6av6372-2dj07-5aa0顺控系统扩展包
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一个密闭---的无尘车间，在使用过程中，主要的漏风途径有以下四种：1门窗缝隙的漏风；开门时的漏风；风淋室、传递室的漏风；室内工艺排风。下面介绍各种漏风量计算方法：1缝隙漏风量的计算方式一：v=1.29*(p)1/2v=s*vp：无尘车间内外压力差pv：从缝隙处流过的风速(m/s)s：缝隙面积m2v：通过缝隙的泄漏风量m3/h例：假设条件：房间正压2pa，门缝长度3.6m，窗缝长度4m，假设缝隙宽度.2m门缝隙面积s1=.2*3.6=.72m2，窗缝隙面积s2=.2*4=.8m2泄漏风量v=s*v=(s1 s2)*36*1.29*(p)1/2=(.72 .8)*36*1.29*1/2=3157m3/hr方法二：l=.827ap)1/21.25=1.3375ap)1/2压差法计算方式式中l正压漏风量(m3/s)；.827漏风系数；a总有效漏风面积(m2)；p压力差(pa)；不严密处附加系数2开门的泄漏风量假设条件：房间正压p=2pa，门面积s=.9*2.=1.8m2，风速v=1.29*(p)1/2=5.77m/s,开启次数n=1次/hr，开启时间t=5s泄漏风量q=s*v*t*n=1.8*5.77*5*1*=51.93m3/h每小时开一次门，开5秒，泄漏空气量51.93m3/h3风淋室与传递窗的空气泄漏量假设条件：风淋室体积15m3，密闭无缝隙开闭顺序为开关开关分析：以风淋室为例开时a/s内压力为常压11325pa关时a/s内压力为常压11325pa，保持不变开时a/s内压力为常压11325pa，保持不变，但稳定后压力变为与房间内同样压力11325 2pa关时a/s内压力为常压11325 2pa,保持不变结论：所以需要补充的空气量为15m3的空间压力变化为2pa时的密闭空间的量根据理想气体方程式pv=mrt，p气压，单位pa；v体积，单位m3；m气体单位kg；r气体常数等于287；t气体开氏温度，单位k，假设室内温度25℃，开氏温度为298k需补充空气m=p*v/r*t=(2*15)/(287*298)=.35kg=3l相对来说风淋室及传递箱的泄漏量较小密闭---的情况下，可忽略不计。
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