This document discusses different types of gates and siphons used in open channel flow. It begins by describing the geometry of sluice gates, including definitions of terms like gate opening and vena contracta. It then shows diagrams and definitions for other gate types like radial gates and drum gates. The document also explains the concepts of siphon flow, including diagrams of a basic siphon and definitions of terms. It provides equations for calculating flow rates, volume flow rate, mass flow rate, and weight flow rate. Examples are given for calculating time of emptying/filling tanks and pipes, as well as worked examples for siphon problems.
Mechanics of fluids is extremely important in many areas of engineering and science. Examples are:
Mechanical engineering:
Pipeline projects.
Design of tanks.
Design of pumps, turbines, air-conditioning equipment.
Petroleum Engineering
Mud logging, cementing.
Chemical Engineering
Design of chemical processing equipment.
Mechanics of fluids is extremely important in many areas of engineering and science. Examples are:
Mechanical engineering:
Pipeline projects.
Design of tanks.
Design of pumps, turbines, air-conditioning equipment.
Petroleum Engineering
Mud logging, cementing.
Chemical Engineering
Design of chemical processing equipment.
Flow Equations for sluice gate.Introduces different flow equations to students which are widely utilized for the design of sluice gates connected to open channel.This tutorial will help to understand and articulate the basic flow equation utilized by designers all over the world.
Unit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flow
Flow Equations for sluice gate.Introduces different flow equations to students which are widely utilized for the design of sluice gates connected to open channel.This tutorial will help to understand and articulate the basic flow equation utilized by designers all over the world.
Unit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flowUnit 5 Open Channel flow
Hierarchical Digital Twin of a Naval Power SystemKerry Sado
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxR&R Consult
CFD analysis is incredibly effective at solving mysteries and improving the performance of complex systems!
Here's a great example: At a large natural gas-fired power plant, where they use waste heat to generate steam and energy, they were puzzled that their boiler wasn't producing as much steam as expected.
R&R and Tetra Engineering Group Inc. were asked to solve the issue with reduced steam production.
An inspection had shown that a significant amount of hot flue gas was bypassing the boiler tubes, where the heat was supposed to be transferred.
R&R Consult conducted a CFD analysis, which revealed that 6.3% of the flue gas was bypassing the boiler tubes without transferring heat. The analysis also showed that the flue gas was instead being directed along the sides of the boiler and between the modules that were supposed to capture the heat. This was the cause of the reduced performance.
Based on our results, Tetra Engineering installed covering plates to reduce the bypass flow. This improved the boiler's performance and increased electricity production.
It is always satisfying when we can help solve complex challenges like this. Do your systems also need a check-up or optimization? Give us a call!
Work done in cooperation with James Malloy and David Moelling from Tetra Engineering.
More examples of our work https://www.r-r-consult.dk/en/cases-en/
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
About
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
Technical Specifications
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
Key Features
Indigenized remote control interface card suitable for MAFI system CCR equipment. Compatible for IDM8000 CCR. Backplane mounted serial and TCP/Ethernet communication module for CCR remote access. IDM 8000 CCR remote control on serial and TCP protocol.
• Remote control: Parallel or serial interface
• Compatible with MAFI CCR system
• Copatiable with IDM8000 CCR
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
Application
• Remote control: Parallel or serial interface.
• Compatible with MAFI CCR system.
• Compatible with IDM8000 CCR.
• Compatible with Backplane mount serial communication.
• Compatible with commercial and Defence aviation CCR system.
• Remote control system for accessing CCR and allied system over serial or TCP.
• Indigenized local Support/presence in India.
• Easy in configuration using DIP switches.
5. (2)
Sluice Gate Geometry
Sluice gate of width “W”
(1)
y1
y2
HGL
EGL
yG
Sluice gates are often used to
regulate and measure the flow
rate in open channel
yG is the gate
opening
(1)
Sectional Elevation Through a Sluice Gate
6. (2)
Sluice gate of width “W”
(1)
y1
y2
HGL
EGL
yG
yG is the
gate opening
(1)
Vena-contracta
Sectional Elevation Through a Sluice Gate
Taking the bed of the channel as a reference datum
16. Flow Rates
₪ Volume flow rate: is the volume of fluid flowing past a
section per unit time.
