Discussion Prepared By: Dr.Khalil M. Al-Astal Eng.Ahmed S. Al-Agha Eng.Ruba M. Awad 2014-2015
Chapter (1) Fluids and their Properties
Fluids and their Properties Fluids (Liquids or gases) which a substance deforms continuously, or flows, when subjected to shearing forces. If a fluid is at rest, there are no shearing forces acting. In fluids we usually deal with continuous streams of fluid without beginning or end. Shear Stress in Moving Particles If fluid is in motion, shear stress are developed this occur if the fluid particles move relative to each other with different velocities. However, if the fluid velocity is the same at every point (fluid particles are at rest relative to each other), no shear stress will be produced. The following figure exhibit the velocity profile in a circular pipe: y v=f(y) Note that fluid next to the pipe wall has zero velocity (fluid sticks to wall), But if the fluid moved away from the wall, velocity increases to maximum. Change in velocity (v) with distance(y) is (velocity gradient): Velocity gradient = dv dy This also called (rate of shear strain) v Page (1) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Newton s Law of Viscosity: τ = μ dv dy (Pa = N/m2 ) τ = shear stress. μ = dynamic viscosity (will be discussed later). dv dy = velocity gradient Newtonian & Non-Newtonian Fluid Newtonian Fluids: Fluids obey (تتبع) Newton s law of viscosity are Newtonian fluids. For this type of fluids, there is a linear relationship between shear stress and the velocity gradient. Dynamic viscosity (μ) is the slope of the line. Dynamic viscosity (μ) is constant for a fluid at the same temperature. As temperature increase (μ) decreases slope decreases. Most common fluids are Newtonian, for example: Air, Water, Oil, etc The following graph explains the linear variation of shear stress with rate of shear strain (velocity gradient) for common fluids: Slope = μ Page (2) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Non-Newtonian Fluids: Fluids don t obey Newton s law of viscosity are Non-Newtonian fluids. For this type of fluids, there is no linear relationship between shear stress and the velocity gradient and the slope of the curves is varies. There are several types of Non-Newtonian fluids based on the relationship between shear stress and the velocity gradient. The general relationship is τ = A + B ( dv dy )n, the values of A, B, and n depends on the type of Non-Newtonians fluid. For Newtonian fluids A = 0.0, B = μ, n = 1 Newtonian fluids is a special case from the above equation. The following graph explains the variation of shear stress with rate of shear strain (velocity gradient) for Different types of Non-Newtonian Fluids: Page (3) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Properties of Fluids Density: There are many ways of expressing density: 1. Mass Density (ρ): Mass per unit volume ρ = M V (Kg/m3 ) Where, M = mass of fluid (Kg), V = volume of fluid (m 3 ) Typical values: Water= 1000 Kg/m 3, Mercury = 13546 Kg/m 3, Air = 1.23Kg/m 3 2. Specific Weight (γ): also called (unit weight): is the weight per unit volume. γ = ρ g = W V (N/m3 ) Where, W = weight of fluid(n) = M g, V = volume of fluid (m 3 ) Typical values: Water= 9814 N/m 3, Mercury = 132943 N/m 3, Air = 12.07Kg/m 3 3. Relative Density (σ): also called (specific gravity SG ): σ = SG = ρ substance ρ water at 4 (Unitless) Typical values: Water = 1, Mercury = 1.3 4. Specific Volume (ν): is the reciprocal of mass density. ν = 1 ρ (m3 /Kg) Page (4) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Viscosity: it s the ability of fluid to resist shear deformation due to cohesion and interaction between molecules of fluid. There are two types of viscosity: 1. Coefficient of Dynamic Viscosity (μ): τ = μ dv dy μ = τ Force Force = Area = Area dv Velocity Distance dy Distance Distance Time = Force Time Area But Force = Mass Lenght Time 2 Units: Kg m s = kg m 1 s 1 μ = Mass length Time Length 2 Time 2 = Mass Length Time Another Unit: N = kg m S 2 Kg = N S2 m Kg m S = N S2 m m S = N m 2. S = Pa. S Note that (μ) is often expressed in Poise (P) where 10P = 1Pa. S 2. Coefficient of Kinematic Viscosity (ν): ν = μ ρ Units: Kg m S Kg m 3 = Kg m3 = m2 m S Kg S Note that (ν) is often expressed in Stokes (St) where 10 4 St = 1 m2 Surface Tension(σ): is a force per unit length (N/m) pπr 2 S σ Page (5) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties R is the radius of the droplet, σ is the surface tension, p is the pressure difference between the inside and outside pressure. The force developed around the edge due to surface tension along the line: F σ = σ Perimeter = σ 2πR The force produced from the pressure difference p: F p = p Area = p πr 2 (Inside direction, because inside pressure largere thanoutside pressure) Now, these two forces are in equilibrium so we equating them: σ 2πR = p πr 2 p = 2σ R = p i p o Capillarity Capillary effect: is the rise or fall of a liquid in a small-diameter tube caused by surface tension. h: is the height of water rises. R: is the radius of the tube (d=2r). θ: is the angle of contact between liquid and solid 2πRσ = πdσ θ The weight of the fluid is balanced with the vertical force caused by surface tension. W = m g = ρ V g πdσcos(θ) = ρ π 4 d2 h g h = 4σ cos(θ) (only for tube) ρgd if any thing else you should derive the equation with the same concept (see p. 12) Notes: If θ < 90 water is rise, If θ > 90 water is fall, If θ = 90 no rise or fall (h = 0.0). If d is too small θ is too small and may be neglected. For clean glass in contact with Water θ = 0.0 (rise). For clean glass in contact with Mercury θ = 130 (fall). Page (6) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Compressibility & Bulk Modulus All materials are compressible under the application of an external force. The compressibility of a fluid is expressed by its bulk modulus (K) such Change in Pressure that: K = Volumetric Strain Change in Pressure = p = P final P initial Volumetric Strain = V final V V K = p V/V = V V The ve sign refer to the volume decrease as the pressure increase. K can be expressed by another form as following: K = ρ p ρ Large values of bulk modulus refer to incompressibility which indicates the large pressures are needed to compress volume slightly. Famous Unit Conversion 1m=0.3048 ft 1ft=12 inch 1inch=2.54 cm 1Ib = 4.45 N γ w = 9810 N m 3 g = 9.81 m S 2 Ib (0.3048)3 m 3 4.45N ft 0.3048m = 32.2ft/S2 To transform from rev rad to min S v ( m S ) = ω r 1slug = 14.59 Kg. ft 3 = 62.4 Ib/ft 3 multiply by 2π 60 rad where, ω = velocity in )ثلث حجم االسطوانة( Volume of cone (V cone ) = π 3 R2 h 1m 3 = 10 3 Liter, 1Liter = 10 3 ml, 1m 3 = 10 6 cm 3. S, r = radius of rotation (m) 1m 3 = 10 6 ml. Page (7) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Problems 1. If 9 m3 of oil weighs 70.5 KN, Calculate its: A. Specific Weight: γ = Weight Volume = 70.5 9 = 7.833 KN/m3 = 7.833 10 3 N/m 3. B. Density: γ(newton s) = ρ g ρ = γ 7.833 103 = g 9.81 C. Specific Volume: ν = 1 ρ = 1 798.47 = 1.25 10 3 m 3 /Kg. = 798.47 Kg/m 3. 2. A cylindrical tube of volume 335 ml is filled with soda. If the Mass of tube when it s filled by soda is 0.37 Kg and the weight of tube when it s empty is 0.155 N. Determine the specific weight, density, and specific gravity for soda. Solution γ soda = W soda V soda V soda = V tube = 335ml = 335 10 6 m 3 M soda+tube = 0.37Kg W soda+tube = M g = 0.37 9.81 = 3.63 N W tube = 0.153N W soda = W soda+tube W tube = 3.63 0.153 = 3.47 N 3.47 γ soda = 335 10 6 = 10358.2 N/m3. γ soda = ρ soda g ρ soda = γ soda g = 10358.2 9.81 S. G = ρ soda ρ water@4 = 1055.88 1000 = 1.0558. = 1055.88 Kg/m 3. Page (8) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties 3. A quantity of soda of mass M is filled in a container having volume V. The density of soda in this container is 1005 Kg/m 3. If the soda is distributed from this container to three similar containers having the same volume V. Calculate specific gravity and specific weight of the soda in each one of three containers after distribution occurs. Solution The mass of soda (M) is the same before and after distribution M 1 = M 2 (Solution Key) ρ = M V M = ρv ρ 1 V 1 = ρ 2 V 2 (ρ 1 = 1005, V 1 = V, ρ 2 =??, V 2 = 3V) 1005V = ρ 2 3V 1005 = 3ρ 2 ρ 2 = 335 Kg/m 3 ρ 2 = 335 Kg/m 3 (For soda in each container) S. G = ρ soda = 335 = 0.335 (For soda in each container). ρ water@4 1000 γ soda = ρ soda g = 335 9.81 = 3286.35 N/m 3. 4. A 10-kg block slides down a smooth inclined surface as shown in Figure below. Determine the terminal velocity of the block if the 0.1-mm gap between the block and the surface contains oil having dynamic viscosity of 0.38 Pa.s. Assume the velocity distribution in the gap is linear, and the area of the block in contact with the oil is 0.1 m 2. Page (9) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Solution Important Notes: In all problems, the direction of shear force is in reverse direction of movement, because it s the resistance of fluid against moving object. So the shear force resists the movement of the object exactly like friction force (if the object moves above rough surface). The summation of forces equal zero in all directions, because the velocity is constant (we assume it constant) and so the acceleration is zero. Return to the above problem, the free body diagram of the block is: Page (10) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties F x = 0.0 Wsin(20) = τa τ = Wsin(20) A W = Mg = 10 9.81 = 98.1 N, A = 0.1 m 2 (given) 98.1 sin (20) τ = = 335.5 Pa. 0.1 τ = μ v y v = τ y μ 335.5 0.1 10 3 = 0.38 = 0.0883 m/s. 5. A cylinder of 40 cm length and 10 cm diameter rotates about a vertical axis inside a fixed cylindrical tube of 105 mm diameter, and 0.4 m length. If the space between the tube and the cylinder is filled with liquid of dynamic viscosity of 0.2 N.s/m 2. Determine the external torque that led the cylinder to rotate by speed of 700 rev. /min. Solution Torque = Shear force radius of rotating cylinder Shear force (F) = τ Side Area of rotating cylinder (A s ) τ = μ du dy μ = 0.2N. s/m 2 du = velocity of rotating cylinder in (m/s). ω = 700rev/min = 700 2π 60 = 73.3 rad/s. v = ω radius of rotating cylinder = 73.3 0.1 2 v = 3.66 m/s. dy = ( always is: distance perpendicular to the direction of shear force) 105 100 dy = = 2.5mm = 0.0025m 2 Page (11) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
τ = 0.2 3.66 = 292.8 N/m2 0.0025 Fluids and their Properties 105 mm A s = πdl = π 0.1 0.4 = 0.1256 m 2 Shear force (F) = 292.8 0.1256 = 36.7 N. Torque = 36.7 0.05 = 1.838 N. m. 100 mm 6. A cylindrical body of 70 mm diameter and 150 mm length falls freely in a 80 mm diameter circular tube as shown in the figure below. If the space between the cylindrical body and the tube is filled with oil of viscosity 0.9 poise. Determine the weight of the body when it falls at a speed of 1.5 m/s. Page (12) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Solution The weight of cylindrical body (downward) equals the resist shear force (upward) due to the equilibrium. F = 0.0 W = τ A The shear stress caused by cylindrical tube is: τ = μ du dy μ = 0.9 pois, but 10pois = 1pa. s 0.9poise = 0.9 10 = 0.09 pa. s. du = velocity of freely falling = 1.5m/s. 80 70 dy = = 5mm = 0.005m 2 τ = 0.09 1.5 = 27 pa. 0.005 Shear force = side area of cylindrical body τ F shear = π d L τ = 3.14 0.07 0.15 27 F shear = 0.89 N The weight of cylindrical body = F shear = 0.89 N. 7. A movable plate of 0.5m 2 area (for each face) is located between two large fixed plates as shown in the figure below. Two different Newtonian fluids having the viscosities indicated are contained between the plates. Determine the magnitude of the force acting on the movable plate when it moves at a speed of 4.0 m/s. Page (13) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Solution The free body diagram of the plate is: τ 1 A τ 2 A A F F = 0.0 F = τ 1 A + τ 2 A F = A (τ 1 + τ 2 ) u 1 = u 2 = 4m/s. u 1 τ 1 = μ 1 = 0.01 4 = 13.33 Pa. y 1 0.003 u 2 τ 2 = μ 2 = 0.02 4 = 13.33 Pa. y 2 0.006 F = 0.5 (13.33 + 13.33) = 13.33 N. Page (14) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties 8. Fluid flow through a circular pipe is one-dimensional, and the velocity profile for laminar flow is given by u(r) = u max (1 r2 R2) where R is the radius of the pipe, r is the radial distance from the center of the pipe, and u max is the maximum flow velocity, which occurs at the center. Determine: (a). a relation for the drag force applied by the fluid on a section of the pipe of length L. (b). the value of the drag force for water flow at 20 C with R = 0.08m, L = 15m, u max = 3m/s, μ = 0.001kg/m. s. Solution a) The shear stress at the surface of the pipe (at r = R) is given by: τ = μ du (Negative sign is because u decresed with r increased) dr τ = μ d dr (u max (1 r2 R 2)) (But, u max is constant) τ = μu max d r2 (1 dr R 2) = μu max (0 2r R2) (because R is constant) τ = 2rμu max R 2 (at the surface of the pipe r = R) τ = 2μu max R. Page (15) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Note: Always we calculate the shear stress at the fixed surface (like pipe surface in the above problem) because it gives maximum shear stress. The drag force that causes shear stress on the pipe surface is: F = 0.0 F D = τ A (A = side area of tube = 2πRL) F D = 2μu max 2πRL R F D = 4πLμu max. b) F D = 4πLμu max F D = 4π 15 0.001 3 = 0.565 N. 9. A layer of water flows down an inclined fixed surface with the velocity profile shown in Figure below. Determine the magnitude of shearing stress that the water exerts on the fixed surface for U= 2m/s and h=0.1m μ = 1.12 10 3 pa. s Solution The shear stress at the fixed surface (at y = 0.0) is given by: τ = +μ du (Positive sign is because u incresed with y increased) dy u U = 2 y h y2 h 2 u = U (2 y h y2 h2) (But, U = 2m/s and h = 0.1m) Page (16) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties u = 2 ( 2y 0.1 y2 0.12) u = 40y 200y2 du dy = 40 400y (But, at the fixed surface y = 0.0) du dy = 40 τ = 1.12 10 3 40 = 44.8 10 3 N/m 2. 