Solution Manual For Introduction To Flight, 7th Edition

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SOLUTIONS MANUAL TO ACCOMPANYINTRODUCTION TO FLIGHT7thEditionByJohn D. Anderson, Jr.

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2Chapter 22.1523/(1.2)(1.0110 )/(287)(300)1.41 kg/m1/1/1.410.71 m /kgρρρ×== p RT =v ===2.2Mean kinetic energy of each atom232033 (1.3810) (500)1.03510J22−−=××k T ==One kg-mole, which has a mass of 4 kg, has 6.02 × 1026atoms. Hence 1 kg has26261 (6.0210)1.505104=××atoms.20266Totalinternal energy(energy per atom)(number of atoms)(1.03510)(1.50510)1.55810 J-==´´=´2.332116slug0.00237(1716)(46059)ftρ ===+pRT3Volume of the room(20)(15)(8)2400 ftTotal mass in the room(2400)(0.00237)5.688slugWeight(5.688)(32.2)183lb======2.432116slug0.00274(1716)(46010)ftpRTρ===-Since the volume of the room is the same, we can simply compare densities between the twoproblems.3slug0.002740.002370.00037 ftρΔ=-=0.00037% change(100)15.6% increase0.00237ρρΔ==´=2.5First, calculate the density from the known mass and volume,3m1500 9001.67 lbftρ//==In consistent units,31.67/32.20.052slug ft .ρ/==Also,T= 70F= 70 + 460 = 530R.Hence,(0.52)(1716)(530)ρpRT==247, 290lb/ftp=or47, 290 / 211622.3 atmp==

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32.6ρpRT=npnpnRnT=++Differentiating with respect to time,111ρρ=+dpddTp dtdtT dtor,Tρρ=+dpp dp dTdtdtdtor,ρρ=+dpddTRTRdtdtdt(1)At the instant there is 1000 lbmof air in the tank, the density is3m1000 / 9001.11lb /ftρ==31.11/32.20.0345slug/ftρ==Also, in consistent units, is given thatT= 50 + 460 = 510Rand that1 /min1 /min0.016/sec===dTFRRdtFrom the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have3mm30.5 lb /sec0.000556 lb /(ft )(sec)900 ftρ==ddt530.0005561.7310slug/(ft )(sec)32.2ρ−==×ddtThus, from equation (1) above,52(1716)(510)(1.7310)(0.0345)(1716)(0.0167)16.115.10.9916.1 lb/(ft )(sec)21160.0076 atm/secρ−=×+=+===ddt2.7In consistent units,10273263 K= −+=TThus,43/(1.710 )/(287)(263)0.225 kg/mρρ==×=p RT2.8533/0.510 /(287)(240)0.726 kg/m1/1/0.7261.38 m /kgρρ==×====p RTv

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42.93300230Force due to pressure(211610 )[21165]6303 lb perpendicular to wall.pFp dxx dxxx===-=-=òò33τ1002132090Force due to shear stress =τ(9)[180 (9)]623.554083.5 lb tangential to wall.Fdxdxxx==+=+=-=òòMagnitude of the resultant aerodynamic force =22(6303)(835)6303.6 lb83.5Arc Tan0.76º6303θR=+=æö÷ç==÷ç÷÷çèø2.103sin2θ∞VV=Minimum velocity occurs when sinθ=0, i.e., whenθ=0° and 180°.Vmin= 0 atθ= 0° and 180°, i.e., at its most forward and rearward points.Maximum velocity occurs when sinθ=1, i.e., whenθ=90°. Hence,max3 (85)(1)127.5 mph at90 ,2θ°V===i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction.

