Hydrostatics

Presentation on Fluid Mechanics I : Hydrostatics

1. Welcome To MY Presentation

2. Group - 2 Serial Name ID 01 Morshedul Hasan 14107006 Sec: Day Sub: MEC 267 Prog : BSME

3. Contents: Introduction Total pressure Total Pressure on an Immersed Surface Total Pressure on a Horizontally Immersed Surface Total Pressure on a Vertically Immersed Surface Total Pressure on an Inclined Immersed Surface Centre of Pressure Centre of Pressure on a Vertically Immersed Surface Centre of Pressure on an Inclined Immersed Surface Centre of Pressure of a Composite Section Hydrostatic Force on a Curved Surface Application Hydrostatics

4. Hydrostatics  is the branch of  fluid mechanics  that studies incompressible  fluids  at rest. Hydrostatics is fundamental to  hydraulics , the  engineering  of equipment for storing, transporting and using fluids. Figure : Hydrostatic Pressure

5. Total pressure : The 'total pressure' is the sum of the static pressure, the dynamic pressure, and the gravitational potential energy per unit volume. It is therefore the sum of the mechanical energy per unit volume in a fluid. P total = P 1 + P 2+ P 3 ………….. Total Pressure on an Immersed Surface: The total pressure exerted by a liquid on an immersed surface. The position of an immersed surface may be:- Horizontal Vertical and Inclined

6. Total Pressure on a Horizontally Immersed Surface: Consider a plane horizontal surface immersed in a liquid as shown in figure: Let, W= Specific weight of the liquid. A= Area of the liquid. X= Depth of the horizontal surface from the liquid level in meters. Now, Total pressure on the surface, P =Weight of the liquid above the immersed surface = Sp. Wt. of liquid x Volume of liquid = Sp. Wt . Of liquid x Area of surface x Depth of liquid = wAx KN Where w is the specific weight of the liquid in KN/m 3

7. Total Pressure on a Vertically Immersed Surface: Consider a plane vertical surface immersed in a liquid as shown in figure: First of all, let us divide the whole immersed surface into a number of small parallel strips as shown in figure. Let, w= specific weight of the liquid. A= Total area of the immersed surface x= Depth of center of gravity of the immersed surface from the liquid surface. Let us consider a strip of thickness dx , width b and at a depth x from the free surface of the liquid as shown in figure. Now, The intensity of pressure on the strip = wx And area of the strip = b.dx

8. Total Pressure on a Vertically Immersed Surface: Now, Total pressure on the surface, P = ʃ wx . bdx = w ʃ x . bdx But, ʃ x . bdx = Moment of the surface area about of the liquid level. = Ax So, P = wAx So, Pressure on the strip, P = Intensity of pressure x Area = wx.bdx

9. Total Pressure on an Inclined Immersed Surface: Consider a plane inclined surface immersed in a liquid as shown in figure: First of all, let us divide the whole immersed surface into a number of small parallel strips as shown in figure. Let, w= specific weight of the liquid. A= Total area of the surface x= Depth of center of gravity of the immersed surface from the liquid surface. θ = Angle at which the immersed surface is inclined with the liquid surface. Let us consider a strip of thickness dx , width b and at a distaance l from O. Now, The intensity of pressure on the strip = wl sin θ And area of the strip = b.dx So, Pressure on the strip, P = Intensity of pressure x Area = wl sin θ . bdx

10. Total Pressure on an Inclined Immersed Surface: Now, Total pressure on the surface, P = ʃ wl sin θ . bdx = w sin θ ʃ l . bdx But, ʃ l . bdx = Moment of the surface about O. = So, P = w sin θ x =

11. Center of pressure: The  center of pressure  is the point where the total sum of a  pressure  field acts on a body, causing a  force  to act through that point. The total force vector acting at the center of pressure is the value of the integrated vectorial pressure field. The resultant force and center of pressure location produce equivalent force and moment on the body as the original pressure field. Pressure fields occur in both static and dynamic fluid mechanics. Specification of the center of pressure, the reference point from which the center of pressure is referenced, and the associated force vector allows the moment generated about any point to be computed by a translation from the reference point to the desired new point. It is common for the center of pressure to be located on the body, but in fluid flows it is possible for the pressure field to exert a  moment  on the body of such magnitude that the center of pressure is located outside the body.

12. Centre of Pressure on a Vertically Immersed Surface: Consider a plane surface immersed vertically in a liquid as shown in figure: First of all, let us divide the whole immersed surface into a number of small parallel strips as shown in figure. Let, w= specific weight of the liquid. A= Total area of the immersed surface x= Depth of center of gravity of the immersed surface from the liquid surface. Let us consider a strip of thickness dx , width b and at a depth x from the free surface of the liquid as shown in figure. Now, The intensity of pressure on the strip = wx And area of the strip = b.dx h G p

13. Centre of Pressure on a Vertically Immersed Surface: h G p So, Pressure on the strip, P = Intensity of pressure x Area = wx.bdx Moment of this pressure about the liquid surface, = ( wx . bdx ) x = wx 2 . bdx Now, The Some of moments of all such pressures about the liquid surface , M = ʃ wx 2 . bdx = w ʃ x 2 . bdx But, ʃ x 2 . bdx = I (i.e. moment of inertia of the surface about the liquid level) So, M= w I ..................( i ) We know that, the sum of the moments of the pressure = P x h..............(ii) Where, p= Total pressure on the surface and h= Depth of center of pressure from the liquid surface.