10. When a 2-mm-diameter tube is inserted into a liquid in an open tank, the liquid is observed to rise 10 mm above the free surface of the liquid. The contact angle between the liquid and the tube is zero, and the specific weight of the liquid is 1.2 x 10 4 N/m 3. Determine the value of the surface tension for this liquid. Solution For tube the capillary rise can be expressed by the relation tha we previously derived: 4σ cos(θ) h = σ = ρgd hρgd 4 cos(θ) h = 0.01m, ρg = γ = 1.2 10 4 N/m 3, d = 0.002m, θ = 0 σ = 0.01 1.2 104 0.002 4 cos(0) = 0.06 N/m. 11. Derive an expression for the capillary height change h for a fluid of surface tension σ and contact angle θ between two vertical parallel plates a distance W apart, as in figure. Page (17) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties Solution Assume the side length of each plate is (b). F = 0.0 F σ = Weight of water column F σ = 2 σ b cos(θ) (2 because there are two plates) Weight = M g = ρ V g (V = W h b) Weight = W h b ρ g 2 σ b cos(θ) = W h b ρ g h = 2σ cos(θ) ρgw. 12. Assume that the surface tension of 7.34x10-2 N/m act at an angle θ relative to the water surface as shown in Figure below. a. If the mass of the double-edge blade is 0.64 x 10-3 Kg, and the total length of its sides is 206 mm. Determine the value of θ required to maintain equilibrium between the blade weight and the resultant surface tension force. b. If the mass of the single-edge blade is 2.61 x 10-3 Kg, and the total length of its sides is 154 mm. Explain why this blade sinks. Support your answer with the necessary calculations. σ σ Solution Page (18) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties a. Note: The surface tension σ exists about all sides of the blade so to calculate the surface tension force F σ we should multiply the value of σ by the total length of blade sides as following: F σ = σ total length = 7.34 10 2 0.206 = 0.0152 N F σ = 0.0152 N with inclination of θ with horizontal The weight of the blade is: W = Mg = 0.64 10 3 9.81 = 6.278 10 3 N (downward) By Equilibrium F vertical = 0.0 0.0152 sin(θ) = 6.278 10 3 θ = 24.4. b. F σ = σ total length = 7.34 10 2 0.154 = 0.0113 N F σ,vertical = 0.0113 sin(θ) W = Mg = 2.61 10 3 9.81 = 0.0256N Now, the maximum value of sin(θ) = 1 so, the maximum value of F σ,vertical = 0.0113 N. Note that the value of W = 0.0256N (downward) is greater than the value of F σ,vertical = 0.0113 N so, the blade will sink downward. 13. A rigid cylinder (15 mm inside diameter) contains column of water 500mm length. If the bulk modulus of water is K water = 2.05 x 10 9 N/m 2. What will be the column length if a 2 KN force is applied it s end by frictionless plunger? Assume no leakage. Page (19) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Fluids and their Properties K = p V/V Solution K = 2.05 10 9 Pa. p = P f P i (P i = 0.0 because no applied forces in first case) P f = Force Area = 2 103 π 4 = 11.317 106 Pa. 0.0152 p = 11.317 10 6 0 = 11.317 10 6 Pa. V V = V f V = L A 0.5 A = L 0.5 V 0.5 A 0.5 K = p V V 2.05 10 9 = 11.317 106 L 0.5 0.5 L = 0.497 m. Page (20) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Chapter (2) Pressure and Head
Pressure and Head Pressure A fluid will exerts a normal force on any boundary it is in contact with, and these boundaries may be varies and the force may differ from place to place. So, it is convenient to work in terms of Pressure ( P ) which is the force per unit area. Force Pressure = Area over which the force is applied P = F A Unit: N m2 = pascal (Pa) Also there is a famous unit used for expressing pressure which is bar such that (1bar = 10 5 Pa). Pascal s Law for Pressure at a point This law applies to fluid at rest and it says Pressure at any point is the same in all directions The following figure clarifies the concept of Pascal s law: As shown in figure above, a tank contains water and the vertical pressure at depth (h) is P V and it s applied on the bottom wall of the tank. Also, the horizontal pressure at the same depth (i.e. same point) is P H and it s applied on the side wall of the tank. The value of P V equal the value of P H at the same point (same depth h) and the values of pressure in all directions are the same. (This is Pascal s Law) Page (22) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head Variation of Pressure Vertically in a Fluid under Gravity Consider the vertical cylinder immersed in fluid of mass density ρ as shown in figure above. The fluid is at rest and in equilibrium so, all forces in the vertical direction sum to zero. These forces are: Force due to pressure P 1 at the bottom of the cylinder at level Z 1 on area A and directed upward = P 1 A. Force due to pressure P 2 at the top of the cylinder at level Z 2 on area A and directed downward = P 2 A Force due to weight of element (cylinder) directed downward = mg But, m = ρv = ρ A (Z 2 Z 1 ) mg = ρ g A (Z 2 Z 1 ). From equilibrium F vertical = 0.0 P 1 A P 2 A mg = 0.0 P 1 A P 2 A ρ g A (Z 2 Z 1 ) = 0.0 (Divided by A) P 1 P 2 ρ g (Z 2 Z 1 ) = 0.0 P = P 1 P 2 = +ρg(z 2 Z 1 ) P 2 P 1 = ρg(z 2 Z 1 ) Page (23) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head From the above derivation we note the following: P 1 > P 2 because the column height of fluid at 1 is larger than at 2 So, as the depth (column height)increase, the pressure will also increase. Pressure difference ( P) always equals (ρgh) such that the distance h called (Pressure Head) and it s the vertical distance (difference in elevation with respect to a specific datum) between the two points that s we want to calculate the pressure difference between them. For example the pressure difference between points 2 and 3 is: P 2 P 3 = ρgl such that L = Z 3 Z 2 and so on. Equality of Pressure at the Same Level in a Static Fluid The fluid is at rest and in equilibrium so, all forces in the horizontal direction sum to zero. F horizontal = 0.0 P L A P R A = 0.0 P L A = P R A (Divided by A) P L = P R By another way, P L P R = P = ρgh (but h = 0) P L P R = 0 P L = P R Page (24) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head From the above derivation we note that the pressure difference between any points having the same elevation is the same but in two conditions: The flow is continuous through these points. The fluid is the same fluid at theses points (same mass density, ρ,). Note: If these two conditions are satisfied the pressure at any points at the same level is equal even if these points don t have the same area because we deal with pressure which not related to area. Note that the conditions are satisfied (continuous fluid and same fluid), so the pressure at points 1 and 2 is equal. But the forces at points 1 and 2 are not equal because the area at point 1 doesn t equal the area at point 2. Page (25) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head General Equation for Variation of Pressure in a Static Fluid (P + P). A θ P. P. A A The fluid is at rest and in equilibrium so, all forces in the all directions sum to zero. F oblique plane = 0.0 PA (P + P)A mg cos(θ) = 0.0 mg = ρvg = ρ. A. S. g PA (P + P)A ρ. A. S. g cos(θ) = 0.0 (Divided by A) P (P + P) ρ. g. S. cos(θ) = 0.0 But, S. cos(θ) = (Z 2 Z 1 ) = vertical distance (difference in elevation) P P P = ρ. g. (Z 2 Z 1 ) P = ρg(z 2 Z 1 ) ve Sign is because the pressure at elevation Z 1 is larger than the pressure at elevation Z 2. Important Note: For all types of surfaces (vertical, horizontal, and oblique) always the difference in pressure between any two points is measured by the relation ( P = ρgh) such that the distance (h) is the vertical distance between these two points. Page (26) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head Pressure and Head For a static fluid of constant density (ρ) the relation of vertical pressure can be expressed as: dp = ρg dp = ρg. dz by integrating both sides dz dp = ρg dz P = ρgz + constant As shown in figure above, distance z is measured from the free surface z = h P = ρgh + constant. The constant in the above equation is the pressure at the free surface of fluid which normally is atmospheric pressure P atmospheric P = ρgh + P atm Atmospheric Pressure: Is the pressure which we live constantly under it, and everything else exists under this pressure. So, it is convenient to take atmospheric pressure as the datum. Thus, we measure pressure above or below atmospheric pressure (i.e. atmospheric pressure is a datum). This measured pressure is known as gauge pressure for example For the above figure, the atmospheric pressure is at the free surface of fluid and the pressure measured from this surface (datum) is P = ρgh which is the gauge pressure(p gauge ). Gauge pressure can be positive (above atmospheric) or negative (lower the atmospheric). Also negative gauge pressure is known as vacuum pressure. Atmospheric pressure at any free surface exposed to atmosphere is zero gauge pressure (because the atmospheric pressure is a datum for gauge pressure). Page (27) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head Absolute Pressure: is the sum of gauge and atmospheric pressures and it is always positive. P absolute = P gauge + P atmospheric Notes: In most cases we deal with gauge pressure (i.e. taking atmospheric pressure as a datum having zero gauge pressure). By taking P atm = 0.0 P = ρgh if the density of fluid is known, we can measure pressure by head such that: Pressure head (h) = ρgh h (m). The Hydrostatic Paradox The pressure exerted by the fluid depends only on: Vertical head of fluid (h). Type of fluid (mass density, ρ, ) Not on the weight of the fluid present (because the weight is already considered in h, ρ, and g). Whatever the shape or size of the containers that contains the same fluid with same head (h), the pressure at the bottom of all containers is the same (this called Pascal s Paradox). But the force at the bottom of each container depends on the base area of the container. Pressure Measurement by Manometers Manometers are devices used to measure the pressure of fluid using the relationship between pressure and head. Different types of manometers will be discussed in problems of this chapter. Page (28) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head Problems 1. Calculate gauge pressure at point A if (ρ mercury = 13600 Kg/m 3 ) Solution Pressure exerted by metal cylinder: P metal cylinder = ρgh = 1.7 1000 9.81 0.025 = 416.925 Pa. Important Note: Since the density of air is very small, we can neglect the pressure exerted by air inside the tube. P 1 = P 2 = P 3 (solution key) 1 2 3 416.935 + 1000 9.81 0.05 + 13600 9.81 0.1 = P A P A = 14249 Pa = 14.249 kpa. Page (29) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 2. Calculate the gauge pressure difference between A and B. 1 2 P 1 = P 2 (Solution Key) Solution Distance from point (2) to gauge (B) = 1.8-1.5= 0.3m P A 1000 9.81 1 0.9 1000 9.81 0.8 +1000 9.81 0.3 = P B P A P B = 13,8930.2 Pa = 13.93 kpa. Page (30) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 3. In the figure shown, determine the reading of the gauge A. Solution Applying pressure equation from point (1) to gauge (A). 