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ENGG 2230 (29)
Lecture

FluidMechWhite5ech02part2b.pdf

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Department
Engineering
Course
ENGG 2230
Professor
Bahram Gharabaghi
Semester
Winter

Description
Chapter 2Pressure Distribution in a Fluid 105 champagne 6 inches above the bottom 24 p096624 ft1356624 ftp0gageAAatmosphere12122orP272 lbfft gage AAThen the force on the bottom end cap is vertical only due to symmetry and equals the force at section AA plus the weight of the champagne below AA FFp A r e a WWVAAAA6in cylinder2in hemisphere22327241209662421261209662423212 42374261058Ans258 lbf 288 Circulararc Tainter gate ABC pivots about point O For the position shown determine a the hydrostatic force on the gate per meter of width into the paper and b its line of action Does the force pass through point O Solution The horizontal hydrostatic force is based on vertical projectionFig P288 FhA9790361176220 Nat 4 m below CHC G v e r tThe vertical force is upward and equal to the weight of the missing water in the segment ABC shown shaded below Reference to a good handbook will give you the geometric properties of a circular segment and you may compute that the segment area is 2and its centroid is 55196 m from 3261 m point O or 03235 m from vertical line AC as shown in the figure The vertical upward hydrostatic force on gate ABC is thusFAunit width979032611VA B C31926 Nat04804 m from B106 Solutions ManualFluid Mechanics Fifth Edition 22 12The net force is thus FFF179100 N per meter of width acting upward to HVthe right at an angle of 1027 and passing through a point 10 m below and 04804 m to the right of point B This force passes as expected right through point O289 The tank in the figure contains benzene and is pressurized to 200 kPa gage in the air gap Determine the vertical hydrostatic force on circulararc section AB and its line of action Solution Assume unit depth into the paper The vertical force is the weight of benzene plus the force due to the airpressure Fig P289 N206108819812000000610FA n s122400 Vm4Most of this 120000 Nm is due to the air pressure whose line of action is in the middle of the horizontal line through B The vertical benzene force is 2400 Nm and has a 255 cm to the right or A line of action see Fig 213 of the text at 4R3The moment of these two forces about A must equal to moment of the combined 122400 Nm force times a distance X to the right of A X299 cm12000030 cm2400255 cm122400 XsolveforAns The vertical force is 122400 Nm down acting at 299 cm to the right of A290 A 1ftdiameter hole in the bottom of the tank in Fig P290 is closed by a 45 conical plug Neglecting plug weight compute the force F required to keep the plug in the hole Solution The part of the cone that is inside the water is 05 ft in radius and h 05tan2251207 ft high The force F Fig P290 equals the air gage pressure times the hole Chapter 2Pressure Distribution in a Fluid 107 area plus the weight of the water above the plug FpAWWgagehole3ftcylinder1207ftcone122231441 ft6241362411207 443 4Ans467 lbf33931470197 291 The hemispherical dome in Fig P291 weighs 30 kN and is filled with water and attached to the floor by six equallyspaced bolts What is the force in each bolt required to hold the dome downSolution Assuming no leakage the Fig P291 hydrostatic force required equals the weight of missing water that is the water in a 4mdiameter cylinder 6 m high minus the hemisphere and the small pipe FWWWtotal2mcylinder2mhemisphere3cmpipe2329790269790232979040034 73814916403328574088 N The dome material helps with 30 kN of weight thus the bolts must supply 57408830000 90700 N Ans or 544088 N The force in each of 6 bolts is 5440886 or Fbolt 292 A 4mdiameter water tank consists of two halfcylinders each weighing 45 kNm bolted together as in Fig P292 If the end caps are neglected compute the force in each bolt Solution Consider a 25cm width of upper cylinder as at right The waterpressure in the bolt plane is Fig P292 ph9790439160 Pa 1
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