1) The document discusses steady, incompressible fluid flow through pipes and channels. It covers topics like shear stress distribution in pipes and laminar flow velocity profiles. 2) Equations are derived relating shear stress at the wall to pressure drop for both the entire cross-section and a definite length. The Hagen-Poiseuille equation is also presented. 3) Details are provided on velocity profiles for fully developed laminar flow in pipes, including the parabolic profile and equations for mean flow velocity. Constants are derived for the laminar flow profile equations.

1. Introduction
The Chapter deals with the steady flow of incompressible fluids
through closed pipes and channels.
Flow of incompressible fluids in pipes
Shear stress distribution in pipes
Flow
rw
r
y
dL
τ
-(p + dp)
p
τ

2. We have
g
w
b
b
a
a
a
a
b
b F
F
S
p
S
p
V
V
m
F 





 )
( 


here 0
, 

 g
a
b
a
b F
and
V
V


Substituting the values we have



 )
2
(
)
(
0 2
2
dL
r
dp
p
r
r
p
F 





0
2 
 
dL
rdp
Dividing by π r we have
𝑑𝑝
𝑑𝐿
= −
2𝜏
𝑟

3. 𝑑𝑝
𝑑𝐿
= −
2𝜏𝑤
𝑟𝑤
For entire cross section
From the two equations we have
r
rw
w 


rw
r
Shear stress, τ
τw
C of pipe

4. Relation between skin friction and wall shear
pa > pb Δps= pa – pb pb = pa - Δps
fs
c
b
b
b
c
b
P
c
a
a
a
c
a
h
g
V
Z
g
g
p
W
g
V
Z
g
g
p







2
2
2
2





fs
s
a
a
h
p
p
p






0
fs
s
h
p




5. w
w
r
dL
dp 
2

w
w
s
r
L
p 
2


L
D
L
r
p
h w
w
w
s
fs





4
2




Δ𝑝𝑠 =
2𝜏𝑤
𝑟𝑤
𝐿
For a definite length L, dp/dL becomes

6. 2
4
2 2
V
D
L
f
L
r
p
h
w
w
fs
s







2
2 V
L
D
p
f
s



Δ𝑝𝑠
𝐿
=
2𝑓𝜌𝑉2
𝐷
Relations between skin friction parameters
2
2
2
2
/ V
V
f w
w






Friction factor

7. Laminar flow in pipes
Velocity distribution for fully developed laminar flow for
Newtonian fluids
dr
r
dS 
2

dr
du /

 

dr
du
r
r w
w
/
/

 

r
r
dr
du
w
w




r
rw
w 



8. rdr
r
du
r
r
w
w
u
w

 



0
)
(
2
2
2
r
r
r
u w
w
w






2
max
w
wr
u 
2
max
1 









w
r
r
u
u
𝑢 = −
𝜏𝑤
𝑟𝑤𝜇
[
1
2
𝑟2
− 𝑟𝑤
2
]

9. udS
S
V
S


1
)
2
(
)
(
2
1 2
2
0
2
dr
r
r
r
r
r
V w
w
w
r
w
w



 








 
  rdr
r
r
r
V w
r
w
w
w
)
( 2
2
0
3

 










4
4
0
0
4
2
4
2
4
3
4
4
3
0
4
2
2
3
w
w
w
w
w
w
w
w
w
r
w
w
w r
r
r
r
r
r
r
r
r
r
V
w
































5
.
0
2
1
2
4
max







w
w
w
w
r
r
u
V
3
4
2 
 
 and

10. 𝛽 =
1
𝜋 𝑟𝑤
2
0
𝑟𝑤 𝜏𝑤
2𝑟𝑤𝜇 (𝑟𝑤
2 − 𝑟2)
𝜏𝑤𝑟𝑤
4𝜇
2
(2𝜋 𝑟 𝑑𝑟)
dS
V
u
S S
2
1






 

𝛽 =
1
𝑟𝑤
2
0
𝑟𝑤
8
𝑟𝑤
4
𝑟𝑤
2 − 𝑟2 2 𝑟 𝑑𝑟
𝛽 =
8
𝑟𝑤
6
0
𝑟𝑤
𝑟𝑤
2 − 𝑟2 2 𝑟 𝑑𝑟
𝛽 =
8
𝑟𝑤
6
0
𝑟𝑤
𝑟𝑤
4 + 𝑟4 − 2𝑟𝑤
2𝑟2 𝑟 𝑑𝑟

11. 𝛽 =
8
𝑟𝑤
6
𝑟𝑤
4
𝑟2
2
+
𝑟6
6
−
2𝑟𝑤
2
𝑟4
4 0
𝑟𝑤
=
8
𝑟𝑤
6
𝑟𝑤
6
2
+
𝑟𝑤
6
6
−
𝑟𝑤
6
2
− 0 − 0 + 0
𝛽 =
8
𝑟𝑤
6
0
𝑟𝑤
𝑟𝑤
4
𝑟 + 𝑟5
− 2𝑟𝑤
2
𝑟3
𝑑𝑟
𝛽 =
8
𝑟𝑤
6
𝑟𝑤
6
6
=
4
3

