您好,欢迎访问三七文档
船舶阻力与推进(2)参考答案第三章兴波阻力1.某长江双桨客货船水线长108m,方形系数0.594,中横剖面系数0.97。试用圆圈P理论判别航速15.5kn、16.7kn、19.5kn时阻力是处于峰值还是谷值。解:依题意得:612.097.0594.0MBPCCC当m/s973.7kn5.15V时,767.0108793.7LV根据767.0612.0LVCP,查图3-14知:阻力处于谷值当m/s590.8kn7.16V时,827.0108590.8LV根据827.0612.0LVCP,查图3-14知:阻力处于谷值当m/s031.10kn5.19V时,965.0108031.10LV根据965.0612.0LVCP,查图3-14知:阻力处于峰值2.某沿海货船垂线间长86m,服务航速10.8kn,最大航速11.62kn,棱形系数0.757.试问该船在上述两个航速下圆圈P值为多少?是否属于有利范围。解:当m/s556.5kn8.10V时551.014.3286757.08.9556.52LgCVPP根据599.086556.5757.0LVCP,查图3-14知:有利干扰当m/s977.5kn62.11V时593.014.3286757.08.9977.52LgCVPP根据645.086977.5757.0LVCP,查图3-14知:不利干扰3.已知某船船长100m,船宽12m,吃水6m,横剖面面积曲线是关于船长的一个分段函数501570)15(352152070205070)20(90722xxxxxAs,当航速为15kn时,求船体首尾横波相位差。解:该船的型排水体积为:315205015220502m33.548370)15(3527070)20(907dxxdxdxxdxAs中横剖面面积:270mAM故有:783.01007033.5483LACMPmgV152.388.95144.01514.32222根据43)(qnLCP得:43LCqnP(其中n为正整数,q为正分数)05.2152.38100783.0LCP故:54280.275.005.2qn所以:542qn,此时船体首尾横波相位差为582q第四章附加阻力1.已知某海洋客货船在中横剖面上主船体的投影面积25㎡,上层建筑的投影面积68㎡,当船速为30km/h,逆风风向角20°,风力为4级时,试估算船的空气阻力。解:空气阻力:221ataaaaVACR其中3/226.1mkga风力为4级时,查表4-2知:smuw/9.7~5.5取中间值smuw/7.6船速:smhkmV/33.8/30故有:smuVVwa/63.1420cos7.633.802936825mAt参考课本220p空气阻力系数平均值:若按客船取31009.0aaC则:NVACRataaaa098.163.1493226.1211009.021232若按普通货船取3101.0aaC则:NVACRataaaa220.163.1493226.121101.021232第五章船模阻力试验1.某海船船模长度6m,沾湿面积10m2,航速0.8m/s,测得阻力为48N。若缩尺比为25,求实船在相当速度时的总阻力。其中,摩擦阻力系数按普朗特—许立汀公式计算,粗糙度补贴系数取为0.410-3。解:假设℃15t,查表可得:smvmkgss/10188.1/91.1025262,smvmkgmm/10139.1/04.999263,当缩尺比25时,有:mmLLms150625222262501025mmSSmssmsmVVms/4/8.0256610214.410139.168.0RemmmmvLV8610051.510188.11504RessssvLV0035.010214.4lg455.0Relg455.058.2658.2mfmC0017.010051.5lg455.0Relg455.058.2858.2sfsCNVSCRmmmfmfm19.118.01004.999210035.02122NVSCCRsssffsfs523210077.14625091.102521104.00017.021)(实船在相当速度下的总阻力为:NRRRRmsfmtmfsts535310983.62504.99991.102519.114810077.12.设船模长度为2m,沾湿面积为1m2,当速度为1.5m/s时测得总阻力为10N。假设由于测量误差的存在使得船模总阻力增大了1%,求当缩尺比分别为50和100时,实船的总阻力将产生多大的相对误差。解:依题意可知船模实际总阻力:NRtm9.9%1110假设℃15t,查表可得:smvmkgss/10188.1/91.1025262,smvmkgmm/10139.1/04.999263,6610634.210139.125.1RemmmmvLV0038.0210634.2lg075.02Relg075.0262mfmCNVSCRmmmfmfm27.45.1104.999210038.02122当50时:mmLLms10025022222500150mmSSmssmsmVVms/6.10/5.1508610923.810188.11006.10RessssvLV0016.0210923.8lg075.02Relg075.0282sfsCNVSCCRsssffsfs523210882.26.10250091.102521104.00016.021)(当NRtm10时,有:NRRRRmsfmtmfsts635310024.15004.99991.102527.41010882.2当NRtm9.9时,有:NRRRRmsfmtmfsts635310011.15004.99991.102527.49.910882.2相对误差:%29.1%10010011.110011.1024.1%100%66tstststsRRRR当100时,有:mmLLms20021002222100001100mmSSmssmsmVVms/15/5.11009610525.210188.120015RessssvLV0014.0210252.2lg075.02Relg075.0292sfsCNVSCCRsssffsfs623210077.2151000091.102521104.00014.021)(当NRtm10时,有:NRRRRmsfmtmfsts636310961.710004.99991.102527.41010077.2当NRtm9.9时,有:NRRRRmsfmtmfsts636310858.710004.99991.102527.49.910077.2相对误差:%31.1%10010858.710858.7961.7%100%66tstststsRRRR3.某海船船模长度为5m,沾湿面积为10m2,缩尺比为25,当速度为1.5m/s时测得船模总阻力为40N。取粗糙度补贴系数0.410-3,摩擦阻力按普朗特—许立汀公式计算,用二因次法求实船在相当速度下的总阻力。若取形状系数0.022,用三因次法求实船阻力,并与二因次法相比较。解:假设℃15t,查表可得:smvmkgss/10188.1/91.1025262,smvmkgmm/10139.1/04.999263,当缩尺比25时,有:mmLLms125525222262501025mmSSmssmsmVVms/5.7/5.1256610585.610139.155.1RemmmmvLV8610891.710188.11255.7RessssvLV0032.010585.6lg455.0Relg455.058.2658.2mfmC0016.010891.7lg455.0Relg455.058.2858.2sfsCNVSCRmmmfmfm97.355.11004.999210032.02122NVSCCRsssffsfs523210607.35.7625091.102521104.00016.021)(0036.05.11004.99921402122mmmtmtmVSRC二因次法:NRRRRmsfmtmfsts535310254.42504.99991.102597.354010607.3三因次法:0024.0104.0)0016.00032.0()022.01(0036.0)1(3ffsfmtmtsCCCkCCNNVSCRssststs5210328.45.7625091.1025210024.021比较可得:%74.1%10010254.410)254.4328.4(%55tsR(注:以上答案仅供参考,如有错误,望批评!)
本文标题:船舶阻力习题答案
链接地址:https://www.777doc.com/doc-2120165 .html