₪ Mass flow rate: is the mass of fluid flowing past a
section per unit time.
₪ Weight flow rate: is the weight of fluid flowing past a
section per unit time.
20. Atank
H1
H2
dh
g
2
V
D
L
f
h
2
)
f
(
L
dh
A
dt
Q tank
.
dh
Q
A
dt
t H
H
tank
.
0
2
1
dh
.
h
g
2
A
D
L
f
5
.
1
A
dt
t
0
H
H
5
.
0
pipe
tank 2
1
0
Qin
g
2
V2
g
V
2
5
0
2
.
D
L
f
5
.
1
g
2
V
g
2
V
g
2
V
D
L
f
g
2
V
5
.
0
h
2
2
2
2
5
.
0
h
.
D
L
f
5
.
1
g
2
V
5
.
0
.
5
.
1
2
.
. h
D
L
f
g
A
V
A
Q pipe
pipe
Q
F, L, D
V : the velocity through the pipe
22. 5 cm
5 cm
5 cm 0.4 cm dia.
Water flows from a 10-cm diameter container that contains three holes as
shown in the figure. The diameter of each fluid stream is 0.4cm, and the
distance between holes is 5cm. If viscous effects are negligible and quasi-
steady conditions are assumed.
Determine the time at which the water stops draining from the top hole.
Compare it with the time required to stop draining from the container.
Quasi-steady flow: means that at any
moment of time, we can solve the
problem as if the flow were steady.
29. Barometer: instruments used for measuring
atmospheric pressure.
At sea level the average atmospheric pressure
is:
101.3 kPa = 760 mm of mercury
= 14.7 psi = 2117 Ib/ft2
Atmospheric Pressure
30. Absolute Pressure: is the total pressure measured
above zero (perfect vacuum),
Gage Pressure: is the pressure measured above
atmospheric pressure.
Absolute Pressure = gage pressure + atmospheric
pressure
Suppose that the tire gage measure measures 30 psi
(gage),
Absolute pressure = 30 psi + 14.7 psi = 44.7psi,
Absolute and Gage Pressure
31. 4.6 m
A
50mm dia.
25mm dia.
0.9 m
B
water
» The volume flow-rate of
water from the nozzle.
» The pressure at points “A”
and “B”.
For the siphon shown in the
figure, calculate:
Worked Example
32. 4.6 m
A
50mm dia.
25mm dia.
0.9 m
B
water
» Because p1 = p2 = patm. = 0,
the remaining terms are
2
1
2
2
1
2
2
2
g
V
g
p
Z
g
V
g
p
Z
» Applying Bernoulli’s Equation
between points “1” and “2”
s
m
Z
Z
g
V
/
5
.
9
6
.
4
0
81
.
9
2
2 2
1
2
Reference
Datum
» The volume flow rate of water, Q
s
L
V
D
V
A
Q
/
66
.
4
5
.
9
4
025
.
0
4
2
2
2
2
2
2
Worked Example
33. The continuity equation gives:
water
of
m
Z
g
p
g
p A
B
19
.
1
9
.
0
29
.
0
B
A
g
V
g
p
Z
g
V
g
p
Z
2
2
2
2
s
m
D
D
V
V
siph
Siph /
39
.
2
25
50
5
.
9
2
2
.
2
2
.
» To determine the pressure at point “A”, apply Bernoulli’s Equation
between points “1” and “A” gives,
water
of
m
g
V
g
p siph
A
29
.
0
81
.
9
2
39
.
2
2
0
2
2
» Similarly, since the siphon cross section is constant (VA =VB), the
pressure at point “B” can be determined as,
2
3
3
/
85
.
2
10
81
.
9
10
29
.
0
m
kN
2
3
3
/
67
.
11
10
81
.
9
10
19
.
1
m
kN
Worked Example
34. 1.0 m
25mm dia.
h
water
» Does changing the
elevation, h, of the siphon
centerline above the water
level in the tank vary the
volume flow-rate through the
siphon? Explain.
» What is the maximum
allowable value of h ?
A siphon is used to draw water
from a large container as
indicated in the figure.
