L:
M
c 1
4
1
F L L
F t L
3 1
t
1
t
1
2
5
2
2
2
t
L
L
L F t
F t
Π1 f Π2 Π3 Π4
H
Π4
D
L
t
t
1
L
1
L
1
1
L
P
3
5
ρ ω D
f
μ
V
H
2 ω D D
ρ ω D
Problem 7.57
[Difficulty: 2]
Given:
A model is to be subjected to the same Reynolds number in air flow and water flow
Find:
(a) Which flow will require the higher flow speed
(b) How much higher the flow speed needs to be
Solution:
For dynamic similarity:
ρw Vw Lw
μw
From Tables A.8 and A.10 at 20 deg C: νw 1.00 10
Va
Vw
1.51 10
1.00 10
ρa Va La
μa
2
6 m
s
and
We know that Lw La
2
5 m
νa 1.51 10
Thus:
Va
Vw
ρw μa
νa
ρa μw
νw
Therefore:
s
5
6
15.1
Air speed must be higher than
water speed.
To match Reynolds number:
Va 15.1 Vw
Problem 7.56
Given:
[Difficulty: 3]
Find:
Airship is to operate at 20 m/s in air at standard conditions. A 1/20 scale model is to be tested in a wind tunnel at
the same temperature to determine drag.
(a) Criterion needed to obtain dynamic similarity
(b) Air pressure required if air speed in wind tunnel is 75 m/s
(c) Prototype drag if the drag on the model is 250 N
Solution:
Dimensional analysis predicts:
F
2
ρ V L
2
f
ρ V L
Therefore, for dynamic similarity, it would follow that:
μ
ρm Vm Lm
μm
ρp Vp Lp
μp
Since the tests are performed at the same temperature, the viscosities are the same. Solving for the ratio of densities:
ρm
ρp
Vp Lp μm
20
p
Thus:
20 1 5.333 Now from the ideal gas equation of state: ρ
Vm Lm μp
75
R T
ρm Tp
pm pp
ρp Tm
5
p m 101 kPa 5.333 1
Fp
From the force ratios:
2
ρp Vp Lp
Substituting known values:
2
p m 5.39 10 Pa
Fm
2
Thus:
2
ρm Vm Lm
2
2
20
2
Fp 250 N
( 20)
5.333 75
1
2
Vp Lp
Fp Fm
ρm Vm
Lm
ρp
Fp 1.333 kN
Problem 7.55
[Difficulty: 3]
Given: Model scale for on balloon
Find: Required water model water speed; drag on protype based on model drag
Solution:
From Appendix A (inc. Fig. A.2)
The given data is
For dynamic similarity we assume
μair 1.8 10
m
Vair 5
s
Lratio 20
ρw Vw Lw
μw
Then
5 Ns
kg
ρair 1.24
3
m
2
m
3 Ns
kg
ρw 999
3
m
μw 10
ρair Vair Lair
μair
μw ρair Lair
μw ρair
m
Vw Vair
Vair
Lratio 5
μair ρw Lw
μair ρw
s
10 3
1.8 10 5
1.24 20
999
m
Vw 6.90
s
Fair
1
2
ρair Aair Vair
2
Fw
1
2
2
ρw Aw Vw
2
Lair
2
Lratio
Aw
Lw
Aair
Hence the prototype drag is
2
Vair
Fair Fw
Lratio
2000 N
ρw
Vw
ρair
2
1.24 202 5
999
6.9
2
m
Fw 2 kN
For the same Reynolds numbers, the drag coefficients will be the same so we have
where
2
Fair 522 N
Problem 7.54
Functional relationship between the maximum pressure experienced in a water hammer wave and other physical
parameters
(a) The number of Π terms that characterize this phenomenon
(b) The functional relationship between the Π terms
Given:
Find:
We will use the Buckingham pi-theorem.
Solution:
1
p max
ρ
2
U0
EV
Select primary dimensions M, L, t:
3
p max
ρ
U0
M
L
3
t
M
L t
4
5
ρ
2
L
n = 4 parameters
EV
M
L t
2
r = 3 dimensions
m = 2 repeating parameters because p max and Ev have the same dimensions.
We have n - m = 2 dimensionless groups.
U0
Setting up dimensional equations:
a
Π1 p max ρ U0
b
M
2
1 3 a b 0
t:
2 b 0
a
Π2 Ev ρ U0
b
M
t:
2 b 0
L
b
Check using F, L, t dimensions:
2
M
a
L
p max
ρ U0
2
b
The solution to this system is:
a 1
F
L
The functional relationship is:
b 2
Π1
0 0 0
3 t M L t
L t L
Thus:
M: 1 a 0
1 3 a b 0
a
a 1
Summing exponents:
L:
M
The solution to this system is:
M: 1 a 0
L:
0 0 0
3 t M L t
L t L
Thus:
Summing exponents:
6
[Difficulty: 4]
2
L
4
F t
2
Π1 f Π2
t
b 2
2
L
2
1
Thus:
F
L
2
L
4
F t
2
t
Π2
Ev
ρ U0
2
2
L
2
1
p max
ρ U0
2
Ev
ρ U 2
0
f
Problem 7.53
[Difficulty: 2]
Given:
Boundary layer profile
Find:
Two groups by inspection; One that is a standard fluid mechanics group; Dimensionless groups
Solution:
Two obvious groups are u/U and y/. A dimensionless group common in fluid mechanics is U (Reynolds number)
Apply the Buckingham procedure
u
y
U
dU/dx
n = 6 parameters
Select primary dimensions M, L, t
u
L
t
U
y U
L
L
t
dU dx
2
1
t
L
t
L
m = r = 3 primary dimensions
m = r = 2 repeat parameters
Then n – m = 4 dimensionless groups will result. We can easily do these by inspection
1
u
U
2
y
3
dU
dy
U
Check using F, L, t as primary dimensions, is not really needed here
Note: Any combination of ’s can be used; they are not unique!
4
U
Problem 7.52
[Difficulty: 3]
Given:
Find:
Functional relationship between the power to drive a marine propeller and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
(a) The number of Π terms that characterize this phenomenon
(b) The Π terms
1
P
2
Select primary dimensions F, L, t:
3
P
t
5
D
ρ
M L
4
D
ρ
2
M
3
ρ
L
V
L
3
V
c
ω
μ
V
c
ω
μ
L
L
1
M
t
t
t
L t
n = 7 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 P ρ V D
Thus:
M L
t
Summing exponents:
2 3 a b c 0
t:
3 b 0
c
Π2 c ρ V D
Thus:
1 3 a b c 0
t:
1 b 0
Summing exponents:
M: a 0
L:
3 a b c 0
t:
1 b 0
b
a
c 2
Thus:
Π1
P
3
1
M
b 1
a
L
c0
c
Π2
V
b
c
0 0 0
L M L t
3
t
t
L
The solution to this system is:
a0
b 1
c1
2
ρ V D
b
The solution to this system is:
L:
c
L
b 3
a0
b
L
M
c
0 0 0
L M L t
3
t
t
L
M: a 0
Π3 ω ρ V D
a
c
0 0 0
3 t L M L t
L
L
Summing exponents:
a
M
a 1
L:
b
The solution to this system is:
M: 1 a 0
a
3
2
ω D
Π3
V
a
b
c
Π4 μ ρ V D
Thus:
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 b 0
M
L t
M
Check using F, L, t dimensions:
L
b
c
0 0 0
3 t L M L t
L
The solution to this system is:
a 1
4
6
a
b 1
3
t 1
F L L
1
2 3 2
t
F t L L
μ
Π4
ρ V D
c 1
L t
1
t L
1
t
L
t
L
4
1
t 1
F t L
1
2
2 L L
L F t
Problem 7.51
[Difficulty: 3]
Given:
That the cooling rate depends on rice properties and air properties
Find:
The groups
Solution:
Apply the Buckingham procedure
dT/dt
c
k
L
cp
V
n = 8 parameters
Select primary dimensions M, L, t and T (temperature)
dT dt
c
k
L
T
t
L2
ML
t 2T
t 2T
cp
V
L2
M
t 2T
L3
M
Lt
L
t
r = 4 primary dimensions
V
L
L
cp
m = r = 4 repeat parameters
Then n – m = 4 dimensionless groups will result. By inspection, one group is c/cp. Setting up a dimensional equation,
d
2
dT L M
c L T
1 V L c
3 L 2
T 0 M 0 L0t 0
dt t L
t T t
a
a
b
b c d
p
Summing exponents,
T:
d 1 0
d 1
M:
b0
b0
L:
t:
Hence
1
a 3b c 2d 0 a c 2 c 1
a 2d 1 0
a 3
dT Lc p
dt V 3
By a similar process, we find
2
k
L2 c p
and
3
LV
Hence
c
dT Lc p
, k ,
f
c p L2 c LV
dt V 3
p
Problem 7.50
Functional relationship between the thrust of a marine propeller and other physical parameters
Given:
Find:
Solution:
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
FT
2
Select primary dimensions F, L, t:
3
FT
ρ
M L
M
t
4
5
D
ρ
2
ρ
[Difficulty: 3]
L
V
D
L
3
V
g
ω
p
μ
V
g
ω
p
μ
L
L
1
2
t
t
t
M
L t
n = 8 parameters
M
L t
2
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 5 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 FT ρ V D
Thus:
M L
t
Summing exponents:
1 3 a b c 0
t:
2 b 0
c
Π2 g ρ V D
Thus:
L:
1 3 a b c 0
t:
2 b 0
c
Summing exponents:
M: a 0
L:
3 a b c 0
t:
1 b 0
b
b 2
a
a0
b
L
c 2
Thus:
Π1
FT
2
1
M
b 2
a
L
c1
Π2
g D
2
V
b
c
0 0 0
L M L t
3
t
t
L
The solution to this system is:
a0
b 1
c1
2
ρ V D
b
The solution to this system is:
M: a 0
a
L
M
c
0 0 0
L M L t
2
3
t
t L
L
Summing exponents:
Π3 ω ρ V D
a
c
0 0 0
3 t L M L t
L
a 1
L:
b
M
The solution to this system is:
M: 1 a 0
a
2
ω D
Π3
V
a
b
c
Π4 p ρ V D
a 1
L:
1 3 a b c 0
t:
2 b 0
b
c
M
b
L
1 3 a b c 0
t:
1 b 0
Check using F, L, t dimensions: F
M
a
L
c0
p
2
ρ V
b
The solution to this system is:
M: 1 a 0
L:
b 2
Π4
c
0 0 0
L M L t
3
t
L t
L
Thus:
Summing exponents:
6
The solution to this system is:
M: 1 a 0
a
a
M
c
0 0 0
3 t L M L t
L t L
2
Summing exponents:
Π5 μ ρ V D
M
Thus:
a 1
L
4
F t
2
t
2
L
2
1
L
2
b 1
1
L
t
2
μ
Π5
ρ V D
c 1
L
t
2
L
2
1
1
t
L
t
L
1
F
L
2
L
4
F t
2
t
2
L
2
4
1
t 1
F t L
1
2
2 L L
L F t
Problem 7.49
Functional relationship between the heat transfer rate in a convection oven and other physical parameters
Given:
Find:
Solution:
The number of Π terms that characterize this phenomenon and the Π terms
We will use the Buckingham pi-theorem.
1
Q
2
Select primary dimensions F, L, t, T (temperature):
3
Q
cp
L
t
5
ρ
Θ
cp
F L
4
[Difficulty: 4]
L
Θ
L
T
L
2
2
t T
V
L
ρ
ρ
F t
L
2
4
μ
V
F t
L
2
t
L
b
c
Π1 Q ρ V L Θ
d
Thus:
a
F t2 L b c d 0 0 0 0
L T F L t T
t 4 t
L
The solution to this system is:
F:
1a0
L:
1 4 a b c 0
t:
1 2 a b 0
T:
d0
b
a 1
c
Π2 cp ρ V L Θ
We have n - m = 3 dimensionless
groups.
F L
Summing exponents:
a
r = 4 dimensions
m = r = 4 repeating parameters
Θ
Setting up dimensional equations:
a
n = 7 parameters
V
μ
d
Thus:
Summing exponents:
b 3
c 2 d 0
a0
L:
2 4 a b c 0
t:
2 2 a b 0
T:
1 d 0
Q
3
ρ V L
a
F t2 L b c d 0 0 0 0
L T F L t T
2
4 t
t T L
L
2
The solution to this system is:
F:
Π1
a0
b 2
c0
d1
Π2
cp Θ
2
V
a
a
b
c
Π3 μ ρ V L Θ
d
Thus:
Summing exponents:
F:
1a0
L:
2 4 a b c 0
t:
1 2 a b 0
T:
d0
F t2 L b c d 0 0 0 0
L T F L t T
2 4 t
L L
F t
The solution to this system is:
a 1
b 1
c 1 d 0
μ
Π3
ρ V L
2
6
Check using M, L, t, T dimensions:
M L
t
The functional relationship is:
3
2
L
3
M
Π1 f Π2 Π3
t
2
L
2
1
L
3
1
L
2
2
t
2
t T L
Q
3
ρ V L
2
2
3
T 1
M L t 1
1
L t M L L
cp Θ μ
V2 ρ V L
f
cp Θ μ
V2 ρ V L
Q ρ V L f
3
2
Problem 7.48
Functional relationship between the power loss in a journal bearing and other physical parameters
Given:
Find:
The Π terms that characterize this phenomenon and the function form of the dependence of P on these
parameters
We will use the Buckingham pi-theorem.
Solution:
1
P
2
Select primary dimensions F, L, t:
3
P
l
F L
t
4
5
[Difficulty: 3]
D
D
c
l
D
c
L
L
L
μ
p
ω
μ
p
1
F t
t
L
n = 7 parameters
F
2
L
2
r = 3 dimensions
m = r = 3 repeating parameters
p
ω
ω
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b c
Π1 P D ω p
Thus:
Summing exponents:
F:
1c0
L:
1 a 2 c 0
t:
1 b 0
a
b c
Π2 l D ω p
c0
L:
1 a 2 c 0
t:
b 0
a
t
a
1
a 3
b c
Summing exponents:
F:
c0
L:
1 a 2 c 0
t:
b 0
c
Thus:
1
L L
t
a
b 1
b
c 1
L L
a
1
P
3
D ω p
c
The solution to this system is:
Thus:
Π1
F
0 0 0
F L t
2
L
a 1
Π3 c D ω p
b
0 0 0
F
2 F L t
t L
L
The solution to this system is:
Summing exponents:
F:
F L
b0
b
c0
l
Π2
D
c
0 0 0
F
2 F L t
t L
The solution to this system is:
a 1
b0
c0
c
Π3
D
a
b c
Π4 μ D ω p
Thus:
F t
L
2
Summing exponents:
6
F:
1c0
L:
2 a 2 c 0
t:
1b0
a
1
b
c
0 0 0
F
2 F L t
t L
The solution to this system is:
a0
Check using M, L, t dimensions:
M L
t
The functional relationship is:
L
3
2
1
L
3
b1
t
L t
Π1 f Π2 Π3 Π4
c 1
2
M
Π4
1 L
1
L
P
3
ω p D
1
L
1
L
l
p
2
1
c μ ω
D D p
f
μ ω
M 1 L t
1
L t t M
c μ ω
D D p
P ω p D f
3
l
Problem 7.47
[Difficulty: 3]
Given:
Find:
Functional relationship between the mass burning rate of a combustible mixture and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
The dependence of mass burning rate
(Mathcad can't render dots!)
1
m
2
Select primary dimensions M, L, t:
3
m
δ
δ
M
5
δ
D
α
M
L
3
t
L
ρ
α
ρ
L
t
4
ρ
n = 5 parameters
D
2
L
2
r = 3 dimensions
t
m = r = 3 repeating parameters
α
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
c
a b
c
Π1 m δ ρ α
Thus:
b 2
M
L
0 0 0
L M L t
3 t
t
L
M
Summing exponents:
a
The solution to this system is:
M: 1 b 0
L:
a 3 b 2 c 0
t:
1 c 0
a b
c
Π2 D δ ρ α
a 1
Thus:
Summing exponents:
M: b 0
L:
2 a 3 b 2 c 0
t:
1 c 0
b 1
2
b
Check using F, L, t dimensions:
The functional relationship is:
c 1
c
M0 L0 t0
The solution to this system is:
a0
b0
4
6
2
L
a M
L
t
3
t
L
L
m
Π1
δ ρ α
t
F t 1 L
1
2 2
L L
F t L
Π1 f Π2
D
Π2
α
c 1
L
2
t
t
L
2
1
m
δ ρ α
f
D
α
Problem 7.46
Given:
Ventilation system of cruise ship clubhouse
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham procedure
c
N
p
D
Select primary dimensions M, L, t
c
1
L3
D
p
N
p
D
p
g
1
M
Lt 2
1
t
M
L3
M
L3
L
t2
L
g
n = 9 parameters
M
Lt
r = 3 primary dimensions
m = r = 3 repeat parameters
Then n – m = 6 dimensionless groups will result. Setting up a dimensional equation,
a
c
M
b1 M
M 0 L0t 0
1 D Δp 3 L
2
L
t
Lt
M:
a 1 0
a 1
Hence
L : 3a b 1 0 b 2
c20
t:
c 2
a
Summing exponents,
b
c
M
2 D 3
L
M:
a 1 0
L : 3a b 1 0
a
Summing exponents,
t:
b
c
c 1 0
The other groups can be found by inspection:
a
1
c
b1 M
M 0 L0t 0
L
t
Lt
a 1
Hence
b 2
2
c 1
3 cD 3
p
D 2 2
p
D 2
6
g
D 2
4 N
5
1
3 4 5 6 1
Check using F, L, t as primary dimensions
1
F
L2
Ft 2 2 1
L 2
L4
t
1
Note: Any combination of ’s is a group, e.g.,
2
Ft
L2
Ft 2 2 1
L
L4
t
p
1
, so the ’s are not unique!
2
Problem 7.45
Functional relationship between the aerodynamic torque on a spinning ball and other physical parameters
Given:
Find:
Solution:
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
T
2
Select primary dimensions M, L, t:
3
T
V
M L
t
4
5
μ
ρ
L
M
M
t
3
L t
2
V
ρ
V
2
ρ
[Difficulty: 3]
L
μ
D
d
ω
d
D
ω
1
L
n = 7 parameters
r = 3 dimensions
L
t
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 T ρ V D
Thus:
M L
t
Summing exponents:
2
2
M
a
L
b
c
0 0 0
3 t L M L t
L
The solution to this system is:
a 1
M: 1 a 0
L:
2 3 a b c 0
t:
2 b 0
b 2
c 3
Check using F, L, t dimensions:
F L
L
4
F t
a
b
c
Π2 μ ρ V D
Thus:
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 b 0
M
M
a
L
T
Π1
2
t
2
2
L
3
ρ V D
2
1
L
3
1
b
c
0 0 0
L M L t
3
t
L t
L
The solution to this system is:
a 1
b 1
c 1
μ
Π2
ρ V D
4
Check using F, L, t dimensions:
t 1
F t L
1
2
2 L L
L F t
a
b
c
Π3 ω ρ V D
Thus:
Summing exponents:
M
a
L
a0
L:
3 a b c 0
t:
1 b 0
b
c
Π4 d ρ V D
b
c
0 0 0
L M L t
3
t
t
L
ω D
Π3
V
The solution to this system is:
M: a 0
a
1
b 1
c1
Check using F, L, t dimensions:
Thus:
L
t:
b 0
The functional relationship is:
L
t
L
1
b
The solution to this system is:
a0
M: a 0
1 3 a b c 0
a
t
L
c
0 0 0
3 t L M L t
L
Summing exponents:
L:
M
1
b0
c 1
Check using F, L, t dimensions:
Π1 f Π2 Π3 Π4
d
Π4
D
1
L
L 1
T
2
3
ρ V D
ω D d
ρ V D V D
f
μ
Problem 7.44
[Difficulty: 3]
Find:
Functional relationship between the mass flow rate of liquid from a pressurized tank through a contoured nozzle
and other physical parameters
(a) How many independent Π terms that characterize this phenomenon
(b) Find the Π terms
(c) State the functional relationship for the mass flow rate in terms of the Π terms
Solution:
We will use the Buckingham pi-theorem.
Given:
1
m
2
Select primary dimensions M, L, t:
3
m
A
A
M
L
t
4
5
ρ
h
ρ
M
2
L
A
h
ρ
L
3
Δp
g
M
L
L t
n = 6 parameters
g
Δp
2
t
r = 3 dimensions
2
m = r = 3 repeating parameters
g
We have n - m = 3 dimensionless groups.
Setting up dimensional equations:
a
b c
Π1 m ρ A g
Thus:
Summing exponents:
M
t
M
3
L
a
2 L2
b
L
c
0
0 0
M L t
t
The solution to this system is:
5
1
a 1
b
c
4
2
M: 1 a 0
L:
3 a 2 b c 0
t:
1 2 c 0
a
b c
5
1
4
2
ρ A g
4
Check using F, L, t dimensions:
1
t
F t L
1
2
5
1
L
F t
L
Π2 h ρ A g
m
Π1
Thus:
Summing exponents:
M: a 0
L:
1 3 a 2 b c 0
t:
2 c 0
L
M
3
L
a
2 L2
b
L
c
0
2
L
0 0
M L t
t
The solution to this system is:
1
a0
b
c0
2
Check using F, L, t dimensions:
Π2
L
1
L
1
h
A
2
a
b c
Π3 Δp ρ A g
Thus:
M
M
2 3
L t L
Summing exponents:
a
2 L2
b
L
c
0
0 0
M L t
t
The solution to this system is:
1
a 1
b
c 1
2
M: 1 a 0
L:
1 3 a 2 b c 0
t:
2 2 c 0
Π3
F
Check using F, L, t dimensions:
L
The functional relationship is:
Π1 f Π2 Π3
m
5
1
4
2
ρ A g
f
h
Δp
2
A ρ g A
L
4
F t
t
Δp
ρ g A
2
1
2 L L
1
So the mass flow rate is:
5
1
m ρ A g f
4
2
h
Δp
A ρ g A
Problem 7.43 (In Excel)
[Difficulty: 3]
Given: Time to speed up depends on inertia, speed, torque, oil viscosity and geometry
Find: groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of groups is:
=8
=3
=3
=5
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , D , T
D
T
M
L
1
1
2
t
-1
-2
GROUPS:
Two groups can be obtained by inspection: /D and L /D . The others are obtained below
t
1:
I
3:
M
0
L
0
a =
b =
c =
1
0
0
M
1
L
2
a =
b =
c =
2
0
-1
Hence the groups are
D 3
I 2
D
T
T
Note that the 1 group can also be easily obtained by inspection
t
L
D
t
1
2:
t
0
4:
M
1
L
-1
a =
b =
c =
1
3
-1
M
0
L
0
a =
b =
c =
0
0
0
t
-1
t
0
Problem 7.42
[Difficulty: 3]
Find:
Functional relationship between the mass flow rate of gas through a choked-flow nozzle and other physical
parameters
(a) How many independent Π terms that characterize this phenomenon
(b) Find the Π terms
(c) State the functional relationship for the mass flow rate in terms of the Π terms
Solution:
We will use the Buckingham pi-theorem.
Given:
1
m
2
Select primary dimensions M, L, t:
3
m
A
A
M
L
t
4
5
p
p
T
p
T
M
2
L t
A
T
n = 5 parameters
R
L
T
2
(Mathcad can't render dots!)
R
2
r = 4 dimensions
2
t T
m = r = 4 repeating parameters
R
We have n - m = 1 dimensionless group.
Setting up dimensional equations:
a
b
c
Π1 m p A T R
d
Thus:
Summing exponents:
The solution to this system is:
1
a 1
b 1 c
2
M: 1 a 0
L:
d
a
2
b
M
2
c L
0 0 0 0
L T
M L t T
2
t 2
L t
t T
M
d
m
Π1
R T
p A
1
2
a 2 b 2 d 0
t:
1 2 a 2 d 0
T:
cd0
1
2
Check using F, L, t dimensions:
L
F t L 1
2
T 1
2
1
L F
L
t T
The functional relationship is:
Π1 C
m
p A
R T C
So the mass flow rate is:
2
m C
p A
R T
Problem 7.41
[Difficulty: 2]
Given:
That the power of a washing machine agitator depends on various parameters
Find:
Dimensionless groups
Solution:
Apply the Buckingham procedure
P
H
D
max
h
f
n = 8 parameters
Select primary dimensions M, L, t
P
ML2
3
t
D
H
D
h max
L
L
L
1
t
max
f
1
t
M
L3
r = 3 primary dimensions
M
Lt
m = r = 3 repeat parameters
Then n – m = 5 dimensionless groups will result. Setting up a dimensional equation,
a
c
2
M
b 1 ML
1 D P 3 L 3 M 0 L0t 0
L
t t
M:
a 1 0
a 1
a
L:
Summing exponents,
b
c
max
3a b 2 0 b 5
c30
t:
Hence
c 3
a
1
P
3
D max
5
c
M
b1 M
M 0 L0t 0
2 D 3 L
L
t
Lt
M:
a 1 0
a 1
Summing exponents,
2
Hence
L : 3a b 1 0 b 2
2
D max
c 1 0
t:
c 1
h
f
H
4
The other groups can be found by inspection:
5
3
D
D
max
a
b
c
max
Check using F, L, t as primary dimensions
1
FL
t
Ft 2 5 1
L 3
L4
t
1
Note: Any combination of ’s is a group, e.g.,
Ft
L2
3 4 5 1
1
Ft 2 2 1
L
L4
t
1
P
, so the ’s are not unique!
2
2 D 3max
2
Problem 7.40
[Difficulty: 2]
Given:
Functional relationship between the length of a wake behind an airfoil and other physical parameters
Find:
Solution:
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
w
2
Select primary dimensions M, L, t:
3
w
V
V
L
L
4
5
ρ
L
t
V
t
L
t
L
L
ρ
μ
ρ
n = 6 parameters
μ
M
M
3
L t
L
r = 3 dimensions
m = r = 3 repeating parameters
L
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
Π1 w ρ V L
c
Thus:
Summing exponents:
L
a0
1 3 a b c 0
t:
b 0
b
Π2 t ρ V L
c
Thus:
L
M
a
L
c 1
L
b0
c
Thus:
Summing exponents:
M: 1 a 0
M
M
a
L
L
1
L
1
b
c
0 0 0
L M L t
3
t
L t
L
The solution to this system is:
a 1
b 1
μ
Π3
ρ V L
c 1
1 3 a b c 0
1 b 0
1
t
Π2
L
c 1
Check using F, L, t dimensions:
b 0
b
L
b
1 3 a b c 0
a
1
The solution to this system is:
a0
Π3 μ ρ V L
t:
b0
w
Π1
L
c
0 0 0
3 t L M L t
L
M: a 0
L:
b
Check using F, L, t dimensions:
Summing exponents:
t:
L
The solution to this system is:
L:
L:
a
c
0 0 0
3 t L M L t
L
M: a 0
a
M
4
Check using F, L, t dimensions:
t 1
F t L
1
2
2 L L
L F t
Problem 7.39 (In Excel)
[Difficulty: 3]
Given: Speed depends on mass, area, gravity, slope, and air viscosity and thickness
Find: groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of groups is:
=7
=3
=3
=4
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose g , , m
M
L
1
1
t
-2
t
-1
g
m
1
V
M
0
L
1
a =
b =
c =
-0.5
-0.5
0
M
0
L
0
a =
b =
c =
0
0
0
GROUPS:
1 :
3 :
Hence
V
1
g
1 1
2 2
V2
g
2
g
2 :
t
0
3
2
1
2m
2 3
2
m g
Note that the 1 , 3 and 4 groups can be obtained by inspection
L
-1
a =
b =
c =
-0.5
1.5
-1
M
0
L
2
a =
b =
c =
0
-2
0
A
4 :
M
1
3
4
A
2
t
-1
t
0
Problem 7.38 (In Excel)
[Difficulty: 3]
Given: Bubble size depends on viscosity, density, surface tension, geometry and pressure
Find: groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of groups is:
=6
=3
=3
=3
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , p , D
p
D
M
1
1
L
-3
-1
1
M
0
L
1
a =
b =
c =
0
0
-1
M
1
L
0
a =
b =
c =
0
-1
-1
t
-2
GROUPS:
d
1:
3:
Hence
1
d
D
2
1
1
2 p 2 D
2
pD 2
Note that the 1 group can be obtained by inspection
t
0
2:
t
-2
4:
3
Dp
M
1
L
-1
a =
b =
c =
-0.5
-0.5
-1
M
0
L
0
a =
b =
c =
0
0
0
t
-1
t
0
Problem 7.37 (In Excel)
[Difficulty: 3]
Given: That dot size depends on ink viscosity, density, and surface tension, and geometry
Find: groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of groups is:
=7
=3
=3
=4
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , V , D
M
1
V
D
L
-3
1
1
t
-1
GROUPS:
d
1:
3:
Hence
1
d
D
2
M
0
L
1
a =
b =
c =
0
0
-1
M
1
L
0
a =
b =
c =
-1
-2
-1
VD
VD
Note that groups 1 and 4 can be obtained by inspection
t
0
2:
t
-2
L
4:
3
V 2 D
4
L
D
M
1
L
-1
a =
b =
c =
-1
-1
-1
M
0
L
1
a =
b =
c =
0
0
-1
t
-1
t
0
Problem 7.36
Functional relationship between the height of a ball suported by a vertical air jet and other physical parameters
Given:
Find:
Solution:
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
h
2
Select primary dimensions M, L, t:
3
h
D
d
L
L
L
4
5
ρ
[Difficulty: 3]
D
V
d
V
ρ
μ
W
μ
W
n = 7 parameters
V
ρ
L
M
M
M L
t
3
L t
2
L
t
r = 3 dimensions
m = r = 3 repeating parameters
d
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b c
Π1 h ρ V d
Thus:
Summing exponents:
1 3 a b c 0
t:
b 0
b c
Π2 D ρ V d
L
b
c
0 0 0
3 t L M L t
L
b0
Thus:
M: a 0
L
M
a
L
h
Π1
d
c 1
Check using F, L, t dimensions:
Summing exponents:
t:
a
a0
L:
L:
M
The solution to this system is:
M: a 0
a
L
L
L
1
b
c
0 0 0
3 t L M L t
L
The solution to this system is:
a0
b0
D
Π2
d
c 1
1 3 a b c 0
b 0
1
Check using F, L, t dimensions:
L
1
L
1
a
b c
Π3 μ ρ V d
Thus:
M
L t
Summing exponents:
t:
L
b
c
0 0 0
3 t L M L t
L
a 1
b 1
b c
Thus:
Summing exponents:
M: 1 a 0
μ
Π3
ρ V d
c 1
4
Check using F, L, t dimensions:
1 b 0
a
t:
a
1 3 a b c 0
Π4 W ρ V d
L:
M
The solution to this system is:
M: 1 a 0
L:
M L
t
2
M
a
L
t 1
F t L
1
2
2 L L
L F t
b
c
0 0 0
3 t L M L t
L
The solution to this system is:
a 1
b 2
Π4
c 2
1 3 a b c 0
2 b 0
Check using F, L, t dimensions:
F
L
4
F t
2
t
2
L
2
W
2 2
ρ V d
1
L
2
1
Problem 7.35
Functional relationship between the diameter of droplets formed during jet breakup and other physical
parameters
(a) The number of dimensionless parameters needed to characterize the process
(b) The ratios (Π-terms)
Given:
Find:
We will use the Buckingham pi-theorem.
