空压机出气点压缩空气温度13度,湿度9%,露点-19度,请问含水量是多少,如何计算?

2025-01-23 06:31:11
推荐回答(1个)
回答1:

还要知道排气量和压力才可以算的出的,列举一个例子,你参照计算吧
1.在大气温度30℃,相对湿度70%的条件下,14.3m3/min的空压机:
24小时吸入水量=g1*70%*14.3*60*24=30.38*70%*14.3*60*24=437.91kg。
( 由大气压力露点/水份含量表查出30℃下含水量g1为30.38g/ m3)
2.通过冷冻式干燥机后的压力露点大概为15℃,在压力0.7MPa下:
通过冷干机后24小时含水量= g2*14.3*60*24=1.876*14.3*60*24=38.63kg
(在此温度下大气露点为-13℃,由大气露点/水份含量表查出g2为1.8764g/ m3。.)
3.通过吸附式干燥机后压力露点为-35℃,在压力0.7 MPa下:
通过吸干机后24小时含水量=g3*14.3*60*24=0.04*14.3*60*24=0.824kg
(在此压力露点下大气露点为-53℃,由大气露点/水份含量表查出g3为0.04g/m3。.)
以上计算的是压缩空气中的饱和含水量,除了以上38.63Kg的水通过冷冻式干燥机进入后压缩空气管道外,其余378.93Kg水中除了一部分被过滤器、冷干机、贮气罐的排水阀排除外,还有相当一部分也进入了后压缩空气管道,经过温差的不断变化,冷冻式干燥机后除了潮湿的压缩空气以外,还有大量的液态水出现,对设备及生产带来了极大的危害。因此只有通过吸附式干燥机才能从根本上将压缩空气中的水份吸附排除,从而从根本上解决压缩空气中的水份对设备及生产的危害。

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