Solution Manual For Gas Dynamics, 3rd Edition

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1CChhaapptteerr OOnneeBBAASSIICC EEQQUUAATTIIOONNSSOOFF CCOOMMPPRREESSSSIIBBLLEE FFLLOOWWProblem 1. – Air is stored in a pressurized tank at a pressure of 120 kPa (gage) and a temperatureof 27°C.The tank volume is 1 m3.Atmospheric pressure is 101 kPa and the local accelerationof gravity is 9.81 m/s2.(a) Determine the density and weight of the air in the tank, and (b)determine the density and weight of the air if the tank was located on the Moon where theacceleration of gravity is one sixth that on the Earth.Kkg/kJ728.0Rs/m81.9gm1C30027327Tkpa122101120PPP23atmgageabs⋅===∀°=+==+=+=a)3mkg5668.2)300)(287.0(221RTP===ρN1801.25)81.9)(1)(5668.2(gmgW==∀ρ==b)3earthmoonmkg5668.2=ρ=ρN1967.4W61WggWearthearthearthmoonmoon===Problem 2. – (a) Show that p/ρhas units of velocity squared. (b) Show that p/ρhas the sameunits as h (kJ/kg). (c) Determine the units conversion factor that must be applied to kineticenergy, V2/2, (m2/s2) in order to add this term to specific enthalpy h (kJ/kg).Air

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2a)22223232VsmsNmkg1kgmNkgmmNpmkg,mNp≈=⎟⎟⎠⎞⎜⎜⎝⎛⋅⋅−=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛≈ρ⎟⎟⎠⎞⎜⎜⎝⎛≈ρ⎟⎟⎠⎞⎜⎜⎝⎛≈b)kgkJ10001J0100kJ1mNJ1kgmNP=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛⋅⋅≈ρc)c2222g10001factorkgkJJ0100kJ1mNJ1mkgsN1sm2V=∴≈⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛⋅⎟⎟⎠⎞⎜⎜⎝⎛⋅⋅≈Problem 3. – Air flows steadily through a circular jet ejector, refer to Figure 1.15. The primaryjet flows through a 10 cm diameter tube with a velocity of 20 m/s. The secondary flow is throughthe annular region that surrounds the primary jet.The outer diameter of the annular duct is 30cm and the velocity entering the annulus is 5 m/s. If the flows at both the inlet and exit areuniform, determine the exit velocity.Assume the air speeds are small enough so that the flowmay be treated as an incompressible flow, i.e., one in which the density is constant.eimm&&=ssppspiVAVAmmmρ+ρ=+=&&&eeeVAmρ=&eessppVAVAVA=+∴SoessppeAVAVAV+=pseAAA+=2ppD4Aπ=2p2osD4D4Aπ−π=2oeD4Aπ=iesp

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3()()sp2o2ps2os2p2op2pessppeVVDDVDVDDVDAVAVAV−+=−+=+=()s/m6667.65203010522=−+=Problem 4. –A slow leak develops in a storage bottle and oxygen slowly leaks out. The volumeof the bottle is 0.1 m3and the diameter of the hole is 0.1 mm. The initial pressure is 10 MPa andthe temperature is 20˚C. The oxygen escapes through the hole according to the relationeeATp04248.0m=&where p is the tank pressure and T is the tank temperature. The constant 0.04248 is based on thegas constant and the ratio of specific heats of oxygen. The units are: pressure N/m2, temperatureK, area m2and mass flow rate kg/s. Assuming that the temperature of the oxygen in the bottledoes not change with time, determine the time it takes to reduce the pressure to one half of itsinitial value.3m1.0=∀MPa01p1=21TK293T==MPa5p2=KkgJ8219.259323.314,8R⋅==From the continuity equationemdtdm&−=butRTpm∀=sopTA80424.0mdtdpRTdtdmee−=−=∀=&Integrating we get,O2m(t)d = 0.1mmeeATp04248.0m=&

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4tATR80424.0pplne12⎟⎟⎠⎞⎜⎜⎝⎛∀−=hrs5979.12sec4076.713,4621ln293)8219.259(mm0100mmm1.04)04248.0(1.0pplnTRA)04248.0(t212e==⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛ π−=∀−=Problem 5. – A normal shock wave occurs in a nozzle in which air is steadily flowing. Becausethe shock has a very small thickness, changes in flow variables across the shock may be assumedto occur without change of cross-sectional area. The velocity just upstream of the shock is 500m/s, the static pressure is 50 kPa and the static temperature is 250 K. On the downstream side ofthe shock the pressure is 137 kPa and the temperature is 343.3 K. Determine the velocity of theair just downstream of the shock.s/m050V1=?V2=kPa05p1=kPa713p2=K025T1=K3.343T2=21AA=From the continuity equation21mm&&=So222111VAVAρ=ρs/m5839.250)500(2503.34313750VTTppVRT/pRT/pVV11221122111212=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛===ρρ=21