Solution:
1
d
2
Select primary dimensions M, L, t:
3
d
ρ
5
V
μ
σ
V
σ
V
M
M
M
L
3
L t
2
t
L
ρ
μ
ρ
L
4
[Difficulty: 3]
t
n = 6 parameters
D
D
r = 3 dimensions
L
m = r = 3 repeating parameters
D
We have n - m = 3 dimensionless groups.
Setting up dimensional equations:
a
b
c
Π1 d ρ V D
Thus:
L
L
a0
L:
1 3 a b c 0
t:
b 0
b
b
b0
c
Thus:
a
a 1
b 1
c 1
1 3 a b c 0
c
Summing exponents:
M: 1 a 0
L:
3 a b c 0
t:
2 b 0
Thus:
L
1
μ
Π2
ρ V D
4
Check using F, L, t dimensions:
1 b 0
b
1
b
The solution to this system is:
M: 1 a 0
a
L
L
M
c
0 0 0
L M L t
3
t
L t
L
M
Summing exponents:
Π3 σ ρ V D
d
Π1
D
c 1
Check using F, L, t dimensions:
Π2 μ ρ V D
t:
The solution to this system is:
M: a 0
L:
a
c
0 0 0
3 t L M L t
L
Summing exponents:
a
M
M
M
a
L
b
t 1
F t L
1
2
2 L L
L F t
c
0 0 0
3 t L M L t
t L
2
The solution to this system is:
a 1
b 2
c 1
Π3
4
Check using F, L, t dimensions:
2
σ
2
ρ V D
t 1
F L
1
2 2 L
L
F t L
Problem 7.34
[Difficulty: 3]
Given:
Functional relationship between the deflection of the bottom of a cylindrical tank and other physical parameters
Find:
Functional relationship between these parameters using dimensionless groups.
Solution:
We will use the Buckingham pi-theorem.
1
δ
2
Select primary dimensions F, L, t:
3
δ
D
h
d
L
L
L
L
D
h
d
γ
E
γ
E
F
L
4
5
D
n = 6 parameters
F
3
L
r = 2 dimensions
2
m = r = 2 repeating parameters
γ
We have n - m = 4 dimensionless groups.
Setting up dimensional equations:
a b
Π1 δ D γ
Thus:
a
Summing exponents:
F:
b0
L:
1 a 3 b 0
b
0 0
3 F L
L
L L
F
The solution to this system is:
a 1
b0
Check using M, L, t dimensions:
Now since h and d have the same dimensions as δ, it
would follow that the the next two pi terms would be:
a b
Π4 E D γ
Thus:
Summing exponents:
F:
1b0
L:
2 a 3 b 0
L
F
L
a
2
δ
Π1
D
h
Π2
D
L
1
L
d
Π3
D
b
0 0
3 F L
L
F
The solution to this system is:
a 1
E
Π4
D γ
b 1
Check using M, L, t dimensions:
Π1 f Π2 Π3 Π4
2 2
1 L t
1
2 L M
M
L t
The functional relationship is:
1
δ
D
h d E
D D D γ
f
(For further reading, one should consult an appropriate text, such as Advanced Strength of Materials by Cook and Young)
Problem 7.33
[Difficulty: 2]
Given:
Functional relationship between the mass flow rate exiting a tank through a rounded drain hole and other
physical parameters
Find:
(a) Number of dimensionless parameters that will result
(b) Number of repeating parameters
(c) The Π term that contains the viscosity
Solution:
We will use the Buckingham pi-theorem.
1
m
2
Select primary dimensions M, L, t:
3
m
h0
M
t
4
5
ρ
D
d
h0
D
d
L
L
L
g
μ
g
ρ
μ
L
M
M
3
L t
t
d
n = 7 parameters
ρ
2
L
r = 3 dimensions
We have n - r = 4 dimensionless groups.
m = r = 3 repeating parameters
g
Setting up dimensional equation involving the viscosity:
a b c
Π1 μ ρ d g
Thus:
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 2 c 0
M
M
L t 3
L
a
L
b
L
0 0 0
2 M L t
t
The solution to this system is:
3
1
a 1
b
c
2
2
4
Check using F, L, t dimensions:
c
1
t
F t L
1
2
2
3
1
L F t
L
2
L
2
Π1
μ
3
ρ d g
Problem 7.32
[Difficulty: 2]
Given:
Functional relationship between the power required to drive a fan and other physical parameters
Find:
Expression for P in terms of the other variables
Solution:
We will use the Buckingham pi-theorem.
1
P
2
Select primary dimensions M, L, t:
3
P
ρ
M L
t
4
5
Q
M
L
3
t
2
L
D
n = 5 parameters
ω
D
ρ
3
ρ
Q
ω
D
3
1
L
r = 3 dimensions
t
m = r = 3 repeating parameters
ω
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 P ρ D ω
Thus:
M L
2 3 a b 0
t:
3 c 0
M
3
L
a
L
b
1
t
c
0 0 0
M L t
The solution to this system is:
a 1
M: 1 a 0
L:
3
t
Summing exponents:
2
b 5
Π1
c 3
P
5
3
ρ D ω
4
Check using F, L, t dimensions:
a
b
c
Π2 Q ρ D ω
Thus:
1 3
F L L
t 1
2 5
t
F t L
L
3
3 3 a b 0
t:
1 c 0
Check using F, L, t
dimensions:
The functional relationship is:
M
3
L
t
Summing
exponents:
M: a 0
L:
a
L
b
1
t
c
0 0 0
M L t
The solution to this system is:
a0
1
t
L
Π1 f Π2
t
L
b 3
Π2
c 1
Q
3
D ω
1
P
5
3
ρ D ω
f
Q
3
D ω
P ρ D ω f
5
3
Q
3
D ω
Problem 7.31
Given:
[Difficulty: 3]
Find:
Functional relationship between the flow rate of viscous liquid dragged out of a bath and other physical
parameters
Expression for Q in terms of the other variables
Solution:
We will use the Buckingham pi-theorem.
1
Q
2
Select primary dimensions M, L, t:
3
Q
L
μ
μ
3
t
4
5
ρ
ρ
V
g
ρ
g
M
M
L
L t
3
2
L
t
h
n = 6 parameters
V
h
V
L
L
r = 3 dimensions
t
m = r = 3 repeating parameters
h
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b c
Π1 Q ρ V h
Thus:
L
3
a
a0
L:
3 3 a b c 0
t:
1 b 0
Check using F, L, t dimensions:
b c
b
The solution to this system is:
M: a 0
Π2 μ ρ V h
L
c
0 0 0
L M L t
t
3
t
L
Summing exponents:
a
M
Thus:
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 b 0
c 2
Q
V h
2
3
t 1
1
t L 2
L
L
M
M
a
L
b
c
0 0 0
L M L t
3
t
L t
L
The solution to this system is:
a 1
4
Check using F, L, t dimensions:
b 1
Π1
b 1
t 1
F t L
1
2
2 L L
L F t
c 1
μ
Π2
ρ V h
a
b c
Π3 g ρ V h
Thus:
L
t:
2 b 0
Check using F, L, t dimensions:
L
a0
L
t
The functional relationship is:
a
b
The solution to this system is:
M: a 0
1 3 a b c 0
M
c
0 0 0
3 t L M L t
t L
2
Summing exponents:
L:
2
L
t
c1
g h
2
V
2
L
b 2
Π3
2
1
Π1 f Π2 Π3
Q
V h
2
ρ V h V2
μ g h
f
ρ V h V2
μ g h
Q V h f
2
Problem 7.30
[Difficulty: 2]
Given:
Functional relationship between the time needed to drain a tank through an orifice plate and other physical
parameters
Find:
(a) the number of dimensionless parameters
(b) the number of repeating variables
(c) the Π term which contains the viscosity
Solution:
We will use the Buckingham pi-theorem.
1
τ
2
Select primary dimensions M, L, t:
3
τ
h0
D
d
T
L
L
L
h0
D
d
g
μ
g
ρ
μ
L
M
M
2
3
L t
t
4
5
ρ
d
n = 7 parameters
ρ
L
r = 3 dimensions
m = r = 3 repeating parameters
g
We have n - m = 4 dimensionless groups.
Setting up dimensional equation including the viscosity:
a b c
Π1 μ ρ d g
Thus:
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 2 c 0
M
M
L t 3
L
a
L
b
L
0 0 0
2 M L t
t
The solution to this system is:
3
1
a 1
b
c
2
2
4
Check using F, L, t dimensions:
c
1
t
F t L
1
2
2
3
1
L F t
L
2
L
2
Π1
μ
3
1
2
2
ρ d g
Problem 7.29
[Difficulty: 2]
Given:
Find:
Functional relationship between the power transmited by a sound wave and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
Expression for E in terms of the other variables
1
E
2
Select primary dimensions M, L, t:
3
E
V
ρ
M
L
M
3
t
3
t
4
5
ρ
V
ρ
V
L
r
n = 5 parameters
n
r
n
1
L
r = 3 dimensions
t
m = r = 3 repeating parameters
r
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b c
Π1 E ρ V r
Thus:
t:
3 b 0
a 1
Check using F, L, t dimensions:
a
b c
Π2 n ρ V r
Thus:
F
L t
1
t
Summing
exponents:
M: a 0
L:
3 a b c 0
t:
1 b 0
Check using F, L, t
dimensions:
The functional relationship is:
L
b
The solution to this system is:
M: 1 a 0
3 a b c 0
c
0 0 0
3 t L M L t
t L
3
Summing exponents:
L:
a
M
M
L
4
F t
2
b 3
t
c0
3
a
L
b
c
0 0 0
3 t L M L t
L
a0
t
3
ρ V
1
n r
Π2
V
The solution to this system is:
1
E
3
L
M
Π1
L
Π1 f Π2
t
L
b 1
c1
1
E
3
ρ V
f
n r
V
E ρ V f
3
n r
V
Problem 7.28 (In Excel)
[Difficulty: 2]
Given: That drain time depends on fluid viscosity and density, orifice diameter, and gravity
Find: Functional dependence of t on other variables
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of groups is:
=5
=3
=3
=2
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , g , d
M
1
g
d
L
-3
1
1
t
-2
GROUPS:
t
1:
M
0
L
0
t
1
a =
b =
c =
0
0.5
-0.5
2:
M
1
L
-1
a =
b =
c =
-1
-0.5
-1.5
M
0
L
0
a =
b =
c =
0
0
0
t
-1
The following groups from Example 7.1 are not used:
3:
Hence
1 t
The final result is
g
d
t
and
M
0
L
0
a =
b =
c =
0
0
0
2
g
d
g
2
f 2 3
gd
1 3
2d 2
t
0
4:
2
gd 3
2
with 1 f 2
t
0
Problem 7.27
Given:
Find:
Solution:
Functional relationship between the load bearing capacity of a journal bearing and other physical parameters
Dimensionless parameters that characterize the problem.
We will use the Buckingham pi-theorem.
1
W
2
Select primary dimensions F, L, t:
3
W
D
l
c
F
L
L
L
4
5
D
[Difficulty: 2]
D
ω
l
c
n = 6 parameters
ω
μ
ω
μ
1
F t
t
2
L
r = 3 dimensions
m = r = 3 repeating parameters
μ
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 W D ω μ
Thus:
a
Summing exponents:
F:
1c0
L:
a 2 c 0
t:
b c 0
1
c
The solution to this system is:
a 2
Check using M, L, t dimensions:
The functional relationship is:
b
0 0 0
F t
2 F L t
t L
F L
b 1
L t
M L 1
t
1
2
2
M
t
L
Π1 f Π2 Π3
c 1
Π1
W
2
D ω μ
By inspection, we can see that:
l
Π2
D
W
2
D ω μ
c
Π3
D
f
D D
l
c
Problem 7.26
Given:
That the power of a vacuum depends on various parameters
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham procedure
P
p
D
d
di
do
n = 8 parameters
Select primary dimensions M, L, t
P
ML2
3
t
D
p
M
Lt 2
D d
L
L
1
t
do
L
di
M
L3
L
r = 3 primary dimensions
m = r = 3 repeat parameters
Then n – m = 5 dimensionless groups will result. Setting up a dimensional equation,
M
1 D P 3
L
M:
a 1 0
L : 3a b 2 0
c30
t:
a
Summing exponents,
b
c
a
c
2
b 1 ML
0 0 0
L 3 M Lt
t t
a 1
Hence
b 5
c 3
a
1
P
D 5 3
c
M
b 1 M
2 D Δp 3 L
M 0 L0t 0
2
L
t Lt
M:
a 1 0
a 1
p
Summing exponents,
Hence
L : 3a b 1 0 b 2
2
D 2 2
c20
t:
c 2
d
d
d
3
The other groups can be found by inspection:
4 i
5 o
D
D
D
a
b
c
Check using F, L, t as primary dimensions
1
FL
t
1
2
F
L2
1
3 4 5
Ft 2 1
L 2
L4
t
1
P
Note: Any combination of 1, 2 and 3 is a group, e.g.,
, so the ’s are not unique!
2 pD 3
Ft 5 1
L 3
L4
t
2
2
L
1
L
Problem 7.25
Given:
That automobile buffer depends on several parameters
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham procedure
T
F
e
n = 6 parameters
Select primary dimensions M, L, t
T
ML2
2
t
F
e
1
t
F
ML
t2
e
L
M
Lt
M
t2
r = 3 primary dimensions
m = r = 3 repeat parameters
Then n – m = 3 dimensionless groups will result. Setting up a dimensional equation,
a
c
2
ML
b 1 ML
1 F e T 2 L 2 M 0 L0t 0
t t
t
a b
c
Summing exponents,
M:
L:
a 1 0
ab20
a 1
b 1
t:
2a c 2 0
c0
a
Summing exponents,
a
T
Fe
2
e2
F
c
ML
b1 M
3 F e 2 L 2 M 0 L0t 0
t
t t
M:
a 1 0
a 1
Hence
L:
ab0
b 1
t : 2a c 2 0 c 0
Check using F, L, t as primary dimensions
1
c
ML
b1 M
2 F a eb c 2 L
M 0 L0t 0
t
t Lt
M:
a 1 0
a 1
Hence
L:
a b 1 0
b2
t : 2a c 1 0 c 1
a b
Summing exponents,
Hence
c
3
e
F
Ft 2 1
F
L
L
2
FL
L
t
L
1
1
2
1
3
1
F
FL
F
T
1
Note: Any combination of 1, 2 and 3 is a group, e.g.,
, so 1, 2 and 3 are not unique!
2 e3
Problem 7.24
Given:
Find:
Solution:
Functional relationship between the flow rate over a weir and physical parameters
An expression for Q based on the other variables
We will use the Buckingham pi-theorem.
1
Q
2
Select primary dimensions L, t:
3
Q
L
h
5
g
h
g
3
h
L
L
t
4
[Difficulty: 2]
t
2
n = 5 parameters
b
b
r = 2 dimensions
L
m = r = 2 repeating parameters
g
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
Π1 Q h g
Thus:
L
3
t
L
a
L
0 0
2 L t
t
Summing exponents:
L:
3ab0
t:
1 2 b 0
Check:
The solution to this system is:
5
1
a
b
2
2
Π1
Q
2
h g h
L3 1 2 t
1
t L L
a b
Π2 b h g
Thus:
L L
a
L
L:
1ab0
t:
2 b 0
L
1
L
b
0 0
2 L t
t
Summing exponents:
Check:
b
The solution to this system is:
a 1
b0
b
Π2
h
1
The functional relationship is:
Π1 f Π2
Q
2
h g D
f
b
h
Therefore the flow rate is:
Q h g h f
2
b
h
Problem 7.23
Given:
Find:
Solution:
Functional relationship between the speed of a capillary wave and other physical parameters
An expression for V based on the other variables
We will use the Buckingham pi-theorem.
1
V
2
Select primary dimensions M, L, t:
3
V
σ
L
M
t
2
4
5
σ
σ
t
λ
[Difficulty: 2]
λ
λ
L
n = 4 parameters
ρ
ρ
M
L
r = 3 dimensions
3
m = r = 3 repeating parameters
ρ
We have n - m = 1 dimensionless group. Setting up dimensional equations:
a b c
Π1 V σ λ ρ
Thus:
t
Summing exponents:
1 b 3 c 0
t:
1 2 a 0
Check using F, L, t dimensions:
The functional relationship is:
M
2
t
a
L
b
M
c
0 0
3 L t
L
The solution to this system is:
1
1
1
a
b
c
2
2
2
M: a c 0
L:
L
Π1 C
Π1 V
λ ρ
σ
2
L L F t L 1
4 F
t
L
V
λ ρ
σ
C
Therefore the velocity is:
V C
σ
λ ρ
Problem 7.22
[Difficulty: 2]
(The solution to this problem was first devised by G.I. Taylor
in the paper "The formation of a blast wave by a very intense
explosion. I. Theoretical discussion," Proceedings of the Royal
Society of London. Series A, Mathematical and Physical
Sciences, Vol. 201, No. 1065, pages 159 - 174 (22 March 1950).)
Given:
Find:
Functional relationship between the energy released by an explosion and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
Expression for E in terms of the other variables
1
E
2
Select primary dimensions M, L, t:
3
E
t
M L
t
4
5
t
R
t
L
ρ
p
M
L t
t
n = 5 parameters
ρ
p
2
2
ρ
R
2
M
L
r = 3 dimensions
3
m = r = 3 repeating parameters
R
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
Π1 E ρ t R
c
Thus:
M L
2
t
Summing exponents:
2 3 a c 0
t:
2 b 0
a b
c
Thus:
M
t:
2 b 0
2 1
t
F t
2
M
L
5
c 5
E t
2
ρ R
5
1
a
The solution to this system is:
a 1
Check using F, L, t dimensions:
F
L
The functional relationship is:
4
b2
Π1
b c
0 0 0
3 t L M L t
L t L
2
M: 1 a 0
1 3 a c 0
a
b c
0 0 0
3 t L M L t
L
L
F L
Summing exponents:
L:
M
a 1
Check using F, L, t dimensions:
Π2 p ρ t R
The solution to this system is:
M: 1 a 0
L:
2
2
L
4
F t
Π1 f Π2
2
b2
2 1
t
L
2
c 2
Π2
p t
2
ρ R
2
1
p t2
f
5
2
ρ R
ρ R
E t
2
E
ρ R
t
2
5
p t2
f
ρ R2
Problem 7.21
[Difficulty: 2]
Given:
Functional relationship between mean velocity for turbulent flow in a pipe or boundary layer and physical
parameters
Find:
(a) Appropriate dimensionless parameters containing mean velocity and one containing the distance from the
wall that are suitable for organizing experimental data.
(b) Show that the result may be written as:
u
yu
f * where u* w is the friction velocity
u*
Solution:
We will use the Buckingham pi-theorem.
w
1
u
2
Select primary dimensions M, L, t:
3
u
w
L
t
M
Lt 2
4
5
y
M
L3
M
Lt
y
L
w
y
n = 5 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
1 u y
a
b c
w
Thus:
L
t
M
3
L
a
L
b
c
0 0 0
2 M L t
L t
M
Summing exponents:
M: a c 0
L:
1 3 a b c 0
t:
1 2 c 0
Check using F, L, t dimensions:
2 a y b wc
Thus:
The solution to this system is:
1
1
a
b0 c
2
2
L t 1
t L
M
L t
M
3
L
a
L
b
c
0 0 0
2 M L t
L t
M
1 u
u
w u*
Summing exponents:
M: 1 a c 0
L:
1 3 a b c 0
t:
1 2 c 0
The solution to this system is:
1
1
a
b 1 c
2
2
2
y w y w yu* yu*
Π2 is the reciprocal of the Reynolds number, so we know that it checks out.
The functional relationship
is:
Π1 g Π2
u
g
u*
yu*
which may be rewritten as:
u
yu
f *
u*
Problem 7.20
Given:
Find:
Solution:
Functional relationship between the speed of a free-surface gravity wave in deep water and physical parameters
The dependence of the speed on the other variables
We will use the Buckingham pi-theorem.
1
V
2
Select primary dimensions M, L, t:
3
V
λ
L
t
4
5
D
D
λ
D
L
L
ρ
n = 5 parameters
g
ρ
g
M
L
3
2
L
ρ
[Difficulty: 2]
t
r = 3 dimensions
m = r = 3 repeating parameters
g
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b c
Π1 V D ρ g
L
Thus:
t
L
a
M
b
L
c
0 0 0
3 2 M L t
L t
Summing exponents:
M: b 0
L:
1 a 3 b c 0
t:
1 2 c 0
The solution to this system is:
1
1
a b0 c
2
2
Check using F, L, t dimensions:
a b c
Π2 λ D ρ g
Thus:
Π1
V
g D
L t 1
t L
L L
a
M
b
L
c
0 0 0
3 2 M L t
L t
Summing exponents:
M: b 0
L:
1 a 3 b c 0
t:
2 c 0
The solution to this system is:
a 1 b 0
Check using F, L, t dimensions: L
The functional relationship is:
1
L
c0
λ
Π2
D
1
Π1 f Π2
V
g D
f
λ
D
Therefore the velocity is:
V
g D f
λ
D
Problem 7.19
Given:
That light objects can be supported by surface tension
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham procedure
W
p
g
n = 5 parameters
Select primary dimensions M, L, t
W
ML
t 2
g
p
g
L
M
L3
L
t2
M
t 2
p
r = 3 primary dimensions
m = r = 3 repeat parameters
Then n – m = 2 dimensionless groups will result. Setting up a dimensional equation,
L
1 g p W 2
t
M:
b 1 0
Summing exponents,
L:
a
b
c
2a 2 0
t:
b
W
g p 3
2
gp 2
b
L M
c M
2 g p 2 3 L 2 M 0 L0t 0
t
t L
M:
b 1 0
b 1
Hence
L : a 3b c 0 c 2
2 a 2 0 a 1
t:
a
1
a 1
a
Summing exponents,
b
M
c ML
0 0 0
3 L 2 M L t
t
L
b 1
Hence
a 3b c 1 0 c 3
a
c
F
F
L
Check using F, L, t as primary dimensions
1
1
2
1
L Ft 2 2
L Ft 2 3
L
L
t 2 L4
t 2 L4
1 Wp
Note: Any combination of 1 and 2 is a group, e.g.,
, so 1 and 2 are not unique!
2
Problem 7.18
Given:
Find:
Solution:
Functional relationship between boundary layer thickness and physical parameters
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.
1
2
Select primary dimensions M, L, t:
3
x
U
L
L
M
L3
M
Lt
L
t
x
U
4
5
x
[Difficulty: 2]
U
n = 5 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
c
Π1 δ ρ x U
Thus:
L
M
a
3
L
L
b
L
c
0 0 0
M L t
t
Summing exponents:
M: 0 a 0
The solution to this system is:
L:
1 3 a b c 0
t:
0c0
a0
Check using F, L, t dimensions: ( L)
b 1 c 0
δ
Π1
x
1
1
L
a b
c
Π2 μ ρ x U
a
Thus:
c
M M Lb L M0 L0 t0
3
L t L
t
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 c 0
Check using F, L, t dimensions:
The solution to this system is:
a 1 b 1 c 1
4
2
F t L 1 t 1
2 2 L L
L F t
μ
Π2
ρ x U
The functional relationship is:
Π1 f Π2
Problem 7.17
Given:
Find:
Solution:
Functional relationship between wall shear stress in a boundary layer and physical parameters
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.
1
w
2
Select primary dimensions M, L, t:
3
w
x
U
M
Lt 2
L
M
L3
M
Lt
L
t
x
U
4
5
[Difficulty: 2]
x
U
n = 5 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
c
Π1 τw ρ x U
a
Thus:
c
M M Lb L M0 L0 t0
2 3
t
L t L
Summing exponents:
M: 1 a 0
The solution to this system is:
L:
1 3 a b c 0
t:
2 c 0
Check using F, L, t dimensions:
a b
c
Π2 μ ρ x U
a 1 b 0
τw
2
ρ U
4
2
F L t 1
2 2 L
L F t
a
Thus:
c 2
Π1
c
M M Lb L M0 L0 t0
3
L t L
t
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 c 0
Check using F, L, t dimensions:
The solution to this system is:
a 1 b 1 c 1
4
2
F t L 1 t 1
2 2 L L
L F t
μ
Π2
ρ x U
The functional relationship is:
Π1 f Π2
Problem 7.16
[Difficulty: 2]
Given:
That speed of shallow waves depends on depth, density, gravity and surface tension
Find:
Dimensionless groups; Simplest form of V
Solution:
Apply the Buckingham procedure
V
D
g
n = 5 parameters
Select primary dimensions M, L, t
V
L
t
g
D
g
L
M
L3
L
t2
M
t 2
D
r = 3 primary dimensions
m = r = 3 repeat parameters
Then n – m = 2 dimensionless groups will result. Setting up a dimensional equation,
a
b
L M
c L
1 g a b D cV 2 3 L M 0 L0t 0
t
t L
M:
b0
b0
1
Summing exponents,
L : a 3b c 1 0 c
Hence
2
1
t:
2a 1 0
a
2
a
b
L M
c M
2 g a b D c 2 3 L 2 M 0 L0t 0
t
t L
M:
b 1 0
b 1
Summing exponents,
Hence
L : a 3b c 0 c 2
2 a 2 0 a 1
t:
L
t
Check using F, L, t as primary dimensions
1
1
The relation between drag force speed V is
L
2
t
1
2
L
1 f 2
V
f
2
gD
gD
1
V
gD
2
gD 2
F
L
2
1
L Ft 2 2
L
t 2 L4
V gD f
2
gD
Problem 7.15
[Difficulty: 2]
Given:
Find:
Functional relationship between drag on an object in a supersonic flow and physical parameters
Solution:
We will use the Buckingham pi-theorem.
Functional relationship for this problem using dimensionless parameters
1
FD
2
Select primary dimensions M, L, t:
3
FD
V
ρ
M L
L
M
2
t
V
ρ
t
4
5
V
A
ρ
A
L
3
L
n = 5 parameters
c
c
L
2
r = 3 dimensions
t
m = r = 3 repeating parameters
A
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
c
Π1 FD V ρ A
Thus:
M L
t
Summing exponents:
2
M: 1 b 0
1 a 3 b 2 c 0
t:
2 a 0
t
2
L
c
c
Thus:
Π1
a 2 b 1 c 1
Check using F, L, t dimensions: F
a b
b
The solution to this system is:
L:
Π2 c V ρ A
L
a
2
0 0 0
M
3 L M L t
t L
L
t
2
L
4
F t
L
2
a
1
L
2
M
L3
t
FD
2
V ρ A
1
b
2
L
c
0
0 0
M L t
Summing exponents:
M: b 0
The solution to this system is:
L:
1 a 3 b 2 c 0
t:
1 a 0
Check using F, L, t dimensions:
The functional relationship is:
a 1 b 0
c0
c
Π2
V
(The reciprocal of Π 2 is also referred
to as the Mach number.)
L t
1
t L
Π1 g Π2
FD
2
V ρ A
f
V
c
Problem 7.14
Functional relationship between buoyant force of a fluid and physical parameters
Given:
Find:
Solution:
Buoyant force is proportional to the specific weight as demonstrated in Chapter 3.
We will use the Buckingham pi-theorem.
1
FB
2
Select primary dimensions F, L, t:
3
FB
V
F
L
V
5
V
n = 3 parameters
γ
γ
F
3
L
4
[Difficulty: 2]
r = 2 dimensions
3
m = r = 2 repeating parameters
γ
We have n - m = 1 dimensionless group. Setting up dimensional equations:
a b
Π1 FB V γ
3 F3
a
Thus:
b
F L
0
F L
0
L
Summing exponents:
F:
1b0
The solution to this system is:
L:
3 a 3 b 0
a 1 b 1
2
FB
Π1
V γ
2
Check using M, L, t dimensions:
M L 1 t L
1
2
3 M
t
L
The functional relationship is:
Π1 C
FB
V γ
C
Solving for the buoyant force:
FB C V γ
Buoyant force is
proportional to γ
(Q.E.D.)
Problem 7.13
[Difficulty: 2]
Given:
Find:
Functional relationship between the drag on a satellite and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
1
Expression for FD in terms of the other variables
2
FD
λ
ρ
L
c
Select primary dimensions M, L, t:
3
FD
λ
M L
t
4
5
L
2
ρ
M
L
L
L
ρ
c
L
L
3
n = 5 parameters
r = 3 dimensions
t
m = r = 3 repeating parameters
c
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b d
Π1 D ρ L c
Thus:
M L
t
Summing exponents:
2
a
L
3
L
b
a 1
L:
1 3 a b d 0
t:
2 d 0
Check using F, L, t dimensions:
F
L
4
F t
a b d
M
L
d
0 0 0
M L t
t
The solution to this system is:
M: 1 a 0
Π2 λ ρ L c
Thus:
L
t:
d0
a
a0
Check using F, L, t dimensions:
The functional relationship is:
M
2
t
L
1
L
FD
2 2
ρ L c
2
L
2
L
b
d 2
1
L
d
0 0 0
M L t
t
The solution to this system is:
M: a 0
1 3 a b d 0
L
3
L
Summing exponents:
L:
2
1
b 2
Π1
b 1
λ
Π2
L
d0
(Π2 is sometimes referred to
as the Knudsen number.)
1
Π1 f Π2
FD
2 2
ρ L c
f
λ
L
λ
2 2
FD ρ L c f
L
Problem 7.12
[Difficulty: 2]
At low speeds, drag F on a sphere is only dependent upon speed V, viscosity μ, and diameter D
Given:
Find:
Solution:
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.