12. 𝛽 =
8
𝑟𝑤
6
0
𝑟𝑤
𝑟𝑤
2
− 𝑟2 2
𝑟 𝑑𝑟
𝛽 =
8
𝑟𝑤
6
𝑟𝑤
2
0
𝑦2
𝑟(−
1
2𝑟
)𝑑𝑦
𝑟𝑤
2 − 𝑟2 = 𝑦 −2𝑟 𝑑𝑟 = 𝑑𝑦 𝑑𝑟 = −
1
2𝑟
𝑑𝑦
At, r = 0, y = 𝑟𝑤
2
and at r = 𝑟𝑤 , y = 0
𝛽 = −
4
𝑟𝑤
6
𝑟𝑤
2
0
𝑦2 𝑑𝑦

13. 𝛽 = −
4
𝑟𝑤
6
𝑦3
3 𝑟𝑤
2
0
= −
4
𝑟𝑤
6 0 −
𝑟𝑤
6
3
=
𝛽 = −
4
𝑟𝑤
6 −
𝑟𝑤
6
3
=
4
3

14. 𝛼 =
1
𝜋 𝑟𝑤
2
0
𝑟𝑤 𝜏𝑤
2𝑟𝑤𝜇 (𝑟𝑤
2 − 𝑟2)
𝜏𝑤𝑟𝑤
4𝜇
3
(2𝜋 𝑟 𝑑𝑟)
𝛼 =
1
𝑆
𝑆
𝑢
𝑉
3
𝑑𝑆
𝛼 =
1
𝑟𝑤
2
0
𝑟𝑤
16
𝑟𝑤
6
𝑟𝑤
2 − 𝑟2 3 𝑟 𝑑𝑟
𝛼 =
16
𝑟𝑤
8
0
𝑟𝑤
𝑟𝑤
2 − 𝑟2 3 𝑟 𝑑𝑟
𝑟𝑤
2 − 𝑟2 = 𝑦 −2𝑟 𝑑𝑟 = 𝑑𝑦 𝑑𝑟 = −
1
2𝑟
𝑑𝑦

15. 𝛼 =
16
𝑟𝑤
8
𝑟𝑤
2
0
𝑦3
𝑟(−
1
2𝑟
)𝑑𝑦
At, r = 0, y = 𝑟𝑤
2 and at r = 𝑟𝑤 , y = 0
𝛼 = −
8
𝑟𝑤
8
𝑟𝑤
2
0
𝑦3 𝑑𝑦
𝛼 = −
8
𝑟𝑤
8
𝑦4
4 𝑟𝑤
2
0
= −
8
𝑟𝑤
8 0 −
𝑟𝑤
8
4
=
𝛼 = −
8
𝑟𝑤
8 −
𝑟𝑤
8
4
= 2

16. L
r
p
h
w
w
fs



2


 w
w
s
L
r
p



2 

4
w
wr
V 


 L
D
p
L
r
p
r
L
r
p
V s
w
s
w
w
s
32
8
4
2
2
2





 2
32
D
V
L
ps



2
32
4
D
V
L
D
L
p w
s





D
V
w


8

V
D
V
D
V
V
f w





 16
8
2
2
2
2


 Re
16

f
Hagen-poiseuille equation

17. 5.1 Prove that the flow of a liquid in laminar flow between infinite
parallel plates is given by
𝑝𝑎 − 𝑝𝑏 =
12𝜇𝐿𝑉
𝑏2
b
L
2x
2𝑥𝑝𝑎 2𝑥𝑝𝑏
𝜏 L
𝜏 L
Where L = length of plate in direction of flow
b = distance between plates
Neglect end effects

18. We have
g
w
b
b
a
a
a
a
b
b F
F
S
p
S
p
V
V
m
F 





 )
( 


here 𝛽𝑏 = 𝛽𝑎 , 𝑉𝑏 = 𝑉
𝑎 , 𝑏𝑟𝑒𝑎𝑑𝑡ℎ = 1𝑚 𝑎𝑛𝑑 𝐹
𝑔 = 0
Substituting the values we have
∑𝐹 = 0 = 2𝑥𝑝𝑎 − 2𝑥𝑝𝑏 − (2𝐿)𝜏
Dividing by 2 and rearranging we have
𝑝𝑎 − 𝑝𝑏
𝐿
=
𝜏
𝑥

