Fluid Statics

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Fluid Statics Fluid Statics and Fluid Dynamics form the two constituents of Fluid Mechanics. Fluid Statics deals with fluids at rest while Fluid Dynamics studies fluids in motion. In this chapter we discuss Fluid Statics. A fluid at rest has no shear stress. Consequently, any force developed is only due to normal stresses i.e, pressure. Such a condition is termed the hydrostatic condition. In fact, the analysis of hydrostatic systems is greatly simplified when compared to that for fluids in motion. Though fluid in motion gives rise to many interesting phenomena, fluid at rest is by no means less important. Its importance becomes apparent when we note that the atmosphere around us can be considered to be at rest and so are the oceans. The simple theory developed here finds its application in determining pressures at different levels of atmosphere and in many pressure-measuring devices. Further, the theory is employed to calculate force on submerged objects such as ships, parts of ships and submarines. The other application of the theory is in the calculation of forces on dams and other hydraulic systems.

Fluid Forces In Fluid Mechanics we consider forces upon fluid elements. It is necessary to discuss the type of forces that could act on fluid elements. These Forces could be divided into two categories - Surface Forces, Fs and Body Forces, FB.

Figure 1 : Classification of forces Surface forces are brought about by contact of fluid with another fluid or a solid body. The best example of this is pressure. The surface forces depend upon surface area of contact and do not depend upon the volume of fluid. On the other hand, body forces depend upon the volume of the substance and are distributed through the fluid element. Examples are weight of any substance, electromagnetic forces etc.

Pressure at a point within a fluid Consider a fluid at rest as shown in Fig. 2. From around the point of interest, P in the fluid let us pull out a small wedge of dimensions dx x dz x ds . Let the depth normal to the plane of paper be b. In some of the derivations we chose z to be the vertical coordinate. This is consistent with the use of z as the elevation or height in many applications involving atmosphere or an ocean. Let us now mark the surface and body forces acting upon the wedge. The surface forces acting on the three faces of the wedge are due to the pressures, Px, Pz and Pn as shown. These forces are normal to the surface upon which they act. We follow the usual convention that compression pressure is positive in sign. We again remind ourselves that since the fluid is at rest there is no shear force acting.

Figure 2 : Pressure at a point In addition we have a body force, the weight, W of the fluid within the wedge acting vertically downwards. Summing the horizontal and the vertical forces we have,

 F x =0 ie. Px b dx−Pn b ds sin=0  F z=0 ie. Pz b dx−Pn b ds cos −W =0 1 or Pz b dx−Pn d ds cos−  g b dx dz=0 2 Noting that

ds sin =dz and ds cos =dx

we have after simplification,

Px =P n and

1 Pz =Pn  g dz . 2

(1)

(2)

(3)

We note that the pressure in the horizontal direction does not change, which is a consequence of the fact that there is shear in a fluid at rest. In the vertical direction there is a change in pressure proportional to density of the fluid, acceleration due to gravity and difference in elevation. Now if we take the limit as the wedge volume decreases to zero, i.e., the wedge collapses to the point P, we have, Px =P z =P n=P (4) This equation is known as Pascal's Law. It is important to note that it is valid only for a fluid at rest. In the case of a moving fluid, pressures in different directions could be different depending upon fluid accelerations in different directions. Hence, for a moving fluid pressure is defined as an average of the three normal stresses acting upon the fluid element.

Equation for Pressure Field We have shown that for a fluid at rest pressure at any point is invariant with direction. But it does not prevent pressure itself varying from point to point within the fluid. In this section we try to establish a relationship for this variation of pressure. Consider a rectangular element of fluid of dimensions dx x dy x dz with its centre at the point P(x,y,z) as shown in Figure 2.3. Let the pressure acting at the point P be equal to P. It is usually assumed that pressure varies continuously across the element.