1
F
2
Select primary dimensions M, L, t:
3
F
V
ML
t2
L
t
M
Lt
V
D
n = 4 parameters
D
n = 4 parameters
L
r = 3 dimensions
D
4
V
5
We have n - m = 1 dimensionless group. Setting up a dimensional equation:
a b
c
Π1 F V μ D
m = r = 3 repeating parameters
a
Thus:
b
M L L M Lc M0 L0 t0
2 t L t
t
Summing exponents:
M: 1 b 0
L:
1abc0
t:
2 a b 0
The solution to this system is:
a 1 b 1 c 1
F
Π1
μ V D
2
Check using F, L, t primary dimensions:
t L 1
F
1 Checks out.
L F t L
Since the procedure produces only one dimensionless group, it must be a constant. Therefore:
F
μ V D
constant
Problem 7.11
Given:
That drag depends on speed, air density and frontal area
Find:
How drag force depend on speed
Solution:
Apply the Buckingham procedure
F
V
A
n = 4 parameters
Select primary dimensions M, L, t
F
V
ML
L
t
M
A
r = 3 primary dimensions
t2
V
L3
L2
A
m = r = 3 repeat parameters
Then n – m = 1 dimensionless groups will result. Setting up a dimensional equation,
1 V a b Ac F
b
a
L M
3 L2
t L
c
ML
t2
M 0 L0 t 0
Summing exponents,
b 1 0
M:
L:
t:
b 1
a 3b 2c 1 0 c 1
a20
a 2
Hence
1
F
V 2 A
Check using F, L, t as primary dimensions
1
1
F
Ft
2
L2
L4 t 2
L2
The relation between drag force F and speed V must then be
F V 2 A V 2
The drag is proportional to the square of the speed.
[Difficulty: 2]
Problem 7.10
Given:
Find:
Solution:
Functional relationship between pressure drop through orifice plate and physical parameters
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.
1
p
2
Select primary dimensions M, L, t:
3
p
V
D
d
M
Lt 2
M
L3
M
Lt
L
t
L
L
V
D
4
5
[Difficulty: 2]
V
D
d
n = 6 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1 Δp ρ V D
a
Thus:
b
M M L Lc M0 L0 t0
2 3 t
L t L
Summing exponents:
M: 1 a 0
The solution to this system is:
L:
1 3 a b c 0
t:
2 b 0
a 1 b 2 c 0
Check using F, L, t primary dimensions:
F
L
a
b
c
Π2 μ ρ V D
2
L
4
F t
a
Thus:
2
t
Π1
Δp
2
ρ V
2
1 Checks
out.
L
2
b
M M L Lc M0 L0 t0
3
L t L t
Summing exponents:
M: 1 a 0
L:
1 3 a b c 0
t:
1 b 0
The solution to this system is:
a 1 b 1 c 1
μ
Π2
ρ V D
(This is the Reynolds number, so it checks out)
a
b
c
Π3 d ρ V D
Thus:
L
M
a
L
b
c
0 0 0
3 t L M L t
L
Summing exponents:
M: a 0
L:
1c0
t:
b0
The solution to this system is:
a0
b0
c 1
d
Π3
D
(This checks out)
Problem 7.9
Given:
Equations Describing pipe flow
Find:
Non-dimensionalized equation; Dimensionless groups
[Difficulty: 2]
Solution:
Nondimensionalizing the velocity, pressure, spatial measures, and time:
u*
u
V
p*
p
p
x*
x
L
r*
r
L
t* t
V
L
Hence
u V u*
p p p *
x Lx*
r Dr*
t
L
t*
V
Substituting into the governing equation
V u *
u
1
1 p *
1 2 u * 1 u *
V 2
p
V
L t *
L x *
t
D r *2 r * r *
The final dimensionless equation is
p p * L 2 u * 1 u *
u *
t *
V 2 x * DV D r *2 r * r *
The dimensionless groups are
p
V
2
DV
L
D
Problem 7.8
Given:
Equation for unsteady, 2D compressible, inviscid flow
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Denoting nondimensional quantities by an asterisk
x*
x
L
y*
y
L
u*
u
c0
v*
v
c0
c*
c
c0
t*
t c0
L
*
L c0
Note that the stream function indicates volume flow rate/unit depth!
Hence
x Lx*
y Ly*
u c0 u *
v c0 v *
c c0 c *
t
Lt *
c0
L c0 *
Substituting into the governing equation
2
2
c03 2 * c03 u *2 v *2 c03 2
c03 2
c03
2 *
2 *
2 *
u
c
v
c
u
v
*
*
*
*
2
*
*
0
2
L
L
t
x *2 L
y *2 L
x * y *
L t *
The final dimensionless equation is
2
2
2 *
2 * u *2 v *2
2
2 *
2
2 *
0
2
*
*
*
*
*
*
u
c
v
c
u
v
x * y *
y *2
x *2
t
t *2
No dimensionless group is needed for this equation!
Problem 7.7
[Difficulty: 4]
Given:
Find:
The Prandtl boundary-layer equations for steady, incompressible, two-dimensional flow neglecting gravity
Solution:
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
Nondimensionalization for the equation using length scale L and velocity scale V 0. Obtain the dimensionless
groups that characterize the flow.
x*
Substituting into the continuity equation:
x
L
y*
y
L
u*
u
V0
v*
v
V0
*
*
u *V0 v *V0
Simplifying this expression: V0 u V0 v 0
0
L x * L y *
x* L
y*L
u * v *
0
x * y *
We expand out the second derivative in the momentum equation by writing it as the derivative of the derivative. Upon substitution:
u *V0
u *V0
1 p
u *V0
*
u V0
v V0
x* L
x* L
y*L
y*L y*L
*
u*
Simplifying this expression yields:
*
1 p
2 u * Now every term in this equation has been non-dimensionalized except the
u *
* u
v
V02 x * V0 L y * 2 pressure gradient. We define a dimensionless pressure as:
y *
x *
p*
p
V02
Substituting this into the momentum equation:
u*
Simplifying this expression yields:
The dimensionless group is
ν
V0 L
*
1 p * V02
2u *
u *
* u
u
v
V0 L y * 2
V02
x *
y *
x *
*
which is the reciprocal of the Reynolds number.
*
2u *
u *
p *
* u
v
x *
y *
x * V0 L y * 2
Problem 7.6
Given:
Equations for modeling atmospheric motion
Find:
Non-dimensionalized equation; Dimensionless groups
[Difficulty: 2]
Solution:
Recall that the total acceleration is
DV V
V V
t
Dt
Nondimensionalizing the velocity vector, pressure, angular velocity, spatial measure, and time, (using a typical velocity magnitude V
and angular velocity magnitude ):
V
V*
V
Hence
V VV *
*
p
p*
p
x*
*
p p p *
x
L
t* t
x Lx*
V
L
t
L
t*
V
Substituting into the governing equation
V V *
V
1 p
V
V V * * V * 2V * V *
p *
L t *
L
L
The final dimensionless equation is
The dimensionless groups are
p
V *
L
p *
V * * V * 2
* V
V
t *
V2
p
V
2
L
V
The second term on the left of the governing equation is the Coriolis force due to a rotating coordinate system. This is a very
significant term in atmospheric studies, leading to such phenomena as geostrophic flow.
Problem 7.5
[Difficulty: 2]
Given:
Find:
Equation describing two-dimensional steady flow in a liquid
Solution:
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
Nondimensionalization for the equation using length scale L and velocity scale V 0. Obtain the dimensionless
groups that characterize the flow.
x*
x
L
h*
h
L
u*
u
V0
y*
y
L
Substituting into the governing equation:
u *V0
2 u *V0
h * L 2 u *V0
u V0
g
x* L
x * L x * L x * L y * L y * L
*
Simplifying this expression:
V02 * u *
h * V0
u
g * 2
L
x *
x
L
2u * 2u *
x * 2 y * 2
Thus:
The dimensionless groups are
g L
V0
reciprocal of the Reynolds number.
2
u*
gL h *
2 u * 2 u *
u *
V02 x * V0 L x * 2 y * 2
x *
which is the reciprocal of the square of the Froude number, and
μ
V0 ρ L
which is the
Problem 7.4
[Difficulty: 2]
Given:
Find:
Equation describing one-dimensional unsteady flow in a thin liquid layer
Solution:
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
Nondimensionalization for the equation using length scale L and velocity scale Vo. Obtain the dimensionless
groups that characterize the flow.
x*
x
L
h*
h
L
u*
u
V0
t*
t
L V0
u *V0
u *V0
h* L
*
u
V
g
Substituting into the governing equation:
0
t * L V0
x* L
x* L
V02 u * V02 * u *
h *
u
g
L t *
L
x *
x *
The dimensionless group is
g L
V0
2
Simplifying this expression:
Thus:
which is the reciprocal of the square of the Froude number.
*
gL h *
u *
* u
u
t *
x *
V02 x *
Problem 7.3
[Difficulty: 2]
Given:
Find:
Equation describing the slope of a steady wave in a shallow liquid layer
Solution:
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
Nondimensionalization for the equation using length scale L and velocity scale V o. Obtain the dimensionless
groups that characterize the flow.
h*
Substituting into the governing equation:
The dimensionless group is
V0
h
L
x*
x
L
u*
u
V0
h* L
u*V0 u *V0
x* L
g x* L
2
g L
which is the square of the Froude number.
h*
V02 * u *
u
x*
gL x*
Problem 7.2
Given:
Equation for beam
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Denoting nondimensional quantities by an asterisk
A*
Hence
A
L2
A L2 A *
Substituting into the governing equation
The final dimensionless equation is
The dimensionless group is
y*
y
L
t* t
y Ly*
t
I*
t*
I
L4
I L4 I *
x*
x
L
x Lx*
2 y *
4 y *
4 1
L L A *
EL 4 LI *
0
t *2
L
x *4
2 y * E 4 y *
I *
A*
0
t *2 L2 2 x *4
2
2
E
2 2
L
Problem 7.1
[Difficulty: 2]
Given:
Find:
Equation describing the propagation speed of surface waves in a region of uniform depth
Solution:
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
Nondimensionalization for the equation using length scale L and velocity scale Vo. Obtain the dimensionless
groups that characterize the flow.
*
Substituting into the governing equation:
h*
L
c V
2
*
0
h
L
c*
c
V0
2 g* L
2h* L
tanh
*
* L
2
L
Simplifying this expression:
2 gL *
2
c*
2
2
*
LV0 V0 2
The dimensionless group is
g L
V0
which is the reciprocal of the square of the Froude number, and
2
which is the inverse of the Weber number.
σ
ρ L V0
2
2h*
tanh *
Problem 6.127
[Difficulty: 4] Part 1/2
Problem 6.127
[Difficulty: 4] Part 2/2
Problem 6.126
[Difficulty: 3] Part 1/2
Problem 6.126
[Difficulty: 3] Part 2/2
Problem 6.125
[Difficulty: 4]
Problem 6.124
[Difficulty: 3] Part 1/2
Problem 6.124
[Difficulty: 3] Part 2/2
Problem 6.123
[Difficulty: 3]
Problem 6.122
[Difficulty: 3]
Problem 6.121
[Difficulty: 3]
Problem 6.120
[Difficulty: 2]
Problem 6.119
[Difficulty: 3]
Open-Ended Problem Statement: Consider flow around a circular cylinder with
freestream velocity from right to left and a counterclockwise free vortex. Show that the
lift force on the cylinder can be expressed as FL = −ρUΓ, as illustrated in Example 6.12.
Discussion: The only change in this flow from the flow of Example 6.12 is that the
directions of the freestream velocity and the vortex are changed. This changes the sign of
the freestream velocity from U to −U and the sign of the vortex strength from K to −K.
Consequently the signs of both terms in the equation for lift are changed. Therefore the
direction of the lift force remains unchanged.
The analysis of Example 6.12 shows that only the term involving the vortex strength
contributes to the lift force. Therefore the expression for lift obtained with the changed
freestream velocity and vortex strength is identical to that derived in Example 6.12. Thus
the general solution of Example 6.12 holds for any orientation of the freestream and
vortex velocities. For the present case, FL = −ρUΓ, as shown for the general case in
Example 6.12.
Problem 6.118
[Difficulty: 3] Part 1/2
Problem 6.118
[Difficulty: 3] Part 2/2
Problem 6.117
[Difficulty: 3]
Problem 6.116
[Difficulty: 3]
Problem 6.115
[Difficulty: 2] Part 1/2
Problem 6.115
[Difficulty: 2] Part 2/2
Problem 6.114
[Difficulty: 2]
Problem 6.113
[Difficulty: 4]
Given:
Complex function
Find:
Show it leads to velocity potential and stream function of irrotational incompressible flow; Show that df/dz
leads to u and v
Solution:
Basic equations: u = ∂ ψ
∂y
First consider
∂
∂x
f =
2
Hence
∂
2
∂x
2
Combining
∂
∂x
2
∂
∂
v=− ψ
∂x
∂
u=− φ
∂x
d
d
d
f = 1⋅ f = f
dz
dz
∂x dz
f =
f +
z⋅
(1)
and also
⎛ ∂ ⎞ d ⎛ d ⎞ d2
⎜ f = ⎜ f = 2f
∂x ⎝ ∂x ⎠ dz ⎝ dz ⎠
dz
2
∂y
2
f =
d
2
dz
2
f −
d
2
dz
2
f =0
and
∂y
f =
∂
∂y
2
∂
d
d
d
f = i⋅ f = i⋅ f
dz
dz
∂y dz
f =
z⋅
(2)
2
⎛∂ ⎞
d ⎛ d ⎞
d
⎜ f = i⋅ ⎜ i⋅ f = − 2 f
dz ⎝ dz ⎠
∂y ⎝ ∂y ⎠
dz
∂
Any differentiable function f(z) automatically satisfies the Laplace
Equation; so do its real and imaginary parts!
We demonstrate derivation of velocities u and v
From Eq 1
d
∂
∂
∂
∂
f = f = ( φ − i⋅ ψ) = φ − i⋅ ψ = −u + i⋅ v
dz
∂x
∂x
∂x
∂x
From Eq 2
1 ∂
1 ∂
d
∂
∂
f = ⋅ f = ⋅ ( φ − i⋅ ψ) = −i⋅ φ − ψ = i⋅ v − u
i ∂y
i ∂y
dz
∂y
∂y
Hence we have demonstrated that
∂
2
∂
∂
∂
v=− φ
∂y
df
dz
= −u + i⋅ v
Problem 6.112
[Difficulty: 4]
Given:
Complex function
Find:
Show it leads to velocity potential and stream function of irrotational incompressible flow; Show that df/dz leads
to u and v
Solution:
u=
Basic equations: Irrotationality because φ exists
∂
Incompressibility
6
f ( z) = z = ( x + i ⋅ y )
6
u +
∂x
∂
∂y
v =0
∂
∂y
∂
v=− ψ
∂x
ψ
Irrotationality
6
∂
∂x
4 2
v −
∂
∂y
2 4
∂
u=− φ
∂x
∂
v=− φ
∂y
u =0
(
6
5
5
We are thus to check the following
6
4 2
2 4
φ( x , y ) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
∂
u ( x , y ) = − φ( x , y )
∂x
so
∂
v ( x , y ) = − φ( x , y )
∂y
so
6
(
)
5
5
3 3
3 2
5
4
4
2 3
5
3 2
5
4
4
2 3
5
ψ( x , y ) = − 6 ⋅ x ⋅ y + 6 ⋅ x ⋅ y − 20⋅ x ⋅ y
u ( x , y ) = 60⋅ x ⋅ y − 6 ⋅ x − 30⋅ x ⋅ y
v ( x , y ) = 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6 ⋅ y
An alternative derivation of u and v is
u( x , y) =
∂
∂y
ψ( x , y )
u ( x , y ) = 60⋅ x ⋅ y − 6 ⋅ x − 30⋅ x ⋅ y
∂
v ( x , y ) = − ψ( x , y )
∂x
Hence
Hence
Next we find
Hence we see
∂
∂x
∂
∂x
v( x , y) −
u( x , y) +
df
dz
df
dz
=
∂
∂y
∂
∂y
v ( x , y ) = 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6 ⋅ y
u( x , y) = 0
Hence flow is IRROTATIONAL
v( x , y) = 0
Hence flow is INCOMPRESSIBLE
( 6) = 6⋅z5 = 6⋅(x + i⋅y)5 = (6⋅x5 − 60⋅x3⋅y2 + 30⋅x⋅y4) + i⋅(30⋅x4⋅y + 6⋅y5 − 60⋅x2⋅y3)
dz
d z
= −u + i⋅ v
Hence the results are verified;
These interesting results are explained in Problem 6.113!
u = −Re⎛⎜
⎞
dz
⎝ ⎠
df
and
)
3 3
f ( z) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y + i⋅ 6 ⋅ x ⋅ y + 6 ⋅ x ⋅ y − 20⋅ x ⋅ y
Expanding
v = Im⎛⎜
df
⎞
⎝ dz ⎠
Problem 6.111
Given:
Velocity potential
Find:
Show flow is incompressible; Stream function
[Difficulty: 2]
Solution:
u=
Basic equations: Irrotationality because φ exists
∂
Incompressibility
6
∂x
u +
4 2
∂
∂y
Hence
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
2 4
6
∂
u ( x , y ) = − φ( x , y )
∂x
u ( x, y ) = 60⋅ x ⋅ y − 6⋅ x − 30⋅ x⋅ y
∂
v ( x , y ) = − φ( x , y )
∂y
v ( x, y ) = 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6⋅ y
∂
Hence flow is INCOMPRESSIBLE
∂x
u ( x, y ) +
u=
∂
∂y
ψ
∂
v=− ψ
∂x
∂
∂y
v ( x, y ) = 0
∂
v=− φ
∂y
v =0
φ( x, y ) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
Hence
∂
3 2
5
4
4
2 3
5
so
⌠
3 3
5
5
ψ( x , y ) = ⎮ u ( x , y ) dy + f ( x ) = 20⋅ x ⋅ y − 6 ⋅ x ⋅ y − 6 ⋅ x ⋅ y + f ( x )
⌡
so
⌠
3 3
5
5
ψ( x , y ) = −⎮ v ( x , y ) dx + g ( y ) = 20⋅ x ⋅ y − 6 ⋅ x ⋅ y − 6 ⋅ x ⋅ y + g ( y )
⌡
Comparing, the simplest stream function is then
3 3
5
ψ( x , y ) = 20⋅ x ⋅ y − 6 ⋅ x ⋅ y − 6 ⋅ x ⋅ y
5
Problem 6.110
Given:
Velocity potential
Find:
Show flow is incompressible; Stream function
[Difficulty: 2]
Solution:
u=
Basic equations: Irrotationality because φ exists
∂
Incompressibility
We have
Hence
Hence
5
∂x
u +
3 2
∂
∂y
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
v =0
4
2
φ( x , y ) = x − 10⋅ x ⋅ y + 5 ⋅ x ⋅ y − x + y
2
∂
u ( x , y ) = − φ( x , y )
∂x
u ( x , y ) = 30⋅ x ⋅ y − 5 ⋅ x + 2 ⋅ x − 5 ⋅ y
∂
v ( x , y ) = − φ( x , y )
∂y
v ( x , y ) = 20⋅ x ⋅ y − 20⋅ x ⋅ y − 2 ⋅ y
∂
Hence flow is INCOMPRESSIBLE
∂x
u( x , y) +
u=
∂
∂y
ψ
∂
v=− ψ
∂x
∂
∂y
2 2
4
3
v( x , y) = 0
∂
v=− φ
∂y
4
3
so
⌠
2 3
4
5
ψ( x , y ) = ⎮ u ( x , y ) dy + f ( x ) = 10⋅ x ⋅ y − 5 ⋅ x ⋅ y + 2 ⋅ x ⋅ y − y + f ( x )
⌡
so
⌠
2 3
4
ψ( x , y ) = −⎮ v ( x , y ) dx + g ( y ) = 10⋅ x ⋅ y − 5 ⋅ x ⋅ y + 2 ⋅ x ⋅ y + g ( y )
⌡
Comparing, the simplest stream function is then
2 3
4
ψ( x , y ) = 10⋅ x ⋅ y − 5 ⋅ x ⋅ y + 2 ⋅ x ⋅ y − y
5
Problem 6.109
Given:
Velocity potential
Find:
Show flow is incompressible; Stream function
[Difficulty: 2]
Solution:
u=
Basic equations: Irrotationality because φ exists
Incompressibility
We have
Hence
Hence
∂
∂x
u +
2
∂
∂y
φ( x , y ) = A⋅ x + B⋅ x ⋅ y − A⋅ y
∂
∂y
ψ
∂
v=− ψ
∂x
∂
u=− φ
∂x
v =0
2
∂
u ( x , y ) = − φ( x , y )
∂x
u ( x , y ) = −2 ⋅ A⋅ x − B⋅ y
∂
v ( x , y ) = − φ( x , y )
∂y
v ( x , y ) = 2 ⋅ A⋅ y − B⋅ x
∂
Hence flow is INCOMPRESSIBLE
∂x
u( x , y) +
u=
∂
∂y
ψ
∂
v=− ψ
∂x
∂
∂y
v( x , y) = 0
∂
v=− φ
∂y
so
⌠
1
2
ψ( x , y ) = ⎮ u ( x , y ) dy + f ( x ) = −2 ⋅ A⋅ x ⋅ y − ⋅ B⋅ y + f ( x )
2
⌡
so
⌠
1
2
ψ( x , y ) = −⎮ v ( x , y ) dx + g ( y ) = −2 ⋅ A⋅ x ⋅ y + ⋅ B⋅ x + g ( y )
2
⌡
Comparing, the simplest stream function is then
ψ( x , y ) = −2 ⋅ A⋅ x ⋅ y +
1
2
2
⋅ B⋅ x −
1
2
⋅ B⋅ y
2
Problem 6.108
[Difficulty: 2]
Given:
Stream function
Find:
Velocity field; Show flow is irrotational; Velocity potential
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
We have
6
v −
∂y
4 2
∂
∂y
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
2 4
6
2 3
ψ( x , y )
4
u ( x , y ) = 60⋅ x ⋅ y − 30⋅ x ⋅ y − 6 ⋅ y
∂
v ( x , y ) = − ψ( x , y )
∂x
v ( x , y ) = 60⋅ x ⋅ y − 6 ⋅ x − 30⋅ x ⋅ y
∂
Hence flow is IRROTATIONAL
∂x
v( x , y) −
∂
∂y
3 2
u( x , y) = 0
∂
v=− φ
∂y
u =0
ψ( x , y ) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
u( x , y) =
Hence
∂x
∂
u=
5
5
4
∂
u=− φ
∂x
so
⌠
5
3 3
5
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) = 6 ⋅ x ⋅ y − 20⋅ x ⋅ y + 6 ⋅ x ⋅ y + f ( y )
⌡
∂
v=− φ
∂y
so
⌠
5
3 3
5
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) = 6 ⋅ x ⋅ y − 20⋅ x ⋅ y + 6 ⋅ x ⋅ y + g ( x )
⌡
Comparing, the simplest velocity potential is then
5
3 3
φ( x , y ) = 6 ⋅ x ⋅ y − 20⋅ x ⋅ y + 6 ⋅ x ⋅ y
5
Problem 6.107
[Difficulty: 2]
Given:
Stream function
Find:
Find A vs B if flow is irrotational; Velocity potential
Solution:
u=
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
We have
3
(
v −
∂y
2
2
∂
∂y
∂y
ψ
∂
v=− ψ
∂x
∂
u=− φ
∂x
2
)
ψ( x , y )
u ( x , y ) = −B⋅ ( 2 ⋅ y − 2 ⋅ x ⋅ y )
(2
∂
v ( x , y ) = − ψ( x , y )
∂x
v ( x , y ) = −3 ⋅ A⋅ x − B⋅ y + 2 ⋅ x
∂
Hence flow is IRROTATIONAL if
∂x
v( x , y) −
∂
∂y
∂
v=− φ
∂y
u =0
ψ( x , y ) = A⋅ x + B⋅ x ⋅ y + x − y
u( x , y) =
Hence
∂x
∂
∂
2
u ( x , y ) = −2 ⋅ x ⋅ ( 3 ⋅ A + B)
)
B = −3 ⋅ A
∂
u=− φ
∂x
so
⌠
2
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) = 2 ⋅ B⋅ y ⋅ x − B⋅ y ⋅ x + f ( y )
⌡
∂
v=− φ
∂y
so
3
⌠
B⋅ y
2
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) = 3 ⋅ A⋅ x ⋅ y + 2 ⋅ B⋅ x ⋅ y +
+ g( x)
3
⌡
Comparing, the simplest velocity potential is then
2
φ( x , y ) = 2 ⋅ B⋅ y ⋅ x − B⋅ y ⋅ x + B⋅
y
3
3
Problem 6.106
Given:
Stream function
Find:
Velocity potential
[Difficulty: 2]
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
3
ψ( x , y ) = A⋅ x − B⋅ x ⋅ y
We have
Then
∂x
u( x , y) =
∂
∂y
ψ( x , y )
Hence
∂
∂x
v( x , y) −
∂
∂y
∂
∂y
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
∂
v=− φ
∂y
u =0
2
u ( x , y ) = −2 ⋅ B⋅ x ⋅ y
∂
v ( x , y ) = − ψ( x , y )
∂x
Then
v −
u=
2
v ( x , y ) = B⋅ y − 3 ⋅ A⋅ x
u ( x , y ) = 2 ⋅ B⋅ x − 6 ⋅ A⋅ x
but
2
2⋅ B − 6⋅ A = 0
1
m⋅ s
hence flow is IRROTATIONAL
∂
u=− φ
∂x
so
⌠
2
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) → φ( x , y ) = B⋅ y ⋅ x + f ( y )
⌡
∂
v=− φ
∂y
so
3
⌠
B⋅ y
2
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) → φ( x , y ) = g ( x ) −
+ 3 ⋅ A⋅ x ⋅ y
3
⌡
Comparing, the simplest velocity potential is then
2
φ( x , y ) = 3 ⋅ A⋅ x ⋅ y −
B⋅ y
3
3
Problem 6.105
[Difficulty: 2]
Given:
Stream function
Find:
Velocity field; Show flow is irrotational; Velocity potential
Solution:
u=
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
∂x
5
v −
∂
∂y
3 2
∂y
ψ( x , y )
3
3
2 2
4
∂
Hence flow is IRROTATIONAL
∂
∂y
∂
v=− φ
∂y
u ( x , y ) = 20⋅ x ⋅ y − 20⋅ x ⋅ y
v ( x , y ) = 30⋅ x ⋅ y − 5 ⋅ x − 5 ⋅ y
v( x , y) −
∂
u=− φ
∂x
4
∂
v ( x , y ) = − ψ( x , y )
∂x
∂x
Hence
∂
∂y
∂
v=− ψ
∂x
ψ
u =0
ψ( x , y ) = x − 10⋅ x ⋅ y + 5 ⋅ x ⋅ y
u( x , y) =
∂
u( x , y) = 0
4
∂
u=− φ
∂x
so
⌠
4
2 3
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) = 5 ⋅ x ⋅ y − 10⋅ x ⋅ y + f ( y )
⌡
∂
v=− φ
∂y
so
⌠
4
2 3
5
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) = 5 ⋅ x ⋅ y − 10⋅ x ⋅ y + y + g ( x )
⌡
Comparing, the simplest velocity potential is then
4
2 3
φ( x , y ) = 5 ⋅ x ⋅ y − 10⋅ x ⋅ y + y
5
Problem 6.104
Given:
Stream function
Find:
Velocity potential
[Difficulty: 2]
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
2
ψ ( x, y ) = A ⋅ x ⋅ y − B⋅ y
We have
Then
∂x
u ( x, y ) =
∂
∂y
Hence
∂
∂x
v ( x, y ) −
∂
∂y
∂
∂y
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
∂
v=− φ
∂y
u =0
3
2
ψ ( x, y )
u ( x, y ) = A ⋅ x − 3⋅ B⋅ y
∂
v ( x, y ) = − ψ ( x, y )
∂x
Then
v −
u=
2
v ( x, y ) = −2⋅ A ⋅ x⋅ y
u ( x, y ) = 6⋅ B⋅ y − 2⋅ A ⋅ y
but
6⋅ B − 2⋅ A = 0
1
m⋅ s
hence flow is IRROTATIONAL
∂
u=− φ
∂x
so
3
⌠
A⋅ x
2
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) → φ( x , y ) = f ( y ) −
+ 3⋅ B⋅ x⋅ y
3
⌡
∂
v=− φ
∂y
so
⌠
2
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) → φ( x , y ) = A⋅ x ⋅ y + g ( x )
⌡
Comparing, the simplest velocity potential is then
2
φ( x , y ) = A⋅ x ⋅ y −
A⋅ x
3
3
Problem 6.103
[Difficulty: 3]
Given:
Data from Table 6.2
Find:
Stream function and velocity potential for a vortex in a corner; plot; velocity along one plane
Solution:
From Table 6.2, for a vortex at the origin
ϕ( r , θ) =
Expressed in Cartesian coordinates
ϕ( x , y ) =
K
2⋅ π
⋅θ
q
2⋅ π
ψ( r , θ) = −
⋅ atan⎛⎜
y⎞
K
2⋅ π
ψ( x , y ) = −
⎝x⎠
⋅ ln( r)
q
4⋅ π
(2
⋅ ln x + y
)
2
To build flow in a corner, we need image vortices at three locations so that there is symmetry about both axes. We need vortices
at (h,h), (h,- h), (- h,h), and (- h,- h). Note that some of them must have strengths of - K!
