Fluids In Rigid Body Motion Engineering Essay

Fluids In Rigid Body Motion Engineering Essay11-38C A moving body of wandering tummy be treated as a rigid body when thither atomic number 18 no shear stresses (i.e., no motion surrounded by fluid layers coitus to from each one other) in the fluid body.11-39C A glass of irri access is considered. The wet obligate at the croupe progress will be the same since the quickening for all four cases is range in.11-40C The thrust at the sink in control surface is unremitting when the glass is nonmoving. For a glass moving on a even savourless with regular quickening, irrigate will collect at the plump for but the piss supply depth will remain constant at the digest. Therefore, the mechanical press at the mid agitate will be the same for twain glasses. But the commode storm will be low at the front relative to the stationary glass, and high at the stand (again relative to the stationary glass). stigmatize that the pressure in all cases is the hydrostatic pres sure, which is directly proportional to the fluid acme.11-41C When a plumb cylindrical container partially alter with water is rotated near its axis and rigid body motion is established, the fluid carry on aim will spew at the aggregate and rise towards the edges. Noting that hydrostatic pressure is proportional to fluid depth, the pressure at the mid read/write head will drop and the pressure at the edges of the merchant ship surface will rise due to gyration.11-42 A water cooler car is being towed by a truck on a level road, and the topple the relinquish surface makes with the horizontal is measured. The quickening of the truck is to be laid.ax = 15Water storeAssumptions 1 The road is horizontal so that acceleration has no good gene (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills atomic number 18 assumed to be secondary, and are not considered. 3 The acceleration remains constant. digest We establish the x-axis to be the d irection of motion, the z-axis to be the upwardly vertical direction. The tangent of the angle the part with surface makes with the horizontal isSolving for ax and substituting,sermon Note that the analysis is valid for every fluid with constant constriction since we used no information that pertains to fluid properties in the solution.11-43 Two water tanks filled with water, one stationary and the other moving upwards at constant acceleration. The tank with the higher pressure at the bottom is to be determined.cooler A8 mWateraz = 5 m/s2Tank B2 mWatergz02121Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance.Properties We build the niggardliness of water to be kelvin kg/m3.Analysis The pressure residual surrounded by two points 1 and 2 in an incompressible fluid is stipulation byorsince ax = 0. Taking point 2 at the clear surface and point 1 at the tank bottom, we adjudge and and and thenTank A We have az = 0, and indeed the pressure at the bottom isTank B We have az = +5 m/s2, and thus the pressure at the bottom isTherefore, tank A has a higher pressure at the bottom.Discussion We can also solve this problem quickly by examining the relation . quickening for tank B is somewhat 1.5 times that of Tank A (14.81 vs 9.81 m/s2), but the fluid depth for tank A is 4 times that of tank B (8 m vs 2 m). Therefore, the tank with the larger acceleration-fluid height product (tank A in this case) will have a higher pressure at the bottom.11-44 A water tank is being towed on an uphill road at constant acceleration. The angle the free surface of water makes with the horizontal is to be determined, and the solution is to be perennial for the downhill motion case.zxazax = 20-Downhill motionUphill motionzxaxFreesurfaceazWater tank = 20HorizontalAssumptions 1 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 2 The acceleration remains constant.Analysis W e induce the x- and z-axes as shown in the figure. From geometrical considerations, the horizontal and vertical constituents of acceleration areThe tangent of the angle the free surface makes with the horizontal is = 22.2When the direction of motion is reversed, some(prenominal) ax and az are in negative x- and z-direction, respectively, and thus become negative quantities,Then the tangent of the angle the free surface makes with the horizontal becomes = 30.1Discussion Note that the analysis is valid for each fluid with constant minginess, not just water, since we used no information that pertains to water in the solution.11-45E A vertical cylindrical tank open to the atmosphere is rotated about the centerline. The angular velocity at which the bottom of the tank will first be exposed, and the maximum water height at this moment are to be determined.2 ftzr0Assumptions 1 The increase in the rotational speed is very behindhand so that the transparent in the container always acts as a rigid body. 2 Water is an incompressible fluid.Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the still is precondition aswhere h0 = 1 ft is the original height of the liquid before rotation. undecomposed before run dry spot appear at the center of bottom surface, the height of the liquid at the center equals zero, and thus zs(0) = 0. Solving the equation to a higher place for and substituting,Noting that one complete revolution corresponds to 2 radians, the rotational speed of the container can also be expressed in terms of revolutions per minute (rpm) asTherefore, the rotational speed of this container should be limited to 108 rpm to avoid both dry spots at the bottom surface of the tank.The maximum vertical height of the liquid go bys a the edges of the tank (r = R = 1 ft), and it isDiscussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property.11-46 A cylindrical tank is being transported on a level road at constant acceleration. The allowable water height to avoid spill of water during acceleration is to be determinedD=40 cmax = 4 m/s2htank =60 cmzWater tankAssumptions 1 The road is horizontal during acceleration so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant.Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction, and the origin to be the midpoint of the tank bottom. The tangent of the angle the free surface makes with the horizontal is(and thus = 22.2)The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midplane experiences no rise or drop during acceleration. Then the maximum vertical rise at the back of the tank relative to the midplane isTherefore, the maximum initial water height in the tank to avoid spilling isDiscussion Note that the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution.11-47 A vertical cylindrical container partially filled with a liquid is rotated at constant speed. The drop in the liquid level at the center of the cylinder is to be determined.zrzsR = 20 cmFreesurfaceho = 60 cmgAssumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 The bottom surface of the container remains covered with liquid during rotation (no dry spots).Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is inclined aswhere h0 = 0.6 m is the original height of the liquid before rotation, andThen the vertical height of the li quid at the center of the container where r = 0 becomesTherefore, the drop in the liquid level at the center of the cylinder isDiscussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property. Also, our assumption of no dry spots is validated since z0(0) is positive.11-48 The motion of a fish tank in the confine of an elevator is considered. The pressure at the bottom of the tank when the elevator is stationary, moving up with a specified acceleration, and moving down with a specified acceleration is to be determined.Fish Tank2az = 3 m/s2h = 40 cmgzWater10Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance.Properties We take the density of water to be 1000 kg/m3.Analysis The pressure variance between two points 1 and 2 in an incompressible fluid is given byorsince ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have and and thus(a) Tank stationary We ha ve az = 0, and thus the gage pressure at the tank bottom is(b) Tank moving up We have az = +3 m/s2, and thus the gage pressure at the tank bottom is(c) Tank moving down We have az = -3 m/s2, and thus the gage pressure at the tank bottom isDiscussion Note that the pressure at the tank bottom art object moving up in an elevator is almost twice that while moving down, and thus the tank is under much greater stress during upward acceleration.11-49 vertical cylindrical milk tank is rotated at constant speed, and the pressure at the center of the bottom surface is measured. The pressure at the edge of the bottom surface is to be determined.zrzsR = 1.50 mFreesurfaceg0hoAssumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 draw is an incompressible substance.Properties The density of the milk is given to be 1030 kg/m3.Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given aswhere R = 1.5 m is the radius, andThe fluid rise at the edge relative to the center of the tank isThe pressure struggle corresponding to this fluid height difference isThen the pressure at the edge of the bottom surface becomesDiscussion Note that the pressure is 14% higher at the edge relative to the center of the tank, and there is a fluid level difference of nearly 2 m between the edge and center of the tank, and these large differences should be considered when designing rotating fluid tanks.11-50 Milk is transported in a completely filled horizontal cylindrical tank accelerating at a specified rate. The maximum pressure difference in the oiler is to be determined. -EESax = 3 m/s21zx0g2Assumptions 1 The acceleration remains constant. 2 Milk is an incompressible substance.Properties The density of the milk is given to be 1020 kg/m3.Analysis We take the x- and z- axes as shown. The horizontal accelerati on is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in elongated rigid body motion is given byThe first term is due to acceleration in the horizontal direction and the resulting condensate effect towards the back of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank issince x1 = 0, x2 = 7 m, z1 = 3 m, and z2 = 0.Discussion Note that the variation of pressure on a horizontal line is due to acceleration in the horizontal direction while the variation of pressure in the vertical direction is due to the effects of gravity and acceleration in the vertical direction (whi ch is zero in this case).11-51 Milk is transported in a completely filled horizontal cylindrical tank decelerating at a specified rate. The maximum pressure difference in the tanker is to be determined. -EESzx21gax = 3 m/s2Assumptions 1 The acceleration remains constant. 2 Milk is an incompressible substance.Properties The density of the milk is given to be 1020 kg/m3.Analysis We take the x- and z- axes as shown. The horizontal deceleration is in the x direction, and thus ax is positive. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given byThe first term is due to deceleration in the horizontal direction and the resulting compression effect towards the front of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank issince x1 = 7 m, x2 = 0, z1 = 3 m, and z2 = 0.Discussion Note that the variation of pressure along a horizontal line is due to acceleration in the horizontal direction while the variation of pressure in the vertical direction is due to the effects of gravity and acceleration in the vertical direction (which is zero in this case).11-52 A vertical U- thermionic tube partially filled with alcohol is rotated at a specified rate about one of its arms. The elevation difference between the fluid levels in the two arms is to be determined.zr0h0 = 20 cmR = 25 cmAssumptions 1 Alcohol is an incompressible fluid.Analysis Taking the base of the left arm of the U-tube as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given aswhere h0 = 0.20 m is the original height of the liquid before rotation, and = 4.2 rad/s. The fluid