Fluid mechanics white 7th edition solution manual free download

Right from manual guideline for all kinds of products and appliance from different brand name or even an expensive university or college book, school academic journal from many different topic for your research study. This listing are filled with by far the most related and relevant subject identical to your current term and even organize into a compact list for your comfort by our program. Solution: a The vertical elevation of the water surface in the slanted tube is 1.

Stay with BG units. Some teachers 6 ft say it is more instructive to calculate these by direct integration of the pressure forces. Find the water force on the panel and its line of action. From Eq. The center of pressure is thus 3. The three bottom shapes and the fluids are the same. This is called the hydrostatic paradox. Explain why it is true and sketch a freebody of each of the liquid columns.

In b side pressures are horizontal. In c upward side pressure helps reduce a heavy W. For what water depth h is this condition reached? Neglecting atmospheric pressure, find the resultant hydrostatic force on panel BC, a from a single formula; b by computing horizontal and vertical forces separately, in the spirit of curved surfaces.

What water level h will dislodge the gate? Neglect Fig. The CG would be For what water depth h will the force at point B be zero? The centroid of Ans.

This weight acts downward at the CG of the full gate as shown not the CG of the submerged portion. Thus, W is 7. For Ans. If the water level is high enough, the gate will open. Compute the depth h for which this happens. The forces on AB and BC are shown in the freebody at right.

This solution is independent of both the water density and the gate width b into the paper. Determine the required force P for equilibrium. Thus it is 3. Find the hydrostatic force on surface AB and its moment about C. Could this force tip the dam over? Would fluid seepage under the dam change your argument? As shown in the figure, the line of action of F is 2. This moment is counterclockwise, hence it cannot tip over the dam. If there were seepage under the dam, the main support force at the bottom of the dam would shift to the left of point C and might indeed cause the dam to tip over.

The Fig. What horizontal force P is required at point B for equilibrium? The gate weight of N is assumed at the centroid of the plate, with moment arm 0. If not, Fig. Neglect the atmosphere. Because the actual plate force is not vertical.

The hinge 15 cm hinge h is 15 cm from the centerline, as shown. Then the moment about the hinge is 0. Just iterate once or twice. What sphere diameter is just right to close the gate? What is the water depth h which will first cause the gate to open? The hinge at A is 2 ft above the freshwater level. Find h when the gate opens. Solution: Find the force on each panel and set them equal: Fig.

Compute a the hydrostatic force of the water on the panel; b its center of pressure; and c the moment of this force about point B. Neglect atmospheric pressure. CD is longer than AB, but its centroid is not as deep. If you have a great insight, let me know. Neglecting atmospheric pressure, determine the lowest level h for which the gate will open.

Is your result independent of the liquid density? The width into the paper is b. Solution: The critical angle is when the hydrostatic force F causes a clockwise moment equal to the counterclockwise moment of the dam weight W. The negative sign occurs because the sign convention for dF was a downward force. Determine the horizontal and vertical components of hydrostatic force against the dam and the point CP where the resultant strikes the dam. Find the force F just sufficient to keep the gate from opening.

The gate is uniform and weighs lbf. It is cm wide into the paper. Find a the vertical and b horizontal water forces on the panel. This is a rectangle, 75 cm by cm, and its centroid is This is in two parts 1 the weight of the rectangular portion above the line AC; and 2 the little curvy piece above the parabola and below line AC. Recall from Ex.

Find the horizontal force P required to hold the gate stationary. The width b into the paper is 3 m. PDF You also possible to search a book or topic on the digital platform.

Munson, Young, Okiishi Solutions Manual Fundamentals of Fluid Mechanics, 5th Edition, Munson, References : Advanced Engineering Mathematics, E. Kreyszig, 7th Ed. Fundamentals of Fluid Mechanics, Munson, Young Fundamentals of Fluid Mechanics R. Introduction to Fluid Mechanics 73 [Solutions Manual] Munson u Young u Okiishi.

Solution:Thebuoyancyisduetothedifferencebetweenhotandcoldairdensity: p kg kg cold 1. Checktomakesuretheballoonmaterialisnotexcessivelyheavy: W balloon 0. Determine a the stringtension; and b the specific gravity ofthewood. Is italso possibletodeterminetheinclinationangle?

SummingmomentsaboutCgives MC 0 W 2. Thus W 0. Letthe Fig. Howmanypoundsofsteel SG 7. Wsteel or: 1. Whatisthespecificgravityoftherod material? What is peculiar about the rest angle30? If the rock exerts no moments on the spar, computetheangleofinclination. We see that the buoyancy exceeds the weight by 19 kN, or more than 2 tons.