Worked Example
35. A
7.5cm dia.
water
» Determine the flow rate
involved?
A water siphon having a
constant inside diameter of
7.5cm is arranged as shown in
the figure. If the friction losses
between A and B is 0.8 V2/2g,
where V is the velocity of flow
in the siphon.
1.0 m
1.0 m
3.0 m
B
Worked Example
37. What throat diameter is needed for a Venturi-meter in a 200cm
horizontal pipe carrying water with a discharge of 10m3/s if the
differential pressure between the throat and upstream section is to
be limited to 200kPa at this discharge? (take Cd = 0.98)
Solution
4
1
2
2
1
2
2
1
2
4 D
D
g
p
p
g
D
C
Q d
act
The basic discharge equation for a Venturi-meter is:
m
g
p
39
.
20
81
.
9
1000
10
200 3
D1 =2.0m
= 0.98
10 m3/s
………….. (1)
Worked Example
38. Eliminating Eq. (1) gives: 2
2
2
16
8
.
6400
13
D
D
Solving by trial and error: D2 = 0.80m
………….. (2)
39. Water flows through a Venturi-meter that has a 30cm throat. The
Venturi-meter is installed in a 60cm pipe. What deflection occur in a
water-mercury manometer connected between the upstream and
throat sections if the discharge of 0.57m3/s. (Cd = 0.98)
Solution
4
1
2
2
1
2
2
1
2
4 D
D
g
p
p
g
D
C
Q d
act
The basic discharge equation for a Venturi-meter is:
D1 =0.60 m
= 0.98
0.57 m3/s
………….. (1)
1
/
2
1
i
h
g
p
p
D2 =0.30 m
Worked Example
40. For the given case,
6
.
13
0625
.
0
16
1
60
.
0
30
.
0
0706
.
0
4
3
.
0
4
4
4
1
2
2
2
2
2
2
w
m
L
i
D
D
m
D
A
Substituting these values into Eq. (1) gives
0625
.
0
1
1
6
.
13
81
.
9
2
0706
.
0
98
.
0
57
.
0
h
………….. (2)
Solving for h, h (the deflection in the manometer) = 25.70cm
Worked Example
44. What is Pitot tube:
Pitot Tube is a device used for measuring the velocity
of flow at any point in a pipe or a channel. The Pitot
tube was invented by the French Engineer Henri Pitot.
The principle of Pitot tube is based on the Bernoulli's
equation, where each term can be interpreted as a
form of pressure.
45. Pitot Tube
It is used to measure flow rate at a particular point in a flowing
fluid. Such devices are frequently used as air speed indicators in
aircraft
The Pitot tube consists of a tube with a small opening at the
measuring end. This small hole faces the flowing fluid,
When the fluid contacts the Pitot tube, the fluid velocity is zero
and the pressure is maximum.
This small hole or impact opening, provide the higher pressure for
pressure measurement.
The purpose of the static ports is to sense the true static pressure
of the free airstream.
47. Note: the gray circles outlined in dotted red boxes are Pitot-static ports,
which, combined with an aircraft's Pitot tubes
48. With no flow, the pressure at
both ports is the same and the
level in each “leg” of the U-tube
is the same.
Flow Sensors are typically a Form of Pitot Tube.
49. The Pitot tube is used to find the Velocity of a fluid. The
parameters for the model include density of fluid “L”,
density of manometer liquid (i), differential head (h).
50.
51. Laminar flow: slow, organized, parallel to vessel
walls, parabolic profile.
Vortex flow: swirling, often countercurrent eddies
seen at bifurcations and after stenosis
Turbulent flow: disorganized, random, with high
velocities.
52. Types of Flow
Laminar Flow
Low velocity
Particles flow in straight
lines
No Mixing
Rare (
اندرة
) in water system
Turbulent Flow
High velocity
Particles move in irregular
paths
Mixing
Most common type of flow.
56. The weir is one of the oldest, cheapest, most straight
forward and reliable structures for the determination of
flow in channels where sufficient water depth is available.
A simple weir consists of a structure in wood, metal or
concrete placed perpendicular to the flow in a channel.
The structure has a sharp-edged opening or notch of
specific shape and dimension through which the water
can flow. Weirs are identified by shape as indicated
below.
