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# FluidMechanics CIEG305

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This 13 page Class Notes was uploaded by Jackeline Parker on Saturday September 19, 2015. The Class Notes belongs to CIEG305 at University of Delaware taught by JackPuleo in Fall. Since its upload, it has received 37 views. For similar materials see /class/207163/cieg305-university-of-delaware in Civil and Environmental Engineering at University of Delaware.

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Date Created: 09/19/15

CIEG 305 RESPONSE TO MUDDIEST POINT October 1 2009 I broke down all the muddiest points into broad categories Several specific questions for some are addressed at the end of each section WARNING LONG 1 Moments If you did not have them in physics or statics a moment is something that is related to torque or the tendency to rotate A moment is de ned as MrgtltF Where M is the moment r is the radius or distance to the force F from some origin and the X is not atimes sign but it stands for the cross product Simplifying the Moment is equal to M T F 31719 where the angle theta is the angle between the moment arm the r and the applied force Typically in our problems this angle is 90 degrees For instance in most of the planar gate problems all forces were normal to the surface and the r value was the distance from some hinge on the gate in the plane of the gate So the angle parts drops out really sin90l Note that for an angled gate if the weight of the gate is important the angle in the moment equation does not drop out because gravity acts down not along gate Most of us use the right hand rule to determine directions etc Use your right hand and point your fingers in the direction of the moment arm Then curl your finger in the direction of the force The angle your fingers swept through is the angle you use in the moment equation supplied above Conceptually you know about moments because when you want to tighten a bolt down really hard or conversely need to loosen a bolt that has become corroded in place you do not search for your Leatherman tool that is 3 inches long You reach for the longest wrench you can find so even though you are applying the same force you are applying a much larger torque due simply to the increased length of the moment arm 2 FBD Free body diagrams should have been covered in Physics and for sure in Statics They are a way to isolate the forces of interest on the object of interest in the problem They are used to develop equilibrium equations usually 2 EC 0 or Z Fy 0 and to look at moment summation 3 Gage Pressure Absolute pressure is pressure referenced to a vacuum Gage pressure is pressure referenced to local atmospheric pressure Thus if local atmospheric pressure is Pam and the absolute pressure at a point A is called PA PAgage PAPaml just subtract off local atmospheric pressure When to incorporate Pam If the problem is open to the atmosphere and the problem asks for absolute pressure you should incorporate a numerical value if possible for Pam Else you can give an answer as whatever you equated Pam Warning if there is a way to determine Palm You should That is if the temperature is given and a table is supplied you should nd a real value for Pam If doing manometer problems and both sides are open to the atmosphere you do not need to include Palm because is acts on both sides of the manometer If doing a manometer problem with one side closed and they ask for Gage pressure or give gage pressure you need to consider that that means they either want an answer relative to atmospheric or you need to subtract atmospheric pressure from your answer respectively All the submerged gate stuff we have done was gage pressure without explicitly stating s0 4 Forces 0n Planar Gates and Surfaces We went through several examples of nding forces on surfaces and surfaces containing gates All of them started out with three steps 1 Find direction acting always normal to surface 2 Determine the force magnitude resultant It is due to pressure and we went through a derivation to show it is F R pghCA where he is the depth to the centroid This is the vertical distance from the top of the water to the centroid of the object NOTE all the derivations we did in class had the coordinate origin at the free surface and NOT at the centroid like the book does 3 Determine where the resultant force acts how far down the planar surface from the free surface I gave you the equation and pointed you to the handout online if you were interested further The equation for the distance from the origin at the free surface is I y R yc 72 where IXX is the moment of inertia around XaXlS through the centr01d of ya