Figure 3: Pressure at a point Consequently the pressures at the different faces of the element are calculated by expanding pressure in a Taylor series about the point P. Second and higher order terms are neglected. Accordingly,

∂ P dx ∂ P dx , PR=P ∂x 2 ∂x 2 ∂ P dy ∂ P dy PG=P− , PF=P ∂y 2 ∂y 2 ∂ P dz ∂ P dz PB=P− , PT=P ∂z 2 ∂z 2 PL=P−

A surface force balance in the x-direction gives

 F x =PL−PR ∂ P dx ∂ P dx = P− dy dz− P dy dz ∂x 2 ∂x 2 −∂ P = dx dy dz ∂x









(5)

Similar force balancing is carried out in each of the other directions. Upon collecting terms we have for surface forces acting on the fluid element,





∂P  ∂P  ∂P  F s=− i j k dx dy dz ∂x ∂y ∂z The terms within the parenthesis is called the pressure gradient. i.e.,



Thus,



(6)



∂P  ∂ P  ∂P ∂  ∂  ∂ grad p=∇ P= i j k = i j k P ∂x ∂y ∂z ∂x ∂y ∂z

(7)

d F s =−grad P dx dy dz=−∇ P dx dy dz

(8)

Thus it is seen that the net surface force upon the element is given by the pressure gradient and is not dependent upon the pressure level itself. Body forces The only body force that we consider is the weight of the fluid element or the gravity force. If this is designated dFB, we have

d F B= g dx dy dz

(9)

Total Force Adding the surface and body forces acting on the fluid element, we have total force,

dF=d F sd F B=−∇ P g dx dy dz

(10)

On a unit volume basis this equation becomes

dF =−∇ P g dV

(11)

Thus we have obtained an expression for force upon a unit volume of a fluid element at rest. To extend this to the case of a moving fluid, one has to include normal and shear stresses due to viscosity in addition to the one given above. They are together balanced by inertia forces. Coming back to fluid at rest, the net force given by Eq.11 should be equal to zero. Accordingly,

−∇ P g=0

(12)

The above equation consists of three separate equations, one for each direction and each of them must be equal to zero. Thus for x, y and z directions we have,

−∂ P −∂ P −∂ P  g x =0 ,  g y =0 ,  g z =0 ∂x ∂y ∂z

(13)

If the coordinate system be so selected as to align one of x, y or z with acceleration due to gravity, g the equations simplify considerably. The natural selection is to have z-direction align with -g, such that gz = - g. Consequently, gx = gy = 0. Then we have,

∂P ∂P =0 , =0 and ∂x ∂y

∂P =− g ∂z

(14)

The above equation shows that pressure in a static fluid does not vary in x or y direction. It varies only in the z-direction. This enables one to write,

∂P =− g=− ∂z

(15)

The above equation is a fundamental equation in Fluid Statics. It defines the manner in which pressure varies with height or elevation and finds many applications. Mainly it enables one to determine atmospheric pressures at different elevations above the sealevel. Then we employ the same equation to determine pressure at various depths of an ocean. The other application is in Manometry , which forms the basis of a class of pressure measuring instruments. A close look at the equation reveals that the pressure gradient is a function of density,  and acceleration due to gravity, g . The latter one, g is almost a constant and therefore it is the variation in density with elevation that influences the pressure values. Density is constant for incompressible fluids and varies with pressure and temperature for compressible fluids. Therefore it is necessary to consider these two types of fluids separately. Incompressible Fluids For incompressible fluids density is a constant. In addition as stated before, for most applications of practical interest acceleration due to gravity is also a constant. As a consequence the pressure equation is greatly simplified and Equ (15) is readily integrated. Thus for incompressible fluids, P2

∫P which on integration yields

1

z2

z2

1

1

dP=− g∫z dz=−∫z dz

(16)

P2−P1=− z 2−z 1 

(17)

A convenient form of the above equation is

P1=P2 h

(18)

where h is the difference in elevation, z2 - z1 . By rewriting the above equation, we have,

h=

 P1−P2  

(19)

These equations demonstrate that pressure difference between two points in an incompressible fluid is proportional to the difference in elevation or height between the two points. The term h is sometimes defined as the pressure head and is the height of the fluid column of density  (and specific weight,  ) that supports a pressure difference of (P1 – P2). The above is used to determine pressures in atmosphere and ocean depths. For this, it is advisable to choose a convenient datum or reference. Depending upon the application as shown in the figure, sealevel (for atmospheric pressure) or free surface (for measurements in oceans and lakes) seems to be ideally suited. For an ocean or a lake, if Pa is the pressure acting on the free surface pressure at any depth h2 is given by

P=Pa water h2

(20)

On the other hand for the atmosphere with Pa being the sealevel pressure, we have,

P=Pa −air h1

(21)

Figure 4 : Measurement of pressure in atmosphere, oceans and lakes. However care should be taken only to apply this to small distances in air as air is incompressible and over large heights, other effects dominate.