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5Problem 6. – A gas flows steadily in a 2.0 cm diameter circular tube with a uniform velocity of1.0 cm/s and a densityρo. At a cross section farther down the tube, the velocity distribution isgiven by V = Uo[1-( r/R)2], with r in centimeters. Find Uo,assuming the gas density to beρo[1+( r/R)2].s/cm1V1=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛−=2o2Rr1UVo1ρ=ρ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+ρ=ρ2o2Rr121mm&&=o221oRo111Ro11RRVrdr2VdAVmρπ=πρ=πρ=ρ=∫∫&()oo22oo1o522oo22o22RoRoo222UR326121RU2RrwheredR2Urdr2Rr1URr1dAVmρπ=⎟⎠⎞⎜⎝⎛−πρ==ξξξ−ξπρ=π⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛+ρ=ρ=∫∫∫&oo2o2UR32Rρπ=ρπ∴sos/cm23Vo=Problem 7. – For the rocket shown in Figure 1.6, determine the thrust. Assume that exit planepressure is equal to ambient pressure.()()()ee2oHeeoHoHeeeatmeAmmVmmmm0VmAppρ+=⎟⎟⎠⎞⎜⎜⎝⎛ρ+++=+−=&&&&&&&Tr12

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6Problem 8. – Determine the force F required to push the flat plate of Figure Pl.8 against theround air jet with a velocity of 10 cm/s. The air jet velocity is 100 cm/s, with a jet diameter of 5.0cm. Air density is 1.2 kg/m3.Figure P1.8To obtain steady state add + Vpto all velocitiesVmF&=()() ()1.015.042.1AVm2+⎟⎠⎞⎜⎝⎛ π=ρ=&s/kg200259.0=()()N100285.01.1002592.0F==Problem 9. – A jet engine (Figure P1.9) is traveling through the air with a forward velocity of300 m/s. The exhaust gases leave the nozzle with an exit velocity of 800 m/s with respect to thenozzle. If the mass flow rate through the engine is 10 kg/s, determine the jet engine thrust. Exitplane static pressure is 80 kPa, inlet plane static pressure is 20 kPa, ambient pressure surroundingthe engine is 20 kPa, and the exit plane area is 4.0 m2.FxV = -10scmVj= 100scmVj= 110scmxV = 0F

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7Figure P1.9()()()( )()()kN24552403008001042080VVmAppieeatme=+=−+−=−+−=&TProblem 10. – A high-pressure oxygen cylinder, typically found in most welding shops,accidentally is knocked over and the valve on top of the cylinder breaks off. This creates a holewith a cross-sectional area of 6.5 x 10-4m2.Prior to the accident, the internal pressure of theoxygen is 14 MPa and the temperature is 27˚C. Based on critical flow calculations, the velocityof the oxygen exiting the cylinder is estimated to be 300 m/s, the exit pressure 7.4 MPa and theexit temperature 250 K.How much thrust does the oxygen being expelled from the cylindergenerate? What percentage is due to the pressure difference? What percentage due to the exitingmomentum? Atmospheric pressure is 101 kPa. Also note that 0.2248 lbf= 1 N.FigureP1.10s/m030Ve=24em105.6A−×=MPa4.7pe=MPa110.0kPa110patm==k025Te=eeeeeeeVARTpVAm=ρ=&kkgJ82.259R⋅=()() ()()s/kg2.22300105.625082.259mN104.7426=×⎟⎟⎠⎞⎜⎜⎝⎛×=−300 m/s800 m/s

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8()()()()()f2243eeatmelb3.548,2N0.336,11N6.66644.4671mkgsN1smkg3002.22N104.6101017400VmApp==+=⎟⎟⎠⎞⎜⎜⎝⎛⋅⋅⎟⎟⎠⎞⎜⎜⎝⎛⋅+×××−=+−=−&TThe thrust due to the pressure is 41% of total and that due to momentum 59%.Problem 11. – Air enters a hand held hair dryer with a velocity of 3 m/s at a temperature of 20°Cand a pressure of 101 kPa. Internal resistance heaters warm the air and it exits through an area of20 cm2with a velocity of 10 m/s at a temperature of 80°C. Assume that internal obstructions donotappreciablyaffectthepressurebetween inletandexitandthatheattransfertothesurroundings are negligible.Determine the power in kW needed to operate the hair dryer atsteady state.()()()()() () ()()skg019939.010100m2035328710101VARTpVAm2232222222==⎟⎟⎠⎞⎜⎜⎝⎛=ρ=&()()kgkJ3.602080KkgkJ005.1TTchh12p12=−⋅=−=−()()kgkJ0455.02000310310g010012VVc2122=−+=⎟⎟⎠⎞⎜⎜⎝⎛−ie

