第一章 绪论1-1.20℃的水2.5m3,当温度升至80℃时,其体积增加多少?[解] 温度变化前后质量守恒,即(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)又20℃时,水的密度(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)80℃时,水的密度(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)则增加的体积为(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)1-2.当空气温度从0℃增加至20℃时,运动粘度(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)增加15%,重度(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)减少10%,问此时动力粘度(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)增加多少(百分数)?[解] (rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)此时动力粘度(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)增加了3.5%1-3.有一矩形断面的宽渠道,其水流速度分布为(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2),式中(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)、(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)分别为水的密度和动力粘度,(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)为水深。试求(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)时渠底(y=0)处的切应力。[解] (rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)当(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)=0.5m,y=0时(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)1-4.一底面积为45×50cm2,高为1cm的木块,质量为5kg,沿涂有润滑油的斜面向下作等速运动,木块运动速度u=1m/s,油层厚1cm,斜坡角22.62 (见图示),求油的粘度。(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)[解] 木块重量沿斜坡分力F与切力T平衡时,等速下滑1-5.已知液体中流速沿y方向分布如图示三种情况,试根据牛顿内摩擦定律,定性绘出切应力沿y方向的分布图。(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)[解]1-6.为导线表面红绝缘,将导线从充满绝缘涂料的模具中拉过。已知导线直径0.9mm,长度20mm,涂料的粘度(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)=0.02Pa.s。若导线以速率50m/s拉过模具,试求所需牵拉力。(1.O1N)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)[解] (rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)1-7.两平行平板相距0.5mm,其间充满流体,下板固定,上板在2Pa的压强作用下以0.25m/s匀速移动,求该流体的动力粘度。[解] 根据牛顿内摩擦定律,得1-8.一圆锥体绕其中心轴作等角速度旋转。锥体与固定壁面间的距离=1mm,用的润滑油充满间隙。锥体半径R=0.3m,高H=0.5m。求作用于圆锥体的阻力矩。(39.6N·m) [解] 取微元体如图所示微元面积:切应力:阻力:阻力矩:(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)1-9.一封闭容器盛有水或油,在地球上静止时,其单位质量力为若干?当封闭容器从空中自由下落时,其单位质量力又为若干?[解] 在地球上静止时:因为 代入柯列勃洛克公式(5-35)得㏒ = -2㏒()所以 = 检验: (rho )_(1)(V)_(1)=(rho )_(2)(V)_(2) (rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)因为,属于过渡区,故假定正确,计算有效。5-16 混凝土排水管的水力半径(rho )_(1)(V)_(1)=(rho )_(2)(V)_(2)。水均匀流动1km的水头损失为1 m,粗糙系数,试计算管中流速。