Weir
60. Each type of weir has certain advantages under specific
conditions. Generally, for accurate readings, a standard V-
notch or a rectangular suppressed (parallel sides with no
side contractions) weir should be used. The Cipolletti and
rectangular weirs with full end contractions are especially
suited for water division.
Normally the observer has a reasonable idea of the
quantities to be measured and taking the following into
account, a choice can be made of the suitable weir for
particular circumstances:
Choice of weir
61. The maximum expected water height above the weir must be
at least 60 mm to prevent the nape from adhering to the weir
crest. Furthermore it is difficult to take accurate readings on
the measuring scale if (h) is too low.
The length of the rectangular and Cipolletti weirs must be at
least equal to three times the water height above the weir
crest.
The V-notch is the most suitable for measurements smaller
than 100 m³/h.
62. The V-notch is as accurate as the other weir types for
flows between 100 and 1 000 m³/h provided that
submergence does not occur.
The weir crest must be as high above the channel floor
as possible so that free overflow (of the nape) will take
place.
The flow depth over the weir should not exceed 600 mm.
63. When the channel feeding the water to the weir is wider than
the weir crest, the sides of the stream will narrow where it
crosses the weir. The width of the stream flowing over the
weir will be slightly narrower than crest width. The
phenomenon is known as end contraction.
To make allowance for end contractions, the overflow
width in the basic equation must be modified as follows:
L' = L – 0.1 n h (8.4)
where:
L' = effective length of weir [m]
L = measured length of weir [m]
n = number of end contractions
h = overflow depth [m].
End Contractions
64. A rectangular weir that does not span the whole channel is called a weir
with end contractions. The effective length of the weir will be less than the
actual weir length due to contraction of the flow jet caused by the
sidewalls.
where,
N = the number of contractions, and
L‘ = measured length of the crest.
L = L' – 0.1 n h
L'
65. A lined channel is provided with a lining of impervious material
on its bed and banks to prevent the seepage of water.
Lined Canal
68. Advantages of Canal Lining:
It reduces the loss of water due to seepage and hence the
duty is enhanced,
It controls the water logging and hence the bad effects of
water-logging are eliminated,
It provides smooth surface and hence the velocity of flow
can be increased.
Advantage of Canal Lining
69. Disadvantages of Canal Lining:
Higher initial investment,
Repair is costly,
Shifting of outlet is costly because it involve dismantling
and relaying of lining,
(
يلهاصتوو نةاالبط تفكيك يشمل نهل ملكف اخملرج نقل نا
)
Longer construction period,
Sophisticated construction equipment and labor is
needed. { هناك
حاجة
ىلا
معدات
بناءلا
املتطورة
العامةلو }
Dis - advantage of Canal Lining
70.
71.
72.
73.
74. Pumps in Series
Volume Flow- rate V
.
Performance Curve “one pump”
Performance Curve “two pump”
System Curve
A
B
Head
H
p
When two pumps are placed in series, the resulting pump performance curve is
obtained by adding heads at the same flow-rate. As shown in the figure, for two
identical pumps in series, both the actual head gained by the fluid and the flow-
rate increased, but neither will be doubled if the system curve remains the
same. The operating point is at “A” for one pump and moves to “B” for two
pumps in series.
For two pumps in series,
added heads
75. Pumps in Parallel
Volume Flow- rate V
.
Performance Curve “one pump”
Performance Curve “two pump”
System Curve
A
B
Head
H
p
For two identical pumps in parallel, the combined performance curve is
obtained by adding flow-rates at the same head. As shown in the figure above,
the flow-rate for the system will not be doubled with the addition of two pumps
in parallel (if the same system curve applies). However, for a relatively flat
system curve (see the shown figure), a significant increase in flow-rate can be
obtained as the operating point moves from point “A” to point “B”.
