Answers assignment 2 fluid statics-fluid mechanics


Published on

Published in: Business, Technology
  • Be the first to comment

  • Be the first to like this

No Downloads
Total Views
On Slideshare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Answers assignment 2 fluid statics-fluid mechanics

  1. 1. PHYS207 Assignment 2 Fluid StaticsQ1. A 6 cm diameter rubber ball weighing 5 g is used to plug a 4 cm hole in the base of a tank. Thetank is in use and is gradually emptying. At approximately what depth will the ball plug fail byfloating up and out of the hole? (hint: keep it simple)Let’s simplify this somewhat. If we replace the ball with a flat plug over the hole we’d have a netvertical pressure force FH acting on the hole area.The combination of curved upper and lower surfaces on the ball produce hydrostatic forces equivalentto the simple flat plug (if you can’t see that from the theory, just treat it as a simplifying assumption):P = gH = 9800H where H is the unknown depthAnd FH = PA Where A is the area of the hole: A = D2/4 = (0.04)2/4 = 1.257 x10-3m2So FH = PA = 9800(1.257 x10-3)HThe ball itself adds a small amount to the downwards force: FM = mg = 0.005(9.8) = 0.049 NSo FH + FM = 9800(1.257 x10-3)H + 0.049The only upwards force on the ball is its buoyancy, which is the displaced weight of water. This isrelated to the submerged volume of the ball. To keep it simple, let’s use the total volume of the ball:V = 4 R3/3 = 4 D3/24 = 4 (0.06)3/24 = 113.1 x 10-6m3The buoyant weight force is then FB = wgV = 9800(113.1 x 10-6) = 1.108 NAt the point where the ball just begins to float, all forces are in balance:FH + FM = FB9800(1.257 x10-3)H + 0.049 = 1.108So H = (1.108 – 0.049)/(9800(1.257 x10-3)) = 0.086 m = 8.6 cm.This is only just above the top of the ball, so the simple plug is reasonably effective.A more accurate model would only allow for the displacement of water not directly above the hole,which is close to the difference between the spherical volume and the volume of a cylinder 6 cm talland 4 cm in diameter: V = 113.1 x 10-6- 75 x 10-6= 38.1 m3so the buoyant force isFb = 1.108 x 38.1/113.1 = 0.37 NAnd 9800(1.257 x10-3)H + 0.049 = 0.37Or H = (0.37 – 0.049)/(9800(1.257 x10-3)) = 2.6 cmEven better, and the plug may not fail at all as the water level drops below the top of the ball.
  2. 2. Q2. Rooms in the lower level of a cruise ship have 30 cm diameter circular windows, or portholes. Inrough weather the lower portholes can be momentarily submerged up to 4 m by waves. Determine theinstantaneous hydrostatic force in terms of magnitude and position at one of these moments.This can be handled by assuming static and horizontal forces:The mean pressure on the porthole is the pressure at it’s centroid: Assuming the depth is 4 m to thecentre of the porthole, we have the working pressure:P = gH = 9800(4) = 39.2 kPaThe area of the porthole is A = D2/4 = (.3)2/4 = 0.0707 m2The force is thus F = PA = 39.2(0.0707) = 2.77 kNThis acts at the line of the pressure centre, which is located at a depth hp:hp = hc + I/ hcA where hc is the depth to the centroid of the porthole, hc = 4mI = D4/64 = (0.3)4/64 = 397.6 x 10-6m4so hp = 4 + 397.6 x 10-6/(4(0.0707)) = 4.001 mA hydrostatic force of 2.8 kN is located 1 mm below the centroid of the porthole.Q3. A 2 x 2 m square gate opens clockwise on a horizontal hinge located at point A, which is 100 mmbelow the centroid of the gate. The gate sits against solid rubber blocks which create an effectivepressure seal. The gate is designed to open from the top when the water exceeds a certain depth. Towhat depth h can the water rise without the gate opening at C?The gate will open as soon as the pressure centre, reflecting the position of the net resultanthydrostatic force, is higher than the hinge at A, since at that point the moment action of thehydrostatic force will have no resistance to opening.The pressure centre is located at:hp = hc + I/ hcABACh
  3. 3. hc is the depth to the centroid of the gate, which we can express as a function of the unknown depth hand the gate dimension CB = 2 m:hc = h – CB/2 = h – 1We’re told that the hinge is 100 mm below the centroid:hhinge = hc + 0.1Thus the gate opens when the pressure centre depth equals the hinge depth:hp = hc + I/ hcA = hc + 0.1or I/ hcA = 0.1that is I/((h – 1)A) = 0.1I = bd3/12 = 2(2)3/12 = 1.33 m4A = bd = 4 m2So I/ ((h – 1)A) = 1.33/(4(h – 1)) = 0.1Thus h – 1 = 1.33/(4(0.1))or h = 1.33/0.4 + 1 = 4.33 mMaximum depth h = 4.33 mQ4. Navy divers are called to rescue a group of office workers trapped in a basement underfloodwater in Leeds, UK. The only access to the basement is via a 1.5 x 1.5 m horizontal hatch