0 + (13.6 1000 9.81 0.7) (13.6 1000 9.81 0.5) (1000 9.81 2.7) (0.8 1000 9.81 2.4) = P A P A = 18,639 Pa. = 18.639 kpa. Negative sign means lower than atmospheric or (vacuum pressure) 1 Page (31) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 4. In the figure shown, the left reservoir contains carbon tetrachloride (SG=1.6) and it is closed and pressurized to 55 kpa. The right reservoir contains water and is open to atmosphere. Determine the depth of water h in the right reservoir. 0.9 m 55 kpa 1 0.1m 2 3 0.2 m 4 0.3 m SG=0.8 5 0.3 m Solution Important Notes: The best way to solve problems in this chapter is to make one equation starting and ending by points of known pressure. And you should always remember (the pressure for two points at the same level is the same, if the fluid is continuous) because this is the key for solving all problems in this chapter, for example P 1 = P 2 = P 3 (in figure) Also there is a general rule in solving all problems which is (if you directed downward you will sum the pressure, however if you directed upward you will subtract the pressure) because pressure increase with depth. Distance from surface to point (1) = 0.9 (0.2 + 0.3 + 0.3) = 0.1m 55 10 3 + 1.6 1000 9.81 (0.1) + 0.8 1000 9.81 0.2 1000 9.81 (h (0.3 + 0.3)) = 0.0 (P atm = zero gauge pressure) Solve the equation for h h = 6.52 m Page (32) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 5. A closed cylindrical tank filled with water has a hemispherical dome and is connected to an inverted piping system as shown in figure below. The liquid in the top part of the piping system has a specific gravity of 0.8, and the remaining parts of the system are filled with water. If the pressure gage reading at A is 60 kpa, determine: 1. The pressure at point B. 2. The pressure head at point C in term of mercury (ρ = 13600 Kg/m 3 ) 1 2 3 Solution 1. P B =??? (60 = P 1 = P 2 = P 3 ) solution key 60 10 3 + 0.8 1000 9.81 3 + 1000 9.81 2 = P B P B = 103164 Pa = 103.164 kpa. 2. h C (in term of mercury) =??? 60 10 3 1000 9.81 3 = P C P C = 30570 Pa = 30.57 kpa. h C (in terms of mercury) = P C ρ mercury g = 30570 = 0.229 m 13600 9.81 Page (33) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 6. An open tube is attached to a tank shown in figure below, if the water rises to a height of 800 mm in the tube, what are the pressures P A and P B of the air above the water? 1 2 Solution To calculate P B, apply the pressure equation from point (1) to point (B): 0 + 1000 9.81 0.8 1000 9.81 0.3 = P B P B = 4905 Pa = 4.905 kpa. To calculate P A, apply the pressure equation from point (B) to point (A): P B = P 2 = 4905 Pa (because same fluid and continuous) 4905 1000 9.81 0.1 = P A P A = 3924 Pa = 3.924 kpa. Page (34) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 7. Calculate the gauge pressure difference between A and B. 2 3 2 4 2 1 2 2 Solution Note that the pressure at A does not equal the pressure at B although the two points have the same elevation, but the fluid does not continuous between them. P 1 = P 2 and P 3 = P 4 (Solution Key) P A 0.91 1000 9.81 y 13.6 1000 9.81 0.1 +0.91 1000 9.81 (0.1 + y) = P B P A 8927.1y 13341.6 + 892.71 + 8927.1 = P B P A P B = 13341.6 892.71 = 12448.89 Pa = 12.448 kpa. Page (35) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 8. For the figure shown below, if the pressure reading at point A is 4.2 kpa. Calculate the pressure at point B. 7.5 cm 20 cm 7.5 cm Solution As we discussed previously, always we deal with vertical distances between the points that s we want to calculate the pressure between them. Vertical distance for the fluid of SG=2.6 = 0.2 sin(30) = 0.1m. 4.2 10 3 + 1000 9.81 0.075 2.6 1000 9.81 0.1 P A 1000 9.81 0.075 = P B P B = 1649.4 Pa = 1.649 kpa. Page (36) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 9. The cylindrical tank with hemispherical ends shown in figure below contains a volatile liquid and its vapor. The liquid density is 800 Kg/m 3 and its vapor density is negligible. The pressure in the vapor is 120 kpa (abs), and the atmospheric pressure is 101 kpa. Determine: (a). the gage pressure reading on the pressure gage. (b). the height, h, of the mercury manometer. Solution (a). P abs = P gauge + P atm P gauge = P abs P atm For vapor: 1 2 3 4 P gauge,vapor = 120 101 = 19 kpa. To find gage reading: 19 10 3 + 800 9.81 1 = P gage P gage = 26848 Pa = 26.848 kpa. (b). P gage = P 1 = P 2 = P 3 = P 4 = 26848 Pa. 26848 13600 9.81 h = 0.0 (zero gauge pressure) h = 0.2012 m. Page (37) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 10. Determine the elevation difference ( h) between the water levels in the two open tanks shown in figure below. 3 4 d 1 2 Solution P 1 = P 2 = 0.0 and P 3 = P 4 (Solution Key) 0.0 1000 9.81 (d + 0.4) + 0.9 1000 9.81 0.4 +1000 9.81 d + 1000 9.81 h = 0.0 9810 d 3924 + 3531.6 + 9810 d + 9810 h = 0.0 h = 0.04 m. Page (38) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 11. A piston having a cross-sectional area of 0.07m 2 is located in a cylinder containing water as shown in figure below. What is the value of the applied force?, F, acting on the piston? The weight of the piston is negligible. F 60 mm 100 mm 1 2 3 4 Solution The force applied on the piston exerts pressure on the piston and this pressure will transfer to water. P piston = F Area = P 1 = P 2 = P 3 = P 4 F 0.07 = 14.28 F 14.28 F + 1000 9.81 0.06 13600 9.81 0.1 = 0.0 F = 893.067 N. Page (39) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 12. If the absolute pressure of air is 76 kpa and the atmospheric pressure is 100 kpa. Determine the specific gravity (SG 2 ) for the right fluid. P gauge,air = 76 100 ( 24 10 3 = 24 10 3 Pa +13.55 1000 9.81 0.22 Solution SG 2 1000 9.81 0.4 = 0.0) SG 2 = 1.34. 22cm Page (40) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Pressure and Head 13. A gas is contained in a vertical, frictionless piston cylinder device. The piston has a mass of 4 kg and a cross sectional area of 35 cm2. A compressed spring above the piston exerts a force of 60 N on the piston. If the atmospheric pressure is 95 kpa, determine the pressure inside the cylinder. Solution The pressure inside the cylinder equals the total applied pressure by piston weight and spring force in addition to atmospheric pressure to get absolute pressure. P abs = 60 + (4 9.81) 35 10 4 + 95 10 3 P abs = 123354.28 Pa = 123.35 KPa. If we want to calculate gauge pressure: P gauge = P abs P atm P gauge = 123.35 95 = 28.35 kpa. Page (41) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Chapter (3) Static Force on Surfaces (Buoyancy)
Static Forces on Surfaces-Buoyancy Introduction When a surface is submerged in a fluid, forces develop on the surface due to the fluid. The determination of these forces is important in the design of storage tanks, ships, dams, and other hydraulic structures. For fluids at rest we know that the force must be perpendicular to the surface since there are no shearing stresses present. We also know that (from previous chapter) the pressure will vary linearly with depth as shown in figures below. The following figures exhibits the pressure distribution and the resultant force of this pressure for different planes: Note that the resultant force(f R ) always acting on the center of pressure (c.p) of the surface. Page (43) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Resultant Force and Center of Pressure on a Plane Surface Immersed in a Liquid (General Case) D y da c.p c.g 1. Determining the resultant force (F R ): F R = df but, df = dp da and dp = ρg h F R = ρg h da but, ρg is constant F R = ρg h da The term h da is the first moment of area about the free surface h da = Object Area Distance from centroide to the free surface h da = A y Page (44) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy F R = ρgay 2. Determining the location of resultant force: The resultant force act perpendicular to the immersed surface at the center of pressure (c.p) at some depth (D) as shown in figure above, the following is the derivation of the formula that used to calculate (D): Moment about free surface = 0.0 What is moment!!! Moment = Force Arm Arm = is the distance perpendicular to the force Moment about free surface = 0.0 F R y R = df y but, df = dp da and dp = ρg h F R y R = ρg h da y but h = y sin(θ) F R y R = ρg y sin(θ) da y but, ρg sin(θ) is constant F R y R = ρg sin(θ) y 2 da The term y 2 da is the second moment of area about tha free surface and it is also called (moment of inertia "I") F R y R = ρg sin(θ) I o but F R = ρgy A (ρgy A) y R = ρg sin(θ) I o A y y R = I o sin(θ) y R = I o sin(θ) A y D = y R sin(θ) D = sin 2 (θ) I o = I cg + Ad 2 I o A y (Parallel axis theorem) I cg = Moment of inertia about the center of gravity of the object A = Area of the object Page (45) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy d = distance from center of gravity to point O d = y c (as shown in figure above) d = y c = y Substitute in D sin(θ) 2 y I cg + A ( D = sin 2 sin(θ) ) (θ) A y [ ] D = sin 2 (θ) I cg y 2 A y + sin2 (θ) A A y sin 2 (θ) D = sin 2 (θ) I cg A y + y I advise you to use this equation when calculating (D). The following figure shows the moment of inertia (I cg ) for different shapes: Page (46) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Pressure Diagram For vertical or inclined walls of constant width it is possible to find the resultant force and the location of center of pressure using a Pressure Diagram. Pressure Diagram: Is the graphical representation of the gauge pressure change with vertical depth along the wall and this change is increasing linearly from zero (at free surface) to the maximum at the base of the wall. As we know, the relation of calculating gauge pressure at any depth is: P = ρgh How to calculate resultant force and center of pressure using pressure diagram: Case I: Always, resultant force equals the volume of pressure diagram: F R = volume of pressure diagram F R = 1 2 γh h b = 1 2 γ h2 b (N) (γ = ρg) If the width (b) is not known, we can represent the resultant force as force per unit area: F R = 1 2 γ h2 (N/m) Page (47) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Location of F R : In the pressure diagram above is a triangle and the resultant force will be in the center of this triangle: D = h h 3 = 2h 3 Case II: Note that the object is fully immersed in fluid at distance h 1 from the free surface. Assume the width (b) is 1m inside the page: F R = F 1 + F 2 Let the vertical length of the object is L = (h 2 h 1 ) F 1 = Volume of the rectangle = γ h 1 L 1 F 2 = Volume of the triangle = 1 2 γ L L 1 Location of (F R ): D = h 1 + y A h 1 is given, but how we can find y A Take Moment about point A: F 1 at the center of rectangle ( L ) from point A 2 Page (48) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy F 2 at the center of Triangle ( 2L ) from point A 3 F R y A = F 1 ( L 2 ) + F 2 ( 2L 3 ) y A = D = h 1 + y A = Case III: If the object is the same at the second case above, but is inclined in angle (θ) with the horizontal, do the same in case two, but in finding the location of center of pressure, find the inclined distance firstly (by taking moment about a specified point ) and then calculate the vertical distance. Case IV: If there are two fluids or more, the pressure diagram will be as following: F R = F 1 + F 2 + F 3 Assume the width (b) is 1m inside the page F 1 = Volume of upper triangle = 1 2 ρ oil g h 1 h 1 1 F 2 = Volume of rectangle = ρ oil g h 1 h 2 1 F 3 = Volume of lower triangle = 1 2 ρ w g h 2 h 2 1 Page (49) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy To find the location D, take moment about the free surface: F R D = F 1 2h 1 3 + F 2 (h 1 + h 2 2 ) + F 3 (h 1 + 2h 2 3 ) D = Important Note: Before solving any problem, if the width of the object is uniform (i.e. the object is rectangle), it is preferable to use pressure diagram method. However, if the object is not rectangle (width is not uniform) like triangle, trapezoidal and circle you are strongly advise to use equations derived in the beginning of this chapter. Problems 1. An open tank has a vertical partition and on one side contains gasoline with a density 700 kg/m 3 at a depth of 4 m, as shown in the figure below. A rectangular gate that is 4m high and 2 m wide is hinged at one end and is located in the partition. Water is slowly added to the empty side of the tank. At what depth (h) will the gate start to open? Solution Page (50) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy What is Hinge??? Is a type of supports