河南城建学院专科毕业设计(论文) 工艺设计
2.1.13 氨分离器物料计算
进器物料：氨分离进器总物料等于水冷器出器气液混合物总物料。即V11 = V11
出
+L 11出=9074.864+907.486=9982.35 m3
出器物料：气液混合物在器内进行分离，分别得到气体和液体。 出器气体：
V12 = V11出=9074.864 m3 ， 出器液体：
L 15=L 11出=907.486 m3 ，
氨分离器出口气体放空V13=130.76 m3 其中： NH3 V13 NH3 =130.76?0.092=12.030 m3
CH4 V13CH4 =130.76?0.1705=22.295 m3 Ar V13AR =130.76?0.0353=4.616 m3 H2 V13H2 =130.76?0.5367=70.179 m3 N2 V13N2 =130.76?0.1655=21.640 m3
2.1.14 冷交换器物料计算
进器物料：进器物料等于氨分离器出口气体物料减去放空气量：
V14 = V12–V13=9074.864-130.76=8944.104 m3 其中：
NH3 V14 NH3 =8944.104?0.0920=822.858 m3
CH4 V14CH4 =8944.104?0.1705=1972.175 m3 Ar V14AR =8944.104?0.0353=584.050 m3 H2 V14H2 =8944.104?0.5367=4174.213 m3 N2 V14N2 =8944.104?0.1655=1390.808 m3 出口物料（热气）：
设热气出口温度17℃，查t=17℃，p=30Mpa，气相中平衡氨含量y*=5.5﹪，计算热气出口冷凝液氨量时，忽略溶解在液氨中的气体。取过饱和度10﹪，故V17NH3=5.5﹪?1.1=6.05﹪.
设热气出口氨体积为a，则：
a 5 解得a=522.975 m3 ?0.0608944.1?0452?2.a858冷交换器热气冷凝液氨量为：
L 17NH3=V14NH3 –a=822.858-522.975=299.883 m3
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河南城建学院专科毕业设计(论文) 工艺设计
冷交换器热气出口气量及组分 其中：
NH3 V17NH3=V14NH3–L17NH3 =522.975 m3
CH4 V17CH4= V14CH4?L17CH=1972.175 m3
4Ar V17Ar = V14Ar?L17Ar=584.050 m3
H2 V17H2 = V14H2?L17H=4174.213 m3
2N2 V17N2 = V14N2?L17N2=1390.808m3
出口总气量：
V17=V14–L17NH3=8944.104-299.883=8644.221 m3 出口气体各组分： NH3 V17NH3 /V17=
CH4 V17CH4 /V17= Ar V17AR /V17 = H2 V17H2 /V17= N2 V17N2 /V17=
522.975?100%?6.05%
8644.2211972.175?100%?22.81%
8644.221584.050?100%?6.76%
8644.2214174.213?100%?48.28%
8644.2211390.808?100%?16.10%
8644.2212.1.15 氨冷器物料计算
进器物料：氨冷器进器物料等于冷交换器出器物料加上补充新鲜气物料
V1=2980.23 m3
其中：CH4 V1CH4 =2980.23?0.0129=38.44m3
Ar V1AR =2980.23?0.0038=11.32m3 H2 V1H2 =2980.23?0.7316=2180.34m3 N2 V1N2 =2980.23?0.2517=750.34m3
V18（进器气体物料）=V1+V17=2980.23+8644.22=11624.45m3 进器气体组分含量V18i=V1i+V17i
其中：NH3 V18 NH3 = V17NH3=522.975m3
CH4 V18CH4 =38.44+1972.175=2010.615m3 Ar V18Ar =11.32+584.050=595.37m3 H2 V18H2 =2180.34+4174.213=6354.553m3
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河南城建学院专科毕业设计(论文) 工艺设计
N2 V18N2 =750.12+1390.808=2140.924m3 各组分百分含量y18i=V18i /V18
522.975其中：NH3 y18 NH3 =?100%?4.50%
11624.45CH4 y18CH4 =Ar y18AR =H2 y18H2 =N2 y18N2 =
2010.615?100%?17.30%
11624.45595.37?100%?5.12%
11624.456354.553?100%?54.67%
