Pressure Fluid Statics Variation of Pressure with Position in a Fluid Measurement of Pressure Hydrostatic Thrusts on Submerged Surfaces Plane Surfaces Curved Surfaces ddendum First and Second Moment of rea Pressure Variation of Pressure with Position in a Fluid Consider a small cylinder of fluid PQ as shown in the figure (refer to notes from class for its location in the bulk fluid). If the fluid is at rest, the cylinder must be in equilibrium and the only forces acting on it are those on its various faces due to pressure, and gravity. The cross-sectional area δ is very small and the variation of pressure over it is therefore negligible. Figure 1 Q (p+δp) δ z+ δz δl z θ pδ P ρgδδl Taking the pressure at the end P to be p, and the pressure at the other end Q to be p + δp, where δp could be negative or positive. The weight of the fluid acting vertically downward is the product of its mass and gravity, which is ρgδδl, δ is the area of the cylinder ends, δl is its length, ρ is the density of the fluid and g is gravity. 1
Because the fluid is at rest, no shear forces act along the sides of the cylinder. So the forces acting on its long sides act perpendicular to its faces and hence have no component along the length of the cylinder. For equilibrium, the algebraic sum of the forces in any direction should sum to zero. Resolving along the length of the cylinder, we get ( p + p) δ pδ + ρgδδl cos θ = 0 δ (1) Take the elevation of P above some datum as z. Therefore, the elevation of Q is z + δz. From this, δz = δl cos θ. Inserting this in the above equation, we get, δ p + ρgδz = 0 () and as the limit δ z 0 p z = ρg (3) The sign convention takes up as positive, so the above equation says that pressure decreases in the direction of increasing z. Note the use of the partial derivative because we have not as yet determine whether pressure also varies in the x and y direction (that is along a horizontal plane.) Pressure is constant along a horizontal plane in the same fluid Consider the case when δz is zero, that is P and Q are on the same horizontal plane. From the equation above, this implies that δp is zero and so the pressure at P and Q will be the same. Further consider a series of cylinders, say, PQ, QR, RS, etc., (illustrated in class) on the same horizontal plane. Since δz is zero from one point to the other, then using the above argument, the pressure at each point will be the same. So, for any fluid in equilibrium, the pressure is the same at any two points in the same horizontal plane, provided that the points are in the same continuous body of fluid.
Density is constant along a horizontal plane Consider the case for another horizontal plane in the same fluid an elevation δz from the plane. The pressure according to Equation 3 above on that plane is given p by p + δz. The pressure everywhere on this horizontal plane must be the same. z p That is to say, does not vary within that horizontal plane (illustrated in class). z p If does not vary then from Equation 3, the product of g and ρ remains constant. z ssuming that g does not change (and this is reasonable as g does not vary horizontally) then it means that ρ also does not vary horizontally. So, for equilibrium of any fluid, compressible or incompressible, the density as well as the pressure must be constant over any horizontal plane. Since pressure varies only in the vertical direction, then the partial derivative in Equation 3 can be replaced by the full derivative to give: dp dz = ρg (4) The pressure at any point can be found by integrating the above equation to get: p = ρ gdz (5) For fluids with constant density, the above becomes p + ρ gz = cons tan t (6) Note that this equation applies throughout a continuous expanse of the same fluid since, in deriving Equation 3, no restriction was place on the value of θ. To specify the pressure at any point, use is made of Equation 6 once the value of the constant can be specified. To do so, we need to know the pressure at a particular value of z. If the fluid is a liquid with a free surface, then the pressure at its surface is atmospheric pressure, say p a. Take the elevation of this free surface to be z = 0. 3