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9⎟⎟⎠⎞⎜⎜⎝⎛+−⎟⎟⎠⎞⎜⎜⎝⎛+=−2Vhm2VhmWQ211222&&&&()()()W2051.203,1kgkJ203205.10455.03.60019939.02VVhhmW212212==+=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−+−=−&&Problem 12. – Air is expanded isentropically in a horizontal nozzle from an initial pressure of1.0 MPa, of a temperature of 800 K, to an exhaust pressure of 101 kPa. If the air enters thenozzle with a velocity of 100 m/s, determine the air exhaust velocity. Assume the air behaves asa perfect gas, with R = 0.287 kJ/kg · K andγ= 1.4. Repeat for a vertical nozzle with exhaustplane 2.0 m above the intake plane.(a)Horizontal nozzle2Vh2Vhh222211o+=+=()()212121p212TT1R2VTTc2VV−−γγ+=−+=kkgJ287R⋅=5194.01000101ppTT4.14.11212=⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=γ−γK5.415T2=()( )()() ()s/m856.8845.460,772000,105.4158004.02874.12100V22=+=−+=(b)Vertical nozzle12p2= 101kPap1= 1MPaT1= 800KV1= 100m/s

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1022221211gz2Vhgz2Vh++=++()()( )()( )281.925.460,782zzg2TT1R2VV2121212+=−+−−γγ+=s/m059.88474.499,782==Problem 13. – Nitrogen is expanded isentropically in a nozzle from a pressure of 2000 kPa, at atemperature of 1000 K, to a pressure of 101 kPa. If the velocity of the nitrogen entering thenozzle is negligible, determine the exit nozzle area required for a nitrogen flow of 0.5 kg/s.Assume the nitrogen to behave as a perfect gas with constant specific heats, mean molecularmass of 28.0, andγ= 1.4.2Vh2Vhh222211o+=+=()()21p212TTc2hh2V−=−=()K1.42620001011000PPTT4.14.011212=⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=γ−γKkgJ9.296283.8314R⋅==()()( )()()s/m2.10921.42610009.2967TTR7TT1R2V21212=−=−=−−γγ=()()3222mkg798.01.4269.296000,101RTp===ρp1= 2000kPaT1= 1000KV1~ small12p2= 101kPam = 0.5kg/s =ρ2A2V2•A2= ?V2= ?

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11()()22222cm473.5m4000573.02.1092798.05.0VmA===ρ=&Problem 14. – Air enters a compressor with a pressure of 100 kPa and a temperature 20°C; themass flow rate is 0.25 kg/s.Compressed air is discharged from the compressor at 800 kPa and50°C. Inlet and exit pipe diameters are 4.0 cm. Determine the exit velocity of the air at thecompressor outlet and the compressor power required. Assume an adiabatic, steady, flow andthat the air behaves as a perfect gas with constant specific heats; cp= 1.005 kJ/kg · K andR = 0.287kJ/kg·K.kgkJ005.1cp=kkgkJ287.0R⋅=s/kg52.0mmm21===&&&()22221m60012.004.04d4AA=π=π==()()3111mkg189.12930.287kPa0.10RTp===ρ()()3122mkg630.83230.287800RTp===ρ()()sm3.16700126.189.125.0AmV111==ρ=&()()sm1.2300126.63.825.0AmV222==ρ=&p1= 100kPaT1= 293Kd = 4.cm12W = ?•m = 0.25 kg/s•p2= 800KPaT2= 323KV2= ?

