第二章流体的运动2-1.一水平圆管，粗处的直径为8cm，流速为1m/s^{-1},粗处的直径为细处的2倍，求细处的流速和水在管中的体积流量.解:(1)已知:d_{1}=8cm,v_{1}=1ms^{-1},d_{1}=2d_{2}.求:v_{2}=?根据连续性方程S_{1}v_{1}=S_{2}v_{2},\frac { \pi }{4}d_{1}^{2}v_{1}= \frac { \pi }{4}d_{2}^{2}v_{2},代入已知条件得v_{2}=4v_{1}=4(ms^{-1})(2)水的体积流量为Q=S_{1}v_{1}=S_{2}v_{2}= \frac { \pi }{4}d_{1}^{2}v_{1}= \frac { \pi }{4} \{ 8 \times 10^{-2})2-2.将半径为2cm的引水管连接到草坪的洒水器上,洒水器装一个有20个小孔的莲蓬头,每个小孔直径为0.5cm.如果水在引水管中的流速为1m/s^{-1},试求由各小孔喷出的水流速度是多少?解:已知:总管的半径r_{1}=2cm水的流速v_{1}=1ms^{-1}支管的半径为r_{2}=0.25cm,支管数目为20.求:V2=?根据连续性方程S_{1}v_{1}=nS_{2}v_{2},有\pi r_{1}^{2}v_{1}=n \pi r_{2}^{2}v_{2},代入数据,得(2 \times 10^{-2})^{2} \times 1=20 \times(0.25 \times 10^{-2})^{2}v_{2}从而,解得小孔喷出的水流速度v_{2}=3.2(ms^{-1}).2-3.一粗细不均匀的水平管，粗处的截面积为30cm^{2},细处的截面积为10cm^{2}.用此水平管排水