For equilibrium of the surface, the pressure immediately below it must be equal to the pressure immediately above it because the surface is at rest. That is, the pressure just below the surface also equals p a. Here, the value of the constant in the equation is p a (illustrated in class). For a point a depth h below the surface, h = -z (since z is positive in the upward direction). Therefore, the pressure at that point is: p = pa + ρgz (7) The pressure therefore increases linearly with depth, whatever the shape of any solid boundary may be. The equation says that the pressure at a point in a liquid in equilibrium is due to atmospheric pressure at the free surface and to the density of the liquid. Note that the atmospheric pressure is acting on all surfaces. For the differences of height encountered in this course, the value of atmospheric pressure is effectively constant. Normally atmospheric pressure is regarded as the zero pressure of the pressure scale. Pressures expressed relative to atmospheric pressure as zero is called gauge pressure. Equation 7 then become p = ρgz This says that the gauge pressure in a liquid in equilibrium is given as the product of ρg and the depth vertically below the plane free surface in contact with the atmosphere. The gauge pressure is directly proportional to the depth for a liquid of constant density. Pressure therefore can be visualized simply as the vertical distance h = p. This is known as pressure head, (or sometimes simply as head) ρg corresponding to p. 4
The sum of the terms on the left hand side of Equation 6 is known as the piezometric pressure. In turn, we can define the piezometric head, which is the p pizometric pressure divided by ρg, which is + z. ρg For a compressible fluid the law connecting ρ with p must be used. For instance, many gases obey the perfect gas law over a wide range of pressures and temperatures. That is pv = RT, where V is the volume per unit mass of gas at pressure p and absolute temperature T. Since ρ (mass per unit volume) is V -1 the gas law can be rewritten as: p = RT or ρ = ρ p RT (8) and substituted into Equation 4 gives dp dz or dp p g = p (9) RT g = dz (10) RT If the temperature is constant at all heights, an integration can be performed between the pressure limits p 1 and p, and height limits z 1 and z to give ln p p g = (11) ( z ) z1 1 RT Note that although the temperature does vary with elevation (at about 6.5 C to every 1000 m) errors would be small if the average temperature between z 1 and z were to be used. The Measurement of Pressure (i) Piezometer, Figure (a) (given in class) vertical transparent tube connected at the bottom end to the fluid under pressure and open at the top. The liquid rises up the tube to a height, h, when the pressure is said to be equal to a head h of the liquid. This device cannot be used if the fluid inside P is a gas; it also is not suitable if h is large (above m) as the 5
required length of the tube makes the device impractical; it also cannot be used if h is small (below about 7 cm) when the accuracy is low. (ii) The U-tube manometer, Figure (b) (given in class) This is a transparent tube having a bend filled with a fluid heavier than the fluid being measured. The fluid is water or alcohol if air is being gauged; it is mercury if water is being gauged. The difference of level, h, of the two surfaces of the heavier fluid gives the gauge pressure at the level of the lower surface B, that is, ρ 1 gh where ρ 1 is the density of the heavy fluid. The determination of the pressure at P will be treated in class. The class will also treat use of the manometer for measuring difference between two unknown pressures (Figure (c)). Modifications to the standard U-tube manometer (Refer to text book) (iii) Bordon Gauge (Refer to Text book) Figure (a). The piezometer (from class) 6
Figure (b). The Manometer (from class) Figure (c) Measurement of pressure differences with manometer (from class) 7
Hydrostatic Thrusts on Submerged Surfaces Relevance: Designing dams, storage tanks, gates, viewing galleries in storage tanks Required are: 1. force due to pressure of the fluid;. the direction of the resultant force; and 3. the position of the resultant. Some examples are given in class. We are going to develop expressions for determining forces, direction and location of the resultant forces on plane areas (eg., the circular viewing window in the aquarium) and on curved surfaces (such as a light shield at the side of a swimming pool). To do so, we take advantage of some properties of plane areas the first and second moments of area as they are needed in determining the forces and their resultant positions. Refer to the addendum on 1 