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52.11The mass of air displaced is3(2.2)(0.002377)5.2310slugM-==´The weight of this air is3air(5.2310)(32.2)0.168 lbW-=´=This is the lifting force on the balloon due to the outside air. However, the helium inside theballoon has weight, acting in the downward direction. The weight of the helium is less than thatof air by the ratio of the molecular weights4(0.168)0.0233 lb.28.8cHW==Hence, the maximum weight that can be lifted by the balloon is0.168−0.0233 = 0.145 lb.2.12Letp3,ρ3, andT3denote the conditions at the beginning of combustion, andp4,ρ4, andT4denote conditions at the end of combustion. Since the volume is constant, and the mass of thegas is constant, thenp4=ρ3= 11.3 kg/m3.Thus, from the equation of state,72444(11.3)(287)(4000)1.310N/mρpRT===´or,7451.310129 atm1.0110p´==´2.13The area of the piston face, where the diameter is 9 cm = 0.09 m, is232(0.09)6.3610m4πA-==´(a)The pressure of the gas mixture at the beginning of combustion is6233311.3(287)(625)2.0210N/mρpRT===´The force on the piston is63433(2.0210 )(6.3610)1.2810 NFp A-==´´=´Since 4.45 N = l lbf,431.28102876 lb4.45F´==(b)72444(11.3)(287)(4000)1.310 N/mρpRT===´The force on the piston is73444= (1/310 ) (6.3610) = 8.2710 NFp A-=´´´448.271018,579 lb4.45F´==

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62.14Letp3andT3denote conditions at the inlet to the combustor, andT4denote the temperature atthe exit. Note:6234410 N/mpp==´(a)63333410= 15.49 kg/m(287)(900)ρpRT´==(b)63444410= 9.29 kg/m(287)(1500)ρpRT´==2.151 mile = 5280 ft, and 1 hour = 3600 sec.So:miles5280 ft1 hour6088 ft/sec.hourmile3600 secæöæöæö÷÷÷ççç=÷÷÷ççç÷÷÷÷÷÷çççèøèøèøA very useful conversion to remember is that60 mph88 ft/sec=also,1 ft = 0.3048 mft0.3048 mm8826.82sec1 ftsecæöæö÷ç÷ç÷ =÷çç÷÷ç÷ç÷÷çèøèøThusftm8826.82secsec=2.16miles88 ft/sec6921015 ft/sechour60 mph=miles26.82 m/sec692309.3 m/sechour60 mph=2.17On the front face55(1.071510 )(2)2.14310NffFp A==×=×On the back face55(1.0110 )(2)2.0210NbbFp A==×=×The net force on the plate is55(2.1432.02)100.12310NfbFFF=−=−×=×From Appendix C,14.448 N.flb=So,50.123102765 lb4.448F×==This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)

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72.18210,100Wing loading43.35 lb/ft233Ws===In SI units:222lb4.448 N1 ft43.351 lb0.3048 mftN2075.5 mWsWs  =   =In terms of kilogram force,221 kkgN2075.5211.89.8 NmmffWs==  2.195miles5280 ft0.3048 mmkm4377.03310703.3hrmile1 fthrhrV  ==×=    0.3048 mAltitude(25, 000 ft)7620 m7.62 km1 ft===2.203ft0.3048 mmkm26, 0007.925107.925sec1 ftsecsecV ==×=  2.21From Fig. 2.16,length of fuselage = 33 ft, 4.125 inches = 33.34 ft0.3048 m33.34 ft10.16 mft==wing span = 40 ft, 11.726 inches = 40.98 ft0.3048 m40.98 ft12.49 mft==

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8Chapter 33.1An examination of the standard temperature distribution through the atmosphere given inFigure 3.3 of the text shows that both 12 km and 18 km are in the same constant temperatureregion. Hence, the equations that apply are Eqs. (3.9) and (3.10) in the text. Since we are inthe same isothermal region with therefore the same base values ofpandρ, these equationscan be written as021(/)()2211ρρgRThhpep--==where points 1 and 2 are any two arbitrary points in the region. Hence, withg0= 9.8 m/sec2andR= 287 joule/kgK, and letting points 1 and 2 correspond to 12 km and 18 km altitudes,respectively,9.8(287)(216.66)(6000)22110.3884ρρpep===Hence:4322132(0.3884)(1.939910 )7.5310N/m(0.3884)(3.119410)0.121 kg/mρp-=´=´=´=and, of course,2216.66 KT=These answers check the results listed in Appendix A of the text within round-off error.3.2From Appendix A of the text, we see immediately thatp= 2.65×104N/m2corresponds to10,000 m, or 10 km, in the standard atmosphere. Hence,pressure altitude = 10 kmThe outside air density is432.65100.419 kg/m(287)(220)ρpRT´===From Appendix A, this value ofρcorresponds to 9.88 km in the standard atmosphere.Hence,density altitude = 9.88 km3.3At 35,000 ft, from Appendix B, we find thatp= 4.99 × 102= 499 lb/ft2.3.4From Appendix B in the text,33,500 ft corresponds top= 535.89 lb/ft232,000 ft corresponds toρ= 8.2704 × 10−4slug/ft3Hence,4535.89378(8.270410)(1716)ρpTRR-===´