19. 𝜇 = −
𝜏
𝑑𝑢/𝑑𝑥
𝜇 = −
𝑥(𝑝𝑎 − 𝑝𝑏)/𝐿
𝑑𝑢/𝑑𝑥
𝑑𝑢
𝑑𝑥
= −
(𝑝𝑎 − 𝑝𝑏)
𝐿𝜇
𝑥
𝑑𝑢 = −
𝑝𝑎 − 𝑝𝑏
𝐿𝜇
𝑥 𝑑𝑥

20. 0
𝑢
𝑑𝑢 = −
𝑝𝑎 − 𝑝𝑏
𝐿𝜇
𝑏/2
𝑥
𝑥𝑑𝑥
𝑢 =
𝑝𝑎 − 𝑝𝑏
2𝐿𝜇
𝑏2
4
− 𝑥2
𝑢 = −
𝑝𝑎 − 𝑝𝑏
𝐿𝜇
[
1
2
𝑥2 −
𝑏2
4
]
udS
S
V
S


1
𝑉 =
1
𝑏/2
0
𝑏/2
𝑝𝑎 − 𝑝𝑏
2𝐿𝜇
𝑏2
4
− 𝑥2
𝑑𝑥

21. 𝑉 =
1
𝑏
𝑝𝑎 − 𝑝𝑏
𝐿𝜇
𝑏2𝑥
4
−
𝑥3
3 0
𝑏/2
=
1
𝑏
𝑝𝑎 − 𝑝𝑏
𝐿𝜇
𝑏3
4
−
𝑏3
3
− 0 + 0
𝑉 =
2
𝑏
𝑝𝑎 − 𝑝𝑏
2𝐿𝜇
0
𝑏/2
𝑏2
4
− 𝑥2 𝑑𝑥
𝑉 =
1
𝑏
𝑝𝑎 − 𝑝𝑏
𝐿𝜇
𝑏3
12
𝑝𝑎 − 𝑝𝑏 =
12𝜇𝐿𝑉
𝑏2

22. 5.3. A Newtonian fluid is in laminar flow in a rectangular channel
with a large aspect ratio . Derive the relationship between local and
maximum velocity, and determine the ratio 𝑢𝑚𝑎𝑥/𝑉.
𝑢 =
𝑝𝑎 − 𝑝𝑏
2𝐿𝜇
𝑏2
4
− 𝑥2
𝑢𝑚𝑎𝑥 =
𝑝𝑎 − 𝑝𝑏
2𝐿𝜇
𝑏2
4
𝑉 =
𝑝𝑎 − 𝑝𝑏
𝐿𝜇
𝑏2
12
𝑢𝑚𝑎𝑥
𝑉
=
12
8
= 1.5

23. 5.4. Calculate the power required per meter of width of stream to force
lubricating oil through the gap between two horizontal flat plates
under the following conditions.
Distance between the plates, 6 mm
Flow rate of oil per meter of width, 100 m3/hr
Viscosity of Oil, 25 cP
Density of Oil, 0.88 g/cm3
Length of plate, 3m
Assume that the plates are wide in comparison with the distance in
between them and the end effects are neglected.

24. Distance between the plates b = 6 mm = 0.006 m
Flow rate of oil per meter of width Q = 100 m3/hr = 0.027778 m3/s
Viscosity of Oil 𝜇 = 25 cP = 0.025 Pa-s
Density of Oil ρ = 0.88 g/cm3 = 880 kg/m3
Length of plate L = 3m
𝑝𝑎 − 𝑝𝑏 =
12𝜇𝐿𝑉
𝑏2
=
12𝑥0.025𝑥3𝑥4.6296
0.0062
= 115,740 𝑃𝑎
𝑉 =
𝑄
𝑆
=
0.027778
0.006
= 4.6296 𝑚/𝑠
𝑃𝑜𝑤𝑒𝑟, 𝑃 = 115740𝑥0.027778 = 3215.03 𝑤𝑎𝑡𝑡𝑠

25. Flow in noncircular channels
The equivalent diameter is defined as 4 times hydraulic radius
p
H
L
S
r 
Where S is cross-sectional area of channel
Lp is perimeter of channel in contact with fluid
4
4
/
2
D
D
D
rH 



D
D
r
D H
eq 


4
4
4

26. 4
4
/
4
/ 2
2
i
o
i
o
i
o
H
D
D
D
D
D
D
r









i
o
i
o
H
eq D
D
D
D
r
D 




4
4
4
4
4
2
b
b
b
rH 

b
b
r
D H
eq 


4
4
4
2
2
b
a
ba
rH 
 b
b
r
D H
eq 2
2
4
4 



27. Laminar flow for non-newtonian fluids
For fluids following the power law model, the velocity variation with
radius follows the formula
𝑢 =
𝜏𝑤
𝑟𝑤𝐾′
1/𝑛′
𝑟𝑤
1+1/𝑛′
− 𝑟
1+1/𝑛′
1 + 1/𝑛′
∆𝑝𝑠 = 2𝐾′
3𝑛′ + 1
𝑛′
𝑛′
𝑉𝑛′
𝑟𝑛′+1
𝐿
For fluids through annulus the friction factor is given by
𝑓 =
16
𝑅𝑒
𝝓𝑎

