Philadelphia University
Faculty Engineering & Technology
Mechanical Engineering Department
Fluid Mechanics Lab.
Experiment Title: Losses in pipes Straight & Expansion-Contraction
(exp#5)
Student Name: Mohammed jehad Abdullah alia
Supervisor Name: Eng .Esraa AL-Hyasat
 Objectives:
The object of this experiment To make the student acquainted with fluid flows
through pipes. And To study and recognize the pressure drop along straight
pipes/expanded-contracted cross sections . One of the most common problems
in fluid mechanics is the estimation of pressure loss. It is the objective of this
experiment to enable pressure loss measurements to be made on several small
bore pipe circuit components such as pipe bends valves and sudden changes in
area of flow .
 Theory:
A Straight Pipe 13.7 mm Bore B
90o Sharp Bend
C Proprietary 90o Elbow D
Gate Valve
E Sudden Enlargement - 13.7 mm / 26.4 mm
F Sudden Contraction - 26.4 mm / 13.7 mm
G Smooth 90o Bend 52 mm Radius
B
C
H Smooth 90o Bend 102 mm Radius J
Smooth 90o Bend 152 mm Radius K
Globe Valve
H
G
L Straight Pipe 26.4 mm Bore.
J
D
K
E
L
F
A
P.V.C. Manometer Tubes
Figure 1. Diagrammatic Arrangement of Apparatus Apparatus
The state of a fluid being transported through pipes is the most common case
which encounters engineers .Such flow is found in sanitary systems to oil lines
and even within machinery in hydraulic and pneumatic circuits that is used to
actuate and control the machines.
As in contacting solid bodies where relative motion creates friction forces,
there is also friction between flowing fluid particles and its container, and
among the fluid particles themselves.
This friction is expressed in terms of fluid layers shear stress, which is a
function of fluid viscosity (µ) and fluid velocity gradient
Figure 2 .Fluid Layer Shear Stress.
 Major Losses (Straight Pipe)
Major losses are that pressure losses that are created by fluid flow
through constant section and straight pipes. Looking at figure 3, if we
took any element of the fluid and drew its free body diagram, we can see
that the pressure decreases along the pipe due to frictional forces
Figure 3 .Fluid Element Force Analysis
This friction in straight pipes is expressed by a coefficient called friction
coefficient (f).This coefficient depends on pipe surface smoothness and
Reynolds number of the flow. It’s found empirically and represented in a
graph called Moody Diagram Our job is to find this coefficient for two
different pipes at different flow rates using these equations
𝒇=
𝟐𝒈∆𝒉
𝒗𝟐 (𝑳⁄𝑫)
𝑹𝒆 =
𝝆𝑽𝑫
𝝁
𝒇 :
Friction factor
g :
acceleration due to gravity, (9.81 m/𝑚2 )
D :
Diameter of pipe, (m)
V :
mean velocity, (m/s)
L :
Length of pipe, (m)
∆𝒉 ∶ Average height (m)
Re :
Reynolds Number
𝝆 ∶ density, (kg/𝑚3 )
𝝁 ∶ dynamic viscosity (N. s/𝑚2 )
 Sudden Expanded-Contracted Pipe loss
The changes in cross-section available on the experimental unit take the
form of discontinuous expansion or constriction. For a continuous change
of cross-section, the coefficient of resistance values can be taken from
special diagrams. For a discontinuous change in cross-section, the
coefficient of resistance can be derived from Bernoulli’s equation and the
principle of linear momentum.
A1
V1
A2
V2
Figure 4 . A Sudden Expansion.