st and nd Moment of rea. 1. Hydrostatic Thrusts on Plane Surfaces Refer to Figure 3 (to be completed in class) and note the two perpendicular axes with origin O. OX runs into the page, OY runs along the plane. Consider an elemental area δ at depth h below the free surface. Then the gauge pressure, p is: p = ρgh (1) So the force δf on this elemental area is δ F = pδ = ρghδ = ρgy sinθδ (13) Now fluid is at rest and therefore there is no movement relative to the plane. This implies that there are no shear forces. If there are no shear forces, the forces on the plane do not have any component parallel to it. That is to say that the forces act perpendicular to the plane. Since it is a plane surface, then all elemental forces are parallel to each other. 8
Figure 3 Free surface (atmosphere) θ O F δf X C ( x, y ) P (x, y ) δ O Edge view of plane View normal to plane Y 9
Total force F on one side of the plane is: F = gy sinθd = ρg sinθ yd ρ (14) Now recall that yd is the 1 st Moment of area about the x-axis, and is given by y, where is the total area and ( x,y ) is the position of the centroid C. Therefore, F = ρ g sinθy = ρgh (15) But ρgh is the pressure at the centroid. So whatever the slope of the plane, the total force exerted on it by the static fluid is given by the product of the area and the pressure at the centroid. Note that above applies for a fluid of uniform density in equilibrium. Centre of Pressure for a Plane Surface So far we have obtained an expression for determining the resultant force due to pressure on a plane surface. We have also reasoned that the direction of this resultant is perpendicular to the plane. However, we still do not know where this resultant acts is it at the centre of the plane, is it at the lowest elevation, is it at the top? Define the centre of pressure as the position at which the line of action of the force meets the plane. For a resultant force to be equivalent to all the individual forces, its moment about any axis must be the same as the sum of the moments of the individual forces about the axis. Taking moments about OX then the moment of δ is δ M = ρgy sinθδ (16) Let the centre of pressure be at P (x, y ). Then, Fy' = ρgy sinθd (17) 10
But recall that F = ρg sinθy (15) Therefore, y' = ρg sinθ y d ρg sinθy = ( k ) y OX (18) where (k ) OX is the nd moment of area about OX. K k + y Now ( ) ( ) OX = (19) C So, y' ( k ) + y ( k ) C C = = + (0) y y y Since a nd moment of area is always positive, then, y ' > y Example 1: sewer discharges to a river as shown in the figure 4 below. t the end of the sewer is a circular gate with a diameter (D) of 0.6 m. The gate is inclined at an angle of 45º to the water surface. The top edge of the gate is 1.0 m below the surface. Calculate: (a) (b) the resultant force on the gate caused by the water in the river; the vertical depth from the water surface to the centre of pressure. River 45 1.0 m Sewer Gate 0.6 m 11
Insert solution (presented in class) to sewer discharge example on this page. 1
Hydrostatic Thrusts on Curved Surfaces Unlike the case for plane surfaces, the forces pδ on elemental areas do not all act in the same direction, and so we cannot do a simple summation of all the elemental forces. The approach to defining the resultant force on a submerged curved surface involves examining the forces required to keep a body of fluid in contact with the surface in equilibrium. The surface exerts some resultant force on the fluid which, by Newton s Third Law, is exactly equal and opposite by the force exerted on the surface by the fluid. We attempt to find the horizontal and vertical components of this resultant force, after which we can find the resultant force and its direction. Refer to Figure 5 (given in class) The vertical solid wall at the left exerts horizontal forces on the fluid in contact with it in reaction to the forces due to the fluid pressure. The resultant force F 1 acts at a distance h/3 from the bottom of the wall. Horizontal Component, F H : The force F a on the right side of the upper part to a depth of h is equal to F 1 in magnitude and acts in the opposite direction. Then they have no effect on the curved surface. By summing forces in the horizontal direction, you can see that F H must be equal to F b acting on the lower part of the right side. The area on which F b acts is the projection of the curved surface onto a vertical plane. The magnitude of F b and its location can be found using the procedures developed for plane surfaces and given above. That is, F b = ρgh (1) where h is the depth to the centroid of the projected area. For the type of surface shown in the figure below, the projected area is a rectangle. Letting the height of the rectangle be L, you can see that h = h + L lso the area is LB where B is the width of the curved surface. Then, F = F = ρ g ( h L )LB () H + b 13