Compressible Fluids, Properties of Atmosphere

The most common compressible fluid we know is air. Assuming air to behave like a perfect gas, Equ 15 becomes,

dP −P =− g= g dz RT or

dP −g dz = P R T

(22)

By integrating the above equation we obtain, P2

∫P

1

 

P dP −g z dz =ln 2 = ∫ P P1 R z T 2

(23)

1

To solve the above equation we need to know how temperature T varies with altitude. For this we rely on the concept of Standard Atmosphere described in a following chapter.

International Standard Atmosphere

Measurement of Pressure One of the direct applications of the equation of Fluid Statics we have derived is in the devices used to measure pressure. Now it is necessary to recall that we have an Absolute Pressure and a Gauge Pressure. We note that pressure is always measured as a difference or with respect to a datum or reference. Absolute pressure is measured relative to a perfect vacuum, whereas gauge pressure is measured relative to atmospheric pressure. Further, Absolute pressure is the sum of atmospheric pressure and the gauge pressure. See Fig. 8 below.

Figure 8 : Definition of gauge and absolute pressures Manometry We saw in previous sections that pressure is proportional to the height of a column of fluid. Manometry exploits this to measure fluid pressure. In other words we measure the height of a column of liquid supported by the pressure (actually the pressure difference). Barometer, piezometer and Utube Manometer are some of the members of this class.

Mercury Barometer

Figure 9 : Mercury Barometer Mercury Barometer (Fig.9) is the simplest device to measure atmospheric pressure at a location. It consists of a glass tube closed at one end immersed in a container filled with mercury. Because of the atmospheric pressure mercury rises in the tube as shown. If h is the height of mercury above the fluid level in the container, we have Patm−P A = h (24) Where PA is the pressure at A and will be equal to the vapour pressure of mercury, Pvap, which is around 0.16pa at a temperature of 20oC. It is usual to neglect PA when the atmospheric pressure is given as

Patm= h= g h

Sometimes atmospheric pressure is expressed as "mms of mercury" being equal to h. At sealevel conditions where the pressure value is 101,327 Pascals and the specific weight of mercury is 133,100 N/m3, the barometric height is 761 mm Hg. Water could be used as the barometer fluid, but in that case the height of water will be around 10.36 m! Piezometer Tube

Figure 10 : Piezometer

Piezometer tube (Fig. 10) is perhaps the simplest of the pressure measuring devices and consists of a vertical tube. In its application one end is connected to the pressure to be measured while the other end is open to the atmosphere as shown. Application of Equ 18 gives P A −Patm= h (25) or simply for the gauge pressure at A,

P A = h

(26)

U-tube Manometer

Figure 11 : U tube Manometer Although the piezometer tube is simple in structure, it has a few practical disadvantages. In this regard U-tube Manometer (Fig. 11) seems to be a better alternative. It is a U-tube filled with what is called a gauge fluid. As before one end of the tube is exposed to the pressure to be measured while the other end is open to atmosphere. We have,

P A 1 h1=P2=P3=P atm 2 h2 P A =2 h2−1 h1

Considerable simplification is possible if the fluid A is a gas when 1 g h1

(27) is negligible, giving

P A =2 h2

Differential U-tube Manometer Differential U-tube manometer (Fig. 12) is very handy to measure the pressure difference directly and is basically similar to the U-tube manometer discussed above. What was the open end before is now connected to a different pressure, PB so that we measure the difference PA – PB. Now we have,

P A 1 h1−2 h2−3 h 3=PB so that

P A −PB =2 h23 h3−1 h 1 .