Hence the composite velocity potential and stream function are
ϕ( x , y ) =
K
2⋅ π
ψ( x , y ) = −
y − h⎞
⋅ ⎛⎜ atan⎛⎜
⎝
K
4⋅ π
⎝x − h⎠
− atan⎛⎜
y + h⎞
⎝x − h⎠
+ atan⎛⎜
y + h⎞
⎝x + h⎠
− atan⎛⎜
y − h ⎞⎞
⎝ x + h ⎠⎠
⎡ ( x − h) 2 + ( y − h) 2 ( x + h) 2 + ( y + h) 2⎤
⎥
⋅
⎢ ( x − h) 2 + ( y + h) 2 ( x + h) 2 + ( y − h) 2⎥
⎣
⎦
⋅ ln⎢
By a similar reasoning the horizontal velocity is given by
u=−
K⋅ ( y − h )
2 ⋅ π⎡⎣( x − h ) + ( y − h )
2
2⎤
+
⎦
K⋅ ( y + h )
2 ⋅ π⎡⎣( x − h ) + ( y + h )
2
2⎤
⎦
−
K⋅ ( y + h )
2 ⋅ π⎡⎣( x + h ) + ( y + h )
2
2⎤
⎦
+
K⋅ ( y − h )
2 ⋅ π⎡⎣( x + h ) + ( y − h )
2
Along the horizontal wall (y = 0)
u=
or
K⋅ h
2 ⋅ π⎡⎣( x − h ) + h
u(x) =
2
K⋅ h
⋅⎡
2⎤
⎦
+
K⋅ h
2 ⋅ π⎡⎣( x − h ) + h
2
2⎤
⎦
1
1
⎤
−
⎢
2
2
2
2⎥
π
( x + h) + h ⎦
⎣ ( x − h) + h
−
K⋅ h
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
−
K⋅ h
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
2⎤
⎦
In Excel:
y
Stream Function
x
Velocity Potential
y
x
Problem 6.102
[Difficulty: 3]
Given:
Velocity field of irrotational and incompressible flow
Find:
Stream function and velocity potential; plot
Solution:
The velocity field is
u=
q⋅ x
2 ⋅ π⎡⎣x + ( y − h )
2
∂
2⎤
+
⎦
q⋅ x
2 ⋅ π⎡⎣x + ( y + h )
2
∂
v=− ψ
∂x
2⎤
v=
⎦
∂
u=− ϕ
∂x
q⋅ ( y − h)
2 ⋅ π⎡⎣x + ( y − h )
2
∂
v=− ϕ
∂y
The basic equations are
u=
Hence for the stream function
⌠
q ⎛
y − h⎞
y + h ⎞⎞
ψ = ⎮ u ( x , y ) dy =
⋅ ⎜ atan⎛⎜
+ atan⎛⎜
+ f ( x)
2⋅ π ⎝
⌡
⎝ x ⎠
⎝ x ⎠⎠
∂y
ψ
⌠
q ⎛
y − h⎞
y + h ⎞⎞
ψ = −⎮ v ( x , y ) dx =
⋅ ⎜ atan⎛⎜
+ atan⎛⎜
+ g( y)
2⋅ π ⎝
⌡
⎝ x ⎠
⎝ x ⎠⎠
q
⋅ ⎛⎜ atan⎛⎜
y − h⎞
+ atan⎛⎜
y + h ⎞⎞
The simplest expression for ψ is
ψ( x , y ) =
For the stream function
⌠
q
2
2
2
2
ϕ = −⎮ u ( x , y ) dx = −
⋅ ln⎡⎣⎡⎣x + ( y − h ) ⎤⎦ ⋅ ⎡⎣x + ( y + h ) ⎤⎦⎤⎦ + f ( y )
4⋅ π
⌡
2⋅ π
⎝
⎝ x ⎠
⎝ x ⎠⎠
⌠
q
2
2
2
2
ϕ = −⎮ v ( x , y ) dy = −
⋅ ln⎡⎣⎡⎣x + ( y − h ) ⎤⎦ ⋅ ⎡⎣x + ( y + h ) ⎤⎦⎤⎦ + g ( x )
4⋅ π
⌡
The simplest expression for φ is
ϕ( x , y ) = −
q
4⋅ π
⋅ ln⎡⎣⎡⎣x + ( y − h ) ⎤⎦ ⋅ ⎡⎣x + ( y + h )
2
2
2
2⎤⎤
⎦⎦
2⎤
⎦
+
q⋅ ( y + h)
2 ⋅ π⎡⎣x + ( y + h )
2
2
In Excel:
Stream Function
Velocity Potential
Problem 6.101
[Difficulty: 3]
Given:
Data from Table 6.2
Find:
Stream function and velocity potential for a source in a corner; plot; velocity along one plane
Solution:
From Table 6.2, for a source at the origin
ψ( r , θ) =
Expressed in Cartesian coordinates
ψ( x , y ) =
q
2⋅ π
⋅θ
q
2⋅ π
ϕ( r , θ) = −
⋅ atan⎛⎜
y⎞
q
2⋅ π
ϕ( x , y ) = −
⎝x⎠
⋅ ln( r)
q
4⋅ π
(2
⋅ ln x + y
)
2
To build flow in a corner, we need image sources at three locations so that there is symmetry about both axes. We need sources at
(h,h), (h,- h), (- h,h), and (- h,- h)
Hence the composite stream function and velocity potential are
ψ( x , y ) =
q
2⋅ π
ϕ( x , y ) = −
⋅ ⎛⎜ atan⎛⎜
⎝
q
4⋅ π
y − h⎞
⎝x − h⎠
+ atan⎛⎜
y + h⎞
+ atan⎛⎜
⎝x − h⎠
y + h⎞
⎝x + h⎠
⋅ ln⎡⎣⎡⎣( x − h ) + ( y − h ) ⎤⎦ ⋅ ⎡⎣( x − h ) + ( y + h )
2
+ atan⎛⎜
2
2
y − h ⎞⎞
⎝ x + h ⎠⎠
2⎤⎤
⎦⎦ −
q
⋅ ⎡⎣( x + h ) + ( y + h ) ⎤⎦ ⋅ ⎡⎣( x + h ) + ( y − h )
2
4⋅ π
2
2
2⎤
⎦
By a similar reasoning the horizontal velocity is given by
u=
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + ( y − h )
2
2⎤
⎦
+
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + ( y + h )
2
2⎤
⎦
+
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + ( y + h )
2
2⎤
+
⎦
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + ( y + h )
2
Along the horizontal wall (y = 0)
u=
or
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + h
u(x) =
2
q
π
2⎤
⎦
+
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + h
2
x−h
x+h
⎤
+
⎢
2
2
2
2⎥
( x + h) + h ⎦
⎣ ( x − h) + h
⋅⎡
2⎤
⎦
+
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
+
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
2⎤
⎦
The results in Excel are:
Velocity Potential
y
x
Stream Function
y
x
Problem 6.100
[Difficulty: 2]
Problem 6.99
[Difficulty: 3]
Given:
Stream function
Find:
If the flow is irrotational; Pressure difference between points (1,4) and (2,1)
Solution:
u=
Basic equations: Incompressibility because ψ exists
2
∂
∂y
∂
v=− ψ
∂x
ψ
∂
Irrotationality
∂x
v −
∂
∂y
u =0
ψ( x , y ) = A⋅ x ⋅ y
u( x , y) =
∂
∂y
ψ( x , y ) =
∂
∂y
(A⋅x2⋅y)
(
2
∂
∂
v ( x , y ) = − ψ( x , y ) = − A⋅ x ⋅ y
∂x
∂x
Hence
∂
∂x
v( x , y) −
∂
∂y
2
u ( x , y ) = A⋅ x
)
v ( x , y ) = −2 ⋅ A⋅ x ⋅ y
∂
u ( x , y ) = −2 ⋅ A⋅ y
∂x
v −
∂
∂y
u ≠0
so flow is NOT IRROTATIONAL
Since flow is rotational, we must be on same streamline to be able to use Bernoulli
At point (1,4)
ψ( 1 , 4 ) = 4 A
ψ( 2 , 1 ) = 4 A
and at point (2,1)
Hence these points are on same streamline so Bernoulli can be used. The velocity at a point is
V( x , y ) =
Hence at (1,4)
V1 =
2
2
⎡ 2.5 × ( 1 ⋅ m) 2⎤ + ⎛ −2 × 2.5 × 1⋅ m × 4⋅ m⎞
⎢
⎥ ⎜
m⋅ s
⎣ m⋅ s
⎦ ⎝
⎠
m
V1 = 20.2
s
Hence at (2,1)
2
2
2.5
2.5
2⎤
⎡
⎛
⎞
V2 = ⎢
× ( 2 ⋅ m) ⎥ + ⎜ −2 ×
× 2⋅ m × 1⋅ m
m⋅ s
⎣ m⋅ s
⎦ ⎝
⎠
m
V2 = 14.1
s
Using Bernoulli
p1
ρ
+
∆p =
1
2
2
⋅ V1 =
1
2
p2
+
ρ
× 1200⋅
kg
3
m
1
2
⋅ V2
(
2
∆p =
)
× 14.1 − 20.2 ⋅ ⎛⎜
2
2
m⎞
⎝s⎠
2
2
u( x , y) + v( x , y)
⋅ ⎛ V − V1
2 ⎝ 2
ρ
2
2
×
N⋅ s
kg⋅ m
∆p = −126 ⋅ kPa
2⎞
⎠
2
Problem 6.98
[Difficulty: 3]
Given:
2D incompressible, inviscid flow field
Find:
Relationships among constants; stream function; velocity potential
Solution:
Basic equations
u=
Incompressible
∂
∂
v=− ψ
∂x
ψ
∂y
u ( x , y ) = a⋅ x + b ⋅ y
Check incompressibility
∂
∂x
∂
u( x , y) +
v( x , y) −
∂
∂y
∂
Irrotational
∂
v=− ϕ
∂y
v ( x , y ) = c⋅ x + d ⋅ y
v( x , y) = a + d
Hence must have
d = −a
u( x , y) = c − b
Hence must have
c = b
Check irrotational
∂x
Hence for the streamfunction
⌠
1
2
ψ ( x , y ) = ⎮ u ( x , y ) dy = a ⋅ x ⋅ y + ⋅ b ⋅ y + f ( x )
2
⌡
∂y
∂
u=− ϕ
∂x
⌠
1
2
ψ( x , y ) = −⎮ v ( x , y ) dx = − ⋅ c⋅ x − d ⋅ x ⋅ y + g ( y )
2
⌡
Comparing
Check
Hence for the velocity potential
ψ( x , y ) =
u( x , y) =
1
2
2
⋅ b⋅ y −
∂
∂y
1
2
2
⋅ c⋅ x + a⋅ x ⋅ y
ψ( x , y ) = a ⋅ x + b ⋅ y
ψ( x , y ) = a ⋅ x ⋅ y +
1
2
(2
⋅ b⋅ y − x
)
2
∂
v ( x , y ) = − ψ( x , y ) = b ⋅ x − a ⋅ y
∂x
⌠
1
2
ϕ( x , y ) = −⎮ u ( x , y ) dx = − ⋅ a⋅ x − b ⋅ x ⋅ y + f ( y )
2
⌡
⌠
1
2
ψ( x , y ) = −⎮ v ( x , y ) dy = −c⋅ x ⋅ y − ⋅ d ⋅ y + g ( x )
2
⌡
Comparing
Check
1
(2
)
1
2 1
2
ϕ( x , y ) = − ⋅ a⋅ x − ⋅ d ⋅ y − b ⋅ x ⋅ y
2
2
ϕ( x , y ) = −b ⋅ x ⋅ y −
∂
u ( x , y ) = − ϕ( x , y ) = a⋅ x + b ⋅ y
∂x
∂
v ( x , y ) = − ϕ( x , y ) = b ⋅ x − a⋅ y
∂y
2
⋅ a⋅ x − y
2
Problem 6.97
[Difficulty: 2]
Problem 6.96
(a)
[Difficulty: 2]
Note that the effect of friction would be that the EGL would tend to drop from
right to left: steeply in the small pipe, gradually in the large pipe, and
suddenly at the expansion. The HGL would then “hang” below the HGL in a
manner similar to that shown.
EGL
Flow
Pump
HGL
(b)
Note that the effect of friction would be that the EGL would tend to drop from
right to left: steeply in the small pipe, gradually in the large pipe, and
suddenly at the expansion. The HGL would then “hang” below the HGL in a
manner similar to that shown.
EGL
Flow
HGL
Pump
Problem 6.95
(a)
[Difficulty: 2]
Note that the effect of friction would be that the EGL would tend to drop:
suddenly at the contraction, gradually in the large pipe, more steeply in the
small pipe. The HGL would then “hang” below the HGL in a manner similar
to that shown.
EGL
Turbine
HGL
(b)
Note that the effect of friction would be that the EGL would tend to drop:
suddenly at the contraction, gradually in the large pipe, more steeply in the
small pipe. The HGL would then “hang” below the HGL in a manner similar
to that shown.
EGL
Turbine
HGL
Problem 6.94
[Difficulty: 5]
Problem 6.93
[Difficulty: 5] Part 1/2
Problem 6.93
4.48
[Difficulty: 5] Part 2/2
.04
Problem 6.92
[Difficulty: 4]
Given:
Unsteady water flow out of tube
Find:
Differential equation for velocity; Integrate; Plot v versus time
Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0
Applying unsteady Bernoulli between reservoir and tube exit
2
⌠
2
2
2
⎮ ∂
V
V
p
dV ⌠
dV
+⎮
+ g⋅ h =
V ds =
+
⋅ ⎮ 1 ds =
+
⋅L
⌡
2
2
2
ρ
dt
dt
⎮ ∂t
1
⌡
2
V
where we work in gage pressure
1
dV
Hence
dt
2
+
V
2⋅ L
=
1
L
L⋅ dV
Separating variables
p
ρ
2
+ g⋅ h −
⋅ ⎛⎜
p
⎝ρ
+ g⋅ h ⎞
is the differential equation for the flow
⎠
= dt
V
2
Integrating and using limits V(0) = 0 and V(t) = V
⎛⎜ p + g ⋅ h ⎞
ρ
V( t) = 2 ⋅ ⎜ + g ⋅ h⎞ ⋅ tanh⎜
⋅ t⎟
2
ρ
⎜
⎝
⎠
⎝ 2⋅ L
⎠
⎛p
25
V (ft/s)
20
15
10
5
0
1
2
3
t (s)
This graph is suitable for plotting in Excel
For large times
V =
2 ⋅ ⎛⎜
p
⎝ρ
+ g ⋅ h⎞
⎠
V = 22.6⋅
ft
s
4
5
Problem 6.91
[Difficulty: 4]
Problem 6.90
Given:
Unsteady water flow out of tube
Find:
Initial acceleration
[Difficulty: 3]
Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0
Applying unsteady Bernoulli between reservoir and tube exit
2
2
⌠
dV ⌠
∂
+ g⋅ h = ⎮
V ds =
⋅ ⎮ 1 ds = a x ⋅ L
⎮ ∂t
ρ
dt ⌡1
⌡
p
where we work in gage pressure
1
Hence
ax =
Hence
ax =
1
L
⋅ ⎛⎜
p
⎝ρ
1
35⋅ ft
+ g⋅ h⎞
⎠
3
⎤
⎡ lbf ⎛ 12⋅ in ⎞ 2
ft
slug⋅ ft
ft
×⎜
×
×
+ 32.2⋅ × 4.5⋅ ft⎥
⎢ in2 ⎝ 1 ⋅ ft ⎠
1.94⋅ slug
2
2
⎥
s ⋅ lbf
s
⎣
⎦
× ⎢3 ⋅
ax = 10.5⋅
ft
2
s
Note that we obtain the same result if we treat the water in the pipe as a single body at rest with gage pressure p + ρgh at the left end!
Problem 6.89
Given:
Unsteady water flow out of tube
Find:
Pressure in the tank
[Difficulty: 3]
Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0
Applying unsteady Bernoulli between reservoir and tube exit
2
2
2
⌠
dV ⌠
V
∂
+ g⋅ h =
V ds =
+
⋅ ⎮ 1 ds
+⎮
⎮ ∂t
2
2
ρ
dt ⌡1
⌡
2
V
p
where we work in gage pressure
1
Hence
Hence
⎛ V2
p = ρ⋅ ⎜
⎝ 2
p = 1.94⋅
− g⋅ h +
slug
ft
3
×
dV
dt
⎞
⋅L
⎠
⎞ ft 2 lbf s2
⎛ 62
⎜ − 32.2 × 4.5 + 7.5 × 35 ⋅ ⎛⎜ ⎞ × ⋅
slug⋅ ft
⎝2
⎠ ⎝s⎠
p = 263 ⋅
lbf
ft
2
p = 1.83⋅ psi
(gage)
Problem 6.88
[Difficulty: 3]
Problem 6.87
[Difficulty: 5]
Problem 6.86
[Difficulty: 5]
Open-Ended Problem Statement: An aspirator provides suction by using a stream of
water flowing through a venturi. Analyze the shape and dimensions of such a device.
Comment on any limitations on its use.
Discussion: The basic shape of the aspirator channel should be a converging nozzle
section to reduce pressure followed by a diverging diffuser section to promote pressure
recovery. The basic shape is that of a venturi flow meter.
If the diffuser exhausts to atmosphere, the exit pressure will be atmospheric. The pressure
rise in the diffuser will cause the pressure at the diffuser inlet (venturi throat) to be below
atmospheric.
A small tube can be brought in from the side of the throat to aspirate another liquid or gas
into the throat as a result of the reduced pressure there.
The following comments can be made about limitations on the aspirator:
1. It is desirable to minimize the area of the aspirator tube compared to the flow area
of the venturi throat. This minimizes the disturbance of the main flow through the
venturi and promotes the best possible pressure recovery in the diffuser.
2. It is desirable to avoid cavitation in the throat of the venturi. Cavitation alters the
effective shape of the flow channel and destroys the pressure recovery in the
diffuser. To avoid cavitation, the reduced pressure must always be above the
vapor pressure of the driver liquid.
3. It is desirable to limit the flow rate of gas into the venturi throat. A large amount
of gas can alter the flow pattern and adversely affect pressure recovery in the
diffuser.
The best combination of specific dimensions could be determined experimentally by a
systematic study of aspirator performance. A good starting point probably would be to
use dimensions similar to those of a commercially available venturi flow meter.
Problem 6.85
[Difficulty: 5]
Open-Ended Problem Statement: Imagine a garden hose with a stream of water
flowing out through a nozzle. Explain why the end of the hose may be unstable when
held a half meter or so from the nozzle end.
Discussion: Water flowing out of the nozzle tends to exert a thrust force on the end of the
hose. The thrust force is aligned with the flow from the nozzle and is directed toward the
hose.
Any misalignment of the hose will lead to a tendency for the thrust force to bend the hose
further. This will quickly become unstable, with the result that the free end of the hose
will “flail” about, spraying water from the nozzle in all directions.
This instability phenomenon can be demonstrated easily in the backyard. However, it will
tend to do least damage when the person demonstrating it is wearing a bathing suit!
Problem 6.84
[Difficulty: 5]
Open-Ended Problem Statement: Describe the pressure distribution on the exterior of a
multistory building in a steady wind. Identify the locations of the maximum and
minimum pressures on the outside of the building. Discuss the effect of these pressures
on infiltration of outside air into the building.
Discussion: A multi-story building acts as a bluff-body obstruction in a thick
atmospheric boundary layer. The boundary-layer velocity profile causes the air speed
near the top of the building to be highest and that toward the ground to be lower.
Obstruction of air flow by the building causes regions of stagnation pressure on upwind
surfaces. The stagnation pressure is highest where the air speed is highest. Therefore the
maximum surface pressure occurs near the roof on the upwind side of the building.
Minimum pressure on the upwind surface of the building occurs near the ground where
the air speed is lowest.
The minimum pressure on the entire building will likely be in the low-speed, lowpressure wake region on the downwind side of the building.
Static pressure inside the building will tend to be an average of all the surface pressures
that act on the outside of the building. It is never possible to seal all openings completely.
Therefore air will tend to infiltrate into the building in regions where the outside surface
pressure is above the interior pressure, and will tend to pass out of the building in regions
where the outside surface pressure is below the interior pressure. Thus generally air will
tend to move through the building from the upper floors toward the lower floors, and
from the upwind side to the downwind side.
Problem 6.83
[Difficulty: 4] Part 1/2
Problem 6.83
[Difficulty: 4] Part 2/2
Problem 6.82
[Difficulty: 4]
Given:
Water flow out of tube
Find:
Pressure indicated by gage; force to hold body in place
Solution:
Basic equations: Bernoulli, and momentum flux in x direction
p
ρ
2
+
V
+ g ⋅ z = constant
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0)
Applying Bernoulli between jet exit and stagnation point
p1
ρ
+
p1 =
V1
2
2
=
p2
V2
+
ρ
2
⋅ ⎛ V − V1
2 ⎝ 2
ρ
2
V2
=
2
2⎞
2
⎠
2
A1
D
V2 = V1 ⋅
= V1 ⋅
A2
2
2
D −d
But from continuity Q = V1 ⋅ A1 = V2 ⋅ A2
⎞
ft ⎛
2
V2 = 20⋅ ⋅ ⎜
2
s ⎜ 2
⎝ 2 − 1.5 ⎠
2
p1 =
Hence
The x mometum is
1
2
× 1.94⋅
slug
ft
where we work in gage pressure
2
ft
V2 = 45.7⋅
s
2
2 ft
× ( 45.7 − 20 ) ⋅ ⎛⎜ ⎞
s
2
⎝ ⎠
3
(
)
F = p 1 ⋅ A1 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞
⎝
⎠
F = 11.4⋅
lbf
2
in
×
2
×
(
lbf ⋅ s
π⋅ ( 2 ⋅ in)
4
2
+ 1.94⋅
F = 14.1⋅ lbf
slug
ft
3
×
2
p 1 = 1638⋅
slug⋅ ft
−F + p 1 ⋅ A1 − p 2 ⋅ A2 = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2
2
where D = 2 in and d = 1.5 in
lbf
ft
2
p 1 = 11.4⋅ psi
(gage)
)
using gage pressures
2
2
2
2
2
⎡⎛ ft ⎞ 2 π⋅ ( 2⋅ in) 2 ⎛
ft
lbf ⋅ s
π⋅ ⎡⎣( 2 ⋅ in) − ( 1.5⋅ in) ⎤⎦ ⎤ ⎛ 1 ⋅ ft ⎞
⎢⎜ 20⋅
⎥×⎜
×
− ⎜ 45.7⋅ ⎞ ×
×
s⎠
4
4
slug⋅ ft
⎣⎝ s ⎠
⎝
⎦ ⎝ 12⋅ in ⎠
in the direction shown
Problem 6.81
[Difficulty: 4] Part 1/2
Problem 6.81
[Difficulty: 4] Part 2/2
Problem 6.80
Given:
Air flow over "bubble" structure
Find:
Net vertical force
Solution:
The net force is given by
L = 50⋅ ft
Available data
[Difficulty: 3]
→ ⌠
→
F = ⎮ p dA
⌡
R = 25ft
∆p = ρ⋅ g ⋅ ∆h
also
V = 35⋅ mph
∆h = 0.75⋅ in
ρ = 1.94⋅
slug
ft
The internal pressure is
∆p = ρ⋅ g ⋅ ∆h
⌠
FV = ⎮
⌡
π
where pi is the internal pressure and p the external
π
ft
(pi − p)⋅ sin(θ)⋅ R⋅ L dθ
0
⌠
FV = ⎮
⎮
⌡
slug
∆p = 187 Pa
Through symmetry only the vertical component of force is no-zero
Hence
3
ρair = 0.00238 ⋅
(
(
)
1
2
2
p = p atm − ⋅ ρair⋅ V ⋅ 1 − 4 ⋅ sin( θ)
2
p i = p atm + ∆p
)
⎡∆p − 1 ⋅ ρ ⋅ V2⋅ 1 − 4 ⋅ sin( θ) 2 ⎤ ⋅ sin( θ) ⋅ R⋅ L dθ
⎢
⎥
2 air
⎣
⎦
0
⌠
FV = R⋅ L⋅ ∆p⋅ ⎮
⌡
π
1
2⌠
π
sin( θ) dθ − R⋅ L⋅ ⋅ ρair⋅ V ⋅ ⎮
⌡
2
0
0
But
⌠
⎮
⎮
⌡
(sin(θ) − 4⋅sin(θ)3) dθ = −cos(θ) + 4⋅⎛⎜ cos(θ) − 13 ⋅cos(θ)3⎞
⎝
⌠
⎮ sin( θ) dθ = −cos( θ)
⌡
Combining results
5
2
FV = R⋅ L⋅ ⎛⎜ 2 ⋅ ∆p + ⋅ ρair⋅ V ⎞
3
⎝
⎠
⎠
(1 − 4⋅sin(θ)2)⋅sin(θ) dθ
so
⌠
⎮
⌡
π
0
so
⌠
⎮
⌡
(sin(θ) − 4⋅sin(θ)3) dθ = − 103
π
sin( θ) dθ = 2
0
4
FV = 2.28 × 10 ⋅ lbf
FV = 22.8⋅ kip
3
Problem 6.79
[Difficulty: 4]
Problem 6.78
[Difficulty: 4] Part 1/2
Problem 6.78
[Difficulty: 4] Part 2/2
Problem 6.77
[Difficulty: 4] Part 1/2
Problem 6.77
[Difficulty: 4] Part 2/2
Problem 6.76
[Difficulty: 4]
Given:
Air jet striking disk
Find:
Manometer deflection; Force to hold disk; Force assuming p 0 on entire disk; plot pressure distribution
Solution:
Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction
p
∆p = SG ⋅ ρ⋅ g ⋅ ∆h
ρ
2
V
+
2
+ g ⋅ z = constant
Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0)
Applying Bernoulli between jet exit and stagnation point
p atm
ρair
2
+
V
2
=
p0
1
2
p 0 − p atm = ⋅ ρair⋅ V
2
+0
ρair
1
But from hydrostatics
p 0 − p atm = SG ⋅ ρ⋅ g ⋅ ∆h
∆h = 0.002377⋅
slug
ft
For x momentum
3
× ⎛⎜ 225 ⋅
⎝
Rx = −0.002377⋅
2
s⎠
slug
3
× ⎛⎜ 225 ⋅
The force of the jet on the plate is then F = −Rx
1
2
p 0 = p atm + ⋅ ρair⋅ V
2
⎝
ft ⎞
s⎠
1
×
)
ft
The stagnation pressure is
ft ⎞
2 π⋅ d
Rx = V⋅ −ρair⋅ A⋅ V = −ρair⋅ V ⋅
4
(
2
∆h =
so
2 ⋅ 1.75
×
2
⋅ ρair⋅ V
SG ⋅ ρ⋅ g
ft
3
2
=
ρair⋅ V
2 ⋅ SG ⋅ ρ⋅ g
2
1.94⋅ slug
s
×
32.2⋅ ft
∆h = 0.55⋅ ft
2
π⋅ ⎛⎜
2
×
0.4
⋅ ft⎞
2
2
⎝ 12 ⎠ × lbf ⋅ s
4
slug⋅ ft
Rx = −0.105 ⋅ lbf
F = 0.105 ⋅ lbf
∆h = 6.60⋅ in
The force on the plate, assuming stagnation pressure on the front face, is
2
1
2 π⋅ D
F = p 0 − p ⋅ A = ⋅ ρair⋅ V ⋅
2
4
(
F =
)
π
8
× 0.002377⋅
slug
ft
3
× ⎛⎜ 225 ⋅
⎝
ft ⎞
2
s⎠
2
×
2
⎛ 7.5 ⋅ ft⎞ × lbf ⋅ s F = 18.5⋅ lbf
⎜
slug⋅ ft
⎝ 12 ⎠
Obviously this is a huge overestimate!
For the pressure distribution on the disk, we use Bernoulli between the disk outside edge any radius r for radial flow
p atm
ρair
+
1
2
p
2
⋅ v edge =
ρair
+
1
2
⋅v
2
We need to obtain the speed v as a function of radius. If we assume the flow remains constant thickness h, then
Q = v ⋅ 2 ⋅ π⋅ r⋅ h = V⋅
π⋅ d
2
v ( r) = V⋅
4
d
2
8⋅ h⋅ r
We need an estimate for h. As an approximation, we assume that h = d (this assumption will change the scale of p(r) but not the
basic shape)
d
Hence
v ( r) = V⋅
Using this in Bernoulli
ρair⋅ V ⋅ d
4
1
2
2
p ( r) = p atm + ⋅ ρair⋅ ⎛ v edge − v ( r) ⎞ = p atm +
⋅⎛
− ⎞
⎝
⎠
⎜
128
2
2
2
r ⎠
⎝D
8⋅ r
2 2
1
2 2
Expressed as a gage pressure
0
p ( r) =
ρair⋅ V ⋅ d
128
1
4
1⎞
⎜ 2− 2
r ⎠
⎝D
⋅⎛
2
p (psi)
− 0.1
− 0.2
− 0.3
r (in)
3
4
Problem 6.75
[Difficulty: 4]
Open-Ended Problem Statement: An old magic trick uses an empty thread spool and a
playing card. The playing card is placed against the bottom of the spool. Contrary to
intuition, when one blows downward through the central hole in the spool, the card is not
blown away. Instead it is ‘‘sucked’’ up against the spool. Explain.
Discussion: The secret to this “parlor trick” lies in the velocity distribution, and hence
the pressure distribution, that exists between the spool and the playing cards.
Neglect viscous effects for the purposes of discussion. Consider the space between the
end of the spool and the playing card as a pair of parallel disks. Air from the hole in the
spool enters the annular space surrounding the hole, and then flows radially outward
between the parallel disks. For a given flow rate of air the edge of the hole is the crosssection of minimum flow area and therefore the location of maximum air speed.
After entering the space between the parallel disks, air flows radially outward. The flow
area becomes larger as the radius increases. Thus the air slows and its pressure increases.
The largest flow area, slowest air speed, and highest pressure between the disks occur at
the outer periphery of the spool where the air is discharged from an annular area.
The air leaving the annular space between the disk and card must be at atmospheric
pressure. This is the location of the highest pressure in the space between the parallel
disks. Therefore pressure at smaller radii between the disks must be lower, and hence the
pressure between the disks is sub-atmospheric. Pressure above the card is less than
atmospheric pressure; pressure beneath the card is atmospheric. Each portion of the card
experiences a pressure difference acting upward. This causes a net pressure force to act
upward on the whole card. The upward pressure force acting on the card tends to keep it
from blowing off the spool when air is introduced through the central hole in the spool.
Viscous effects are present in the narrow space between the disk and card. However, they
only reduce the pressure rise as the air flows outward, they do not dominate the flow
behavior.
Problem 6.74
[Difficulty: 3]
c
V
H
CS
W
y
x
Ry
Given:
Flow through kitchen faucet
Find:
Area variation with height; force to hold plate as function of height
Solution:
2
p
Basic equation
ρ
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the faucet (1) and any height y
V1
2
2
2
+ g⋅ H =
2
+ g⋅ y
where we assume the water is at patm
2
V( y ) =
Hence
V
V1 + 2 ⋅ g ⋅ ( H − y )
m
V1 = 0.815
s
The problem doesn't require a plot, but it looks like
V( 0 ⋅ m) = 3.08
m
s
5
V (m/s)
4
3
2
1
0
5
10
15
20
25
30
35
40
45
y (cm)
The speed increases as y decreases because the fluid particles "trade" potential energy for kinetic, just as a falling solid particle does!