The sphere would float nicely. W 9. Slightly less than doubling the wall thickness would create neutral buoyancy for the sphere. Estimate a thetensioninthe mooring line, and b the height in the standardatmospheretowhichtheballoon willriseifthemooringlineiscut.

Theyestimatetheweightofthe boxandtreasure inair atlbf. Theirplanistoattachtheboxtoasturdyballoon, inflatedwithairto3atmpressure. Theboxis2ftwide, 5ftlong,and18inhigh. Theboxvolumeis 2ft 5ft 1. Thusthe balloonmustdevelopanetupwardforceof1. Theairweightin theballoonisnegligible,butwecancomputeitanyway.

Fnet 64 0. Determine a ; b wood. A Summoments abouttheleftcorner,point uniformheavyspheretiedtotheleftcorner C: causes the beam to float exactly on its diagonal.

Thentheverticalforcebalanceon 7. Whatistheratio ofthedistancesaandbforthiscondition? Intermsofseawater,thistotalvolumewouldbe equivalentto Iffilledwithhydrogenat18psia and 60F and released, at what U. Solution:Assumethatitremainsat18psiaand60 F. Thedensityofthehydrogenintheballoonisthus p 18 H 2 0. Ifso, b howmanycubiccentimetersareexposed, and c how high will a spherical cap protrude above the surface?

From Dr. Math, or a good math book, the sphere formulas relatetothedistances r,h,R showninthefigureatright. Oryoucould useEESandgettheanswerdirectly. Then itsvolumeissubmergedinfluidX.

Estimate thegagepressureoftheairinthetank. Solution:In order to apply the hydro staticrelationfortheairpressurecalcula tion,thedensityofFluidXmustbefound. Neglectthe Fig. If seawaterisdenotedasS1,theiceberghasS0. The metacenterpositionisdeterminedbyEq. SupposethattheheightisLandthedepth intothepaperisLbutthewidthdecreases to H L.

Again with S 0. Wouldthiscylinderbestableifplacedto floatwithitsaxisverticalinoil SG0. Solution:Averticalforcebalancegives 0. Itispiledsohighwith gravelthatitscenterofgravityis3ftabove thewaterline,asshown. Solution:Example 2. ThecenterofbuoyancyBisatthecentroidofafrustrumofa submerged cone: 0. Thebase radius is R and the cone height is H, as shown. Which position is more stable?

Assume largebodywidthintothepaper. Solution:The calculations are similar to the floating cone of Prob. Let the trianglebeLbyLbyL.

Equation 2. Solution:For the given SG 0. Then,forr4 ft,. Then,forsteadyrotation,thewatersurfaceintheglasswillslopeattheangle ax 6. Solution: a Theslopeoftheliquidgivesustheacceleration: a x 28 15cm tan 0. Letthetankbepulled alongahorizontalroadinrigidbodymotion.

Findthe accelerationanddirectionfor which a aconstantpressuresurfaceextendsfromthetopofthefrontendtothebottom ofthebackend;and b thetopofthebackendisatapressure0.

Solution: a Wearegiventhattheisobarorconstantpressurelinereachesfrompoint Cto pointBin thefigure above, is negative,hence the tankis decelerating. The ellipticalshapeisimmaterial,onlythe2mheight. Theisobarslopegivestheacceleration: 2m a tanC B 0. Thus,again,theisobarmustslopeupwardthroughBbutnotnecessarily pass through point C. The pressure difference along line AB gives the correct deceleration: kg p A B 0. Thisismorethanpart a ,sotheisobaranglemustbesteeper: 5. Assumingrigid bodymotion,compute a theaccelerationa, b whethertheaccelerationisupordown, and c thepressureatpointAifthefluid ismercuryat20C.

Solution:Thefreesurfaceistiltedattheangle Thisangle mustsatisfyEq. The topsurfaceishorizontal. Determinetherigid bodyaccelerationsforwhichthewaterat oppositetopcornerBwillcavitate,for a horizontal,and b verticalmotion. Solution:From Table A5 the vapor pressure of the water is Pa. Assuming rigidbody motion, estimate the maximum water depth to avoidspilling. Whichisthebestwayto alignthetank? Solution:Thebestwayistoalignthe16inchwidthwiththecarsdirectionofmotion, tominimizetheverticalsurfacechangez.

Insideisa 10cm balloon filled with helium at kPa. Willitleantotheleftorto theright? Solution:The acceleration sets up pressureisobarswhichslantdownandto the right, in both the water and in the helium. This means there will be a buoyancyforceontheballoonupandto the right, as shown at right.

