1 / 55

ME804306-2 Fluid Mechanics

Chapter 2 Fluid Statics. ME804306-2 Fluid Mechanics. 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/kmguedri Email: kmguedri@uq.edu.sa. Objectives.

hilde
Download Presentation

ME804306-2 Fluid Mechanics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 2 Fluid Statics ME804306-2Fluid Mechanics 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: kmguedri@uq.edu.sa

  2. 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.

  3. OUTLINE 1. Absolute and gauge pressure 2. Basic Equation of Fluid Statics 3. Pressure - Depth Relationships a) Constant density fluids b) Ideal gases 4. Pressure measurement devices a) The barometre b) The manometers b) Other pressure measurement devices 5. Hydrostatic forces on submerged plane surfaces 6. Hydrostatic forces on submerged curved surfaces 7. Buoyancy and stability 8. Fluid in rigid-body in motion

  4. 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)

  5. 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’.

  6. 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

  7. 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.

  8. 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.

  9. Example 2.2 • 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.

  10. Solution: • At point A, the oil is exposed to the atmosphere Thus 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).

  11. 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

  12. 2.4. Pressure measurement devices

  13. Example 2.3

  14. Example 2.4

  15. Example 2.5

  16. 2.5. Hydrostatic forces on submerged plane surfaces

  17. Example 2.6

  18. 2.6. Hydrostatic forces on submerged curved surfaces

  19. Example 2.7 below. 2.7

  20. 2.7. Buoyancy and stability

  21. Example 2.8

  22. 2.8. Fluid in rigid-body motion

More Related