Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition

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Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition - Page 1 preview image

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Chapter 11-1(a) 98 Btu/(hr-ft-F) x1.7307 = 170 W/(m-K)(b) 0.24 Btu/(Ibm-F) x4186.8=1.0 kJ/kg-K(c)0.04 Ibm/(ft-hr)3600 sec/hrx1.488 = 16.52Nsmμ(d) 1050BtuIbmx419.48x10−JBtux2.20462 Ibmkg= 2.44MJkg(e) 12,000BtuIbmx13.412= 3.52 kW(f) 14.72Ibfinx 6894.76 = 101 kPa1-2(a) 120 kPa x2lbf / in6.89476kPa= 17.4 lbf/in2(b) 100WmK−x 0.5778 = 57.8 Btu/hr-ft-F(c) 0.82WmK−x 0.1761 = 0.14 Btu/hr-ft2-F(d) 10-6N-s/m2x11.488= 6.7 x 10-7lbmftsec−(e) 1200 kW x 3412 = 4.1 x 10-6Btu/hr

Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition - Page 2 preview image

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Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition - Page 3 preview image

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2(f) 1000kJkgx1 Btu1.055 kJx1 kg2.2046 lbm= 430Btulbm1-3Hp = 50 (ft) x 0.3048 (mft) = 15.2 m∆P =15.2 m1000 Pa/kPax9.8071(Nkg) x 1000 (kg/m3) = 149 kPa1-4P =∆412(ft) x 0.3048 (mft) x9.8071(Nkg) x 1000 (3kgm)∆P = 996 Pa1.0 kPa≈1-5TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE+ METER CHARGE()()()()96,000kw - hrs0.045 $ / kwhr +624 kw 1150 $ / kw−−+ $68 = $4,320 + $7,176 + $68 = $11,5641-67 AM to 6 PM11 hrs/day, 5 days/wkhrsdays(11)(22)242 hrs / monthdaymonths=students enrolled in courses for which the textbook has been adopted.Any other reproduction or translation of this work beyond that

Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition - Page 4 preview image

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3()()()624 kwratio =1.5796,000 kwhr242 hr=⎛⎞−⎜⎟⎝⎠1-7This is a trial and error solution since eq. 1-1 cannot be solvedexplicitly for i.Answer converges at just over 4.2% using eq. 1-11-8Determine present worth of savings using eq. 1-1()()()12120.012$10001-1+12P =0.01212P$134,000−⎡⎤⎛⎞⎢⎥⎜⎟⎝⎠⎢⎥⎣⎦⎛⎞⎜⎟⎝⎠=1-9(a)QVA== 2 x 3.08 x 10-3= 6.16 x 10-3m3/sm6.16 x 10Qρ==-3x 998 = 6.15 kg/s(b) A=4π(0.3)2= 7.07 x 10-2m2Q7.07x10=-2x 4 = 0.283 m3/ s;ρ=1.255 kq/m3m= 1.225 x 0.283 = 0.347 kg/s1-10V = 3x10x20 = 600m3students enrolled in courses for which the textbook has been adopted.Any other reproduction or translation of this work beyond that

Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition - Page 5 preview image

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4= 600 xiQ14x13600= 4.17 x 10-2m3/s1-11pp3q = mcTc = 4.183 kJ/(kg-K)= 983.2 kg/mρ∆1-11 (cont’d)( )()()()3c3mkgkJq =1983.24.183520,564skgKmq = 20,564 kw=−kJsstudents enrolled in courses for which the textbook has been adopted.Any other reproduction or translation of this work beyond that permitted1-12=watqairq−11,200(1)(10) ==25000x60x14.7x144x0.24(t50)(53.35x510)−11,200 = 5601.5 (t2-50); t2= (11,200/5601.5) + 50 = 70 F1-13Diagram as in 1-12 above.qwat= -qair1.5 (4186)(90-t2) = 2.4 (1.225)(1.0)(30-20)(1000)6279(90-t2) = 29,400

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-l purposes only tostudents enrolled in courses for which the textbook has been adopted.this work beyond that permitted5t2= 90 -29,4006279= 85.3 C1-14qhA(t=s-)t∞A=π(1/12) x 10 = 2.618 ft2st= tsur212 F≈= 10x2.618x(212-50) = 4241 Btu/hrq1-15A=πx 0.25x4 = 3.14 16 m2hA(tq=s-)t∞h=sqA(t -t)∞=12503.1416(10010)−; h = 4.42 W/(m2– C)1-16(tpqmc=2-t1) ;mQxρ=ρ=P/RT = 14.7x144/53.35(76+460)ρ= 0.074 lbm/ft3m= 5000x0.074x60 = 22,208 lbm/hr= 0.24 Btu/lbm-Fpcq= 22,208x0.24(58-76) = -95,939 Btu/hrNegative sign indicates cooling1-17(t1 pm c3-t1) +for-profit basis for testing or instructionaAny other reproduction or translation of

Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition - Page 7 preview image

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6(t2p2m c3-t2) = 0=p1cp2ct3=1 12 212(m tm t )(mm )++12mQ1ρ== 1000x14.7x14453.35(46050)+= 73.5 lbm/min1-17 (cont’d)22mQ2ρ== 600x14.7x14453.35(46050)+= 46.7 lbm/min3(73.5x80)(46.7 x 50)t68.3 F(73.546.7)+==+students enrolled in courses for which the textbook has been adopted.Any other reproduction or translation of this work beyond that

Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition - Page 8 preview image

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Problem 1.1The force,F, of the wind blowing against a building is givenby22DFCVA, whereVisthe wind speed,the density of the air,Athe cross-sectional area of the building, andDCisa constant termed the drag coefficient. Determine the dimensions of the drag coefficient.Solution1.122DAFCVor22 DFCVA, where2FMLT,3ML,1VLT,2ALThus, -20002-2-12DMLTCML TMLLTLHence,DCis dimensionless.

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Problem 1.2The Mach number is a dimensionless ratio of the velocity of an object in a fluid to the speed ofsound in the fluid. For an airplane flying at velocityVin air at absolute temperatureT, theMach numberMais,MaVkRT,wherekis a dimensionless constant andRis the specific gas constant for air. Show thatMaisdimensionless.Solution 1.2We denote the dimension of temperature byand use Newton’s second law to get2MLFT.Then 2221LLTTMFLMLLMT FTor 1M.

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Problem 1.3Verify the dimensions, in both theFLTandMLTsystems, of the following quantities whichappearinTable B.1Physical Properties of Water (BG/EE Units).(a)Volume,(b)acceleration,(c)mass,(d)moment of inertia (area), and(e)work.Solution1.3a)3volumeLb)12accelerationtime rate of change of velocityLTLTTc)massM212or withmassFMLTFLTd)224moment of inertia areasecond moment of areaLLLe)workforcedistanceFL222or withworkFMLTML T

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Problem 1.4Verify the dimensions, in both theFLTandMLTsystems, of the following quantities whichappearinTable B.1Physical Properties of Water (BG/EE Units).(a)Angular velocity,(b)energy,(c)moment of inertia (area),(d)power, and(e)pressure.Solution1.4a)1angular displacementangular velocitytimeTb)energy ~ capacity of body to do worksingle workforcedistanceenergyFL2or withFMLT222energyMLTLML Tc)224second mommoment of ineent of arertiaa ear aLLLd)12123powerrate of doing workFLFLTMLTLTML TTe)222122forcepressureareaFFLMLTLML TL

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Problem 1.5Verify the dimensions, in both theFLTsystem and theMLTsystem, of the following quanti-ties which appearinTable B.1Physical Properties of Water (BG/EE Units).(a)Frequency,(b)stress,(c)strain,(d)torque, and(e)work.Solution1.5a)-1cyclesfrequency=Ttimeb)-22-2-2-1-22forceFstress=FLareaLSince FMLT,MLTstressML TLc)0change in lengthLstrain=LdimensionlesslengthLd)-22-2torque=forcedistanceFLMLTLML Te)-22-2work=forcedistanceFLMLTLML T

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Problem 1.6Ifuis a velocity,xa length, andta time, what are the dimensions (in theMLTsystem) of(a)/ut,(b)2/ ux t, and(c)(/)ut dx?Solution1.6a)12uLTLTtTb)212()() uLTTx tLTc)122(L)LTuxL TtT

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Problem 1.7Verify the dimensions, in both theFLTsystem and theMLTsystem, of the following quantitieswhich appearinTable B.1Physical Properties of Water (BG/EE Units).(a)Acceleration,(b)stress,(c)moment of a force,(d)volume, and(e)work.Solution1.7a)22velocityaccelerationtimeLLTTb)2222122forcestressareaSincestressFFLLFMLTMLTMLTLc)222moment of a forceforcedistanceFLMLTLML Td)33volume(length)Le)222workforcedistanceFLMLTLML T

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Problem 1.8Ifpis a pressure,Va velocity, anda fluid density, what are the dimensions (in theMLTsystem) of(a)/p,(b)pV, and(c)2/pV?Solution1.8a)2221222333pFLMLTLML TL TMLMLMLb)1213233pVML TLTMLM LTc)120002231dimensionlesspMLTM L TVMLLT

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Problem 1.9IfPis a force andxa length, what are the dimensions (in theFLTsystem) of(a)/dPdx,(b)33/d Pdx, and(c)P dx?Solution1.9a)2dPFFLdxLb)3333d PFFLdxLc)PdxFL

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