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ENGG 2230 (29)
Lecture

Ch 2 Solutions.pdf

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Department
Engineering
Course
ENGG 2230
Professor
Bahram Gharabaghi
Semester
Winter

Description
Pressure Distribution Chapter 2 in a Fluid 21 For the twodimensional stress field in Fig P21 let3000 psf2000 psfxxyy500 psf xyFind the shear and normal stresses on plane AA cutting through at 30 Solution Make cut AA so that it justhits the bottom right corner of the element Fig P21 This gives the freebody shown at right Now sum forces normal and tangential to side AA Denote side length AA as L LF0AAnAA3000sin30500cos30Lsin30 2000cos30500sin30Lcos30 2Solve for aAns2683 lbfft AAF0L3 0 0 0 c o s 3 0 5 0 0 s i n 3 0L s i n 3 05 0 0 c o s 3 0 2 0 0 0 s i n 3 0L c o s 3 0tAAAA 2Solve for bAns683 lbfft AA 22 For the stress field of Fig P21 change the known data to 2000 psf 3000 xxyyAA2500 psf Computeand the shear stress on plane AA psf and nxySolution Sum forces normal to and tangential to AA in the element freebody above AA known andunknown with nxyF2500Lcos302000sin30Lsin30nAAxy s i n 3 03 0 0 0 c o s 3 0L c o s 3 00xy72 Solutions ManualFluid Mechanics Fifth Edition 2Solve for250050022500866 aAns289 lbfft xy In like manner solve for the shear stress on plane AA using our result for xyFL2 0 0 0 c o s 3 0 2 8 9 s i n 3 0L s i n 3 0tAAAA 289cos303000sin30Lcos3002Solve for9381515 bAns577 lbfft AAThis problem and Prob 21 can also be solved using Mohrs circle23 A vertical clean glass piezometer tube has an inside diameter of 1 mm When a pressure is applied water at 20C rises into the tube to a height of 25 cm After correcting for surface tension estimate the applied pressure in Pa 3Solution For water let Y0073 Nm contact angle 0 and 9790 Nm The capillary rise in the tube from Example 19 of the text is 2Ycos20073 cos0Nm0030 mh cap3RNmm9790 00005 025 m003 m022 m Then the rise due to applied pressure is less by that amount hpress3The applied pressure is estimated to be ph9790 Nm022 m2160 Pa Ans press BourdonWP24 Pressure gages such as the Bourdon gage gage in Fig P24 are calibrated with a deadweight piston If the Bourdon gage is designed to rotate the pointer 2 cmOil diameter 10 degrees for every 2 psig of internal pressure how many degrees does the pointer rotate if the piston and Fig P24weight together total 44 newtonsSolutionThe deadweight divided by the piston area should equal the pressure applied to the Bourdon gageStay in SI units for the momentN44Flbfa68948pP203140060Bourdon22Apiston002min4 Chapter 2Pressure Distribution in a Fluid 73At 10 degrees for every 2 psig the pointer should move approximately100 degrees AnsDenver Colorado has an average altitude of 5300 ft On a US standard day pres25sure gage A reads 83 kPa and gage B reads 105 kPa Express these readings in gage or vacuum pressure whichever is appropriateSolution We can find atmospheric pressure by either interpolating in Appendix Table A6 or more accurately evaluate Eq 227 at 5300 ft1615 m gRB526Bz00065 Km1615 m pp110135 kPa1834 kPaaoT28816 KoTherefore Gage A83 kPa834 kPa04 kPa gagekPa vacuum48321Gage B105 kPa4 kPa kPa gageAns6 26 Express standard atmospheric pressure as a head hpg in a feet of glycerin b inches of mercury c meters of water and d mm of ethanol Solution Take the specific weights g from Table A3 divide p byatm23787 lbfft269 ft Ans a a Glycerin h2116 lbfft23846 lbfft250 ft300 inches Ans b b Mercury h2116 lbfft239790 Nm1035 m Ans c c Water h101350 Nm237740 Nm131 m13100 mm Ans d d Ethanol h101350 Nm P27La Paz Bolivia is at an altitude of approximately 12000 ftAssume a standard atmosphereHow high would the liquid rise in a methanol barometer assumed at 20C HINTDont forget the vapor pressureSolutionConvert 12000 ft to 3658 meters and Table A6 or Eq 220 give Bz000653658526gRBppPa1101350164400LaPazoT28816o
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