Fluid Mechanics Tuts and Answers

ME 215.3 Fluid Mechanics I Example Problems c James D. Bugg January 2009 Department of Mechanical Engineering University of Saskatchewan 1. The tank shown holds oil of specific gravity 0.89. The top of the tank is closed and the space in the tank above the oil contains air. The U-tube manometer contains water and the displacements are as indicated. Atmospheric pressure is 101.3 kPa. What is the pressure of the air in the tank? (Solution: page 49) 6 c m 4 c m 1 5 c m Air Air Water Oil Open to atmosphere 2. A flat, vertical gate holds back a pool of water as shown. Find the force that the water exerts on the gate. (Solution: page 51) 3 m 3 m ρ = 1000 kg/m 3 hinge gate (5 m wide) stop P atm = 100 kPa 1 ME 215.3 Example Problems 3. A Tainter gate is constructed from a quarter-cylinder and is used to hold back a pool of water. The radius of the gate is 1.22 m and it is 2.44 m long. Calculate the hydrostatic force on the gate. (Solution: page 53) R = 1 . 2 2 m gate ρ = 1000 kg/m 3 length=2.44 m 4. A flat, vertical gate holds back a pool of water as shown. Find the force that the stop exerts on the gate. (Solution: page 56) P atm = 100 kPa ρ = 1000 kg/m 3 hinge gate (5 m wide) stop 3 m 3 m ME 215.3 Example Problems 2 5. A circular hatch on a submarine is hinged as shown. The radius of the hatch is 35 cm above the centre of the hatch. Determine how hard a sailor has to push on the centre of the hatch to open it. The seawater density is 1025 kg/m 3 and the centre of the hatch is 2 m below the surface of the ocean. The air pressure inside the submarine is equal to atmospheric pressure. hinge opening hatch hatch 70 cm A A View A-A g submarine outside of submarine inside of force 6. A rectangular gate is hinged along the top edge as shown. The gate is 4 m long, 15 cm thick, and is made of concrete (s.g.=2.3). Determine the water depth H if the gate is just about to open. What force does the gate exert on the hinge? The water density is 1000 kg/m 3 . H gate 45 o 2 m hinge g 3 ME 215.3 Example Problems 7. A pool of fluid has a density that varies linearly from 1000 kg/m 3 at the surface to 1600 kg/m 3 at a depth of 4 m. A 2 m by 2 m square gate is hinged along its bottom edge and held in place by a force F at its top edge. Find the force F. F 2 m 2 m g hinge 8. A long, square wooden block is pivoted along one edge. The block is in equilibrium when immersed in water (ρ = 1000 kg/m 3 ) to the depth shown. Evaluate the specific gravity of the wood. hinge wood 0.6 m water 1.2 m 1.2 m ME 215.3 Example Problems 4 9. A cylindrical gate with a radius of curvature of √ 2 m holds back a pool of water with a layer of oil floating on the surface. The gate is 8 m long. What force does the pinned connection at A exert on the gate? 1 m g A √ 2 m Water (ρ = 1000 kg/m 3 ) Oil (s.g. = 0.8) 1 m 10. A parabolic gate 4 m wide (in the z direction) holds a pool of water as shown. Find the tension in the cable AB required to hold the gate in the position shown. Find the reaction force which the hinge at C exerts on the gate. Note that point A is directly above point C. C A B y = x 2 Parabolic gate 0.5 m 2 m 1.5 m g ρ = 1000 kg/m 3 water y x Cable Hinge 5 ME 215.3 Example Problems 11. A partially submerged pipe rests against a frictionless wall at B as shown. The specific gravity of the pipe material is 2.0 and the pipe has an outside diameter of 2 m. The fluid is water with ρ = 1000 kg/m 3 . Find the inside diameter of the pipe and the force which the wall exerts on the pipe. The ends of the pipe are closed. g s.g. = 2.0 0.5 m B 12. A gate of mass 2000 kg is mounted on a frictionless hinge along its lower edge. The width of the gate (perpendicular to the plane of view) is 8 m. For the equilibrium position shown, calculate the length of the gate, b. b Water hinge 1 m 30 o ME 215.3 Example Problems 6 13. A vertical, plane wall holds back a pool of water (ρ = 1000 kg/m 3 ) which is 1.5 m deep. The wall has a triangular gate in it that is hinged along the bottom edge and held closed by a horizontal force F applied at the top corner. Calculate the force F required to hold it closed. 2 m 1 m g gate gate View A-A A A 1 . 5 m hinge F 14. A Tainter gate holds back a pool of water (ρ = 1000 kg/m 3 ) which is 2 m deep. The radius of curvature of the gate is 2 m and it is 6 m long. The mass of the gate is 1000 kg and its centre of gravity is at the position indicated on the diagram. Assume that the point where the gate contacts the bottom of the pool is frictionless. Calculate the reaction force on the hinge. gate c. of g. g 1.8 m 45 o 45 o 2 m hinge 2 m 7 ME 215.3 Example Problems 15. Determine H when the L-shaped gate shown is just about to open. Neglect the weight of the gate and let the density of the fluid be ρ. hinge L H gate g 16. What force F is needed to hold the 4 m wide gate closed? The fluid is water and it has a density of 1000 kg/m 3 . hinge 3 m F 9 m ME 215.3 Example Problems 8 17. A semi-circular gate is hinged at the bottom as shown. The density of the fluid varies linearly from 1000 kg/m 3 at the surface of the reservoir to 1500 kg/m 3 at the bottom of the reservoir. Find the force F required to hold the gate in place? 1 m A A g hinge Section A-A gate hinge F 3 m 18. A circular cylinder holding back a pool of water is held in place by a stop as shown. The height of the stop is 0.5 m and the water depth is 1.5 m. If the water depth were increased beyond 1.5 m the cylinder would roll over the stop. Assuming that the water contacts the cylinder surface right up to point A, what is the specific gravity of the cylinder. Diameter = 2 m A 0.5 m 1.5 m 9 ME 215.3 Example Problems 19. A gate composed of a quarter circle portion and a straight portion holds back a pool of water. The gate is hinged at A and held in place by a force F applied as shown. Find the magnitude of the force F required to hold the gate in the position shown. The width of the gate is 8 m. F g A Width of Gate is 8 m Water (ρ = 1000 kg/m 3 ) Hinge 1.5 m 0.5 m R = 1 m ME 215.3 Example Problems 10 20. A circular cylinder of radius R = 50 mm, length b = 100 mm, and density ρ c = 800 kg/m 3 blocks a slot in the bottom of a water tank as shown. The line joining the centre of the cylinder and the point where the cylinder contacts the edge of the slot subtends an angle of α = 30 o with the vertical. If h = 150 mm what is the force exerted on the cylinder by the tank? h R α = 30 o g 11 ME 215.3 Example Problems 21. Water at 20 o C is retained in a pool by a triangular gate which is hinged along its top edge and held in place by a stop at its lowest point. What force does the stop exert on the gate? What force does the hinge exert on the gate? stop 3 . 5 m 7 m g A 1 m Section A-A A hinge 22. A 35 kg, 10 cm cube of material is suspended from a wire in a fluid of unknown den- sity. The tension in the wire is 335.5 N. Determine the specific gravity of the fluid. (Solution: page 59) 23. A U-tube is used as a crude way to measure linear acceleration. Determine the mag- nitude of the acceleration as a function of the geometry of the tube, the acceleration due to gravity, and the displacement of the fluid in the tube. (Solution: page 61) ME 215.3 Example Problems 12 24. An open-top cart half full of water (ρ = 1000 kg/m 3 ) is shown at rest in the figure below. The cart begins to accelerate to the right at a constant 3 m/s 2 . After some time, the fluid reaches a hydrostatic state. Determine the net hydrostatic force on the rear, vertical end of the cart. The cart is 0.8 m wide. 1 m 3 m/s 2 2 m g 0 . 5 m 25. A container of water (ρ = 1000 kg/m 3 ) accelerates on a 30 o slope. It is completely closed except for a small hole in the position indicated. If the gauge pressure at point A is 25 kPa, what is the acceleration a? If the width of the container (perpendicular to the page) is 0.5 m, what is the net hydrostatic force on the top of the container? 0.75 m g hole A 30 o 1 m a 13 ME 215.3 Example Problems 26. A partially full can with an open top spins around an axis as shown. The diameter of the can is 50 cm and is spinning at 50 rpm. If the depth of the fluid at the outer edge of the can is 30 cm, what is the depth of the fluid on the axis of rotation? (Solution: page 63) 30 cm 50 cm ω = 50 rpm 27. A cylindrical container is rotated about its axis. Derive a general relationship for the shape of the free surface as a function of the rotation rate, container radius, and the depth of the fluid when it is not spinning. (Solution: page 65) ω R H o ME 215.3 Example Problems 14 28. A cylindrical can of radius 4 cm and height 12 cm has an open top. It is initially at rest and completely full of liquid. It is rotated about its axis at 250 rev/min until the fluid inside it achieves solid body rotation. The rotation is then stopped and the fluid within the container is allowed to come to rest. How deep will the fluid in the container be? 4 cm g 12 cm 29. A U-tube manometer contains two fluids with different densities as shown. The fluid positions shown in the diagram are for the case when the tube is not spinning. It is then spun around the axis shown until the liquid level in both legs is equal? Find ω. 1000 kg/m 3 1200 kg/m 3 g 20 cm ω 1 0 c m 15 ME 215.3 Example Problems 30. The U-tube shown is rotated about the vertical axis indicated on the diagram at 60 rev/min. Determine the displacement of the water in each leg from its rest position. Perform this calculation on the centreline of the tubes. What is the pressure at point A? Diameter, 8 mm 5 c m A g 3 cm 10 cm Axis of rotation Diameter, 5 mm Rest level 31. A 80 cm diameter cylindrical can has a closed top except for a small vent hole at the centre. If the density of the fluid is 1000 kg/m 3 and the can spins at 60 rpm, what is the force on the top of the can? (Solution: page 67) 60 rpm cylindrical can ρ = 1000 kg/m 3 vent ME 215.3 Example Problems 16 32. An upright, 10-cm diameter, cylindrical paint can 20 cm deep spins around its axis of symmetry at 500 rpm. It is completely full of mineral oil (sg = 0.87) and there is a very small hole in the lid at the rim of the can. Determine the net hydrostatic force on the lid. 33. A quarter-circle, cylindrical gate 5 m long and with a 1.5 m radius is hinged at the point indicated below. It holds back a pool of water (ρ = 998 kg/m 3 ) with a horizontal force F applied as shown. Determine the force F and the reaction at the hinge. g hinge 1 . 5 m F 34. You are driving down the road at 100 km/hr with a cup of coffee in your drink holder. The cup has no lid and is 8 cm in diameter. The coffee is initially 1 cm from the top of the cup. You come to a curve in the highway which is not banked. Determine the minimum radius of the curve for which the coffee will not spill. If you go around a curve whose radius is 75% of the minimum, how much coffee will spill (in cm 3 )? 35. A square gate (1.25 m x 1.25 m) is hinged along a line 0.25 m from its top edge. A force F 1 applied at the top of the gate holds it closed. The pool of water (ρ = 998 kg/m 3 ) is 1.75 m deep. Calculate the force F 1 required to hold the gate closed. 1 . 7 5 m g gate hinge 1 . 2 5 m 1 m F 1 17 ME 215.3 Example Problems 36. A cylindrical container is 40 cm deep and 30 cm in diameter and is open at the top. Water is put in the container to a depth of 20 cm. The container is then spun around its axis at an angular speed of ω until the water reaches a hydrostatic state. The spinning is stopped and the water is allowed to come to rest. After coming to rest, the water is 10 cm deep. Determine ω in revolutions per minute (rpm). 4 0 c m 30 cm g 2 0 c m initial water level 37. The police are using a fire hose to move a flat barricade. What is the horizontal force on the barricade due to the stream of water? (Solution: page 69) A j = 0.01 m 2 V = 15 m/s ME 215.3 Example Problems 18 38. A jet of water from issues from a 0.01 m 2 nozzle at 15 m/s. It impinges on a vane mounted on a movable cart and is deflected through 45 o . If the cart is moving away from the nozzle at 5 m/s, what is the force which the water exerts on the cart? (Solution: page 72) 45 o A j = 0.01 m 2 cart 39. A 50 mm diameter pipe carries water at 1 m/s. A nozzle with an exit diameter of 25 mm is attached at the end of the pipe. If the pressure just upstream of the nozzle is 10 kPa, what is the force on the bolts at the flange connection attaching the nozzle to the pipe? (Solution: page 79) 40. Consider the entrance region of a circular pipe for laminar flow. What is the frictional drag on the fluid between axial locations 1 and 2 in terms of the pressure at those locations, the density of the fluid, the mean velocity of the fluid, and the pipe radius. (Solution: page 81) 2 1 z u(r) = U max (1 −(r/R) 2 ) radius = R circular pipe U o r 19 ME 215.3 Example Problems 41. A 50 mm diameter pipe carries water at 1 m/s. A nozzle with an exit diameter of 25 mm is attached at the end of the pipe. Assuming frictionless flow, find the force on the flange connection. (Solution: page 84) 42. If the nozzle from the previous problem has a pressure of 10 kPa at its entrance, what is the loss coefficient? (Solution: page 86) 43. A transition piece turns 45 o and expands from 50 mm to 80 mm. The pressure at the entrance is 20 kPa and the loss coefficient is 0.50 based on the discharge velocity. Water flows through the transition at 4 kg/s. Find the force required to hold the transition in place. (Solution: page 87) 4 kg/s K m = 0.50 (based on discharge velocity) transition piece 20 kPa D 1 = 50 mm D 2 = 80 mm 45 o ME 215.3 Example Problems 20 44. Air (ρ = 1.2 kg/m 3 ) blows parallel to a flat plate upon which a boundary layer grows. The plate is 10 m long in the streamwise direction and 4 m wide (normal to the page). At the leading edge of the plate the air speed is uniform at U ∞ = 10 m/s. The velocity profile is measured at the end of the plate and is found to vary with distance y from the plate according to u = U ∞ (y/δ) 1/7 where δ = 170 mm is the boundary layer thickness. Determine the viscous force on the plate. Consider only the flow over the top surface of the plate. The dashed line labeled “A” is a streamline in this flow. A 170 mm 10 m 10 m/s y flat plate 45. An object is placed in a 1 m diameter wind tunnel and the air velocity downstream of the object is found to linearly vary from zero at the centreline of the wind tunnel to a maximum at the wind tunnel wall. The air velocity upstream of the object is a uniform 20 m/s. A mercury manometer indicates a 10 mm Hg pressure difference as shown. Assume the pressure is uniform across the wind tunnel at any axial location. Neglect shear at the wind tunnel walls and let ρ = 1.2 kg/m 3 . Determine the drag force on the object. 10 mm g 1 m diameter 20 m/s mercury (s.g.=13.56) 21 ME 215.3 Example Problems 46. A tank on wheels contains water (ρ = 1000 kg/m 3 ) and is held in place by a cable as shown below. A pump mounted on top of the tank draws water from the tank and discharges the water through a horizontal, 1 cm diameter nozzle at a mass flow rate of 5 kg/s. The cable breaks and the cart begins to move. At some instant in time later, the total mass of the tank, pump assembly, and the water remaining in the tank is 100 kg and the velocity of the system is 20 m/s. Determine the acceleration of the system at this instant in time. The flow rate supplied by the pump remains constant. Pump 5 kg/s g Cable 1 cm dia. 47. A new type of lawn sprinkler is developed that has two arms of different lengths. The arms lie in a horizontal plane and rotate about a vertical axis. Both nozzles are aimed upward at 20 o from the horizontal plane and have an exit diameter of 2 mm. The flow to each arm is equal. One arm is 10 cm from the pivot and the other is 20 cm from the pivot. If the pivot is assumed to be frictionless and the sprinkler delivers 2 L/min, find the angular speed of the sprinkler? 10 cm D = 2 mm 20 cm ME 215.3 Example Problems 22 48. An aluminium block weighing 10 N is supported by a jet of water issuing from a 2.5 cm diameter nozzle. What jet exit velocity is required to hold the block 10 cm above the nozzle exit? The jet of water is deflected through 180 o when it strikes the block and the guides holding the block in place are frictionless. 10 cm guides g aluminum block 2.5 cm 49. An elbow connected to a 5 cm diameter pipe discharges water (ρ = 1000 kg/m 3 ) as shown. The difference between the stagnation and static pressures ∆P measured by the pitot-static tube is 500 Pa. The mass of the elbow and the water which it contains is 1 kg. The elbow has a minor loss coefficient of 0.75 based on the inlet velocity. What force does the pipe exert on the elbow at the flange connection? D 1 = 5 cm ∆P ˙ m Pipe Flange g Elbow D 2 = 3 cm 23 ME 215.3 Example Problems 50. A pipe 4 cm in diameter supplies water to a 2 cm diameter nozzle. The gauge pressure just upstream of the nozzle is 60 kPa and the head loss in the nozzle is given by h L = 0.2V 2 e /2g where V e is the velocity at the nozzle exit. The jet of water (not shown) hits a frictionless splitter plate which is inclined at 30 o to the axis of the pipe. Half of the water is deflected downwards along this plate. Find the force required to hold the plate in this position. water 30 o ρ = 1000 kg/m 3 60 kPa 51. Water (ρ = 1000 kg/m 3 ) is pumped at volume flowrate Q through the nozzle shown. If the flowrate is high enough, a force F will be required to keep the cart stationary. Assume that the depth of the water in the cart remains at 0.8 m. Derive an expression for F as a function of Q and indicate the range of Q for which it is valid. Sketch the function. 