that prevent objects to move in two perpendicular directions (usually horizontal and vertical directions) so, this support having two reactions against movement as shown in the following figure: As shown the hinge support can t resist any moments, thus we take summation moments about hinge in most cases to avoid reactions of hinge and because moment at hinge is zero. Because the width is uniform, we can solve the problem using pressure diagram as following: the free body diagram for the gate is shown below: Width = 2m F 1 = Volume of right triangle = 1 (700 9.81 4) 4 2 = 109872 N 2 F 2 = Volume of left triangle = 1 2 (1000 9.81 h) h 2 = 9810h2 N M @hinge = 0.0 109872 4 3 = 9810h2 h 3 h3 = 44.8 h = 3.55m. Page (51) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 2. The 4-m-diameter circular gate in the shown figure below is located in the inclined wall of a large reservoir containing water. The gate is mounted on a shaft along its horizontal diameter, and the water depth is 10 m above the shaft. Determine: a) the magnitude and location of the resultant force exerted on the gate by the water. b) the moment that would have to be applied to the shaft to open the gate. Solution Since the object is circle, (width is not constant), we use equations. The pressure distribution on the gate is shown in the following figure: h 1 = 10 2 sin(60) = 8.268 m h 2 = 10 + 2 sin(60) = 11.732 m From Equation: F R = ρgay A = π 4 D2 = π 4 42 = 12.56 m 2, y = 10m F R = 1000 9.81 12.56 10 = 1232136 N. Page (52) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Find the location of resultant force from the free surface: I cg D = sin 2 (θ) A y + y θ = 60, A = 12.56 m 2, y = 10m I cg (for circle) = π 4 R4 = π 4 24 = 12.56m 4 D = sin 2 (60) 12.56 + 10 = 10.075 m. 12.56 10 To find the moment about the shaft we want to find the distance between centroid and center of pressure: Inclined distance = (D y ) sin 60 = 10.075 10 sin 60 = 0.0866 m M about shaft = F R 0.0866 = 1232136 0.0866 = 106702.97 N. m. 3. Two sides of the container are filled with two different liquids. Find the resultant force exerted on the gate AB and the position of center of pressure. Page (53) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Solution Since the object is not rectangle, (width is not constant), we use equations. The stress distribution along the gate AB is shown in figure below: It is clear that the force F 1 is larger than force F 2 F R = F 1 F 2 F 1 = ρ 1 gay 1 ρ 1 g = γ 1 = 9810 N/m 3 A = Area of the gate A = (rectangle area + triangle area ) A = 3 3 + 1 3 1 = 10.5 m2 2 To calculate y, firstly we should calculate the distance (h) from the centroid of the gate to point A by taking moment of area about point A. A h = A rectangle h rectangle + A triangle h triangle h rectangle = distance from center of rectangle to point A h triangle = distance from center of triangle to point A 10.5 h = (3 3) 1.5 + ( 1 2 3 1) (3 + 1 1) h = 1.76 m 3 y 1 = 1.76 + 5 = 6.76 m F 1 = 9810 10.5 6.76 = 696313.8 N F 2 = ρ 2 gay 2 ρ 2 g = γ 2 = 7847 N/m 3, y 2 = 1.76 + 2 = 3.76 m F 2 = 7847 10.5 3.76 = 309799.56 N F R = 696313.8 309799.56 = 386514.24 N. To calculate the location of center of pressure firstly we should calculate the values of D 1 and D 2 for each force: I cg D 1 = sin 2 (θ) + y 1 A y 1 θ = 90, A = 10.5 m 2, y 1 = 6.76 m, I cg =??? Now we want to calculate the moment of inertia about the centroid of the gate. Page (54) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy I cg = (I rectangle + A rectangle d 2 ) + (I triangle + A triangle d 2 ) I cg = 3 33 12 + 3 13 36 I cg = 11.14 m 4 + (3 3) 0.262 + (0.5 3 1) 1.572 D 1 = sin 2 (90) 11.14 + 6.76 = 6.9 m 10.5 6.76 I cg D 2 = sin 2 (θ) + y 2 y 2 = 3.76 m A y 2 D 1 = sin 2 (90) 11.14 + 3.76 = 4.042 m 10.5 3.76 Now to find the location of center of pressure from point A, take the moment about point A. Distance from F 1 to point A = 6.9 5 = 1.9 m Distance from F 2 to point A = 4.042 2 = 2.042 m M @hinge = 0.0 F R D c.p = F 1 1.9 F 2 2.042 386514.24 D c.p = 696313.8 1.9 309799.56 2.042 D c.p = 1.78 m from point A. Page (55) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 4. Calculate the minimum force F to keep the cover AB closed (the cover is rectangular with 1.5m wide). Solution Important Note: In problems similar to this problem, if you are not given any information related to weight of the object, neglect the weight of the object, but if the weight is given you must consider it. Since the cover is rectangular, we can use pressure diagram method. Firstly we want to calculate the pressure at points A and B using principals that we learned in chapter (2). P A = 1.6 1000 9.81 1.85 0.825 1000 9.81 0.86 = 22077.4 Pa. P B = 22077.4 0.825 1000 9.81 1.04 = 13660.42 Pa. Now we can draw the free body diagram for the cover: Page (56) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Note that if we want to calculate the resultant force (F R ): F R = F 1 + F 2 But the required is (F) so we divide the pressure diagram into rectangle (F 1 ) and triangle (F 1 ) to find the location of each force from the hinge easily. Now, calculate the length of the cover: L = 0.6 2 + 1.04 2 = 1.2 m The distance (perpendicular to F 1 ) to the hinge is 1.2 = 0.6 m. 2 The distance (perpendicular to F 2 ) to the hinge is 2 1.2 3 = 0.8 m. The distance (perpendicular to F) to the hinge is 0.3m (given). F 1 = Volume of rectangle = area of rectangle cover width = 13660.4 1.2 1.5 = 24588.72 N. F 2 = Volume of triangle = area of traingle cover width = 1 (22077.4 13660.4) 1.2 1.5 = 7575.3 N. 2 To calculate force F, take summation moments about hinge: M @hinge = 0.0 F 0.3 = 24588.72 0.6 + 7575.3 0.8 F = 69378.24 N. Page (57) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 5. For the system shown, water tank is pressurized to 90 cm Hg and 35 kpa.air pressure. Determine the resultant hydrostatic force per meter width on panel AB. Solution Since the required is force per meter length assume the width is constant and it is equal 1m, so we can solve the problem using pressure diagram method. To draw the pressure diagram along panel AB we must calculate the pressure at points A and B from both sides (Right and Left). From right side, the fluid is air and as we know the air pressure is constant so the pressure at A equal the pressure at B = 35000 Pa. From left side, we calculate the pressure at points A and B as we learned in Chapter 2: P A = 13600 9.81 0.9 + 1000 9.81 5 = 169124.4 Pa. P B = 169124.4 + 1000 9.81 3 = 198554.4 Pa. The free body diagram for panel AB is shown below: Page (58) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy F R = F 1 + F 2 F 3 F 1 = volume of left rectangle F 1 = 169124.4 3 1 F 1 = 507373.2 N/m F 2 = volume of triangle F 2 = 1 2 (198554.4 169124.4) 3 1 F 2 = 44145 N/m F 3 = volume of left rectangle F 3 = 35000 3 1 = 105000 N/m F R = 507373.2 + 44145 105000 = 446518.2 N = 446.52 kn. Page (59) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 6. A 1.2 m wide gate of negligible weight shown in the figure pivots about the frictionless hinge O. the gate is held in position shown by the 1000 kg counter mass M. determine the water depth h. 1m 0.6m Width = 1.2 m M = 1000 kg Solution Since the gate having constant width (1.2m), we can use pressure diagram method. P free surface = 0.0 P O = 1000 9.81 h = 9810 h The free body diagram of the system is shown below: Page (60) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy F R = Volume of triangle F R = 1 9810h h 1.2 = 5886 h2 2 To calculate the value of (h) we take summation moment at hinge (O): M @O = 0.0 5886 h 2 h = (1000 9.81) 1 3 1962 h 3 = 9810 h = 1.71 m. 7. The gate in figure below is 1.5 m wide, is hinged at point B, and rests against a smooth wall at point A. Compute: (a) The force on the gate due to water pressure. (b) The horizontal force P exerted by the wall at point A. (c) The reactions at the hinge B. Water 4.5m 1.8 m 2.4m Solution Since the width of the gate is constant (1.5m), we can use pressure diagram method. Page (61) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy P A = 0 + 1000 9.81 (4.5 1.8) = 26487 Pa. P B = 26487 + 1000 9.81 1.8 = 44145 Pa. The weight of the gate is not given, so we neglect it. The free body diagram of the gate is shown below: Length of the gate (L) = 1.8 2 + 2.4 2 = 3m F R = F 1 + F 2 F 1 = volume of the rectangle = (26487 3) 1.5 = 119191.5 N F 2 = volume of the triangle F 2 = ( 1 (44145 26487) 3) 1.5 = 39730.5 N 2 F R = 119191.5 + 39730.5 = 158922 N. To calculate the force (P) take the moment at hinge (B): M @B = 0.0 P 1.8 = 119191.5 1.5 + 39730.5 1 P = 121398.75 N. F x = 0.0 B X + F 1 sin(θ) + F 2 sin(θ) = P sin(θ) = 1.8 3 = 0.6 B X + 119191.5 0.6 + 39730.5 0.6 = 121398.75 B X = 26045.5 N. Page (62) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy F y = 0.0 B y = F 1 cos(θ) + F 2 cos(θ) cos(θ) = 2.4 3 = 0.8 B y = 119191.5 0.8 + 39730.5 0.8 = 127137.6 N. Page (63) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Force on a Curved Surface due to Hydrostatic Pressure If the surface is curved, the forces on each element of the surface will not be parallel (normal to the surface at each point) and must be combined using some vectorial method. The most significant method to solve these types of problems is to calculate the vertical and horizontal components, and then combine these two forces to obtain the resultant force and its direction. There are two cases: Case I: if the fluid is above the curved surface: Horizontal Component (R h ): The resultant horizontal force of a fluid above a curved surface is: R h = Resultant force on the projection of the curved surface onto a vertical plane (i.e. along line AC in the above figure). We know that the force must be normal to the plane, so if we take the vertical plane, the force will act horizontally through the center of pressure of the projected vertical plane as shown in figure below, and we can use pressure diagram method. Page (64) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy R h cp Vertical Component (R v ): Because the fluid is at rest, there are no shear forces on the vertical edges, so the vertical component can only be due to the weight of the fluid. The resultant vertical force of a fluid above a curved surface is: R v = Weight of fluid directly above the curved surface and will act vertically downward through the center of gravity of the mass of fluid as shown in figure below. R v = Weight of fluid above the curved surface = ρgv = γv Page (65) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Resultant Force (R): The overall resultant force is found by combining the vertical and horizontal components vectorialy: R = R h 2 + R v 2 This resultant force acts through point O at an angle (θ) with R h The position of O is the point of intersection of the horizontal line of action of R h and the vertical line of action of R v as shown in figure below. θ = tan 1 ( R v R h ) Case II: if the fluid is below the curved surface: Page (66) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy The calculation of horizontal force R h is the same as case I, but calculation of vertical force R v will differ from case I. Vertical force component in case of fluid below curved surface: If the curved surface AB is removed, the area ABDE will replaced by the fluid and the whole system would be in equilibrium. Thus the force required by the curved surface to maintain equilibrium is equal to that force which the fluid above the surface would exert (weight of fluid above the curved surface). I.e. The resultant vertical force of a fluid below a curved surface is: R v = Weight of the imaginary volume of fluid vertically above the curved surface. Problems 1. The gate shown in figure is 5.4 m wide and it is a quarter circle hinged at H. Determine the horizontal force P required to hold the gate in place. Solution 1. Calculate the horizontal component of hydrostatic force (R h ): As stated above, the value R h is calculated on the projection on vertical plane as following: R h = Volume of pressure diagram = Area width R h = 1 (1000 9.81 1.8) 1.8 5.4 2 Page (67) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy R h = 65817.88 N 2. Calculate the vertical component of hydrostatic force (R v ): R v = Weight of fluid above the curved surface = ρgv V = Area of quarter circle Width V = π 1.82 5.4 = 13.74 m 3 4 R v = 1000 9.81 13.74 = 134789.4 N R v acts at distance 4R 3π = 4 1.8 = 0.764m from the circle center 3 π The weight of the gate is not given, so neglect it. The free body diagram of the whole system