ME804306-2Fluid Mechanics
Chapter 2
Fluid Statics
Dr. Kamel Mohamed Guedri
Mechanical Engineering Department,The College of Engineering and Islamic Architecture,
Umm Al-Qura University, Room H1091
Website: https://uqu.edu.sa/kmguedriEmail: [email protected]
Objectives
• Determine the variation of pressure in a fluid at rest
• Calculate pressure using various kinds of manometers
• Calculate the forces exerted by a fluid at rest on plane
or curved submerged surfaces.
• Analyze the stability of floating and submerged
bodies.
• Analyze the rigid-body motion of fluids in containers
during linear acceleration or rotation.
OUTLINE
1. Absolute and gauge pressure2. Basic Equation of Fluid Statics3. Pressure - Depth Relationships
a) Constant density fluidsb) Ideal gases
4. Pressure measurement devicesa) The barometreb) The manometersb) Other pressure measurement devices
5. Hydrostatic forces on submerged plane surfaces6. Hydrostatic forces on submerged curved surfaces7. Buoyancy and stability8. Fluid in rigid-body in motion
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2.1 Absolute and Gauge Pressure• Pressure measurements are generally indicated as
being either absolute or gauge pressure.
• Gauge pressure
• is the pressure measured above or below the atmospheric pressure (i.e. taking the atmospheric as datum).
• can be positive or negative. • A negative gauge pressure is also known as vacuum
pressure.
• Absolute pressure
• uses absolute zero, which is the lowest possible pressure.
• Therefore, an absolute pressure will always be positive.
• A simple equation relating the two pressure measuring system can be written as:
• Pabs = Pgauge + Patm (2.2)
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•Atmospheric pressure• refers to the prevailing pressure in the air around
us.
• It varies somewhat with changing weather conditions, and it decreases with increasing altitude.
• At sea level, average atmospheric pressure is 101.3 kPa (abs), 14.7 psi (abs), or 1 atmosphere (1 bar = 1x105 Pa).
• This is commonly referred to as ‘standard atmospheric pressure’.
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Example 2.1 Express a pressure of 155 kPa (gauge) as an absolute
pressure. Express a pressure of –31 kPa (gauge) as an absolute
pressure. The local atmospheric pressure is 101 kPa (abs).
• Solution:
Pabs = Pgauge + Patm
Pabs = 155 + 101 = 256 kPa
Pabs = -31 + 101 = 70 kPa
2.2 Basic Equation of Fluid Statics
The pressure at a point in a static fluid is the same in all directions. What this means that the pressure on the small cube in Figure 2.1 is the numerically the same on each face as the cube shrinks to zero volume. At the surface, the pressure is zero gage. Since depth increases in the downward z-direction, the sign on the specific weight in equation 2.1 is negative.
2.3 Pressure - Depth Relationships
2.3.1 Constant Density Fluids
To establish the relationship between pressure and depth for a constant density fluid, the pressure - position relationship must be separated and integrated
Equation 2.3 is one of the most important and useful equations in fluid statics.
Figure below shows a tank
with one side open to the
atmosphere and the other
side sealed with air above
the oil (SG=0.90).
Calculate the gauge
pressure at points
A,B,C,D,E.
Example 2.2
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Solution:
• At point A, the oil is exposed to the atmosphereThus PA=Patm = 0 (gauge)
• Point B is 3 m below point A, Thus PB = PA + oilgh
= 0 + 0.9x1000x9.81x3= 26.5 kPa (gauge)
• Point C is 5 m below point A, Thus PC = PA + oilgh
= 0 + 0.9x1000x9.81x5= 44.15 kPa (gauge)
• Point D is at the same level of point B,Thus PD = PB
= 26.5 kPa (gauge)• Point E is higher by 1 m from point A,
Thus PE = PA - oilgh= 0 - 0.9x1000x9.81x1= -8.83 kPa (gauge).
2.3.2 Variable Density Fluids
Most fluids are relatively incompressible, but gases are not. This means that the density increases with depth so we cannot use the specific weight as a constant in determining the pressure. If we assume an ideal gas, then the density is given by
where M is the molecular weight, T is the absolute temperature, and R is the gas constant in appropriate units.Replacing in the differential equation (2.1) we get