2 m D = 5 cm 45 o F g Q 0.8 m 1 m 2 m ME 215.3 Example Problems 24 52. An engineer is measuring the lift and drag on an aerofoil section mounted in a two– dimensional wind tunnel. The wind tunnel is 0.5 m high and 0.5 m deep (into the paper). The upstream air velocity is uniform at 10 m/s. The downstream velocity is uniform at 12 m/s in the lower half of the wind tunnel. The downstream velocity in the upper half is uniform. The vertical component of velocity is zero at the beginning and end of the test section. The test section is 1 m long. The engineer measures the pressure distribution in the tunnel along the upper and lower walls and finds P u = 100 −10x −20x(1 −x) [Pa, gauge] P l = 100 −10x + 20x(1 −x) [Pa, gauge] where x is the distance in metres measured from the beginning of the test section. The air density is constant at 1.2 kg/m 3 . Find the lift and drag forces acting on the aerofoil. Neglect shear on the walls of the wind tunnel. of Test section of Beginning 1 m g 12 m/s x 10 m/s 0.25 m 0.25 m End Test section 53. A cart mounted on straight, level rails is used to test rocket engines. A 400 kg cart has a rocket mounted on it which has an initial mass of 500 kg. Eighty percent of the mass of the rocket is fuel. The products of combustion exhaust at a speed of 1000 m/s relative to the rocket nozzle. Neglect friction in the wheels of the cart and aerodynamic drag. Determine the speed of the rocket when the fuel is all used. g 25 ME 215.3 Example Problems 54. Water (ρ = 1000 kg/m 3 ) exits from a circular pipe 10 cm in diameter with a mass flow rate of 4 kg/s and strikes a flat plate at 90 o . The velocity at the exit varies linearly from a maximum at the pipe centreline to zero at the pipe wall as shown. Determine the force exerted on the flat plate. Compare this to the force that would be exerted at the same mass flow rate if the pipe exit velocity had been uniform. ˙ m = 4 kg/s 10 cm diameter 55. A two-arm sprinkler is constructed as shown below. The total mass flow rate of water (ρ = 1000 kg/m 3 ) is 1 kg/s and it is divided equally between the two arms. At the end of one arm a 12 mm diameter nozzle is oriented perpendicular to the arm and is in the same horizontal plane as the arm. The other arm has a 1 mm wide slot that also emits water in the horizontal plane. Find the rotational speed of the sprinkler in revolutions per minute assuming that the pivot is frictionless. Section A-A 15 cm 15 cm A A Plan View 12 mm dia. axis of rotation 1 mm 2 cm (enlarged) ME 215.3 Example Problems 26 56. Two circular coaxial jets of incompressible liquid with speed V collide as shown. The interaction region is open to atmosphere. Liquid leaves the interaction region as a conical sheet. Obtain an expression for the angle θ of the resulting flow in terms of d 1 and d 2 . d 1 V V V d 2 θ V 57. A tank of water sitting on a weigh scale is being filled with water (ρ = 1000 kg/m 3 ) from a 2 cm diameter pipe at a rate of 5 kg/s. The tank is circular and has a diameter of 0.5 m. The empty tank has a mass of 2 kg. At the instant when the water is 0.5 m deep in the tank, what force will the scale read? Carefully explain each of your assumptions. 5 kg/s g 0.5 m 0.5 m diameter 0.5 m 2 cm Scale 27 ME 215.3 Example Problems 58. Calculate the force of the water (ρ = 1000 kg/m 3 ) on the frictionless vane if (a) the blade is stationary, (b) the blade moves to the right at 20 m/s, and (c) the blade moves to the left at 20 m/s. 5 cm 60 o 40 m/s dia. 59. A jet of water (ρ = 998 kg/m 3 ) strikes a frictionless splitter vane as shown. The flowrate is 1 kg/s while the diameter of the nozzle is 1 cm. The position of the splitter vane is adjusted in the z direction so that there is no reaction in the z direction. What is the reaction in the x direction? vane 1 kg/s D = 1 cm splitter x z 45 o ME 215.3 Example Problems 28 60. Water at 20 o C flows from a nozzle of diameter D = 5 mm at speed V = 10 m/s and strikes a vane which splits the flow and deflects it as shown. Thirty-five percent of the mass flow deflects through 90 o (to the left). What force does the fluid exert on the vane? 45 o V D 61. Water at 20 o C flows through an elbow/nozzle arrangement and exits to atmosphere as shown below. The inlet pipe diameter is 12 cm while the nozzle exit diameter is 3 cm. The mass flowrate is 1 kg/s. If the pressure at the flange connection is 200 kPa(gauge), what is the force and moment on the flanged connection? Neglect the weight of the elbow and water. Water flanged 45 o 30 cm g 200 kPa connection 29 ME 215.3 Example Problems 62. A nozzle for a spray system is designed to produce a flat radial sheet of water. The sheet leaves the nozzle at V 2 = 10 m/s, covers 180 o of arc and has thickness t = 1.5 mm. The nozzle discharge radius is R = 50 mm. The water supply pipe is 35 mm in diameter and the inlet pressure P 1 is 50 kPa above atmospheric. Calculate the force exerted by the spray nozzle on the supply pipe through the flanged connection. R 35 mm supply pipe spray nozzle P 1 diameter Water thickness, t connection flanged V 2 63. A ride at an amusement park consists of a wheeled cart that zooms down an inclined plane onto a straight and level track where it decelerates to rest by means of a high- speed jet of air projected directly forward through a nozzle. The jet is supplied by a compressed air cylinder aboard the cart. The initial gross mass of the cart and its occupants is 500 kg. Air escapes from the braking jet at a constant mass flow rate of 20 kg/s and a constant velocity (relative to the nozzle) of 150 m/s. At the instant when the braking jet is activated, the speed of the cart is 40 m/s. Determine how much air must escape in order to stop the cart. What is the stopping distance? ME 215.3 Example Problems 30 64. Fluid enters a 5 cm diameter pipe with a uniform velocity of U o . The fluid exits the pipe at two locations. One is a 5 cm diameter exit with a turbulent velocity profile described by u z = U c 1 _ 1 − r R 1 _ 1/7 . where u z is the axial component of velocity, U c1 is the centreline velocity, r is the radial coordinate, and R 1 is the radius of the pipe. The second exit is 1.5 cm in diameter and has a laminar, parabolic velocity profile described by u z = U c 2 _ 1 − _ r R 2 _ 2 _ . If U c 1 is measured to be 2 m/s and U c 2 is measured to be 1 m/s, what is U o ? 1.5 cm diameter laminar velocity profile turbulent velocity profile 5 cm diameter uniform velocity 65. A 6 mm diameter angled nozzle is attached to the end of a 2 cm diameter pipe with a flanged connection as shown below. The angle between the nozzle and the supply pipe is 45 o . The pressure just upstream of the flanged connection is 200 kPa (gauge). Water (ρ = 998 kg/m 3 ) exits the nozzle at 20 m/s. Determine the force and torque in the flanged connection. 4 cm 200 kPa 2 cm diameter 6 mm diameter flanged connection 31 ME 215.3 Example Problems 66. A two-arm lawn sprinkler, as viewed from above, is shown in the sketch. Water is delivered to the sprinkler at a volume flow rate of 5 L/min and it can be assumed that this flow is split evenly between the two nozzles. The nozzles both have a diameter of 2 mm and are both angled upwards from the horizontal plane at 30 o . The density of the water is 998 kg/m 3 . Assume the pivot is frictionless. Calculate the rotational speed of the sprinkler. 15 cm 45 o 15 cm 67. Fluid (ρ = 850 kg/m 3 ) enters a 5-cm diameter pipe with a uniform velocity of 3 m/s at location A. At location B, the fluid has a turbulent velocity profile described by u z = U c _ 1 − r R _ 1/5 where u z is the axial component of velocity, U c is the centreline velocity, r is the radial coordinate, and R is the radius of the pipe. The pressure at location A is 4 kPa larger than at location B. Determine the viscous force on the wall of the pipe between location A and B. B A 5 cm diameter uniform velocity 68. A 2”ID steel pipe 50 m long carries water at a rate of 0.04 m 3 /s. There are two 90 o regular flanged elbows and an open flanged globe valve. The net elevation change is 30 m. Calculate the pressure difference between the ends of the pipe. (Solution: page 90) ME 215.3 Example Problems 32 69. Two reservoirs are connected by a pipe as shown. The elevation change between the two reservoirs is 20 m. Find the volume flowrate between the reservoirs. (Solution: page 92) 20 m ρ = 1000 kg/m 3 L = 5 m D = 2 cm ǫ = 0.05 mm D = 4 cm L = 5 m b a µ = 0.001 kg/(m· s) 70. Calculate the frictional pressure drop in 100 m of 3-cm diameter smooth pipe with water (ρ = 1000 kg/s, µ = 0.001 Pa · s) flowing at 1 kg/s. What would the frictional pressure loss be if the roughness were 0.5 mm? 71. The frictional pressure drop in 100 m of 3-cm diameter smooth pipe with water (ρ = 1000 kg/s, µ = 0.001 Pa · s) flowing is 1 MPa. Determine the volume flow rate. 72. A piping system contains a valve (K m = 6.5) and discharges water (ρ = 998 kg/m 3 , ν = 10 −6 m 2 /s) to atmosphere. A mercury manometer 20 m from the end of the pipe reads as shown. The right leg of the manometer is open to atmosphere. The pipe is smooth and its diameter is 5 cm. Determine the volume flowrate through the pipe in L/min. g 20 cm 20 cm mercury (s.g.=13.56) 33 ME 215.3 Example Problems 73. A pump has a characteristic curve that can be approximated by a parabola as shown. It pumps water through 100 m of 20 cm diameter cast iron pipe. What is the flowrate? (Solution: page 94) Flowrate(m 3 /s) P u m p H e a d ( m ) 2.5 2.0 1.5 1.0 0.5 0.0 100 80 60 40 20 0 74. Two large water (ν = 10 −6 m 2 /s) reservoirs are joined by two equal-length pipes as shown. The 25 mm diameter pipe has a roughness of 0.5 mm while the 20 mm diameter pipe is smooth. The flow rates through the two pipes are equal. What is the total flow rate between the two reservoirs? Neglect minor losses. 25 mm, ǫ = 0.5 mm g 20 mm, smooth ME 215.3 Example Problems 34 75. Water (ρ = 1000 kg/m 3 , µ = 0.001 Pa · s) flows through a horizontal section of 4 cm diameter pipe. The pipe has a roughness of 0.2 mm. A stagnation pressure tap and a static pressure tap are mounted 1 m apart as shown. The pressure difference between these two taps is measured with a mercury (s.g.=13.56) manometer which shows a displacement of 5 cm. The manometer tubes are otherwise full of water. What is the volume flow rate of the water? Stagnation tap g Q mercury (s.g.=13.56) 5 cm 1 m Static tap 35 ME 215.3 Example Problems 76. A vertical section of 4 cm diameter pipe with ǫ = 0.2 mm has two pressure gauges mounted 5 m apart which, at the current flow rate, read the same (P A = P B ) . A fluid with a kinematic viscosity of 4 ×10 −6 m 2 /s flows through the pipe. Does the fluid flow up or down? What is the volume flow rate? 5 m D = 4 cm g P B P A ME 215.3 Example Problems 36 77. A pump draws water (ρ = 1000 kg/m 3 , µ = 0.001 Pa · s) from a reservoir and discharges it into 100 m of 10 cm diameter pipe which has a roughness of 0.1 mm. The discharge of the pipe is 20 m lower than the surface of the reservoir. Neglect minor losses. If the pump characteristics are represented by the pump curve shown below, estimate the flowrate. Flow Rate (m 3 /min) H e a d ( m ) 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 60 55 50 45 40 35 30 25 20 15 10 5 0 37 ME 215.3 Example Problems 78. An engineer needs to measure the minor loss coefficient of a reducing elbow which has already been installed in a piping system. Because of limited access to the piping system, the only available pressure taps are 1 m upstream and 0.75 m downstream of the elbow. A mercury manometer is attached at these locations and reads 7.5 cm Hg when the mass flowrate is 1 kg/s. What is the minor loss coefficient for the elbow? Assume that all pipes are smooth. The flowing fluid is water with ρ = 1000 kg/m 3 and µ = 0.001 Pa · s. 2 cm diameter 1 m 7.5 cm 0.75 m Reducing elbow not to scale mercury (s.g.=13.56) g ˙ m = 1 kg/s 4 cm diameter ME 215.3 Example Problems 38 79. A centrifugal pump has a pump curve that can be described by h p = 4 −10Q 2 where h p is the pump head in metres of water and Q is the volume flowrate in L/min. This pump is used to supply water (ρ = 1000 kg/m 3 , ν = 10 −6 m 2 /s) to the system shown below. The sum of all minor loss coefficients is 12 and the total length of tube is 8 m. The tube has an inside diameter of 5 mm and roughness ǫ = 0.025 mm. What is the flowrate achieved? 2.0 m Diameter D = 5 mm Total length L = 8 m Roughness ǫ = 0.05 mm atmosphere Discharge to 80. A centrifugal pump draws water (ρ = 1000 kg/m 3 , µ = 0.001 Pa · s) from a large reservoir and pumps it through 1335 m of 30 cm inside diameter pipe with roughness ǫ = 0.5 mm. Over the range of interest the pump head can be expressed by h p = A −BQ 2 where h p is the head produced by the pump, Q is the volume flowrate, and A and B are constants. When the pipe exit is at the same elevation as the reservoir surface the flowrate is 17.0 m 3 /min. However, when the exit of the same pipe is raised to an elevation of 50 m the flow reduces to 13.3 m 3 /min. How high can the exit be raised before the flow will be zero? For all conditions the pipe discharges to atmosphere. Neglect minor losses in this problem. 39 ME 215.3 Example Problems 81. A 6 mm internal diameter, thin-walled, smooth rubber hose 12 m long is used to siphon water (ρ = 1000 kg/m 3 , ν = 10 −6 m 2 /s) from a large tank. The outlet of the hose is 6 m below the water surface in the tank. What will the volume flowrate be? Neglect minor losses due to bends in the hose. 6 m 6 mm I.D., 12 m long 82. The piping system shown is fitted with a centrifugal pump whose characteristics can be approximated by h p = 46−2.5Q 2 where h p is the head produced by the pump in m of water and Q is the volume flowrate in m 3 /min. What is the flowrate through this piping system? Gate valve All pipes have ǫ = 0.15 mm All elbows shown are 90 o , regular, flanged. Pump L = 340 m D = 200 mm L = 170 m D = 100 mm Sudden contraction expansion Sudden Fully open ME 215.3 Example Problems 40 83. You are going to measure the minor loss coefficient of a new valve design using the apparatus shown. Assume that the water tank is large and the pipe discharges to atmosphere. The 2-cm diameter smooth pipe is 4 m long. The free surface in the tank is 5 m above the pipe exit. If you measure the flowrate to be 60 L/min what is the minor loss coefficient of the valve? Let ν = 10 −6 m 2 /s. New valve D = 2 cm 5 m L = 4 m 84. A centrifugal pump is connected to a piping system as shown below. The pipe is 5 cm in diameter and has a roughness of ǫ = 0.5 mm. The two valves are identical and have minor loss coefficients of 1.0 when fully open and 20 when 50% open. With both valves fully open, the pressure guage reads P A = 250 kPa. When both valves are 50% open, the flow is 355 L/min. The pump curve can be represented by a parabola. The free surface in the tank is at the same elevation as the pipe exit. Let ν = 10 −6 m 2 /s and ρ = 998 kg/m 3 . Determine P A if the first valve is fully open and the second valve is fully closed. Neglect losses which occur on the suction side of the pump. 25 m 25 m 25 m 25 m Valve Valve Pump P A 41 ME 215.3 Example Problems 85. A portion of a piping system is shown below. At point B, the piping system discharges to the atmosphere. The two elbows each have a minor loss coefficient of 2.4. The valve has a minor loss coefficient of 6.9. The pipe is 25 mm in diameter and has a roughness of 0.05 mm. There is a total of 15 m of pipe between point A and point B. Point B is 3 m above point A. The pressure P A is 200 kPa guage. Determine the volume flow rate (in L/min) if the fluid is water with ρ = 998 kg/m 3 and µ = 0.001 Pa · s. B A P A Valve ME 215.3 Example Problems 42 86. Flow around a certain bridge pier can be modelled by a freestream and a single source. If the freestream velocity is 5 m/s and the pressure a long distance upstream is 50 kPa, what is the pressure at point A on the pier surface? The stagnation point will be 1 m upstream of the source. (Solution: page 96) A y x 2 m 43 ME 215.3 Example Problems 87. A pair of doublets of equal strength are placed in a freestream as indicated below. The streamline which passes through the origin also passes through the point (4 m, 1 m). What is the doublet strength? Referenced to P ∞ , what is the pressure at the origin? (Solution: page 99) x U ∞ = 10 m/s doublets 30 o 2 m 2 m 2 m y 88. The landing gear strut on a small aircraft has a cross–section as indicated below. This shape can be modelled as a Rankine Oval with a source and a sink placed as shown. The aircraft is flying at 45 m/s and the air density is 1.2 kg/m 3 . What is the difference in static pressure between points A and B? (Solution: page 101) A U ∞ 12 cm 12 cm B sink source 30 cm ME 215.3 Example Problems 44 89. Consider the irrotational flow around a circular cylinder which is creating no lift. (a) Derive an expression for the velocity along the positive y axis in terms of U ∞ , a, and y. (b) Sketch this function. (c) Consider the streamline labelled A. Far upstream of the cylinder this streamline is a distance L from the horizontal plane of symmetry. At what distance from the origin does the streamline cross the y axis? (Solution: page 104) a U ∞ g L A x y 45 ME 215.3 Example