Here are the solutions to the assignment problems:
1. Given: Pressure = 40 m of oil of specific gravity 0.6
Density of oil (ρo) = 0.6 x 1000 = 600 kg/m3
Density of mercury (ρm) = 13,600 kg/m3
Using the formula: ρ1h1 = ρ2h2
ρoh = ρmh
600 x 40 = 13,600 h
h = 40 x 13,600/600 = 68 m of mercury
2. Given pressures:
(a) 150 kPa
(b) 1800 millibars
(c) 20 m of water
(d) 1240 mm
Engineering Thermodynamics: Properties of Pure SubstancesMAYURDESAI42
Engineering thermodynamics is a branch of thermodynamics that deals with the practical application of thermodynamic principles and concepts. One of the fundamental topics in engineering thermodynamics is the properties of pure substances.
A pure substance is a material that has a fixed and constant chemical composition, regardless of its physical state. This means that a pure substance cannot be separated into two or more different substances by physical means. Examples of pure substances include water, oxygen, and carbon dioxide.
The properties of a pure substance are critical in thermodynamics because they are used to calculate important thermodynamic parameters, such as enthalpy, entropy, and internal energy. These parameters are used to understand the behavior of systems and predict their response to changes in temperature, pressure, and other conditions.
One of the key properties of a pure substance is its temperature-pressure phase diagram, which provides information about the physical state of the substance under different conditions. For example, water can exist as a solid (ice), a liquid (water), or a gas (steam) depending on the temperature and pressure conditions. This information is critical for understanding the behavior of a substance in different thermodynamic systems, such as power plants, refrigeration and air-conditioning systems, and chemical processes.
Another important property of a pure substance is its enthalpy of vaporization, which is the amount of energy required to convert a unit mass of the substance from a liquid to a gas at a constant temperature. This property is critical in many applications, such as the design of steam power plants, which use the energy stored in steam to generate electricity.
The specific heat capacity of a pure substance is another critical property. It represents the amount of energy required to raise the temperature of a unit mass of the substance by a unit temperature. This property is used to calculate the heat transfer in thermodynamic systems, such as refrigeration and air-conditioning systems.
Another important property of a pure substance is its thermal conductivity, which represents its ability to transfer heat. This property is critical in the design of heat exchangers, where heat is transferred from one fluid to another.
In conclusion, the properties of pure substances play a critical role in engineering thermodynamics. They provide valuable information about the behavior of a substance under different conditions and are used to calculate important thermodynamic parameters, such as enthalpy, entropy, and internal energy. This information is critical for the design and operation of a wide range of thermodynamic systems, such as power plants, refrigeration and air-conditioning systems, and chemical processes.
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.
Engineering Thermodynamics: Properties of Pure SubstancesMAYURDESAI42
Engineering thermodynamics is a branch of thermodynamics that deals with the practical application of thermodynamic principles and concepts. One of the fundamental topics in engineering thermodynamics is the properties of pure substances.
A pure substance is a material that has a fixed and constant chemical composition, regardless of its physical state. This means that a pure substance cannot be separated into two or more different substances by physical means. Examples of pure substances include water, oxygen, and carbon dioxide.
The properties of a pure substance are critical in thermodynamics because they are used to calculate important thermodynamic parameters, such as enthalpy, entropy, and internal energy. These parameters are used to understand the behavior of systems and predict their response to changes in temperature, pressure, and other conditions.
One of the key properties of a pure substance is its temperature-pressure phase diagram, which provides information about the physical state of the substance under different conditions. For example, water can exist as a solid (ice), a liquid (water), or a gas (steam) depending on the temperature and pressure conditions. This information is critical for understanding the behavior of a substance in different thermodynamic systems, such as power plants, refrigeration and air-conditioning systems, and chemical processes.
Another important property of a pure substance is its enthalpy of vaporization, which is the amount of energy required to convert a unit mass of the substance from a liquid to a gas at a constant temperature. This property is critical in many applications, such as the design of steam power plants, which use the energy stored in steam to generate electricity.
The specific heat capacity of a pure substance is another critical property. It represents the amount of energy required to raise the temperature of a unit mass of the substance by a unit temperature. This property is used to calculate the heat transfer in thermodynamic systems, such as refrigeration and air-conditioning systems.