28. Turbulent flow in pipes

29. 𝑢+ ≡
𝑢
𝑢∗
𝑢+ = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡
Velocity distribution for turbulent flow
𝑢∗ ≡ 𝑉
𝑓
2
= 𝑉
𝜏𝑤
2𝜌 𝑉2 2
=
𝜏𝑤
𝜌
𝑢∗ = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑦+ ≡
𝑦𝑢∗𝜌
𝜇
=
𝑦
𝜇
𝜏𝑤𝜌 𝑦+ = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠
𝑟𝑤 = 𝑟 + 𝑦
Equations relating u+ and y+ are called universal distribution laws

30. 

 y
u
05
.
3
ln
00
.
5 
 

y
u
5
.
5
ln
5
.
2 
 

y
u
5


y
30
5 
 
y

 y
30
Laminar layer
Buffer layer
Turbulent core
𝑑𝑢
𝑑𝑦
=
𝜏
𝜇
=
𝜏𝑤
𝜇
𝑑(𝑢+𝑢∗)
𝑑(𝑦+𝜇/𝜌𝑢∗)
=
𝜏𝑤
𝜇
𝑢∗2
𝜇/𝜌
𝑑(𝑢+
)
𝑑(𝑦+)
=
𝜏𝑤
𝜇
𝑑(𝑢+)
𝑑(𝑦+)
=
1
𝑢∗2
𝜇
𝜌
𝜏𝑤
𝜇
𝑑(𝑢+
)
𝑑(𝑦+)
= 1
𝑑(𝑢+
)
𝑑(𝑦+)
=
𝜌
𝜏𝑤
𝜏𝑤
𝜌

32. 𝑢+ = 2.5 ln 𝑦+ + 5.5 
 y
30
Turbulent core
Average Velocity:
For center line above equation becomes
𝑢𝑐
+
= 2.5 ln 𝑦𝑐
+
+ 5.5







w
w
y
yu
y
u
u
u
f
V
u







*
*
*
2
𝑢𝑐
+ =
𝑢𝑚𝑎𝑥
𝑢∗
𝑦𝑐
+ =
𝑟𝑤𝑢∗
𝜗
𝑢+
= 𝑢𝑐
+
+ 2.5 ln
𝑦+
𝑦𝑐
+

33. Turbulent core
For center line above equation becomes
udS
S
V
S


1
𝑉 =
1
𝜋𝑟𝑤
2
0
𝑟𝑤
𝑢 2𝜋𝑟𝑑𝑟 =
2𝜋
𝜋𝑟𝑤
2
0
𝑟𝑤
𝑢 𝑟𝑑𝑟 =
2
𝑟𝑤
2
0
𝑟𝑤
𝑢 𝑟𝑑𝑟
𝑉 =
2
𝑟𝑤
2
𝑟𝑤
0
𝑢 𝑟𝑤 − 𝑦 −𝑑𝑦 =
2
𝑟𝑤
2
0
𝑟𝑤
𝑢 𝑟𝑤 − 𝑦 𝑑𝑦
𝑉 =
2
𝑟𝑤
2
0
𝑦𝑐
+
𝑢∗𝑢+
𝜗
𝑢∗
𝑦𝑐
+ −
𝜗
𝑢∗
𝑦+
𝜗
𝑢∗
𝑑𝑦+

34. 𝑉 =
2
𝑟𝑤
2
𝜗2
𝑢∗ 2
0
𝑦𝑐
+
𝑢∗ 𝑢𝑐
+ + 2.5 ln
𝑦+
𝑦𝑐
+ 𝑦𝑐
+ − 𝑦+ 𝑑𝑦+
𝑉 =
5
𝑟𝑤
2
𝜗2
𝑢∗
0
𝑦𝑐
+
0.4𝑢𝑐
+ + ln
𝑦+
𝑦𝑐
+ 𝑦𝑐
+ − 𝑦+ 𝑑𝑦+
𝑉 =
5
𝑟𝑤
2
𝜗2
𝑢∗
0
𝑦𝑐
+
0.4𝑢𝑐
+𝑦𝑐
+ + 𝑦𝑐
+ln
𝑦+
𝑦𝑐
+ − 0.4𝑢𝑐
+𝑦+ − 𝑦+ln
𝑦+
𝑦𝑐
+ 𝑑𝑦+
𝑉 =
5
𝑟𝑤
2
𝜗2
𝑢∗ 0
𝑦𝑐
+
0.4𝑢𝑐
+
𝑦𝑐
+
− 0.4𝑢𝑐
+
𝑦+
𝑑𝑦+
+
5
𝑟𝑤
2
𝜗2
𝑢∗ 0
𝑦𝑐
+
𝑦𝑐
+
ln
𝑦+
𝑦𝑐
+ 𝑑𝑦+
− 0
𝑦𝑐
+
𝑦+
ln
𝑦+
𝑦𝑐
+ 𝑑𝑦+