rise at the right a rm relative to the fluid level in the left arm (the center of rotation) isDiscussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property.11-53 A vertical cylindrical tank is completely filled with gasoline, and the tank is rotated about its vertical axis at a specified rate. The pressures difference between the centers of the bottom and top surfaces, and the pressures difference between the center and the edge of the bottom surface are to be determined. -EESh = 3 mD = 1.20 mzr0Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Gasoline is an incompressible substance.Properties The density of the gasoline is given to be 740 kg/m3.Analysis The pressure difference between two points 1 and 2 in an incompressible fluid rotating in rigid body motion is given bywhere R = 0.60 m is the radius, and(a) Taking points 1 and 2 to be the centers of the bottom and top surfaces, respectively, we have and . Then,(b) Taking points 1 and 2 to be the center and edge of the bottom surface, respectively, we have , , and . Then,Discussion Note that the rotation of the tank does not affect the pressure difference along the axis of the tank. But the pressure difference between the edge and the center of the bottom surface (or any other horizontal plane) is due entirely to the rotation of the tank.11-54 Problem 11-53 is reconsidered. The effect of rotational speed on the pressure difference between the center and the edge of the bottom surface of the cylinder as the rotational speed varies from 0 to 500 rpm in increments of 50 rpm is to be investigated.g=9.81 m/s2rho=740 kg/m3R=0.6 mh=3 momega=2*pi*n_dot/60 rad/sDeltaP_axis=rho*g*h/1000 kPaDeltaP_bottom=rho*omega2*R2/2000 kPaRotation rate, rpmAngular speed, rad/sPcenter-edgekPa0501001502002503003504004505000.05.210.515.720.926.231.436.741.947.152.40.03.714.632.958.491.3131.5178.9233.7295.8365 .211-55E A water tank partially filled with water is being towed by a truck on a level road. The maximum acceleration (or deceleration) of the truck to avoid spilling is to be determined.axh = 2 ftWater tankhw = 6 ftzx0L=20 ftAssumptions 1 The road is horizontal so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant.Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in figure. The tangent of the angle the free surface makes with the horizontal is given bywhere az = 0 and, from geometric considerations, tan isSubstituting,The solution can be repeated for deceleration by replacing ax by ax. We obtain ax = -6.44 m/s2.Discussion Note that the analysis is valid for any fluid with constant density since we used n o information that pertains to fluid properties in the solution.11-56E A water tank partially filled with water is being towed by a truck on a level road. The maximum acceleration (or deceleration) of the truck to avoid spilling is to be determined.axh = 0.5 ftWater tankhw = 3 ftzx0L= 8 ftAssumptions 1 The road is horizontal so that deceleration has no vertical component (az = 0). 2 Effects of splashing and driving over bumps are assumed to be secondary, and are not considered. 3 The deceleration remains constant.Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in figure. The tangent of the angle the free surface makes with the horizontal is given bywhere az = 0 and, from geometric considerations, tan isSubstituting,Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solu tion.11-57 Water is transported in a completely filled horizontal cylindrical tanker accelerating at a specified rate. The pressure difference between the front and back ends of the tank along a horizontal line when the truck accelerates and decelerates at specified rates. -EESax = 3 m/s2zx012gAssumptions 1 The acceleration remains constant. 2 Water is an incompressible substance.Properties We take the density of the water to be 1000 kg/m3.Analysis (a) We take the x- and z- axes as shown. The horizontal acceleration is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given bysince z2 z1 = 0 along a horizontal line. Therefore, the pressure difference between the front and back of the tank is due to acceleration in the horizontal direction and the resulting compression effect towards the back of t he tank. Then the pressure difference along a horizontal line becomessince x1 = 0 and x2 = 7 m.(b) The pressure difference during deceleration is determined the way, but ax = 4 m/s2 in this case,Discussion Note that the pressure is higher at the back end of the tank during acceleration, but at the front end during deceleration (during breaking, for example) as expected.Review Problems11-58 The density of a wood log is to be measured by tying lead weights to it until both the log and the weights are completely submerged, and then weighing them distributively in air. The average out density of a given log is to be determined by this approach.Properties The density of lead weights is given to be 11,300 kg/m3. We take the density of water to be 1000 kg/m3.Analysis The weight of a body is equal to the buoyant pierce when the body is floating in a fluid while being completely submerged in it (a consequence of vertical strength balance from static equilibrium). In this case the average density of the body must be equal to the density of the fluid sinceLead, 34 kgLog, 1540 NFBWaterTherefore,whereSubstituting, the volume and density of the log are determined to beDiscussion Note that the log must be completely submerged for this analysis to be valid. Ideally, the lead weights must also be completely submerged, but this is not very critical because of the small volume of the lead weights.11-59 A rectangular gate that leans against the floor with an angle of 45 with the horizontal is to be opened from its lower edge by applying a pattern absorb at its center. The minimum force F required to open the water gate is to be determined.Assumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the attach is negligible.Properties We take the density of water to be 1000 kg/m3 