(28)

Figure 12 : Differential U-tube manometer Hydrostatic Force on a submerged surface The other important utility of the hydrostatic equation is in the determination of force acting upon submerged bodies. Among the innumerable applications of this are the force calculation in storage tanks, ships, dams etc.

Figure 13 : Force upon a submerged object First consider a planar arbitrary shape submerged in a liquid as shown in the figure. The plane makes an angle  with the liquid surface, which is a free surface. The depth of water over the plane varies linearly. This configuration is efficiently handled by prescribing a coordinate frame such that the y-axis is aligned with the submerged plane. Consider an infinitesimally small area dA=dx dy at (x,y). Let this small area be located at a depth h from the free surface. From Equ 18 we know that P=P A h (29) Where PA is the pressure acting on the free surface. The hydrostatic force on the plane is given by,

F=∫ P dA=∫A P A hdA F=P A A∫A h dA F=P A A∫A  y sin dA F=P A A sin∫A y dA

(30)

The integral,

∫A y dA

is the first moment of surface area about x axis. If yc is the y-coordinate of

the centroid of the area we have,

∫A y dA= y c A

(31)

F=P A A sin y C A=P A hC  A=PC A

(32)

Consequently, Eq. 30 is rewritten as

where hC = y C sin 

and PC is the pressure acting at the centroid.

Centre of Pressure Force, F given by Equ 32.is the resultant force acting on the plane due to the liquid and acts at what is called the Center of Pressure (CP). It does not act at the centroid of the plane as it may seem. Let the coordinates of CP be (xp,yp). Noting that the moment of the resultant force is equal to the moment of the distributed force about the same axis, we have

x P F=∫ A x P dA , y P F=∫A y P dA

(33)

Before substituting for F in the above equation we note that the atmospheric pressure PA acting at the free surface also acts everywhere within the fluid and also on both sides of the plane. As such it does not contribute to the net force upon the plane. So we drop term PA from the equation for F . Equ.33 becomes

y P sin  y c A=∫A y  h dA= sin∫ A y 2 dA The term

∫A y

2

(34)

dA is the well-known second moment of area about the x-axis denoted by Ixx

leading to

yP =

I xx A yc

(35)

Ixx is related to that about the x-axis passing through the centroid of the area, Ixc through the Parallel Axes Theorem given by (36) I xx =I xc  A y 2c Consequently, we have

y P = y c

I xc A yc

(37)

Similarly, taking moments about the y-axis, we obtain,

x P F=∫ A sin  x y dA I xy Leading to x P = A yc

(38)

Ixy is the product of inertia with respect to x and y axes. Again on the application of the parallel axis theorem we have

x P =x c

I xyc yc A

(39)

Where Ixyc is the product of inertia about the axes passing through the centroid. The coordinates of the Centre of Pressure are thus given by (xP, yP) (Eqns. 39 and 2.37). The resulting force upon the immersed surface is therefore given by

F= hc A

(40)

The centre of pressure is given by

x P =x c

I xyc I xc , yP = yc yc A yc A

(41)

Expressions for the moments Ixc, Ixyc etc for some of the common shapes are given in the next section. Geometrical Properties of Common Shapes

Figure 14 : Properties of some common shapes

Table 2 Properties of Common Shapes Shape

A

Ixc

A) Circle

R

B) Rectangle

bh

b h3 12

C) Triangle

bh 2

bh 36

D) Semicircle

R 2

2



Iyc

R4 4

2

0.1098 R

R4 4

0

b3 h 12

0

-

b h2 b−2d 72



3

4

Ixyc

0.3927 R

4

0

Note that the determination of the resultant force FR hinges on the knowledge of the position of the centroid for the given shape. The location of CP , the Center of Pressure depends upon the moment of inertia and the product of inertia. These are functions of the geometry only and can be calculated once the shape is given. Table 2 along with Figure 14 gives these properties for some of the common shapes.