14. Centre of Pressure on a Vertically Immersed Surface: Now, equating equation ( i ) and (ii) We know the theorem of parallel axis that, I =I G + Ah 2 Where, I G = Moment of inertia of the figure, about horizontal axis through its center of gravity h = Distance between the liquid surface and the center of gravity of the figure. ( in this case) Now rearranging the equation (iii),

15. Centre of Pressure on an Inclined Immersed Surface: Consider a plane inclined surface immersed in a liquid as shown in figure: First of all, let us divide the whole immersed surface into a number of small parallel strips as shown in figure. Let, w= specific weight of the liquid. A= Total area of the immersed surface x= Depth of center of gravity of the immersed surface from the liquid surface. θ = Angle at which the immersed surface is inclined with the liquid surface Let us consider a strip of thickness dx , width b and at a depth x from the free surface of the liquid as shown in figure. Now, The intensity of pressure on the strip = wl sin θ And area of the strip = b.dx G p l 1

16. Centre of Pressure on an Inclined Immersed Surface: So, Pressure on the strip, P = Intensity of pressure × Area = wl sin θ . bdx and moment of this pressure about 0, = ( wl sin θ . bdx ) l = wl 2 sin θ . bdx So , M= ʃ wl 2 sin θ . bdx = wsin θ ʃ l 2 .bdx But, ʃ l 2 . bdx = I (i.e. moment of inertia of the surface about 0) so, M = wsin θ I .....................( i ) We know that the sum of the moments of all such pressure about 0, Where, p= Total pressure on the surface and h= Depth of center of pressure from the liquid surface.

17. Centre of Pressure on an Inclined Immersed Surface: Now equating equation ( i ) and (ii) We know the theorem of parallel axis that, I =I G + Ah 2 Where, I G =Moment of inertia of the figure, about horizontal axis through its center of gravity h = Distance between the liquid surface and the center of gravity of the figure. ( l 1 in this case) Now rearranging the equation (iii),

18. Hydrostatic Force on a Curved Surface: On a curved surface the forces pδA on individual elements differ in direction, so a simple summation of them may not be made. Instead, the resultant forces in certain directions may be determined, and these forces may then be combined vectorially . It is simplest to calculate horizontal and vertical components of the total force. Horizontal component of hydrostatic force: Any curved surface may be projected on to a vertical plane. Take, for example, the curved surface illustrated in figure.

19. Its projection on to the vertical plane shown is represented by the trace AC. Let Fx represent the component in this direction of the total force exerted by the fluid on the curved surface. Fx must act through the center of pressure of the vertical projection and is equal in magnitude to the force F on the fluid at the vertical plane. In any given direction, therefore, the horizontal force on any surface equals the force on the projection of that surface on a vertical plane perpendicular to the given direction. The line of action of the horizontal force on the curved surface is the same as that of the force on the vertical projection. Vertical component of hydrostatic force: The vertical component of the force on a curved surface may be determined by considering the fluid enclosed by the curved surface and vertical projection lines extending to the free surface. Thus F H = F 2 = ρgz s F V = F 1 = ρgV where V is the volume of the liquid between the free surface liquid and solid curved surface. The magnitude of the resultant is obtained form the equation

20. Some Application Hydrostatics : The Hydrostatics Pressure is either utilized in the working of a Hydraulic Structure, or a Structure is checked to withstand the Hydrostatic Pressure exerted on it. 01)Water Pressure on Sluice Gate:-

21. To regulate the flow of Water, in the path of a River, a Sluice Gate is provided. The Sluice Gate is made to move up and down with the help of Rollers fixed on Skin Plate, which travel on Vertical Rails called Guide. These Rails are fixed on Piers or Vertical Walls. In between these two Skin Plates, a number of I-beams are provided horizontally to withstand the Water Pressure. The spacing between the I-beam is lesser at the bottom than that at top of Sluice Gate, as the Water Pressure varies in the depth.

22. 02)Water Pressure on Lock Gate:- The Water Level on both sides of the Dam will be different. If it is desired to have Boating in such river, the Chamber, known as Lock, is constructed between these two different Water Levels. Two sets of Lock Gates are provided in order to transfer a Boat from Higher Water Level to Lower Water Level. The Upstream gates are opened, and the Water Level in the Chamber is rises up to the Upstream Water level. The Boat is then shifted in the chamber, then Upstream Gates are closed and Downstream Gates are opened and the Water Level in the Chamber is lowered to the Downstream Water level. The procedure is reversed for the transfer of Boat from Downstream to Upstream.

23. 03)Water Pressure on Masonry Walls:- When Water on one side of Masonry Walls, the Water Pressure will act perpendicular to the Wall. A little consideration will show, that the intensity of Pressure at Water Level will be Zero and will be increase by a Straight Line Law to  wH  at the bottom. Thus the Pressure of the Water on a Vertical Wall will act through a point at a distance H/3 from the bottom, where H is the depth of Water.

24. Conditions for Stability of Dam. 01)The Resultant must pass within the base to safeguard Dam against Overturning. 02)The Resultant must pass through the middle third of the base, to avoid tension at the base. 03)To prevent sliding, the maximum Frictional Force should be more than the Horizontal Force. 04)The Maximum Stress developed at the bottom of the Dam should be within the permissible Stress of the site.