35. 𝑉 =
5
𝑟𝑤
2
𝜗2
𝑢∗ 0.4𝑢𝑐
+𝑦𝑐
+2
−
0.4𝑢𝑐
+𝑦𝑐
+2
2
+
5
𝑟𝑤
2
𝜗2
𝑢∗ 0
𝑦𝑐
+
𝑦𝑐
+ln
𝑦+
𝑦𝑐
+ 𝑑𝑦+ − 0
𝑦𝑐
+
𝑦+ln
𝑦+
𝑦𝑐
+ 𝑑𝑦+
𝑉 =
5
𝑟𝑤
2
𝜗2
𝑢∗
0.4𝑢𝑐
+𝑦𝑐
+2
2
+
5
𝑟𝑤
2
𝜗2
𝑢∗ 0
𝑦𝑐
+
𝑦𝑐
+2
ln 𝑥 𝑑𝑥 −
𝑉 =
5
𝑟𝑤
2
𝜗2
𝑢∗
0.4𝑢𝑐
+𝑦𝑐
+2
2
+
5
𝑟𝑤
2
𝜗2
𝑢∗ 𝑦𝑐
+2
−
𝑦𝑐
+2
4

36. 𝑢+ = 𝑢𝑐
+ + 2.5 ln
𝑦+
𝑦𝑐
+
𝑉 =
5
𝑟𝑤
2
𝜗2
𝑢∗
𝑦𝑐
+2 0.4𝑢𝑐
+
2
−
3
4
𝑉 =
1
𝑟𝑤
2
𝜗2
𝑢∗
𝑦𝑐
+2
𝑢𝑐
+ − 3.75
𝑉 = 𝑢∗ 1
𝑟𝑤
2
𝜗2
𝑢∗2 𝑦𝑐
+2
𝑢𝑐
+
− 3.75 = 𝑢∗ 1
𝑦𝑐
+2 𝑦𝑐
+2
𝑢𝑐
+
− 3.75
𝑉
𝑢∗
= 𝑢𝑐
+
− 3.75 =
1
𝑓/2

37. 𝑉
𝑢∗
= 𝑢𝑐
+ − 3.75 =
1
𝑓/2
𝑆𝑖𝑛𝑐𝑒 𝑢∗ ≡ 𝑉 𝑓/2
𝑢𝑐
+ 𝑓/2 − 3.75 𝑓/2 = 1
𝑢𝑐
+ 𝑓/2 = 1 + 3.75 𝑓/2
𝑢𝑚𝑎𝑥
𝑢∗
𝑓/2 = 1 + 3.75 𝑓/2 𝑆𝑖𝑛𝑐𝑒 𝑢𝑐
+ =
𝑢𝑚𝑎𝑥
𝑢∗
𝑢𝑚𝑎𝑥
𝑉
= 1 + 3.75 𝑓/2 𝑆𝑖𝑛𝑐𝑒 𝑢∗ ≡ 𝑉 𝑓/2

38. 𝑉
𝑢𝑚𝑎𝑥
=
1
1 + 3.75 𝑓/2
The Reynolds number – friction factor law for smooth tubes
𝑦𝑐
+ =
𝑟𝑤𝑢∗
𝜗
=
𝑟𝑤𝑉
𝜗
𝑓
2
=
𝐷𝑉
2𝜗
𝑓
2
= 𝑅𝑒
𝑓
8
𝑢𝑐
+ =
1
𝑓/2
+ 3.75
𝑉
𝑢∗
= 𝑢𝑐
+ − 3.75 =
1
𝑓/2
𝑢𝑐
+
= 2.5 ln 𝑦𝑐
+
+ 5.5

39. 75
.
1
8
Re
ln
5
.
2
2
/
1










f
f
6
4
10
10 
 
y
f
f
f
91
.
3
1
)
9
.
15
15
(
78
.
0
1







031
.
1
084
.
1




011
.
1
032
.
1




Von Karman equation
The kinetic energy and momentum correction factors
Re =104
f= 0.0079
Re =106
f= 0.0029
1
𝑓/2
+ 3.75 = 2.5 ln Re
𝑓
8
+ 5.5