Problems 90. Consider the flow of air (ρ = 1.2 kg/m 3 ) over a 20 cm diameter circular cylinder. The freestream velocity is 20 m/s. The volume flowrate per unit length between the two streamlines labelled A and B is 1.19 m 2 /s. The streamline labelled B passes through the point (x, y) = (0, 0.12 m). What is the lift force per unit length on the cylinder? (Solution: page 107) −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 B A y x (x, y) = (0, 0.12 m) 91. An inviscid flow of air (ρ = 1.2 kg/m 3 ) is produced by a freestream (U ∞ = 50 m/s) aligned with the x axis, a source at the origin, and a second source of equal strength located at (x, y) = (0.1 m, 0). The stagnation streamline crosses the y axis at y = 9 cm. Determine the gauge pressure at the point where the stagnation streamline crosses the y axis. Determine the acceleration 2 cm upstream of the first stagnation point. ME 215.3 Example Problems 46 92. An inviscid flow is produced by a freestream (U ∞ = 10 m/s) aligned with the x axis, three sources placed as shown, and a single sink with a strength appropriate to form a closed streamline in the flow. The streamline shown on the diagram is a distance a from the x axis far upstream of the origin and passes through the point (x, y) = (0, 2a). Determine the thickness of the object at x = 0 if a = 0.2 m. a sink sources x 2a y a a U ∞ = 10 m/s 93. The maximum pressure difference between any two points on the surface of the Rankine oval shown below is measured to be 500 Pa. The source and the sink are placed 40 cm apart. The maximum thickness of the oval is 20 cm. Determine the freestream velocity if the density of the air is 0.8 kg/m 3 . U ∞ source sink 40 cm 2 0 c m 47 ME 215.3 Example Problems Solution: Problem # 1 Given: • Air trapped over oil in a closed tank as shown g z 1 B A P atm = 101.3 kPa h 2 = 6 c m h 3 = 4 c m h 1 = 1 5 c m Air Air Water s.g = 0.89 Oil Open to atmosphere Find: • Calculate the air pressure in the tank. Assumptions: • fluid is in a hydrostatic state • no acceleration • density of the air is negligible • ρ w = 998 kg/m 3 (constant) • ρ oil is constant • g is constant Analysis: ∇P = ρ(g −a) g = −g ˆ k ME 215.3 Example Problems 48 a = 0 ∇P = −ρg ˆ k dP dz = −ρg dP = −ρgdz Since ρ and g are constant, ∆P = −ρg∆z Now, P 1 = (P 1 −P B ) + (P B −P A ) + (P A −P atm ) + P atm P 1 = −ρ oil g(z 1 −z B ) −ρ air g(z B −z A ) −ρ w g(z A −z atm ) + P atm z 1 −z B = h 3 z A −z atm = −h 1 P 1 = −ρ oil gh 3 + ρ w gh 1 + P atm P 1 = ρ w g(h 1 −sg oil h 3 ) + P atm P 1 = 998 _ kg m 3 _ 9.81 _ m s 2 _ (0.15[m] −0.89(0.04)[m]) + 101, 300[Pa] P 1 = 102.4 kPa(abs) The pressure in the air chamber is 102.4 kPa (abs). 49 ME 215.3 Example Problems Solution: Problem # 2 Given: • A rectangular gate holding a pool of water. g x z gate (5 m wide) hinge ρ = 1000 kg/m 3 P atm = 100 kPa 3 m 3 m stop Find: • The force which the water exerts on the gate Assumptions: • fluid is in a hydrostatic state • no acceleration • g is constant • ρ is constant ME 215.3 Example Problems 50 Analysis: Establish a coordinate system (see diagram). F s = − __ S Pˆ ndA ∇P = ρ(g −a) g = g ˆ k a = 0 ∇P = ρg ˆ k dP dz = ρg dP = ρgdz _ dP = ρg _ dz P = ρgz + C at z = 0, P = P atm . Therefore, C = P atm . P = ρgz + P atm ˆ n = ˆı F s = − _ z 2 z 1 _ w 0 (ρgz + P atm )ˆıdydz F s = −ˆı _ z 2 z 1 (ρgz + P atm )dz _ w 0 dy F s = −ˆı w _ ρgz 2 2 + P atm z _ z 2 z 1 F s = −ˆı 5[m] _ 1000 _ kg m 3 _ 9.81 _ m s 2 _ (6 2 −3 2 ) 2 [m 2 ] + 100, 000[Pa](6 −3)[m] _ F s = −2.16ˆı MN The water will exert a horizontal force of 2.16 MN on the gate. 51 ME 215.3 Example Problems Solution: Problem # 3 Given: • A Tainter gate holding back a pool of water R = 1 . 2 2 m g z x ρ = 1000 kg/m 3 gate length=2.44 m Find: • The net hydrostatic force on the gate Assumptions: • atmosphere is constant pressure • fluid is in a hydrostatic state • density of water is constant • g is constant • no acceleration Analysis: Establish coordinate system (see diagram). F s = − __ S Pˆ ndA ∇P = ρ(g −a) g = −g ˆ k a = 0 ME 215.3 Example Problems 52 dP dz = −ρg P = −ρgz + C at z = R/ √ 2, P = 0 (gauge). Therefore, C = ρgR/ √ 2. P = ρg(R/ √ 2 −z) ˆ n = cos θˆı + sin θ ˆ k θ ˆ n dA = Rdθdy F s = − _ w 0 _ θ 2 θ 1 ρg(R/ √ 2 −z)(cos θˆı + sin θ ˆ k)Rdθdy z = Rsin θ F s = −ρgwR 2 _ θ 2 θ 1 _ cos θˆı √ 2 + sin θ ˆ k √ 2 −sin θ cos θˆı −sin 2 θ ˆ k _ dθ Recall that sin 2 θ = 1 −cos 2θ 2 F s = −ρgwR 2 _ sin θˆı √ 2 − cos θ ˆ k √ 2 − sin 2 θ 2 ˆı − 1 2 _ θ − sin 2θ 2 _ ˆ k _ θ 2 θ 1 F s = −ρgwR 2 _ ˆı _ sin θ √ 2 − sin 2 θ 2 _ − ˆ k _ cos θ √ 2 + θ 2 − sin 2θ 4 __ π/4 −π/4 F s = − 1000 _ kg m 3 _ 9.81 _ m s 2 _ 2.44[m]1.22 2 [m 2 ] _ ˆı __ 1 2 − 1 4 _ − _ − 1 2 − 1 4 __ − ˆ k __ 1 2 + π 8 − 1 4 _ − _ 1 2 − π 8 + 1 4 ___ 53 ME 215.3 Example Problems F s = −35630[N] _ ˆı − ˆ k _ π 4 − 1 2 __ F s = _ −35.6ˆı + 10.2 ˆ k _ kN With respect to the coordinate system shown on the diagram, the net hydrostatic force on the gate is (−35.6ˆı + 10.2 ˆ k) kN. ME 215.3 Example Problems 54 Solution: Problem # 4 Given: • A rectangular gate hinged at the top holding back a pool of water g x z gate (5 m wide) hinge ρ = 1000 kg/m 3 3 m 3 m P atm = 100 kPa stop Find: • The force which the stop exerts on the gate Assumptions: • fluid is in a hydrostatic state • no acceleration • ρ is constant • g is constant • hinge is frictionless Analysis: Establish coordinate system (see diagram) Draw a free body diagram of the gate. 55 ME 215.3 Example Problems x F s F b F t z M o = 0 (H ˆ k × F b ) + M s = 0 M s = − __ S P(r × ˆ n)dA r = z ˆ k ˆ n = ˆı r × ˆ n = z ˆ k ׈ı = zˆ  ∇P = ρ(g −a) a = 0 g = g ˆ k dP dz = ρg P = ρgz + C at z = 0, P = ρgh 0 . Therefore, C = ρgh 0 . P = ρg(z + h 0 ) dA = dydz M s = − _ w 0 _ z 2 z 1 (zˆ )ρg(z + h 0 )dzdy M s = −ρgwˆ  _ z 2 z 1 (z 2 + h 0 z)dz M s = −ρgwˆ  _ z 3 3 + h 0 z 2 2 _ z 2 z 1 M s = −1000 _ kg m 3 _ 9.81 _ m s 2 _ 5[m]ˆ  _ 3 3 3 [m 3 ] + 3[m]3 2 2 [m 2 ] _ ME 215.3 Example Problems 56 M s = −1104ˆ  kN · m Recall that, (H ˆ k × F b ) + M s = 0 ˆ k × F b = 1104ˆ  3[m] [kN · m] F b = 368 kNˆı The stop exerts a force of 368 kN on the gate. 57 ME 215.3 Example Problems Solution: Problem # 22 Given: • An object suspended in a liquid by a wire g z 10 cm cube, m = 35 kg Find: • The specific gravity of the fluid Assumptions: • fluid is in a hydrostatic state • no acceleration • block is in equilibrium • mass and volume of wire are zero • ρ is constant • g is constant Analysis: Establish coordinate system (see diagram). F z = 0 F b + T −W = 0 ME 215.3 Example Problems 58 ρgV + T −mg = 0 ρ9.81 _ m s 2 _ (0.1[m]) 3 + 335.5[N] −35[kg]9.81 _ m s 2 _ = 0 ρ = 800 kg/m 3 sg = 800/998 The specific gravity of the liquid is 0.801. 59 ME 215.3 Example Problems Solution: Problem # 23 Given: • A U–tube manometer used to measure linear acceleration g L z x a 2 1 d h Find: • The magnitude of a in terms of geometry, g, and h Assumptions: • a constant • fluid is in a hydrostatic state • ρ and g are constant Analysis: Establish coordinate system (see diagram). Points 1 and 2 are at the same pressure. Therefore, _ 2 1 dP = _ 2 1 dR · ∇P = 0 dR = dxˆı + dz ˆ k ∇P = ρg −ρa ME 215.3 Example Problems 60 g = −g ˆ k a = aˆı ∇P = −ρg ˆ k −ρaˆı dR · ∇P = −ρadx −ρgdz 0 = _ 2 1 (ρadx + ρgdz) = ρa(x 2 −x 1 ) + ρg(z 2 −z 1 ) 0 = ρaL + ρg(−h) a = gh/L The acceleration of the tube is gh/L. 61 ME 215.3 Example Problems Solution: Problem # 26 Given: • A partially full container rotating around its axis. 1 2 z r g ω = 50 rpm 50 cm 30 cm Find: • The depth of the fluid on the axis Assumptions: • ω is constant • fluid is in a hydrostatic state • ρ is constant • g is constant Analysis: Establish coordinate system (see diagram). The pressure at 1 and 2 are equal. Therefore, _ 2 1 dP = 0 = _ 2 1 dR · ∇P dR = drˆ r + dz ˆ k ME 215.3 Example Problems 62 ∇P = ρ(g −a) g = −g ˆ k a = −ω 2 rˆ r ∇P = −ρg ˆ k + ρω 2 rˆ r 0 = _ 2 1 (−ρgdz + ρω 2 rdr) ρg(z 2 −z 1 ) = ρω 2 2 (r 2 2 −r 2 1 ) 0.3[m] −z 1 = (5π/3[1/s]) 2 (0.25[m]) 2 2(9.81)[m/s 2 ] z 1 = 21.3 cm The fluid is 21.3 cm deep on the centreline. 63 ME 215.3 Example Problems Solution: Problem # 27 Given: • A cylindrical container rotating about its axis. g r z ω R H o Find: • Derive an equation for the shape of the liquid surface if the container spins at ω about the z axis. Assumptions: • ω is constant • fluid is in a hydrostatic state • g is constant • ρ is constant Analysis: Establish coordinate system (see diagram). The surface is a line of constant pressure. Therefore, dP = 0 along the surface. 0 = dR · ∇P ME 215.3 Example Problems 64 dR = drˆ r + dz ˆ k a = −ω 2 rˆ r ∇P = ρ(ω 2 rˆ r −g ˆ k) 0 = ρω 2 rdr −ρgdz _ ω 2 r g dr = _ dz z = ω 2 r 2 2g + C Equate the initial and final volumes. _ R 0 z2πrdr = H 0 πR 2 H 0 πR 2 = 2π _ R 0 _ ω 2 r 3 2g + Cr _ dr H 0 R 2 = 2 _ ω 2 r 4 8g + Cr 2 2 _ R 0 H 0 R 2 = ω 2 R 4 4g + CR 2 C = H 0 − ω 2 R 2 4g The surface is defined by z = H 0 + ω 2 2g _ r 2 − R 2 2 _ 65 ME 215.3 Example Problems Solution: Problem # 31 Given: • A closed cylindrical container rotating about its axis. r 60 rpm z g cylindrical can ρ = 1000 kg/m 3 vent Find: • The force on the top of the can (net hydrostatic force) Assumptions: • fluid is in a hydrostatic state • ρ is constant • g is constant Analysis: Establish a coordinate system (see diagram). F s = − __ S Pˆ ndA ∇P = ρ(g −a) g = −g ˆ k ME 215.3 Example Problems 66 a = −ω 2 rˆ r ∇P = −ρg ˆ k + ρω 2 rˆ r Need P as a function of r. Therefore, ∂P ∂r = ρω 2 r P = ρω 2 r 2 2 + C at r = 0, P = 0. Therefore, C = 0. ˆ n = − ˆ k dA = rdrdθ F s = − _ 2π 0 _ R 0 ρω 2 r 2 2 (− ˆ k)rdrdθ F s = 2πρω 2 2 ˆ k _ R 0 r 3 dr = πρω 2 ˆ k R 4 4 F s = π1000 _ kg m 3 _ _ 2π _ 1 s __ 2 0.4 4 [m 4 ] 4 ˆ k F s = 794 ˆ k N The net hydrostatic force on the lid of the can is 794 N upwards. 67 ME 215.3 Example Problems Solution: Problem # 35 Given: • A jet of water hitting a flat barricade. A j = 0.01 m 2 V = 15 m/s 2 2 1 y x g Find: • The force that the water exerts on the barricade. Assumptions: • steady state • jet is horizontal • jet deflects to vertical plane • pressure uniform on control surface • neglect body forces • neglect fluid shear • uniform flow ME 215.3 Example Problems 68 Analysis: Since we are asked to find a force we should probably consider the linear momentum equation. ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = d dt ___ V − (ρ V )dV − + __ S ρ V ( V r · ˆ n)dA Consider the control volume shown. Each term of the momentum equation will now be discussed. Body Force: ___ V − ρgdV − = −W ˆ k W is the weight of EVERYTHING in the control volume, water, air, barricade etc. Pressure Force: __ S Pˆ ndA = 0 P is uniform over the entire control surface. Therefore, the net pressure force is ZERO. Viscous Force: F v = 0 The flow is perpendicular to the control surface everywhere. Reaction: R = 0 We have chosen a control surface which “cuts” the support so we can find R. 69 ME 215.3 Example Problems Unsteady term: d dt ___ V − (ρ V )dV − = 0 Steady state is assumed. Momentum transport term: __ S ρ V ( V r · ˆ n)dA = d ( ˙ m V ) d − i ( ˙ m V ) i The properties at all inlets and discharges are assumed to be uniform. With these simplifications, the momentum equation becomes −W ˆ k + R = (ρ w A 2 V 2 ) V 2 + (ρ w A 3 V 3 ) V 3 −(ρ w A 1 V 1 ) V 1 Consider only the x direction R x = −ρ w A 1 V 2 1 R x = −998 _ kg m 3 _ 0.01[m 2 ] _ 15 _ m s __ 2 = −2.25 kN This is the force exerted on the control volume by the “mounting strut”. The force exerted on the barricade is in the opposite direction. The water exerts a force of 2.25kN to the right on the barricade. ME 215.3 Example Problems 70 Solution: Problem # 36 Given: • A jet of water deflecting off a vane mounted on a moving cart 45 o V c = 5 m/s constant A j = 0.01 m 2 V j = 15 m/s cart Find: • Force which the water exerts on the cart Assumptions: • neglect body forces • neglect friction on vane • uniform flow • cart travelling at constant velocity Analysis: Since we are asked to find a force we should probably consider the linear momentum equa- tion. ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = d dt ___ V − (ρ V )dV − + __ S ρ V ( V r · ˆ n)dA We must choose a control volume. 71 ME 215.3 Example Problems Should it be stationary or moving? (we will try both) What should it look like? First consider what the control volume should look like. It is important to consider what we are being asked to solve for. In this case we are looking for the force that the water exerts on the cart. Therefore, one possible choice for a control volume is . plate (gap shown only for clarity) C.S. between water and Consider the left-hand side of the momentum equation for this choice. ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = ... • The body force term is assumed to be zero. We are neglecting the weight of the water on the vane. • The pressure term is unknown. – the pressure everywhere except on the surface of the plate is zero – the integrated effect of the pressure acting on the plate is what we were asked to find in this problem – this term gives the force ON the CV. Therefore, it is equal and opposite to the unknown in this problem • The viscous force term is zero since we are assuming a frictionless surface. • The reaction force is zero since this CS does not cut through any solid objects. ME 215.3 Example Problems 72 Another choice for a CV may be as follows. Consider the LHS of the momentum equation now. ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = ... • The body force term is zero as before. • The pressure is now zero everywhere on the control surface. • The viscous force term is zero as before. • Now the control surface cuts through the cart. Therefore, the reaction force is NOT zero. If we consider a free–body–diagram of the cart, we realise that R is opposite to the force which the water exerts on the cart. Moral of the Story!! Either choice for a CV is fine. It only changes which term on the LHS gives us the information we want. We will continue the problem with the second option. 1. Stationary Control Volume 73 ME 215.3 Example Problems Attach the coordinate system AND the control volume to the cart. z x • This is considered a stationary control volume because it is not moving with respect to the coordinate system. • The coordinate system is inertial because it is moving at constant velocity. We have already simplified the LHS so now, R = d dt ___ ρ V dV− + __ ρ V ( V r · ˆ n)dA Since nothing in the CV is changing with time, it is a steady problem. Also, assume uniform flow. R = ( ˙ m V ) out −( ˙ m V ) in or, because ˙ m out = ˙ m in R = ˙ m( V out − V in ) ˙ m = ρ(V j −V c )A j V in = (V j −V c )ˆı Assuming that the vane is frictionless, the magnitude of this velocity will not change at the outlet. V out = (V j −V c ) _ √ 2 2 ˆı + √ 2 2 ˆ k _ ME 215.3 Example Problems 74 R = ρ(V j −V c )A j _ (V j −V c ) _ √ 2 2 _ (ˆı + ˆ k) −(V j −V c )ˆı _ R = ρ(V j −V c ) 2 A j __ √ 2 2 −1 _ ˆı + √ 2 2 ˆ k _ R = 998 _ kg m 3 _ _ 10 _ m s __ 2 0.01[m 2 ] __ √ 2 2 −1 _ ˆı + √ 2 2 ˆ k _ R = (−292ˆı + 705 ˆ k) N Remember, this is opposite to the force the water exerts on the cart. The water exerts a force of (292ˆı − 705 ˆ k) N on the cart. The coordinate system is shown on the diagram. 2. Moving Control Volume Now, attach the coordinate system to the pipe but attach the control volume to the cart. The flow inside the control volume is still steady. Therefore, R = ˙ m _ V out − V in _ is still true and ˙ m = ρ(V j −V c )A j is still true. However, V out and V in are different. Now, V in = V j ˆı V out still has a magnitude of (V j − V c ) when viewed with respect to the vane. However, in our coordinate system, the forward velocity of the cart must be added to this. ( V j − V c ) √ 2 / 2 ( ˆ i + ˆ k ) V out V c ˆ i V out = (V j −V c ) √ 2 2 (ˆı + ˆ k) + V c ˆı 75 ME 215.3 Example Problems Now, R = ρ(V j −V c )A j _ (V j −V c ) √ 2 2 (ˆı + ˆ k) + V c ˆı −V j ˆı _ R = ρ(V j −V c )A j _ (V j −V c ) √ 2 2 (ˆı + ˆ k) −(V j −V c )ˆı _ This intermediate result is identical to the stationary CV. Therefore, the answer will be identical. 3. Stationary Control Volume Try again with a different stationary control volume. Attach the coordinate system to the supply pipe. Draw a stationary CV such that the cart is in it at a certain instant in time. How is the analysis different now? Again, the LHS is the same. R = d dt ___ V − ρ V dV − −+ ( ˙ m V ) out −( ˙ m V ) in The unsteady term is NOT ZERO anymore. The uniform flow assumption is still valid. Consider a sketch showing the CV at two times. The total momentum ( ___ ρ V dV −) in (ii) is greater than in (i) because the column of liquid going at V j ˆı is longer. This column is getting longer at speed V c . Therefore, d dt ___ ρ V dV − = ρV j ˆıA j V c Now, at the inlet ( ˙ m V ) in = (ρV j A j )V j ˆı However, note that ˙ m in = ˙ m out for this analysis because mass is accumulating in the CV. ˙ m out = ˙ m in −ρA j V c ˙ m out = ρV j A j −ρV c A j = ρA j (V j −V c ) Back to the momentum equation. R = ρV j V c A j ˆı + ρA j (V j −V c ) __ √ 2 2 _ (V j −V c )(ˆı + ˆ k) + V c ˆı _ −ρV 2 j A j ˆı ME 215.3 Example Problems 76 (i) (ii) z x z x R = ρ(V j −V c )A j _ (V j −V c ) √ 2 2 (ˆı + ˆ k) −(V j −V c )ˆı _ Again, this is the same intermediate result so the answer will be the same. 77 ME 215.3 Example Problems Solution: Problem # 37 Given: • A nozzle which discharges water from a pipe 1 x z D 2 = 25 mm V 1 = 1 m/s D 1 = 50 mm 10 kPa 2 Find: • The force on the bolts at the flange connection Assumptions: • neglect body forces • no viscous force on CV • uniform flow • steady state Analysis: Since we are asked to find a force, consider the linear momentum equation. Choose a control volume that will help us find what we are after. Therefore, cut the bolts with the control surface. A coordinate system is also shown on the diagram. Consider the linear momentum equation. ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = d dt ___ V − (ρ V )dV − + __ S ρ V ( V r · ˆ n)dA ME 215.3 Example Problems 78 Consider each term in this equation. 1. Neglect the body force. We have no information about weight of the nozzle. 2. Use gauge pressures so that the only non–zero pressure is at 1. P 1 = 10 kPa ˆ n = −ˆı Since P 1 is uniform and ˆ n doesn’t vary, − __ P ˆ ndA = −P 1 (−ˆı)A 1 = P 1 A 1 ˆı 3. Neglect the viscous force term. There is no flow parallel to the CS. 4. Since our CS cuts the bolts, the reaction force will give us what we want to find. 5. The unsteady term on the right-hand-side is zero because this is a steady problem. 6. Assume uniform flow to simplify the last term in the equation. So, the momentum equation becomes P 1 A 1 ˆı + R = ˙ m( V out − V in ) = ˙ m( V 2 − V 1 ) V 1 = 1ˆı m/s Find the magnitude of V 2 from conservation of mass d dt _ _ V − _ (ρ)dV − + _ S _ ρ( V · ˆ n)dA = 0 For steady, uniform flow this becomes ρV 1 A 1 = ρV 2 A 2 V 2 = V 1 A 1 A 2 = V 1 _ 50 25 _ 2 V 2 = 4ˆı m/s R = −10 ×10 3 [Pa] π 4 (0.05) 2 [m 2 ]ˆı + 998 _ kg m 3 _ 1 _ m s _ π 4 (0.05) 2 [m 2 ](4ˆı −1ˆı) _ m s _ R = −13.8ˆı N This means the bolts are exerting a force ON the CV in the −ˆı direction. Therefore, the bolts are in tension. There is a tensile force on the bolts of 13.8 N. 79 ME 215.3 Example Problems Solution: Problem # 38 Given: • The inlet section of a laminar pipe flow 1 u(r) = U max (1 −(r/R) 2 ) radius= R circular pipe U o r z 2 Find: • Frictional drag on the fluid between 1 and 2 in terms of P 1 , P 2 , ρ, U o , and R Assumptions: • neglect body forces • uniform flow at 1 • steady state • ρ is constant Analysis: Since we are asked for a force (drag) on the fluid we should consider the linear momentum equation. A coordinate system has already been given in the problem. ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = d dt ___ V − (ρ V )dV − + __ S ρ V ( V r · ˆ n)dA The dashed line on the diagram is chosen as the CV. Consider each term of the momentum equation. ME 215.3 Example Problems 80 1. Since no information has been give about the orientation of the pipe and we are only interested in frictional drag, ignore the body force term. 2. The only significant pressure forces on our CS are at 1 and 2. Although pressure is certainly acting on the rest of the CS, it is equal around the circumference and therefore yields no net force. 3. The viscous force term is the unknown in this problem. We have chosen a CV including just the water on the pipe so that the friction at the pipe walls enters the momentum equation through this term. 4. R = 0 since no solid member penetrates the CS. 5. Since this is apparently a steady flow, assume the unsteady term is zero. 6. The final term in the equation must be handled carefully. Although we do have uniform flow at the inlet, we certainly don’t at the outlet. Therefore, we can apply the uniform flow assumption at 1, but not at 2. (P 1 −P 2 )πR 2 ˆ k + F v = __ 2 ρ V ( V r · ˆ n)dA−( ˙ m V ) 1 At 2, ( V r · ˆ n) = u(r) V 2 = U max _ 1 − _ r R _ 2 _ ˆ k dA = rdθdr F v = (P 2 −P 1 )πR 2 ˆ k + _ R 0 _ 2π 0 ρ _ U max _ 1 − _ r R _ 2 __ 2 ˆ krdrdθ −U o πR 2 ρU o ˆ k F v = (P 2 −P 1 )πR 2 ˆ k + 2πρU 2 max ˆ k _ R 0 _ 1 − 2r 2 R 4 + r 4 R 4 _ rdr −U 2 o πR 2 ρ ˆ k F v = (P 2 −P 1 )πR 2 ˆ k + 2πρU 2 max ˆ k _ r 2 2 − 2r 4 4R 4 + r 6 6R 4 _ R 0 −U 2 o πR 2 ρ ˆ k F v = πR 2 _ (P 2 −P 1 ) + ρU 2 max 3 −ρU 2 o _ ˆ k Eliminate U max with conservation of mass (steady). d dt _ _ V − _ (ρ)dV − + _ S _ ρ( V · ˆ n)dA = 0 0 = __ 2 ρU max _ 1 − _ r R _ 2 _ rdrdθ −U o ρπR 2 81 ME 215.3 Example Problems 0 = ρU max 2π _ R 0 _ r − r 3 R 2 _ dr −U o ρπR 2 0 = ρU max 2π _ r 2 2 − r 4 4R 2 _ R 0 −U o ρπR 2 0 = ρU max 2π R 2 4 −U o ρπR 2 U max = 2U o F v = πR 2 _ (P 2 −P 1 ) + ρ(2U o ) 2 3 −ρU 2 o _ ˆ k The frictional drag on the fluid is πR 2 _ (P 2 −P 1 ) + ρU 2 o 3 _ ME 215.3 Example Problems 82 Solution: Problem # 39 Given: • A nozzle mounted at a pipe exit D 2 = 25 mm x z ? V 1 = 1 m/s D 1 = 50 mm 2 1 Find: • Force on the bolts assuming frictionless flow Assumptions: • frictionless flow • uniform inlet and outlet • steady, incompressible flow • no shaft work or heat transfer • no elevation change Analysis: The control volume and coordinate system are shown above. Since we are asked for a force, consider the momentum equation. ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = d dt ___ V − (ρ V )dV − + __ S ρ V ( V r · ˆ n)dA With the assumptions listed this becomes (see solution on page 79 for more details) R = ρV 1 A 1 __ V 1 A 1 A 2 _ ˆı −V 1 ˆı _ −P 1 A 1 ˆı 83 ME 215.3 Example Problems Bernoulli’s equation is P 1 + ρV 2 1 2 + ρgz 1 = P 2 + ρV 2 2 2 + ρgz 2 Since z 1 = z 2 P 1 = ρ 2 (V 2 2 −V 2 1 ) = 998 2 _ kg m 3 _ (4 2 −1 2 ) _ m 2 s 2 _ P 1 = 7485 Pa R = 1000 _ kg m 3 _ 1 _ m s _ π 4 (0.05) 2 [m 2 ] _ 1 _ m s _ _ 50 25 _ 2 ˆı −1 _ m s _ ˆı _ −7485[Pa] π 4 (0.05) 2 [m 2 ] = −8.82ˆı N The tensile force in the bolts is 8.82 N. ME 215.3 Example Problems 84 Solution: Problem # 40 Given: • A nozzle mounted at the exit of a pipe D 2 = 25 mm V 1 = 1 m/s D 1 = 50 mm 10 kPa z x 2 1 Find: • The loss coefficient for the nozzle Assumptions: • steady, incompressible flow • no shaft work, no heat transfer • no elevation change Analysis: Bernoulli’s equation P 1 + ρV 2 1 2 + ρgz 1 = P 2 + ρV 2 2 2 + ρgz 2 + ∆P L ∆P L = P 1 + ρ 2 _ V 2 1 −V 2 2 _ ∆P L = 100000[Pa] + 998 2 _ kg m 3 _ _ 1 2 _ m 2 s 2 _ −4 2 _ m 2 s 2 __ ∆P L = 2515 Pa ∆P v = K m ρV 2 1 2 K m = 5.04 The loss coefficient of this nozzle based on the inlet velocity is 5.04. 85 ME 215.3 Example Problems Solution: Problem # 41 Given: • A transition piece as shown K m = 0.50 (based on . discharge velocity) y x 20 kPa D 1 = 50 mm D 2 = 80 mm 45 o 4 kg/s 2 1 Find: • Force required to hold transition in place Assumptions: • neglect body forces • uniform inlet and outlet • steady, incompressible flow • no shaft work or heat transfer Analysis: Since we are asked to find a force, consider the linear momentum equation. Begin with the complete equation ___ V − ρgdV − − __ S Pˆ ndA+ F v + R = d dt ___ V − (ρ V )dV − + __ S ρ V ( V r · ˆ n)dA • Since we have no information about the mass of the transition, or even the direction of g, we will neglect body forces. ME 215.3 Example Problems 86 • Choose a CV which isolates the transition piece. Therefore, R becomes the unknown in the problem. • A uniform pressure exists at 1 and 2. • The flow is steady so d dt ___ ρ V dV − = 0 • Use the uniform flow assumption for the last term since we have no information about how the velocity varies across the pipe. Now, we have − __ Pˆ ndA + R = ˙ m( V 2 − V 1 ) V 1 = ˙ m ρA 1 ˆı = 4[kg/s] 998[kg/m 3 ] 4 π(0.05) 2 [m 2 ] = 2.04ˆı m/s V 2 = ˙ m ρA 2 _ √ 2 2 _ (ˆı + ˆ ) = √ 2 2 4[kg/s] 998[kg/m 3 ] 4 π(0.08) 2 [m 2 ] (ˆı + ˆ ) = 0.5638(ˆı + ˆ ) m/s Now, let’s work on the pressure force term. At 1, P 1 = 20 kPa ˆ n = −ˆı At 2 ˆ n = _ √ 2 2 _ (ˆı + ˆ ) P 2 =? We must find P 2 using Bernoulli’s equation. Write Bernoulli’s equation (with pressure losses) from 1 to 2. P 1 + ρV 2 1 2 = P 2 + ρV 2 2 2 + ∆P L ∆P L = K m ρV 2 2 2 P 2 = P 1 + ρ 2 (V 2 1 −V 2 2 ) −K m ρV 2 2 2 87 ME 215.3 Example Problems P 2 = 20, 000[Pa] + 998 2 _ kg m 3 _ _ (2.04) 2 −2(0.5638) 2 _ _ m 2 s 2 _ − 0.5 998 2 _ kg m 3 _ 2(0.5638) 2 _ m 2 s 2 _ P 2 = 21.6 kPa Now, return to the momentum equation R = −P 1 A 1 ˆı + P 2 A 2 √ 2 2 (ˆı + ˆ ) + ˙ m( V 2 − V 1 ) R = − 20, 000[Pa] π 4 (0.05) 2 [m 2 ]ˆı + 21, 600[Pa] π 4 (0.08) 2 [m 2 ] √ 2 2 (ˆı + ˆ ) + 4 _ kg s _ (0.5638ˆı + 0.5638ˆ  −2.04ˆı) _ m s _ R = (31.6ˆı + 79.0ˆ ) N The force required to hold the transition in place is (31.6ˆı +79.0ˆ ) N in the coordinate system shown. ME 215.3 Example Problems 88 Solution: Problem # 64 Given: • A piping system as shown below 90 o , regular, flanged elbows g Q = 0.04 m 3 /s water, 20 o C 30 m open, flanged, globe valve 1 2 50 m, 2” I.D. steel pipe Find: • P 1 −P 2 Assumptions: • no shaft work or heat transfer • uniform flow at 1 and 2 • no diameter change • steady, incompressible flow Analysis: Write Bernoulli’s equation with losses from 1 to 2. P 1 + ρV 2 1 2 + ρgz 1 = P 2 + ρV 2 2 2 + ρgz 2 + f L D ρV 2 2 + K m ρV 2 2 89 ME 215.3 Example Problems Since there is no diameter change, V 1 = V 2 . Also, let z 1 = 0. P 1 −P 2 = ρgz 2 + f L D ρV 2 2 + K m ρV 2 2 P 1 −P 2 = ρgz 2 + ρV 2 2 _ f L D + K m _ V = Q A = 0.04[m 3 /s] (2(0.0254)) 2 π/4[m 2 ] = 19.74 m/s We need to find f. It is a function of roughness (ǫ) and Reynolds number (Re). For steel pipe, ǫ = 0.046 mm (Table 6.1). Therefore, ǫ D = 0.046 2(25.4) = 0.000906 Re = ρV D µ = V D ν = 19.74[m/s](2(0.0254)[m]) 1 ×10 −6 [m 2 /s] = 1 ×10 6 From the Colebrook equation with this (ǫ/D) and Re we get f = 0.0195. Now find the minor loss coefficients from table 6–5. K valve = 8.5 K elbow = 0.39 P 1 −P 2 = 998 _ kg m 3 _ 9.81 _ m s 2 _ 30[m] + 998 2 _ kg m 3 _ _ 19.74 _ m s __ 2 _ 0.0195 50 2(0.0254) + 8.5 + 2(0.39) _ P 1 −P 2 = 5.83 MPa The pressure drop between 1 and 2 is 5.83 MPa at this flowrate. ME 215.3 Example Problems 90 Solution: Problem # 65 Given: • A pipe connecting two reservoirs as shown 20 m ρ = 1000 kg/m 3 L = 5 m D = 2 cm 2 1 ǫ = 0.05 mm D = 4 cm L = 5 m b a µ = 0.001 kg/(m· s) Find: • Volume flowrate between the reservoirs Assumptions: • V 1 = V 2 = 0 • P 1 = P 2 • steady, incompressible flow • no heat transfer or shaft work Analysis: Write Bernoulli’s equation (with losses) from 1 to 2. P 1 + ρV 2 1 2 + ρgz 1 = P 2 + ρV 2 2 2 + ρgz 2 + f L D ρV 2 2 + K m ρV 2 2 ρgz 1 = f a _ L D _ a ρV 2 a 2 + f b _ L D _ b ρV 2 b 2 + K m ρV 2 2 91 ME 215.3 Example Problems Minor losses: K m source entrance 0.5 Figure 6.21 expansion 0.56 Figure 6.22 exit 1.0 Figure 6.21 ρgz 1 = f a _ L D _ a ρV 2 a 2 + f b _ L D _ b ρV 2 b 2 + ρV 2 a 2 (K ent. + K exp. ) + ρV 2 b 2 (K exit ) From conservation of mass, V b = V a A a A b = V a _ D a D b _ 2 ρgz 1 = ρV 2 a 2 _ _ f L D _ a + _ f L D _ b _ D a D b _ 4 + K ent. + K exp. + K exit _ D a D b _ 4 _ _ ǫ D _ a = 0.05[mm] 20[mm] = 0.0025 _ ǫ D _ b = 0.05[mm] 40[mm] = 0.00125 392.4 = V 2 a _ 250f a + 125 16 f b + 1.1225 _ Initial guess, fully rough zone f a = 0.025, f b = 0.021. Procedure: • calculate V a from the previous equation • use V a to get Re a and Re b • use Re a and Re b to get new f a and f b • repeat if necessary V a Re a f a Re b f b 7.22 144,300 0.026 72,200 0.024 7.09 141,770 0.026 70,900 0.024 This result is converged because the f’s have stopped changing. The volume flowrate between the reservoirs is 0.00223 m 3 /s. ME 215.3 Example Problems 92 Solution: Problem # 69 Given: • The head versus flowrate characteristics of a pump • Water is pumped through 100 m of 20 cm diameter cast iron pipe • Water temperature is 20 o C Flowrate, m 3 /s Parabola P u m p h e a d , m 2.5 0 100 80 60 40 20 0 1 1.5 2 0.5 Find: • Flowrate Assumptions: • no elevation change • no minor losses • pump inlet the same diameter as exit pipe • pump inlet pressure same as pipe exit pressure • no heat transfer • steady, incompressible flow 93 ME 215.3 Example Problems Analysis: Writing Bernoulli’s equation from the inlet of the pump (1) to the exit of the pipe (2) gives, P 1 + ρV 2 1 2 + ρgz 1 = P 2 + ρV 2 2 2 + ρgz 2 + ∆P f + ∆P m −∆P p With the assumptions stated above this reduces to ∆P f = ∆P p or h f = h p The pump curve can be described by, h p = 80 −20Q 2 where Q is in m 3 /s and h p is in m. The frictional head loss is h f = f L D Q 2 2gA 2 . (80 −20Q 2 )[m] = f _ 100[m] 0.2[m] _ Q 2 2(9.81[m/s 2 ]) _ 4 π(0.2[m]) 2 _ 2 80 = 25, 821fQ 2 + 20Q 2 Q = _ 80 25, 821f + 20 For cast iron pipe, ǫ = 0.26 mm so ǫ/D = 0.0013. The fully rough friction factor for this ǫ/D is f = 0.0210. Therefore, Q = 0.3772 m 3 /s Check the Reynolds number. Re = ρV D µ = V D ν = QD Aν = 2.4 ×10 6 This is fully rough so we can accept the answer. The flow rate is 0.377 m 3 /s. ME 215.3 Example Problems 94 Solution: Problem # 81 Given: • Flow around a bridge pier • U ∞ = 5 m/s • P ∞ = 50 kPa • stagnation point is 1 m upstream of source 2 m x 1 m A y Find: • Pressure at point A Assumptions: • two-dimensional flow • incompressible, steady flow • inviscid, irrotational flow 95 ME 215.3 Example Problems Analysis: Must find the source strength required to give the stagnation point 1 m upstream of the source. Place the source at the origin. Therefore, for the source v r = m r At the stagnation point, u = 0. Therefore, 0 = U ∞ − m r 0 = 5 _ m s _ − m 1[m] Therefore, m = 5 m 2 /s. Now we need the velocity at point A. ψ = ψ freestream + ψ source ψ = U ∞ r sin θ + mθ Therefore, v θ = −∂ψ ∂r = −U ∞ sin θ v r = 1 r ∂ψ ∂θ = U ∞ cos θ + m r We need the (r, θ) coordinates of point A to find these velocities. At the stagnation point θ = π, r = 1 m/s. Therefore, ψ sp = 0 +mπ = mπ Since A is on the same streamline ψ A = mπ In (x, y) coordinates. ψ = U ∞ y + mtan −1 _ y x _ at A, 5[m 2 /s]π = 5[m/s]2[m] + 5[m 2 /s] tan −1 _ 2 x _ x A = 0.9153 ME 215.3 Example Problems 96 r A = _ (0.9153) 2 + (2) 2 = 2.200 m θ A = tan −1 _ 2 0.9152 _ = 1.142 rad Therefore, V 2 A = v 2 θ + v 2 r = (−5 sin(1.142)) 2 + (5 cos(1.142) + 5/2.200) 2 V 2 A = 39.61 m 2 /s 2 P A = P ∞ + 1 2 ρ(V 2 ∞ −V 2 A ) = 50, 000[Pa] + 1000 2 _ kg m 3 _ (5 2 −39.61) _ m s _ P A = 42.7 kPa The pressure at A is 42.7 kPa. 97 ME 215.3 Example Problems Solution: Problem # 82 Given: • A pair of doublets as shown • U ∞ = 10 m/s x U ∞ = 10 m/s doublets 30 o 2 m 2 m 2 m y Find: • Doublet strength • Pressure at the origin Assumptions: • two-dimensional flow • incompressible, steady flow • inviscid, irrotational flow ME 215.3 Example Problems 98 Analysis: ψ = ψ fs + ψ d1 + ψ d2 ψ = U ∞ (y cos α −xsin α) − λ sin θ 1 r 1 − λ sin θ 2 r 2 ψ = U ∞ (y cos α −xsin α) − λ(y −2) (y −2) 2 + (x −2) 2 − λ(y −2) (y −2) 2 + (x + 2) 2 ψ 0,0 = ψ 4,1 ψ 0,0 = 0 − λ(−2) (−2) 2 + (−2) 2 − λ(−2) (−2) 2 + (2) 2 ψ 0,0 = 2λ 8 + 2λ 8 = λ 2 ψ 4,1 = 10(cos 30 o −4 sin 30 o ) − λ(−1) (−1) 2 + (2) 2 − λ(−1) (−1) 2 + (6) 2 ψ 4,1 = −11.34 + λ 5 + λ 37 = λ 2 λ _ 1 2 − 1 5 − 1 37 _ = −11.34 λ = −41.54 m 3 /s The doublet strength must be −41.54 m 3 /s. By symmetry, only the freestream contributes to the velocity at the origin. Therefore, the pressure there is the same as the pressure far away from the doublets. The pressure at the origin is P ∞ . 99 ME 215.3 Example Problems Solution: Problem # 83 Given: • A landing gear strut • U ∞ = 45 m/s • ρ = 1.2 kg/m 3 A U ∞ 12 cm 12 cm B sink source 30 cm Find: • P A −P B Assumptions: • two-dimensional flow • incompressible, steady flow • inviscid, irrotational flow ME 215.3 Example Problems 100 Analysis: A Rankine oval can be modelled by superimposing a source and a sink with the freestream. ψ = ψ fs + ψ source + ψ sink ψ = U ∞ r sin θ + mθ source −mθ sink Find m. At the stagnation point, u = 0. 0 = U ∞ − m (0.15 −0.12) + m (0.15 + 0.12) m = 1.519 m 2 /s Need V B . At B, v = 0. Therefore, we only need u at B. ψ = U ∞ y + mtan −1 _ y x + a _ −mtan −1 _ y x −a _ u = ∂ψ ∂y = U ∞ + m 1 + _ y x+a _ 2 _ 1 x + a _ − m 1 + _ y x−a _ 2 _ 1 x −a _ PROBLEM!!!!! Don’t know y at B. ψ B = ψ sp = 0 0 = 45y + 1.519 tan −1 _ y 0.12 _ −1.519 tan −1 _ y −0.12 _ Note that tan −1 _ y −0.12 _ = π −tan −1 _ y 0.12 _ 0 = 45y + 2(1.519) tan −1 _ y 0.12 _ −π(1.519) 45y = 1.519π −2(1.519) tan −1 _ y 0.12 _ y = 0.07027 m Now, u B = 45 + 1.519 _ 1 1 + ( 0.07027 0.12 ) 2 _ 1 0.12 __ 2 101 ME 215.3 Example Problems u B = 63.85 m/s P A + 1 2 ρU 2 A = P B = 1 2 ρU 2 B P A −P B = 1 2 ρU 2 B = 1.2 2 (63.85) 2 P A −P B = 2.45 kPa The pressure difference between points A and B is 2.45 kPa. ME 215.3 Example Problems 102 Solution: Problem # 84 Given: • Flow over a circular cylinder creating no lift a U ∞ g L A x y Find: • Derive an expression for the velocity along the positive y axis in terms of U ∞ , a, and y. • Sketch this function. • Consider the streamline labelled A. Far upstream of the cylinder this streamline is a distance L from the horizontal plane of symmetry. At what distance from the origin does the streamline cross the y axis? Assumptions: • two-dimensional flow • incompressible, steady flow • inviscid, irrotational flow Analysis: Non-lifting flow over a cylinder can be modelled by a doublet and a freestream. ψ = ψ fs + ψ doublet 103 ME 215.3 Example Problems ψ = U ∞ r sin θ − λ sinθ r ψ = sin θ(U ∞ r − λ r ) At the stagnation point, θ = π and r = a. Therefore, ψ sp = 0 Therefore, ψ = 0 on the circle. Therefore, U ∞ a − λ a = 0 λ = U ∞ a 2 ψ = U ∞ sin θ(r − a 2 r ) Along the y axis, v r = 0 v θ = − ∂ψ ∂r = −U ∞ sin θ _ 1 + a 2 r 2 _ θ = π 2 v θπ/2 = −U ∞ _ 1 + a 2 r 2 _ Along the positive y axis, u = −v θ . Therefore u U ∞ = 1 + a 2 r 2 The velocity along the positive y axis is given by u U ∞ = 1 + a 2 r 2 . ψ A −ψ sp = U ∞ L ψ A = U ∞ L U ∞ L = U ∞ sin _ π 2 _ _ r − a 2 r _ L = r − a 2 r r 2 −Lr −a 2 = 0 ME 215.3 Example Problems 104 1 u/U ∞ r/a 1 2 r = L 2 ± ¸ _ L 2 + a 2 _ Since r > a, we want the positive root. r = L 2 + ¸ _ L 2 + a 2 _ The streamline labelled A crosses the y axis at a distance of L 2 + ¸ _ L 2 + a 2 _ from the origin. 