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1、【精品文档】如有侵权，请联系网站删除，仅供学习与交流螺旋板式换热器2013年5月计算过程结果已知数据物性参数传热量 M1= 2000 kg/h = 0.556 kg/s t 1 = 140 C t 1 = 40 Ct m1 = 40 + 0.3(140-40) = 70 C参考2：表7-8 表7-10 表7-111 = 778.52510- 6 kg/(ms) 1 = 0.1085 W /(mK)Cp1 = 2.23kJ/(kgK) 1 = 788 kg/m3Q = M1Cp1( t 1 - t 1) = 123.988kJ用试算法确定t2 t 2 =25 C t m2 = (t2+t2 )/
2、2(1) 假定水的出口温度 t 2 = 35 C 则 t m2 = 30 C1 = 995.7 kg/m3 Cp2 =4.18kJ/(kgK) t 2 = 123.988/( 995.74.1815/3600 ) = 32.14 C 相差太大，重新取值。(2) 假定水的出口温度 t2 = 33 C t m2 = 29 C1 = 995.98 kg/m3 Cp2 =4.18kJ/(kgK) t 2 = 123.988/( 995.984.1815/3600 ) = 32.15 C 相差太大，重新取值。Q =123.99kJ设计内容计算过程结果出口温度 物性参数选型 当量直径(3) 假定水的出口温
3、度 t2 = 32.15 C tm2 = 28.6 C1 = 996.09 kg/m3 Cp2 =4.18kJ/(kgK) t2= 123.988/( 996.094.1815/3600 ) = 32.147 C 相差0.003 C，合适。tm2 = ( t1+t2 )/2 = ( 25+32.15 )/2 = 28.6 C参考2：表2-12 = 794.6910-6 kg/(ms) 2 = 0.609 W/(mK)Cp2 =4.18kJ/(kgK) 2 = 996.09 kg/m3换热器为液液热交换器，煤油和水相对清洁，选用型参考1附录F.1冷却水 w2 = 0.5 m/s 煤油w1 = 0
4、.4 m/sA2 = m2 /( 3600w22 ) = 15/(36000.5 ) =0.0083 m2A1 = M1/( 3600w11 ) = 2000/(36000.4788 ) =0.0018 m2参考1表3.1H = 0.6m He = H-2 = 0.6-20.01 = 0.58 mb2 = A2/He = 0.0083/0.58 = 0.014 mb1 = A1/He =0.0018/0.58 = 0.003 mt2=32.15 C设计内容计算过程结果 参考1表3.1查看产品样本：取b2 = 15 mm b1 = 5 mmde2 = 2Heb2/( He+b2 )= 20.58
5、 0.015/(0.58+0.015 ) =0.0292 mde1 = 2Heb1/( He+b1 ) = 20.580.005/(0.58 +0.005 )=0.0099 m2 = m2 / ( 3600A22 ) = 15 / ( 3600 0.0083 ) = 0.50 m / sRe2 = 2de22 /2 = 0.50.0292996.09 / ( 794.6910 -6 ) = 18300Pr2 =2Cp2 / 2 = 794.6910 -64.18103 =5.451 = M1 / ( 3600A11 ) = 2000 / ( 36000.0018788 ) = 0.39 m /
6、 sRe1 = 1de11 /1 = 0.390.0099788 / ( 778.52510 -6 ) = 3908de2 = =0.0292 mde1 =0.0099 m设计内容计算过程结果换热系数传热系数平均温差 Pr1 =1 Cp1 /1=778.52510 -62.23103 =162 = 0.0397( 2 /de2 )Re20.784Pr20.4 = 0.0397( 0.609 /0.0292 )183000.784 5.450.4 = 3583 W/( m2 C)1 = 0.0397( 1 /de1 )Re10.784Pr10.4 = 0.0397( 0.1085 /0.0099
7、 )39080.784 160.4 = 655 W/( m2 C)查询网络资料，介质为水和煤油，选用Q235卷筒钢板，厚度4mm，导热系数为= 46.5 W/（ m C )参考1附录C取r1,s = r2,s = 0.00017 m2 C /W1 /K = 1 / 1 + 1/2 + / + r1,s + r2,s =1 /3583 + 1 /655 + 0.004 /46.5+0.000172=2.2310 -3K = 448 W / ( m2 / C )附图1在题后。t m = ( t1 - t2 ) - ( t1 - t2 ) / ln ( t1 2 = 3583W/(m2 C)1 =6
8、55W/(m2 C)K = 448 W /( m2 / C )设计内容计算过程结果 传热面积有效长度内测螺旋板 外侧螺旋板外径- t2 ) /( t1 - t2 ) =(140 - 32.75 ) - (40 - 25) / ln ( 140 - 32.15 ) /( 40 - 25 ) = 47 CF =Q /( Kt m ) = 123988 /( 44847 ) = 5.89 m2L e = 0.5F / he = 0.55.98 /0.58 = 5.08 m设 d2 = 200 mm C = b1 + b2 + 2 = 28 mmd1 = d2 - ( b2 -b1 ) = 200 -
9、 ( 15 - 5 ) = 190 mmNe =( 2b2 - d1 - d2 ) + ( d1 + d2 - 2b2 ) 2 +16Cl e / 3.1415 ) / ( 4C) =( 20.015 - 0.2 - 0.19)+ ( 0.028 + 0.19 - 20.015 ) 2 +160.285.08 / 3.1415 ) / ( 40.28) = 5.03 取有效圈数 n = 5li = l1 = = 5.03 mlo = l2 = = 5.62 mD = d2 + 2nC + 2 = 200 + 2528 + 24t m =47 Cn = 5li = 5.03 mlo = 5.62 m设计内容计算过程结果 压降 参考书目 存在问题老师批示 = 488 mmP1 = = = 0.025 M Pa 符合设计要求。P2 = = = 0.033M Pa 符合设计要求。1 热交换器原理与设计，东南大学出版社2 工程物质的热物理性质手册，新时代出版社P1 所用的公式是雷诺数在500044000之间时适用，但Re1不在此范围内，已查阅相关论文均用此公式直接运算，并未考虑适用范围，由于没有找到相关运算公式，按课本计算。D = 488 mm老师请写出存在问题，谢谢设计内容计算过程结果 .精品文档.设计内容