如图所示,输送液体只计沿程水头损失,当H、l一定时,试证明hf=H/3条件下,管路输送功率最大。若已知H=127.4m,l=500m,管路末端可用水头为,管路末端可用功率1000kW,λ=0.024,试求管路的输送流量与直径。
根据气力输送的特点,降低动力消耗,节约能源是一个重要的研究课题,从经验中知,管道内径是关于能耗的重要参数。直观上看,当输送量一定时,管径过小,输送易阻滞,管径过大,虽输送畅通,但又造成能量的浪费。根据经验,把管径分成三组,各组的试验结果如下表所示,试用方差分析法比较各组的效果.
管径(mm) | 单位功耗 |
230 | 0.0308,0.0476,0.0504 |
250~260 | 0.0532,0.032,0.0218,0.028,0.028, 0.042,0.0336,0.042,0.042,0.028 |
280~320 | 0.07,0.07,0.0644,0.0312,0.0756,0.0756, 0.07,0.0588,0.0588,0.042,0.0308,0.0364, 0.0448,0.21,0.154,0.1064,0.1288,0.112, 0.1064,0.1288,0.0756,0.0644,0.0504,0.0644, 0.0504,0.0308 |
从水箱下部引一管道,管道水平段上有一收缩段。从收缩段引出的玻璃管插入容器A的水中(图3—58)。已知管径d1=4cm,收缩段直径d2=3cm。水箱至收缩段的水头损失hω1=3v2/2g,收缩段至管道出口的水头损失hω2=v2/2g(v为管道流速)。当水流通过管道流出时,玻璃管中水柱高h=0.35m,求作用水头H(取动校正系数为1)。
(扬州大学2004—2005学年第2学期期末考试试题)如图6—44所示,有一长度L=10m的有压管道,管径d=0.1m,作用水头H=2.5m,沿程阻力系数λ=0.02,不计行进流速,管道进口及阀门的局部阻力系数分别为ζ1=0.5、ζ2=1.5。求: (1)阀门全开时管道中通过的流量。 (2)阀门由全开突然关闭时管道中的水击压强水头。(水击波波速为1000m/s)