文档介绍：工程流体力学第二版课后****题答案工程流体力学第二版课后****题答案【篇一:工程流体力学****题答案】图3-35****题3-2示意图【解】p1gas??h2ogh1?p1airp2gas??h2ogh2?p2airp1gas??gasgh?p2gasp1air??aigh?p2airr?h2ogh1?p1air??gasgh?p2gasp2gas?p1air??h2ogh2??airgh?h2ogh1??gasgh??h2ogh2??airgh?hoh1??hoh2??airh??gash22h202o?gas??hh1?h2??air?1000?????3?图3-43****题3-10示意图【解】p1??h2og?h?h1??pp1??hgg?h2?h1??p2p3??h2og?h2?h3??p2p3??hgg?h4?h3??pa?h?h?h2o2og?h?h1??p??hgg?h2?h1??p22og?h2?h3??p2??hgg?h4?h3??pag?h?h1?h2?h3??p??hgg?h2?h1?h4?h3??pap??hgg?h4?h3?h2?h1???h2og?h?h1?h2?h3??pa?13600????????1000???3?????101325??pa?p??hgg?h4?h3?h2?h1???h2og?h?h1?h2?h3??pa?13600????????1000???3?????101325??pa?【3-15】图3-48所示为一等加速向下运动的盛水容器,水深h=2m,加速度a=。试确定:(1)容器底部的流体绝对静压强;(2)加速度为何值时容器底部所受压强为大气压强?(3)加速度为何值时容器底部的绝对静压强等于零?图3-48****题3-15示意图【解】fx?0,fy?0,fz?a?g压强差公式dp???fxdx?fydy?fzdz?dp???fxdx?fydy?fzdz????a?g?dz?ppadp???h??a?g?dz?a??p?pa???a?g???h????a?g?0??h?g?a???gh?1???g???a??p?pa??gh?1???g??p?pa??h?g?a?a?g?p?pa?h(1)p?pa??h?g?a??101325?1000?2??????pa?(2)a?g?p?pa?hp?pa?g?pa?pa?h?g??2???(3)a?g??h??0?1013252?1000?2【3-16】图3-49所示为一圆柱形容器,直径d=300mm,高h=500mm,容器内装水,水深h1=300mm,使容器绕铅直轴作等角速旋转。(1)试确定水正好不溢出时的转速n1;(2)求刚好露出容器底面时的转速n2;这时容器停止旋转,水静止后的深度h2等于多少?图3-49****题3-16示意图【解】初始状态圆筒中没有水的那部分空间体积的大小为12v??d?h?h1?(1)4圆筒以转速n1旋转后,将形成如图所示的旋转抛物面的等压面。令h为抛物面顶点到容器边缘的高度。空体积旋转后形成的旋转抛物体的体积等于具有相同底面等高的圆柱体的体积的一半v?112??dh(2)24?h1??112??dh(3)24由(1)(2),得14?d2?h即h?2?h?h1?(4)等角速度旋转容器中液体相对平衡时等压面的方程为?r222?gz?c(5)d2对于自由液面,c=0。圆筒以转速n1旋转时,自由液面上,边缘处,r?2,z?h,则?d?????2??gh?0(6)22得??由于22ghd(7)??2?30?30n160(8)602ghn1?????22ghd??d(9)(1)水正好不溢出时,由式(4)(9),得n1?602g?2h?h1?120gh?h1?d?d(10)即n1?120?????rmin???(2)求刚好露出容器底面时,h=h,则n1?602gh?602gh?60?2????rmin??d?d??(3)旋转时,旋转抛物体的体积等于圆柱形容器体积的一半v?112??dh(11)24这时容器停止旋转,水静止后的深度h2,无水部分的体积为12v??d?h?h2?(12)4由(11)(12),得11122??dh??d?h?h2?(13)244得??m?22h2??h另一侧的水位h=,闸门自动开启,试求铰链至水闸下端的距离x。图3-54****题3-21示意图【解】设水闸宽度为b,水闸左侧水淹没的闸门长度为l1,水闸右侧水淹没的闸门长度为l2。作用在水闸左侧压力为fp1??ghc1a1(1)其中hc1?h2l1?hsin?a1?bl1?bhsin?则fp1??gh2bhsin???ghb2sin?2(2)作用在水闸右侧压力为fp2??ghc2a2(3)其中hc2?h2l2?hsin?