whichis currently under 1.8 m of water. The hatch is hinged on one edge, with a solidly built brass ringhandle at the other end. The divers try to pull it open, and soon realise they will need a crane. Whatminimum lifting load should they ask for? What mistake are they making?The pressure on the hatch is:P = gH = 9800(1.8) = 17.6 kPaThe pressure force isF = P.A = 17.6 x 103(1.5)2= 39.69 kNThis acts from the centre of the hatch to the hinge, an offset of d = 1.5/2 = 0.75 m. This creates aclosing moment:Mc = F.d = 39.69x103(0.75) = 29.77 kN.mThe lifting force on the handle (positioned at the other edge, 1.5 m from the hinge) acts over the fullwidth of the hatch, creating an opening moment. To get the hatch open, the opening moment mustequal the closing moment:Mo = Fo(1.5) = 29.77x103
  4. 4. Fo = 29.77x103/1.5 = 19.85 kNConvert that to mass for purposes of ordering a crane:Crane Lift = 19.85x103/9.8 = 2.025 Tonnes.A 2.5 tonne crane should do the job.But will the door handle be able to hold this? Be careful not to drown the workers when you lift thehatch!Q5. A long cylinder of radius 2 m hinged at point A is used as an automatic gate. When the waterlevel reaches 15 m the gate automatically opens. Determine the magnitude of the net hydrostaticmoment on the cylinder at the moment the gate opens. What is the mass per unit length of thecylinder?Work this out in horizontal and vertical components:Horizontal: The mean pressure on the rectangular projection of the gate is the pressure at thecentroid of the rectangle, at depth hc:hc = 15 – R/2 = 15 – 1 = 14 mMean horizontal pressure: PH = ghc = 9800(14) = 137.2 kPaThe gate is described as “long” and we are not told how long. A useful technique is to express theproblem in terms of force per unit length (into the page), so we don’t have to factor in the length – itcancels. The area on which the pressure applies is just the height of the gate multiplied by unit length,that isA per unit length = 2 x 1 = 2.0 m2/mThe horizontal pressure force per unit length of gate is thus:FH = PHA = 137.2(2.0) = 274.4 kN/mRectangular projection of gate faceR = 2m
  5. 5. This force acts at the pressure centre:hp = hc + I/ hcAWhere I = bh3/12 = h3/12 and A = bh = h for a unit rectangleI = 23/12 = 0.67 m4A = 2 m2thus hp = hc + I/ hcA = 14 + 0.67/((14)2) = 14.024 mThis is a d=1.024 m offset from the hinge, creating a horizontal moment ofMH = FH d = 274.4 (1.024) = 281.0 kN.m / m widthVertical: The vertical force on the gate can be visualised as the weight of water missing (displaced)above the curving face of the gate. This is a rectangle with a hemispherical end:Volume per unit width = area A = 13(2) + 22/4 = 29.14 m3/mFV = Weight per unit width = A g = 29.14(9800) = 285.6 kN/mThis force acts upwards through the combined centroid of the rectangle-hemisphere. Measuring fromthe centre of the circle,AAxxi 1Xp = ((1)26 + (4R/3 )3.14)/29.14 = (26 + 8/3)/29.14 = 0.984 mThis is an offset d = 2 – 0.984 = 1.016 m from the hingeThis creates a moment per unit width of MV = FV d = 285.6 (1.016) = 290.2 kNm /mThe total hydrostatic moment about the hinge is thus MT = MV + MH = 290.2 + 281.0 = 571.2 kNm /mThis is balanced by the weight per unit width of the gate acting on the offset of the hinge from thecentre of the circle:MT = MgR where M is the mass per unit length of the gate:M = MT/gR = 571.2 x103/(9.8(2)) = 29.14 x 103kg / m widthQ6. A 1.2 m diameter steel pipe carries oil of relative density 0.822 under a head of 70 m of oil. Whatminimum thickness of 120 MPa (yield stress) steel would be required for a safety factor of 1.5?The safety factor reduces the allowable stress to a level significantly less than the yield stress. Theallowable stress in the thin pipe walls is S = 120/1.5 = 80 MPa.Taking a unit length of pipe and considering forces on the cross-section:
  6. 6. The pressure is worked out from the head and fluid density:P = gH = 822(9.8)70 = 563.9 kPaThe pressure force is the pressure times the area of cross-section, Fp = P.D for a unit length:Fp = P.D = 563.9(1.2) = 676.7 kN/mThe tensile force in the pipe wall is equal to the tensile stress times the area of the wall: T = S.t. for aunit length. There are two tensile forces to account for as the cross-section has two walls. The twotensile forces thus equal the pressure force:Fp = 2S.tS is the allowable stress, 80 MPa = 80 x106PaThus 676.7 x103= 2(80 x106).tAnd t = 676.7 x103/(2(80 x106)) = 0.0042 m4.2 mm is a reasonable thickness for a tensile steel medium pressure tank.As a check, use the longitudinal stress formula given in the study guide:tPD2= 563.9x103(1.2)/(2(0.0042)) = 80x106OKDL = 1t = wall thicknessFp = P.DT = S.t T = S.t