2
But we have
Hence
π⋅ D
Q = V1 ⋅ A1 = V1 ⋅
= V⋅ A
4
A=
V1 ⋅ A1
V
2
A( y ) =
π⋅ D1 ⋅ V1
2
4 ⋅ V1 + 2 ⋅ g ⋅ ( H − y )
45
A( H) = 1.23⋅ cm
y (cm)
The problem doesn't require a plot, but it looks like
2
A( 0 ) = 0.325 ⋅ cm
30
15
2
0
0.5
1
A (cm2)
The area decreases as the speed increases. If the stream falls far enough the flow will change to turbulent.
For the CV above
(
)
Ry − W = u in⋅ −ρ⋅ Vin⋅ Ain = −V⋅ ( −ρ⋅ Q)
2
2
Ry = W + ρ⋅ V ⋅ A = W + ρ⋅ Q⋅ V1 + 2 ⋅ g ⋅ ( H − y )
Hence Ry increases in the same way as V as the height y varies; the maximum force is when y = H
2
Rymax = W + ρ⋅ Q⋅ V1 + 2 ⋅ g ⋅ H
1.5
Problem 6.73
[Difficulty: 3]
Problem 6.72
[Difficulty: 3] Part 1/2
Problem 6.72
[Difficulty: 3] Part 2/2
Problem 6.71
[Difficulty: 4]
Given:
Flow through branching blood vessel
Find:
Blood pressure in each branch; force at branch
Solution:
Basic equations
p
ρ
2
+
V
∑ Q= 0
+ g ⋅ z = const
2
Q = V⋅ A
∆p = ρ⋅ g ⋅ ∆h
CV
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Given data
L
Q1 = 4.5⋅
min
L
Q2 = 2 ⋅
min
SG Hg = 13.6
ρ = 999 ⋅
kg
3
m
For Q 3 we have
∑ Q = −Q1 + Q2 + Q3 = 0
D1 = 10⋅ mm
D2 = 5 ⋅ mm
D3 = 3 ⋅ mm
kg
ρb = 1060⋅
3
m
h 1 = 140 ⋅ mm
(pressure in in. Hg)
so
Q3 = Q1 − Q2
L
Q3 = 2.50⋅
min
CV
Q1
V1 =
A1
We will need each velocity
Similarly
V2 =
V1 =
4 ⋅ Q2
π⋅ D2
4 ⋅ Q1
π⋅ D1
2
m
V2 = 1.70
s
2
m
V1 = 0.955
s
V3 =
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ
+
V1
2
2
=
p2
ρ
+
V2
2
2
where we ignore gravity effects
4 ⋅ Q3
π⋅ D3
2
m
V3 = 5.89
s
ρ
2
2
p 2 = p 1 + ⋅ ⎛ V1 − V2 ⎞
⎝
⎠
2
and
Hence
ρb
2
2
p2 = p1 +
⋅ ⎛ V − V2 ⎞
⎠
2 ⎝ 1
p 2 = 17.6⋅ kPa
In mm Hg
h2 =
Similarly for exit (3)
ρ
2
2
p 3 = p 1 + ⋅ ⎛ V1 − V3 ⎞
⎝
⎠
2
In mm Hg
h3 =
p2
p 1 = SGHg⋅ ρ⋅ g ⋅ h 1
p 1 = 18.7⋅ kPa
h 2 = 132 ⋅ mm
SGHg⋅ ρ⋅ g
p3
p 3 = 1.75⋅ kPa
h 3 = 13.2⋅ mm
SGHg⋅ ρ⋅ g
Note that all pressures are gage.
(
)
(
Rx + p 3 ⋅ A3 ⋅ cos( 60⋅ deg) − p 2 ⋅ A2 ⋅ cos( 45⋅ deg) = u 3 ⋅ ρ⋅ Q3 + u 2 ⋅ ρ⋅ Q2
For x momentum
)
(
Rx = p 2 ⋅ A2 ⋅ cos( 45⋅ deg) − p 3 ⋅ A3 ⋅ cos( 60⋅ deg) + ρ⋅ Q2 ⋅ V2 ⋅ cos( 45⋅ deg) − Q3 ⋅ V3 ⋅ cos( 60⋅ deg)
Rx = p2 ⋅
π⋅ D2
2
4
⋅ cos( 45⋅ deg) − p 3 ⋅
π⋅ D3
2
4
(
⋅ cos( 60⋅ deg) + ρ⋅ Q2 ⋅ V2 ⋅ cos( 45⋅ deg) − Q3 ⋅ V3 ⋅ cos( 60⋅ deg)
(
)
(
Ry − p 3 ⋅ A3 ⋅ sin( 60⋅ deg) − p 2 ⋅ A2 ⋅ sin( 45⋅ deg) + p 1 ⋅ A1 = v 3 ⋅ ρ⋅ Q3 + v 2 ⋅ ρ⋅ Q2
For y momentum
)
)
Rx = 0.156 N
)
(
Ry = p 2 ⋅ A2 ⋅ sin( 45⋅ deg) + p 3 ⋅ A3 ⋅ sin( 60⋅ deg) − p 1 ⋅ A1 + ρ⋅ Q2 ⋅ V2 ⋅ sin( 45⋅ deg) + Q3 ⋅ V3 ⋅ sin( 60⋅ deg)
Ry = p2 ⋅
π⋅ D2
4
2
⋅ sin( 45⋅ deg) + p 3 ⋅
π⋅ D3
4
2
⋅ sin( 60⋅ deg) − p 1 ⋅
π⋅ D1
4
)
2
(
+ ρ⋅ Q2 ⋅ V2 ⋅ sin( 45⋅ deg) + Q3 ⋅ V3 ⋅ sin( 60⋅ deg)
)
Ry = −0.957 N
Problem 6.70
[Difficulty: 3]
Given:
Flow nozzle
Find:
Mass flow rate in terms of ∆p, T1 and D 1 and D 2
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
+
ρ
But we have
V1
2
=
2
p2
ρ
Q = V1 ⋅ A1 = V1 ⋅
+
V2
2
where we ignore gravity effects
2
π⋅ D1
4
2
= V2 ⋅
π⋅ D2
2
⎛ D2 ⎞
V1 = V2 ⋅ ⎜
⎝ D1 ⎠
so
4
2
Note that we assume the flow at D 2 is at the same pressure as the entire section 2; this will be true if there is turbulent mixing
2⋅ (p2 − p1)
⎛ D2 ⎞
V2 − V2 ⋅ ⎜
=
ρ
⎝ D1 ⎠
(
4
Hence
Then the mass flow rate is
Using
For a flow nozzle
2
2
mflow = ρ⋅ V2 ⋅ A2 = ρ⋅
p = ρ⋅ R⋅ T
mflow = k ⋅ ∆p where
π⋅ D2
4
2
⋅
(
2⋅ p1 − p2
⎡
⎢
ρ⋅ 1 −
⎢
⎣
k=
)
⎛ D2 ⎞
⎜
⎝ D1 ⎠
mflow =
4⎤
2⋅ 2
⋅
π⋅ D2
⎡
⎢
ρ⋅ 1 −
⎢
⎣
2
2⋅ 2
2
2⋅ 2
2
=
⎥
⎥
⎦
π⋅ D2
π⋅ D2
2⋅ p1 − p2
V2 =
or
⋅
⋅
)
⎛ D2 ⎞
⎜
⎝ D1 ⎠
4⎤
⎥
⎥
⎦
∆p⋅ ρ
⎡
⎢
⎢1 −
⎣
⎛ D2 ⎞
⎜
⎝ D1 ⎠
4⎤
⎥
⎥
⎦
∆p⋅ p 1
⎡
⎢
R ⋅ T1 ⋅ 1 −
⎢
⎣
⎛ D2 ⎞
⎜
⎝ D1 ⎠
4⎤
⎥
⎥
⎦
p1
⎡
⎢
R ⋅ T1 ⋅ 1 −
⎢
⎣
4⎤
⎛ D2 ⎞ ⎥
⎜
⎥
⎝ D1 ⎠ ⎦
We can expect the actual flow will be less because there is actually significant loss in the device. Also the flow will experience a
vena contracta so that the minimum diameter is actually smaller than D 2. We will discuss this device in Chapter 8.
Problem 6.69
[Difficulty: 3]
Given:
Flow through reducing elbow
Find:
Gage pressure at location 1; x component of force
Solution:
Basic equations:
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline 5) Ignore elevation change 6) p 2 = p atm
Available data:
From contnuity
Q = 2.5⋅
V1 =
L
Q = 2.5 × 10
s
Q
3
−3m
D = 45⋅ mm
s
V2 =
p1
Hence, applying Bernoulli between the inlet (1) and exit (2)
ρ
or, in gage pressures
From x-momentum
p 1g =
ρ
2
⋅ ⎛ V2 − V1
⎝
2
(
ρ = 999 ⋅
kg
3
m
m
V1 = 1.57
s
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
d = 25⋅ mm
2⎞
+
Q
m
V2 = 5.09
s
⎛ π⋅ d 2 ⎞
⎜
⎝ 4 ⎠
V1
2
2
=
p2
ρ
+
V2
2
2
p 1g = 11.7⋅ kPa
⎠
)
(
)
Rx + p 1g⋅ A1 = u 1 ⋅ −mrate + u 2 ⋅ mrate = −mrate⋅ V1 = −ρ⋅ Q⋅ V1
because
u 1 = V1
u2 = 0
2
π⋅ D
Rx = −p 1g⋅
− ρ⋅ Q⋅ V1
4
The force on the supply pipe is then
Rx = −22.6 N
Kx = −Rx
Kx = 22.6 N
on the pipe to the right
Problem 6.68
[Difficulty: 3]
Problem 6.67
[Difficulty: 3]
c
d
Rx
Given:
Flow through fire nozzle
Find:
Maximum flow rate
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
+
ρ
V1
2
=
2
p2
ρ
V2
+
2
where we ignore gravity effects
2
2
But we have
π⋅ D
π⋅ d
Q = V1 ⋅ A1 = V1 ⋅
= V2 ⋅
4
4
Hence in Bernoulli
4 2⋅ p − p
2
1
2
2 d
V2 − V2 ⋅ ⎛⎜ ⎞ =
ρ
⎝ D⎠
(
2
)
so
d
V1 = V2 ⋅ ⎛⎜ ⎞
D
or
V2 =
⎝ ⎠
(
2⋅ p1 − p2
⎡
ρ⋅ ⎢1 −
⎣
3
V2 =
2×
π⋅ d
Q = V2 ⋅
4
Then
m
3 N
1000⋅ kg
× ( 700 − 0 ) × 10 ⋅
1
×
2
4
2
Q =
(
π
× 37.6⋅
× ( 0.025 ⋅ m)
2
4
)
1−
m
s
×
⎛ 25 ⎞
⎜ 75
⎝ ⎠
m
(
)
2
3
Q = 0.0185⋅
Hence
⎡
π⋅ D
π⋅ D
Rx = −p 1 ⋅
+ ρ⋅ Q⋅ V2 − V1 = −p 1 ⋅
+ ρ⋅ Q⋅ V2 ⋅ ⎢1 −
4
4
⎣
3 N
Rx = −700 × 10 ⋅
2
m
×
π
4
2
⋅ ( 0.075 ⋅ m) + 1000⋅
kg
3
m
3
× 0.0185⋅
m
s
m
s
Q = 18.5⋅
L
s
using gage pressures
2
)
m
V2 = 37.6
s
N⋅ s
Rx + p 1 ⋅ A1 = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2
(
)
4⎤
⎛d⎞ ⎥
⎜
⎝ D⎠ ⎦
kg⋅ m
From x momentum
2
2
× 37.6⋅
m
s
⎡
2⎤
⎛d⎞ ⎥
⎜
⎝ D⎠ ⎦
× ⎢1 −
⎣
3
2
⎛ 25 ⎞ ⎤⎥ × N⋅ s
⎜
⎝ 75 ⎠ ⎦ kg⋅ m
Rx = −2423 N
This is the force of the nozzle on the fluid; hence the force of the fluid on the nozzle is 2400 N to the right; the nozzle is in tension
Problem 6.66
[Difficulty: 3]
Problem 6.65
[Difficulty: 3]
Given:
Velocity field for plane doublet
Find:
Pressure distribution along x axis; plot distribution
Solution:
p
The governing equation is the Bernoulli equation
ρ
+
1
2
2
⋅ V + g ⋅ z = const
3
Λ = 3⋅
The given data is
m
s
2
u +v
2
p 0 = 100 ⋅ kPa
3
m
Λ
Vr = − ⋅ cos( θ)
2
r
From Table 6.1
kg
ρ = 1000⋅
V=
where
Λ
Vθ = − ⋅ sin( θ)
2
r
where Vr and Vθ are the velocity components in cylindrical coordinates (r,θ). For points along the x axis, r = x, θ = 0, Vr = u and Vθ
=v=0
Λ
u=−
v = 0
2
x
p
so (neglecting gravity)
ρ
+
1
2
2
⋅ u = const
Apply this to point arbitrary point (x,0) on the x axis and at infinity
x →0
At
Hence the Bernoulli equation becomes
u→0
p0
ρ
=
p
ρ
+
p → p0
Λ
2
2⋅ x
4
or
u=−
At point (x,0)
p(x) = p0 −
ρ⋅ Λ
2⋅ x
Λ
x
2
2
4
The plot of pressure can be done in Excel:
x (m) p (Pa)
99.892
99.948
99.972
99.984
99.990
99.993
99.995
99.997
99.998
99.998
99.999
99.999
99.999
99.999
99.999
100.000
Pressure Distribution Along x axis
100.0
p (kPa)
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
100.0
99.9
99.9
99.8
0.0
0.2
0.4
0.6
0.8
1.0
x (m)
1.2
1.4
1.6
1.8
2.0
Problem 6.64
[Difficulty: 3]
Given:
Velocity field
Find:
Pressure distribution along wall; plot distribution; net force on wall
Solution:
3
m
q = 2⋅
The given data is
u=
s
h = 1⋅ m
m
kg
ρ = 1000⋅
3
m
q⋅ x
2 ⋅ π⎡⎣x + ( y − h )
2
+
2⎤
⎦
q⋅ x
2 ⋅ π⎡⎣x + ( y + h )
2
v=
2⎤
⎦
q⋅ ( y − h)
2 ⋅ π⎡⎣x + ( y − h )
2
+
2⎤
⎦
q⋅ ( y + h)
2 ⋅ π⎡⎣x + ( y + h )
2
2⎤
⎦
The governing equation is the Bernoulli equation
p
ρ
+
1
2
2
⋅ V + g ⋅ z = const
V=
where
2
u +v
2
Apply this to point arbitrary point (x,0) on the wall and at infinity (neglecting gravity)
x →0
At
u=
At point (x,0)
u→0
q⋅ x
(2
π⋅ x + h
v→0
V→0
v=0
)
2
V=
q⋅ x
(2
π⋅ x + h
)
2
2
q⋅ x
⎡
⎤
=
+ ⋅
ρ
2
2 ⎥
ρ
2 ⎢
⎣ π⋅ x + h ⎦
p atm
Hence the Bernoulli equation becomes
p
1
(
)
ρ
p(x) = − ⋅ ⎡
2 ⎢
q⋅ x
⎤
2
2 ⎥
⎣ π⋅ x + h ⎦
or (with pressure expressed as gage pressure)
(
2
)
(Alternatively, the pressure distribution could have been obtained from Problem 6.8, where the momentum equation was used to find
the pressure gradient
∂
∂x
2
p =
(2
ρ⋅ q ⋅ x ⋅ x − h
2
(2
π ⋅ x +h
)
2
) along the wall. Integration of this with respect to x leads to the same result for p(x))
2
3
The plot of pressure can be done in Excel (see below). From the plot it is clear that the wall experiences a negative gage pressure
on the upper surface (and zero gage pressure on the lower), so the net force on the wall is upwards, towards the source
10⋅ h
⌠
F=⎮
⌡
The force per width on the wall is given by
(pupper − plower) dx
F=−
− 10⋅ h
ρ⋅ q
2 ⌠
⎮
⋅⎮
2
2⋅ π ⎮
⎮
⌡
10⋅ h
− 10⋅ h
⌠
⎮
⎮
⎮
⎮
⌡
The integral is
x
(x2 + h2)
F=−
so
atan⎛⎜
2
ρ⋅ q
2
2
⋅ ⎛⎜ −
10
101
2⋅ π ⋅ h ⎝
2
(x
2
2
+h
)
2
2
x⎞
⎝h⎠ −
dx =
x
2⋅ h
x
2
2⋅ h + 2⋅ x
2
+ atan( 10) ⎞
⎠
2
2
⎛ m2 ⎞
1
10
N⋅ s
F = −
× 1000⋅
× ⎜ 2⋅
×
× ⎛⎜ −
+ atan( 10) ⎞ ×
2
3 ⎝
s ⎠
1 ⋅ m ⎝ 101
⎠ kg⋅ m
2⋅ π
m
1
kg
F = −278 ⋅
N
m
In Excel:
q =
h =
2
1
m3/s/m
m
ℵ= 1000 kg/m 3
x (m) p (Pa)
0.00
-50.66
-32.42
-18.24
-11.22
-7.49
-5.33
-3.97
-3.07
-2.44
-1.99
Pressure Distribution Along Wall
0
0
1
2
3
4
5
-10
p (Pa)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
-20
-30
-40
-50
-60
x (m)
6
7
8
9
10
dx
Problem 6.63
[Difficulty: 2]
Problem 6.62
[Difficulty: 2]
Given:
Race car on straightaway
Find:
Air inlet where speed is 60 mph; static pressure; pressure rise
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline 5) Standard atmosphere
Available data
p atm = 101 ⋅ kPa
slug
ρ = 0.002377⋅
ft
3
V1 = 235⋅ mph
V2 = 60⋅ mph
Between location 1 (the upstream flow at 235 mph with respect to the car), and point 2 (on the car where V = 60 mph),
Bernoulli becomes
p1
ρ
Hence
2
+
V1
2
=
ρ
2
V1
+
2
=
⎡
⎢
p 2 = p atm + ⋅ ρ⋅ V1 ⋅ 1 −
⎢
2
⎣
Note that the pressure rise is
1
2
1
p2
ρ
2
+
⎛ V2 ⎞
⎜
⎝ V1 ⎠
⎡
⎢
∆p = ⋅ ρ⋅ V1 ⋅ 1 −
⎢
2
⎣
The freestream dynamic pressure is
Then
p atm
2
V2
2
2⎤
⎥
⎥
⎦
⎛ V2 ⎞
⎜
⎝ V1 ⎠
p 2 = 15.6 psi
2⎤
⎥
⎥
⎦
∆p = 0.917 ⋅ psi
2
q = 0.980 ⋅ psi
q =
1
∆p
= 93.5⋅ %
q
2
⋅ ρ⋅ V1
Note that at this speed the flow is borderline compressible!
Problem 6.61
Given:
Flow through fire nozzle
Find:
Maximum flow rate
[Difficulty: 2]
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ
+
V1
2
=
2
p2
ρ
+
V2
2
where we ignore gravity effects
2
2
But we have
π⋅ D
π⋅ d
Q = V1 ⋅ A1 = V1 ⋅
= V2 ⋅ A2 =
4
4
(
4 2⋅ p − p
2
1
2
2 d
V2 − V2 ⋅ ⎛⎜ ⎞ =
ρ
⎝ D⎠
Hence
V2 =
(
2⋅ p1 − p2
⎡
ρ⋅ ⎢1 −
⎣
Then
2
2
⎝ ⎠
)
)
4⎤
⎛d⎞ ⎥
⎜
⎝ D⎠ ⎦
3
2
12⋅ in ⎞
V2 = 2 ×
× ( 100 − 0 ) ⋅
× ⎛⎜
×
1.94⋅ slug
2 ⎝ 1 ⋅ ft ⎠
in
ft
π⋅ d
Q = V2 ⋅
4
d
V1 = V2 ⋅ ⎛⎜ ⎞
D
so
lbf
2
Q =
π
4
× 124 ⋅
ft
s
×
⎛ 1 ⋅ ft⎞
⎜
⎝ 12 ⎠
1
1−
⎛1⎞
⎜3
⎝ ⎠
3
×
slug⋅ ft
ft
V2 = 124 ⋅
s
2
lbf ⋅ s
2
Q = 0.676 ⋅
ft
3
s
Q = 304 ⋅
gal
min
Problem 6.60
[Difficulty: 3]
Problem 6.59
[Difficulty: 3]
Given:
Water flow over a hydrofoil
Find:
Stagnation pressure; water speed relative to airfoil at a point; absolute value
Solution:
Basic equations
p
ρ
2
+
V
+ g ⋅ z = const
2
∆p = ρ⋅ g ⋅ h
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
m
V1 = 20⋅
s
ρ = 999 ⋅
kg
3
h = 3⋅ m
p 2 = −75⋅ kPa
p 1 = ρ⋅ g ⋅ h
p 1 = 29.4⋅ kPa
(gage)
m
Using coordinates fixed to the hydrofoil, the pressure at depth h is
Applying Bernoulli between the upstream (1) and the stagnation point (at the front of the hydrofoil)
p1
ρ
+
V1
2
=
2
p0
or
ρ
1
2
p 0 = p 1 + ⋅ ρ⋅ V1
2
p 0 = 229 ⋅ kPa
Applying Bernoulli between the upstream point (1) and the point on the hydrofoil (2)
p1
ρ
Hence
+
V2 =
V1
2
2
=
2
p2
ρ
V1 + 2 ⋅
+
V2
2
2
(p1 − p2)
ρ
m
V2 = 24.7
s
This is the speed of the water relative to the hydrofoil; in absolute coordinates
V2abs = V2 + V1
m
V2abs = 44.7
s
Problem 6.58
[Difficulty: 3]
Given:
Air flow over a wing
Find:
Air speed relative to wing at a point; absolute air speed
Solution:
Basic equation
p
ρ
2
V
+
+ g ⋅ z = const
2
p = ρ⋅ R⋅ T
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
T = −10 °C
For air
ρ=
km
V1 = 200 ⋅
hr
p 1 = 65⋅ kPa
p1
3 N
ρ = ( 65) × 10 ⋅
R⋅ T
2
m
×
kg⋅ K
286.9 ⋅ N⋅ m
×
p 2 = 60⋅ kPa
1
( −10 + 273 ) ⋅ K
R = 286.9 ⋅
ρ = 0.861
N⋅ m
kg⋅ K
kg
3
m
Hence, applying Bernoulli between the upstream point (1) and the point on the wing (2)
p1
ρ
Hence
+
V1
2
=
2
2
p2
ρ
+
2
where we ignore gravity effects
2
( p1 − p2)
V2 =
V1 + 2 ⋅
V2 =
kg⋅ m
m
3 N
⎛ 200 ⋅ km ⎞ × ⎛ 1000⋅ m ⎞ × ⎛ 1⋅ hr ⎞ + 2 × m
× ( 65 − 60) × 10 ⋅
×
V2 = 121
⎜
⎜
⎜
0.861 ⋅ kg
2
2
hr ⎠
s
⎝
⎝ 1 ⋅ km ⎠ ⎝ 3600⋅ s ⎠
m
N⋅ s
ρ
2
Then
V2
2
2
3
NOTE: At this speed, significant density changes will occur, so this result is not very realistic
The absolute velocity is
V2abs = V2 − V1
m
V2abs = 65.7
s
Problem 6.57
[Difficulty: 2]
Problem 6.56
Given:
Flow through tank-pipe system
Find:
Velocity in pipe; Rate of discharge
[Difficulty: 2]
Solution:
Basic equations
p
ρ
2
+
V
+ g ⋅ z = const
2
∆p = ρ⋅ g ⋅ ∆h
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the free surface and the manometer location
p atm
ρ
=
p
ρ
2
− g⋅ H +
V
where we assume VSurface <<, and H = 4 m
2
2
Hence
V
p = p atm + ρ⋅ g ⋅ H − ρ⋅
2
For the manometer
p − p atm = SGHg⋅ ρ⋅ g ⋅ h 2 − ρ⋅ g ⋅ h 1
Combining equations
ρ⋅ g ⋅ H − ρ⋅
Note that we have water on one side and mercury on
the other of the manometer
2
Hence
V =
V
2
= SGHg⋅ ρ⋅ g ⋅ h 2 − ρ⋅ g ⋅ h 1
2 × 9.81⋅
m
2
or
V=
(
2 ⋅ g ⋅ H − SG Hg⋅ h 2 + h 2
× ( 4 − 13.6 × 0.15 + 0.75) ⋅ m
V = 7.29
s
2
The flow rate is
Q = V⋅
π⋅ D
4
)
Q =
π
4
× 7.29⋅
m
s
× ( 0.05⋅ m)
2
m
s
3
Q = 0.0143
m
s
Problem 6.55
[Difficulty: 2]
Problem 6.54
Given:
Flow rate through siphon
Find:
Maximum height h to avoid cavitation
[Difficulty: 3]
Solution:
Basic equation
p
ρ
2
V
+
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
From continuity
Q = 5⋅
V=
3
−3m
L
Q
A
Q = 5 × 10
s
D = 25⋅ mm
s
p atm = 101⋅ kPa
3
m
4⋅ Q
=
kg
ρ = 999⋅
V=
2
π⋅ D
4
π
3
× 0.005⋅
m
s
×
⎛ 1 ⎞
⎜
⎝ .025⋅ m ⎠
2
V = 10.2
m
s
Hence, applying Bernoulli between the free surface and point A
p atm
ρ
=
pA
ρ
2
+ g⋅ h +
V
where we assume VSurface <<
2
2
Hence
V
p A = p atm − ρ⋅ g ⋅ h − ρ⋅
2
p v = 2.358 ⋅ kPa
From the steam tables, at 20oC the vapor pressure is
This is the lowest permissible value of pA
2
p atm − p v
2
Hence
V
p A = p v = p atm − ρ⋅ g ⋅ h − ρ⋅
2
Hence
m
s
h = ( 101 − 2.358 ) × 10 ⋅
×
⋅
×
×
− × ⎛⎜ 10.2 ⎞ ×
h = 4.76 m
2
9.81⋅ m
2
9.81⋅ m
999 kg
2 ⎝
s⎠
m
N⋅ s
3 N
h=
or
1
3
m
2
s
ρ⋅ g
kg⋅ m
−
V
2⋅ g
1
2
2
Problem 6.53
Given:
Ruptured Coke can
Find:
Pressure in can
[Difficulty: 2]
Solution:
Basic equation
p
ρCoke
2
+
V
2
+ g ⋅ z = const
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
kg
ρw = 999 ⋅
3
m
h = 0.5⋅ m
From a web search
SG DietCoke = 1
SG RegularCoke = 1.11
Hence, applying Bernoulli between the coke can and the rise height of the coke
p can
ρCoke
=
p atm
ρCoke
+ g⋅ h
where we assume VCoke <<, and h is the rise height
Hence
p Coke = ρCoke ⋅ g ⋅ h = SG Coke ⋅ ρw⋅ g ⋅ h where p Coke is now the gage pressure
Hence
p Diet = SG DietCoke⋅ ρw⋅ g ⋅ h
p Diet = 4.90⋅ kPa
(gage)
and
p Regular = SGRegularCoke ⋅ ρw⋅ g ⋅ h
p Regular = 5.44⋅ kPa
(gage)
Problem 6.52
Given:
Ruptured pipe
Find:
Height benzene rises from tank
[Difficulty: 2]
Solution:
Basic equation
p
ρben
2
+
V
+ g ⋅ z = const
2
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
ρ = 999 ⋅
kg
p ben = 50⋅ kPa
3
(gage)
From Table A.2
m
Hence, applying Bernoulli between the pipe and the rise height of the benzene
p ben
ρben
Hence
h =
=
p atm
ρben
+ g⋅ h
p ben
SG ben⋅ ρ⋅ g
h = 5.81 m
where we assume Vpipe <<, and h is the rise height
where p ben is now the gage pressure
SG ben = 0.879
Problem 6.51
Given:
Siphoning of wort
Find:
Flow rate; plot; height for a flow of 2 L/min
[Difficulty: 2]
Solution:
Basic equation
p
ρwort
2
V
+
+ g ⋅ z = const
2
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the open surface of the full tank and tube exit to atmosphere
p atm
ρgas
=
2
p atm
+
ρgas
Hence
V=
The flow rate is then
Q = V⋅ A =
V
2
− g⋅ h
where we assume the tank free surface is slowly changing so V tank <<,
and h is the difference in levels
2⋅ g⋅ h
2
π⋅ D
4
⋅ 2⋅ g⋅ h
D = 5 ⋅ mm
For
Q (L/min)
3
2
1
0
50
100
150
200
250
h (mm)
For a flow rate of
Q = 2⋅
L
min
3
−5m
Q = 3.33 × 10
s
2
solving for h
h =
8⋅ Q
2
4
π ⋅D ⋅g
h = 147 ⋅ mm
Problem 6.50
Given:
Siphoning of gasoline
Find:
Flow rate
[Difficulty: 2]
Solution:
Basic equation
p
ρgas
2
+
V
2
+ g ⋅ z = const
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the gas tank free surface and the siphon exit
p atm
ρgas
=
p atm
ρgas
Hence
V=
The flow rate is then
Q = V⋅ A =
2
V
+
2
− g⋅ h
where we assume the tank free surface is slowly changing so V tank <<,
and h is the difference in levels
2⋅ g⋅ h
2
Q =
π
4
π⋅ D
4
⋅ 2⋅ g ⋅ h
2
× ( .5⋅ in) ×
1 ⋅ ft
2
2
144 ⋅ in
×
2 × 32.2
ft
2
s
× 1 ⋅ ft
Q = 0.0109⋅
ft
3
s
Q = 4.91⋅
gal
min
Problem 6.49
[Difficulty: 2]
Problem 6.48
[Difficulty: 2]
Given:
Flow in pipe/nozzle device
Find:
Gage pressure needed for flow rate; repeat for inverted
Solution:
Basic equations
p
Q = V⋅ A
ρ
2
+
V
2
+ g ⋅ z = const
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
From continuity
D1 = 1 ⋅ in
D2 = 0.5⋅ in
Q = V1 ⋅ A1 = V2 ⋅ A2
ft
V2 = 30⋅
s
z2 = 10⋅ ft
A2
⎛ D2 ⎞
V1 = V2 ⋅ ⎜
⎝ D1 ⎠
V1 = V2 ⋅
A1
ρ = 1.94⋅
ft
or
2
ft
V1 = 7.50
s
Hence, applying Bernoulli between locations 1 and 2
p1
ρ
+
V1
2
2
+0=
p2
ρ
+
V2
2
2
Solving for p 1 (gage)
⎛⎜ V 2 − V 2
⎞
1
2
p 1 = ρ⋅ ⎜
+ g ⋅ z2
2
⎝
⎠
When it is inverted
z2 = −10⋅ ft
⎛⎜ V 2 − V 2
⎞
1
2
p 2 = ρ⋅ ⎜
+ g ⋅ z2
2
⎝
⎠
+ g ⋅ z2 = 0 +
V2
2
slug
2
+ g ⋅ z2working in gage pressures
p 1 = 10.0⋅ psi
p 2 = 1.35⋅ psi
3
Problem 6.47
4.123
[Difficulty: 4]
Problem 6.46
[Difficulty: 2]
Problem 6.45
[Difficulty: 2]
Problem 6.44
[Difficulty: 2]
Given:
Wind tunnel with inlet section
Find:
Dynamic and static pressures on centerline; compare with Speed of air at two locations
Solution:
Basic equations
p dyn =
1
2
2
⋅ ρair⋅ U
p 0 = p s + p dyn
p
ρair =
Rair⋅ T
∆p = ρw⋅ g ⋅ ∆h
p atm = 101⋅ kPa
kg
h 0 = −10⋅ mm ρw = 999⋅
3
m
p s = −1.738 kPa
hs =
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
T = −5 °C
m
J
U = 50R
⋅ = 287⋅
s
kg⋅ K
For air
p atm
ρair =
R⋅ T
ρair = 1.31
p dyn =
1
2
kg
3
m
2
⋅ ρair⋅ U
Also
p 0 = ρw⋅ g ⋅ h 0
and
p 0 = p s + p dyn
p dyn = 1.64⋅ kPa
p 0 = −98.0 Pa
so
(gage)
p s = p 0 − p dyn
(gage)
∂
Streamlines in the test section are straight so
In the curved section
∂
∂n
p =0
and
2
V
p = ρair⋅
R
∂n
so
p w < p centerline
p w = p centerline
ps
ρw⋅ g
h s = −177 mm
Problem 6.43
[Difficulty: 2]
Problem 6.42
Given:
Air jet hitting wall generating pressures
Find:
Speed of air at two locations
[Difficulty: 2]
Solution:
Basic equations
p
ρair
2
V
+
p
ρair =
Rair⋅ T
+ g ⋅ z = const
2
∆p = ρHg⋅ g ⋅ ∆h = SG Hg⋅ ρ⋅ g ⋅ ∆h
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
For the air
R = 287 ⋅
J
T = −10 °C
kg⋅ K
kg
ρ = 999 ⋅
p = 200 ⋅ kPa
3
SG Hg = 13.6
m
p
ρair =
R⋅ T
ρair = 2.65
kg
3
m
Hence, applying Bernoulli between the jet and where it hits the wall directly
p atm
ρair
Hence
∆h = 25⋅ mm
+
Vj
2
=
2
p wall
p wall =
ρair
p wall = SGHg⋅ ρ⋅ g ⋅ ∆h =
Vj =
hence
ρair⋅ Vj
ρair⋅ Vj
2
(working in gage pressures)
2
2
Vj =
so
2
2 × 13.6 × 999 ⋅
kg
3
m
×
1
2 ⋅ SGHg⋅ ρ⋅ g ⋅ ∆h
ρair
3
⋅
m
2.65 kg
× 9.81⋅
m
2
× 25⋅ mm ×
s
1⋅ m
m
Vj = 50.1
s
1000⋅ mm
Repeating the analysis for the second point
∆h = 5 ⋅ mm
p atm
ρair
+
Vj
2
2
=
p wall
ρair
2
Hence
V =
2
+
V
2
V=
2
Vj −
2 ⋅ p wall
ρair
=
2
Vj −
2 ⋅ SG Hg⋅ ρ⋅ g ⋅ ∆h
ρair
3
⎛ 50.1⋅ m ⎞ − 2 × 13.6 × 999 ⋅ kg × 1 ⋅ m × 9.81⋅ m × 5 ⋅ mm × 1 ⋅ m
⎜
3
2
s⎠
2.65 kg
1000⋅ mm
⎝
m
s
V = 44.8
m
s
Problem 6.41
Given:
Velocity of automobile
Find:
Estimates of aerodynamic force on hand
[Difficulty: 2]
Solution:
The basic equation is the Bernoulli equation (in coordinates attached to the vehicle)
p atm +
1
2
2
⋅ ρ⋅ V = p stag
where V is the free stream velocity
For air
ρ = 0.00238 ⋅
slug
ft
3
We need an estimate of the area of a typical hand. Personal inspection indicates that a good approximation is a square of sides 9
cm and 17 cm
A = 9 ⋅ cm × 17⋅ cm
A = 153 ⋅ cm
2
Hence, for p stag on the front side of the hand, and p atm on the rear, by assumption,
(
)
1
2
F = p stag − p atm ⋅ A = ⋅ ρ⋅ V ⋅ A
2
(a)
V = 30⋅ mph
2
ft ⎞
⎛
22⋅
⎜
1
1
slug
s
2
2
F = ⋅ ρ⋅ V ⋅ A = × 0.00238 ⋅
× ⎜ 30⋅ mph⋅
× 153 ⋅ cm ×
15⋅ mph ⎠
3
2
2
⎝
ft
(b)
⎛ 1 ⋅ ft ⎞
⎜ 12
⎜
⎝ 2.54⋅ cm ⎠
2
⎛ 1 ⋅ ft ⎞
⎜ 12
⎜
⎝ 2.54⋅ cm ⎠
2
F = 0.379 ⋅ lbf
V = 60⋅ mph
2
ft ⎞
⎛
22⋅
⎜
1
1
slug
s
2
2
F = ⋅ ρ⋅ V ⋅ A = × 0.00238 ⋅
× ⎜ 60⋅ mph⋅
× 153 ⋅ cm ×
15
⋅
mph
3
2
2
⎝
⎠
ft
F = 1.52⋅ lbf
These values pretty much agree with experience. However, they overestimate a bit as the entire front of the hand is not at
stagnation pressure - there is flow around the had - so the pressure is less than stagnation over most of the surface.