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93.50.021GGhhhhh-==-From Eq. (3.6), the above equation becomes1110.02GGrhhrræö+÷ç-=--=÷ç÷÷çèø650.020.02(6.35710 )1.2710 m127 kmGGhrh==´=´=3.6T= 15−0.0065h= 15−0.0065(5000) =−17.5°C = 255.5°K0.0065dTadh== -From Eq. (3.12)0/(2.8) /(0.0065)(287)11255.50.533288gaRpTpT---æöæö÷ç÷ç÷===ç÷ç÷÷ç÷ç÷÷çèøèø54210.5330.533 (1.0110 )5.3810 N/mpp==´=´3.711()pgnhhpRT= --1111(4157)(150)0.524.9phhRTnngp-= -= -Lettingh1= 0 (the surface)h = 17,358 m = 17.358 km

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103.8A standard altitude of 25,000 ft falls within the first gradient region in the standard atmosphere.Hence, the variation of pressure and temperature are given by:11gaRpTpT-æö÷ç÷= ç÷ç÷÷çèø(1)andT=T1+a(h–h1)(2)Differentiating Eq. (1) with respect to time:11111ggaRaRdpgdTTpdtTARdtæö-÷ç-- ÷ç÷÷ççèøæöæö÷ç÷ç÷=-ç÷ç÷÷ç÷ç÷÷çèøèø(3)Differentiating Eq. (2) with respect to time:dTdhadtdt=(4)Substitute Eq. (4) into (3)111()ggaRaRdpgdhp TTdtRdtæö÷ç-+ ÷ç÷÷ççèøæö÷ç= -÷ç÷÷çèø(5)In Eq. (5),dh/dtis the rate-of-climb, given bydh/dt= 500 ft/sec. Also, in the first gradientregion, the lapse rate can be calculated from the tabulations in Appendix B. For example,take 0 ft and 10,000 ft, we find2121483.04518.690.0035710, 0000ft°TTRahh--=== ---Also from Appendix B,p1= 2116.2 lb/ft2at sea level, andT= 429.64 °Rat 25,000 ft.Thus,32.25.256(0.00357)(1716)gaR== --Hence, from Eq. (5)5.2564.25632.2(2116.2)(518.69)(429.64)(500)1716dpdt-æö÷ç= -÷ç÷÷çèø2lb17.17 ft secdpdt= -

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113.9From the hydrostatic equation, Eq. (3.2) or (3.3),0ρdpg dh= -or0ρdpdhgdtdt= -The upward speed of the elevator isdh/dt, which is0/ρdpdp dtdtg= -At sea level,ρ= 1.225 kg/m3. Also, a one-percent change in pressure per minute starting fromsea level is532(1.0110 )(0.01)1.0110N/mper minutedpdt= -´= -´Hence,31.011084.1meter per minute(1.225)(9.8)dhdt-´==3.10From Appendix B:At 35,500 ft:p= 535.89 lb/ft2At 34,000 ft:p= 523.47 lb/ft2For a pressure of 530 lb/ft2, the pressure altitude is535.8953033,50050033737 ft535.89523.47æö-÷ç+=÷ç÷÷çèø-The density at the altitude at which the airplane is flying is435307.91910slug/ft(1716)(390)ρρRT-===´From Appendix B:4343At 33,000 ft:7.965610slug/ftAt 33,500 ft:7.816510slug/ftρρ--=´=´Hence, the density altitude is7.96567.91933, 00050033,156 ft7.96567.8165æö-÷ç+=÷ç÷÷çèø-