105 ME 215.3 Example Problems Solution: Problem # 85 Given: • Flow around a circular cylinder • U ∞ = 20 m/s • ρ = 1.2 kg/m 3 • Volume flowrate per unit length between streamlines A and B is 1.19 m 2 /s. • Streamline B passes through point (x, y) = (0, 0.12m) −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 B A y x (x, y) = (0, 0.12 m) Find: • Lift force per unit length ME 215.3 Example Problems 106 Assumptions: • two-dimensional flow • incompressible, steady flow • inviscid, irrotational flow Analysis: This flow can be modelled by combining a freestream, a doublet, and a vortex. ψ = ψ fs + ψ doublet + ψ vortex ψ = U ∞ r sin θ − λ sinθ r −K ln _ r a _ For ψ = 0 on r = a, 0 = U ∞ a sin θ − λ sin θ a λ = U ∞ a 2 ψ B −ψ A = Q = 1.19 m 2 /s Therefore, ψ B = 1.19 m 2 /s Now, at (x, y) = (0, 0.12 m) 1.19 _ m 2 s _ = 20 _ m s _ (0.12[m]) − 20[m/s](0.1) 2 [m 2 ] 0.12[m] −K ln _ 0.12[m] 0.1[m] _ K = −2.505 m 2 /s Γ = 2πK = −15.74 m 2 /s Kutta-Joukowski Theorem F L = −ρU ∞ Γ F L = −1.2 _ kg m 3 _ 20 _ m s _ _ −15.74 _ m 2 s __ F L = 378 N/m The lift force per unit length is 378 N/m. 107 ME 215.3 Example Problems 1. The tank shown holds oil of specific gravity 0.89. The top of the tank is closed and the space in the tank above the oil contains air. The U-tube manometer contains water and the displacements are as indicated. Atmospheric pressure is 101.3 kPa. What is the pressure of the air in the tank? (Solution: page 49) Open to atmosphere Air 6 cm Air 4 cm 15 cm Water Oil 2. A flat, vertical gate holds back a pool of water as shown. Find the force that the water exerts on the gate. (Solution: page 51) Patm = 100 kPa ρ = 1000 kg/m3 3m hinge 3m gate (5 m wide) stop 1 ME 215.3 Example Problems 22 m and it is 2.44 m 1. A Tainter gate is constructed from a quarter-cylinder and is used to hold back a pool of water.3. Calculate the hydrostatic force on the gate. A flat. 22 m R gate ρ = 1000 kg/m3 4. (Solution: page 53) length=2. vertical gate holds back a pool of water as shown. (Solution: page 56) Patm = 100 kPa = ρ = 1000 kg/m3 3m hinge 3m gate (5 m wide) stop ME 215.3 Example Problems 2 . Find the force that the stop exerts on the gate. The radius of the gate is 1.44 m long. What force does the gate exert on the hinge? The water density is 1000 kg/m3 . The gate is 4 m long. The air pressure inside the submarine is equal to atmospheric pressure.5. 15 cm thick.3). and is made of concrete (s. Determine how hard a sailor has to push on the centre of the hatch to open it. hinge A g opening force hatch inside of submarine outside of submarine A hatch View A-A 70 cm 6. The radius of the hatch is 35 cm above the centre of the hatch. A rectangular gate is hinged along the top edge as shown.=2.g.3 Example Problems . Determine the water depth H if the gate is just about to open. The seawater density is 1025 kg/m3 and the centre of the hatch is 2 m below the surface of the ocean. g 2 m H hinge gate 45 o 3 ME 215. A circular hatch on a submarine is hinged as shown. g 2m F 2m hinge 8.7. A pool of fluid has a density that varies linearly from 1000 kg/m3 at the surface to 1600 kg/m3 at a depth of 4 m. 1.2 m ME 215. The block is in equilibrium when immersed in water (ρ = 1000 kg/m3 ) to the depth shown.6 m water hinge 1. A 2 m by 2 m square gate is hinged along its bottom edge and held in place by a force F at its top edge. A long. square wooden block is pivoted along one edge.2 m wood 0. Find the force F .3 Example Problems 4 . Evaluate the specific gravity of the wood. 5 m A Cable B 0.5 m Parabolic gate y = x2 water ρ = 1000 kg/m3 2m y C x Hinge 5 ME 215.g. = 0. Note that point A is directly above point C.√ 9. The gate is 8 m long. A parabolic gate 4 m wide (in the z direction) holds a pool of water as shown. Find the reaction force which the hinge at C exerts on the gate. A cylindrical gate with a radius of curvature of 2 m holds back a pool of water with a layer of oil floating on the surface. What force does the pinned connection at A exert on the gate? √ 2m g 1m Oil (s.8) A 1m Water (ρ = 1000 kg/m3) 10. g 1.3 Example Problems . Find the tension in the cable AB required to hold the gate in the position shown. 5 m B g 12. The ends of the pipe are closed.3 Example Problems 6 . A gate of mass 2000 kg is mounted on a frictionless hinge along its lower edge. b. Find the inside diameter of the pipe and the force which the wall exerts on the pipe.g. For the equilibrium position shown.0 and the pipe has an outside diameter of 2 m. = 2. b 1m Water 30o hinge ME 215. The width of the gate (perpendicular to the plane of view) is 8 m. The fluid is water with ρ = 1000 kg/m3 .0 0. calculate the length of the gate. s. A partially submerged pipe rests against a frictionless wall at B as shown. The specific gravity of the pipe material is 2.11. The radius of curvature of the gate is 2 m and it is 6 m long. A F gate g 1.5 m deep. plane wall holds back a pool of water (ρ = 1000 kg/m3 ) which is 1.5 m 2m gate A hinge 1m View A-A 14. of g. Assume that the point where the gate contacts the bottom of the pool is frictionless. gate c.3 Example Problems . A vertical. Calculate the force F required to hold it closed. 2m 45o g 2m 45o 1. The wall has a triangular gate in it that is hinged along the bottom edge and held closed by a horizontal force F applied at the top corner.13. A Tainter gate holds back a pool of water (ρ = 1000 kg/m3 ) which is 2 m deep. Calculate the reaction force on the hinge.8 m hinge 7 ME 215. The mass of the gate is 1000 kg and its centre of gravity is at the position indicated on the diagram. 15. Determine H when the L-shaped gate shown is just about to open. Neglect the weight of the gate and let the density of the fluid be ρ. gate H g hinge L 16. What force F is needed to hold the 4 m wide gate closed? The fluid is water and it has a density of 1000 kg/m3 . 9m hinge 3m F ME 215.3 Example Problems 8 17. A semi-circular gate is hinged at the bottom as shown. The density of the fluid varies linearly from 1000 kg/m3 at the surface of the reservoir to 1500 kg/m3 at the bottom of the reservoir. Find the force F required to hold the gate in place? A g 3m F gate 1m hinge hinge Section A-A A 18. A circular cylinder holding back a pool of water is held in place by a stop as shown. The height of the stop is 0.5 m and the water depth is 1.5 m. If the water depth were increased beyond 1.5 m the cylinder would roll over the stop. Assuming that the water contacts the cylinder surface right up to point A, what is the specific gravity of the cylinder. Diameter = 2 m 1.5 m A 0.5 m 9 ME 215.3 Example Problems 19. A gate composed of a quarter circle portion and a straight portion holds back a pool of water. The gate is hinged at A and held in place by a force F applied as shown. Find the magnitude of the force F required to hold the gate in the position shown. The width of the gate is 8 m. F 0.5 m g 1.5 m Water (ρ = 1000 kg/m3 ) Width of Gate is 8 m R = 1m A Hinge ME 215.3 Example Problems 10 length b = 100 mm. and density ρc = 800 kg/m3 blocks a slot in the bottom of a water tank as shown. A circular cylinder of radius R = 50 mm. The line joining the centre of the cylinder and the point where the cylinder contacts the edge of the slot subtends an angle of α = 30o with the vertical.3 Example Problems . If h = 150 mm what is the force exerted on the cylinder by the tank? h g R α = 30o 11 ME 215.20. Water at 20o C is retained in a pool by a triangular gate which is hinged along its top edge and held in place by a stop at its lowest point.5 m 12 . and the displacement of the fluid in the tube. The tension in the wire is 335. 10 cm cube of material is suspended from a wire in a fluid of unknown density.5 N. Determine the magnitude of the acceleration as a function of the geometry of the tube. What force does the stop exert on the gate? What force does the hinge exert on the gate? A g hinge 1m 7m stop A Section A-A 22. (Solution: page 61) ME 215.3 Example Problems 3. (Solution: page 59) 23. A U-tube is used as a crude way to measure linear acceleration.21. the acceleration due to gravity. Determine the specific gravity of the fluid. A 35 kg. After some time.24. what is the net hydrostatic force on the top of the container? hole 1m a g 0. A container of water (ρ = 1000 kg/m3 ) accelerates on a 30o slope. 2m 3 m/s2 1m 0. vertical end of the cart.75 m A 30o 13 ME 215. The cart begins to accelerate to the right at a constant 3 m/s2 .3 Example Problems . It is completely closed except for a small hole in the position indicated.5 m g 25. The cart is 0. Determine the net hydrostatic force on the rear.8 m wide.5 m. what is the acceleration a? If the width of the container (perpendicular to the page) is 0. the fluid reaches a hydrostatic state. If the gauge pressure at point A is 25 kPa. An open-top cart half full of water (ρ = 1000 kg/m3) is shown at rest in the figure below. (Solution: page 65) ω Ho R ME 215. container radius. The diameter of the can is 50 cm and is spinning at 50 rpm. A partially full can with an open top spins around an axis as shown. what is the depth of the fluid on the axis of rotation? (Solution: page 63) ω = 50 rpm 30 cm 50 cm 27. Derive a general relationship for the shape of the free surface as a function of the rotation rate. and the depth of the fluid when it is not spinning. A cylindrical container is rotated about its axis. If the depth of the fluid at the outer edge of the can is 30 cm.26.3 Example Problems 14 . 28. A cylindrical can of radius 4 cm and height 12 cm has an open top. It is initially at rest and completely full of liquid. It is rotated about its axis at 250 rev/min until the fluid inside it achieves solid body rotation. The rotation is then stopped and the fluid within the container is allowed to come to rest. How deep will the fluid in the container be? g 12 cm 4 cm 29. A U-tube manometer contains two fluids with different densities as shown. The fluid positions shown in the diagram are for the case when the tube is not spinning. It is then spun around the axis shown until the liquid level in both legs is equal? Find ω. 1000 kg/m3 1200 kg/m3 g 10 cm 20 cm ω 15 ME 215.3 Example Problems 30. The U-tube shown is rotated about the vertical axis indicated on the diagram at 60 rev/min. Determine the displacement of the water in each leg from its rest position. Perform this calculation on the centreline of the tubes. What is the pressure at point A? Axis of rotation Diameter, 5 mm Diameter, 8 mm Rest level 5 cm 10 cm 60 rpm vent ρ = 1000 kg/m3 cylindrical can A g 3 cm 31. A 80 cm diameter cylindrical can has a closed top except for a small vent hole at the centre. If the density of the fluid is 1000 kg/m3 and the can spins at 60 rpm, what is the force on the top of the can? (Solution: page 67) ME 215.3 Example Problems 16 32. An upright, 10-cm diameter, cylindrical paint can 20 cm deep spins around its axis of symmetry at 500 rpm. It is completely full of mineral oil (sg = 0.87) and there is a very small hole in the lid at the rim of the can. Determine the net hydrostatic force on the lid. 33. A quarter-circle, cylindrical gate 5 m long and with a 1.5 m radius is hinged at the point indicated below. It holds back a pool of water (ρ = 998 kg/m3 ) with a horizontal force F applied as shown. Determine the force F and the reaction at the hinge. F g 1. 5m hinge 34. You are driving down the road at 100 km/hr with a cup of coffee in your drink holder. The cup has no lid and is 8 cm in diameter. The coffee is initially 1 cm from the top of the cup. You come to a curve in the highway which is not banked. Determine the minimum radius of the curve for which the coffee will not spill. If you go around a curve whose radius is 75% of the minimum, how much coffee will spill (in cm3 )? 35. A square gate (1.25 m x 1.25 m) is hinged along a line 0.25 m from its top edge. A force F1 applied at the top of the gate holds it closed. The pool of water (ρ = 998 kg/m3 ) is 1.75 m deep. Calculate the force F1 required to hold the gate closed. hinge g 1.75 m F1 gate 1.25 m 1m 17 ME 215.3 Example Problems the water is 10 cm deep. What is the horizontal force on the barricade due to the stream of water? (Solution: page 69) V = 15 m/s Aj = 0. Water is put in the container to a depth of 20 cm. After coming to rest.01 m2 ME 215. A cylindrical container is 40 cm deep and 30 cm in diameter and is open at the top. Determine ω in revolutions per minute (rpm).36. The police are using a fire hose to move a flat barricade. initial water level 40 cm g 20 cm 30 cm 37. The container is then spun around its axis at an angular speed of ω until the water reaches a hydrostatic state. The spinning is stopped and the water is allowed to come to rest.3 Example Problems 18 . 01 m2 cart 39. A nozzle with an exit diameter of 25 mm is attached at the end of the pipe. A jet of water from issues from a 0. A 50 mm diameter pipe carries water at 1 m/s. If the cart is moving away from the nozzle at 5 m/s.01 m2 nozzle at 15 m/s. the density of the fluid. (Solution: page 81) Uo r 1 u(r) = Umax (1 − (r/R)2) z 2 circular pipe radius = R 19 ME 215. It impinges on a vane mounted on a movable cart and is deflected through 45o . what is the force which the water exerts on the cart? (Solution: page 72) 45o Aj = 0. If the pressure just upstream of the nozzle is 10 kPa. What is the frictional drag on the fluid between axial locations 1 and 2 in terms of the pressure at those locations. and the pipe radius. Consider the entrance region of a circular pipe for laminar flow.3 Example Problems . what is the force on the bolts at the flange connection attaching the nozzle to the pipe? (Solution: page 79) 40. the mean velocity of the fluid.38. The pressure at the entrance is 20 kPa and the loss coefficient is 0. A transition piece turns 45o and expands from 50 mm to 80 mm.50 based on the discharge velocity.3 Example Problems 20 . A 50 mm diameter pipe carries water at 1 m/s. A nozzle with an exit diameter of 25 mm is attached at the end of the pipe. Find the force required to hold the transition in place. what is the loss coefficient? (Solution: page 86) 43. Water flows through the transition at 4 kg/s. (Solution: page 84) 42. find the force on the flange connection. (Solution: page 87) D2 = 80 mm 20 kPa 45o 4 kg/s D1 = 50 mm transition piece Km = 0. If the nozzle from the previous problem has a pressure of 10 kPa at its entrance. Assuming frictionless flow.41.50 (based on discharge velocity) ME 215. 20 m/s 1 m diameter mercury (s. Assume the pressure is uniform across the wind tunnel at any axial location. Neglect shear at the wind tunnel walls and let ρ = 1.56) 10 mm g 21 ME 215.44. The dashed line labeled “A” is a streamline in this flow. The velocity profile is measured at the end of the plate and is found to vary with distance y from the plate according to u = U∞ (y/δ)1/7 where δ = 170 mm is the boundary layer thickness.3 Example Problems . Consider only the flow over the top surface of the plate. The plate is 10 m long in the streamwise direction and 4 m wide (normal to the page). At the leading edge of the plate the air speed is uniform at U∞ = 10 m/s. Determine the viscous force on the plate.2 kg/m3 ) blows parallel to a flat plate upon which a boundary layer grows. A mercury manometer indicates a 10 mmHg pressure difference as shown. The air velocity upstream of the object is a uniform 20 m/s. A 170 mm 10 m/s flat plate y 10 m 45. An object is placed in a 1 m diameter wind tunnel and the air velocity downstream of the object is found to linearly vary from zero at the centreline of the wind tunnel to a maximum at the wind tunnel wall. Air (ρ = 1.g.2 kg/m3 . Determine the drag force on the object.=13. 46. 1 cm diameter nozzle at a mass flow rate of 5 kg/s. The cable breaks and the cart begins to move. Determine the acceleration of the system at this instant in time. Cable 47. Pump 5 kg/s g 1 cm dia. The flow to each arm is equal. pump assembly. A pump mounted on top of the tank draws water from the tank and discharges the water through a horizontal. find the angular speed of the sprinkler? 10 cm 20 cm D = 2 mm ME 215. The flow rate supplied by the pump remains constant. and the water remaining in the tank is 100 kg and the velocity of the system is 20 m/s. The arms lie in a horizontal plane and rotate about a vertical axis. A new type of lawn sprinkler is developed that has two arms of different lengths.3 Example Problems 22 . One arm is 10 cm from the pivot and the other is 20 cm from the pivot. Both nozzles are aimed upward at 20o from the horizontal plane and have an exit diameter of 2 mm. If the pivot is assumed to be frictionless and the sprinkler delivers 2 L/min. At some instant in time later. the total mass of the tank. A tank on wheels contains water (ρ = 1000 kg/m3 ) and is held in place by a cable as shown below. An aluminium block weighing 10 N is supported by a jet of water issuing from a 2.3 Example Problems . The elbow has a minor loss coefficient of 0.75 based on the inlet velocity. What force does the pipe exert on the elbow at the flange connection? ∆P D1 = 5 cm Elbow m ˙ g Pipe Flange D2 = 3 cm 23 ME 215.48. The difference between the stagnation and static pressures ∆P measured by the pitot-static tube is 500 Pa.5 cm diameter nozzle.5 cm 49. guides g aluminum block 10 cm 2. What jet exit velocity is required to hold the block 10 cm above the nozzle exit? The jet of water is deflected through 180o when it strikes the block and the guides holding the block in place are frictionless. The mass of the elbow and the water which it contains is 1 kg. An elbow connected to a 5 cm diameter pipe discharges water (ρ = 1000 kg/m3) as shown. If the flowrate is high enough. Derive an expression for F as a function of Q and indicate the range of Q for which it is valid. 