空气动力学课后答案(北航)

钱第一章1.1解：）（ksm84.259mkR22328315RTp36mkg63.5063032.5984105RTP气瓶中氧气的重量为354.938.915.0506.63Gvg1.2解：建立坐标系根据两圆盘之间的液体速度分布量呈线性分布则离圆盘中心r，距底面为h处的速度为0uknu当n=0时u=0推出0u0当n=h时u=wr推出hwrk则摩擦应力为hwrudnduu上圆盘半径为r处的微元对中心的转矩为drdhwrurrdrdhwrurdAd3则2D0332032Dudrdhru1.4解：在高为10000米处T=288.15-0.006510000=288.15-65=223.15压强为TaTPaP5.2588MKN43.26TaTpap2588.5密度为2588.5TaTamkg4127.0TaTa2588.51-7解:2MKG24.464RTPRTp空气的质量为kg98.662vm第二章2-2解流线的微分方程为yxvdyvdx将vx和vy的表达式代入得ydyxdxyx2dyxy2dx22，将上式积分得y2-x2=c，将（1，7）点代入得c=7因此过点（1，7）的流线方程为y2-x2=482-3解:将y2+2xy=常数两边微分2ydy+2xdx+2ydx=0整理得ydx+（x+y）dy=0（1）将曲线的微分方程yxVdyVdy代入上式得yVx+（x+y）Vy=0由22y2xy2xV得Vx2+Vy2=x2+2xy+y2（（2）由（1）（2）得yvyxvyx，2-5解:直角坐标系与柱坐标系的转换关系如图所示速度之间的转换关系为cosvsinvvsinvcosvvryrx由cosr1yvsinyrsinr1xvcosxrrsinyrcosxsinr1sinVcosVcossinVcosVrxvvxrrvxvrrxxxsincosVsinVsinVcosVr1cossinrVcosrVrrrcossinVr1sinVr1sinVr1cossinVr1cossinrVcosrV22rr2rcosr1cosVsinVsincosVsinVryvvVyrVVVVrryxyxycosr1sinVcosVcosVsinVsincosrVsinrVrrrcossinVr1cosVr1cosVr1cossinvVr1cossinrVsinrV22rr2rzVVVr1rVzVyVxVdivzrrzyx2-6解：（1）sinyx3xV2xsinyx3yV2y0yVxVyx此流动满足质量守恒定律（2）sinyx3xV2xsinyx3yV2y0sinyx6yVxV2yx此流动不满足质量守恒定律（3）Vx=2rsinrxy2Vy=-2rsin2ry2233ry2xVx332yr2yyx4yV0ryx4yVxV32yx此流动不满足质量守恒方程（4)对方程x2+y2=常数取微分，得xdydydx由流线方程yxvdyvdx(1)由）（得2rkvvrkv422y2x由（1）（2）得方程3xrkyv3yrkxv25xrkxy3xV25yrkxy3yV0yVxVyx此流动满足质量守恒方程2—7解：0xVzV0ryz23ryz23zVyVzx2727yz同样0yVxVxy该流场无旋2322222223222zyxzyxzyxd21zyxzdzydyxdxdzvdyvdxvdczyx12222—8解：（1）axVxxayVyyazVzz021v;021v;021vzyxyVxVxVzVzVxVxxzxyz（2）0yVxV210xVzV210zVyV21xyzzxyyzx；；位该流线无旋，存在速度（3）azdz2aydyaxdxdzvdyvdxvdzyxcazay21ax212222—9解：曲线x2y=-4，04yxyxf2，切向单位向量22422422y2x2y2xyx4xxy2iyx4xxjfffxifffytttvvvt切向速度分量把x=2，y=-1代入得jx2xiyx2xjyixv2j21i21jyx4x2xyiyx4xxt224224223tvvtj23i23j21i2123tvvtt2—14解：v=180hkm=50sm根据伯努利方程22V21V21ppap驻点处v=0，表示为1531.25pa501.22