A Note that yC is the distance down the yaXis from the origin at the free surface to the centroid yR is the distance down the yaXis from the origin at the free surface to where the resultant force FR acts The distance yR is also sometimes referred to as ycp where CP stands for center of pressure 3 L For most of our problems it was either a rectangle shaped surface and I M 2 where the one that gets cubed is the one that lies along the y axis or a circular surface where 4 727 I If you are asked to calculate forces and they do not act at the same location then a good bet is that you will have to determine that force through a moment calculation EXAMPLE Calculate the minimum force F necessary to hold a uniform 1th square gate weighing 5001b closed on a tank of water Assume air pressure to left of gate is zero 0 Q hinge o N 4 P The picture on the left shows the problem set up and the one on the right shows the coordinate system all the way up to the free surface l The force direction supplied by the water is normal to the gate see free body diagram in the moment calculation step below 2 Find the force magnitude on the gate Use the equation F R pghCA But we need to know the depth to the centroid Ifthe length dawn the y axis is 6 ft to the centroid yo then the depth to the centroid is ha 6 sin45 he 424 slugs So FR 194322S 2424 144 2 38140121 3 Determine the resultant force location It must lie below the centroid I 7 yR ya yCA 3 yR 6ft W 8 The force acts 8 ft down the yaXis Draw a Free body diagram to determine the force F Calculate the sum of the moments about the hinge at point 2M0 0 8 FR 12 P 6 sin45V P 2 amp 6ftsin45W 12 P 8 381401b 6 sin455001b 12 P 255201 I note again that all the stuff presented in class used an origin at the free surface The book puts the origin of the coordinate system at the centroid with y positive towards the surface If you use the books approach use the books equation for the location of the resultant force In their coordinate system it is negative so that it still lies below the centroid How to nd centroid The centroid is always at the center of a rectangular or circular surface For a triangle it is 13 up from the base These are in the textbook on page 82 The centroid is where an object would balance if you rested it on your finger This is NOT the center of pressure where the resultant force acts The resultant force acts below the centroid unless horizontal surface How to calculate IXX Use the table on page 82 For use it is almost always a rectangle or circle For a rectangle the part that gets cubed is the part that lies along the yaXis bL3 XX E rectangle 1W XX T ClT Cle What is center of pressure The center of pressure is the location where the pressure force acts I called this yR in class to indicate it is where the resultant force FR acts Gate problems with moments If a problem asks to nd a force and there are multiple forces that are not collinear on same line but your bottom dollar a moment calculation is needed The easiest way to do it right is to draw a free body diagram and label it with the forces correct direction and the distance from the point of interest to where the force acts Then just be sure to maintain positive and negative direction based on where your thumb points when using right hand rule Don t forget to include sine of angle your ngers sweep through The reference point for these calculations is almost always wherever the hinge is located 5 Forces 0n Curved Surfaces There are horizontal and vertical forces on curved surfaces that we calculated in class by proceeding by example We came up with the horizontal force on a curved submerged surface equals F H pghcypm Pm where subscript projected means you have to project the curved surface to a vertical wall The vertical force was determined to be the weight of the uid contained above the surface For a quarter circle it is the weight of uid contained in the quarter circle The Resultant force comes from adding the horizontal and vertical forces vectorally as F R 4F FV2 This only tells me the magnitude of the resultant not direction The direction comes from the following figure as Ta an mnmemcalculman wnh hs we have m kmw Wm m mdwldul fumes mm w cange hnse fmmkmwmg thath hamamnl farce acts thmugh m cemmldafthz da aura sheenhsxsZ athmydnwnfmmthzfnesnrfacefax 13m shnwnbelnw mwmmm ran quantum pmblem m mm WAX3m s ndmsafqumxcucle m a 1m af uevemcaldashzd hm afthz qmexcxrclz 5 Shawn mummy 221mm This xsbecmlse m mum farce must 5 m cemmxdaf u Wm af mdsmce ms a m weight EXAMPLE A 2mm gm 5 a mumquot urn and 11mgde Determine m hnmamal farce p nqmdm mam gm mphce Neglect a weight afthz gum and fnconnm m 11mg p Fustwe medm rm m 1man farce suppllzdbythz water FL WowAWN 