60 kPa water ρ = 1000 kg/m3 30o 51. Water (ρ = 1000 kg/m3 ) is pumped at volume flowrate Q through the nozzle shown.2Ve2 /2g where Ve is the velocity at the nozzle exit. A pipe 4 cm in diameter supplies water to a 2 cm diameter nozzle. The jet of water (not shown) hits a frictionless splitter plate which is inclined at 30o to the axis of the pipe. Half of the water is deflected downwards along this plate. Find the force required to hold the plate in this position. g F D = 5 cm 45o 0.8 m. Sketch the function. The gauge pressure just upstream of the nozzle is 60 kPa and the head loss in the nozzle is given by hL = 0. Assume that the depth of the water in the cart remains at 0. a force F will be required to keep the cart stationary.50.8 m 1m 2m Q 2m ME 215.3 Example Problems 24 . The wind tunnel is 0. The downstream velocity in the upper half is uniform.25 m 12 m/s Pl = 100 − 10x + 20x(1 − x) [Pa. Determine the speed of the rocket when the fuel is all used.5 m deep (into the paper). The air density is constant at 1.2 kg/m3 . An engineer is measuring the lift and drag on an aerofoil section mounted in a two– dimensional wind tunnel. Neglect friction in the wheels of the cart and aerodynamic drag. level rails is used to test rocket engines. The test section is 1 m long. The vertical component of velocity is zero at the beginning and end of the test section. The engineer measures the pressure distribution in the tunnel along the upper and lower walls and finds Pu = 100 − 10x − 20x(1 − x) [Pa. gauge] where x is the distance in metres measured from the beginning of the test section. The upstream air velocity is uniform at 10 m/s.52. Beginning of Test section End of Test section g 0. Eighty percent of the mass of the rocket is fuel. Neglect shear on the walls of the wind tunnel.25 m 10 m/s 0. Find the lift and drag forces acting on the aerofoil. A 400 kg cart has a rocket mounted on it which has an initial mass of 500 kg. The downstream velocity is uniform at 12 m/s in the lower half of the wind tunnel. A cart mounted on straight. gauge] 1m x 53. The products of combustion exhaust at a speed of 1000 m/s relative to the rocket nozzle. g 25 ME 215.5 m high and 0.3 Example Problems . 54. The velocity at the exit varies linearly from a maximum at the pipe centreline to zero at the pipe wall as shown. 15 cm 12 mm dia. The other arm has a 1 mm wide slot that also emits water in the horizontal plane.3 Example Problems 26 . The total mass flow rate of water (ρ = 1000 kg/m3 ) is 1 kg/s and it is divided equally between the two arms. Find the rotational speed of the sprinkler in revolutions per minute assuming that the pivot is frictionless. At the end of one arm a 12 mm diameter nozzle is oriented perpendicular to the arm and is in the same horizontal plane as the arm. Water (ρ = 1000 kg/m3 ) exits from a circular pipe 10 cm in diameter with a mass flow rate of 4 kg/s and strikes a flat plate at 90o . A two-arm sprinkler is constructed as shown below. Determine the force exerted on the flat plate. 2 cm A 1 mm axis of rotation Plan View A 15 cm Section A-A (enlarged) ME 215. Compare this to the force that would be exerted at the same mass flow rate if the pipe exit velocity had been uniform. 10 cm diameter m = 4 kg/s ˙ 55. 5 m diameter g 0.5 m. Two circular coaxial jets of incompressible liquid with speed V collide as shown.5 m Scale 27 ME 215. A tank of water sitting on a weigh scale is being filled with water (ρ = 1000 kg/m3 ) from a 2 cm diameter pipe at a rate of 5 kg/s.3 Example Problems . The interaction region is open to atmosphere. The empty tank has a mass of 2 kg.56. The tank is circular and has a diameter of 0. V θ d1 V V d2 V 57. Liquid leaves the interaction region as a conical sheet.5 m 0. what force will the scale read? Carefully explain each of your assumptions. Obtain an expression for the angle θ of the resulting flow in terms of d1 and d2 . At the instant when the water is 0. 5 kg/s 2 cm 0.5 m deep in the tank. (b) the blade moves to the right at 20 m/s. and (c) the blade moves to the left at 20 m/s. Calculate the force of the water (ρ = 1000 kg/m3 ) on the frictionless vane if (a) the blade is stationary. 59.58. 40 m/s 60o 5 cm dia. What is the reaction in the x direction? z 1 kg/s 45 o x splitter vane D = 1 cm ME 215. The flowrate is 1 kg/s while the diameter of the nozzle is 1 cm. A jet of water (ρ = 998 kg/m3 ) strikes a frictionless splitter vane as shown. The position of the splitter vane is adjusted in the z direction so that there is no reaction in the z direction.3 Example Problems 28 . The inlet pipe diameter is 12 cm while the nozzle exit diameter is 3 cm. The mass flowrate is 1 kg/s.60. what is the force and moment on the flanged connection? Neglect the weight of the elbow and water. Thirty-five percent of the mass flow deflects through 90o (to the left). 200 kPa Water g flanged connection 30 cm 45o 29 ME 215.3 Example Problems . Water at 20o C flows from a nozzle of diameter D = 5 mm at speed V = 10 m/s and strikes a vane which splits the flow and deflects it as shown. What force does the fluid exert on the vane? 45o D V 61. If the pressure at the flange connection is 200 kPa(gauge). Water at 20o C flows through an elbow/nozzle arrangement and exits to atmosphere as shown below. t flanged connection spray nozzle 63. At the instant when the braking jet is activated. What is the stopping distance? ME 215. P1 supply pipe Water 35 mm diameter R V2 thickness. Determine how much air must escape in order to stop the cart. The initial gross mass of the cart and its occupants is 500 kg.3 Example Problems 30 . The sheet leaves the nozzle at V2 = 10 m/s. The water supply pipe is 35 mm in diameter and the inlet pressure P1 is 50 kPa above atmospheric. The jet is supplied by a compressed air cylinder aboard the cart. the speed of the cart is 40 m/s. covers 180o of arc and has thickness t = 1. A ride at an amusement park consists of a wheeled cart that zooms down an inclined plane onto a straight and level track where it decelerates to rest by means of a highspeed jet of air projected directly forward through a nozzle.62. A nozzle for a spray system is designed to produce a flat radial sheet of water. Calculate the force exerted by the spray nozzle on the supply pipe through the flanged connection. The nozzle discharge radius is R = 50 mm.5 mm. Air escapes from the braking jet at a constant mass flow rate of 20 kg/s and a constant velocity (relative to the nozzle) of 150 m/s. what is Uo ? uniform velocity 5 cm diameter turbulent velocity profile 1. If Uc1 is measured to be 2 m/s and Uc2 is measured to be 1 m/s.5 cm diameter laminar velocity profile 65. Uc1 is the centreline velocity. r is the radial coordinate. Fluid enters a 5 cm diameter pipe with a uniform velocity of Uo . The second exit is 1. Water (ρ = 998 kg/m3 ) exits the nozzle at 20 m/s.5 cm in diameter and has a laminar. One is a 5 cm diameter exit with a turbulent velocity profile described by uz = Uc1 1 − r R1 1/7 . The pressure just upstream of the flanged connection is 200 kPa (gauge). A 6 mm diameter angled nozzle is attached to the end of a 2 cm diameter pipe with a flanged connection as shown below. where uz is the axial component of velocity. The fluid exits the pipe at two locations.3 Example Problems . parabolic velocity profile described by uz = Uc2 1− r R2 2 . The angle between the nozzle and the supply pipe is 45o . and R1 is the radius of the pipe. Determine the force and torque in the flanged connection.64. 200 kPa flanged connection 6 mm diameter 2 cm diameter 4 cm 31 ME 215. The density of the water is 998 kg/m3 . the fluid has a turbulent velocity profile described by uz = Uc 1 − r R 1/5 where uz is the axial component of velocity. is shown in the sketch.3 Example Problems 32 . Fluid (ρ = 850 kg/m3 ) enters a 5-cm diameter pipe with a uniform velocity of 3 m/s at location A. Determine the viscous force on the wall of the pipe between location A and B. and R is the radius of the pipe. The pressure at location A is 4 kPa larger than at location B. as viewed from above. A two-arm lawn sprinkler. 15 cm 15 cm 45o 67. Uc is the centreline velocity. Assume the pivot is frictionless. There are two 90o regular flanged elbows and an open flanged globe valve. Water is delivered to the sprinkler at a volume flow rate of 5 L/min and it can be assumed that this flow is split evenly between the two nozzles. (Solution: page 90) ME 215. The nozzles both have a diameter of 2 mm and are both angled upwards from the horizontal plane at 30o . At location B.66. uniform velocity 5 cm diameter A B 68. Calculate the pressure difference between the ends of the pipe. Calculate the rotational speed of the sprinkler. The net elevation change is 30 m.04 m3 /s. A 2”ID steel pipe 50 m long carries water at a rate of 0. r is the radial coordinate. ν = 10−6 m2 /s) to atmosphere.=13. A mercury manometer 20 m from the end of the pipe reads as shown. µ = 0. Two reservoirs are connected by a pipe as shown.g. The right leg of the manometer is open to atmosphere. Calculate the frictional pressure drop in 100 m of 3-cm diameter smooth pipe with water (ρ = 1000 kg/s.5) and discharges water (ρ = 998 kg/m3 . (Solution: page 92) ρ = 1000 kg/m3 µ = 0.001 Pa · s) flowing at 1 kg/s. A piping system contains a valve (Km = 6.56) 20 cm 33 ME 215. Determine the volume flow rate. The frictional pressure drop in 100 m of 3-cm diameter smooth pipe with water (ρ = 1000 kg/s.69. What would the frictional pressure loss be if the roughness were 0.001 Pa · s) flowing is 1 MPa. g 20 cm mercury (s. µ = 0.001 kg/(m · s) 20 m a D = 2 cm L = 5m b D = 4 cm L = 5m ǫ = 0. The pipe is smooth and its diameter is 5 cm. 72.5 mm? 71. Find the volume flowrate between the reservoirs. Determine the volume flowrate through the pipe in L/min. The elevation change between the two reservoirs is 20 m.3 Example Problems .05 mm 70. It pumps water through 100 m of 20 cm diameter cast iron pipe.5 3 Flowrate(m /s) 2. The 25 mm diameter pipe has a roughness of 0. What is the total flow rate between the two reservoirs? Neglect minor losses. Two large water (ν = 10−6 m2 /s) reservoirs are joined by two equal-length pipes as shown. What is the flowrate? (Solution: page 94) 100 80 Pump Head (m) 60 40 20 0 0.3 Example Problems 34 .5 mm g ME 215. smooth 25 mm. A pump has a characteristic curve that can be approximated by a parabola as shown. The flow rates through the two pipes are equal.0 0. 20 mm.73.0 2.0 1.5 mm while the 20 mm diameter pipe is smooth.5 74. ǫ = 0.5 1. 56) 35 ME 215. µ = 0. Water (ρ = 1000 kg/m3 .=13. The pipe has a roughness of 0. The pressure difference between these two taps is measured with a mercury (s.001 Pa · s) flows through a horizontal section of 4 cm diameter pipe. The manometer tubes are otherwise full of water. A stagnation pressure tap and a static pressure tap are mounted 1 m apart as shown.=13. What is the volume flow rate of the water? Stagnation tap Static tap Q g 5 cm 1m mercury (s.75.g.3 Example Problems .g.56) manometer which shows a displacement of 5 cm.2 mm. A fluid with a kinematic viscosity of 4 × 10−6 m2 /s flows through the pipe.3 Example Problems 36 . A vertical section of 4 cm diameter pipe with ǫ = 0.76. Does the fluid flow up or down? What is the volume flow rate? PA 5m g PB D = 4 cm ME 215. read the same (PA = PB ) .2 mm has two pressure gauges mounted 5 m apart which. at the current flow rate. 0 Flow Rate (m3 /min) 37 ME 215. Neglect minor losses.0 1.0 3. If the pump characteristics are represented by the pump curve shown below.1 mm. 60 55 50 45 40 35 30 25 20 15 10 5 0 0.5 4.001 Pa · s) from a reservoir and discharges it into 100 m of 10 cm diameter pipe which has a roughness of 0.3 Example Problems .0 2.5 2.77.5 3. µ = 0. estimate the flowrate. A pump draws water (ρ = 1000 kg/m3 .0 Head (m) 0.5 1.0 4. The discharge of the pipe is 20 m lower than the surface of the reservoir.5 5. What is the minor loss coefficient for the elbow? Assume that all pipes are smooth.g.5 cmHg when the mass flowrate is 1 kg/s.75 m 2 cm diameter not to scale 1m g 4 cm diameter m = 1 kg/s ˙ 7.3 Example Problems 38 .001 Pa · s.=13.5 cm mercury (s. Reducing elbow 0.56) ME 215. A mercury manometer is attached at these locations and reads 7.75 m downstream of the elbow. the only available pressure taps are 1 m upstream and 0. The flowing fluid is water with ρ = 1000 kg/m3 and µ = 0.78. Because of limited access to the piping system. An engineer needs to measure the minor loss coefficient of a reducing elbow which has already been installed in a piping system. The sum of all minor loss coefficients is 12 and the total length of tube is 8 m.5 mm. Over the range of interest the pump head can be expressed by hp = A − BQ2 where hp is the head produced by the pump. Q is the volume flowrate. ν = 10−6 m2 /s) to the system shown below.0 m Total length L = 8 m Diameter D = 5 mm Roughness ǫ = 0. and A and B are constants. Neglect minor losses in this problem.05 mm 80. A centrifugal pump has a pump curve that can be described by hp = 4 − 10Q2 where hp is the pump head in metres of water and Q is the volume flowrate in L/min.0 m3 /min.001 Pa · s) from a large reservoir and pumps it through 1335 m of 30 cm inside diameter pipe with roughness ǫ = 0. What is the flowrate achieved? Discharge to atmosphere 2.025 mm. The tube has an inside diameter of 5 mm and roughness ǫ = 0.3 m3 /min. How high can the exit be raised before the flow will be zero? For all conditions the pipe discharges to atmosphere. A centrifugal pump draws water (ρ = 1000 kg/m3 . 39 ME 215. when the exit of the same pipe is raised to an elevation of 50 m the flow reduces to 13. However. When the pipe exit is at the same elevation as the reservoir surface the flowrate is 17.79. This pump is used to supply water (ρ = 1000 kg/m3 .3 Example Problems . µ = 0. 3 Example Problems 40 .. Sudden expansion All pipes have ǫ = 0. A 6 mm internal diameter. regular. What is the flowrate through this piping system? All elbows shown are 90o . What will the volume flowrate be? Neglect minor losses due to bends in the hose. The piping system shown is fitted with a centrifugal pump whose characteristics can be approximated by hp = 46 − 2. thin-walled.81.15 mm Pump D = 100 mm L = 170 m D = 200 mm L = 340 m Gate valve Fully open Sudden contraction ME 215. smooth rubber hose 12 m long is used to siphon water (ρ = 1000 kg/m3 . 6m 6 mm I.5Q2 where hp is the head produced by the pump in m of water and Q is the volume flowrate in m3 /min. ν = 10−6 m2 /s) from a large tank. The outlet of the hose is 6 m below the water surface in the tank. 12 m long 82.D. flanged. 83. The pipe is 5 cm in diameter and has a roughness of ǫ = 0. Determine PA if the first valve is fully open and the second valve is fully closed. If you measure the flowrate to be 60 L/min what is the minor loss coefficient of the valve? Let ν = 10−6 m2 /s. With both valves fully open. Neglect losses which occur on the suction side of the pump. You are going to measure the minor loss coefficient of a new valve design using the apparatus shown.5 mm. Let ν = 10−6 m2 /s and ρ = 998 kg/m3 . The 2-cm diameter smooth pipe is 4 m long. The pump curve can be represented by a parabola. The free surface in the tank is 5 m above the pipe exit.0 when fully open and 20 when 50% open. the flow is 355 L/min. Assume that the water tank is large and the pipe discharges to atmosphere. When both valves are 50% open. PA Pump Valve Valve 25 m 25 m 25 m 25 m 41 ME 215. A centrifugal pump is connected to a piping system as shown below. The two valves are identical and have minor loss coefficients of 1. The free surface in the tank is at the same elevation as the pipe exit. 5m L = 4m D = 2 cm New valve 84. the pressure guage reads PA = 250 kPa.3 Example Problems . Point B is 3 m above point A.05 mm.9. At point B.001 Pa · s. The two elbows each have a minor loss coefficient of 2. Valve B PA A ME 215. the piping system discharges to the atmosphere. The pipe is 25 mm in diameter and has a roughness of 0.85. The valve has a minor loss coefficient of 6. There is a total of 15 m of pipe between point A and point B.4.3 Example Problems 42 . Determine the volume flow rate (in L/min) if the fluid is water with ρ = 998 kg/m3 and µ = 0. The pressure PA is 200 kPa guage. A portion of a piping system is shown below. (Solution: page 96) y x 2m A 43 ME 215. If the freestream velocity is 5 m/s and the pressure a long distance upstream is 50 kPa.3 Example Problems . Flow around a certain bridge pier can be modelled by a freestream and a single source. what is the pressure at point A on the pier surface? The stagnation point will be 1 m upstream of the source.86. 2 kg/m3 . This shape can be modelled as a Rankine Oval with a source and a sink placed as shown.87. The streamline which passes through the origin also passes through the point (4 m. The aircraft is flying at 45 m/s and the air density is 1. What is the doublet strength? Referenced to P∞ . 1 m). A pair of doublets of equal strength are placed in a freestream as indicated below. The landing gear strut on a small aircraft has a cross–section as indicated below. what is the pressure at the origin? (Solution: page 99) 2m 2m doublets 2m y x 30 o U∞ = 10 m/s 88. What is the difference in static pressure between points A and B? (Solution: page 101) B A source 12 cm sink 12 cm U∞ 30 cm ME 215.3 Example Problems 44 . 