For two pumps in parallel,
added flow-rates
76.
y
D
sin
8
,
2
D
A
Area
2
,
D
P
perimeter
Wetted
4
sin
,
D
P
A
R
raduis
hydraulic
The h
Where is in radians
If the section is laid on a constant slope “So”, and the Manning coefficient “n”
3
2
3
5
3
2
3
8
2
1 sin
4
8
,
D
S
n
k
Q
discharge
The o
The above equation can be written in terms of the flow depth, y, using
2
sin
1
2
D
y
Uniform Flow in a Circular Section
77. V2 = 6m/s
V1 = 4m/s
V3
10 cm
12 cm
90o
The water jets collide and form one
homogenous jet as shown in the
figure. Determine:
» The speed, V, and direction,
, of the combined jet.
» The loss for a fluid particle
flowing from “1” to “3” and
from “2” to “3”. Gravity is
negligible.
Linear Momentum Equation
78. V2 = 6m/s
V1 = 4m/s
V3
10 cm
12 cm
90o
V3 cos
V3 sin
s
m
Q /
0314
.
0
4
1
.
0
4
3
2
1
s
m
Q /
0679
.
0
6
12
.
0
4
3
2
2
s
m
Q
Q
Q /
1
.
0
0679
.
0
0314
.
0 3
2
1
3
Applying momentum equation in x-direction gives,
2
2
3
3
3
3
2
2
)
( cos
cos
0 V
Q
V
Q
A
p
A
p
F x
Substituting p2 = p3 = patm (gage) and incompressible flow ( is constant),
….. (1)
2
2
3
3 cos V
Q
V
Q
2
2
2
3
2
3
cos
A
Q
A
Q
or
For Q3 = 0.1 m3/s , Q2 = 0.0679 m3/s
2
2
3
2
2
2
3
2
3
2
2
4
1
.
0
0679
.
0
cos
D
A
A
A
Q
Q
(2)
Linear Momentum Equation
79. Applying momentum equation in y-direction gives,
2
1
3
2
2
1
3
2
3
2
1
4
1
.
0
0314
.
0
sin
D
A
A
A
Q
Q
….. (4)
1
1
3
3
3
3
1
1
)
( sin
sin
0 V
Q
V
Q
A
p
A
p
F x
….. (3)
Similarly,
Solving Eq. (2) and Eq. (4) for ,
2
2
2
1
2
2
10
.
0
12
.
0
0679
.
0
0314
.
0
0679
.
0
0314
.
0
tan
D
D
308
.
0
10
.
0
12
.
0
0679
.
0
0314
.
0
0679
.
0
0314
.
0
tan
2
2
2
1
2
2
D
D
and
12
.
17
Linear Momentum Equation
82. 20m
o
5m 35m
20m
5m
Hydrostatic Forces “HTI Dec. 2012”
F1
2
cp )
y
(
2
CG )
y
(
CG
cp
5 m
m
77
25
20
25
62
20
y 2
CG .
.
)
(
m
14
27
62
20
200
77
25
12
62
20
200
77
25
y
3
2
cp .
)
.
(
.
)
/
.
(
.
)
(
o
1
96
.
75
5
20
tan
83. Hydrostatic Forces “HTI Dec. 2012”
N
10
525
.
24
)
5
200
(
)
2
/
5
(
81
.
9
10
A
h
g
F 6
3
1
1
1
The hydrostatic force on the vertical face of the dam, F1
It acts at a vertical distance = ⅔ 5 =3.33 m below the water surface
The hydrostatic force on the inclined face of the dam, F2
N
10
85
.
606
)
62
.
20
200
(
)
2
/
20
5
(
81
.
9
10
A
h
g
F 6
3
2
2
2
cos
F
F
2
F
F
F 2
1
2
2
2
1
t
= 165.96
86. 45o
500 mm dia-
pU.S = 1000 KN/m2
2
0
0
m
m
d
i
a
-
Linear Momentum Equation “HTI Dec. 2012)
A reducing pipe bend is shown in the figure. If the discharge is 500L/s and the
upstream pressure is1000kN/m2, find the magnitude and direction of the force
on the bend.
Linear Momentum Equation “HTI Dec. 2012”
87. 45o
500 mm dia-
P inlet = 103 KN/m2
V1
V2
p1
p2
P2 cos 45o
P2 sin 45o
C V
Linear Momentum Equation “HTI Dec. 2012”
88. Section “2”
Section “1”
P2
in KN
P1
in KN A2 in m2
D2 ( m)
A1 in m2
D1 ( m)
0.0314
0.20
0.196
0.50
???