Problem 6.40
Given:
Air speed
Find:
Plot dynamic pressure in mm Hg
[Difficulty: 2]
Solution:
p dynamic =
Basic equations
1
2
2
⋅ ρair⋅ V
kg
ρw = 999 ⋅
3
m
Available data
1
Hence
2
kg
ρair = 1.23⋅
3
m
SG Hg = 13.6
2
⋅ ρair⋅ V = SGHg⋅ ρw⋅ g ⋅ ∆h
V( ∆h) =
Solving for V
p = ρHg⋅ g ⋅ ∆h = SGHg⋅ ρw⋅ g ⋅ ∆h
2 ⋅ SG Hg⋅ ρw⋅ g ⋅ ∆h
ρair
250
V (m/s
200
150
100
50
0
50
100
150
Dh (mm)
200
250
Problem 6.39
Given:
Air speed of 100 km/hr
Find:
Dynamic pressure in mm water
[Difficulty: 1]
Solution:
Basic equations
Hence
p dynamic =
∆h =
1
2
⋅ ρair⋅ V
2
p = ρw⋅ g ⋅ ∆h
ρair V2
⋅
ρw 2 ⋅ g
1.23⋅
kg
3
m
∆h =
999 ⋅
kg
3
m
×
1
2
× ⎛⎜ 100 ⋅
⎝
km ⎞
hr
⎠
2
2
×
2
2
⎛ 1000⋅ m ⎞ × ⎛ 1 ⋅ hr ⎞ × s
⎜
⎜
9.81⋅ m
⎝ 1⋅ km ⎠ ⎝ 3600⋅ s ⎠
∆h = 48.4⋅ mm
Problem 6.38
Given:
Water at speed 25 ft/s
Find:
Dynamic pressure in in. Hg
[Difficulty: 1]
Solution:
Basic equations
p dynamic =
1
2
2
⋅ ρ⋅ V
p = ρHg⋅ g ⋅ ∆h = SGHg⋅ ρ⋅ g ⋅ ∆h
2
Hence
∆h =
∆h =
2
ρ⋅ V
2⋅ SGHg⋅ ρ⋅ g
1
2
× ⎛⎜ 25⋅
⎝
ft ⎞
s
⎠
=
V
2⋅ SG Hg⋅ g
2
×
1
13.6
×
s
2
32.2⋅ ft
×
12⋅ in
1⋅ ft
∆h = 8.56⋅ in
Problem 6.37
[Difficulty: 4] Part 1/2
Problem 6.37
[Difficulty: 4 ] Part 2/2
Problem 6.36
[Difficulty: 4]
Given:
x component of velocity field
Find:
y component of velocity field; acceleration at several points; estimate radius of curvature; plot streamlines
Solution:
3
Λ = 2⋅
The given data is
The basic equation (continuity) is
∂
∂x
u +
m
u=−
s
∂
∂y
(2
Λ⋅ x − y
)
2
(x2 + y2)
2
v =0
The basic equation for acceleration is
⌠
v = −⎮
⎮
⌡
Hence
Integrating (using an integrating factor)
v=−
⌠
⎮
dy = −⎮
dx
⎮
⎮
⌡
du
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
(x2 + y2)
3
2
) dy
2⋅ Λ⋅ x⋅ y
(x2 + y2)
2
Alternatively, we could check that the given velocities u and v satisfy continuity
u=−
so
∂
∂x
(2
Λ⋅ x − y
u +
(x2 + y2)
∂
∂y
v =0
)
2
2
∂
∂x
u =
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
(x2 + y2)
3
2
)
v=−
2⋅ Λ⋅ x⋅ y
(x2 + y2)
∂
2
∂y
v =−
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
(x2 + y2)
3
2
)
For steady, 2D flow the acceleration components reduce to (after considerable math!):
x - component
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
(
u
)
(
⎡ Λ⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ x⋅ x2 − 3⋅ y2
⎥ ⋅⎢
3
⎢ 2 2 2⎥ ⎢
2
2
x +y
⎣ x +y ⎦⎣
ax = ⎢−
y - component
ay = u ⋅
(
∂
∂x
)
v + v⋅
∂
∂y
(
(
)
Evaluating at point (0,1)
Evaluating at point (0,2)
Evaluating at point (0,3)
u = 2⋅
(
)
)
m
(
(
)
v = 0⋅
s
u = 0.5⋅
m
v = 0⋅
s
u = 0.222 ⋅
(
)
2
2
2
2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x − y ⎤
2Λ
⎥ax = − ⋅ ⋅ x
⋅⎢
⎢
⎥
3
3
⎥
⎢ x 2 2 2⎥ ⎢
2
2
2
2
x +y
x +y
+y ⎦⎣
⎣
⎦
⎥
⎦
(
)
(
)
m
s
v = 0⋅
)⎤⎥ + ⎡−
y = 1m
)
(
)
2
2
2
2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ y − x ⎤
2Λ
⎥ay = − ⋅ ⋅ y
⋅⎢
⎢
⎥
3
3
⎥
⎢ x 2 + y2 2⎥ ⎢
2
2
2
2
x +y
x +y
⎣
⎦⎣
⎦
⎥
⎦
m
(
)
ax = 0 ⋅
s
(
m
2
)
s
m
ax = 0 ⋅
s
m
2
ax = 0 ⋅
s
m
2
⎛ 2⋅ m ⎞
⎜ s
⎝
⎠
r =
or
m
2
ay = −0.25⋅
m
2
s
ay = −0.0333⋅
s
u
aradial = −ay = −
r
)
s
s
m
8⋅
(
ay = −8 ⋅
m
2
s
2
The instantaneous radius of curvature is obtained from
For the three points
(
v
⎡ Λ⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x2 − y2
⎥ ⋅⎢
3
⎢ 2 2 2⎥ ⎢
2
2
x +y
⎣ x +y ⎦⎣
ay = ⎢−
)⎤⎥ + ⎡−
r= −
u
2
ay
2
r = 0.5 m
m
2
s
y = 2m
⎛ 0.5⋅ m ⎞
⎜
s⎠
⎝
r =
0.25⋅
2
r = 1m
m
2
s
y = 3m
⎛ 0.2222⋅ m ⎞
⎜
s⎠
⎝
r =
0.03333 ⋅
m
2
r = 1.5⋅ m
2
s
The radius of curvature in each case is 1/2 of the vertical distance from the origin. The streamlines form circles tangent to the x axis
2⋅ Λ⋅ x⋅ y
−
(x2 + y2) = 2⋅x⋅y
=
=
2
2
dx
u
(x2 − y2)
Λ⋅ (x − y )
−
2
(x2 + y2)
dy
The streamlines are given by
2
v
(2
)
2
−2 ⋅ x ⋅ y ⋅ dx + x − y ⋅ dy = 0
so
This is an inexact integral, so an integrating factor is needed
R=
First we try
F=e
Then the integrating factor is
(
)
⎡d 2 2 d ( −2⋅ x ⋅ y)⎤ = − 2
⋅⎢ x − y −
⎥
−2 ⋅ x ⋅ y ⎣dx
y
dy
⎦
1
⌠
⎮
2
⎮ − dy
y
⎮
⌡
=
1
y
(2
2
) ⋅dy = 0
2
The equation becomes an exact integral
x
x −y
−2 ⋅ ⋅ dx +
2
y
y
So
2
⌠
x
x
u = ⎮ −2 ⋅ dx = −
+ f ( y)
⎮
y
y
⌡
ψ=
Comparing solutions
x
and
2
y
+y
(1)
(x2 − y2) dy = − x2 − y + g(x)
⌠
⎮
u=⎮
⎮
⌡
y
2
2
3.50
5.29
4.95
4.64
4.38
4.14
3.95
3.79
3.66
3.57
3.52
3.50
3.75
5.42
5.10
4.82
4.57
4.35
4.17
4.02
3.90
3.82
3.77
3.75
y
2
x + y = ψ⋅ y = const ⋅ y
or
These form circles that are tangential to the x axis, as can be shown in Excel:
x values
The stream function can be evaluated using Eq 1
2.50
2.25
2.00
1.75
1.50
1.25
1.00
0.75
0.50
0.25
0.00
0.10
62.6
50.7
40.1
30.7
22.6
15.7
10.1
5.73
2.60
0.73
0.10
0.25
25.3
20.5
16.3
12.5
9.25
6.50
4.25
2.50
1.25
0.50
0.25
See next page for plot:
0.50
13.0
10.6
8.50
6.63
5.00
3.63
2.50
1.63
1.00
0.63
0.50
0.75
9.08
7.50
6.08
4.83
3.75
2.83
2.08
1.50
1.08
0.83
0.75
1.00
7.25
6.06
5.00
4.06
3.25
2.56
2.00
1.56
1.25
1.06
1.00
1.25
6.25
5.30
4.45
3.70
3.05
2.50
2.05
1.70
1.45
1.30
1.25
1.50
5.67
4.88
4.17
3.54
3.00
2.54
2.17
1.88
1.67
1.54
1.50
1.75
5.32
4.64
4.04
3.50
3.04
2.64
2.32
2.07
1.89
1.79
1.75
2.00
5.13
4.53
4.00
3.53
3.13
2.78
2.50
2.28
2.13
2.03
2.00
2.25
5.03
4.50
4.03
3.61
3.25
2.94
2.69
2.50
2.36
2.28
2.25
y values
2.50
5.00
4.53
4.10
3.73
3.40
3.13
2.90
2.73
2.60
2.53
2.50
2.75
5.02
4.59
4.20
3.86
3.57
3.32
3.11
2.95
2.84
2.77
2.75
3.00
5.08
4.69
4.33
4.02
3.75
3.52
3.33
3.19
3.08
3.02
3.00
3.25
5.17
4.81
4.48
4.19
3.94
3.73
3.56
3.42
3.33
3.27
3.25
4.00
5.56
5.27
5.00
4.77
4.56
4.39
4.25
4.14
4.06
4.02
4.00
4.25
5.72
5.44
5.19
4.97
4.78
4.62
4.49
4.38
4.31
4.26
4.25
4.50
5.89
5.63
5.39
5.18
5.00
4.85
4.72
4.63
4.56
4.51
4.50
4.75
6.07
5.82
5.59
5.39
5.22
5.08
4.96
4.87
4.80
4.76
4.75
5.00
6.25
6.01
5.80
5.61
5.45
5.31
5.20
5.11
5.05
5.01
5.00
Problem 6.35
[Difficulty: 4] Part 1/2
Problem 6.35
[Difficulty: 4] Part 2/2
Problem 6.34
[Difficulty: 2]
Given:
Flow rates in elbow for uniform flow and free vortes
Find:
Plot discrepancy
Solution:
(
)
For Example 6.1 QUniform = V⋅ A = w⋅ r2 − r1 ⋅
⎛ r2 ⎞
For Problem 6.32 Q = w⋅ ln⎜
⎝ r1 ⎠
2
⋅
2 ⋅ r1 ⋅ r2
1
⎛ r2 ⎞
ρ⋅ ln⎜
⎝ r1 ⎠
⋅ ∆p
or
⎛ r2
⎞
−1
⎜
QUniform⋅ ρ
⎝ r1
⎠
=
w⋅ r1 ⋅ ∆p
⎛ r2 ⎞
ln⎜
⎝ r1 ⎠
2
2
2
ρ⋅ ⎛ r2 − r1 ⎞
⎝
⎠
⋅ ∆p
or
Q⋅ ρ
w⋅ r1 ⋅ ∆p
=
(1)
⎛ r2 ⎞ ⎛ r2 ⎞
2
⎜ r ⋅ ln⎜ ⋅
⎝ 1 ⎠ ⎝ r1 ⎠ ⎡⎢⎛ r2 ⎞ 2
−
⎢⎜ r
⎣⎝ 1 ⎠
⎤
⎥
1
⎥
⎦
(2)
It is convenient to plot these as functions of r2/r1
Eq. 1
Eq. 2
Error
1.01
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95
2.00
2.05
2.10
2.15
2.20
2.25
2.30
2.35
2.40
2.45
2.50
0.100
0.226
0.324
0.401
0.468
0.529
0.586
0.639
0.690
0.738
0.785
0.831
0.875
0.919
0.961
1.003
1.043
1.084
1.123
1.162
1.201
1.239
1.277
1.314
1.351
1.388
1.424
1.460
1.496
1.532
1.567
0.100
0.226
0.324
0.400
0.466
0.526
0.581
0.632
0.680
0.726
0.769
0.811
0.851
0.890
0.928
0.964
1.000
1.034
1.068
1.100
1.132
1.163
1.193
1.223
1.252
1.280
1.308
1.335
1.362
1.388
1.414
0.0%
0.0%
0.1%
0.2%
0.4%
0.6%
0.9%
1.1%
1.4%
1.7%
2.1%
2.4%
2.8%
3.2%
3.6%
4.0%
4.4%
4.8%
5.2%
5.7%
6.1%
6.6%
7.0%
7.5%
8.0%
8.4%
8.9%
9.4%
9.9%
10.3%
10.8%
10.0%
7.5%
Error
r2/r1
5.0%
2.5%
0.0%
1.0
1.2
1.4
1 .6
1 .8
r 2 /r 1
2.0
2.2
2.4
2.6
Problem 6.33
[Difficulty: 4] Part 1/2
Problem 6.33
[Difficulty: 4] Part 2/2
Problem 6.32
Given:
Velocity field for free vortex flow in elbow
Find:
Similar solution to Example 6.1; find k (above)
[Difficulty: 3]
Solution:
Basic equation
∂
∂r
2
p =
ρ⋅ V
c
V = Vθ =
r
with
r
Assumptions: 1) Frictionless 2) Incompressible 3) free vortex
2
∂
p =
2
ρ⋅ V
ρ⋅ c
d
=
p =
3
r
dr
r
For this flow
p ≠ p ( θ)
Hence
2⎛ 2
2⎞
⌠ 2
2
2
ρ⋅ c ⎛ 1
1 ⎞ ρ⋅ c ⋅ ⎝ r2 − r1 ⎠
⎮ ρ⋅ c
dr =
∆p = p 2 − p 1 = ⎮
⋅⎜
−
=
3
2
2 2
2 ⎜ 2
r
r1
r2
2⋅ r1 ⋅ r2
⎮
⎝
⎠
⌡r
so
∂r
r
(1)
1
Next we obtain c in terms of Q
⌠ →→
⎮
Q = ⎮ V dA =
⌡
r
r
⌠2
⌠ 2 w⋅ c
⎛ r2 ⎞
dr = w⋅ c⋅ ln⎜
⎮ V⋅ w dr = ⎮
r
⎮
⌡r
⎝ r1 ⎠
⌡r
1
1
Hence
c=
Q
⎛ r2 ⎞
w⋅ ln⎜
⎝ r1 ⎠
ρ⋅ c ⋅ ⎛ r2 − r1
⎝
2
Using this in Eq 1
∆p = p 2 − p 1 =
2
2
2 ⋅ r1 ⋅ r2
2
Solving for Q
2
2
2⎞
⎠ =
2 ⋅ r1 ⋅ r2
⎛ r2 ⎞
Q = w⋅ ln⎜
⋅
⋅ ∆p
⎝ r1 ⎠ ρ⋅ ⎛⎝ r2 2 − r12⎞⎠
ρ⋅ Q ⋅ ⎛ r2 − r1
⎝
2
2
2⎞
⎠
2
⎛ r2 ⎞ 2 2
2 ⋅ w ⋅ ln⎜
⋅ r1 ⋅ r2
⎝ r1 ⎠
2
2
2
2 ⋅ r1 ⋅ r2
⎛ r2 ⎞
k = w⋅ ln⎜
⋅
⎝ r1 ⎠ ρ⋅ ⎛⎝ r2 2 − r12⎞⎠
Problem 6.31
[Difficulty: 2]
Problem 6.30
[Difficulty: 2]
Given:
Flow in a curved section
Find:
Expression for pressure distribution; plot; V for wall static pressure of 35 kPa
Solution:
Basic equation
∂
∂n
2
p = ρ⋅
V
R
Assumptions: Steady; frictionless; no body force; constant speed along streamline
ρ = 999 ⋅
Given data
kg
3
V = 10⋅
m
At the inlet section
p = p( y)
m
L = 75⋅ mm
s
∂
hence
∂n
2
p( y) = pc −
Integrating from y = 0 to y = y
ρ⋅ V
R0⋅ L
⋅y
p =−
2
dp
dy
R0 = 0.2⋅ m
2
= ρ⋅
V
R
p c = 50⋅ kPa
2 2⋅ y
= ρ⋅ V ⋅
2 2⋅ y
dp = −ρ⋅ V ⋅
L⋅ R 0
p ⎛⎜
p ( 0 ) = 50⋅ kPa
(1)
L⎞
⎝2⎠
L⋅ R 0
⋅ dy
= 40.6⋅ kPa
40
y (mm)
30
20
10
40
42
44
46
48
50
p (kPa)
For a new wall pressure
p wall = 35⋅ kPa
solving Eq 1 for V gives
V =
(
)
4 ⋅ R0 ⋅ p c − p wall
ρ⋅ L
V = 12.7
m
s
Problem 6.29
[Diffculty: 4]
Given:
Velocity field for flow over a cylinder
Find:
Expression for pressure gradient; pressure variation; minimum pressure; plot velocity
Solution:
Basic equations
Given data
ρ = 1.23⋅
kg
a = 150 ⋅ mm
3
U = 75⋅
m
For this flow
⎤
⎡ a
Vr = U⋅ ⎢⎛⎜ ⎞ − 1⎥ ⋅ cos( θ)
r
On the surface r = a
Vr = 0
2
Hence on the surface:
⎣⎝ ⎠
⎦
m
s
⎤
⎡ a
Vθ = U⋅ ⎢⎛⎜ ⎞ + 1⎥ ⋅ sin( θ)
r
2
⎣⎝ ⎠
⎦
Vθ = 2 ⋅ U⋅ sin( θ)
For r momentum
2
⎛⎜
⎛⎜ V 2 ⎞
Vθ
Vθ ⎞
θ
∂
∂
∂
ρ⋅ ⎜ Vr⋅ Vr +
= ρ⋅ ⎜ −
⋅ Vr −
=− p
r ⎠
r ∂θ
⎝ a ⎠ ∂r
⎝ ∂r
For θ momentum
Vθ
Vr⋅ Vθ ⎞
⎛ ∂
⎛ Vθ ∂
⎞
2 ⋅ U⋅ sin( θ)
1 ∂
∂
ρ⋅ ⎜ Vr⋅ Vθ +
= ρ⋅ ⎜
⋅ 2 ⋅ U⋅ cos( θ) = − ⋅ p
⋅ Vθ +
⋅ Vθ = ρ⋅
a
r ∂θ
r ∂θ
r ⎠
⎝ ∂r
⎝ a ∂θ ⎠
2
∂
∂θ
p =−
4 ⋅ ρ⋅ U
a
∂
∂r
2
p = ρ⋅
Vθ
2
⋅ sin( θ) ⋅ cos( θ) = −
2 ⋅ ρ⋅ U
a
⋅ sin( 2 ⋅ θ)
For the pressure distribution we integrate from θ = 0 to θ = θ, assuming p(0) = p atm (a stagnation point)
θ
⌠
∂
p ( θ) − p atm = ⎮
p dθ =
⎮ ∂θ
⌡
0
θ
⌠
2
⎮
4 ⋅ ρ⋅ U
⎮ −
⋅ sin( θ) ⋅ cos( θ) dθ
a
⎮
⌡
0
a
2
= ρ⋅ 4 ⋅ U ⋅ sin( θ)
2
2⌠
θ
2
p ( θ) = −4 ⋅ ρ⋅ U ⎮ sin( θ) ⋅ cos( θ) dθ
⌡
p ( θ) = −2 ⋅ U ⋅ ρ⋅ sin( θ)
2
Minimum p:
0
Pressure (kPa)
0
50
100
p ⎛⎜
π⎞
⎝2⎠
= −13.8⋅ kPa
150
−5
− 10
− 15
x (m)
⎤
⎡ a
Vθ( r) = U⋅ ⎢⎛⎜ ⎞ + 1⎥
r
2
For the velocity as a function of radial position at θ = π/2
Vr = 0
so
V = Vθ
⎣⎝ ⎠
⎦
5
r/a
4
3
2
1
75
100
125
150
V(r) (m/s)
The velocity falls off to V = U as directly above the cylinder we have uniform horizontal as the effect of the cylinder decreases
m
Vθ( 100 ⋅ a) = 75
s
Problem 6.28
Given:
Velocity field for doublet
Find:
Expression for pressure gradient
[Difficulty: 2]
Solution:
Basic equations
For this flow
Hence for r momentum
Λ
Vr( r , θ) = − ⋅ cos( θ)
2
r
Λ
Vθ( r , θ) = − ⋅ sin( θ)
2
r
Vz = 0
2
⎛⎜
Vθ
Vθ ⎞
∂
∂
ρ⋅ g r − p = ρ⋅ ⎜ Vr⋅ Vr +
⋅ V −
r ⎠
r ∂θ r
∂r
⎝ ∂r
∂
Ignoring gravity
⎡⎢
⎢
Λ
Λ
∂
∂
p = −ρ⋅ ⎢⎛ − ⋅ cos( θ) ⎞ ⋅ ⎛ − ⋅ cos( θ) ⎞ +
⎜
⎜
2
2
⎢⎣⎝ r
∂r
⎠ ∂r ⎝ r
⎠
For θ momentum
ρ⋅ g θ −
⎛ − Λ ⋅ sin( θ)⎞
⎜ 2
⎝ r
⎠ ⋅ ∂ ⎛ − Λ ⋅ cos( θ) ⎞ −
r
∂θ ⎜ r2
⎝
⎠
⎛ − Λ ⋅ sin( θ) ⎞
⎜ 2
⎝ r
⎠
r
2⎤
⎥
⎥
⎥
⎥⎦
∂
∂r
2
p =
2⋅ Λ ⋅ ρ
5
r
Vθ
Vr⋅ Vθ ⎞
⎛ ∂
1 ∂
∂
⋅ p = ρ⋅ ⎜ Vr⋅ Vθ +
⋅ Vθ +
r ∂θ
r ∂θ
r ⎠
⎝ ∂r
Ignoring gravity
⎡
⎢
Λ
∂
⎢ Λ
∂
p = −r⋅ ρ⋅ ⎛ − ⋅ cos( θ) ⎞ ⋅ ⎛ − ⋅ sin( θ) ⎞ +
⎜
⎜
⎢
2
2
∂θ
⎣⎝ r
⎠ ∂r ⎝ r
⎠
The pressure gradient is purely radial
⎛ − Λ ⋅ sin( θ) ⎞
⎜ 2
⎝ r
⎠ ⋅ ∂ ⎛ − Λ ⋅ sin( θ)⎞ +
r
∂θ ⎜ r2
⎝
⎠
⎛ − Λ ⋅ sin( θ)⎞ ⋅ ⎛ − Λ ⋅ cos( θ) ⎞ ⎤
⎥
⎜ 2
⎜ 2
⎝ r
⎠⎝ r
⎠⎥
⎥
r
⎦
∂
∂θ
p =0
Problem 6.27
[Difficulty: 5]
Given:
Velocity field
Find:
Constant B for incompressible flow; Equation for streamline through (1,2); Acceleration of particle; streamline
curvature
Solution:
Basic equations
(4
2 2
u ( x, y ) = A ⋅ x − 6⋅ x ⋅ y + y
For this flow
∂
∂x
∂
∂x
u ( x, y ) +
u( x , y) +
∂
∂y
v ( x, y ) =
∂
∂y
∂
∂x
)
v ( x, y ) = B⋅ x ⋅ y − x⋅ y
(3
(
)
4
3
)
(
)
⎡⎣A ⋅ x4 − 6⋅ x2⋅ y2 + y 4 ⎤⎦ + ∂ ⎡⎣B⋅ x3⋅ y − x⋅ y 3 ⎤⎦ = 0
(3
∂y
) + A⋅(4⋅x3 − 12⋅x⋅y2) = (4⋅A + B)⋅x⋅(x2 − 3⋅y2) = 0
2
v ( x , y) = B⋅ x − 3⋅ x ⋅ y
B = −4 ⋅ A
Hence
1
B = −8
3
m ⋅s
Hence for ax
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
(4
)
4 ∂
2 2
u = A⋅ x − 6 ⋅ x ⋅ y + y ⋅
2
(2
ax = 4 ⋅ A ⋅ x ⋅ x + y
)
2
∂x
(
)
(
)
(
)
⎡⎣A⋅ x 4 − 6⋅ x 2⋅ y2 + y 4 ⎤⎦ + ⎡⎣−4 ⋅ A⋅ x3⋅ y − x⋅ y 3 ⎤⎦ ⋅ ∂ ⎡⎣A⋅ x 4 − 6⋅ x 2⋅ y2 + y 4 ⎤⎦
∂y
3
For ay
ay = u ⋅
∂
∂x
v + v⋅
∂
∂y
(4
2
(2
ay = 4 ⋅ A ⋅ y ⋅ x + y
For a streamline
dy
dx
Let
)
4 ∂
2 2
v = A⋅ x − 6 ⋅ x ⋅ y + y ⋅
u=
=
v
u
so
)
2
∂x
(
)
(
)
(
)
⎡⎣−4⋅ A⋅ x 3⋅ y − x ⋅ y3 ⎤⎦ + ⎡⎣−4⋅ A⋅ x 3⋅ y − x ⋅ y3 ⎤⎦ ⋅ ∂ ⎡⎣−4 ⋅ A⋅ x3⋅ y − x⋅ y 3 ⎤⎦
∂y
3
dy
dx
y
du
x
dx
=
=
(3
−4 ⋅ A⋅ x ⋅ y − x ⋅ y
)
3
A⋅ x − 6 ⋅ x ⋅ y + y
4
)
=−
d ⎛⎜
d ⎛⎜
1⎞
(4
2 2
y⎞
(3
4⋅ x ⋅ y − x⋅ y
(x4 − 6⋅x2⋅y2 + y4)
⎝ x ⎠ = 1 ⋅ dy + y⋅ ⎝ x ⎠ = 1 ⋅ dy − y
dx
x dx
dx
)
3
x dx
x
2
so
dy
dx
= x⋅
du
dx
+u
dy
Hence
dx
x⋅
du
dx
dx
Separating variables
x
= x⋅
du
(3
4⋅ x ⋅ y − x⋅ y
dx
+u=−
⎡⎢
⎢
⎢⎣
=−u+
(x4 − 6⋅x2⋅y2 + y4)
(
)
4
(
2
(
4⋅ 1 − u
(
⎛ 1 − 6⋅ u + u3⎞
⎜
⎝u
⎠
)
4⋅ 1 − u
)
2
⎛ 1 − 6⋅ u + u3⎞
⎜
⎝u
⎠
)
1
5
3
ln( x ) = − ⋅ ln u − 10⋅ u + 5 ⋅ u + C
5
⋅ du
(u5 − 10⋅u3 + 5⋅u)⋅x5 = c
3 2
)u +
2
(
2
4
2
u ⋅ ( u − 10⋅ u + 5 )
5
=−
4
2
⎤⎥
u ⋅ u − 10⋅ u + 5
=−
4
2
⎛ 1 − 6⋅ u + u3⎞ ⎥
u − 6⋅ u + 1
⎜
⎥
⎝u
⎠⎦
4⋅ 1 − u
u − 6⋅ u + 1
=−
)
3
5
3 2
4
y − 10⋅ y ⋅ x + 5 ⋅ y ⋅ x = const
4
y − 10⋅ y ⋅ x + 5 ⋅ y ⋅ x = −38
For the streamline through (1,2)
Note that it would be MUCH easier to use the stream function method here!