throughout.Analysis The length of the gate and the distance of the upper edge of the gate (point B) from the fre e surface in the plane of the gate areFRF45B0.5 m3 mAThe average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the household area gives the result hydrostatic on the surface,The distance of the pressure center from the free surface of water along the plane of the gate isThe distance of the pressure center from the hinge at point B isTaking the moment about point B and setting it equal to zero givesSolving for F and substituting, the required force is determined to beDiscussion The applied force is mutually proportional to the distance of the point of application from the hinge, and the required force can be rock-bottom by applying the force at a lower point on the gate.11-60 A rectangular gate that leans against the floor with an angle of 45 with the horizontal is to be opened from its lower edge by applying a normal force at its center. The minimum force F required to open the water gate is to be determined.Assumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible.Properties We take the density of water to be 1000 kg/m3 throughout.FRF45B1.2 m3 mAAnalysis The length of the gate and the distance of the upper edge of the gate (point B) from the free surface in the plane of the gate areThe average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the upshot hydrostatic on the surface,The distance of the pressure center from the free surface of water along the plane of the gate isThe distance of the pressure center from the hinge at point B isTaking the moment about point B and setting it equal to zero givesSolving for F and substituting, the required force is determined to beDiscussion The applied force is inversely proportional to the distance of the point of application from the hinge, and the required force can be red uced by applying the force at a lower point on the gate.11-61 A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point B. The force exerted to the plate by the ridge is to be determined.Assumptions The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience.FR3 mA2 mypProperties We take the density of water to be 1000 kg/m3 throughout.Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the gate,The vertical distance of the pressure center from the free surface of water is11-62 A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point B. The force exerted to the plate by the ridge is to be determined.Assumptions The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience.FR3 myPh = 2 mAProperties We take the density of water to be 1000 kg/m3 throughout.Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the wetted plate area gives the resultant hydrostatic force on the gate,The vertical distance of the pressure center from the free surface of water is11-63E A semicircular tunnel is to be built under a lake. The total hydrostatic force acting on the roof of the tunnel is to be determined.Assumptions The atmospheric pressure acts on both sides of the tunnel, and thus it can be ignored in calculations for convenience.Properties We take the density of water to be 62.4 lbm/ft3 throughout.Analysis We consider the free body diagram of the liquid block enclosed by the circular surface of the tunnel and its vertical (on both sides) and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as followsHorizontal force on vertical surface (each side)FyWFxFxVertical force on horizontal surface (downward)R = 15 ftWeight of fluid block on each side within the control volume (downward)Therefore, the net downward vertical force isThis is also the net force acting on the tunnel since the horizontal forces acting on the right and left side of the tunnel cancel each other since they are equal ad opposite.11-64 A hemispherical domed stadium on a level surface filled with water is to be lifted by attaching a long tube to the top and filling it with water. The required height of water in the tube to lift the dome is to be determined.Assumptions 1 The atmospheric pressure acts on both sides of the dome, and thus it can be ignored in calculations for convenience. 2 The weight of the tube and the water in it is negligible.Properties We take the density of water to be 1000 kg/m3 throughout.Analysis We take the dome and the water in it as the system. Wh en the dome is about to rise, the reaction force between the dome and the ground becomes zero. Then the free body diagram of this system involves the weights of the dome and the water, balanced by the hydrostatic pressure force from below. Setting these forces equal to each other givesFVR = 3 mhWSolving for h givesSubstituting,Therefore, this dome can be lifted by attaching a tube which is 2.02 m long.Discussion This problem can also be solved without finding FR by finding the lines of action of the horizontal hydrostatic force and the weight.11-65 The water in a reservoir is restrained by a triangular argue. The total force (hydrostatic + atmospheric) acting on the inner surface of the wall and the horizontal component of this force are to be determined.FRh = 25 mypAssumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible.Properties We take the density of water to be 1000 kg/m3 throughout.Analysis The length of the wall surface underwater isThe average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the surface,Noting thatthe distance of the pressure center from the free surface of water along the wall surface isThe magnitude of the horizontal component of the hydrostatic force is simply FRsin ,Discussion The atmospheric pressure is usually ignored in the analysis for convenience since it acts on both sides of the walls.11-66 A U-tube that contains water in right arm and another liquid in the left is rotated about an axis closer to the left arm. For a known rotation rate at which the liquid levels in both arms are the same, the density of the fluid in the left arm is to be determined.1*1FluidWater