Another important property of a pure substance is its thermal conductivity, which represents its ability to transfer heat. This property is critical in the design of heat exchangers, where heat is transferred from one fluid to another.
In conclusion, the properties of pure substances play a critical role in engineering thermodynamics. They provide valuable information about the behavior of a substance under different conditions and are used to calculate important thermodynamic parameters, such as enthalpy, entropy, and internal energy. This information is critical for the design and operation of a wide range of thermodynamic systems, such as power plants, refrigeration and air-conditioning systems, and chemical processes.
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.
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1. M.405.8 0
DEPARTMENT OF TECHNICAL EDUCATION
ANDHRA PRADESH
Name : Dr. P. Ugandhar
Designation : Sr. Lecturer in Mechanical Engg.
Branch : Mechanical Engineering
Institute : Govt. Polytechnic, SKLM
Year / Semester : IV
Subject : Hydraulics & Hydraulic Machinery
Subject Code : M-405
Topic :Fluid Properties
Duration :50 Minutes
Sub Topic : Problems on Manometers
Teaching Aids : PPT Animations
2. M.405.6 1
OBJECTIVES
On completion of this period, you would be able to
• Know the various pressure measuring devices
• Know the Difference between simple and differential manometers
4. M.405.6 3
Hydrostatic Law:
· The pressure at any depth ‘h’ from the free surface in a liquid of
density is given by P = gh
· If ‘h’ is measured in upward direction P = -gh
The pressure of liquid if expressed in terms of the height (h) of
liquid column, it is known as pressure head of liquid.
Pressure head h = P/ g
5. M.405.6 4
Piezometer is used to measure low liquid pressures
What is Piezometer
Fig. 2 Piezometer
6. M.405.6 5
Manometer
• It is pressure measuring device
• It is used to measure difference of pressures between two
points.
• Its one column of liquid is balanced by another column of
liquid
7. M.405.6 6
Types of Manometers
• Simple U – tube Manometer
• Differential U – tube Manometer
• Simple U-tube manometer is used to measure
pressure at a point (gauge or vacuum)
• Differential U-tube manometer is used to measure
the difference of pressure between two points
8. M.405.6 7
How to find pressure at a point using U-tube Manometer
Fig 3 U-tube Manometer
• The Pressures at two
points P and Q must be equal
(being the same liquid)
9. M.405.6 8
U-tube Manometer:
• Equating pressures at P and Q, we have
P1+Ag(y+x) = patm +Bgx
P1-Patm = (B-A) gx - Agy
• Mercury is generally used as a manometric fluid because of its
high density and low vapour pressure
10. M.405.6 9
To find Vaccum Pressure:
• The Pressures at two points
P and Q must be equal.
• Equating pressures at P and
Q, we have
Fig 4 U-tube Manometer
11. M.405.6 10
Conversion of one fluid column into other fluid
column
• The pressure is given by P = gh where h is the height
in terms of fluid of density
• The same pressure can be expressed in terms of different
fluid columns.
i.e. P = 1gh1 = 2gh2
Where h1 and h2 are heights in terms of fluids of density
1 and 2 respectively.
1h1 = 2h2
Or
s1h1 = s2h2
12. M.405.6 11
SUMMARY
• In this class, we have discussed about
• Hydrostatic Law
• Piezometers
• Simple U-tube manometer to measure positive gauge pressures.
• Simple U-tube manometer to measure Vaccum Pressures
13. M.405.6 12
QUIZ
• A U-tube manometer with both limbs open to atmosphere
contains two immiscible liquids of densities 1 and 2 as shown
in figure. Under equilibrium ‘h’ is given by
(a) h =0
(b) h = L 2/ 1
(c) h = L(1- 1/ 2)
(d) h = L ( 2/ 1 –1)
Fig-5
14. M.405.6 13
• A U-tube manometer is connected to a pipeline conveying
water as shown in figure. Calculate the pressure head of water
in the pipeline is
Assignment
Fig-6
15. M.405.7 14
OBJECTIVES
On the completion of this period, you would be
able to
• Know the principle of differential manometer
• Working principle of Bourdon’s pressure gauge
16. M.405.7 15
Differential Manometer:
1 2
w w m
P y x g P y g x g
1 2 m w
P P gx
• The Pressures at two points P and
Q must be equal.