40. 2
.
0
Re
046
.
0 

f
2
.
0
Re
023
.
0
2


f
32
.
0
Re
125
.
0
0014
.
0 

f
𝑓 = 0.026(𝑘/𝐷)0.24

41. V
L
ps


8
.
1
V
L
ps


2
V
L
ps



42. Effect of roughness

44. 𝑉 = 15 𝑓𝑡 𝑠 = 15𝑥0.3048 = 4.572 𝑚/𝑠
𝐷 = 2 𝑓𝑡 = 2𝑥0.3048 = 0.6096𝑚
𝜇 = 1.31 𝑐𝑃 = 1.31𝑥10−3 𝑃𝑎. 𝑠
ρ = 1000 𝑘𝑔/𝑚3
𝑅𝑒 =
𝜌𝐷𝑉
𝜇
=
1000𝑥0.6096𝑥4.572
1.31𝑥10−3
= 2.1𝑥106
5.8. A steel pipe 2 ft in diameter carries water at 15 ft/s. if the pipe
has a roughness of 0.0003 ft, could the capacity be increased by
inserting a smooth plastic liner that reduces the inside diameter to
1.9 ft? Calculate the change in pressure drop for the same flow and
change in capacity for a fixed pressure drop.

45. For k/d = 0.0003/2 = 0.00015
𝑓1 = 0.026(
𝑘
𝑑
)0.24= 0.00328
For smooth tube
𝑓2 = 0.046(𝑅𝑒)−0.20
= 0.0025
∆𝑝2
∆𝑝1
=
4𝑓2𝐿𝑉2
2
𝜌/2𝐷2
4𝑓1𝐿𝑉1
2
𝜌/2𝐷1
=
𝑓2𝑉2
2
/𝐷2
𝑓1𝑉1
2
/𝐷1
=
𝑓2𝑉2
2
𝐷1
𝑓1𝑉1
2
𝐷2
∆𝑝2
∆𝑝1
=
𝑓2𝐷1
5
𝑓1𝐷2
5 =
0.0025𝑥0.60965
0.00328𝑥0.57915 =
2.1046𝑥10−4
2.1362𝑥10−4 =0.985
For constant volumetric flow
𝑉2
𝑉1
=
𝐷1
2
𝐷2
2

46. For constant volumetric flow
∆𝑝2
∆𝑝1
= 1
1 =
𝑓2𝑉2
2
𝐷1
𝑓1𝑉1
2
𝐷2
𝑉2
2
𝑉1
2 =
𝑓2𝐷1
𝑓1𝐷2
=
0.00328𝑥0.5791
0.0025𝑥0.6096
= 1.2464
𝑄2
𝑄1
=
𝑉2 𝐷2
2
𝑉1 𝐷1
2
= 1.162
0.5791
0.6096
2
= 1.0486
𝑉2
𝑉1
= 1.1642

48. Drag Reduction

50. Friction loss from sudden expansion of cross section

51. 2
2
a
e
fe
V
K
h 
g
w
b
b
a
a
a
a
b
b F
F
S
p
S
p
V
V
m
F 





 )
( 


b
b
a
a
a
a
b
b S
p
S
p
V
V
m 

 )
( 


fe
a
a
b
b
b
a
h
V
V
p
p




2
2
2



1
1




b
a
b
a





52. b
a S
S 
b
a
a
b
b
p
p
V
V
S
m


 )
(

fe
a
b
b
a
a
b
b h
V
V
p
p
V
V
V 

 





2
)
(
2
2
fe
a
b
b
a
b h
V
V
V
V
V 



2
2
2
2
fe
a
b
b
a
b h
V
V
V
V
V 2
)
(
)
(
2 2
2
2





53. )
2
2
(
2 2
2
2
a
b
b
a
b
fe V
V
V
V
V
h 



2
)
( 2
b
a
fe
V
V
h


2
2
2
1
2
2
))
/
(
(












b
a
a
b
a
a
a
fe
S
S
V
S
S
V
V
h
2
1 









b
a
e
S
S
K

54. Friction loss from contraction of cross section

55. 2
2
b
c
fc
V
K
h 
𝐾𝑐 = 0.4 1 −
𝑆𝑏
𝑆𝑎

56. 2
2
a
f
ff
V
K
h 
Friction loss from different fitting and valves
Elbow 45o 0.35
90o 0.75
Elbows 0.19
Tee straight through 0.4
Used as elbow 1.0
Gate Valve Half open 4.5
Wide open 0.17
Globe valve 6.0

57. 2
4
2
V
K
K
K
D
L
f
h f
c
e
f 









2
4
2
2
2
2
2
V
K
K
K
D
L
f
V
g
Z
p
W
V
g
Z
p
f
c
e
b
b
b
b
p
a
a
a
a






















58. 5.13. A centrifugal pump takes brine from the bottom of a supply
tank and delivers it into the bottom of another tank. The brine level
in the discharge tank is 150 ft above that in the supply tank. The
line between the tanks is 600ft of 4in schedule 40 pipe . The flow
rate is 400 gal/min. In the line there are two gate valves, four
standard tees, and four ells. What is the energy cost for running this
pump for one 24-h day? The specific gravity for brine is 1.18 and
the viscosity is 1.2 cP and the energy cost is $400 per horsepower-
year on the basis of 300 d/yr. the overall efficiency of pump is 60
percent.
Given data
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒 𝐿 = 600𝑓𝑡 = 600𝑥0.3048 = 182.88𝑚
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑖𝑝𝑒 𝐷 = 4.026𝑖𝑛 = 0.10226𝑚
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑄 = 400
𝑔𝑎𝑙
𝑚𝑖𝑛
=
400
264.17𝑥60
= 0.02524𝑚3/𝑠
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑡𝑎𝑛𝑘 𝑍 = 150 𝑓𝑡 = 45.72𝑚