Figure 5 (given in class) 14
The location of F b is the centre of pressure of the projected area. gain, using the principles developed earlier, we get h' h + ( k ) C = (3) h 3 But for the rectangular projected area, ( k ) BL C = and = LB 1 L Then h' h + 1h = (4) Vertical component, F V : To find this component we sum the forces exerted by the fluid on the surface in the vertical direction. Only the weight of the fluid acts downward, and only the vertical component F v, acts upward. Then the weight of the fluid is simply ρgv, where V is the volume of the isolated fluid. The volume is the product of the cross-sectional area of the volume shown in the figure and the length of interest, B. So, F V = ρgb (5) The total resultant force F R is F + R = F F (6) H V The resultant force acts at an angle, φ, relative to the horizontal, found from φ = tan 1 (7) ( ) F V F H Forces on a Curved Surface with Fluid Below it Covered in the tutorial 15
ddendum to notes on hydrostatics and buoyancy First and Second Moments of rea First Moment of rea Consider the plane area in Figure 1 at the end of this section. The first moment of the elemental area δ about an axis in the plane is defined as the product of δ and its perpendicular distance from the axis. So, the expression is: xδ The 1 st Moment of the entire area is: Consider now an axis at x = k, xd 1 st Moment about this axis is ( x k ) d = xd k (1) If, k 1 = x = xd () then this first moment is zero. The axis at x is called the centroidal axis. You can repeat above, this time for the OX axis, to get: y 1 = yd (3) Intersection of the two axes occurs at the point known as the centroid. Note that this is an axis of symmetry since for every element on one side of the axis and contributing xδ to the total amount, there is an element on the other side contributing xδ. 16
Position of the centroid of Volume Using a similar approach to that for the plane area, we can specify a centroid of volume. If we consider an elemental volume δv being a perpendicular distance x from a plane yz, then as above, we can write: x 1 = xdv V V (4) Likewise, the centre of mass is given by: x 1 = xdm M M (5) Note that the position of the centre of mass coincides with the centre of gravity for bodies small compared with the earth. Such are the bodies we will encounter in this course. Second Moment of rea Definition: Second moment of area of the plane about the y-axis is: and about the x-axis, it is: x d y d Notation: (about the y-axis) ( ) k OY = xd (6) Dimensional formula: [L 4 ] Typical units: m 4 Parallel xis Theorem To find the nd Moment of area about a particular axis a distance x from a parallel axis. Consider the centroidal axis parallel to the given axis OY. nd Moment of area about the centroid is: 17
( K ) ( x x ) d = x d x xd + x (7) (8) (9) So, OY = = = ( K ) x( x ) + x OY ( K ) x OY ( K ) = ( k ) + x OY C (10) d To state the Parallel xis Theorem: Since OY was arbitrarily chosen, then it follows: The nd Moment of a plane area about any axis equals the sum of the nd moments about a parallel axis through the centroid and the product of the area of and the square of the perpendicular distance between the axes. See Table 1 below for expressions of nd moment of area for some regular figures. Perpendicular xis Theorem: Now by definition: ( K ) + ( k ) = ( x y ) d = r d OY OX + (11) where r is measured from an axis perpendicular to the plane of interest. So the perpendicular axis theorem is: The nd moment of a plane area about an axis meeting the plane perpendicularly at any point P is the sum of the second moments of that area about two axes in the plane that intersects perpendicularly at P. 18
Second Moment of Mass about a particular axis is: M z dm where z is the perpendicular distance of an element from the axis. If the mean value of z is k, then we can write the nd moment of mass as: Mk Recall that the nd moment of mass is also called the moment of inertia, in which k is known as the radius of gyration. 19
Figure 1 x-k Y k x x δ C r y y O X 0