第六章 能量损失及管路计算
第六章 能量损失及管路计算 6-8 一旧铸铁管长l=30m,管径d=0.3m。管中水流速度v=1.5m/s,水温。t,20C
试计算沿程损失。
2lv0.3,,,0.35mh 解:根据谢维列夫公式:, ,,,0.021/0.03df2dg
,626-9 直径d=250mm的铸铁管。当量粗糙度。用它输送,,,1.310m/s,,0.5mm的水。分别计算流动处于水力光滑区的最大输水量和阻力平方区时的最小流量。
d8/7 解:当流动处于水力光滑区时: Re26.98()32778,,max,
vd,2max?,v0.17m/s,, Re,Qdv,,8.34L/smaxmaxmaxmax,4
d0.85当流动处于阻力平方区时: Re4160()454300,,min,2
vd,23min?,v2.36m/s,, Re,Qdv,,0.12m/smaxminminmin,4
6-10 某水管直径d=0.5m,,水温。分别用公式法和查图法确定15C,,0.5mm
333Q,2m/s流量分别为,,时的沿程阻力系数。 Q,0.005m/sQ,0.1m/s,312
,62 解:1)时, t,15C,,,1.13910m/s
3v,0.025m/s公式法:, Q,0.005m/s11
dvd8/71,, Re11178,,26.98()72379.1,,,
d8/7位于水里光滑管区。 ?,,4000Re26.98(),
,0.237 ,,,,0.00320.221Re0.056
查图法: ,,0.06
3v,0.51m/s2)公式法:, Q,0.1m/s22
dvd58/72, Re2.23610,,,26.98()72379.1,,,
d8/7,位于水力光滑区。 ?,,4000Re26.98(),
,0.237 ,,,,0.00320.221Re0.04
查图法: ,,0.03
3Q,2m/sv,10.2m/s3)公式法:, 33
vddd68/70.8553,, Re4.510,,,26.98()72379.1,4160()8.210,,,,,2
dd8/70.85,位于第二过渡区。 ?,,26.98()Re4160(),,2
19.35,,,,1.142lg(), ?,,0.025dRe,,
查图法:,,0.025
6-14 如图6-22所示,用一直径d=20mm、长l=0.5m的管段做沿程阻力实验。当
,33,62Q,,1.210m/s的水以通过时,两侧压管液面高差h=0.6m,,,,0.8910m/s
,试计算。若流动处于阻力平方区,确定当量粗糙度。 ,
2lvQh,, 解:,,,流动处于阻力平方区 v,,3.82m/s?,,0.032fdg2A
d,2, ,,,(1.142lg)?,,0.12mm,
d,225mm,,0.025l,7m6-17 泵送供水管如图6-25所示。已知吸水管,,,111
l,50m,,0.028H,45m,,4d,200mm;排水管,,。。设水泵的扬程H1211c
2HQ,,652500与流量Q的关系为。不计其他局部损失,该管路每昼夜的供水量是多少,
2222lvvlvv2111222 解:由
题
快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题
意可得:, H,,,,,,,,,HHQ,,,652500C112dggdgg222212
,,,2223v,0.045m/sdvdvQ,解得:, ,,?,,Qdv0.081m/s1112211444
3Q,,,2436006975m所以,供水量=