• Equating pressures at P and Q, we
have
Fig-1
17. M.405.7 16
Inverted Differential Manometer
• When a small difference of pressure between two
points is to be measured, then inverted differential manometer
is used
• A liquid of specific gravity less than that of the
flowing liquid is used
18. M.405.7 17
Fig-2 Inverted differential manometer
Pressure at D = Pressure at C
PA - 1gb- mgc = PB - 2g(a+b+c)
PB-PA = 2g(a+b+c) - 1gb - mg c
19. M.405.7 18
Bourdon’s Pressure Gauge:
∙ It is a mechanical gauge and is used for measuring pressure in
steam boilers. This gauge has an elastic metallic tube of elliptical
section bent into circular arc
Fig-3 Bourdon´s Pressure Guage
20. M.405.7 19
Bourdon’s Pressure Gauge:
•When fluid enters the bent tube at one end, it tends to straighten
the tube.
• This causes the other end to move, causing pointer to move on a
graduated scale giving pressure reading
As the tube is surrounded by atmospheric pressure, the movement
of the tube is against atmospheric pressure and hence this gauge
measures pressure above or below atmospheric pressure (gauge
pressure)
21. M.405.7 20
SUMMARY
In this class, we have discussed about
• Differential Manometers
• Differential inverted U-tube Manometers
• Bourdon’s Pressure Gauge
22. M.405.7 21
QUIZ
1. The manometer shown in figure connects
two pipes carrying oil and water respectively.
The difference between oil and water pressure
is
(a) ( Water+ oil)gH
(b) ( Water / oil)gH
(c) ( Water - oil)gH
(d) Oil gH
Fig-4
23. M.405.7 22
2. The manometer shown in figure connects two pipes carrying
oil and water respectively. From this figure it can be concluded
that
(a) pressure in pipes are equal
(b) the pressure in oil pipe is higher
(c) the pressure in water pipe is
higher
(d) The pressure in two pipes can not
be compared (for want of Sufficient
data)
QUIZ
Fig-5
24. M.405.7 23
Frequently Asked Questions
1. What is differential manometer?
2. Why is mercury used as a manometric liquid?
3. How does Bourdon’s pressure gauge work?
25. M.405.8 24
RECAP
In the previous class, we have discussed about
• Principle of manometers
• Finding pressure at a point by using simple manometer
• Finding pressure difference between two points by using differential
manometer
26. M.405.8 25
OBJECTIVES
• On the completion of this period, you would be able to
•solve simple problems on manometers
27. M.405.8 26
Problem –1:
• Find the pressure of a fluid of specific gravity 0.9 at the
point A as shown in figure.
Fig-1
28. M.405.8 27
Solution
• Let x-x be the datum line
Density of oil (0) = 0.9 x 1000 = 900 kg/m3
• Pressure in the left limb at x-x = Pressure in the right limb at XX
0
20 12 20
0
100 100
m
g g
• PA+
30. M.405.8 29
• Problem – 2
A differential manometer is connected as shown in Fig.
Find the pressure difference between A and B in N/m2 and in
cm of water.
• Let us choose xx as datum line.
• Pressure in left limb at xx = Pressure in right limb at xx
Fig-2
31. M.405.8 30
Solution
2 2
0
38 23
(0.53) (0.23)
100
A H o B H o
P g g P g
2 0
0.61 0.53 (0.23)
A B H o
P P g g
2 0
(0.08) (0.23)
A B H o
P P g
9.81 1000(0.08) 810(0.23)
A B
P P
3
0.81 1000 810 /
o kg m
2
1042.8 /
A B
P P N m
.
.
.
.
.
.
.
32. M.405.8 31
• Pressure difference in terms of water
P = gh
• h = -0.1063 m of water
or h = -0.1063 100 cm
h = -10.63 cm of H20
2
A B
P = P P 1042.8 /
N m
.
.
33. M.405.8 32
ASSIGNMENT
1. The pressure at certain point is observed to be 40 m of oil of
specific gravity 0.6. Find the corresponding height in terms of
mercury of specific gravity 13.6 at that point.
2. Pressures have been observed at four different point in different
units of measurements of follows:
(a) 150 kpa (b) 1800milli bar
(c) 20m of water (d) 1240mm of mercury
Arrange the points in the descending order of magnitude of
pressure