3 Example Problems . At what distance from the origin does the streamline cross the y axis? (Solution: page 104) y U∞ A g L a x 45 ME 215. (a) Derive an expression for the velocity along the positive y axis in terms of U∞ . and y. a. (c) Consider the streamline labelled A. Far upstream of the cylinder this streamline is a distance L from the horizontal plane of symmetry. Consider the irrotational flow around a circular cylinder which is creating no lift.89. (b) Sketch this function. y) = (0.3 −0. and a second source of equal strength located at (x. The freestream velocity is 20 m/s.2 (x. 0).2 −0.4 0. The streamline labelled B passes through the point (x.1 0. y) = (0.2 kg/m3 ) is produced by a freestream (U∞ = 50 m/s) aligned with the x axis.12 m) 0. 0. a source at the origin. Consider the flow of air (ρ = 1.5 −0.4 0.90.1 B −0.19 m2 /s. Determine the gauge pressure at the point where the stagnation streamline crosses the y axis.3 −0. What is the lift force per unit length on the cylinder? (Solution: page 107) 0.5 0.3 Example Problems 46 .1 y 0 x −0.3 0. y) = (0.3 0. Determine the acceleration 2 cm upstream of the first stagnation point.12 m). ME 215.2 kg/m3 ) over a 20 cm diameter circular cylinder.4 −0. An inviscid flow of air (ρ = 1.2 0. The stagnation streamline crosses the y axis at y = 9 cm. The volume flowrate per unit length between the two streamlines labelled A and B is 1.5 91.5 −0.4 −0.2 A −0. 0.1 m.1 0 0. source 20 cm U∞ sink 40 cm 47 ME 215. The source and the sink are placed 40 cm apart. The maximum pressure difference between any two points on the surface of the Rankine oval shown below is measured to be 500 Pa. Determine the thickness of the object at x = 0 if a = 0. 2a).92. Determine the freestream velocity if the density of the air is 0. The maximum thickness of the oval is 20 cm. y U∞ = 10 m/s a a sources x sink a 2a 93.2 m. An inviscid flow is produced by a freestream (U∞ = 10 m/s) aligned with the x axis.8 kg/m3 .3 Example Problems . three sources placed as shown. y) = (0. The streamline shown on the diagram is a distance a from the x axis far upstream of the origin and passes through the point (x. and a single sink with a strength appropriate to form a closed streamline in the flow. 3 kPa h2 = 6 cm h1 = 15 cm Air g 1 Oil s.g = 0.3 Example Problems 48 .89 Water z B A Find: • Calculate the air pressure in the tank. Assumptions: • fluid is in a hydrostatic state • no acceleration • density of the air is negligible • ρw = 998 kg/m3 (constant) • ρoil is constant • g is constant Analysis: ∇P = ρ(g − a) ˆ g = −g k ME 215.Solution: Problem # 1 Given: • Air trapped over oil in a closed tank as shown Open to atmosphere Air h3 = 4 cm Patm = 101. 81 2 (0. 300[Pa] 3 m s ˆ ∇P = −ρg k P1 = 102. ∆P = −ρg∆z Now.15[m] − 0.a=0 dP = −ρg dz dP = −ρgdz Since ρ and g are constant. 49 ME 215.3 Example Problems .89(0.4 kPa(abs) The pressure in the air chamber is 102. P1 = (P1 − PB ) + (PB − PA ) + (PA − Patm ) + Patm P1 = −ρoil g(z1 − zB ) − ρair g(zB − zA ) − ρw g(zA − zatm ) + Patm z1 − zB = h3 zA − zatm = −h1 P1 = −ρoil gh3 + ρw gh1 + Patm P1 = ρw g(h1 − sgoil h3 ) + Patm P1 = 998 m kg 9.04)[m]) + 101.4 kPa (abs). Solution: Problem # 2 Given: • A rectangular gate holding a pool of water.3 Example Problems 50 . x Patm = 100 kPa z ρ = 1000 kg/m3 3m hinge g 3m gate (5 m wide) stop Find: • The force which the water exerts on the gate Assumptions: • fluid is in a hydrostatic state • no acceleration • g is constant • ρ is constant ME 215. Fs = − S P ndA ˆ ∇P = ρ(g − a) ˆ g = gk a=0 ˆ ∇P = ρg k dP = ρg dz dP = ρgdz dP = ρg P = ρgz + C at z = 0.Analysis: Establish a coordinate system (see diagram). P = ρgz + Patm n =ˆ ˆ ı z2 w dz Fs = − Fs = −ˆ ı (ρgz + Patm )ˆdydz ı z1 z2 0 w (ρgz + Patm )dz z1 z2 z1 0 dy ρgz 2 + Patm z Fs = −ˆ w ı 2 kg m (62 − 32 ) 2 Fs = −ˆ 5[m] 1000 ı 9.16 ˆ MN ı The water will exert a horizontal force of 2.81 2 [m ] + 100.3 Example Problems . P = Patm .16 MN on the gate. 51 ME 215. 000[Pa](6 − 3)[m] m3 s 2 Fs = −2. Therefore. C = Patm . 3 Example Problems 52 = g . 22 m z R gate x ρ = 1000 kg/m3 Find: • The net hydrostatic force on the gate Assumptions: • atmosphere is constant pressure • fluid is in a hydrostatic state • density of water is constant • g is constant • no acceleration Analysis: Establish coordinate system (see diagram).44 m 1. Fs = − S P ndA ˆ ∇P = ρ(g − a) ˆ g = −g k a=0 ME 215.Solution: Problem # 3 Given: • A Tainter gate holding back a pool of water length=2. 44[m]1. P = 0 (gauge).81 2 2.222 [m2 ] ˆ ı m3 s 1 π 1 1 π 1 + − − − + 2 8 4 2 8 4 1 1 − − − 2 4 ME 215. √ Therefore.dP = −ρg dz P = −ρgz + C √ at z = R/ 2. C = ρgR/ 2. √ P = ρg(R/ 2 − z) ˆ n = cos θˆ + sin θk ˆ ı n ˆ θ dA = Rdθdy w θ2 θ1 Fs = − 0 √ ˆ ı ρg(R/ 2 − z)(cos θˆ + sin θk)Rdθdy θ2 θ1 z = R sin θ Fs = −ρgwR2 Recall that sin2 θ = 1 − cos 2θ 2 2 ˆ cos θˆ sin θk ı ˆ √ + √ − sin θ cos θˆ − sin2 θk dθ ı 2 2 Fs = −ρgwR ˆ sin θˆ cos θk sin2 θ ı 1 √ − √ − ˆ− ı 2 2 2 2 sin θ sin2 θ √ − 2 2 ˆ −k sin 2θ θ− 2 θ2 ˆ k θ1 π/4 Fs = −ρgwR2 ˆ ı cos θ θ sin 2θ √ + − 2 4 2 1 1 − 2 4 −π/4 Fs = − 1000 ˆ − k 53 kg m 9.3 Example Problems . 2k kN ı With respect to the coordinate system shown on the diagram. the net hydrostatic ˆ force on the gate is (−35. ı ME 215.2k) kN.6ˆ + 10.3 Example Problems 54 .6ˆ + 10.Fs = −35630[N] ˆ − k ı ˆ π 1 − 4 2 ˆ Fs = −35. Solution: Problem # 4 Given: • A rectangular gate hinged at the top holding back a pool of water Patm = 100 kPa ρ = 1000 kg/m3 x g 3m z 3m hinge gate (5 m wide) stop Find: • The force which the stop exerts on the gate Assumptions: • fluid is in a hydrostatic state • no acceleration • ρ is constant • g is constant • hinge is frictionless Analysis: Establish coordinate system (see diagram) Draw a free body diagram of the gate. 55 ME 215.3 Example Problems . x Ft z Fs Fb Mo = 0 ˆ (H k × Fb ) + Ms = 0 Ms = − S P (r × n)dA ˆ ˆ r = zk n =ˆ ˆ ı ˆ ı r × n = z k × ˆ = zˆ ˆ  ∇P = ρ(g − a) a=0 ˆ g = gk dP = ρg dz P = ρgz + C at z = 0.81 2 5[m]ˆ  [m ] + [m ] m3 s 3 2 56 ME 215. P = ρgh0 . C = ρgh0 . P = ρg(z + h0 ) dA = dydz w z2 Ms = − (zˆ)ρg(z + h0 )dzdy  0 z1 z2 Ms = −ρgwˆ  Ms = −ρgwˆ  Ms = −1000 (z 2 + h0 z)dz z2 z1 z1 z 3 h0 z 2 + 3 2 kg m 33 3 3[m]32 2 9.3 Example Problems . Therefore. 57 ME 215. ˆ (H k × Fb ) + Ms = 0 1104ˆ  ˆ k × Fb = [kN · m] 3[m] Fb = 368 kNˆ ı The stop exerts a force of 368 kN on the gate.Ms = −1104ˆkN · m  Recall that.3 Example Problems . 3 Example Problems 58 . m = 35 kg z g Find: • The specific gravity of the fluid Assumptions: • fluid is in a hydrostatic state • no acceleration • block is in equilibrium • mass and volume of wire are zero • ρ is constant • g is constant Analysis: Establish coordinate system (see diagram).Solution: Problem # 22 Given: • An object suspended in a liquid by a wire 10 cm cube. Fz = 0 Fb + T − W = 0 ME 215. 81 2 = 0 s s 3 ρ = 800 kg/m sg = 800/998 The specific gravity of the liquid is 0.ρgV + T − mg = 0 m m ρ9.1[m])3 + 335.3 Example Problems .801. 59 ME 215.81 2 (0.5[N] − 35[kg]9. 3 Example Problems 60 . Therefore. 2 2 dP = 1 1 dR · ∇P = 0 ˆ dR = dxˆ + dz k ı ∇P = ρg − ρa ME 215. g.Solution: Problem # 23 Given: • A U–tube manometer used to measure linear acceleration d 1 g h 2 a z x L Find: • The magnitude of a in terms of geometry. Points 1 and 2 are at the same pressure. and h Assumptions: • a constant • fluid is in a hydrostatic state • ρ and g are constant Analysis: Establish coordinate system (see diagram). 61 ME 215.ˆ g = −g k a = aˆ ı ˆ ∇P = −ρg k − ρaˆ ı dR · ∇P = −ρadx − ρgdz 2 0= 1 (ρadx + ρgdz) = ρa(x2 − x1 ) + ρg(z2 − z1 ) 0 = ρaL + ρg(−h) a = gh/L The acceleration of the tube is gh/L.3 Example Problems . Solution: Problem # 26 Given: • A partially full container rotating around its axis.3 Example Problems 62 . Therefore. The pressure at 1 and 2 are equal. z ω = 50 rpm g 1 2 30 cm r 50 cm Find: • The depth of the fluid on the axis Assumptions: • ω is constant • fluid is in a hydrostatic state • ρ is constant • g is constant Analysis: Establish coordinate system (see diagram). 2 2 dP = 0 = 1 1 dR · ∇P ˆ dR = drˆ + dz k r ME 215. ∇P = ρ(g − a) ˆ g = −g k ˆ ∇P = −ρg k + ρω 2 rˆ r 2 a = −ω 2 rˆ r 0= 1 (−ρgdz + ρω 2 rdr) ρg(z2 − z1 ) = ρω 2 2 2 (r2 − r1 ) 2 (5π/3[1/s])2 (0.3[m] − z1 = 2(9. 63 ME 215.3 cm deep on the centreline.3 cm The fluid is 21.3 Example Problems .25[m])2 0.81)[m/s2 ] z1 = 21. Therefore. Assumptions: • ω is constant • fluid is in a hydrostatic state • g is constant • ρ is constant Analysis: Establish coordinate system (see diagram).3 Example Problems 64 .Solution: Problem # 27 Given: • A cylindrical container rotating about its axis. dP = 0 along the surface. The surface is a line of constant pressure. 0 = dR · ∇P ME 215. z ω g Ho r R Find: • Derive an equation for the shape of the liquid surface if the container spins at ω about the z axis. ˆ dR = drˆ + dz k r ˆ ∇P = ρ(ω 2 rˆ − g k) r 0 = ρω 2 rdr − ρgdz ω2r dr = g dz a = −ω 2 rˆ r z= ω 2r2 +C 2g Equate the initial and final volumes.3 Example Problems . R z2πrdr = H0 πR2 0 R H0 πR2 = 2π 0 2 ω 2r3 + Cr dr 2g R 0 ω 2 r 4 Cr 2 + H0 R = 2 8g 2 H0 R 2 = ω R + CR2 4g ω 2 R2 4g 2 4 C = H0 − The surface is defined by ω2 2g R2 2 z = H0 + r2 − 65 ME 215. Solution: Problem # 31 Given: • A closed cylindrical container rotating about its axis.3 Example Problems 66 . Fs = − S P ndA ˆ ∇P = ρ(g − a) ˆ g = −g k ME 215. z 60 rpm vent r ρ = 1000 kg/m3 g cylindrical can Find: • The force on the top of the can (net hydrostatic force) Assumptions: • fluid is in a hydrostatic state • ρ is constant • g is constant Analysis: Establish a coordinate system (see diagram). 67 ME 215. C = 0. Therefore.44 [m4 ] ˆ k 4 The net hydrostatic force on the lid of the can is 794 N upwards.3 Example Problems . Therefore. P = ˆ n = −k ˆ dA = rdrdθ 2π R 0 a = −ω 2 rˆ r Fs = − Fs = 0 ρω 2 r 2 ˆ (−k)rdrdθ 2 R 2πρω 2 ˆ k 2 ˆ r 3 dr = πρω 2 k 2π 1 s 2 0 R4 4 Fs = π1000 ˆ Fs = 794k N kg m3 0. P = 0. ∂P = ρω 2 r ∂r ρω 2 r 2 +C 2 at r = 0.ˆ ∇P = −ρg k + ρω 2 rˆ r Need P as a function of r. 2 V = 15 m/s Aj = 0.3 Example Problems 68 . Assumptions: • steady state • jet is horizontal • jet deflects to vertical plane • pressure uniform on control surface • neglect body forces • neglect fluid shear • uniform flow ME 215.01 m2 y 1 2 x g Find: • The force that the water exerts on the barricade.Solution: Problem # 35 Given: • A jet of water hitting a flat barricade. Therefore. Each term of the momentum equation will now be discussed.3 Example Problems .Analysis: Since we are asked to find a force we should probably consider the linear momentum equation. ρgdV − − − V S P ndA + Fv + R = ˆ d dt − V (ρV )dV + − S ρV (Vr · n)dA ˆ Consider the control volume shown. Reaction: R=0 We have chosen a control surface which “cuts” the support so we can find R. air. water. the net pressure force is ZERO. 69 ME 215. Viscous Force: Fv = 0 The flow is perpendicular to the control surface everywhere. Pressure Force: P ndA = 0 ˆ S P is uniform over the entire control surface. Body Force: ˆ ρgdV = −W k − − V W is the weight of EVERYTHING in the control volume. barricade etc. 01[m2 ] 15 3 m s 2 = −2.25 kN This is the force exerted on the control volume by the “mounting strut”.25kN to the right on the barricade.3 Example Problems 70 . Momentum transport term: ρV (Vr · n)dA = ˆ S d (m V )d − ˙ (mV )i ˙ i The properties at all inlets and discharges are assumed to be uniform. ME 215. The force exerted on the barricade is in the opposite direction. With these simplifications. the momentum equation becomes ˆ −W k + R = (ρw A2 V2 )V2 + (ρw A3 V3 )V3 − (ρw A1 V1 )V1 Consider only the x direction Rx = −ρw A1 V12 Rx = −998 m kg 0. The water exerts a force of 2.Unsteady term: d dt − V (ρV )dV = 0 − Steady state is assumed. ρgdV − − − V S P ndA + Fv + R = ˆ d dt − V (ρV )dV + − S ρV (Vr · n)dA ˆ We must choose a control volume. 71 ME 215.Solution: Problem # 36 Given: • A jet of water deflecting off a vane mounted on a moving cart 45o Aj = 0.3 Example Problems .01 m2 Vj = 15 m/s Vc = 5 m/s constant cart Find: • Force which the water exerts on the cart Assumptions: • neglect body forces • neglect friction on vane • uniform flow • cart travelling at constant velocity Analysis: Since we are asked to find a force we should probably consider the linear momentum equation. – the pressure everywhere except on the surface of the plate is zero – the integrated effect of the pressure acting on the plate is what we were asked to find in this problem – this term gives the force ON the CV. • The pressure term is unknown. • The reaction force is zero since this CS does not cut through any solid objects.. one possible choice for a control volume is . In this case we are looking for the force that the water exerts on the cart. Therefore. It is important to consider what we are being asked to solve for..Should it be stationary or moving? (we will try both) What should it look like? First consider what the control volume should look like. ME 215. it is equal and opposite to the unknown in this problem • The viscous force term is zero since we are assuming a frictionless surface. ˆ • The body force term is assumed to be zero. We are neglecting the weight of the water on the vane.S. ρgdV − − − V S P ndA + Fv + R = .3 Example Problems 72 . between water and plate (gap shown only for clarity) Consider the left-hand side of the momentum equation for this choice. Therefore. C. . It only changes which term on the LHS gives us the information we want.Another choice for a CV may be as follows. • The pressure is now zero everywhere on the control surface. ˆ • The body force term is zero as before. ρgdV − − − V S P ndA + Fv + R = . Stationary Control Volume 73 ME 215.3 Example Problems . we realise that R is opposite to the force which the water exerts on the cart. We will continue the problem with the second option. Moral of the Story!! Either choice for a CV is fine.. Therefore. • The viscous force term is zero as before. If we consider a free–body–diagram of the cart. 1. Consider the LHS of the momentum equation now. the reaction force is NOT zero. • Now the control surface cuts through the cart. the magnitude of this velocity will not change at the outlet. Also.Attach the coordinate system AND the control volume to the cart. z x • This is considered a stationary control volume because it is not moving with respect to the coordinate system. • The coordinate system is inertial because it is moving at constant velocity. because mout = min ˙ ˙ R = m(Vout − Vin ) ˙ m = ρ(Vj − Vc )Aj ˙ Assuming that the vane is frictionless. R= d dt ρV dV + − ρV (Vr · n)dA ˆ Since nothing in the CV is changing with time. it is a steady problem. We have already simplified the LHS so now. √ √ 2 2ˆ Vout = (Vj − Vc ) ˆ+ ı k 2 2 ME 215. R = (mV )out − (mV )in ˙ ˙ or.3 Example Problems 74 Vin = (Vj − Vc )ˆ ı . assume uniform flow. attach the coordinate system to the pipe but attach the control volume to the cart. Vout and Vin are different. However.3 Example Problems (V j − V c) . ˆ The water exerts a force of (292ˆ − 705k) N on the cart. in our coordinate system. 2. R = m Vout − Vin ˙ is still true and m = ρ(Vj − Vc )Aj ˙ is still true. Vcˆ i √ 2/ 2( ˆ i+ kˆ) Vout Vout = (Vj − Vc ) 75 √ 2 (ˆ + k) + Vcˆ ı ˆ ı 2 ME 215. the forward velocity of the cart must be added to this. The coordinate system is ı shown on the diagram.01[m ] 2 √ 2 2ˆ − 1 ˆ+ ı k 2 2 ˆ R = (−292ˆ + 705k) N ı Remember. Therefore. The flow inside the control volume is still steady. Now. Moving Control Volume Now. this is opposite to the force the water exerts on the cart.R = ρ(Vj − Vc )Aj (Vj − Vc ) R = ρ(Vj − Vc )2 Aj kg R = 998 m3 m 10 s √ √ 2 2 (ˆ + k) − (Vj − Vc )ˆ ı ˆ ı √ 2 2ˆ − 1 ˆ+ ı k 2 2 √ 2 0. Vin = Vjˆ ı Vout still has a magnitude of (Vj − Vc ) when viewed with respect to the vane. However. 3. ı d dt ρV dV = ρVjˆAj Vc − ı Now. Therefore. Therefore. the LHS is the same. mout = ρVj Aj − ρVc Aj = ρAj (Vj − Vc ) ˙ √ 2 2 R = ρVj Vc Ajˆ + ρAj (Vj − Vc ) ı ME 215. This column is getting longer at speed Vc . The uniform flow assumption is still valid. Consider a sketch showing the CV at two times. ˙ ˙ mout = min − ρAj Vc ˙ ˙ Back to the momentum equation. Draw a stationary CV such that the cart is in it at a certain instant in time. Stationary Control Volume Try again with a different stationary control volume.Now. at the inlet (mV )in = (ρVj Aj )Vjˆ ˙ ı However.3 Example Problems (Vj − Vc )(ˆ + k) + Vcˆ − ρVj2 Ajˆ ı ˆ ı ı 76 . the answer will be identical. R = ρ(Vj − Vc )Aj R = ρ(Vj − Vc )Aj (Vj − Vc ) (Vj − Vc ) √ 2 (ˆ + k) + Vcˆ − Vjˆ ı ˆ ı ı 2 2 (ˆ + k) − (Vj − Vc )ˆ ı ˆ ı 2 √ This intermediate result is identical to the stationary CV. R= d dt − V ρV dV − + (mV )out − (mV )in − ˙ ˙ The unsteady term is NOT ZERO anymore. The total momentum ( ρV dV ) in (ii) is greater than in (i) because the column of liquid − going at Vjˆ is longer. note that min = mout for this analysis because mass is accumulating in the CV. Attach the coordinate system to the supply pipe. How is the analysis different now? Again. 