Fluid Mechanics FLUID STATICS Pressure Variation with Elevation F 0 l dp dz Variation in pressure with elevation. Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Variation in pressure with elevation. • Hydrostatic equation dp dz From a vertical datum, pressure decreases as z-datum elevation increases. Pressure Variation in Uniform Density Assuming that the density (ρ) and specific weight () of a fluid are uniform through the fluid dp Integrate to get dz p z pz pz : Piezometric pressure p z constant Hydrostatic Equation Piezometric head p1 p2 z1 z2 Pressure Variation in Uniform Density p1 p2 z1 z2 Hydrostatic Equation applies only in a fluid with a constant specific weight. It applies to two point in the same fluid but not across an interface of two fluids having different specific weight. Example: What is the water pressure at a depth of 35 ft in the tank shown? Specific Weight γ = 62.4 lb/ft3 Example 3.3 (p. 37) Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Solution: p1 p2 z1 z2 0 + 250 = (P2/) + 215 35 ft = (P2/) P2 = 35 * 62.4 P2 = 2180 psfg = 15.2 psig Psig = pound force per square inch gage Example: Oil with a specific gravity of 0.80 forms a layer 0.90m deep in an open tank that is otherwise filled with water. The total depth of water and oil is 3 m. What is the gage pressure at the bottom of the tank. γ = 9810 N/m3 p2 = 0.90 x (0.8 x 9810) = 7.06 kPa p3 = 7.06 + 2.1 x 9,810 = 27.7 kPa Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Hydrostatic Equation p1 p2 z1 z2 P h Change in pressure between two points depends on the specific weight of the fluid, and the vertical distance between the two points. Pressure Variation with Elevation p0 p0 p0 p h1 h p0 1 p1=p0+1h1 p=h p=p0+h h2 2 p2=p1+2h2 h h p Pressure Measurements Barometer → to measure atmospheric pressure Absolute pressure pabsolute patm h Gage pressure ph Many pressure-measuring device measure not absolute pressure but only differences in pressure. Manometer • A pressure measuring method that utilizes the change in pressure with elevation to evaluate pressure. Piezometer (manometer ) attached to a pipe P h - Accurate & simple . - The problem is that a piezometer is impractical for high pressure and useless for gasses. If h is 10 m, what is the pressure at the centre of the pipe? U-tube Manometer General manometer equation p2 p1 i hi i hi down 1: Initial point index 2: Final point index Problem: water in pipe, mercury manometer liquid ( mercury specific weight =133 kN/m3) ∆h = 60 cm l =180 cm Find the pressure at the centre of the pipe ? Ans: 62.1 kPa up Example 3.7: Manometer Analysis Question: Pressure of the air? Given: l1 = 40 cm l2 = 100 cm l3 = 80 cm slide 14 Differential Manometers To measure the pressure difference btw two points in a pipe Here, the pressure difference between 1 and 2 is: P ( m f )h (this is for a horizontal pipe… z1 = z2) γm : the specific weight of the manometer liquid, γf : the specific weight of the fluid, Δh : the deflection of this liquid. Differential Manometers • Example 3.8: • Specific gravity of manometer fluid is 3. Δh = 5 cm Δz = 1 m y = 2 cm What is the pressure difference? What is the change in piezometric pressure? Solution: Piezometric pressure p z pz Piezometric difference P ( m f )h Applying the manometer equation between points 1 and 2: P2 P1 w (y h) mh w (y z2 z1 ) ( P2 w z2 ) ( P1 w z1 ) h( w m ) Change in piezometric pressure: Pz 2 Pz1 h( w ym ) slide 18 FLUID STATICS: Hydrostatic Force on Plane Surfaces Distribution of hydrostatic pressure on a plane surface Pressure on the differential area can be computed if the y distance to the point is known dF = p dA = ( y sin) dA Integrating the differential force over the entire area A Hydrostatic Force _ _ _ F y SinA p A Pressure at the centroid F Sin y dA Sin y A A Integral is the first moment of the area Hydrostatic Force Hydrostatic Force Terms • Δh: Vertical distance from centroid to the water surface (This distance determines the pressure at the centroid) • y: Inclined distance from water surface to the centroid • ycp: Inclined distance from water surface to centre of pressure • P¯: the pressure at the centroid Vertical Location of Line of Action of resultant Hydrostatic Force - Considering moments of the pressure about the horizontal axis 0-0: _ _ ycp y I slide 21 _ yA • • • • ycp = (inclined) distance to the centre of pressure y ¯= (inclined) distance to the centroid I ¯= area moment of inertia about horizontal axis passing the centriod A = surface area Restrictions: 1- One liquid involved 2- Gage pressure is zero at the