A1
V1
A2
V2
Figure 5 . A Sudden Contraction
𝑽𝟐
∆𝒉 = 𝑲𝒕𝒉
𝟐𝒈
𝑲𝒆𝒙𝒑
𝒅𝟐𝟐
= [ 𝟐 − 𝟏]
𝒅𝟏
𝑲𝒆𝒙𝒑 :
Loss Coefficient experiment
𝑲𝒕𝒉 :
Loss Coefficient theoretical
𝟐
 Collected Data & Calculations & Results :
Readings:
Straight Pipe
L = 800 mm
D = 17 mm
(Q)L
T
ℎ1 )cm(
ℎ2 )cm(
5
85
80
95
v(m/s)
Re
Q (𝑚3 /s)
0.23
3.91𝑥106
5.26𝑥10−5
fth
0.39
fex(Moody)
0.045
e%
88.46%
Expansion
L1 = 65 mm, L2 = 30 mm D1 = 17 mm , D2= 28.4 mm
(Q)L
T
ℎ1 )cm(
ℎ2 )cm(
5
Q (𝑚3 /s)
4.76𝑥10−5
75
80
v(m/s)
0.46
105
Kth
-0.1237
Kex
3.207
e%
107%
Contraction
L1 = 30 mm, L2 = 65 mm
D2 = 28.4 mm , D0= 17 mm
) Q( L
TIME
ℎ1 )cm(
ℎ2 )cm(
5
87
80
97
Q (m3/s)
v(m/s)
Kth
Kex
e%
−5
0.5
−7.116
145.06%
3.207
5.15𝑥10
Straight Pipe
𝑄=
𝐴=
𝑉
𝑡
=
5𝑥10−3
95
= 5.26𝑥10−5 𝑚3 /s
𝜋 2 𝜋
𝑥𝑑 = 𝑥(0.017)2 = 2.27𝑥10−4 𝑚2
4
4
𝑄
5.26𝑥10−5
𝑣= =
= 0.23
𝐴 2.27𝑥10−4
𝑹𝒆 =
(𝟏𝟎𝟎𝟎)𝒙(𝟎. 𝟐𝟑)𝐱(𝟎. 𝟎𝟏𝟕)
= 𝟑. 𝟔𝒙𝟏𝟎𝟔
𝟏𝒙𝟏𝟎−𝟔
𝝁 = 𝟏𝒙𝟏𝟎−𝟔 𝒇𝒓𝒐𝒎 𝑻𝒂𝒃𝒍𝒆 𝑨. 𝟓 𝒇𝒐𝒓 𝒘𝒂𝒕𝒆𝒓 @𝟐𝟎𝒄
∆𝒉 = ℎ1 − ℎ2 = (85 − 80)𝑚𝑚 = 0.05 𝑚
𝑳
𝒗𝟐
∆𝒉 = 𝒇𝒙 ( ) 𝒙 ( ).
𝑫
𝟐𝒈
𝟎. 𝟎𝟖
(𝟎. 𝟐𝟑)𝟐
𝟎. 𝟎𝟓 = 𝒇𝒙 (
)𝒙(
).
𝟎. 𝟎𝟏𝟕
𝟐𝒙𝟗. 𝟖𝟏
𝒓𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒑𝒊𝒑𝒆 =
𝒇𝒕𝒉𝒆𝒐 = 𝟎. 𝟑𝟗
𝝐 𝟎. 𝟎𝟐𝟓
=
= 𝟎. 𝟎𝟏𝟒𝟕 ≈ 𝟎. 𝟎𝟏𝟓
𝒅
𝟏𝟕
𝒇𝒆𝒙𝒑 = 𝟎. 𝟎𝟒𝟓
𝒆% = |
𝟎. 𝟑𝟗 − 𝟎. 𝟎𝟒𝟓
| 𝒙𝟏𝟎𝟎 = 𝟖𝟖. 𝟒𝟔%
𝟎. 𝟑𝟗
Expansion
𝑉 5𝑥10−3
𝑄= =
= 4.76𝑥10−5 𝑚3
𝑡
105
𝐴=
𝜋
𝜋
𝑥(𝐷2 − 𝐷1 )2 = 𝑥(0.0284 − 0.017)2 = 1.02𝑥10−4 𝑚2
4
4
𝑄
4.76𝑥10−5
𝑣= =
= 0.46
𝐴 1.02𝑥10−4
𝟐
𝑲𝒆𝒙𝒑
𝟐
𝒅𝟐𝟐
𝟎. 𝟎𝟐𝟖𝟒𝟐
= ( 𝟐 − 𝟏) = (