2
an = −
To find the radius of curvature we use
2
V
V
R =
or
R
an
r
V
We need to find the component of acceleration normal to the velocity vector
(3
)
⎤
⎥
4
2 2
4 ⎥
⎣ x − 6⋅ x ⋅ y + y ⎦
⎡
v
θvel = atan⎛⎜ ⎞ = atan⎢−
⎢
⎝u⎠
At (1,2) the velocity vector is at angle
4⋅ x ⋅ y − x⋅ y
(
4⋅ ( 2 − 8) ⎤
θvel = atan⎡⎢−
⎥
⎣ 1 − 24 + 16⎦
3
r
a
)
∆θ
θvel = −73.7⋅ deg
At (1,2) the acceleration vector is at angle
) ⎤⎥ = atan⎛ y ⎞
⎥
⎜
3
x
⎥
)⎦ ⎝ ⎠
⎡ 2
⎢ 4⋅ A ⋅ y⋅ x2 + y2
⎛ ay ⎞
θaccel = atan⎜
= atan⎢
⎝ ax ⎠
⎢ 4 ⋅ A2⋅ x ⋅ x 2 + y2
⎣
(
(
3
∆θ = θaccel − θvel
Hence the angle between the acceleration and velocity vectors is
an = a⋅ sin( ∆θ)
The component of acceleration normal to the velocity is then
At (1,2)
(2
2
a =
2
)
2
ax = 4 ⋅ A ⋅ x ⋅ x + y
3
7
2
a = 4472
2
(4
2 2
2
Then
R =
V
an
2
m
an = a⋅ sin( ∆θ)
2
) = −14⋅ ms
4
(3
⎝
m⎞
s
2
2
(2
3
2
ay = 4 ⋅ A ⋅ y ⋅ x + y
)
2
⎠
2
×
) = 48⋅ ms
3
1
an = 3040
m
2
V=
2
s
3040 m
2
u + v = 50⋅
2
⋅
= 4000⋅
m
2
s
s
v = B⋅ x ⋅ y − x ⋅ y
R = ⎛⎜ 50⋅
ax + ay
a=
s
s
u = A⋅ x − 6 ⋅ x ⋅ y + y
θaccel = 63.4⋅ deg
∆θ = 137 ⋅ deg
where
⎛ 2 ⎞ = 2000⋅ m
⎜ 3
2
s
⎝ m ⋅s ⎠
7
= 500 ⋅ m × A = 500 ⋅ m ×
2 m
2000 + 4000 ⋅
2
θaccel = atan⎛⎜ ⎞
⎝1⎠
R = 0.822 m
m
s
Problem 6.26
[Difficulty: 4] Part 1/2
Problem 6.26
[Difficulty: 4] Part 2/2
Problem 6.25
[Difficulty: 4]
Given:
Velocity field
Find:
Constant B for incompressible flow; Acceleration of particle at (2,1); acceleration normal to velocity at (2,1)
Solution:
Basic equations
For this flow
3
u ( x , y ) = A⋅ x + B⋅ x ⋅ y
∂
∂x
∂
∂x
u( x , y) +
u( x , y) +
∂
∂y
∂
∂y
2
v( x , y) =
3
∂
∂x
Hence for ax
ax = u ⋅
u + v⋅
∂
∂y
(
ay = u ⋅
∂
(2
∂x
v + v⋅
2
∂
∂y
(2
3
3
0.2
2
2
)
2
)
2
∂y
2
(
2
2
∂x
3
) (A⋅y3 − 3⋅A⋅x2⋅y) + (A⋅y3 − 3⋅A⋅x2⋅y)⋅∂ (A⋅y3 − 3⋅A⋅x2⋅y)
2 ∂
∂x
∂y
2
2
2
ax + ay
1
) (A⋅x3 − 3⋅A⋅x⋅y2) + (A⋅y3 − 3⋅A⋅x2⋅y)⋅∂ (A⋅x3 − 3⋅A⋅x⋅y2)
0.2 ⎞
2
2
ay = 3 ⋅ ⎛
× 1 ⋅ m × ⎡⎣( 2 ⋅ m) + ( 1 ⋅ m) ⎤⎦
⎜ 2
⎝ m ⋅s ⎠
a =
B = −0.6
2 ∂
⎞ × 2⋅ m × ⎡( 2⋅ m) 2 + ( 1 ⋅ m) 2⎤
⎣
⎦
⎜ 2
m
⋅
s
⎝
⎠
ax = 3 ⋅ ⎛
B = −3 ⋅ A
Hence
v ( x , y ) = A⋅ y − 3 ⋅ A⋅ x ⋅ y
v = A⋅ x − 3 ⋅ A⋅ x ⋅ y ⋅
ay = 3 ⋅ A ⋅ y ⋅ x + y
Hence at (2,1)
2
u = A⋅ x − 3 ⋅ A⋅ x ⋅ y ⋅
ax = 3 ⋅ A ⋅ x ⋅ x + y
For ay
)=0
2
m ⋅s
3
2
∂y
(2
u ( x , y ) = A⋅ x − 3 ⋅ A⋅ x ⋅ y
∂x
(A⋅x3 + B⋅x⋅y2) + ∂ (A⋅y3 + B⋅x2⋅y) = 0
v ( x , y ) = ( 3 ⋅ A + B) ⋅ x + y
We can write
∂
2
v ( x , y ) = A⋅ y + B⋅ x ⋅ y
2
ax = 6.00⋅
m
2
s
2
ay = 3.00⋅
m
2
s
a = 6.71
m
2
s
We need to find the component of acceleration normal to the velocity vector
At (2,1) the velocity vector is at angle
r
a
⎛ 1 − 3⋅ 2 ⋅ 1 ⎞
θvel = atan⎜
⎜ 3 − 3⋅ 2⋅ 12
⎝2
⎠
θvel = −79.7⋅ deg
⎛ ay ⎞
θaccel = atan⎜
⎝ ax ⎠
1
θaccel = atan⎛⎜ ⎞
⎝2⎠
θaccel = 26.6⋅ deg
∆θ = θaccel − θvel
∆θ = 106 ⋅ deg
3
At (1,2) the acceleration vector is at angle
r
V
⎛ A⋅ y 3 − 3⋅ A⋅ x2⋅ y ⎞
v
θvel = atan⎛⎜ ⎞ = atan⎜
⎜ A⋅ x 3 − 3⋅ A⋅ x⋅ y 2
⎝u⎠
⎝
⎠
Hence the angle between the acceleration and velocity vectors is
The component of acceleration normal to the velocity is then
2
an = a⋅ sin( ∆θ) = 6.71⋅
∆θ
m
2
s
⋅ sin( 106 ⋅ deg)
an = 6.45⋅
m
2
s
Problem 6.24
[Difficulty: 3]
Problem 6.23
[Difficulty: 4]
Given:
Rectangular chip flow
Find:
Velocity field; acceleration; pressure gradient; net force; required flow rate; plot pressure
Solution:
Basic equations
→→
(
∑ V⋅A) = 0
∂
∂x
CS
The given data is
ρ = 1.23⋅
kg
u +
∂
∂y
v =0
p atm = 101 ⋅ kPa
3
h = 0.5⋅ mm
b = 40⋅ mm
M length = 0.005 ⋅
m
Assuming a CV that is from the centerline to any point x, and noting that q is inflow per unit area, continuity leads to
q ⋅ x ⋅ L = U⋅ h ⋅ L
u ( x ) = U( x ) = q ⋅
or
x
h
For acceleration we will need the vertical velocity v; we can use
∂
∂x
Hence
u +
∂
∂y
v =0
q
x
du
∂
d
= − ⎛⎜ q ⋅ ⎞ = −
v =− u =−
h
dx
dx ⎝ h ⎠
∂y
∂x
∂
or
⌠
v ( y = y ) − v ( y = 0 ) = −⎮
⎮
⌡
y
0
But
v( y = 0) = q
For the x acceleration
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
q
h
dy = −q ⋅
y
h
y⎞
so
v ( y ) = q ⋅ ⎛⎜ 1 −
u
x q
y
ax = q ⋅ ⋅ ⎛⎜ ⎞ + q ⋅ ⎛⎜ 1 − ⎞ ⋅ ( 0 )
h ⎝h⎠
h⎠
⎝
⎝
h⎠
ax =
q
h
2
2
⋅x
kg
m
For the y acceleration
ay = u ⋅
∂
∂x
v + v⋅
∂
x
y
q
ay = q ⋅ ⋅ ( 0 ) + q ⋅ ⎛⎜ 1 − ⎞ ⋅ ⎛⎜ − ⎞
h
h⎠ ⎝ h⎠
⎝
v
∂y
∂
Hence
∂x
ρ⋅
Also
p = −ρ⋅
q
h
Dv
Dx
Du
ρ⋅
For the pressure gradient we use x and y momentum (Euler equation)
Dx
ax =
q
2
h
⋅ ⎛⎜
y
⎝h
− 1⎞
⎠
⎛ ∂
∂ ⎞
∂
u + v ⋅ u = ρ⋅ ax = − p
∂y ⎠
∂x
⎝ ∂x
= ρ⋅ ⎜ u ⋅
2
2
⋅x
⎛ ∂
∂ ⎞
∂
v + v ⋅ v = ρ⋅ ay = − p
∂y ⎠
∂y
⎝ ∂x
∂
= ρ⋅ ⎜ u ⋅
p = ρ⋅
∂y
q
2
h
⋅ ⎛⎜ 1 −
⎝
y⎞
h⎠
For the pressure distribution, integrating from the outside edge (x = b/2) to any point x
x
p ( x = x ) − p ⎛⎜ x =
b⎞
⎝
2⎠
x
⌠
2
2
2
⌠
⎮
q
q
q
2
2
⎮ ∂
= p ( x ) − p atm =
p dx = ⎮ −ρ⋅ ⋅ x dx = −ρ⋅
⋅ x + ρ⋅
⋅b
⎮ ∂x
2
2
2
⎮
h
2⋅ h
8⋅ h
⎮
⎮b
⌡b
⌡
2
2 2
p ( x ) = p atm + ρ⋅
q ⋅b
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
⎥
⎝b⎠ ⎦
8⋅ h ⎣
2
x⎞
2
2⎤
For the net force we need to integrate this ... actually the gage pressure, as this pressure is opposed on the outer surface by p atm
2 2
pg( x) =
b
ρ⋅ q ⋅ b
8⋅ h
2
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
⎣
x⎞
2⎤
⎥
⎝b⎠ ⎦
b
⌠2
⌠2
2
2 2
⎮
⎮ ρ⋅ q 2⋅ b 2 ⎡
ρ⋅ q ⋅ b ⋅ L ⎛ b
x ⎤
1 b
Fnet = 2 ⋅ L⋅ ⎮ p g ( x ) dx = 2 ⋅ L⋅ ⎮
⋅ ⎢1 − 4 ⋅ ⎛⎜ ⎞ ⎥ dx =
⋅⎜ − ⋅ ⎞
2 ⎣
2
⌡
⎝b⎠ ⎦
⎝2 3 2⎠
⎮
8⋅ h
4⋅ h
0
⌡
2 3
Fnet =
ρ⋅ q ⋅ b ⋅ L
12⋅ h
2
0
2 3
The weight of the chip must balance this force
M ⋅ g = M length ⋅ L⋅ g = Fnet =
ρ⋅ q ⋅ b ⋅ L
12⋅ h
2
2 3
or
M length ⋅ g =
ρ⋅ q ⋅ b
12⋅ h
3
m
2
q =
Solving for q for the given mass/length
12⋅ h ⋅ g ⋅ M length
ρ⋅ b
q = 0.0432⋅
3
s
2
m
b
The maximum speed
b⎞
b⋅ q
Umax =
2⋅ h
Umax = u ⎛⎜ x =
= q⋅
h
2⎠
⎝
2
2 2
The following plot can be done in Excel
pg( x) =
ρ⋅ q ⋅ b
8⋅ h
2
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
⎣
x⎞
2⎤
⎥
⎝b⎠ ⎦
m
Umax = 1.73
s
2
2
Pressure (Pa)
1.5
1
0.5
− 0.02
− 0.01
0
0.01
0.02
x (m)
The net force is such that the chip is floating on air due to a Bernoulli effect: the speed is maximum at the edges and zero at the
center; pressure has the opposite trend - pressure is minimum (p atm) at the edges and maximum at the center.
Problem 6.20
Problem 6.22
[Difficulty: 3]
Problem 6.21
[Difficulty: 4]
Problem 6.20
[Difficulty: 4]
Problem 6.19
[Difficulty: 3]
Given:
Diffuser geometry
Find:
Acceleration of a fluid particle; plot it; plot pressure gradient; find L such that pressure gradient is less than
25 kPa/m
Solution:
The given data is
Di = 0.25⋅ m
Do = 0.75⋅ m
D( x) = Di +
For a linear increase in diameter
From continuity
Hence
Q = V⋅ A = V⋅
V( x) ⋅
π
4
Do − Di
L
π 2
2
⋅ D = Vi⋅ ⋅ Di
4
4
2
ρ = 1000⋅
4⋅ Q
m
s
Vi
V( x) =
or
Do − Di ⎞
⎛
⋅x
π⋅ ⎜ Di +
L
⎝
⎠
2
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
The governing equation for this flow is
ax = V⋅
or, for steady 1D flow, in the notation of the problem
2
Hence
ax ( x ) = −
(
d
V=
dx
Vi
d
2 dx
Vi
⋅
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
This can be plotted in Excel (see below)
From Eq. 6.2a, pressure gradient is
∂
∂x
3
m
⋅x
Q = 0.245
V( x) =
2 ⋅ Vi ⋅ Do − Di
kg
3
π
⋅ D( x) = Q
m
Vi = 5⋅
s
L = 1⋅ m
p = −ρ⋅ ax
∂
∂x
2
p =
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
2
2
This can also plotted in Excel. Note that the pressure gradient is adverse: separation is likely to occur in the diffuser, and occur
near the entrance
∂
At the inlet
∂x
p = 100 ⋅
kPa
∂
At the
exit
m
∂x
p = 412 ⋅
Pa
m
To find the length L for which the pressure gradient is no more than 25 kPa/m, we need to solve
2
∂
∂x
p ≤ 25⋅
kPa
m
=
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
with x = 0 m (the largest pressure gradient is at the inlet)
2
L≥
Hence
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
Di⋅
∂
∂x
)
L ≥ 4⋅ m
p
This result is also obtained using Goal Seek in Excel.
In Excel:
Di
Do
L
Vi
=
=
=
=
0.25
0.75
1
5
m
m
m
m/s
( =
1000
kg/m 3
x (m) a (m/s 2)
dp /dx (kPa/m)
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.40
0.50
0.60
-100
-62.1
-40.2
-26.9
-18.59
-13.17
-9.54
-5.29
-3.125
-1.940
100
62.1
40.2
26.93
18.59
13.17
9.54
5.29
3.125
1.940
0.70
0.80
0.90
1.00
-1.256
-0.842
-0.581
-0.412
1.256
0.842
0.581
0.412
For the length L required
for the pressure gradient
to be less than 25 kPa/m
use Goal Seek
L =
4.00
x (m)
dp /dx (kPa/m)
0.0
25.0
m
Acceleration Through a Diffuser
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.9
1.0
2
a (m/s )
-20
-40
-60
-80
-100
-120
x (m)
Pressure Gradient Along A Diffuser
dp /dx (kPa/m)
120
100
80
60
40
20
0
0.0
0.1
0.2
0.3
0.4
0.5
x (m)
0.6
0.7
0.8
Problem 6.18
[Difficulty: 3]
Given:
Nozzle geometry
Find:
Acceleration of fluid particle; Plot; Plot pressure gradient; find L such that pressure gradient < 5 MPa/m in
absolute value
Solution:
The given data is
Di = 0.1⋅ m
Do = 0.02⋅ m
D( x ) = Di +
For a linear decrease in diameter
From continuity
Hence
or
Q = V⋅ A = V⋅
V( x ) ⋅
π
4
Do − Di
L
π 2
2
⋅ D = Vi⋅ ⋅ Di
4
4
3
m
⋅x
Q = 0.00785
2
⋅ D( x ) = Q
m
s
4⋅ Q
V( x ) =
Do − Di ⎞
⎛
⋅x
π⋅ ⎜ Di +
L
⎝
⎠
Vi
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
kg
ρ = 1000⋅
3
π
V( x ) =
m
Vi = 1 ⋅
s
L = 0.5⋅ m
2
2
The governing equation for this flow is
or, for steady 1D flow, in the notation of the problem
d
ax = V⋅ V =
dx
Vi
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
d
⋅
2 dx
2
Vi
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
This is plotted in the associated Excel workbook
From Eq. 6.2a, pressure gradient is
∂
∂x
p = −ρ⋅ ax
∂
∂x
2
p =
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
2
ax ( x ) = −
(
2 ⋅ Vi ⋅ Do − Di
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
This is also plotted in the associated Excel workbook. Note that the pressure gradient is always negative: separation is unlikely to
occur in the nozzle
∂
At the inlet
∂x
p = −3.2⋅
kPa
∂
At the exit
∂x
m
p = −10⋅
To find the length L for which the absolute pressure gradient is no more than 5 MPa/m, we need to solve
∂
∂x
2
p ≤ 5⋅
MPa
m
=
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
with x = L m (the largest pressure gradient is at the outlet)
2
L≥
Hence
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
5
L ≥ 1⋅ m
⎛ Do ⎞ ∂
Di⋅ ⎜
⋅
p
Di
x
∂
⎝ ⎠
This result is also obtained using Goal Seek in the Excel workbook
From Excel
x (m) a (m/s 2)
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.420
0.440
0.460
0.470
0.480
0.490
0.500
3.20
4.86
7.65
12.6
22.0
41.2
84.2
194
529
843
1408
2495
3411
4761
6806
10000
dp /dx (kPa/m)
-3.20
-4.86
-7.65
-12.6
-22.0
-41.2
-84.2
-194
-529
-843
-1408
-2495
-3411
-4761
-6806
-10000
For the length L required
for the pressure gradient
to be less than 5 MPa/m (abs)
use Goal Seek
L =
1.00
x (m)
dp /dx (kPa/m)
1.00
-5000
m
MPa
m
Acceleration Through A Nozzle
12000
2
a (m/s )
10000
8000
6000
4000
2000
0
0.0
0.1
0.1
0.2
0.2
0.3
0.3
0.4
0.4
0.5
0.5
0.4
0.5
0.5
x (m)
Pressure Gradient Along A Nozzle
0
dp/dx (kPa/m)
0.0
0.1
0.1
0.2
0.2
0.3
-2000
-4000
-6000
-8000
-10000
-12000
x (m)
0.3
0.4
Problem 6.17
[Difficulty: 3]
Given:
Flow in a pipe with variable area
Find:
Expression for pressure gradient and pressure; Plot them
Solution:
Assumptions: 1) Incompressible flow 2) Flow profile remains unchanged so centerline velocity can represent average velocity
Basic equations
Given data
Q = V⋅ A
ρ = 1250⋅
kg
2
A0 = 0.25⋅ m
3
a = 1.5⋅ m
L = 5⋅ m
m
For this 1D flow
Q = u 0 ⋅ A0 = u ⋅ A
so
A0
u( x) = u0⋅
=
A
⎛
−
⎜
A( x ) = A0 ⋅ ⎝ 1 + e
x
a
−
−e
x
⎞
2⋅ a
⎠
u 0 = 10⋅
m
u0
⎛
−
⎜
⎝1 + e
x
a
−
−e
x
⎞
2⋅ a
⎠
⎛ −
2
2⋅ a ⎜
⋅ ⎝ 2⋅ e
u0 ⋅ e
−
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
u =
∂
∂x
u0
⎡
⎢
x ⎞ ∂x ⎢ ⎛
x
−
−
−
⎜
⎢
2⋅ a
a
−e
−e
⎠ ⎣⎝1 + e
u0
⎛
−
⎜
⎝1 + e
x
a
p = −ρ⋅ ax − ρ⋅ g x = −
⋅
⎛ −
2
2⋅ a ⎜
⋅ ⎝ 2⋅ e
ρ⋅ u 0 ⋅ e
−
For the pressure
p 0 = 300 ⋅ kPa
s
⎛ −
⎜
2 ⋅ a⋅ ⎝ e
x
x
a
−
−e
∂
⎞
x
2⋅ a
x
2⋅ a
− 1⎠
⎞
+ 1⎠
3
⎤
⎥=
x ⎞⎥
⎛ −
2⋅ a ⎥
⎠ ⎦ 2 ⋅ a⋅ ⎜⎝ e
x
x
a
−
−e
⎞
x
2⋅ a
x
2⋅ a
− 1⎠
⎞
+ 1⎠
3
x
dp =
and
∂
∂x
⌠
⎮
−
⎮
x
2
⌠
⋅
⋅
e
ρ
u
⎮
0
∂
p − pi = ⎮
p dx = ⎮ −
⎮ ∂x
⎮
⌡
⎛ −
0
⎮
⎜
⎮
2 ⋅ a⋅ ⎝ e
⌡
p ⋅ dx
⎛ −
2⋅ a ⎜
⋅ ⎝ 2⋅ e
x
x
a
−
−e
⎞
x
2⋅ a
− 1⎠
⎞
x
2⋅ a
3
dx
+ 1⎠
0
∂
This is a tricky integral, so instead consider the following:
x
x
0
0
∂x
p = −ρ⋅ ax = −ρ⋅ u ⋅
( )
∂
1 ∂
2
u
u = − ⋅ ρ⋅
2
∂x
∂x
⌠
⌠
ρ
ρ
2
2
2
∂
∂
p − pi = ⎮
p dx = − ⋅ ⎮
u dx = ⋅ u ( x = 0 ) − u ( x )
⎮ ∂x
⎮
2
2
∂x
⌡
⌡
Hence
( )
ρ
2
2
p(x) = p0 + ⋅ ⎛ u0 − u(x) ⎞
⎝
⎠
2
p( x) = p0 +
ρ⋅ u 0
2
2
⎡
⋅ ⎢1 −
⎢
⎢
⎣
(
)
which we recognise as the Bernoulli equation!
1
⎡⎢
x
⎢⎛
−
−
⎜
⎢⎣ ⎝ 1 + e a − e
⎤⎥
x ⎞⎥
2⋅ a ⎥
⎠⎦
2⎤
⎥
⎥
⎥
⎦
The following plots can be done in Excel
0.26
Area (m2)
0.24
0.22
0.2
0.18
0
1
2
3
x (m)
4
5
Pressure Gradient (kPa/m)
20
0
1
2
3
4
5
− 20
− 40
− 60
x (m)
300
Pressure (kPa)
290
280
270
260
250
0
1
2
3
x (m)
4
5
Problem 6.16
[Difficulty: 3]
Given:
Flow in a pipe with variable area
Find:
Expression for pressure gradient and pressure; Plot them; exit pressure
Solution:
Assumptions: 1) Incompressible flow 2) Flow profile remains unchanged so centerline velocity can represent average velocity
Basic equations
Q = V⋅ A
Given data
ρ = 1.75⋅
slug
ft
For this 1D flow
2
p i = 35⋅ psi
3
Q = u i⋅ Ai = u ⋅ A
Ai = 15⋅ in
(Ai − Ae)
A = Ai −
L
⋅x
so
2
Ae = 2.5⋅ in
L = 10⋅ ft
Ai
u ( x ) = u i⋅
= u i⋅
A
ui = 5⋅
Ai
Ai −
⎡ ( Ai − Ae) ⎤
⎢
⋅ x⎥
⎣ L
⎦
Ai
⎤ Ai ⋅ L ⋅ u i ⋅ ( Ae − Ai)
⎡
⎥=
ax = u ⋅ u + v ⋅ u = u i ⋅
⋅ ⎢u i⋅
∂x
∂y
⎡ ( Ai − Ae) ⎤ ∂x ⎢
⎡ ( Ai − Ae) ⎤ ⎥ ( A ⋅ L + A ⋅ x − A ⋅ x) 3
i
e
i
Ai − ⎢
⋅ x⎥
⋅ x⎥ ⎥
⎢ Ai − ⎢ L
⎣ L
⎦ ⎣
⎣
⎦⎦
∂
For the pressure
∂
∂x
Ai
∂
2
p = −ρ⋅ ax − ρ⋅ g x = −
2
2
2
∂
(
ρ⋅ Ai ⋅ L ⋅ u i ⋅ Ae − Ai
(Ai⋅ L + Ae⋅ x − Ai⋅ x)
dp =
∂
∂x
3
⌠
x
2 2 2
⎮
⌠
ρ⋅ Ai ⋅ L ⋅ u i ⋅ Ae − Ai
∂
⎮
⎮
p − pi =
p dx =
−
dx
⎮ ∂x
⎮
3
Ai⋅ L + Ae⋅ x − Ai⋅ x
⌡
⎮
0
⌡
p ⋅ dx
(
(
This is a tricky integral, so instead consider the following:
x
x
0
0
∂
∂x
p = −ρ⋅ ax = −ρ⋅ u ⋅
( )
∂
1 ∂
2
u
u = − ⋅ ρ⋅
2 ∂x
∂x
⌠
⌠
ρ
ρ
2
2
2
∂
∂
p − pi = ⎮
p dx = − ⋅ ⎮
u dx = ⋅ u ( x = 0 ) − u ( x )
⎮ ∂x
⎮
2
2
∂x
⌡
⌡
( )
(
)
)
)
0
Hence
2
)
x
and
2
ft
s
ρ
2
2
p( x) = pi + ⋅ ⎛ ui − u( x) ⎞
⎠
2 ⎝
p( x) = pi +
Hence
⎡
⎢
⋅ 1−
2 ⎢
⎢
⎣
ρ⋅ u i
2
which we recognise as the Bernoulli equation!