is: By taking summation moment about the hinge: M @hinge = 0.0 P 1.8 = R h 0.6 + R v 0.764 P 1.8 = 65817.88 0.6 + 134789.4 0.764 P = 79149.9 N. Page (68) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 2. The gate shown in figure below consists of a quarter of a circular cylinder and is used to maintain a water depth of 4 m. Determine the weight of the gate per meter of length. Solution Assume the width of whole system is 1m 1. Calculate the horizontal component of hydrostatic force (R h ): R h = Volume of pressure diagram = Area width R h = F 1 + F 2 F 1 = volume of rectangle F 1 = (1000 9.81 3) 1 1 F 1 = 29430 N F 2 = volume of triangle F 2 = 1 [1000 9.81 (4 3)] 1 1 2 F 2 = 4905 N 3. Calculate the vertical component of hydrostatic force (R v ): R v = Weight of fluid above the curved surface = ρgv Note that there is no water above the gate, so the value of R v equals the weight of imaginary fluid above the gate as shown below: Page (69) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy The most important note that the imaginary fluid is above the gate to reach free surface w = w 1 + w 2 w 1 = γ rectangle volume w 1 = 9810 (3 1) 1 = 29430 N w 2 = γ quarter circle volume w 2 = 9810 π 12 4 1 = 7704.7 N R v = w 1 + w 2 R v = 29430 + 7704.7 = 37134.7 N. Location of R v : w 1 located at the center of rectangle at distance ( 1 ) from hinge 2 w 2 located at the center of quarter circle at distance (1 4R ) from hinge 3π = 1 4 1 3 π = 0.576 m To find the location of R v take the moment about hinge: R v x = w 1 0.5 + w 2 0.576 37134.7 x = 29430 0.5 + 7704.7 0.576 x = 0.516 m Now, draw the free body diagram for the entire system: 0.576 0.516 Page (70) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy The weight of the gate is subjected on the center of the gate which is a quarter circle (i.e. at distance 0.576 from hinge) By taking summation moment about the hinge: M @hinge = 0.0 w 0.576 = F 1 0.5 + F 2 0.67 + R v 0.516 w 0.576 = 29430 0.5 + 4905 0.67 + 37134.7 0.516 w = 64490.48 N. 3. The hemispherical dome in below weighs 30 kn and is filled with water and attached to the floor by six equally spaced bolts. What is the force in each bolt required to hold down the dome? Solution The hemispherical dome tries to move up due to the vertical component of hydrostatic pressure (R v ), but the bolts and the weight of the dome resist this movement downward. So for equilibrium, summation of vertical forces should be equal zero: F vertical = 0.0 R v = F bolts + F dome (1) F dome = 30000N Page (71) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Calculation of (R v ): Note that the fluid is below the curved surface, so to calculate R v we should calculate the imaginary weight of water above the dome till reaching the free surface as shown in figure: To maintain equilibrium, the value of R v upward, should equal the imaginary weight of fluid downward (weight of hatch volume). W hatch = V hatch ρg V hatch = V total V dome+tube V total = cylinder volume V total = π 4 42 6 = 75.4 m 3 V dome+tube = V dome + V tube V dome = hemisphere volume V dome = 2π 3 R3 = 2π 3 23 V dome = 16.75 m 3 V tube = π 4 0.032 4 = 0.00283 m 3 V hatch = 75.4 (16.75 + 0.00283) = 58.65 m 3 W hatch = 58.65 9810 = 575356.5 N R v = W hatch = 575356.5 Substitute in Eq.(1): 575356.5 = F bolts + 30000 F bolts = 545356.5 N Force in each bolt = F bolts 6 = 545356.5 6 = 90892.75 N = 90.89kN. Page (72) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Buoyancy When a body is submerged or floating in a static fluid, the resultant force exerted on it by the fluid is called the buoyancy force. Buoyancy Force= weight of fluid displaced by the body and this force will act vertically upward through the centroid of the volume of fluid displaced, known as the center of buoyancy. Archimedes principle Archimedes Principle states that the buoyant force has a magnitude equal to the weight of the fluid displaced by the body and is directed vertically upward. F b = ρ fluid g V displaced by body (Upward ) Problems 1. A 1-m-diameter cylindrical mass, M, is connected to a 2-m-wide rectangular gate as shown in figure below. Determine the required value for M if h = 2.5m. Solution Firstly, we calculate the tension in the cable. The free body diagram for the gate is shown below: Page (73) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy F R = Volume of triangle F R = 1 (9810 2.5) 2.5 2 = 61312.5 N 2 M @hinge = 0.0 T 4 = F R 0.83 T 4 = 61312.5 0.83 T = 12722.34 N Now we take the free body diagram for the cylindrical mass: The force F b is buoyant force F b = ρ fluid g V displaced by body V displaced by body = volume of object immersed in water V displaced by body = π 4 12 (2.5 1) = 1.178 m 3 F b = 1000 9.81 1.178 = 11557.16 N F Y = 0.0 T + F b = Mg Mg = 12722.34 + 11557.16 = 24279.5 N M = 24279.5 9.81 = 2475 kg. Page (74) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 2. A wooden rod weighting 25N is hinged at one end, the rod is 5m long and uniform cross section, and the support is 3m below the free surface. At what angle θ will it come to rest when it is allowed to drop from a vertical position? The cross section of the rod is 10 cm 2. Solution Let X is the length of rod immersed in water, the free body diagram of the rod is shown below: Page (75) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy From the above figure, we note the following: The weight of the rod (25N) act at the centroid of rod (i.e. at distance 2.5 from hinge). The buoyant Force(F b ) act at the centroid of the immersed volume of the rod (i.e. at distance X/2 from the hinge). F b = ρ fluid g V displaced by body V displaced by body = volume of object immersed in water V displaced by body = 10 10 4 X F b = 1000 9.81 10 10 4 X = 9.81 X Now, taking summation moment about hinge: M @hinge = 0.0 25 2.5 cos (θ) = 9.81 X X 2 cos θ 25 2.5 = 9.81 X X 2 X = 3.57m From geometry sin(θ) = 3 X sin(θ) = 3 3.57 θ = 57.17. Page (76) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Equilibrium of Floating Bodies: To be the floating body in equilibrium, two conditions must be satisfied: The buoyant Force (F b ) must equal the weight of the floating body (W). F b and W must act in the same straight line. So, for equilibrium: F b = W object The equilibrium of a body may be: Stable. Unstable. Neutral (could be considered stable) Stability of a Submerge Bodies Stable equilibrium: if when displaced, it returns to its original equilibrium position. Unstable equilibrium: if when displaced, it returns to a new equilibrium position Notes: In this case (body is fully immersed in water) when the body is tilted, the shape of the displaced fluid doesn t change, so the center of buoyancy remains unchanged relative to the body. The weight of the body is located at the center of gravity of the body (G) and the buoyant force located at the center of buoyancy (B). Stable Equilibrium: A small angular displacement υ from the equilibrium position will generate a moment equals: (W x BG x υ). The immersed body is considered Stable if G is below B, this will generate righting moment and the body will tend to return to its original equilibrium position. Page (77) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Unstable Equilibrium: The immersed body is considered Unstable if G is above B, this will generate an overturning moment and the body will tend to be in new equilibrium position. Stability of Floating Bodies Here, the volume of the liquid remains unchanged since F b =W, but the shape of this volume changes and thereby its center of buoyancy will differ. When the body is displaced through an angle υ the center of buoyancy move from B to B 1 and a turning moment is produced. Metacenter (M): The point at which the line of action of the buoyant force (F b ) cuts the original vertical line through G. So, Moment Generated is (W x GM x υ). GM is known as a metacentric height. Stability: Stable If M lies above G, a righting moment is produced, equilibrium is stable and GM is regarded as positive. (GM=+VE) Page (78) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Unstable If M lies below G, an overturning moment is produced, equilibrium is unstable and GM is regarded as negative. (GM=.( VE Neutral: If M coincides with G, the body is in neutral equilibrium. Determination of the Position of Metacenter Relative to Centre of Buoyancy: BM = I V displaced I = the smallest moment of inertia of the waterline plane Procedures for Evaluating the Stability of Floating Bodies Page (79) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 1. Determine the position of the floating body (Draft) using the principles of buoyancy (Total Weights = Buoyant Force). 2. Locate the center of buoyance B and compute the distance from some datum to point B (y B ). The bottom of the object is usually taken as a datum. 3. Locate the center of gravity G and compute (y G ) measured from the same datum. 4. Determine the shape of the area at the fluid surface (plane view) and compute I for that shape. 5. Compute the displace volume (V d ) 6. Compute BM distance (BM = I / V d ). 7. Compute (y M = y B +BM) 8. If (y M > y G ) >> the body is stable.(gm = +VE) 9. If (y M < y G ) >> the body is unstable.(gm = ( VE Important Note: If y M = y G (GM = 0), this case is called neutral and the object could be considered stable. Page (80) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Problems 1. A barge 6 m wide and 20 m long is loaded with rocks as shown below. Assume that the center of gravity of the rocks and barge is located along the centerline at the top surface of the barge. If the rocks and the barge weigh 200,000 kg, would the barge be stable? Solution Assume the datum is at the bottom of the barge. Firstly we calculate the draft from principles of buoyancy: F b = Total weight γ w V dis = 200,000 9.81 200,000 9.81 V dis = = 200 m 3 9810 V dis = 6 20 D D = 200 6 20 = 1.667m y B = D 2 = 1.667 = 0.833m (from datum) 2 y G = 2.4m (as given) Calculation of BM: BM = I V dis I = Moment of inertia for the water line ( smallest I for top view) Page (81) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy I = 20 63 12 = 360m 4 BM = 360 200 = 1.8 m y M = y B + BM y M = 0.833 + 1.8 = 2.633 m GM = y M y G = 2.633 2.4 = 0.233 (+VE) Stable. 2. For the shown figure below, a cube of wood of side length (L) is float in water. If the specific gravity of the wood is 0.88. Determine if this cube is stable or not. Solution Assume the datum is at the bottom of the cube. Firstly we calculate the draft (h=d) in terms of L from principles of buoyancy: F b = Total weight F b = γ w V dis = 9810DL 2 Total weight = ρ cube material (wood) g V cube Total weight = 0.88 1000 9.81 L 3 = 8632.8 L 3 9810DL 2 = 8632.8 L 3 D = h = V dis = DL 2 = 0.88L 3 y B = D 2 = 0.88L 2 y G = L 2 = 0.5L = 0.44L (from datum) 8632.8 L3 9810L 2 = 0.88L Page (82) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Calculation of BM: BM = I V dis I = L L3 12 = L4 12 BM = L 4 12 0.88L 3 = 0.0947L y M = y B + BM y M = 0.44L + 0.0947L = 0.534L GM = y M y G = 0.534L 0.5L = 0.034 L (+VE) Stable. 3. A barge is 4.5m wide and 12m long and floats with a draft of 1.2m. It is piled so high with gravel so that its center of gravity became 1m above the waterline. Is it stable? Solution Firstly, we draw a neat sketch for this problem: Assume the datum is at the bottom of the barge. y B = 1.2 2 = 0.6m y G = 2.2m Calculation of BM: BM = I V dis 12 4.53 I = = 91.125m 4, V 12 dis = 4.5 12 1.2 = 64.8m 3 BM = 91.15 64.8 = 1.4m y M = 0.6 + 1.4 = 2m GM = y M y G = 2 2.2 = 0.2 ( VE) Unstable. Page (83) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 4. The figure below shows a cross section of boat. Show if the boat is stable or not. If the boat is stable compute the righting moment if the angle of heel is 10 o. The boat is float in water and its 6m long. Solution Assume the datum is at the bottom of the boat. y B = 1.5 2 = 0.75m 1.5 + 0.3 y G = = 0.9m 2 Calculation of BM: BM = I V dis I = 6 33 12 = 13.5, V dis = 3 6 1.5 = 27m 3 BM = 13.5 27 = 0.5m y M = 0.75 + 0.5 = 1.25m GM = y M y G = 1.25 0.9 = 0.35 (+VE) Stable. Important Note: If the boat is orientated with any angle, the value of GM remains unchanged. Page (84) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Now when the boat is orientated with an angle of 10 o the cross section will be as following: The righting Moment (M)= W X W = F b = γ w V dis = 9810 27 = 264870N X = GM sin(10) = 0.35 sin(10) = 0.0607 m M = 264870 0.0607 = 16077.6 N. m = 16.07KN. m. Page (85) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy 5. A wooden cone floats in water in the position shown in figure below. The specific gravity of the wood is 0.6. Would the cone be stable? Solution Assume the datum is at the tip of the cone. The center of gravity of the cone is located at 3h 4 from the tip of the cone Firstly we calculate the draft from principles of buoyancy: F b = Total weight F b = γ w V dis = 9810V dis Total weight = ρ cone g V cone 2 V cone = πr2 h 3 = π ( 0.18 2 ) 3 = 0.00212m 3 0.25 Total weight = 0.6 1000 9.81 0.00212 = 12.48 N. 9810V dis = 12.48 V dis = 0.001272m 3 (1) Let the diameter at the level of water is X By interpolation 0.18 = X D = 1.388X 0.25 D 2 V dis = π (X 2 ) 1.388X = 0.363X 3 substitute from (1) 3 0. 363X 3 = 0.001272 X = 0.152m D = 1.388 0.152 = 0.21m y B = 3 0.21 = 0.1575m (from tip of cone) 4 y G = 3 0.25 = 0.1875m (from tip of cone) 4 Page (86) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Static Forces on Surfaces-Buoyancy Calculation of BM: BM = I V dis I = Moment of inertia at the plane at the water line The plane (top view) of the cone at the water line is a circle of diameter X I = π 64 D4 = π 64 0.1524 = 2.62 10 5 m 4 V dis = 0.001272m 3 BM = 2.62 10 5 0.001272 = 0.02m y M = 0.1575 + 0.02 = 0.1775m GM = y M y G = 0.1775 0.1875 = 0.01 ( VE) Unstable. Page (87) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Chapter (4) Motion of Fluid Particles and Streams