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12⎟⎟⎠⎞⎜⎜⎝⎛+−⎟⎟⎠⎞⎜⎜⎝⎛+=−2Vhm2VhmWQ21112222&&&&()⎥⎥⎦⎤⎢⎢⎣⎡−+−=−2VVhhmW212212&&() ()()()()( )()⎥⎦⎤⎢⎣⎡−++−=1000216731.231.233.167293323005.125.0()()kW1.4skJ1.473.1315.3025.0==−=Problem 15. – Hot gases enter a jet engine turbine with a velocity of 50 m/s,a temperature of1200 K, and a pressure of 600 kPa. The gases exit the turbine at a pressure of 250 kPa and avelocity of 75 m/s.Assume isentropic steady flow and that the hot gases behave as a perfect gaswith constant specific heats (mean molecular mass 25,γ= 1.37). Find the turbine power outputin kJ/(kg of mass flowing through the turbine).KkgJ6.332253.8314R⋅==37.1=γ()()KkgkJ2314.137.6.33237.11Rcp⋅==−γγ=K3.9476002501200ppTT37.137.011212=⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=γ−γ12W•V1= 50m/sT1= 1200Kp1= 600kPap2= 250 kPaV2= 75 m/s

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13⎟⎟⎠⎞⎜⎜⎝⎛+−⎟⎟⎠⎞⎜⎜⎝⎛+=−2Vhm2VhmWQ21112222&&&&()⎥⎥⎦⎤⎢⎢⎣⎡−+−=2VVhhmW222121&&()()()2000VVVVTTc2VVhhmWW212121p222121−++−=−+−==&&()()()()kgkJ6.309563.118.3112000251253.94712002314.1W=−=⎟⎠⎞⎜⎝⎛−+−=Problem 16. –Hydrogen is stored in a tank at 1000 kPa and 30°C. A valve is opened, whichvents the hydrogen and allows the pressure in the tank to fall to 200 kPa. Assuming that thehydrogen that remains in the tank has undergone an isentropic process, determine the amount ofhydrogen left in the tank. Assume hydrogen is a perfect gas with constant specific heats; the ratioof specific heats is 1.4, and the gas constant is 4.124 kJ/kg · K. The tank volume is 2.0m3.K303TkPa1000p11==kPa200p2=()K3.1911000200303ppTT4.14.011212=⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=γ−γ()( )()()kg507.03.191124.42200RTpm222==∀=

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14Problem 17. – Methane enters a constant-diameter, 3 cm duct at a pressure of 200 kPa, atemperature of 250 K, and a velocity of 20 m/s.At the duct exit, the velocity reaches 25 m/s.Forisothermal steady flow in the duct, determine the exit pressure, mass flow rate, and rate at whichheat is added to the methane. Assume methane behaves as a perfect gas; the ratio of specificheats is 1.32 (constant) and the mean molecular mass is 16.0.222111VAVAρ=ρ222111VRTpVRTp⎟⎟⎠⎞⎜⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛2211VpVp=()22112mN1602520200VVpp=⎟⎠⎞⎜⎝⎛=⎟⎟⎠⎞⎜⎜⎝⎛=()22m000707.003.04A=π=KkgJ519.6163.8314R⋅==()()()() ()skg02176.020250000707.06.519200VARTpVAm1111111==⎟⎟⎠⎞⎜⎜⎝⎛=ρ=&() ()( )W448.225202502176.02VVmQ2122=+=⎟⎟⎠⎞⎜⎜⎝⎛−=&&Problem 18. – Air is adiabatically compressed from a pressure of 300 kPa and a temperature of27 C to a pressure of 600 kPa and a temperature of 327 C. Is this compression actually possible?12T = constantQ•p1= 200kPaT1= 250KV1= 20 m/sp2= ?m = ?•V2= 25m/sQ = ?•d = 3cmγ= 1.32MW = 16

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15kPa300p1=kPa600p2=K30027327T1=+=K600273327T2=+=300600lnR300600lncpplnRTTlncssp1212p12−=−=−()02lnc2lnRcvp>=−=possible∴Problem 19. – Two streams of air mix in a constant-area mixing tube of a jet ejector.Theprimary jet enters the tube with a speed of 600 m/s, a pressure of 200 kPa and a temperature of400˚C.The secondary stream enters with a velocity of 30 m/s, a pressure of 200 kPa and atemperature of 100˚C. The ratio of the area of the secondary flow to the primary jet is 5:1. Theair behaves as a perfect gas with constant specific heats, cp= 1.0045 kJ/kg· K. Using the iterativenumerical procedure described in Example 1.9 determine the velocity, pressure and temperatureof the air leaving the mixing tube.gc1α5γ1.4R287cp1004.5PrimarySecondaryV600.0030.00T673373P200,000200,000A43,122.5078B263,528.7595C706,538.5693nVe(m/s)Pe(Pa)Te(K)10.0000101,000.0293.15002125.1620244,722.8695.57573122.5671245,112.7695.89574122.4284245,133.6695.91265122.4210245,134.7695.91356122.4206245,134.7695.91367122.4206245,134.7695.9136

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