11624.452140.928?100%?18.41%
11624.45进器液体等于冷交换器冷凝液氨量：
L 18= L 18NH3 = L 17NH3 =299.883m3
进器总物料= V18 + L 18 =11624.45+299.883=11924.333m3
出器物料：已知出器气体中氨含量为2.00﹪，设出器气体中氨含量为b m3，
b?0.02
11624.45?522.975解b =222.030m3 则氨冷器中冷凝液氨量：
L18= V18–b =522.975-222.030=300.945m3 氨冷器出器总液氨量：
L 2NH3 = L 18NH3+ L 18NH3 =299.883+300.945=600.828m3 氨冷器出器气体量：
V2=V18 –b=11624.45-300.945=11323.505m3 其中：NH3 V2 NH3 =222.030m3
CH4 V2CH4 =V18CH4=2010.615m3 Ar V2AR =V18AR=595.370m3 H2 V2H2 =V18H2=6354.553m3 N2 V2N2 =V18N2=2140.924m3
各组分百分含量y2i=V2i /V2
222.030其中：NH3 y2 NH3 =?100%?1.96%
11323.505CH4 y2CH4 =
2010.615?100%?17.76%
11323.505 23
河南城建学院专科毕业设计(论文) 工艺设计
Ar y2AR =H2 y2H2 =N2 y2N2 =
595.370?100%?5.25%
11323.5056354.553?100%?56.12%
11323.5052140.924?100%?18.91%
11323.505出器总物料= V2 + L 2NH3 =11323.505+600.828=11924.333m3
2.1.16 冷交换器物料计算
进器物料：冷交换器进器总物料等于氨冷器出器总物料。其中气体入口V2 =11323.505m3，液体入口L 2NH3 =600.828m3，由气液平衡计算得：以1kmol进口物料为计算基准： 即F =1
(1)??V?L?F ?
L?V?F(2)yN3HmN?33H?xNH将yNH3?0.02,xNH3?0.9852代入上式，
V?xNH3?mNH3xNH3?yNH3
mNH30.9652得
0.9852?mNH30.9852?0.02?1.021?
式中的mNH可由物料平衡和氨平衡计算mNH3?3V2NH3V2
?V??V?V?117?2?? V?V?V?L?1781315?V??V?2NH32NH3?L17NH3?L18NH3??式中V2?—冷交入口总物料；
?—冷交热气出口总物料； V17V2?NH—冷交入口总氨物料；
3将V8?9982.35m3，V13?130.46m3， L15?970.486m3 ??9982.35-130.76-970.486=8944.104m3V17
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河南城建学院专科毕业设计(论文) 工艺设计
?∴ V2=2980.23+8944.104=11924.334m3
V2,NH3=222.03+299.883+300.945=822.858m3
'V2NH3822.858??0.069 代入（3）式得 ∴mNH3?'11924.334V2V?1.021?0.069?0.949 L?1?V?0.051
0.9652L0.051??0.054 V0.949?L?由??可求出冷交换器冷凝液体量 。 ?V?L16?L?????0.05 V3?V?冷凝液体量
L=凝液体量。
V3 = V2?–L 16?11924.333?596.217?11328.116m3 其中：NH3 V3 NH3 =11328.116×0.02=226.562m3
CH4 V3CH4 =11328.116?0.1519=1720.741m3 Ar V3AR =11328.116?0.0448=507.499m3 H2 V3H2 =11328.116?0.5877=6657.534m3 N2 V3N2 =11328.116?0.1965=2215.779m3
V?V11328.116?11408.41计算误差：35?100%=??0.17%
V311328.116校核氨分离器液氨百分数
L15x15?100% G分﹪=
L15x15?L16x16=
907.486?0.9777?60.167%
907.486?0.9777?596.217?0.98523=0.05×11924.333=596.217m3
出器物料：冷交换器（冷气）出口气体物料等于进口总物料减去冷
2.1.17 氨贮槽物料计算
进槽物料：氨分离器入槽液体L 15 =8907.486m3 其中：NH3 L15 NH3 =907.486?0.9777=887.249m3
CH4 L15CH4 =907.486?0.0117=10.617m3
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