Motion of Fluid Particles & Streams Read all Theoretical subjects from (slides Dr.K.AlASTAL) Patterns of Flow Reynolds Number (R e ): A dimensionless number used to identify the type of flow. R e = Inertia Forces Viscous Forces = ρ V D = V D μ ν V = mean velocity ( m ) s, D = pipe diameter (m) ρ = fluid density (Kg/m 3 ) μ = Dynamic viscosity (Pa. s), ν = kinematic viscosity (m 2 /s) For flow in pipe: If (R e 2000) The flow is laminar If (2000 < R e 4000) The flow is transitional If (R e > 2000) The flow is turbulent Read the two Examples (in slides) Discharge and Mean Velocity: Discharge: The total quantity of fluid flowing in unit time past any cross section of a stream is called the discharge or flow at that section (flow rate). Discharge can be measured by one of the following two methods: 1. In terms of mass (Mass Flow Rate,m, ): Mass of fluid m = time taken to collect the fluid = dm dt = ρ Q (Kg/s). 2. In terms of volume (Volume Flow Rate or discharge,q,): Volume of Fluid Q = = V Time t (m3 /s). This method is the most commonly used method to represents discharge. There is another important way to represents Q: Q = V t = Area L t = Area Speed Q = A v (m 3 /s). Page (89) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Motion of Fluid Particles & Streams Continuity of Flow Matter cannot be created or destroyed (principle of conservation of mass) Mass entering per = Mass leaving + Increasing of mass in the control unit time per unit time volume per unit time If the flow is steady, no increase in the mass within the control volume. So, Mass entering per unit time = Mass leaving per unit time Continuity Equation for Steady Flow and Incompressible Flow: A 1 v 1 = A 2 v 2 = Q = constant This equation is a very powerful tool in fluid mechanics and will be used repeatedly throughout the rest of this course. The following problems clarify the concept of continuity of flow Problems 1. Water flows through pipe AB 1.3m diameter at speed of 3m/s and passes through a pipe BC of 1.6m diameter. At C the pipe branches. Branch CD is 0.7m in diameter and carries one third of the flow in AB. The velocity in branch CE is 2.7m/s. Find the flow rate in AB, the velocity in BC, the velocity in CD and the diameter of CE. Page (90) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Motion of Fluid Particles & Streams Solution Q AB = A AB v AB = ( π 4 1.32 ) 3 = 3.982 m 3 /s. Q AB = Q BC = 3.982 m 3 /s Q BC = A BC v BC v BC = Q BC = 3.982 A π = 1.98m/s. BC 1.62 4 Q CD = 1 3 Q AB ( as given) Q CD = 1 3 3.982 = 1.327m3 /s Q CD = A CD v CD v CD = Q CD = 1.327 A π = 3.45 m/s. CD 0.72 4 Q CE = 2 3 Q AB = 2 3 3.982 = 2.65 m3 /s Q CE = A CE v CE A CE = Q CE v CE π 4 D AB 2 = 2.65 2.7 D AB = 1.11 m. Page (91) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Motion of Fluid Particles & Streams 2. Pipe flow steadily through the piping junction (as shown in the figure) entering section (1) at a flow rate of 4.5m 3 /hr. The average velocity at section (2) is 2.5 m/s. A portion of the flow is diverted through the showerhead 100 holes of 1-mm diameter. Assuming uniform shower flow, estimate the exit velocity from the showerhead holes. Solution By applying continuity equation for the system given: Q 1 = Q 2 + Q 3 Q 1 = 4.5m 3 /hr = 4.5 60 60 = 1.25 10 3 m 3 /s Q 2 = A 2 v 2 A 2 = π 4 (0.02)2 = 3.14 10 4 m 2, v 2 = 2.5m/s (given) Q 2 = 3.14 10 4 2.5 = 7.854 10 4 m 3 /s Q 3 = Q 1 Q 2 Q 3 = 1.25 10 3 7.854 10 4 = 4.646 10 4 m 3 /s Q 3 = Q in the 100 hole Q hole = Q 3 4.646 10 4 = = 4.646 10 6 m 3 /s 100 100 Q hole = A hole v hole A hole = π 4 (0.001)2 = 7.854 10 7 m 2 v hole = Q hole A hole = 4.646 10 6 = 5.68 m/s. 7.854 10 7 Page (92) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Motion of Fluid Particles & Streams 3. For the shown tank, if the inlet pipe is 40 cm in diameter and the outlet pipes are 20 cm in diameter. A. Is the tank filling or emptying and at what rate? B. What should be the velocity of water at the inlet pipe to keep the water level in the tank constant? A. Solution Q 1 = A 1 v 1 = ( π 4 0.42 ) 3 = 0.377 m 3 /s Q 2 = A 2 v 2 = ( π 4 0.22 ) 1.5 = 0.047 m 3 /s Q 3 = A 3 v 3 = ( π 4 0.22 ) 2 = 0.063 m 3 /s Q 1 (Q 2 + Q 3 ) = 0.377 (0.047 + 0.063) = +0.267m 3 /s Since the inlet flow rate is larger than outlet flow rate, the tank is filling and with flow rate of 0.267 m 3 /s. B. The water level in the tank will be constant if the inlet flow equals the outlet flow. Q 1 = (Q 2 + Q 3 ) ( π 4 0.42 ) v 1 = (0.047 + 0.063) v 1 = 0.875 m/s. Page (93) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Motion of Fluid Particles & Streams 4. In the figure shown is the tank filling or emptying? At what velocity is the water level rising or falling in this circular tank? 1 3 2 Solution Q 1 = Q in = ( π 4 0.12 ) 3 = 0.0235 m 3 /s Q 2 = ( π 4 0.152 ) 2.1 = 0.0371 m 3 /s Q 3 = ( π 4 0.0752 ) 1.2 = 0.0053 m 3 /s Q out = Q 2 + Q 3 = 0.0371 + 0.0053 = 0.0424 m 3 /s Q in Q out = 0.0235 0.0424 = 0.0189 Q in < Q out Emptying. The rate of emptying is Q emptying = 0.0189 m 3 /s Q emptying = A tank v emptying 0.0189 = ( π 4 1.82 ) v emptying v emptying = 0.0074 m/s. Page (94) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Motion of Fluid Particles & Streams 5. In the figure shown below, the diameter of pipes (1) and (2) is 3cm and the diameter of pipe (3) is 4cm. Alcohol (SG = 0.8) enters section (1) at 6m/s while water enters section (2) at 10 m/s. Assuming ideal mixing of incompressible fluids, compute the exist velocity and the density of the mixture in section (3). Solution Calculation of velocity in section (3): Q 1 + Q 2 = Q 3 Q 1 = ( π 4 0.032 ) 6 = 0.00424 m 3 /s Q 2 = ( π 4 0.032 ) 10 = 0.00706 m 3 /s 0.00424 + 0.00706 = ( π 4 0.042 ) v 3 v 3 = 9m/s. Calculation of density of mixture (in section 3): m 1 + m 2 = m 3 We know that m = ρ Q ρ 1 Q 1 + ρ 2 Q 2 = ρ 3 Q 3 800 0.00424 + 1000 0.00706 = ρ 3 (0.00424 + 0.00706) ρ 3 = 925 Kg/m 3. Page (95) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Chapter (5) Momentum Equation and Its Applications
Momentum Equation & Its Applications Momentum and Fluid Flow In fluid Mechanics, the analysis of motion is performed in the same way as in solid mechanics (by use of Newton s Laws of Motion ). From solid mechanics (Newton s Second Law) stated that: Total Force(F T ) = ma, m = mass of the solid body, a = acceleration But, in fluid mechanics, it is not clear what mass of moving fluid, thus we should use a different form of the equation of Newton s Second Law. Newton s 2 nd Law (for fluids) can be written as following: The rate of change of momentum of a body is equal to the total force acting on the body, and takes place in the direction of the force. Rate of change of momentum is: m (v 2 v 1 ) = Mass Flow Rate Change of velocity According Newton s 2 nd law: F T = m (v 2 v 1 ), m = ρq F T = ρq(v 2 v 1 ) This is the total force acting on the fluid in the direction of motion. Page (97) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications Momentum Equation for Two Dimensional Flow along a Streamline (General Case) In this case, momentum and total force can be resolving into components in the x and y directions (since both momentum and force are vector quantities) F T,x = Rate of change of momentume in x direction F T,x = m (v 2,x v 1,x ) = m (v 2 cosθ v 1 cosθ) F T,y = Rate of change of momentume in y direction F T,y = m (v 2,y v 1,y ) = m (v 2 sinθ v 1 sinθ) m = ρq = ρ (A 1 v 1 ) = ρ (A 2 v 2 ) F T = F T,x 2 + F T,y 2 θ = tan 1 ( F T,y F T,x ) Page (98) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications Important Note: The force F T is made up of three components: F 1 = F R = Force exerted in the direction of the fluid by any solid body touching the control volume. F 2 = F B = Force exerted in the direction of the fluid by a body force (gravity force). F B = Fluid weight + object weight = (ρvg) fluid + (ρvg) object F 3 = F P = Force exerted in the direction of the fluid by a fluid pressure outside the control volume F P = F P,1 + F P,2 + F P,n (If the flow is in two dimensional the pressure will be analyze in x and y directions).such that F P = Pressure Area The final relationship for momentum equation will be as following: F T = F R + F B +F P = ρq(v 2 v 1 ) F R = The force exerted by the body on the fluid But, R = F R = The force exerted by the fluid on the body Usually the value of R is required for design purposes F T = R + F B +F P = ρq(v 2 v 1 ) R = F B +F P ρq(v 2 v 1 ) R = F B +F P + ρq(v 1 v 2 ) = Resultant force acting on the body Note: When calculating F P the pressure at the inlet and outlet is always applied on the fluid (if the pressure is positive not vacuum ), however the velocity is always in the direction of fluid. The following figure clarifies this note: Page (99) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications Bernoulli s Equation This equation states the relationship between velocity (v), Pressure (P), and elevation (z) for: steady flow of frictionless fluid of constant density. This equation is one of the most important equations in fluid mechanics and engineering applications (as we will discuss in Ch.6). P + v2 + z = constant = H (Total Head) ρg 2g Bernoulli equation can be written along a streamline between two points (1) and (2) as following: P 1 ρg + v 2 1 2g + z 1 = P 2 ρg + v 2 2 2g + z 2 + h L h L = head loss between 1 and 2 (due to friction in pipes) Assuming no losses between points 1 and 2, the equation will be: P 1 2 ρg + v 1 2g + z 1 = P 2 ρg + v 2 2g + z 2 2 z 1 and z 2 are elevations of points 1 and 2 from a specific datum usually taken at point (1) z 1 = 0 and z 2 = vertical distance between 1 and 2 and if you are not given any information about this distance, assume it zero (i. e. assume the two points in the same level)and caculate the pressure. Applications of the Momentum Equation Momentum equation is used in several engineering applications as we will discuss in the problems of this chapter. Steps in Analysis: Draw the control volume. Decide on coordinate axis system. Calculate the total force F T = ρq(v 2 v 1 ) Calculate the pressure force F P (usually the pressure is known at one point and not known at the other point, so we use Bernoulli equation to find it). Calculate the body force F B (if you are given the volume of the fluid). Calculate the resultant force on the system R (usually it s required). Page (100) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications Problems 1. The figure below shows a smooth curved vane attached to a rigid foundation. The jet of water, rectangular in section, 75 mm wide and 25 mm thick, deflected on the vane with a velocity of 25m/s. Calculate the vertical and horizontal components of the force exerted on the vane and indicate in which direction these components act. Solution R = F B +F P + ρq(v 1 v 2 ), but F B = F P = 0.0 R = ρq(v 1 v 2 ) A 1 = A 2 = 0.075 0.025 = 0.001875 m 2 Since A 1 = A 2 v 1 = v 2 = 25 m/s(as given) Q = A 1 v 1 = 0.001875 25 = 0.0468 m 3 /s ρq = 1000 0.0468 = 46.8 kg/s R x = 46.8(v 1,x v 2,x ) v 1,x = +25 cos45 = 17.67 m/s, v 2,x = +25 cos25 = 22.66 m/s R x = 46.8(17.67 22.66) = 233.5 N (From right to left). R