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123.11Letℓbe the length of one wall of the tank,ℓ=30 ft. Letdbe the depth of the pool,d= 10 ft.At the water surface, the pressure is atmospheric pressure,pa. The water pressure increaseswith increasing depth; the pressure as a function of distance below the surface,h, is givenby the hydrostatic equationdp =ρg dh(1)Note:The hydrostatic equation given by Eq. (3.2) in the text has a minus sign becausehGismeasured positive in the upward direction. In Eq. (1),his measured positive in the downwarddirection, withh= 0 at the surface of the water. Hence, no minus sign appears in Eq. (1); ashincreases (as we go deeper into the water),pincreases. Eq. (1) is consistent with this fact.Integrating Eq. (1) fromh= 0 wherep=pato some local depthhwhere the pressure isp, andnoting thatρis constant for water, we have0ρρahpdpgdh=òòor,p−pa=ρg hor,p=ρg h+pa(2)Eq. (2) gives the water pressure exerted on the wall at an arbitrary depthh.Consider an elementary small sliver of wall surface of lengthℓand heightdh.The water force on this sliver of area isdF=pℓdhTotal force, F, on the wall is00FdFdFpdh==òò(3)wherepis given by Eq. (2). Inserting Eq. (2) into (3),0(+)ρdaFg hpdh=òor,22ρadFgpdæö÷ç÷ç=+÷ç÷ç÷÷çèø(4)In Eq. (4), the productρgis the specific weight (weight per unit volume) of water;ρg= 62.4 lb/ft3. From Eq. (4),2(10)(62.4)(30)(2116)(30)(10)2F=+93, 600634,800728, 400 lbF=+=Note:This force is the combined effect of the force due to the weight of the water, 93,600 lb,and the force due to atmospheric pressure transmitted through the water, 634,800 Ib. In thisexample, the latter is the larger contribution to the force on the wall. If the wall were freestandingwith atmospheric pressure exerted on the opposite side, then the net force exerted on the wallwould be that due to the weight of the water only, i.e., 93,600 Ib. In tons, the force on the sideof the wall in contact with the water is728, 400364.2 tons2000F==In the case of a freestanding wall, the net force, that due only to the weight of the water, is93, 60046.8 tons2000F==

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133.12For the exponential atmosphere model,0/()0(9.8)(45,000) /(287)(240)6.4020ρρρρg hRTeee---===Hence,6.4023330(1225)(1.657510)2.0310kg/mρρe---==´=´From the standard atmosphere, at 45 km,p= 2.02 × 10−3kg/m3. The exponential atmospheremodel gives a remarkably accurate value for the density at 45 km when a value of 240 K is usedfor the temperature.3.13At 3 km, from App. A,T= 268.67 K,p= 7.0121×104N/m2, and= 0.90926 kg/m3. At 3.1 km,from App. A,T= 268.02 K,p= 6.9235×104N/m2, and= 0.89994 kg/m3. Ath= 3.035 km,using linear interpolation:ρ=−−==×−×−×=×=−−=4444230.035268.67(268.67268.02)268.44 K0.10.0357.012110(7.0121106.923510 )0.1N6.9809910m0.035kg0.09026(0.909260.89994)0.9060.1mTp

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143.14The altitude of 3.035 km is in the gradient region. From Example 3.1,a= –0.0065 K/m. FromEq. (3.14),11()(0)288.16(0.0065)(3035)268.43 KsTTa hhTa h=+−=+−=−=From Eq. (3.12),0/1109.85.25328( 0.0065)(28.7)gaRpTpTgaR−=−−==−So5.253285.25328542268.430.688946288.16N0.688946(1.0132510 )6.980710msspTpTp====×=×By comparison the approximate result from the solution of Prob. 3.13 differs from the exactresult above by56.98099006.98074.15100.00415%6.9807−−=×=A very small amount.From Eq. (3.13),ρρρρ−+=+= −+= −===0[(/) 1]1104.25384.2432815.2532814.25328268.430.73958288.16gaRssTTgaRTT3kg0.735980.73958 (1.2250)0.90599 msρρ===

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153.15We want to calculatehGwhen0.011GGGhhhhh−==−From Eq. (3.6)6, where6.35676610mGGrhhrrh==×+Thus,0.01110.010.99GGrrhrrh=−+=−=+0.99()Grhr+=5310.990.0101 (6.35676610 )0.9964.2110m64.21 kmGGhrh−==×=×=3.16From Eq. (3.1),20Grggrh=+wherer= 6.356766×106m,g0= 9.8 m/sec2, and0.3048 m(70,000 ft)21,336 m1 ftGh==Thus,6626.35676610m9.89.7676.3567661021,336secg×==×+9.89.7671000.34%98−=Thus, at 70,000 ft, the acceleration of gravity has decreased only by 0.34% compared to its sealevel value.

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