3 Example Problems .z (i) x z (ii) x R = ρ(Vj − Vc )Aj (Vj − Vc ) √ 2 (ˆ + k) − (Vj − Vc )ˆ ı ˆ ı 2 Again. this is the same intermediate result so the answer will be the same. 77 ME 215. consider the linear momentum equation. A coordinate system is also shown on the diagram. cut the bolts with the control surface.Solution: Problem # 37 Given: • A nozzle which discharges water from a pipe 10 kPa z D2 = 25 mm x 1 D1 = 50 mm V1 = 1 m/s 2 Find: • The force on the bolts at the flange connection Assumptions: • neglect body forces • no viscous force on CV • uniform flow • steady state Analysis: Since we are asked to find a force. Therefore.3 Example Problems . ρgdV − − − V S P ndA + Fv + R = ˆ d dt − V (ρV )dV + − S ρV (Vr · n)dA ˆ 78 ME 215. Choose a control volume that will help us find what we are after. Consider the linear momentum equation. the momentum equation becomes P1 A1ˆ + R = m(Vout − Vin ) = m(V2 − V1 ) ı ˙ ˙ V1 = 1ˆ m/s ı Find the magnitude of V2 from conservation of mass d dt (ρ)dV + − ρ(V · n)dA = 0 ˆ − V S For steady. 5.Consider each term in this equation. 4. P1 = 10 kPa n = −ˆ ˆ ı − Since P1 is uniform and n doesn’t vary. 1. So. 2. 6. ˆ ˆ ı ı P ndA = −P1 (−ˆ)A1 = P1 A1ˆ 3. The unsteady term on the right-hand-side is zero because this is a steady problem. Assume uniform flow to simplify the last term in the equation.8ˆ N ı 79 ME 215. the reaction force will give us what we want to find. the ı bolts are in tension. Use gauge pressures so that the only non–zero pressure is at 1. There is a tensile force on the bolts of 13. There is no flow parallel to the CS. Therefore.8 N. Neglect the viscous force term. Neglect the body force. Since our CS cuts the bolts. R = −13.3 Example Problems .05)2 [m2 ]ˆ + 998 ı 1 (0.05)2 [m2 ](4ˆ − 1ˆ) ı ı 3 4 m s 4 s This means the bolts are exerting a force ON the CV in the −ˆ direction. We have no information about weight of the nozzle. uniform flow this becomes ρV1 A1 = ρV2 A2 V2 = V1 A1 = V1 A2 50 25 2 V2 = 4ˆ m/s ı π kg m π m R = −10 × 103 [Pa] (0. P2 .3 Example Problems 80 . Consider each term of the momentum equation. Uo . ρgdV − − − V S P ndA + Fv + R = ˆ d dt − V (ρV )dV + − S ρV (Vr · n)dA ˆ The dashed line on the diagram is chosen as the CV. ME 215.Solution: Problem # 38 Given: • The inlet section of a laminar pipe flow Uo r 1 z 2 u(r) = Umax (1 − (r/R)2) circular pipe radius= R Find: • Frictional drag on the fluid between 1 and 2 in terms of P1 . A coordinate system has already been given in the problem. ρ. and R Assumptions: • neglect body forces • uniform flow at 1 • steady state • ρ is constant Analysis: Since we are asked for a force (drag) on the fluid we should consider the linear momentum equation. The viscous force term is the unknown in this problem. 4. we can apply the uniform flow assumption at 1. assume the unsteady term is zero.1. (ρ)dV + − ρ(V · n)dA = 0 ˆ r R 2 − V S ρUmax 1 − rdrdθ − Uo ρπR2 ME 215. We have chosen a CV including just the water on the pipe so that the friction at the pipe walls enters the momentum equation through this term. Since this is apparently a steady flow. R = 0 since no solid member penetrates the CS. 2. Therefore. (Vr · n) = u(r) ˆ V2 = Umax 1 − dA = rdθdr Fv = (P2 − P1 )πR k + 2ˆ R 0 0 R 0 2π r R 2 ˆ k ρ Umax r 1− R 2 2 ˆ ˆ krdrdθ − Uo πR2 ρUo k 2 ˆ rdr − Uo πR2 ρk R 0 2 ˆ − Uo πR2 ρk 2 ˆ ˆ Fv = (P2 − P1 )πR2 k + 2πρUmax k 1− r4 2r 2 + 4 R4 R Fv = (P2 − P1 )πR k + 2ˆ 2 ˆ 2πρUmax k 2 ρUmax r2 2r 4 r6 − + 2 4R4 6R4 Fv = πR2 (P2 − P1 ) + d dt 0= 2 3 2 ˆ − ρUo k Eliminate Umax with conservation of mass (steady).3 Example Problems 81 . Although pressure is certainly acting on the rest of the CS. it is equal around the circumference and therefore yields no net force. Since no information has been give about the orientation of the pipe and we are only interested in frictional drag. The only significant pressure forces on our CS are at 1 and 2. ignore the body force term. 3. ˆ (P1 − P2 )πR2 k + Fv = 2 ρV (Vr · n)dA − (mV )1 ˆ ˙ At 2. Although we do have uniform flow at the inlet. The final term in the equation must be handled carefully. 6. 5. we certainly don’t at the outlet. but not at 2. R 0 = ρUmax 2π 0 r− r3 R2 R 0 dr − Uo ρπR2 − Uo ρπR2 0 = ρUmax 2π 0 = ρUmax 2π Umax = 2Uo r2 r4 − 2 4R2 2 R − Uo ρπR2 4 ρ(2Uo )2 2 ˆ − ρUo k 3 Fv = πR2 (P2 − P1 ) + The frictional drag on the fluid is 2 ρUo 3 πR2 (P2 − P1 ) + ME 215.3 Example Problems 82 . ρgdV − − − V S P ndA + Fv + R = ˆ d dt − V (ρV )dV + − S ρV (Vr · n)dA ˆ With the assumptions listed this becomes (see solution on page 79 for more details) R = ρV1 A1 V1 A1 ˆ − V1ˆ − P1 A1ˆ ı ı ı A2 ME 215. Since we are asked for a force. consider the momentum equation.3 Example Problems 83 . incompressible flow • no shaft work or heat transfer • no elevation change Analysis: The control volume and coordinate system are shown above.Solution: Problem # 39 Given: • A nozzle mounted at a pipe exit ? z x D1 = 50 mm V1 = 1 m/s 1 2 D2 = 25 mm Find: • Force on the bolts assuming frictionless flow Assumptions: • frictionless flow • uniform inlet and outlet • steady. ME 215.3 Example Problems 84 .82ˆ N ı 4 1000 50 25 2 R= ˆ− 1 ı m ˆ ı s The tensile force in the bolts is 8.05)2[m2 ] = −8.Bernoulli’s equation is P1 + ρV 2 ρV12 + ρgz1 = P2 + 2 + ρgz2 2 2 Since z1 = z2 ρ 2 998 kg m2 2 2 2 P1 = (V2 − V1 ) = (4 − 1 ) 2 2 2 m3 s P1 = 7485 Pa m π m kg 1 (0.05)2 [m2 ] 1 3 m s 4 s π −7485[Pa] (0.82 N. Solution: Problem # 40 Given: • A nozzle mounted at the exit of a pipe 10 kPa z x D1 = 50 mm V1 = 1 m/s 1 2 D2 = 25 mm Find: • The loss coefficient for the nozzle Assumptions: • steady.04 The loss coefficient of this nozzle based on the inlet velocity is 5. 85 ME 215.04.3 Example Problems . no heat transfer • no elevation change Analysis: Bernoulli’s equation P1 + ρV12 ρV 2 + ρgz1 = P2 + 2 + ρgz2 + ∆PL 2 2 ρ ∆PL = P1 + V 2 − V22 2 1 998 kg m2 m2 ∆PL = 100000[Pa] + 12 2 − 42 2 2 m3 s s ∆PL = 2515 Pa ρV 2 ∆Pv = Km 1 2 Km = 5. incompressible flow • no shaft work. incompressible flow • no shaft work or heat transfer Analysis: Since we are asked to find a force.3 Example Problems 86 .50 (based on D1 = 50 mm discharge velocity) 2 45o . ME 215. Begin with the complete equation ρgdV − − − V S P ndA + Fv + R = ˆ d dt − V (ρV )dV + − S ρV (Vr · n)dA ˆ • Since we have no information about the mass of the transition. consider the linear momentum equation. or even the direction of g. we will neglect body forces.Solution: Problem # 41 Given: • A transition piece as shown D2 = 80 mm 20 kPa y x 4 kg/s 1 Km = 0. Find: • Force required to hold transition in place Assumptions: • neglect body forces • uniform inlet and outlet • steady. • The flow is steady so d dt ρV dV = 0 − • Use the uniform flow assumption for the last term since we have no information about how the velocity varies across the pipe. P1 + ρV12 ρV 2 = P2 + 2 + ∆PL 2 2 √ 2 2 (ˆ + ) ı ˆ ρV22 2 ρ ρV 2 P2 = P1 + (V12 − V22 ) − Km 2 2 2 ∆PL = Km 87 ME 215. R becomes the unknown in the problem. At 1.08)2 [m2 ] ρA2 2 2 998[kg/m Now. let’s work on the pressure force term.04ˆ m/s ı 3 ] π(0. Now.• Choose a CV which isolates the transition piece.05)2 [m2 ] ρA1 998[kg/m √ √ m ˙ 4 2 2 4[kg/s] (ˆ + ) = ı ˆ V2 = (ˆ + ) = 0. we have − V1 = P ndA + R = m(V2 − V1 ) ˆ ˙ 4[kg/s] 4 m ˙ ˆ= ı = 2. Therefore. Write Bernoulli’s equation (with pressure losses) from 1 to 2.3 Example Problems . P1 = 20 kPa n = −ˆ ˆ ı At 2 n= ˆ P2 =? We must find P2 using Bernoulli’s equation.5638(ˆ + ) m/s ı ˆ ı ˆ 3 ] π(0. • A uniform pressure exists at 1 and 2. return to the momentum equation √ 2 (ˆ + ) + m(V2 − V1 ) ı ˆ ˙ R = −P1 A1ˆ + P2 A2 ı 2 √ π π 2 2 2 2 2 R = − 20.04ˆ) ı  ı s s R = (31. 000[Pa] (0.08) [m ] ı (ˆ + ) ı ˆ 4 4 2 kg m + 4 (0.6ˆ + 79.5638ˆ − 2.3 Example Problems 88 .04)2 − 2(0.0ˆ) N in the coordinate ı  system shown.P2 = 20.5638ˆ + 0.6 kPa Now. ME 215. 000[Pa] + 998 kg (2.5 2(0.6ˆ+ 79.5638)2 2 2 m3 s m2 s2 P2 = 21.0ˆ) N ı  The force required to hold the transition in place is (31.5638)2 2 m3 998 kg m2 − 0.05) [m ]ˆ + 21. 600[Pa] (0. 20o C Q = 0. steel pipe Find: • P1 − P2 Assumptions: • no shaft work or heat transfer • uniform flow at 1 and 2 • no diameter change • steady. incompressible flow Analysis: Write Bernoulli’s equation with losses from 1 to 2.3 Example Problems . 2” I. flanged. flanged elbows g 1 water. globe valve 30 m 90o .D. regular.Solution: Problem # 64 Given: • A piping system as shown below 2 open.04 m3 /s 50 m. P1 + ρV 2 L ρV 2 ρV12 + ρgz1 = P2 + 2 + ρgz2 + f + 2 2 D 2 Km ρV 2 2 89 ME 215. 74[m/s](2(0.0254))2π/4[m2 ] We need to find f .3 Example Problems 90 .046 ǫ = = 0.83 MPa at this flowrate.0254) ME 215.1). V1 = V2 .5 + 2(0. Therefore. 0. m s 2 P1 − P2 = 998 19. let z1 = 0.Since there is no diameter change.81 2 30[m] + 3 m s 2 m3 0. P1 − P2 = ρgz2 + f P1 − P2 = ρgz2 + V = L ρV 2 + D 2 f L + D Km ρV 2 2 ρV 2 2 Km 0.83 MPa The pressure drop between 1 and 2 is 5.74 50 + 8.046 mm (Table 6. Now find the minor loss coefficients from table 6–5.5 Kelbow = 0. ǫ = 0.0195.39) 2(0. It is a function of roughness (ǫ) and Reynolds number (Re). Kvalve = 8.04[m3 /s] Q = = 19.0254)[m]) = = = 1 × 106 −6 [m2 /s] µ ν 1 × 10 From the Colebrook equation with this (ǫ/D) and Re we get f = 0.74 m/s A (2(0.0195 P1 − P2 = 5.4) Re = ρV D VD 19. For steel pipe.39 kg m 998 kg 9. Also.000906 D 2(25. incompressible flow • no heat transfer or shaft work Analysis: Write Bernoulli’s equation (with losses) from 1 to 2.05 mm Find: • Volume flowrate between the reservoirs Assumptions: • V1 = V2 = 0 • P1 = P2 • steady.3 Example Problems Km ρV 2 2 ρgz1 = fa 91 a b .001 kg/(m · s) 2 a D = 2 cm L = 5m b D = 4 cm L = 5m ǫ = 0. P1 + ρV 2 L ρV 2 ρV12 + ρgz1 = P2 + 2 + ρgz2 + f + 2 2 D 2 L D ρVa2 + fb 2 L D ρVb2 + 2 ρV 2 Km 2 ME 215.Solution: Problem # 65 Given: • A pipe connecting two reservoirs as shown 1 20 m ρ = 1000 kg/m3 µ = 0. 026 70. The volume flowrate between the reservoirs is 0.3 Example Problems 92 .21 Figure 6.56 1.900 0.22 Figure 6.00223 m3/s.4 = Va2 250fa + Initial guess. + Kexp. fully rough zone fa = 0.Minor losses: entrance expansion exit Km 0.1225 16 392.0 L D source Figure 6. + Kexit Da Db 4 a 0. Aa Vb = Va = Va Ab ρgz1 = ǫ D ǫ D ρVa2 2 = = b Da Db L D 2 f + f a L D b Da Db 4 + Kent.5 0.024 This result is converged because the f ’s have stopped changing.025.300 0.21 ρVa2 + fb 2 L D ρVb2 ρVa2 ρV 2 + (Kent.09 141. + Kexp.021.00125 40[mm] 125 fb + 1.200 0. ME 215.026 72.0025 20[mm] 0.05[mm] = 0. fb = 0.22 144.770 0.05[mm] = 0.024 7. ) + b (Kexit ) 2 2 2 ρgz1 = fa a b From conservation of mass. Procedure: • calculate Va from the previous equation • use Va to get Rea and Reb • use Rea and Reb to get new fa and fb • repeat if necessary Va Rea fa Reb fb 7. Solution: Problem # 69 Given: • The head versus flowrate characteristics of a pump • Water is pumped through 100 m of 20 cm diameter cast iron pipe • Water temperature is 20o C 100 80 Pump head.5 Flowrate. m /s Find: • Flowrate Assumptions: • no elevation change • no minor losses • pump inlet the same diameter as exit pipe • pump inlet pressure same as pipe exit pressure • no heat transfer • steady.5 1 3 1. m 60 Parabola 40 20 0 0 0.5 2 2. incompressible flow 93 ME 215.3 Example Problems . Q = 0. Re = ρV D VD QD = = = 2. hp = 80 − 20Q2 where Q is in m3 /s and hp is in m. The frictional head loss is hf = f L Q2 . D 2gA2 2 (80 − 20Q )[m] = f 80 25. ǫ = 0.26 mm so ǫ/D = 0. 821f Q2 + 20Q2 Q= For cast iron pipe. 821f + 20 100[m] 0.3772 m3 /s Check the Reynolds number.2[m])2 2 80 = 25.4 × 106 µ ν Aν This is fully rough so we can accept the answer. ME 215.3 Example Problems 94 .377 m3 /s.0013. The fully rough friction factor for this ǫ/D is f = 0. P1 + ρV 2 ρV12 + ρgz1 = P2 + 2 + ρgz2 + ∆Pf + ∆Pm − ∆Pp 2 2 With the assumptions stated above this reduces to ∆Pf = ∆Pp or hf = hp The pump curve can be described by.81[m/s2 ]) 4 π(0. The flow rate is 0.Analysis: Writing Bernoulli’s equation from the inlet of the pump (1) to the exit of the pipe (2) gives. Therefore.2[m] Q2 2(9.0210. irrotational flow 95 ME 215. steady flow • inviscid.Solution: Problem # 81 Given: • Flow around a bridge pier • U∞ = 5 m/s • P∞ = 50 kPa • stagnation point is 1 m upstream of source y x 2m A 1m Find: • Pressure at point A Assumptions: • two-dimensional flow • incompressible.3 Example Problems . 3 Example Problems 96 2 x y x . vθ = vr = −∂ψ = −U∞ sin θ ∂r 1 ∂ψ m = U∞ cos θ + r ∂θ r We need the (r. Therefore.Analysis: Must find the source strength required to give the stagnation point 1 m upstream of the source. r = 1 m/s. ψsp = 0 + mπ = mπ Since A is on the same streamline ψA = mπ In (x. 5[m2 /s]π = 5[m/s]2[m] + 5[m2 /s] tan−1 xA = 0. y) coordinates. Place the source at the origin. θ) coordinates of point A to find these velocities.9153 ME 215. u = 0. ψ = U∞ y + m tan−1 at A. Therefore. m = 5 m2 /s. for the source vr = m r m r At the stagnation point. Therefore. Now we need the velocity at point A. At the stagnation point θ = π. 0 = U∞ − 0=5 m m − s 1[m] Therefore. ψ = ψfreestream + ψsource ψ = U∞ r sin θ + mθ Therefore. 142) + 5/2.200)2 2 VA = 39.3 Example Problems .7 kPa The pressure at A is 42. 2 2 2 VA = vθ + vr = (−5 sin(1.142 rad θA = tan−1 Therefore.9152 = 1.7 kPa.61) PA = P∞ + ρ(V∞ − VA ) = 50. 000[Pa] + 3 2 2 m s PA = 42.61 m2 /s2 1000 kg m 1 2 2 (52 − 39.200 m 2 0.9153)2 + (2)2 = 2.142))2 + (5 cos(1.rA = (0. 97 ME 215. 3 Example Problems 98 . steady flow • inviscid.Solution: Problem # 82 Given: • A pair of doublets as shown • U∞ = 10 m/s 2m 2m doublets 2m y x 30o U∞ = 10 m/s Find: • Doublet strength • Pressure at the origin Assumptions: • two-dimensional flow • incompressible. irrotational flow ME 215. 0 = λ sin θ1 λ sin θ2 − r1 r2 λ(y − 2) λ(y − 2) − 2 + (x − 2)2 (y − 2) (y − 2)2 + (x + 2)2 λ(−2) λ(−2) − 2 + (−2)2 (−2) (−2)2 + (2)2 2λ 2λ λ + = 8 8 2 λ(−1) λ(−1) − 2 + (2)2 (−1) (−1)2 + (6)2 ψ4.Analysis: ψ = ψfs + ψd1 + ψd2 ψ = U∞ (y cos α − x sin α) − ψ = U∞ (y cos α − x sin α) − ψ0. Therefore.1 ψ0.0 = 0 − ψ0.34 + λ 1 1 1 − − 2 5 37 λ λ λ + = 5 37 2 = −11.3 Example Problems .54 m3 /s The doublet strength must be −41. 99 ME 215.1 = −11.0 = ψ4. the pressure there is the same as the pressure far away from the doublets.54 m3 /s. The pressure at the origin is P∞ .1 = 10(cos 30o − 4 sin 30o ) − ψ4.34 λ = −41. only the freestream contributes to the velocity at the origin. By symmetry. steady flow • inviscid.3 Example Problems 100 . irrotational flow ME 215.2 kg/m3 B A source 12 cm sink 12 cm U∞ 30 cm Find: • PA − PB Assumptions: • two-dimensional flow • incompressible.Solution: Problem # 83 Given: • A landing gear strut • U∞ = 45 m/s • ρ = 1. 12) m = 1. 0 = U∞ − m m + (0.12 y 45y = 1. ψ = ψfs + ψsource + ψsink ψ = U∞ r sin θ + mθsource − mθsink Find m.07027 )2 0.519 tan−1 0.12 2 101 ME 215.12 y − π(1.519 tan−1 Note that tan−1 y −0.Analysis: A Rankine oval can be modelled by superimposing a source and a sink with the freestream.12 y −0. At B. uB = 45 + 1.519) tan−1 0. we only need u at B.12) (0.519) tan−1 Now.519π − 2(1. ψ = U∞ y + m tan−1 u= y x+a 2 − m tan−1 1 x+a − y x−a m 1+ y x−a 2 ∂ψ m = U∞ + y ∂y 1 + x+a 1 x−a PROBLEM!!!!! Don’t know y at B.07027 m 0 = 45y + 2(1. At the stagnation point.12 = π − tan−1 y 0. v = 0.519 m2 /s Need VB .12 1 0.519 1 1+ ( 0.15 + 0.12 y − 1. Therefore. u = 0.12 y = 0.3 Example Problems .15 − 0.519) 0. ψB = ψsp = 0 0 = 45y + 1. 45 kPa The pressure difference between points A and B is 2.3 Example Problems 102 .85)2 2 2 PA − PB = 2. ME 215.2 PA − PB = ρUB = (63.45 kPa.uB = 63.85 m/s 1 2 1 2 PA + ρUA = PB = ρUB 2 2 1 2 1. a. • Sketch this function. Far upstream of the cylinder this streamline is a distance L from the horizontal plane of symmetry. steady flow • inviscid.3 Example Problems . ψ = ψfs + ψdoublet 103 ME 215. • Consider the streamline labelled A. and y.Solution: Problem # 84 Given: • Flow over a circular cylinder creating no lift y U∞ A g L a x Find: • Derive an expression for the velocity along the positive y axis in terms of U∞ . At what distance from the origin does the streamline cross the y axis? Assumptions: • two-dimensional flow • incompressible. irrotational flow Analysis: Non-lifting flow over a cylinder can be modelled by a doublet and a freestream. ψ = 0 on the circle. vr = 0 ψ = U∞ sin θ(r − ∂ψ a2 vθ = − = −U∞ sin θ 1 + 2 ∂r r θ= π 2 a2 r2 vθπ/2 = −U∞ 1 + u a2 =1+ 2 U∞ r Along the positive y axis. U∞ a − λ =0 a λ = U∞ a2 a2 ) r Along the y axis. Therefore.λ sin θ r λ ψ = sin θ(U∞ r − ) r ψ = U∞ r sin θ − At the stagnation point.3 Example Problems 104 . u = −vθ . Therefore The velocity along the positive y axis is given by u a2 = 1 + 2. ψsp = 0 Therefore. U∞ r ψA − ψsp = U∞ L ψA = U∞ L U∞ L = U∞ sin L=r− π 2 r− a2 r a2 r 2 r − Lr − a2 = 0 ME 215. θ = π and r = a. Therefore. we want the positive root.3 Example Problems . r= L + 2 L + a2 2 The streamline labelled A crosses the y axis at a distance of L + 2 L + a2 2 from the origin. 105 ME 215.u/U∞ 2 1 1 r/a r= L ± 2 L + a2 2 Since r > a. 1 0 0. 0.3 0.3 Example Problems 106 .4 0.2 kg/m3 • Volume flowrate per unit length between streamlines A and B is 1.3 −0.1 y 0 x −0.5 −0.12m) 0.Solution: Problem # 85 Given: • Flow around a circular cylinder • U∞ = 20 m/s • ρ = 1.4 −0.5 −0.2 0.3 0. y) = (0.1 0.1 B −0.3 −0. • Streamline B passes through point (x.2 A −0.12 m) 0.2 −0.4 0.2 (x. 0.4 −0.5 Find: • Lift force per unit length ME 215.19 m2 /s. y) = (0.5 0. 2 kg m 20 3 m s −15.74 m2 s Kutta-Joukowski Theorem FL = 378 N/m The lift force per unit length is 378 N/m.12[m] 0. 0.12[m] 0.19 m2 /s Therefore. at (x.19 m2 m 20[m/s](0.19 m2/s Now. steady flow • inviscid. a doublet. ψ = ψfs + ψdoublet + ψvortex ψ = U∞ r sin θ − For ψ = 0 on r = a.12[m]) − s s 0. and a vortex.12 m) 1. y) = (0.1)2 [m2 ] − K ln = 20 (0.1[m] λ sin θ a λ sin θ r − K ln r a K = −2. irrotational flow Analysis: This flow can be modelled by combining a freestream. 0 = U∞ a sin θ − λ = U∞ a2 ψB − ψA = Q = 1.3 Example Problems . ψB = 1. 107 ME 215.505 m2 /s Γ = 2πK = −15.Assumptions: • two-dimensional flow • incompressible.74 m2 /s FL = −ρU∞ Γ FL = −1.