y = 46.8(v 1,y v 2,y ) v 1,y = 25 sin45 = 17.67m/s, v 2,y = +25 sin25 = 10.56 m/s R y = 46.8( 17.67 10.56) = 1321.16 N (downward). Page (101) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications 2. A square plate of mass 12.7 kg, having uniform thickness and 300 mm length of edge, is hinged so that it can swing freely about its horizontal edge. A horizontal jet, 20mm in diameter, strikes the plate with a velocity of 15 m/s. the centerline of the jet is 150mm below the upper edge of the plate so that when the plate is vertical the jet strikes the plate normally at the center. Find: (a)the force (P) must be applied at the lower edge of the plate to keep it vertical. (b) What inclination to the vertical the plate will assume under the action of the jet if allowed to swing freely. A B Solution (a) Firstly we calculate the force (F) from the water to the jet: F = R x = F B,x +F P,x + ρq(v 1,x v 2,x ) F B,x = 0.0 and F P,x = 0.0 F = R x = ρq(v 1,x v 2,x ) Q = Av = π 4 0.022 15 = 0.0047 m 3 /s v 1,x = 15 m/s (Inlet velocity) v 2,x = 0.0 m/s (velocity at the wall is zero) F = R x = 1000 0.0047(15 0) = 70.5 N Page (102) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications Now by taking summation of moments about hinge equal zero: M @hinge = 0.0 70.5 0.15 = P 0.3 P = 35.25 N. (b) Firstly we calculate the force (F) from the water to the jet: Assume x-direction is in the same direction of the force F. F = R x = F B,x +F P,x + ρq(v 1,x v 2,x ) F B,x = 0.0 and F P,x = 0.0 F = R x = ρq(v 1,x v 2,x ) Q = Av = π 4 0.022 15 = 0.0047 m 3 /s v 1,x = 15 cosθ (Inlet velocity) v 2,x = 0.0 m/s (velocity at the wall is zero) F = R x = 1000 0.0047(15 cosθ 0) = 70.5 cosθ N The weight of the Plate is act at center G and is equal: W plate = mg = 12.7 9.81 = 124.58 N Now by taking summation of moments about hinge equal zero: M @hinge = 0.0 70.5 cosθ 0.15 = 124.58 0.15 sinθ cosθ θ = 34.46. Page (103) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications 3. Water flows through the elbow as shown in figure below and exits to the atmosphere. At a mass flow rate of 15 kg/s, the pressure P 1 is 233 kpa. Neglecting the weight of water and elbow, estimate the horizontal force on the flange bolts at section (1). [D 1 =10cm, D 2 =3cm] Solution The horizontal force on flange bolts equals the horizontal component of the fluid force (R x ). R x = F B,x +F P,x + ρq(v 1,x v 2,x ) but F B = 0.0 (neglect the wieght of water and elbow "as given") R x = F P,x + ρq(v 1,x v 2,x ) Caculation of F T,x = ρq(v 1,x v 2,x ) ρq = 15 kg/s (as given) 1000Q = 15 Q = 0.015 m 3 /s v 1 = Q = 0.015 A π 1 4 = 1.9 m/s, v Q 0.12 2 = = 0.015 A π = 21.22 m/s 2 0.032 4 v 1,x = +1.9 m/s, v 2,x = 21.22 cos40 = 16.25 m/s F T,x = 15(1.9 ( 16.25)) = 272.25 N Page (104) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications Calculation of F P,x : The pressure at section (1) is 233 kpa and the pressure at section (2) is 0.0 because water in section (2) exposed to atmosphere. F P,x = P 1 A 1 + 0.0 = 233 10 3 π 4 0.12 = 1830 N F bolts in flange = R x = 1830 + 272.25 = 2102.25 N. 4. For the shown cart in the figure. Compute the force on the wheels caused be deflecting the jet and the compression of the spring if its stiffness is 1.6 kn/m. Solution R = F B +F P + ρq(v 1 v 2 ), but F B = F P = 0.0 R = ρq(v 1 v 2 ) force from wter to the jet R y = ρq(v 1,y v 2,y ) F y = 0.0 R y + F wheels = 0.0 F wheels = R y = ρq(v 2,y v 1,y ) Q = A v = π 4 0.042 20 = 0.025 m 3 /s Since A 1 = A 2 v 1 = v 2 = 20 m/s v 1,y = 0.0 v 2,y = +20 sin45 = 14.14 m/s F wheels = 1000 0.025(14.14 0) = 353.55 N. Page (105) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications R x = ρq(v 1,x v 2,x ) F x = 0.0 R x + F spring = 0.0 F spring = R x = ρq(v 2,x v 1,x ) Q = 0.025 m 3 /s v 1,x = 20 m/s v 2,x = +20 cos45 = 14.14 m/s F spring = 1000 0.025(14.14 20) = 146.5 N F spring = K X (Compression) F spring = 146.5 = 1.6 1000 X X = 0.0915m = 91.5 mm. 5. The 6-cm-diameter water jet shown in figure below strikes a plate containing a hole of 4-cm diameter. Part of the jet passes through the hole, and part is deflected. Determine the horizontal force required to hold the plate. Solution The most important note in this question is the flow is not the same at 1 and 2 because the flow at 1 divided into three parts (up, down and to the hole). The force on the plate equals the horizontal component of the force of water on the plate (R x ). R x = F B,x +F P,x + ρq(v 1,x v 2,x ) F B,x = 0.0 (No body force in x direction) Page (106) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications F P,x = 0.0 (No given presures) R x = ρq(v 1,x v 2,x ) But!! (Whats wrong ) since Q is not the same at 1 and 2 F Plate = R x = ρq 1 v 1,x ρq 2 v 2,x Q 1 = v 1 A 1 = 25 π 4 0.062 = 0.0706 m 3 /s Q 2 = v 2 A 1 = 25 π 4 0.042 = 0.0314 m 3 /s F Plate = 1000 0.0706 25 1000 0.0314 25 = 980 N. 6. In the figure shown below, what is the resultant force on the jets? 1 3 2 Solution Here the fluid also divided into two components. R x = F B,x +F P,x + ρq(v 1,x v 2,x ) R y = F B,y +F P,y + ρq(v 1,y v 2,y ) No body force is given F B,x = F B,y = 0.0 R x = F P,x + ρq(v 1,x v 2,x ) R y = F P,y + ρq(v 1,y v 2,y ) But, because the value of Q is not constant Page (107) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications R x = F P,x + ρq 1 v 1,x ρq 2 v 2,x ρq 3 v 3,x (1 is inlet, 2 and 3 are outlets) R y = F P,y + ρq 1 v 1,y ρq 2 v 2,y ρq 3 v 3,y Q 2 = A 2 v 2 = 0.02 20 = 0.4 m 3 /s Q 3 = A 3 v 3 = 0.01 20 = 0.2 m 3 /s Q 1 = Q 2 + Q 3 = 0.2 + 0.4 = 0.6m 3 /s v 1 = Q 1 = 0.6 A 1 0.1 = 6 m/s Calculation of R x : F P,x =?? The pressure at 1 is 200 kpa (in x direction), however the pressure at 2 and 3 is 0.0 because they exposed to atmosphere. F P,x = (200 10 3 ) 0.1 + 0 + 0 = 20000 Pa v 1,x = +6 m/s, v 2,x = +20 m/s, v 3,x = +20 cos30 = +17.32 m/s R x = 20000 + 1000 0.6 6 1000 0.4 20 1000 0.2 17.32 R x = 12136 N Calculation of R y : F P,y =?? The pressure at 1 is 0.0 (in y direction), also the pressure at 2 and 3 is 0.0 because they exposed to atmosphere. F P,y = 0 + 0 + 0 = 0 v 1,y = 0, v 2,y = 0, v 3,y = +20 sin30 = 10 m/s R y = 0.0 + 1000 0.6 0 1000 0.4 0 1000 0.2 10 R y = 2000 N = R x 2 + R y 2 = 12136 2 + 2000 2 = 12299.7 N. θ = tan 1 ( R y R x ) = tan 1 ( 2000 12136 ) = 9.36. Page (108) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications 7. The discharge of water through a 140º bend, shown in figure below, is 0.03 m 3 /s. The bend is lying in the horizontal plane and the diameters at the entrance and exit are 200mm and 100mm respectively. The pressure measured at the entrance is 100 kn/m 2, what is the magnitude and direction of the force exerted by the water on the bend? R = F B +F P + ρq(v 1 v 2 ) R x = F B,x +F P,x + ρq(v 1,x v 2,x ) R y = F B,y +F P,y + ρq(v 1,y v 2,y ) Solution Caculation of F T = ρq(v 1 v 2 ) Q = 0.03 m 3 /s v 1 = Q = 0.03 A π = 0.955 m/s 1 0.22 4 v 2 = Q = 0.03 A π = 3.82 m/s 2 0.12 4 F T,x = ρq(v 1,x v 2,x ) ρq = 1000 0.03 = 30 kg/s v 1,x = v 1 = +0.955 m/s, v 2,x = v 2 cos40 = 3.82 cos40 = 2.92 m/s F T,x = 30(0.955 ( 2.92)) = 116.25 N Page (109) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications F T,y = ρq(v 1,y v 2,y ) ρq = 30 kg/s v 1,y = 0.0, v 2,y = +v 2 sin40 = 3.82 sin40 = 2.45 m/s F T,y = 30(0 2.45) = 73.5 N Calculation of Body force (F B ): The volume of the fluid in the pipe is not given, so we can neglect the body force. Calculation of Pressure force (F P ): F P = F P,1 + F P,2 P 1 = 100 kpa = 10000 Pa, P 2 =?? we use bernoulli equation to find P 2 P 1 2 ρg + v 1 2g + z 1 = P 2 ρg + v 2 2g + z 2 2 (Assume no losses) Since the bend is in the horizontal plane, so the elevation of the two points is the same (i.e. canceled each other from Bernoulli equation). P 1 ρg + v 2 1 2g = P 2 ρg + v 2 2 100000 2g 1000 9.81 + 0.9552 2 9.81 = P 2 1000 9.81 + 3.822 2 9.81 P 2 = 93160 Pa F P,x = Pressure force applied on the x direction F P,x = 100000 π 4 0.22 + 93160 cos 40 π 4 0.12 = 3702.1 N F P,y = Pressure force applied on the y direction F P,y = 0.0 93160 sin 40 π 4 0.12 = 470.3 N Now, R x = F B,x +F P,x + ρq(v 1,x v 2,x ) = 0 + 3702.1 + 116.25 = 3818.35 N R y = F B,y +F P,y + ρq(v 1,y v 2,y ) = 0 + ( 470.3) + ( 73.5) = 543.8 N R = R x 2 + R y 2 = 3818.35 2 + 543.8 2 = 3856.9 N. θ = tan 1 ( R y R x ) = tan 1 ( 543.8 3818.35 ) = 8.1. Page (110) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications 8. The outlet pipe from a pump is a bend of 45 rising in the vertical plane. The bend is 150 mm diameter at its inlet (1) and 300 mm diameter at its outlet (2). The pipe axis at the inlet is horizontal and at the outlet is 1m higher. By neglecting friction, calculate the force on the bend and its direction if the inlet pressure is 100 kn/m 2 and the flow of water through the pipe is 0.3 m 3 /s. The volume of the pipe is 0.075 m 3. R x = F B,x +F P,x + ρq(v 1,x v 2,x ) R y = F B,y +F P,y + ρq(v 1,y v 2,y ) Solution Caculation of F T = ρq(v 1 v 2 ) Q = 0.3 m 3 /s v 1 = Q 0.3 = A π = 16.97 m/s 1 0.152 4 v 2 = Q = 0.3 A π = 4.24 m/s 2 0.32 4 F T,x = ρq(v 1,x v 2,x ) ρq = 1000 0.3 = 300 kg/s v 1,x = v 1 = +16.97 m/s, v 2,x = +v 2 cos45 = 4.24 cos45 = 3 m/s F T,x = 300(16.97 3) = 4191 N Page (111) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications F T,y = ρq(v 1,y v 2,y ) ρq = 300 kg/s v 1,y = 0.0, v 2,y = +v 2 sin45 = 4.24 sin45 = 3 m/s F T,y = 300(0 3) = 900 N Calculation of Body force (F B ): The volume of the fluid in the pipe equals 0.075 m 3 F B,x = 0.0 (no weight in x direction) F B,y = ρvg = 1000 0.075 9.81 = 735.75 N Calculation of Pressure force (F P ): F P = F P,1 + F P,2 P 1 = 100 kpa = 100000 Pa, P 2 =?? we use bernoulli equation to find P 2 P 1 2 ρg + v 1 2g + z 1 = P 2 ρg + v 2 2g + z 2 2 (Assume no losses) Assume the datum is at the centerline of the inlet: P 1 2 ρg + v 1 2g + z 1 = P 2 ρg + v 2 2g + z 2 2 100000 1000 9.81 + 16.972 2 9.81 + 0 = P 2 1000 9.81 + 4.242 2 9.81 + 1 P 2 = 225191.65 Pa F P,x = Pressure force applied on the x direction F P,x = 100000 π 4 0.152 225191.65 cos 45 π 4 0.32 = 9488.5 N F P,y = Pressure force applied on the y direction F P,y = 0.0 225191.65 cos 45 π 4 0.32 = 11255.65 N Now, R x = F B,x +F P,x + ρq(v 1,x v 2,x ) = 0 + 9488.5 + 4191 = 5297.5 N R y = F B,y +F P,y + ρq(v 1,y v 2,y ) R y = 735.75 + ( 11255.65) + ( 900) = 12891.4 N Page (112) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Momentum Equation & Its Applications R = R x 2 + R y 2 = 5297.5 2 + 12891.4 2 = 13937.4 N. θ = tan 1 ( R y R x ) = tan 1 ( 12891.4 5297.5 ) = 67.66. Page (113) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Chapter (6) Energy Equation and Its Applications
Energy Equation & Its Applications Bernoulli Equation Bernoulli equation is one of the most useful equations in fluid mechanics and hydraulics. And it s a statement of the principle of conservation of energy along a stream line. Bernoulli Equation can be written as following: P ρg + v2 2g + z = H T = constant All these terms have a unit of length (m) P ρg = pressure energy per unit weight = pressure head We know that P = ρgh pressure h pressure = P ρg (m). v2 = kinetic energy per unit weight = velocity head 2g We know that K. E = 1 2 mv2 divided by weight 1 2 mv2 mg = v2 2g (m). z = potential energy per unit weight = (potential elevation head) We know that P. E = mgz divided by weight mgz mg = z (m). H T = total energy per unit weight = total head (m). By using principle of conservation of energy, we can apply Bernoulli equation between two points (1 and 2) on the streamline: Total head at (1) = Total head at (2) P 1 ρg + v 2 1 2g + z 1 = P 2 ρg + v 2 2 2g + z 2 But!!, this equation no energy losses ( e.g. from friction) or energy gains (e.g. from a pump) along a stream line, so the final form for Bernoulli equation is: P 1 ρg + v 2 1 2g + z 1 + h P = P 2 ρg + v 2 2 2g + z 2 + h L + h T Page (116) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications h P = q = Energy supplied by pump per unit weight (m) h T = w = work done by turbine per unit weight (m) h L = Total friction losses per unit weight (m) Representation of Energy Changes in a Fluid System (HGL and EGL) It is often convenient to plot mechanical energy graphically using heights. Hydraulic Grade Line or (HGL): HGL = P ρg + z It is the line that joins all the points to which water would rise if piezometric tubes were inserted. Energy Grade Line or (Total Energy Line) EGL: EGL = total head = P ρg + v2 2g + z It is the line that joins all the points that represent the total head (i.e. the EGL is always above HGL by a velocity head ( v2 2g ) ). Important