合成氨最后的设计 - 图文(7)

2019-03-15 13:16

河南城建学院毕业论文(设计)

???V?L?F(1)??LxNH3?VyNH3?FmNH3(2) 将yNH3?0.02,xNH3?0.9776代入上式,

V?xNH3?mNH3x

NH3?yNH3得

0.9776?mNH3mNH30.9776?0.02?1.02089?0.9576(3)

式中的mV2NH3NH3可由物料平衡和氨平衡计算mNH3?V

2??V2??V1?V17???V?17??V 8?V13?L15??V2?NH3?V2NH3?L17NH?L18?3NH3式中V2?—冷交入口总物料;

V17?—冷交热气出口总物料; V2?NH3—冷交入口总氨物料;

将V8?15183.47307m3(标),V313?172.39435m(标), L15?1561.46883m3(标)

V17??15183.47307-172.39435-1561.46883 =13449.60989 m3(标)

∴V2?=4199.84482+13449.60989=17649.45471 m3(标)

V2?NH3=334.66191+46.03371+535.1906=915.88622 m3(标)

∴mV2?NH3915.88622NH3?V2??17649.45471?0.05189代入(3)式得

V?0.9667L?1?V?0.0333

??L?0.0333?V???0.9667?0.03445

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河南城建学院毕业论文(设计)

?L?由??可求出冷交换器冷凝液体量 。 ?V?L16?L?????0.03445 V3?V?冷凝液体量

L16?0.03445V3?0.03445?17075.67272?588.2569252m3(标)

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凝液体量。 V3 = V2?–L 16 =17649.45471-588.25693=17061.19779m3(标)

其中 NH3 V3 NH3 =17061.19779?0.02=341.22369m3(标)

CH4 V3CH4 =17061.19779?0.11719=1999.40177m3(标) Ar V3AR =17061.19779?0.03281=559.7779m3(标) H2 V3H2 =17061.19779?0.6225=10620.59562m3(标) N2 V3N2 =17061.19779?0.2075=3540.19854m3(标) 计算误差:

V3?V5?100%=-0.08477% V3校核氨分离器液氨百分数 G分﹪=

1561.46883?0.97054L15x15 ?100%=

1561.46883?0.97054?588.2569252?0.9776L15x15?L16x16 =72.491%

冷交分离液氨百分数:G冷﹪=1–G分﹪=1-72.491=21.509% 计算误差=

G??G分0.71849?0.72491?100%?=-0.88563% ?G0.724913.3.13 氨贮槽物料计算

进槽物料:氨分离器入槽液体L 15 =1561.46883m3(标) 其中 NH3 L15 NH3 =1561.46883?0.97054=1515.46796

CH4 L15CH4 =1561.46883?0.0107=16.70772m3(标) Ar L15AR =1561.46883?0.00136=2.1236m3(标) H2 L15H2 =1561.46883?0.01264=19.73697m3(标)

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河南城建学院毕业论文(设计)

N2 L15N2 =1561.46883?0.00476=7.4326m3(标) 冷交换器入槽液体L 16 =588.2569m3(标) 其中 NH3 L16 NH3 ==575.07995m3(标)

CH4 L16CH4 =588.2569?0.00414=2.43538m3(标) Ar L16AR =588.2569?0.00107=0.62943m3(标) H2 L16H2 =588.2569?0.01193=7.0179m3(标) N2 L16N2 =588.2569?0.00526=3.09423m3(标)

入槽混合物料:L 21 = L 15 + L 16 =1561.46883+588.2569=2149.72573m3(标) 各组分物料含量:L 21i = L 15i + L 16i

其中 NH3 L21 NH3 =1515.46796+575.07995=2090.5464m3(标)

CH4 L21CH4 =16.70772+2.43538=19.1431m3(标) Ar L21AR =2.1236+0.62943=2.75303m3(标) H2 L21H2 =19.73697+7.0179=26.75487m3(标) N2 L21N2 =7.4326+3.09423=10.72683m3(标) 百分含量: x21NH3 = L 21i/ L 21

2090.54764?97.247%

2149.7257319.1431?0.89% CH4 x21CH4 =

2149.725732.75303?0.128% Ar x21AR =

2149.7257326.75487?1.245% H2 x21H2 =

2149.7257310.72683?0.49% N2 x21N2 =

2149.72573其中 NH3 x21 NH3 =

出槽物料:液氨贮槽出口驰放气 V20 =158.32505m3(标) 其中 NH3 V20 NH3 =15832505?0.62641=99.17639m3(标)

CH4 V20CH4 =15832505?0.11639=18.7245m3(标) Ar V20AR =15832505?0.017642.72066m3(标) H2 V20H2 =15832505?0.17184=2.79285m3(标) N2 V20N2 =15832505?0.06772=10.52177m3(标)

出口液氨总物料L 19 = L 21–L 20 =2149.72573-158.32505=1991.40025 m3(标) 其中 NH3 L19 NH3 = L 21NH3–L 20NH3 =2090.54764-99.17639=1991.37125 m3(标)

33

河南城建学院毕业论文(设计)

CH4 L19CH4 = L 21CH4–L 20CH4 =19.1431-18.7245=0.71565 m3(标) Ar L19AR = L 21AR–L 20AR =2.75-2.72066=0.02934 m3(标) H2 L19H2 = L 21H2–L 20H2 =26.75487-20.738=6.01687 m3(标) N2 L19N2 = L 21N2–L 20N2 =10.72683-10.52177=0.20506 m3(标) 各组分百分含量:

其中 NH3 x18 NH3 =

1991.371251991.40025?99.652 %

CH4 x18CH4 =

0.715651991.40025?0.035 % Ar x18AR =

0.029341991.40025?0.001 % H2 x18H2 =

6.016871991.40025?0.302 % N2 x18N2 =

0.205061991.40025?0.01 % 液氨计算核算:NH3=

1991.3712522.4?17?1511.3085kg

34

河南城建学院毕业论文(设计)

第四章 热量衡算

4.1 冷交换器热量计算

(1)热气进口温度 冷交换器热气紧搂温度等于水冷器气体出口温度,由题意知t=35o

(2)冷气进口温度 为了保证合成塔入口氨含量在2.4%,出氨冷气气体的氨含量必须等于或小于2.4%,设过饱和度为10%,则在该冷凝器下的平衡氨含量为: y* =0.02/1.1 ?100%=1.82%

查知平衡氨含量为此时,冷凝温度t2 =-10 o ,故冷气进口温度等于-10 o

(3)热气体带入热量 热气体在器内处于氨饱和区内,计算气体比热容时先求常压下比热容,然后用压力校正的方法计算实际的气体比热容。 查t14=35℃,p=28.4MPa时的各组分比热容并计算得: Cp°=29.806kJ/(kmol.℃) pc=3.246MPa Tc=112.237K Pr=P/ pc =8.755 TR =T/ Tc=2.743

查《小氮肥厂工艺设计手册》附图 ⊿ Cp=4.598 kj/(kmol.℃)

Cp14 = Cp°+⊿ Cp =34.405kj/(kmol.℃)

热气体带入热量

Q14=V14Cp14t14=

13449.13682?34.405?35=722996.17544kJ

22.4(4)冷气体带入热量

查t2=-10℃,p=28.12Mpa时的各组分比热容并计算得: Cp°=31.904kJ/(kmol.℃) pc=2.487Mpa Tc=83.882K Pr=P/ pc =11.326 TR =T/ Tc=3.135

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