− 𝟏) = 𝟑. 𝟐𝟎𝟕
𝟎. 𝟎𝟏𝟕𝟐
𝒅𝟏
𝒗𝟐 𝑳𝟏
𝒗𝟐
𝒗𝟐 𝑳𝟐
∆𝒉 = (𝒇
𝒙 ) + 𝑲𝒕𝒉 𝒙
+ (𝒇
𝒙 )
𝟐𝒈 𝒅𝟏
𝟐𝒈
𝟐𝒈 𝒅𝟐
𝒇=
𝟐𝒈∆𝒉
𝟐𝒙𝟗. 𝟖𝟏𝒙𝟎. 𝟎𝟓
=
= 𝟏. 𝟐𝟏
𝒗𝟐 (𝑳⁄𝑫) 𝟎. 𝟒𝟔𝟐 (𝟎. 𝟎𝟔𝟓⁄𝟎. 𝟎𝟏𝟕)
𝟎. 𝟒𝟔𝟐 𝟎. 𝟎𝟔𝟓
𝟎. 𝟒𝟔𝟐
𝟎. 𝟎𝟓 = (𝟏. 𝟐𝟏𝒙
𝒙
) + 𝑲𝒕𝒉 𝒙
𝟐𝒙𝟗. 𝟖𝟏 𝟎. 𝟎𝟏𝟕
𝟐𝒙𝟗. 𝟖𝟏
𝟎. 𝟒𝟔𝟐
𝟎. 𝟎𝟑
+ (𝟏. 𝟐𝒙
𝒙
)
𝟐𝒙𝟗. 𝟖𝟏 𝟎. 𝟎𝟐𝟖𝟒
𝑲𝒕𝒉 = -0.1237
𝒆% = |
−𝟎. 𝟏𝟐𝟑𝟕 − 𝟏. 𝟐𝟏
| 𝒙𝟏𝟎𝟎 = 𝟏𝟎𝟕. 𝟐%
−𝟎. 𝟏𝟐𝟑𝟕
Contraction
𝑉 5𝑥10−3
𝑄= =
= 5.15𝑥10−5
𝑇
97
𝐴=
𝜋
(𝐷2 − 𝐷1 )2 = 1.02𝑥10−4
4
𝑣=
𝑄
= 0.505
𝐴
∆ℎ = ℎ1 − ℎ2 = 0.87 − 0.80 = 0.07𝑚
𝑓=
∆ℎ = (𝑓
2𝑔∆ℎ
2𝑥9.81𝑥0.07
=
= 3.111
𝑣 2 (𝐿⁄𝐷) 0.52 ( 0.03 )
0.017
𝑣 2 𝐿1
𝑣2
𝑣 2 𝐿2
𝑥 ) + (K 𝑡ℎ
) + (𝑓
𝑥 ) = 0.07
2𝑥𝑔 𝐷1
2𝑥𝑔
2𝑥𝑔 𝐷2
0.52
0.03
0.52
) + (𝐾𝑡ℎ 𝑥
)
= (3.111
𝑥
2 ∗ 9.81 0.017
2𝑥9.81
0.52
0.065
) = 𝐾𝑡ℎ = −7.116
+ (3.111
𝑥
2𝑥9.81 0.0284
2
𝐾𝑒𝑥𝑝
e% = (
2
𝑑22
0.0172
= [ 2 − 1] = [
− 1] = 3.2071
0.02842
𝑑1
𝐾𝑡ℎ − 𝐾𝑒𝑥𝑝
−7.116 − 3.2071
) 𝑥100% = (
) ∗ 100% = 145.06%
𝐾𝑡ℎ
−7.116
 Questions:
1. 1. What are the factors that have an effect on friction coefficient of
pipe?
1_Roughness factor of inner surface of the pipe
2_length of the pipe
3inner diameter of the pipe
4_square velocity of the fluid
5_density of the fluid.
2. Does the expansion have a pressure loss? Why?
 Pressure drop is defined as the difference in total pressure
between two points of a fluid carrying network. A pressure drop
occurs when frictional forces, caused by the resistance to flow,
act on a fluid as it flows through the tube