Ai
⎡⎢
⎤⎥
⎢
⎡ ( Ai − Ae) ⎤ ⎥
⋅ x⎥ ⎥
⎢ Ai − ⎢ L
⎣
⎣
⎦⎦
2⎤
⎥
⎥
⎥
⎦
p ( L) = 29.7 psi
At the exit
Pressure Gradient (psi/ft)
The following plots can be done in Excel
6
4
2
0
2
4
6
8
10
6
8
10
Pressure (psi)
x (ft)
35
30
25
0
2
4
x (ft)
Problem 6.15
[Difficulty: 3]
Problem 6.14
[Difficulty: 3]
Given:
Velocity field
Find:
The acceleration at several points; evaluate pressure gradient
Solution:
The given data is
3
3
m
q = 2⋅
m
s
K = 1⋅
m
s
m
ρ = 1000⋅
kg
q
Vr = −
2 ⋅ π⋅ r
3
m
K
Vθ =
2 ⋅ π⋅ r
The governing equations for this 2D flow are
The total acceleration for this steady flow is then
2
2
Vθ
Vθ
∂
ar = Vr⋅ Vr +
⋅ Vr −
r
r ∂θ
∂r
ar = −
θ - component
Vθ
Vr⋅ Vθ
∂
∂
aθ = Vr⋅ Vθ +
⋅ Vθ +
r ∂θ
r
∂r
aθ = 0
Evaluating at point (1,0)
ar = −0.127
r - component
Evaluating at point (1,π/2)
Evaluating at point (2,0)
From Eq. 6.3, pressure gradient is
∂
ar = −0.127
m
∂
∂r
s
m
aθ = 0
2
s
m
2
s
Evaluating at point (1,π/2)
Evaluating at point (2,0)
∂
∂r
∂
∂r
∂
∂r
p = 127 ⋅
p = 127 ⋅
∂r
Pa
m
Pa
p = 15.8⋅
aθ = 0
∂
p = −ρ⋅ ar
1 ∂
⋅ p = −ρ⋅ aθ
r ∂θ
Evaluating at point (1,0)
m
Pa
m
2 3
4⋅ π ⋅ r
aθ = 0
2
ar = −0.0158
2
q +K
p =
(2
2 3
4⋅ π ⋅ r
1 ∂
⋅ p =0
r ∂θ
1 ∂
⋅ p =0
r ∂θ
1 ∂
⋅ p =0
r ∂θ
1 ∂
⋅ p =0
r ∂θ
)
2
ρ⋅ q + K
Problem 6.13
[Difficulty: 2]
Problem 6.12
[Difficulty: 2]
Given:
Velocity field
Find:
Expression for acceleration and pressure gradient; plot; evaluate pressure at outlet
Solution:
Basic equations
Given data
U = 20⋅
m
L = 2⋅ m
s
u ( x ) = U⋅ e
ρ = 900 ⋅
kg
3
m
−
Here
p in = 50⋅ kPa
x
L
u ( 0 ) = 20
m
u ( L) = 7.36
s
m
s
−
The x component of acceleration is then
The x momentum becomes
The pressure gradient is then
ax ( x ) = u ( x ) ⋅
ρ⋅ u ⋅
dp
dx
2
∂
∂x
ax ( x ) = −
u( x)
=
ρ
L
−
L
2
⋅U ⋅e
2⋅ x
L
⌠
p ( x ) = p in − ρ⋅ ⎮ ax ( x ) dx
⌡
0
2
Hence
L
d
d
u = ρ⋅ aa = − p
dx
dx
x
Integrating momentum
U ⋅e
2⋅ x
p ( L) = p in −
( − 2 − 1)
U ⋅ ρ⋅ e
2
⎛ − 2⋅ x
⎞
2 ⎜
L
U ⋅ ρ⋅ ⎝ e
− 1⎠
p ( x ) = p in −
2
p ( L) = 206 ⋅ kPa
dp/dx (kPa/m)
200
150
100
50
0
0.5
1
1.5
2
x (m)
ax (m/s2)
0
0.5
1
− 50
− 100
− 150
− 200
x (m)
1.5
2
Problem 6.11
[Difficulty: 2]
Given:
Velocity field
Find:
Expression for pressure gradient; plot; evaluate pressure at outlet
Solution:
Basic equations
Given data
U = 15⋅
m
L = 5⋅ m
s
ρ = 1250⋅
kg
3
m
u ( x ) = U⋅ ⎛⎜ 1 −
Here
p in = 100 ⋅ kPa
⎝
x⎞
u ( 0 ) = 15
L⎠
ρ⋅ u ⋅
The x momentum becomes
dp
The pressure gradient is then
dx
s
u ( L) = 0
d
d
u = ρ⋅ aa = − p
dx
dx
ax ( x ) = u ( x ) ⋅
Hence
m
2
= −ρ⋅
∂
∂x
⋅ ⎛⎜
U
x
⎝L
L
m
s
U ⋅ ⎛⎜
2
ax ( x ) =
u( x)
⎝L
⎠
L
− 1⎞
⎠
x
⌠
p ( x ) = p in − ρ⋅ ⎮ ax ( x ) dx
⌡
Integrating momentum
− 1⎞
x
2
p ( x ) = p in −
U ⋅ ρ⋅ x ⋅ ( x − 2 ⋅ L)
0
2⋅ L
2
2
p ( L) =
Hence
ρ⋅ U
2
+ p in
p ( L) = 241 ⋅ kPa
dp/dx (kPa/m)
60
40
20
0
1
2
3
x (m)
4
5
Problem 6.10
Given:
Velocity field
Find:
Expression for pressure field; evaluate at (2,2)
[Difficulty: 2]
Solution:
Basic equations
Given data
A = 4⋅
1
B = 2⋅
s
1
x = 2⋅ m
s
y = 2⋅ m
u ( x , y ) = A⋅ x + B⋅ y
v ( x , y ) = B⋅ x − A⋅ y
Note that
∂
∂
Then
∂x
∂
∂y
v( x , y) = 0
ax ( x , y ) = u ( x , y ) ⋅
ay ( x , y ) = u ( x , y ) ⋅
∂
∂x
∂
∂x
∂x
u( x , y) + v( x , y) ⋅
v( x , y) + v( x , y) ⋅
∂
The momentum equation becomes
∂x
p = −ρ⋅ ax
v( x , y) −
∂
∂y
∂
∂y
∂
∂y
∂
∂y
(
x
y
0
0
p ( x , y ) = 80⋅ kPa
(
2
2
)
2
2
−
(
2
2
)
ax ( x , y ) = 40
m
2
s
(
2
ay ( x , y ) = y ⋅ A + B
v( x , y)
2
)
ay ( x , y ) = 40
m
2
s
p = −ρ⋅ ay
⌠
⌠
p ( x , y ) = p 0 − ρ⋅ ⎮ ax ( x , y ) dx − ρ⋅ ⎮ ay ( x , y ) dy
⌡
⌡
ρ⋅ A + B ⋅ y
p 0 = 200 ⋅ kPa
3
u( x , y) = 0
ax ( x , y ) = x ⋅ A + B
u( x , y)
Integrating
p( x , y) = p0 −
kg
m
For this flow
u( x , y) +
ρ = 1500⋅
2
2
)
ρ⋅ A + B ⋅ x
2
2
and
p = dx⋅
∂
∂x
p + dy⋅
∂
∂y
p
Problem 6.9
[Difficulty: 2]
Problem 6.8
[Difficulty: 3]
Given:
Velocity field
Find:
Expressions for velocity and acceleration along wall; plot; verify vertical components are zero; plot pressure
gradient
Solution:
3
m
q = 2⋅
The given data is
u=
s
h = 1⋅ m
m
ρ = 1000⋅
kg
3
m
q⋅ x
2 ⋅ π⎡⎣x + ( y − h )
2
2⎤
+
⎦
q⋅ x
2 ⋅ π⎡⎣x + ( y + h )
2
v=
2⎤
⎦
q⋅ ( y − h)
2 ⋅ π⎡⎣x + ( y − h )
2
2⎤
+
⎦
q⋅ ( y + h)
2 ⋅ π⎡⎣x + ( y + h )
2
2⎤
⎦
The governing equation for acceleration is
For steady, 2D flow this reduces to (after considerable math!)
x - component
y - component
u=
(
2
π⋅ x + h
∂x
u + v⋅
∂
∂y
2
u =−
⎡
(2
q ⋅ x⋅ ⎣ x + y
)
2
2
(2
− h ⋅ h − 4⋅ y
2
2⎤
)⎦
2
⎡⎣x2 + ( y + h ) 2⎤⎦ ⋅ ⎡⎣x 2 + ( y − h) 2⎤⎦ ⋅ π2
2
2 ⎡ 2
2
2 2
2⎤
− h ⋅ h + 4⋅ x ⎦
q ⋅ y⋅ ⎣ x + y
∂
∂
ay = u ⋅ v + v ⋅ v = −
2
2
∂x
∂y
2 2
2
2
2
π ⋅ ⎡⎣x + ( y + h ) ⎤⎦ ⋅ ⎡⎣x + ( y − h ) ⎤⎦
(
)
2
2
)
2
(
)
y = 0⋅ m
For motion along the wall
q⋅ x
ax = u ⋅
∂
v=0
(No normal velocity)
ax = −
(2
q ⋅ x⋅ x − h
2
(2
π ⋅ x +h
)
2
)
2
3
ay = 0
(No normal acceleration)
The governing equation (assuming inviscid flow) for computing the pressure gradient is
Hence, the component of pressure gradient (neglecting gravity) along the wall is
∂
∂x
p = −ρ⋅
Du
∂
Dt
∂x
(2
2
p =
ρ⋅ q ⋅ x ⋅ x − h
2
(2
π ⋅ x +h
)
2
)
2
3
The plots of velocity, acceleration, and pressure gradient are shown below, done in Excel. From the plots it is clear that the fluid
experiences an adverse pressure gradient from the origin to x = 1 m, then a negative one promoting fluid acceleration. If flow
separates, it will likely be in the region x = 0 to x = h.
q =
h =
2
1
m 3/s/m
m
0.35
∠=
1000
kg/m 3
0.30
0.00000
0.00000
0.01945
0.00973
0.00495
0.00277
0.00168
0.00109
0.00074
0.00053
0.00039
0.00
0.00
-19.45
-9.73
-4.95
-2.77
-1.68
-1.09
-0.74
-0.53
-0.39
u (m/s)
0.00
0.32
0.25
0.19
0.15
0.12
0.10
0.09
0.08
0.07
0.06
dp /dx (Pa/m)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
10
8
9
10
9
10
x (m)
Acceleration Along Wall Near A Source
0.025
0.020
a (m/s 2)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
a (m/s2)
0.015
0.010
0.005
0.000
0
1
2
3
4
5
6
7
-0.005
x (m)
Pressure Gradient Along Wall
5
dp /dx (Pa/m)
x (m) u (m/s)
Velocity Along Wall Near A Source
0
0
1
2
3
4
5
-5
-10
-15
-20
-25
x (m)
6
7
8
Problem 6.7
[Difficulty: 3]
Given:
Velocity field
Find:
Expressions for local, convective and total acceleration; evaluate at several points; evaluate pressure gradient
Solution:
A = 2⋅
The given data is
1
ω = 1⋅
s
∂
Check for incompressible flow
∂x
∂
Hence
∂x
u +
u +
1
ρ = 2⋅
s
∂
∂y
∂
∂y
kg
u = A⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
3
v = −A⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
m
v =0
v = A⋅ sin( 2 ⋅ π⋅ ω⋅ t) − A⋅ sin( 2 ⋅ π⋅ ω⋅ t) = 0
Incompressible flow
The governing equation for acceleration is
The local acceleration is then
∂
x - component
∂t
∂
y - component
∂t
u = 2 ⋅ π⋅ A⋅ ω⋅ x ⋅ cos( 2 ⋅ π⋅ ω⋅ t)
v = −2 ⋅ π⋅ A⋅ ω⋅ y ⋅ cos( 2 ⋅ π⋅ ω⋅ t)
For the present steady, 2D flow, the convective acceleration is
x - component
u⋅
y - component
u⋅
∂
∂x
∂
∂x
u + v⋅
v + v⋅
The total acceleration is then
∂
2
∂y
∂
u = A⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ⋅ ( A⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) + ( −A⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) ⋅ 0 = A ⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
2
∂y
2
v = A⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ⋅ 0 + ( −A⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) ⋅ ( −A⋅ sin( 2 ⋅ π⋅ ω⋅ t) ) = A ⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
x - component
y - component
∂
∂t
∂
∂t
u + u⋅
v + u⋅
∂
∂x
∂
∂x
u + v⋅
v + v⋅
∂
∂y
∂
∂y
2
2
u = 2 ⋅ π⋅ A⋅ ω⋅ x ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
2
2
v = −2 ⋅ π⋅ A⋅ ω⋅ y ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
2
Evaluating at point (1,1) at
t = 0⋅ s
12.6⋅
Local
m
m
−12.6⋅
and
2
2
s
12.6⋅
Total
m
m
12.6⋅
and
2
12.6⋅
and
2
s
t = 1⋅ s
12.6⋅
Local
12.6⋅
Convective
2
0⋅
m
0⋅
and
2
s
∂y
Evaluated at (1,1) and time
2
m
2
m
−12.6⋅
and
2
m
Convective
2
s
m
−12.6⋅
and
2
0⋅
m
0⋅
and
2
s
m
2
s
p = −ρ⋅
Du
∂
Dt
∂x
p = −ρ⋅
Dv
∂
Dt
∂x
(
)
2
2
p = −ρ⋅ −2 ⋅ π⋅ A⋅ ω⋅ y ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ y ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
t = 0⋅ s
x comp.
t = 0.5⋅ s
x comp.
t = 1⋅ s
2
p = −ρ⋅ 2 ⋅ π⋅ A⋅ ω⋅ x ⋅ cos( 2 ⋅ π⋅ ω⋅ t) + A ⋅ x ⋅ sin( 2 ⋅ π⋅ ω⋅ t)
(
x comp.
−25.1⋅
25.1⋅
Pa
y comp.
m
Pa
y comp.
m
−25.1⋅
Pa
m
m
2
s
Hence, the components of pressure gradient (neglecting gravity) are
∂
m
s
The governing equation (assuming inviscid flow) for computing the pressure gradient is
∂x
2
s
s
s
∂
m
2
m
s
Total
2
s
s
m
−12.6⋅
0⋅
and
s
s
Total
m
m
−12.6⋅
and
2
−12.6⋅
Local
0⋅
s
s
t = 0.5⋅ s
Convective
y comp.
25.1⋅
)
Pa
m
−25.1⋅
25.1⋅
2
Pa
m
Pa
m
Problem 6.6
[Difficulty: 2]
Given:
Velocity field
Find:
Simplest y component of velocity; Acceleration of particle and pressure gradient at (2,1); pressure on x axis
Solution:
Basic equations
For this flow
u ( x , y ) = A⋅ x
Hence
v ( x , y ) = −A⋅ y
For acceleration
ax = u ⋅
∂
∂x
ay = u ⋅
⌠
⌠
⎮ ∂
so
v ( x, y ) = −⎮
u dy = −⎮ A dy = −A ⋅ y + c
u +
v =0
⌡
∂x
∂y
⎮ ∂x
⌡
is the simplest y component of velocity
∂
u + v⋅
∂
∂x
∂
u = A ⋅ x⋅
∂y
v + v⋅
∂
∂y
ax =
a =
⎛ 2 ⎞ × 2⋅ m
⎜
⎝ s⎠
2
ax + ay
∂
∂x
v = A⋅ x ⋅
2
Hence at (2,1)
∂
2
( A ⋅ x) + ( −A ⋅ y ) ⋅
∂
∂x
∂
2
∂y
( −A⋅ y ) + ( −A⋅ y ) ⋅
2
( A ⋅ x) = A ⋅ x
∂
∂y
ax = A ⋅ x
2
ay = A ⋅ y
( −A⋅ y )
2
⎛ 2 ⎞ × 1⋅ m
⎜
⎝ s⎠
⎛ ay ⎞
θ = atan⎜
⎝ ax ⎠
ay =
ax = 8
m
ay = 4
2
s
a = 8.94
m
θ = 26.6⋅ deg
2
s
2
m
N⋅ s
kg
× 8⋅ ×
p = ρ⋅ g x − ρ⋅ ax = −1.50⋅
3
2
kg⋅ m
∂x
s
m
∂
∂x
2
m
N⋅ s
kg
× 4⋅ ×
p = ρ⋅ g y − ρ⋅ ay = −1.50⋅
3
2
kg⋅ m
∂y
s
m
∂
∂
∂z
kg
p = ρ⋅ g z − ρ⋅ az = 1.50 ×
For the pressure on the x axis
1
2 2
p ( x ) = p 0 − ⋅ ρ⋅ A ⋅ x
2
3
× ( −9.81) ⋅
dp =
∂x
p
p ( x ) = 190 ⋅ kPa −
⌠
p − p0 = ⎮
⌡
x
0
1
2
⋅ 1.5⋅
2
s
m
∂
m
kg
3
m
(
∂y
N⋅ s
∂
kg⋅ m
∂y
⌠
ρ⋅ g x − ρ⋅ ax dx = ⎮
⌡
)
x
0
2
×
∂
2
×
2
s
For the pressure gradient
∂
m
p = −6 ⋅
Pa
m
Pa
m
p = −14.7⋅
Pa
m
(−ρ⋅A2⋅x) dx = − 12 ⋅ρ⋅A2⋅x2
2
⎛ 2 ⎞ × N⋅ s × x 2
⎜
kg⋅ m
⎝ s⎠
p = −12⋅
p ( x ) = 190 −
3
1000
⋅x
2
(p in kPa, x in m)
Problem 6.5
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,1)
Solution:
Basic equations
ax = u ⋅
∂
∂x
(2
2
(2
2
u ( x , y ) = A⋅ x − y
For this flow
u + v⋅
∂
∂y
u = ⎡⎣A⋅ x − y
) − 3 ⋅ B⋅ x
v ( x , y ) = −2 ⋅ A⋅ x ⋅ y + 3 ⋅ B⋅ y
) − 3⋅B⋅x⎤⎦ ⋅∂ ⎡⎣A⋅(x2 − y2) − 3⋅B⋅x⎤⎦ + (−2⋅A⋅x⋅y + 3⋅B⋅y)⋅∂ ⎡⎣A⋅(x2 − y2) − 3⋅B⋅x⎤⎦
∂x
(
∂y
2
ax = ( 2 ⋅ A⋅ x − 3 ⋅ B) ⋅ A⋅ x − 3 ⋅ B⋅ x + A⋅ y
ay = u ⋅
∂
∂x
v + v⋅
∂
∂y
(2
v = ⎡⎣A⋅ x − y
)
2
) − 3⋅B⋅x⎤⎦ ⋅∂ (−2⋅A⋅x⋅y + 3⋅B⋅y) + (−2⋅A⋅x⋅y + 3⋅B⋅y)⋅∂ (−2⋅A⋅x⋅y + 3⋅B⋅y)
2
∂x
∂y
(2
ay = ( 3 ⋅ B⋅ y − 2 ⋅ A⋅ x ⋅ y ) ⋅ ( 3 ⋅ B − 2 ⋅ A⋅ x ) − 2 ⋅ A⋅ y ⋅ ⎡⎣A⋅ x − y
Hence at (1,1)
ax = ( 2 ⋅ 1 ⋅ 1 − 3 ⋅ 1 ) ⋅
1
s
2
ax + ay
)s
2 ft
2
× 1⋅ 1 − 3⋅ 1⋅ 1 + 1⋅ 1 ⋅
ay = ( 3 ⋅ 1 ⋅ 1 − 2 ⋅ 1 ⋅ 1 ⋅ 1 ) ⋅
a =
(
) − 3⋅B⋅x⎤⎦
2
1
× ( 3⋅ 1 − 2⋅ 1⋅ 1) ⋅
s
ft
ax = 1 ⋅
1
− 2⋅ 1⋅ 1⋅
s
s
⎛ ay ⎞
2
θ = atan⎜
(2
× ⎡⎣1 ⋅ 1 − 1
⎝ ax ⎠
) − 3⋅1⋅1⎤⎦ ⋅ fts
ay = 7 ⋅
∂x
ft
ft
θ = 81.9⋅ deg
2
s
lbf
p = ρ⋅ g x − ρ⋅ ax = −2 ⋅
slug
ft
3
× 1⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂x
2
p = −2 ⋅
ft
ft
= −0.0139⋅
psi
ft
lbf
∂
∂y
p = ρ⋅ g y − ρ⋅ ay = 2 ⋅
slug
ft
3
× ( −32.2 − 7 ) ⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂y
2
s
For the pressure gradient
∂
2
s
2
a = 7.1⋅
ft
2
p = −78.4⋅
ft
ft
= −0.544 ⋅
psi
ft
Problem 6.4
Given:
Velocity field
Find:
Pressure gradient at (1,1) at 1 s
[Difficulty: 2]
Solution:
Basic equations
Given data
A = 2⋅
1
B = 1⋅
2
s
1
x = 1⋅ m
2
y = 1⋅ m
t = 1⋅ s
ρ = 1000⋅
s
kg
3
m
u ( x , y , t) = ( −A⋅ x + B⋅ y ) ⋅ t
v ( x , y , t) = ( A⋅ y + B⋅ x ) ⋅ t
The acceleration components and values are
axt( x , y , t) =
∂
∂t
u ( x , y , t) = B⋅ y − A⋅ x
∂
∂t
m
axt( x , y , t) = −1
2
s
axc( x , y , t) = u ( x , y , t) ⋅
ayt( x , y , t) =
axt( x , y , t) = B⋅ y − A⋅ x
∂
∂x
u ( x , y , t) + v ( x , y , t) ⋅
v ( x , y , t)
∂
∂y
(
2
2
u ( x , y , t) axc( x , y , t) = t ⋅ x ⋅ A + B
2
)
axc( x , y , t) = 5
m
2
s
ayt( x , y , t) = A⋅ y + B⋅ x
m
ayt( x , y , t) = 3
2
s
ayc( x , y , t) = u ( x , y , t) ⋅
∂
∂x
v ( x , y , t) + v ( x , y , t) ⋅
∂
∂y
v( x , y , t)
2
2
2
)
ayc( x , y , t) = 5
m
2
s
2 2
ax ( x , y , t) = axt( x , y , t) + axc( x , y , t)
(
ayc( x , y , t) = t ⋅ y ⋅ A + B
2 2
ax ( x , y , t ) = x ⋅ A ⋅ t − x ⋅ A + x ⋅ B ⋅ t + y ⋅ B
ax ( x , y , t ) = 4
m
2
s
2 2
ay ( x , y , t) = ayt( x , y , t) + ayc( x , y , t)
2 2
ay ( x , y , t ) = y ⋅ A ⋅ t + y ⋅ A + y ⋅ B ⋅ t + x ⋅ B
ay ( x , y , t ) = 8
m
2
s
Hence for the pressure gradient
∂
∂x
∂
∂y
p = −ρ⋅ ax = −1000⋅
kg
3
× 4⋅
kg
3
m
2
2
×
s
m
p = −ρ⋅ ay = −1000⋅
m
× 8⋅
m
2
s
N⋅ s
∂
kg⋅ m
∂x
2
×
N⋅ s
∂
kg⋅ m
∂y
p = −4000⋅
Pa
p = −8000⋅
Pa
m
m
= −4 ⋅
kPa
= −8 ⋅
kPa
m
m
Problem 6.3
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,2)
Solution:
Basic equations
Given data
A = 1⋅
1
B = 2⋅
s
m
2
x = 1⋅ m
y = 2⋅ m
t = 5⋅ s
ρ = 999 ⋅
s
u ( x , y , t) = −A⋅ x + B⋅ t
kg
3
m
v ( x , y , t) = A⋅ y + B⋅ t
The acceleration components and values are
axt( x , y , t) =
∂
∂t
u ( x , y , t) = B
∂
∂t
m
axt( x , y , t) = 2
2
s
axc( x , y , t) = u ( x , y , t) ⋅
ayt( x , y , t) =
axt( x , y , t) = B
∂
∂x
u ( x , y , t) + v ( x , y , t) ⋅
v ( x , y , t)
ayc( x , y , t) = u ( x , y , t) ⋅
∂
∂y
2
u ( x , y , t) axc( x , y , t) = A ⋅ x − A⋅ B⋅ t
axc( x , y , t) = −9
m
2
s
ayt( x , y , t) = B
m
ayt( x , y , t) = 2
2
s
∂
∂x
v ( x , y , t) + v ( x , y , t) ⋅
ax ( x , y , t) = axt( x , y , t) + axc( x , y , t)
∂
∂y
2
ayc( x , y , t) = y ⋅ A + B⋅ t⋅ A
v( x , y , t)
ayc( x , y , t) = 12
2
s
2
ax ( x , y , t ) = x ⋅ A − B ⋅ t ⋅ A + B
ax ( x , y , t) = −7
m
2
s
ay ( x , y , t) = ayt( x , y , t) + ayc( x , y , t)
m
2
ay ( x , y , t ) = y ⋅ A + B ⋅ t ⋅ A + B
ay ( x , y , t) = 14
m
2
s
For overall acceleration
a( x , y , t ) =
2
ax ( x , y , t ) + ay ( x , y , t )
2
(x⋅A2 − B⋅t⋅A + B) + (y⋅A2 + B⋅t⋅A + B)
2
a( x , y , t ) =
2
a( x , y , t) = 15.7
m
2
s
⎛ ay ( x , y , t ) ⎞
θ = atan⎜
⎝
ax ( x , y , t )
⎠
For the pressure gradient we need
θ = −63.4⋅ deg
−ρ⋅ ax ( x , y , t) = 6.99⋅
kPa
m
−ρ⋅ ay ( x , y , t) = −13.99 ⋅
kPa
−ρ⋅ g = −9.80⋅
m
Hence for the pressure gradient
2
m
N⋅ s
kg
× 7⋅ ×
p = ρ⋅ g x − ρ⋅ ax = 999 ⋅
3
2
kg⋅ m
∂x
s
m
∂
∂
∂x
2
m
N⋅ s
kg
× ( −9.81 − 14) ⋅ ×
p = −ρ⋅ g y − ρ⋅ ay = 999 ⋅
3
2
kg⋅ m
∂y
s
m
∂
∂
∂y
p = 6990⋅
Pa
m
p = −23800 ⋅
= 6.99⋅
Pa
m
kPa
m
= −23.8⋅
kPa
m
kPa
m
Problem 6.2
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (2,2)
Solution:
Basic equations
Given data
For this flow
A = 1⋅
B = 3⋅
s
1
s
x = 2⋅ m
y = 2⋅ m
∂
∂x
ay = u ⋅
u + v⋅
∂
∂x
∂y
v + v⋅
ax = ( 1 + 9)
a =
∂
2
1
s
∂
∂y
kg
3
v ( x , y ) = B⋅ x − A⋅ y
u = ( A ⋅ x + B⋅ y ) ⋅
∂
∂x
v = ( A ⋅ x + B⋅ y ) ⋅
( A ⋅ x + B⋅ y ) + ( B⋅ x − A ⋅ y ) ⋅
∂
∂x
∂
∂y
( B⋅ x − A ⋅ y ) + ( B⋅ x − A ⋅ y ) ⋅
m
× 2⋅ m
ax = 20
2
θ = atan ⎜
ax + ay
ρ = 999 ⋅
m
u ( x , y ) = A⋅ x + B⋅ y
ax = u ⋅
Hence at (2,2)
1
ay = ( 1 + 9)
s
⎛ ay ⎞
a = 28.28
⎝ ax ⎠
∂
∂y
1
s
(
2
2
)
(
2
2
( A ⋅ x + B⋅ y )
ax = A + B ⋅ x
( B⋅ x − A ⋅ y )
ay = A + B ⋅ y
× 2⋅ m
ay = 20
m
)
m
s
θ = 45⋅ deg
s
For the pressure gradient
2
m
N⋅s
kg
× 20⋅
×
p = ρ⋅ g x − ρ⋅ ax = −999⋅
3
2
kg⋅ m
∂x
s
m
∂
∂
∂x
2
m
N⋅s
kg
× ( −9.81 − 20) ⋅
×
p = −ρ⋅ g y − ρ⋅ ay = 999⋅
3
2
kg⋅ m
∂y
s
m
∂
∂
∂y
p = −20000⋅
Pa
p = −29800⋅
Pa
m
m
= −20.0⋅
kPa
= −29.8⋅
kPa
m
m
Problem 6.1
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,1)
Solution:
NOTE: Units of B are s-1 not ft-1s-1
Basic equations
(2
u ( x , y ) = A⋅ y − x
For this flow
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
) − B⋅ x
2
v ( x , y ) = 2 ⋅ A⋅ x ⋅ y + B⋅ y
(2
u = ⎡⎣A⋅ y − x
(
) − B⋅x⎤⎦ ⋅∂ ⎡⎣A⋅(y2 − x2) − B⋅x⎤⎦ + (2⋅A⋅x⋅y + B⋅y)⋅∂ ⎡⎣A⋅(y2 − x2) − B⋅x⎤⎦
2
∂x
2
ax = ( B + 2 ⋅ A⋅ x ) ⋅ A⋅ x + B⋅ x + A⋅ y
ay = u ⋅
∂
∂x
v + v⋅
∂
∂y
(2
v = ⎡⎣A⋅ y − x
∂y
)
2
) − B⋅x⎤⎦ ⋅∂ (2⋅A⋅x⋅y + B⋅y) + (2⋅A⋅x⋅y + B⋅y)⋅∂ (2⋅A⋅x⋅y + B⋅y)
2
∂x
∂y
(2
ay = ( B + 2 ⋅ A⋅ x ) ⋅ ( B⋅ y + 2 ⋅ A⋅ x ⋅ y ) − 2 ⋅ A⋅ y ⋅ ⎡⎣B⋅ x + A⋅ x − y
ax = ( 1 + 2 ⋅ 1 ⋅ 1 ) ⋅
Hence at (1,1)
ay = ( 1 + 2 ⋅ 1 ⋅ 1 ) ⋅
a =
2
ax + ay
1
s
1
s
(
)s
)⎦
2⎤
2 ft
2
× 1⋅ 1 + 1⋅ 1 + 1⋅ 1 ⋅
× ( 1⋅ 1 + 2⋅ 1⋅ 1⋅ 1) ⋅
2
ft
s
ft
ax = 9 ⋅
− 2⋅ 1⋅ 1⋅
1
s
(2
× ⎡⎣1 ⋅ 1 + 1 ⋅ 1 − 1
)⎦ ⋅ s
2 ⎤ ft
⎛ ay ⎞
θ = atan⎜
ay = 7 ⋅
2
s
a = 11.4⋅
⎝ ax ⎠
2
s
ft
ft
2
θ = 37.9⋅ deg
s
For the pressure gradient
lbf
∂
∂x
p = ρ⋅ g x − ρ⋅ ax = −2 ⋅
slug
ft
3
× 9⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂x
2
p = −18⋅
ft
= −0.125 ⋅
ft
psi
ft
lbf
∂
∂y
p = ρ⋅ g y − ρ⋅ ay = 2 ⋅
slug
ft
3
× ( −32.2 − 7 ) ⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂y
2
p = −78.4⋅
ft
ft
= −0.544 ⋅
psi
ft
Problem 5.107
[Difficulty: 3]
du
2
k U u
dt
v U u
M
dv du
dv
kv 2
dt
dv
k 2
v 0
dt M
M
t
k =
M =
1.000
0.02
0.3
vi2 2 v g i vi v g2 i
vi vi 1
k
2 v g i vi v g2 i 0
t
M
k
v g i 1
t v g2 i
M
vi
k
1 2
t v g i
M
lbf.s2/ft2
slug
t
Iteration
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
0.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
1.000
25.000
15.385
13.365
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
2.000
25.000
15.385
10.213
8.603
8.477
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
3.000
25.000
15.385
10.213
7.269
6.158
6.043
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
4.000
25.000
15.385
10.213
7.269
5.480
4.715
4.621
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
5.000
25.000
15.385
10.213
7.269
5.480
4.323
3.781
3.706
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
6.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
3.136
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
7.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.668
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
8.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.314
2.274
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
9.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
2.039
2.006
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
10.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.820
1.792
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
11.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.641
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
12.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.493
1.473
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
13.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.457
1.369
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
14.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.457
1.338
1.263
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
15.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.457
1.338
1.237
1.172
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
29
30
31
32
33
34
35
36
37
38
39
40
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