It must be balancedbyastringtensiondownandto the left. If we neglect balloon material weight, the balloon leans up and to the rightatangle a x 5. What istheangleofthefreesurface? Canyou explainthisinterestingresult? Inaframeofreference movingwiththechild,whichwaywilltheballoontilt,forwardorbackward?

Theheliumfilled balloonisnotincontactwithanypartofthecar seats,ceiling,etc. When thetrafficlightturnsgreen,thecaracceleratesforward. Inaframeofreferencemoving withthecarandchild,whichwaywilltheballoontilt,forwardorbackward? Conductascientificexperimenttoseeif yourpredictionsinparts a and b arecorrect. Solution: a Onlythechildandballoonaccelerate,notthesurroundingair.

Thisisnot rigidbodyfluidmotion. Our predictionswerecorrect. What isthewatersurfaceangleatpt. Use an average radius R 6. If L Fig. Whatuniformaccel Fig. Solution:If pressures at A and C are the same, the free surface must join thesepoints: 45, a x g tan g Findthecentral rigidbodyrotationrateforwhich a one thirdofthewaterwillspillout;and b the bottomcenterofthecanwillbeexposed.

Solution: The molten plastic viscosity is a red herring, ignore. The appropriate final rotating surface shape is a paraboloid of radius 7 inches and depth 8 inches. Thus, from Fig. Whenspunsteadily toachieverigidbodyrotation,a4cmdiameterdryspotappearsatthebottomofthevase. Solution:Itisinterestingthattheanswer R hasnothingtodowiththewaterdensity. Thevalueofcubiccentimeterswas chosentomaketherestdepthanicenumber: rest position H cm 3 5cm 2 H , solve H Onewaywouldbetointegrateandfindthevolume Fig.

Theformulasinthetext,concerningtheparaboloidsofair,would,inthewritersopinion, bedifficulttoapplybecauseofthefreesurfaceextendingbelowthebottomofthevase. The fluidismercuryat20C. Solution:Let ho be the height of the freesurfaceatthecenterline. Then,from Eq. If the fluid is wateratFandatmosphericpressureis psfa, at what rotation rate will the fluidbegintovaporize?

Atwhatpointin thetubewillthishappen? WhenspinningaroundDC,thefreesurface comesdownfrom pointAtoaposition below pointD,asshown. Thereforethefluid pressureislowestatpointD Ans.

Thusthedrawingiswildlydistortedandthedashedlinefallsfar belowpointC! Solvefor 2 2 Solution: a IfpressuresareequalatBandC,theymustlieonaconstantpressure paraboloidsurfaceassketchedinthefigure.

Thisfamilyofparaboloidshasthe formula. Thefocusof the mirror is tobe4mfrom themirror, measuredalongthecenterline. Solution:Wehavetoreviewourmathbook,orahandbook,torecallthatthe focusFof aparabolaisthepointforwhichallpointsontheparabolaareequidistantfromboththe focusandasocalleddirectrixline whichisonefocallengthbelowthemirror.

Forthefocallengthhandthezraxesshowninthefigure,theequationoftheparabolais 2 givenbyr 4hz,withh4mforourexample. Alltubesarelongandhaveequal small diameters. Solution: a The freesurface during rotation is visualized as the dashed line in Fig. Thetotaldisplacement betweenouterand 2 2 center menisci is, from Eq. The center meniscusfallstwothirdsofthisamountandfeedstheoutertubes,whicheachrise onethirdofthisamountabovetherestposition:.

Whatisthehydrostaticforceonthe gate? Howhigh abovethewatersurfacedoesthewoodenendoftherodprotrude? Wecanthenneglectmovement ofthereservoir level. Ifthereservoirisnotlarge,itslevelwillmove,asin thefigure. Tubeheighthismeasuredfromthezeropressure level,asshown. WriteanexactExpressionforp1gageasafunction of h, d, D, and gravity g.

Ifweneglectairdensity, thepressuredifferenceisp1pamg hH. Nevertheless,the Utubeisstilltobeusedtomeasurethepressureinthe airtank. Calculate h in cm, ignoringsurfacetensionandairdensityeffects.

Dynamics,ridingthemerrygoroundwithhisson,hasbroughtalong hisUtubemanometer. As shown in Fig. Solution: a Approximate:Theaverageaccelerationof 2 2 the manometer is Ravg 5. Applythistotheleftleg z 1 andrightleg z2. Asabove,the rotationrateis6. Theglassiscylindrical, twiceastallasitiswide,andfilledtothebrim.