liquid surface Lateral Location of Line of Action of resultant Hydrostatic Force - The same principles as above can be used for the lateral location - Starts with taking moments about a line normal to line 0-0 Review of Centroid & Area Moment of Inertia Example: 3.10 • An elliptical gate covers the end of a pipe 4m in diameter. If the gate is hinged at the top, what normal force F is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the atmosphere on the other side? Neglect the weight of gate. Resultant hydrostatic force: Fp p A hA (9810x10)ab (a, b: half of major and minor axes) Fp = 1.54 MN _ _ ycp y I _ yA 1 / 4a 3b y ab 0.125 m y¯(slant distance from surface to centroid): 12.5m Example cont’d Moment about the hinge. Moment arm for the hydrostatic force: 2.5 +0.125 = 2.625m M hinge 0 (1.541x106 N x 2.625m) - (F x 5 m) 0 F 809 kN Normal Force required to open gate Pressure Prism The volume called the pressure prism, that is a geometric representation of the hydrostatic force on a plane surface The resultant force must pass through the centroid of the pressure prism. Pressure Prism • An informative and useful graphical interpretation can be made for the force developed by a fluid acting on a plane area. • Consider the pressure distribution along a vertical wall of a tank of width b, which contains a liquid having a specific weight . • Since the pressure must vary linearly with depth, we can represent the variation as is shown in Figure below, where the pressure is equal to zero at the upper surface and reach to maximum at the bottom. • It is apparent from this diagram that the average pressure occurs at the depth h/2 and therefore the resultant force acting on the rectangular area (A = b h) is FLUID STATICS: Hydrostatic Forces on Curved surfaces Hydrostatic Forces on Curved Surfaces Find the magnitude and line of action of the hydrostatic force acting on surface AB Important Questions to Ask 1. What is the shape of the curve? 2. How deep is the curved surface? 3. Where does the curve intersect straight surfaces? 4. What is the radius of the curve? Hydrostatic Forces on Curved Surfaces A free-body diagram of a suitable volume of fluid can be used to determine the resultant force acting on a curved surface. Hydrostatic forces on Curved surfaces. Find the magnitude and line of action of the hydrostatic force acting on surface AB Forces acting on the fluid element 1. 2. 3. 4. FV : Force on the fluid element due to the weight of water above CB FH : Force on the fluid element due to horizontal hydrostatic forces on AC W : Weight of the water in fluid element ABC F : The force that counters all other forces - F has a horizontal component: Fx - F has a vertical component: Fy Hydrostatic forces on Curved surfaces Find the magnitude and line of action of the hydrostatic force acting on surface AB - Given: Surface AB with a width of 1 m Problem Solving Preparation 1. By inspection, curve is a ¼ circle. 2. The depth to the beginning of the curve (4 m depth to B) 3. The curve radius (2 m horizontal curve projection distance = curve radius) 4. Label relevant points: • BCDE is water above fluid element defined by the curve • ABC is the fluid element defined by the curve Example 3.11: Hydrostatic forces on Curved surfaces Find Fv, FH, W, Fx, Fy, F, Line of action for FH & Fv Given: Surface AB goes 1 m into the paper Fx= FH = (5 x 9810) (2 x 1) = 98.1 kN Pres. at the cenroid AC side area Fy= W + Fv Fv= 9810 x 4 x 2 x 1 = 78.5 kN W= γVABC= 9810 (1/4 x r2) 1 = 30.8 kN Fy= 78.5 + 30.8 = 109.3 kN The hydrostatic force acting on AB is equal and opposite to the force F shown The centroid of the quadrant Location of the resultant force Slide 33 FLUID STATICS: Buoyancy Buoyancy, Flotation & Stability Archimedes’ Principle The resultant fluid force acting on a body that is completely Submerged or floating in a fluid is called the buoyant force. Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Buoyancy: Floating Object Depends on submerged portion of the volume VD is the submerged volume Buoyant force FB VD where g is the specific weight of the fluid and VD is the volume of the body Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Example 3.12: Bouyant force on a metal part - Wood block (1) has dimensions of 10‐mm x 50 mm x 50 mm -Specific gravity of 0.3 - Metal object (2) has volume of 6600 mm3 – Find the tension in the cable and mass of object 2. Steps • Find the buoyant forces. • Find the weight of the block. • Perform force balances on both objects. Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Solution: