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4.3.6 氨冷器II .................................................................................................................
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致 谢 ......................................................................................................................................
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年产30万吨合成氨合成工段工艺设计
摘要:氨是一种重要的化工产品,在国民经济中有重要的作用。对合成氨工艺进行设计研究,并对其过程进行设计优化。氨合成工段包括氨的合成、分离、气体再循环、惰性气体排放等基本过程,其中氨合成是合成氨工艺的中心环节。本设计主要目的是对合成氨的合成工段进行设计,根据已给组成的原料气的组成,进行工艺系统计算,包括物料衡算、热量衡算、设备的数据计算及选型等。合成工段中的主要设备为氨合成塔,结合设计数据及技术现状,本设计选择的氨合成塔的内件为三段绝热冷激--内冷式内件,该内件具有结构合理、氨净值高、产量大等优点。根据物料及热量衡算的数据,计算出内件中绝热床层及换热器的有关尺寸数据,并对一些辅助设备进行设计选型。根据计算数据,绘制出主要设备及带控制点的工艺流程图等。
关键词:合成氨 物料衡算 热量衡算 合成塔
Process Design of the Section Which Synthetizes Liquid
Ammonia of 300000ta
Abstract: Ammonia is an important chemical product,plays an important role in the national economy.It is very necessary to explore and design the process of synthetic ammonia,then to optimize the process and the equipment.Ammonia synthesis process includes the separation of ammonia,the gas recirculation, the ammonia synthesis,the emission of inert gas and so on,and during the process,the ammonia synthesis is the most important link.The main purpose of this design is to devise the synthesis process of ammonia synthesis.According to the composition of raw gas that ,we carry on the calculation of the craft system,including material balance, and selection of the equipment.The ammonia synthesis tower is the crucial equipment of the process,combining with the design data and technical status,this design chooses three adiabatic cold shock -internal cooling type internal parts,the inner parts calculate the relevant size date of the adiabatic bed and and select some auxiliary equipment.
Key Words: Synthetic ammonia; materal balance; =1),首先假定催化剂层温度=390℃
设出第一段气体温度 =480℃ a.查物料数据:
第一段平均温度=(+)÷2=(480+390÷2=435℃
由=435℃,=0.025 查=39.06 =51110KJkmolNH3 由联立方程式 ×(-)
其中表示第n-1段至n段催化剂用量
为入塔氨分解基无惰性气体体积流量 为催化剂活性系数
,为扣除了惰性气体的第n段催化剂进出计算氨含量, 为第n段进出口反应速度倒数
b. 求反应热及生成氨含量:令气体第一段焓升(反应热)
?I1?MK,0?C平,pk,1??TK,1?TK,0? 则第一床层生成氨的量为:
?M1?MK,0?C平,pk,1?TK,1?TK,0??HT1
c.求出口氨含量:
y0NH3,1?M1y0NH3,0由公式 : ?22.4??1?y0NH3,1V0,11?y0NH3,0 = 22.4×723.268÷206021.429+ =0.0588+ 由公式:
=0.025÷(1-0.025)=0.0256, =0.0250 上式=0.0588+0.0250=0.0838, =0.0915
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