521V21pap22相对流速为60sm处得表示为75.63760225.12125.1531V21V21pap222第三章3—1解：根据叠加原理，流动的流函数为xyarctg2QyVyx，速度分量是22y22xyxy2QxVyxx2QVyV；驻点A的位置由VAX=0VAy=0求得0yV2QxAA；过驻点的流线方程为2xyarctg2yxyarctg2yyQVQVAAAsin2rxyarctg2yVVQ或即在半无限体上，垂直方向的速度为-sinvrsin2yxy2v222yQQ线面求极值0-sinv-cossinv2ddv22y当0sin0vvminyy2-tgmaxyyvv用迭代法求解2-tg得取最小值时，y1v2183.1139760315.1取最大值时，y2v7817.2463071538.4由-sinvrsin2yxy2v222yQQ-cossinvrcos2vyxx2vv22xQQ可计算出当v6891574.0vv724611.0vxy1，时，6891514.0vv724611.0vxy2，时，合速度vvv2y2xV3—3解：设点源强度为Q，根据叠加原理，流动的函数为xa3-yarctg2axyarctg2axyarctg2两个速度分量为222222a3-yxxyaxaxyaxax2x222222ya3-yxa3-yyaxyyaxy2v对于驻点，0vvyx，解得a33y0xAA，3—4解：设点源的强度为Q，点涡的强度为T，根据叠加原理得合成流动的位函数为Q2lnr22r1r12r1rrVV；速度与极半径的夹角为QarctgarctgrVV3—5根据叠加原理得合成流动的流函数为yayyaarctgayyaarctgV两个速度分量为1yv2222xyaxaxayaxaxaV2222yyvyaxyyaxyaV由驻点0a30，得驻点位置为yxvv零流线方程为0ayyaarctgayyxaarctgyVV对上式进行改变，得aytanay2ayx222当0x时，数值求解得a03065.1y3—9解：根据叠加原理，得合成流动的流函数为ayyarctg2ayyarctg2yvQQ速度分量为2222xyaxax2yaxax2yvvQQ2222yyaxax2yaxax2vQQ由0vvyx得驻点位置为0vaa2，Q过驻点的流线方程为0ayyarctg2ayyarctg2yvQQ上面的流线方程可改写为ayyarctgayyarctgyv2Q222ayxay2ayyarctgayyarctgtanyv2tanQ容易看出y=0满足上面方程当0y时，包含驻点的流线方程可写为Qyv2tanay2ayx222当12vaQ时，包含驻点的流线方程为tanyy21yx223—10解：偶极子位于原点，正指向和负x轴夹角为，其流函数为22yxxsinycos2M当45时22yxxy222M3—11解：圆柱表面上的速度为a2sinv2v222222a4a2sinv4v222222va4av2sin4sin4vv压强分布函数为222pvasin41sin41vv1C第四章4—1解：查表得标准大气的粘性系数为nkg1078.1u565el1023876.11078.16.030225.1uLVR平板上下两面所受的总得摩擦阻力为NSVLRF789.021e664.02224—2解：沿边阶层的外边界，伯努利方程成立代表逆压梯度代表顺压梯度，时；当时当0m0m00m00mmvvv21p12201002xpxpxvxvxvxxpcmmm4—4解：（a）将2xy21y23vv带入（4—90）中的第二式得28039dyvv1vv0xx由牛顿粘性定律uu23yvu0yxw下面求动量积分关系式，因为是平板附面层0dxdv积分关系式可表示为dxdv2w将上述关系式代入积分关系式，得vdxud14013边界条件为x=0时，0积分上式，得平板边界层的厚度沿板长的变化规律64.428039646.0xxx64.4llRR（b）74.164.483xx83dyvv1l0xR（c）由（a）知64.4xxlR（d）646.0xx646.0v21324xx64.4u23lfl2wflwRCRCR）得—由（；（e）单面平板的摩擦阻力为292.1xx292.1sv21bbdxv