推荐回答（4个）
由理想气体的状态方程PV=nRT,可得V=nRT/P,其中n为物质的量，当n=1时，就是气体的摩尔体积了；R=8.3144J/(mol·K），T=273.15+169.6=442.75 K, 所以V= 5.254 L/mol，又水蒸气的摩尔质量为18 g/mol，所以密度为3.4g/L.
由题意，一小时内流出的水蒸气的体积为4*3.14*300*3600=13564800L,所以质量为密度与体积之积等于46.1吨。
这是用理想气体状态方程计算的，但是因为水分子有极性等原因导致它的误差会很大~最好还是用PRO2 软件模拟下好些~
管径=400mm=0.4m 小时=3600S
每小时体积流量；
0.2平方*3.1416*30*3600=13571.7 米立方
查表在该参数下蒸汽密度是3.7公斤/m3
13571.7*3.7=50215公斤/小时=50.215吨/小时
加热室温度差=壳层压力(真空度)相应温度-加热室料液温度
蒸汽过热度=蒸汽温度-饱和蒸汽压力相应温度
水的饱和蒸汽的温度与压力的公式，
根据克劳佩斯－克劳修斯方程：
dlnP/dT=△vapHm/(RT^2),其中=△vapHm蒸发焓，P为饱和蒸汽压，T为温度
饱和蒸汽压力温度密度对照表
7kg压力是165度、密度3.6
温度 （℃） 密度 （kg/m3) 绝对 压力（Kg)
100 0.6 1.103
105 0.705 1.208
110 0.827 1.433
115 0.965 1.691
120 1.122 1.985
125 1.298 2.321
130 1.497 2.701
135 1.719 3.131
140 1.967 3.614
145 2.242 4.155
150 2.548 4.76
155 2.886 5.433
160 3.26 6.181
165 3.671 7.008
170 4.123 7.92
175 4.618 8.924
180 5.16 10.027
185 5.752 11.233
190 6.397 12.551
195 7.1 13.987
200 7.864 15.549
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