59. 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓 𝑓𝑟𝑜𝑚 𝑔𝑟𝑎𝑝ℎ = 0.0045
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑉 =
𝑄
𝑆
=
0.02524
0.0082126
= 3.0733𝑚/𝑠
Mass flow rate , 𝑚 = 𝑄𝜌 = 0.02524𝑥1180 = 29.7832 𝑘𝑔/𝑠
𝑅𝑒 =
𝜌𝐷𝑉
𝜇
=
1180𝑥0.10226𝑥3.0733
1.2𝑥10−3
= 3.09037 𝑥 105
𝑅𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 =
𝑘
𝐷
=
0.00015
4.026 12
= 0.00045
ℎ𝑓 = 4𝑓
𝐿
𝐷
+ 𝐾𝑒 + 𝐾𝑐 + 𝐾𝑓
𝑉2
2
c/s area 𝑜𝑓 𝑝𝑖𝑝𝑒 𝑆 = 0.0884𝑓𝑡2 = 0.0082126𝑚2

60. ℎ𝑓 = 4𝑥0.0045
182.88
0.10226
+ 1.0 + 0.4 + (2 0.17 + 4 1.0 + 0.4 + 4 0.75 )
3.07332
2
ℎ𝑓 = 32.1909 + 1.0 + 0.4 + 8.94
3.07332
2
= 200.8559 𝐽/𝑘𝑔
𝑝𝑎
𝜌
+ 𝑍𝑎𝑔 +
𝛼𝑎𝑉
𝑎
2
2
+ 𝜂𝑊
𝑝 =
𝑝𝑏
𝜌
+ 𝑍𝑏𝑔 +
𝛼𝑏𝑉𝑏
2
2
+ ℎ𝑓
𝜂𝑊
𝑝 = 45.72𝑥9.80665 +
3.07332
2
+ 200.8559
𝜂𝑊
𝑝 = 448.36 + 4.7226 + 200.8559

61. 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑑𝑎𝑦 = 𝑝𝑜𝑤𝑒𝑟 𝑥 𝑝𝑜𝑤𝑒𝑟 𝑐𝑜𝑠𝑡/𝑑𝑎𝑦𝑠
= 43.4989𝑥400/300 = $60
𝑃 = 𝑚𝑤𝑝 = 29. 7832𝑥 1089.8975 = 32460.64𝑤𝑎𝑡𝑡𝑠
𝐼𝑓 𝑝𝑢𝑚𝑝 𝑤𝑜𝑟𝑘𝑠 𝑓𝑜𝑟 300 𝑑𝑎𝑦𝑠 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 and
the cost is $400 per horse power-year
𝑤𝑝 =
653.9385
0.6
= 1089.8975 J/kg
𝑃 = 32460.64/746.24 = 43.4989 hp

62. 5.14. Cooling water for a chemical plant must be pumped from a
river 2,500 ft from the plant site. Preliminary design calls for a flow
of 600 gal/min and 6in. Steel pipe. Calculate the pressure drop and
the annual pumping cost if power costs 3 cents per kilowatt hour.
Would the use of an 8-in pipe reduce the power cost enough to
offset the increased pipe cost? Use $15/ft of length for the installed
cost of 6 in pipe and $20/ft for 8 in pipe. Annual charges are 20
percent of the installed cost.
Given data
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒 𝐿 = 2500𝑓𝑡 = 2500𝑥0.3048 = 762𝑚
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑖𝑝𝑒 𝐷 = 6 𝑖𝑛. = 6𝑥0.0254 = 0.1524𝑚
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑄 = 600
𝑔𝑎𝑙
𝑚𝑖𝑛
=
600
264.17𝑥60
= 0.03785𝑚3/𝑠
c/s area 𝑜𝑓 𝑝𝑖𝑝𝑒 𝑆 =
𝜋
4
𝐷2 =
𝜋
4
0.15242 = 0.01824𝑚2