Notes (Guidelines) for drawing HGL and EGL: EGL and HGL are falls in the direction of flow due to friction losses (h L ). EGL and HGL are vertically downward due to minor losses such as: [ entrance, exit, elbow, valve, increasing or decreasing diameter] if exist. EGL and HGL are vertically upward due to the head supplied by pump (if exist) and tend vertically downward due to head consumed by turbine (if exist). If there exist a reservoir, the EGL and HGL are coinciding with the surface of fluid in the reservoir because the velocity is zero and the pressure is atmospheric pressure (zero gauge pressure). But if the water exits from a nozzle with a specific velocity the HGL will coincide with the water because the pressure is zero, but the EGL will be above HGL by a velocity head. Another notes will be described in the problems that we will solve. Page (117) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications Problems 1. Water flows through the horizontal branching pipe shown below. Calculate: a) The water speed at section (2). b) The pressure at section (3). c) Flow rate at section (4). Assume no losses and branch in horizontal plane A 2 = 0.0065 m 2 P 2 = 35 kpa Solution a) (v 2 =?? ) v 1 = Q 1 = 0.28 = 3.11 m/s A 1 0.09 By applying Bernoulli equation between sections (1) and (2): P 1 ρg + v 2 1 2g + z 1 = P 2 ρg + v 2 2 2g + z 2 z 1 = z 2 = 0.0 (system is in horizontal plane) 70 10 3 1000 9.81 + 3.112 2 35 103 + 0 = 2 9.81 1000 9.81 + v 2 2 9.81 + 0 v 2 = 8.9 m/s. Page (118) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications b) (P 3 =?? ) By applying Bernoulli equation between sections (1) and (3): P 1 ρg + v 2 1 2g + z 1 = P 3 ρg + v 2 3 2g + z 2 z 1 = z 3 = 0.0 (system is in horizontal plane) 70 10 3 1000 9.81 + 3.112 2 9.81 + 0 = P 3 1000 9.81 + 6.12 2 9.81 + 0 P 3 = 5636.8 Pa = 56.37 kpa. c) (Q 4 =?? ) Q 2 = A 2 v 2 = 0.0065 8.9 = 0.0578 m 3 /s Q 3 = A 3 v 3 = 0.018 6.1 = 0.1098 m 3 /s Q in = Q out (continuity equation) Q 1 = Q 2 + Q 3 + Q 4 0.28 = 0.0578 + 0.1098 + Q 4 Q 4 = 0.1124 m 3 /s. Page (119) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 2. For the shown figure below, if D 1 > D 2. Draw the HGL and EGL. Solution Notes: Since D 1 >D 2 v 1 < v 2 v 1 2 < v 2 2g 2g As the velocity increase, head losses increase and the line will be steeper. Page (120) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 3. For the shown figure below, if D 1 > D 2. Draw the HGL and EGL. Solution Notes: Since D 1 >D 2 v 1 < v 2 v 1 2 < v 2 2g 2g The exit velocity is not zero, so EGL will not enclosed to HGL at the end. Page (121) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 4. For the shown figure below, if D 3 >D 1 > D 2. Draw the HGL and EGL. Solution Notes: The new idea here, that there is enlargement as shown above and with this enlargement, velocity will decrease and thereby head losses decreased as shown. The turbine is considered (power consumed) i.e. losses Page (122) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 5. For the shown figure below. Draw the HGL and EGL. Solution Note that when the diameter is gradually decreased, the HGL and EGL also will gradually decrease as a curved lines as shown. Page (123) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 6. For the system shown below. a) What is the direction of flow? b) What kind of machine is at A? c) Do you think both pipes, AB and CA, are the same diameter? d) Sketch the EGL for the system. Solution a) The flow direction is from right (point B) to left (point C) because the head losses increased from right to left. b) The machine at A is a pump because it s generating energy (rising head) as shown. c) No, pipe CA has a smaller diameter than pipe AB because the slope of the HGL due to pipe CA is steeper than HGL due to pipe AB and this means the velocity in pipe CA is higher than velocity in pipe AB and thereby losses. d) Page (124) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications Applications of Bernoulli Equation There are several applications of Bernoulli equation and will be discussed in the problems of this chapter. Important notes before solving problems: Three equations should be in mind: Bernoulli equation. Continuity Equation. Up down method (Ch.2) When the fluid is flowing from section (1) to section (2), some losses of energy will occur between sections (1) and (2), so we include a coefficient of discharge (C d ) to get the actual discharge: Q actual = C d Q Theoritical Q Theoritical = the calculated discharge = Av This will be clarified in problems later. Pump is a machine that consuming electricity power to produce a head of water. Turbine is a machine that consuming energy from water to produce power elevtricity. If we want to calculate the power: Power = ρqgh (Watt) 746 Watt = 1 horse power Power consumed by pump = ρqgh p Power generated by turbine = ρqgh t If we need the power supplied by any object (machine, reservoir, jet, etc ) just calculate the total head and apply in power equation above. We can also calculate the efficiency of pump or turbine or any system: η = Power out Power in For pump: Power out = ρqgh p Power in = electricity power For turbine: Power out = electricity power Power in = ρqgh t Page (125) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications Problems 1. What is the head that must be supplied by the pump to the water to pump 0.4 m 3 /s from the lower to the upper reservoir? Hint: [The head loss in the pipes is given in the figure as a function of water velocity]. 2 1 Solution Firstly we calculate the velocities of each pipe, assume the pipe in left side is pipe (1) and the pipe in right side is pipe (2). Q = 0.4 m 3 /s v 1 = Q = 0.4 A π 1 4 0.22 = 12.73 m/s v 2 = Q 0.4 = A π = 22.64 m/s 2 0.152 4 Now we calculate the head loss for each pipe (from given formulas): h L,1 = 22.5 V 1 2 12.732 = 22.5 2g 2 9.81 = 185.84m h L,2 = 45 V 2 2 22.642 = 45 2g 2 9.81 = 1175.62m h L,total = 185.84 + 1175.62 = 1361.46m Now apply Bernoulli equation between points (1) and (2): P 1 2 2 ρg + v 1 2g + z 1 + h P = P 2 ρg + v 2 2g + z 2 + h L + h T h T = 0.0 (because there is no turbine between 1 and 2) Page (126) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications P 1 ρg + v 2 1 2g + z 1 + h P = P 2 ρg + v 2 2 2g + z 2 + h L P 1 = P 2 = 0.0 (1 and 2 are exposed to atmosphere) v 1 = v 2 = 0.0 (surface of each reservoir is constant) 0 + 0 + 100 + h P = 0 + 0 + 150 + 1361.46 h P = 1411.46 m. 2. A pipe of a diameter 6 mm connects tank A and open container B as shown in figure below. The liquid having a specific weight of 9780 N/m 3 and a viscosity of 0.0008 kg/m.s. The pressure P A =34.5 kpa and the head loss from pipe CD is h L = 25 v2 a) Which direction will the water flow and what is the flow rate? b) Is the flow in pipe CD laminar or turbulent? 2g 1 2 Solution a) We calculate the head loss for pipe CD: h L = 25 v2 2g = 25 v 2 2 9.81 = 1.274v2 Assume the fluid is moving from container B to tank A. Now apply Bernoulli equation between points (1) and (2): P 1 2 2 ρg + v 1 2g + z 1 + h P = P 2 ρg + v 2 2g + z 2 + h L + h T h P = h T = 0.0 (because there is no pump and turbine between 1 and 2) Page (127) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications P 1 ρg + v 2 1 2g + z 1 = P 2 ρg + v 2 2 2g + z 2 + h L P 1 = 0.0 (1 is exposed to atmosphere) P 2 = 34.5 kpa = 34500 Pa v 1 = v 2 = 0.0 (surface of each reservoir is constant) z 1 (from datum) = 1.4 + 4.3 sin45 = 4.44m z 2 (from datum) = 1m 0 + 0 + 4.44 = 34500 9780 + 0 + 1 + 1.274v2 v 2 = 0.06875 v = 0.06875 = 0.262m/s The negative sign means, the fluid is moving opposite to the assumed direction (i.e. the flow is from tank A to container B). Q = A v = π 4 0.0062 0.262 = 7.40 10 6 m 3 /s. b) R e = ρ V D μ ρ = 9780 = 996.94 kg/m3 9.81 996.94 0.262 0.006 R e = = 1959 < 2000 Laminar. 0.0008 Page (128) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 3. When the pump shown in figure below draws 0.06 m 3 /s of water from the reservoir, the total friction head loss is 5 m. The flow discharges through a nozzle to the atmosphere. Estimate the pump power in kw delivered to the water. Solution We apply Bernoulli Equation between points (1) and (2) Q = 0.06 m 3 /s v exit = Q A e = 0.06 π 4 0.052 = 30.6 m/s v exit = v 2 v 1 = 0.0 (fluid is static in reservoir) P 1 = P 2 = 0.0 (exposed to atm. ) Assume the datum at point (1), so: z 1 = 0.0, z 2 = 2m P 1 ρg + v 2 1 2g + z 1 + h P = P 2 ρg + v 2 2 2g + z 2 + h L + h T h T = 0.0 (because there is no turbine between 1 and 2) P 1 2 ρg + v 1 2g + z 1 + h P = P 2 ρg + v 2 2g + z 2 + h L Apply now: 2 0 + 0 + 0 + h P = 0 + 30.62 2 9.81 + 2 + 5 h P = 54.72m The pump power = ρqgh P = 1000 0.06 9.81 54.72 = 32208.2 Watt The pump power = 32.2 kw. 1 2 Page (129) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 4. For the water system shown, the velocity at section (1) is v 1 =0.6m/s. Calculate the mercury manometer reading, h. Notes: Losses are neglected. Section 2 is opened to atmosphere. ρ mercury = 13600 kg/m 3, ρ water = 1000 kg/m 3 Solution Apply Bernoulli equation between points (1) and (2): P 1 ρg + v 2 1 2g + z 1 + h P = P 2 h P = h T = h L = 0.0 ρg + v 2 2 2g + z 2 + h L + h T P 1 ρg + v 2 1 2g + z 1 = P 2 ρg + v 2 2 2g + z 2 Assume datum is passing through centerline of section (1) z 1 = 0.0, z 2 = 3m v 1 = 0.6m/s (apply continuity equation to find v 2 ) A 1 v 1 = A 2 v 2 v 2 = A 1 v 1 A 2 = π 4 0.0752 0.6 π 4 0.0252 = 5.4 m/s P 2 = 0.0 (opened to atmosphere) To calculate P 1, use up down method: P 1 + 9810 0.6 13600 9.81 h = 0.0 P 1 = 133416 h 5886 Now apply in Bernoulli equation: 133416 h 5886 9810 h = 0.372 m. + 0.62 2 9.81 + 0 = 0 + 5.42 2 9.81 + 3 Page (130) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 5. Kerosine flows through the pump shown in figure below at 0.065 m 3 /s. Head losses between 1 and 2 are 2.4 m, and the pump delivers 8 hp to the flow. What should the mercury manometer reading h be? ρ mercury = 13600 kg/m 3, ρ kerosine = 804 kg/m 3 2 1 y Solution Apply Bernoulli equation between points (1) and (2): P 1 2 2 ρg + v 1 2g + z 1 + h P = P 2 ρg + v 2 2g + z 2 + h L + h T h T = 0.0 h L = 2.4 m Power = 8 hp = 8 746 = 5968 Watt Power = ρqgh P 5968 = 804 9.81 0.065 h P h P = 11.64 m Assume datum is passing through centerline of section (1) z 1 = 0.0, z 2 = 1.5m v 1 = Q = 0.065 A π = 14.7m/s 1 0.0752 4 v 2 = Q = 0.065 A π = 3.68m/s 2 0.152 4 Page (131) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications Apply in Bernoulli equation: P 1 804 9.81 + 14.72 2 9.81 + 0 + 11.64 = P 2 804 9.81 + 3.682 + 1.5 + 2.4 2 9.81 P 2 P 1 = 142471.37 Eq. (1) Now apply up down method between 1 and 2: P 1 + ρ k gy + ρ m gh ρ k g(h + y + 1.5) = P 2 P 2 P 1 = 125528.26 h 11830.86 Substitute from Eq. (1): 125528.26 h 11830.86 = 142471.37 h = 1.23 m. 6. Oil flows through a pipe as shown in figure below. If the coefficient of discharge (C d ) for the orifice in the pipe is 0.63. Calculate the discharge of oil in the pipe. y Solution Firstly we apply Bernoulli equation between points (1) and (2): P 1 2 2 ρg + v 1 2g + z 1 = P 2 ρg + v 2 2g + z 2 z 1 = z 2 = 0.0 (by assuming the datum passing through points 1 and2) P 1 P 2 = v 2 2 2 v 1 ρg 2g Now, by applying continuity equation between (1) and (2): A 1 v 1 = A 2 v 2 Page (132) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications A 1 = π 4 0.252 = 0.049 m 2 A 2 = π 4 0.12 = 0.00785 m 2 0.049 v 1 = 0.00785v 2 v 1 = 0.16 v 2 P 1 P 2 = v 2 2 2 0.16v 2 P 1 P 2 = 0.84v 2 2 ρg 2g ρg 2g Now, by applying up down method between (1) and (2): P 1 + 910 9.81 y + 910 9.81 0.35 13600 9.81 0.35 910 9.81 y = P 2 P 1 P 2 = 43571.115 Pa. 43571.115 = 0.84v 2 2 v ρg 2g 2 = 10.67 m/s Q theo = A 2 v 2 = 0.00785 10.67 = 0.0838 m 3 /s Q actual = Q theo C d = 0.0838 0.63 = 0.053m 3 /s. Page (133) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad
Energy Equation & Its Applications 7. A weir of cross section shown below is used to measure the flow rate. Calculate this value of flow rate (Q). Solution We divide this cross section for rectangle with dimensions of (0.2 x 0.4) and triangle with dimensions of (0.4 x 0.1) and then we sum the two values of Q. For rectangle: The flow rate for the shown element is: Q = A v A = 0.4 h v = 2gh Q = 0.4 2gh h But the value of h is varies from 0 to 0.2m, so the value of v also varies and thereby Q is varies with h, so, to find the total discharge for this rectangle we integrate from 0 to 0.2m to get total amount of discharge: 0.2 0 0.2 0 Q = 0.4 2gh h = 0.4 2g h 0.5 h Page (134) Dr.Khalil Al-astal Eng. Ahmed Al-Agha Eng. Ruba Awad