7 ring F 7 1 94 In 2 23 20 6 F 224221 Nam thecemmxdafpmlectzdmust hz 6n ndnls vvuZ memuchzmsheeL m hummus farce mg 2a afthz mydnwnthz pmjectzdarea Ifymlda mt 1x m Emma m m aid nyanmllA m ge hz 5m Answer So FH acts at 236ft 4ft down Now the vertical force It is due to the weight of the water above the gate FV pg v w i 6 2 FV 194 ff Inasz 4 20 FV 353071 The vertical force acts through the centroid of the uid volume Where is that It is hard to calculate unless one oes back to the ugly formula for a centroid Else use the table in the book on page 81 5 edition or page 80 6th edition or page 82 7th edition Fv acts 4W3n 255 ft to the right of the vertical dashed line Resultant force magnitude FR 4224882 353072 418601b Angle 2 WA 22488113 2 3250 353071 Draw the free body diagram and then sum moments Center of pressure an drawn in scale ZMH 0 Z FH r 255 FV 6 P kw 6 p e W W P 225001b Note you can also use the resultant force lts angle and some more geometry to do the moment calculation as follows Center ufpressure an drawn tn szale ZMH 0 382 FR sin575 6ftP P 382 FR sin575 W P 382 418601bsin575 W P 224781 And the slight difference is due to rounding errors in the moment arm and angles How to determine moments about a curved surface Look at problem above Ithink easiest way is to calculate the horizontal and vertical forces and where they act Then use their lines of action where they act to determine the moment arm for each force How to nd buoyant force on a curved surface For the curved surface type problems we have been doing there is no buoyant force Only the hydrostatic force are of interest The volume is assumed to be small 6 Buoyancy A uid exerts a resultant force due to pressure differences on bodies that are either oating or submerged The resultant force is equivalent to the weight of the displaced uid Thus you must calculate the weight of the uid that is now occupied by the submerged portion of the object of interest Note that if the object is something like a balloon then the uid being displaced is the air the balloon is oating in An example for aforce balance for a helium balloon ride would be F B W pazrgvballumz Whalluurmaterzal phzlmmgvballumt Wpeuple If the left side is greater than the right side the balloon rises else it stays on the ground Example Determine the force required to just push a beach ball under the free surface in your backyard pool Assume the beach ball is 15 ft in diameter The force required to push the ball down must be balanced by the buoyant force trying to push it up There is also the weight of the air in the balloon leading to a free body diagram that looks like FE L X W The force balance looks like the following when the ball is just under the surface 2 F2 0 F3 W F pwatergvdisplazed pairgvball F Note that once ball is fully under water volume displaced and volume of the ball are equal Solve for F F pwate rgvd39uplaced pairgvball Factor out ball volume and g to get and you see the air weight is basically negligible F 194 234x1035 322 rt075ft3 Or F1101b The equations of equilibrium for buoyant force will almost always look like WFB The problem done above is sirni1arto the LawnchairLarry problem For lawnchair larry sthuandall entry contained in the weather balloons Compare those to the buoyant force associated with me quot th uall ii he n39ses else he stays iota on the ground 7 The Concrete Block Example wouldthewaterlevel39 L r 39 reasoning ms given as as in aboat andthen39 all The an e as am Determine displaced wlume Black in nal Black in Paul W FE V V Pow Wow 7 PW Wanton WWW comm commquot pumaquot i Vduplacad 2 V Pwot when block in pool it displaces the volume ofconcrete block When block in boat it i i i c L i i i I an disp1aeesan 1 densities Since the density ofconcrete is much greater than water The block in the boat diniae a So i 39 H quot 39 thetherlevelalongthe pool edge falls 39 ith sides of5 m 39 o Wer 39 L ith amass 0f500kg into the boat Determine the volume displaced FE W PwaiiygVanp 500mg pwatervdlsp 500kg Vt SOOkg1000kgm3 V 05m3 dzsp or the uid level increases that amount divided by the surface area of the pool 3 Elevationidisplaced 0395 002m 25m Newiwater ilevel 4m 002m 402m Instead place the block in the pool It displaces its own volume W 500kg VCUVLCVEZE 7 7 pmmezeg 2300 E f 39 3 3 or concrete dens1ty of 2300 kgm m V 021m3 CUVLCVEZE or the uid level increases that amount divided by the surface area of the pool 3 03921 0008m 25m Newiwater ilevel 4m 0008m 4008m Elevation idisplaced So when the block is in the boat the water level is at 402m When it is in the pool the water level is at 4008 m Thus when the block is in the pool the water level is lower

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