Hydrostatic Force on a Curved Surface We encounter many arbitrarily shaped bodies immersed in liquids such as pipes and walls of containers. Forces on these may be calculated in the same manner as in the previous section. But the required integration of involved terms becomes very tedious. A more simplistic approach is to consider the forces resolved in the three coordinate directions separately. It may be noted that each component of force acts upon a projected area of the body. For example, force in x-direction will act normally on the area projected upon the y-z plane.

Figure 15 : Hydrostatic forces on a curved surface Consider a curved surface as shown in Fig.15, immersed in a liquid. The resultant force FR can be resolved into two components - FH in the horizontal direction and FV in the vertical direction. We are considering a thin body which is two-dimensional and as such there is no force in the direction normal to the paper. The configuration can be split into two parts for discussion purposes- 1:, the part between the body and the free stream , pqrs and 2: the body itself, poq. Consider each part separately. Block of fluid pqrs is in equilibrium. The horizontal forces acting on it cancel out. Thus,

Similarly on the block poq we have,

F HL1=F HR1

(42)

F HL2 =F HR2=F H

(43)

The vertical forces acting upon the fluid are (1) FA due to atmosphere, (2) Wpqrs, the weight of block pqrs and (3) Wpoq, the weight of block poq. Consequently, FV =F AW pqrsW poq (44) Force FA is given by the atmospheric pressure times the projected area normal to it. The resultant force, FRis given by,

F R= F 2H F2V

(45)

Buoyancy and Stability What is the vertical force acting on a body which is completely submerged in a fluid? Answer to such a question can be very well found in the theory developed in the previous section. Archimedes seems

to have discovered the laws concerning submerged bodies as well as floating bodies. What is well known as Archimedes principle states 1. The vertical buoyant force experienced by a body immersed in a fluid is equal to the weight of the fluid displaced. 2. A floating body displaces its own weight of the fluid.

Figure 16 : Forces about a body immersed in a fluid-1

Figure 17 : Forces about a body immersed in a fluid-2 Proof is straight forward. Consider an elemental volume within the immersed body as shown in Fig.2.16 . Now the buoyant force is given by,

F B=∫body  P2−P1 d A H

(46)

where dAH is the area of cross section of the elemental volume chosen. We have,

F B=− ∫body  z2 −z1  d A H F B=volume of body  F B=weight of fluid displaced

(47)

It can be shown that the buoyant force, FB passes through the centroid of the displaced volume as shown in Fig.2.17. The point where this force acts is called "Center of Buoyancy", denoted as CB The above result holds good even in the case of a partially submerged body i.e., a floating body. It is assumed that part of the body above the liquid level is in air. The weight of air displaced as a consequence is ignored. (Fig. 18). For this case as well,

F B=volume of displaced fluid =weight of the body

(48)

Figure 18 : Partially Submerged Body The theory developed so far does hold good in case of a fluid for which specific gravity  is not a constant, a layered fluid for example. However now the buoyant force may not act at the centroid of the displaced volume. The theory developed is also applicable where the fluid involved is a gas, say air. Convection currents established in atmosphere depend upon the buoyant forces generated. Stability of Immersed and Floating Bodies Stability becomes an important consideration when floating bodies such as a boat or ferry is designed. It is an obvious requirement that a floating body such as a boat does not topple when slightly disturbed. We say that a body is in stable equilibrium if it is able to return to its position when slightly disturbed. Failure to do so denotes unstable equilibrium. What equilibrium a body enjoys is decided by the couple formed by the weight of the body and the buoyancy force. Consider the immersed body shown in Fig.2.19. In general, if the center of gravity of the body lies below the center of buoyancy stable equilibrium prevails. An overturning couple leading to unstable equilibrium results if the center of gravity is above the center of buoyancy (Fig.20).

Figure 19 : Stability of an immersed body

Figure 20: Instability of an immersed body It becomes more complicated when floating bodies are considered. Now as the body rotates responding to any disturbance the center of buoyancy can shift. This could render the body stable even though the center of gravity is above the center of buoyancy. This is particularly true of the bodies with a broader base such as a barge (Fig. 21). A slender body as shown in Fig. 22 is very susceptible for instability.

Figure 21 : Stability of a floating body

Figure 22 : Instability of a floating body

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