103
The data given are:
To calculate the pressure at the outlet section P2, we use the
Bernoulli’s equation. The Bernoulli’s equation for a streamline
passing through the centerline of the pipe bend is expressed as
2
2
1
2
g
2
V
g
p
g
2
V
g
p
The pipe bend is in a
horizontal position
Linear Momentum Equation “HTI Dec. 2012”
89. or
The inlet and outlet velocities, V1 and V2, may be determined
using the continuity equation as
g
2
V
g
2
V
g
p
g
p 2
2
2
1
1
2
………… (1)
s
/
m
55
.
2
196
.
0
5
.
0
A
Q
V 1
1
s
/
m
92
.
15
0314
.
0
5
.
0
A
Q
V 2
2
and
Then, substituting the given data in Eq. (1),
water
of
m
35
.
89
92
.
15
55
.
2
81
.
9
2
1
81
.
9
10
10
10
g
p 2
2
3
3
3
2
Linear Momentum Equation “HTI Dec. 2012”
90. or
Hence, the pressure forces on the pipe bend ends are:
2
3
2 m
/
kN
53
.
876
35
.
89
81
.
9
10
35
.
89
g
p
……… (2)
P2 . A2 in KN
P1 .A1 in KN
876.530.0314 = 27.52
1030.196 = 196
The reducing pipe bend, surrounded by a suitable control
volume, is shown in the figure. The forces acting in the bend
are due to the change in direction and the reduction in
diameter.
Linear Momentum Equation “HTI Dec. 2012”
91. The momentum equations along the x- and y- directions are
)
V
45
cos
V
(
Q
45
cos
A
p
A
p
F 1
2
2
2
1
1
)
x
(
…… (3)
Substituting the given values, Eq. (3) gives
)
0
45
sin
V
(
Q
45
sin
A
p
0
F 2
2
2
)
y
(
(where all terms are taken positive with x- and y- axis)
)
55
.
2
45
cos
92
.
15
(
5
.
0
10
45
cos
10
52
.
27
10
196
F 3
3
3
)
x
(
kN
19
.
172
F )
x
(
Similarly,
Linear Momentum Equation “HTI Dec. 2012”
92. Hence
)
0
45
sin
92
.
15
(
5
.
0
10
45
sin
10
52
.
27
0
F 3
3
)
y
(
and
kN
09
.
25
F )
y
(
kN
174
09
.
25
19
.
172
F
F
F 2
2
2
)
y
(
2
)
x
(
t
'
)
x
(
)
y
(
17
8
19
.
172
09
.
25
F
F
From negative x-direction to
negative y-direction
Then the total force of the elbow on the control volume is 174 KN
Fy
Fx
Ft
Linear Momentum Equation “HTI Dec. 2012”
93. kN
174
F
F t
flange
on
t
From Newton’s third law, the force exerts on the bend is the negative
of Ft
Fx
Fy Ft
Linear Momentum Equation “HTI Dec. 2012”
94. Air
S.g. = 0.80
water
0.85 ft
1.60 ft
2.50 ft
4.0 ft
6
.
0
f
t
The figure shows a container holding
oil and water. Air at 710Ibf / ft2 below
atmospheric pressure is above the
oil. Calculate:
₪
The pressure and force at the
bottom of the container in Ibf /ft2
(gage).
B
The pressure at the bottom of the container
2
f
w
w
oil
oil
air
B
ft
/
Ib
946.13
)
5
.
2
6
.
1
8
.
0
(
2
.
32
94
.
1
710
h
g
h
g
P
P
The force at the bottom of the container f
B
B
B Ib
707
22
)
6
4
(
13
.
946
A
P
F
Hydrostatic Pressure Forces “HTI Dec. 2012)
95. o 5m 35m
20m
5m
Hydrostatic Forces “HTI Dec. 2012)
A mass concrete dam has the section shown in the figure and spans a channel
200m wide.
I) Estimate the magnitude of the resultant force,
II) Its angle to the horizontal, and
III) The point at which its line of action passes through the base line.
Hydrostatic Pressure Forces “HTI Dec. 2012)