63. 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓 𝑓𝑟𝑜𝑚 𝑔𝑟𝑎𝑝ℎ = 0.0042
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑉 =
𝑄
𝑆
=
0.03785
0.01824
= 2.0751𝑚/𝑠
Mass flow rate , 𝑚 = 𝑄𝜌 = 0.03785𝑥1000 = 37.85 𝑘𝑔/𝑠
𝑅𝑒, =
𝜌𝐷𝑉
𝜇
=
1000𝑥0.1524𝑥2.0751
1𝑥10−3
= 3.1625 𝑥 105
𝑅𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 =
𝑘
𝐷
=
0.00015
6 12
= 0.0003
∆𝑝𝑠 =
4𝑓𝑙𝑉2𝜌
2𝐷
=
4𝑥0.0042𝑥762𝑥2.07512𝑥1000
2𝑥0.1524
= 180853.68 𝑃𝑎
𝑝𝑎
𝜌
+ 𝑍𝑎𝑔 +
𝛼𝑎𝑉
𝑎
2
2
+ 𝜂𝑊
𝑝 =
𝑝𝑏
𝜌
+ 𝑍𝑏𝑔 +
𝛼𝑏𝑉𝑏
2
2
+ ℎ𝑓

64. 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 = 𝑝𝑜𝑤𝑒𝑟 𝑥 𝑡𝑖𝑚𝑒 𝑥 𝑝𝑜𝑤𝑒𝑟 𝑐𝑜𝑠𝑡
= 8.55664𝑥7200𝑥0.03 = $1848.23
𝜌𝑤𝑝 =
∆𝑝𝑆
𝜌
=
180853.68
1000
= 180.85368 J/kg
𝑃 = 𝑚𝑤𝑝 = 37.85 𝑥 226.06713 = 8556.64𝑤𝑎𝑡𝑡𝑠 = 8.55664 𝑘𝑊
𝐼𝑓 𝑝𝑢𝑚𝑝 𝑤𝑜𝑟𝑘𝑠 𝑓𝑜𝑟 300 𝑑𝑎𝑦𝑠 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 then it means it works
for 300x24 = 7200 hrs
𝑤𝑝 =
180.85368
0.8
= 226.06718 J/kg
Pipe cost Annual Cost Operating cost Total annual cost
15x2500 = 37500 0.2x37500=7500 1848.23 9348.23
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑖𝑝𝑒 𝐷 = 8 𝑖𝑛. = 8𝑥0.0254 = 0.2032𝑚
c/s area 𝑜𝑓 𝑝𝑖𝑝𝑒 𝑆 =
𝜋
4
𝐷2 =
𝜋
4
0.20322 = 0.03243𝑚2

65. 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓 𝑓𝑟𝑜𝑚 𝑔𝑟𝑎𝑝ℎ = 0.0042
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑉 =
𝑄
𝑆
=
0.03785
0.03243
= 1.1671𝑚/𝑠
Mass flow rate , 𝑚 = 𝑄𝜌 = 0.03785𝑥1000 = 37.85 𝑘𝑔/𝑠
𝑅𝑒, =
𝜌𝐷𝑉
𝜇
=
1000𝑥0.2032𝑥1.1671
1𝑥10−3
= 2.3715 𝑥 105
𝑅𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 =
𝑘
𝐷
=
0.00015
8 12
= 0.000225
∆𝑝𝑠 =
4𝑓𝑙𝑉2𝜌
2𝐷
=
4𝑥0.0042𝑥762𝑥1.16712𝑥1000
2𝑥0.2032
= 42906 𝑃𝑎
𝑝𝑎
𝜌
+ 𝑍𝑎𝑔 +
𝛼𝑎𝑉
𝑎
2
2
+ 𝜂𝑊
𝑝 =
𝑝𝑏
𝜌
+ 𝑍𝑏𝑔 +
𝛼𝑏𝑉𝑏
2
2
+ ℎ𝑓

66. 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 = 𝑝𝑜𝑤𝑒𝑟 𝑥 𝑡𝑖𝑚𝑒 𝑥 𝑝𝑜𝑤𝑒𝑟 𝑐𝑜𝑠𝑡
= 2.02842𝑥7200𝑥0.03 = $438.14
𝜌𝑤𝑝 =
∆𝑝𝑆
𝜌
=
42906
1000
= 42.906 J/kg
𝑃 = 𝑚𝑤𝑝 = 37.85 𝑥 53.6336 = 2028.42𝑤𝑎𝑡𝑡𝑠 = 2.02842 𝑘𝑊
𝐼𝑓 𝑝𝑢𝑚𝑝 𝑤𝑜𝑟𝑘𝑠 𝑓𝑜𝑟 300 𝑑𝑎𝑦𝑠 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 then it means it works
for 300x24 = 7200 hrs
𝑤𝑝 =
42.906
0.8
= 53.6336 J/kg
Pipe cost Annual Cost Operating cost Total annual cost
20x2500 = 50000 0.2x50000=10000 438.14 10438.14

67. Pipe
diameter
Pipe cost Annual Cost Operating
cost
Total annual
cost
6 in. 37500 7500 1848.23 9348.23
8 in. 50000 10000 438.14 10438.14
6 in. 37500 3750 1848.23 5598.23
8 in. 50000 5000 438.14 5438.14