气体动力学基础456章答案

14.3解：最大速度和单位长度上的压差定常不可压层流动则其中相应速度分布应为抛物线型分布（在相同的横截面上），若达到最大速度则相应的受力达到平衡则可得到如下方程：drdVrlPPdrdVlPPArlAPAP2sinsin2)sin()sin(0sin)(2221122112121则又4.4解：5631110*88.210*13.110*305*067.1dVRe水3632110*59.110*20510*305*067.1dVRe油4.5解：222044dqVVdVrqVVmqddqVdRVVe512.010*08.6*14.3*232010*67.5*4232042320463则4.7证明：在层流流动中dVdrdrdV在两固定平面之间，则由ylPPblbyPbyP2022121则lPyVylyyPVydylPdVdydVyyV404)(220max22000时，有则)3/(42020000yylPbyVbdybyqVyVmax3/2VV24.11解：沿层损失（层流）222222322322642eeeeefgdVlgVdVdlgVdRlgVdlh流量一定，面积一定则ed最大时沿层损失最小AbaabAde224当a+b最小时ed最大，即a=b时取得minfh4.12解：4222212832322dlqAdlqVdlVVdlpVV流量比为16)(42121ddqqVV湍流时设为光滑的22Vdlp356.6)(75.125.122121ddQQ4.15解：由题意可知，321,,lll三段流量相等由33305.0VAqV21,ll同为粗糙段4.18解：由题意可知，3l段流量为slqVT/102l段流量为slqV/201l段流量为slqqqVVT/301由4.17题可知22222)3(gAldSqqqqShVVVTVTW321gVdl2211112233332WhgVdl=2.09575+2.4839+5.88765=10.46734.19解：smdVAVqV/1)*4//(0708.00708.02段压力损失同理可得得段在粗糙管区，由则可知212333333585.034356333233,?20196.0)7.351.2lg(2110*56.9)2/(416010*6.9)/(8010*23.410*003.115.0*83.2/83.2405.0llgVdlhdRldddVRsmdVwee445.0/75232010*667.13eeRVdR层流paPVdl5210*56.325.3解：由aPP*则aPVP2215210*032.021VapTCVTT22*12*)211(kkaMkPP2*211aMkTTVTAVqm5.4解：绝能等熵流动*T不变，*P不变pCVTT2211*pCVT22222T=322.2KaKKPPTTPP5*111*11*110*5439.1）（由aKKPPTTPP5212*22*210*5008.1）（由由AVqm得1111VATRPg2222VATRPg23210*52.2mA（1）FNAPAPVVqm26.210)(221112（2）8.0111TKRVMga2.0222TKRVMga（3）*T不变，*P不变5.6解：绝能等熵流动*T不变，*P不变5aPP2KTTTPPKK4.210212*22*2）（由由pCVTT22*smV/88.39425.14解：AqTPKqm)(**得11*1*1)(AqTPK=22*2*2)(AqTPK9529.0)(11q查表得由645.08461.0)(22查表得q5.16解：由121*11111)(TKKTT由动量方程799.0)4.0/14.0(1200/600)()(21*2*12zTTz第六章6.6解：扰动被限制在马赫锥内由030/1sin得aM则扰动被限制在以扰动源为顶点半顶角为30度的马赫锥内6.7解：0647.021120133.112*112*2KKaMKPPPPPP）（查气动函数表，得44.22aM查膨胀波表，得028.37)(aM以右伸波为例，则12)(2aM-)(1aM=37.8-27.14=10.666.8解：由题意为左伸波查膨胀波表0219.33)(aM|)(2aM-)(1aM|=019.336012174.9645.2619.33901arcsinaM6.9解：96.102.25)(15)(02.10)(435.12.347/34.498/2.3473002201201111111aaaaaaMMMMMCVMKRTCKT）（则以左伸波，由96.11arcsin15435.11arcsin02159.198-29.247=29.950KTMKMKTTaa68.2402112112222112由aKKaaPPMKMKPP52122211210*4077.0211211）（由6.12解：3//2112VV和121112111222221112)3(31PVPVPPVVPP4.13又3)1(2)1(13.411122121122112aaaMKMKkkMKKPP961.1,22.11aMK由557.0112122212122aaaaMMKKKMM376.1211211222112aaMKMKTT由6.13解：查表得：0020,42.53由smVVV/378)cos(cos2216.14解：smV/07.57426.16解：4.294;10;4.7221511TPV7则：1.2/95.34311111CVMKRTCa则677.1)1(212111aaMKKM则667.125613.0)1()1(22222KKMa由aaPPKKMKKPP52211210*978.41112由KTMKKKMKMTTaaa21.521)211)(21()12(122121221129.2565613.0222KRTCMVa6.21解：同例6.6253.1aM6.22解：8112112121*1*2]1112[])1(2)1([KaKKaaKKMKKMKMKPP