题目
一列管式换热器,管径为 times 2.5mm, 传热面积-|||-为10m^2(按管外径计)。今拟用于使80℃的饱和苯蒸气冷凝、-|||-冷却到50℃。苯走管外,流量为 https:/img.cdnjtzy.com/zyb_cbee51e3b4b149c30b8408e040d11d2c.jpg.25kg/s; 冷却水走管内与苯-|||-逆流,流量为 /s, 进口温度为10℃。现已估算出苯冷凝、冷-|||-却时的对流传热系数分别为 v/((m)^2cdot k) 850-|||-W/(m^2·K); 水的对流传热系数为 v/((m)^2cdot k) 忽略管-|||-两侧污垢热阻和管壁热阻。已知水、苯(液体)的比热容分别为-|||-.18times (10)^3J/(kgcdot R) https:/img.cdnjtzy.com/zyb_cbee51e3b4b149c30b8408e040d11d2c.jpg.76times (10)^3J/(kgcdot R); 苯蒸气在80℃的-|||-冷凝潜热为 times (10)^3J/kg 问此换热器是否合用?一列管式换热器，管径为φ25mm×2.5mm，传热面积为10m2(按管外径计)。今拟用于使80℃的饱和苯蒸气冷凝、冷却到50℃。苯走管外，流量为1.25kg/s；冷却水走管内与苯逆流，流量为6kg/s，进口温度为10℃。现已估算出苯冷凝、冷却时的对流传热系数分别为1600W/(m2·K)、850W/(m2·K)；水的对流传热系数为2500W/(m2·K)。忽略管两侧污垢热阻和管壁热阻。已知水、苯(液体)的比热容分别为4.18×103J/(kg·K)、1.76×103J/(kg·K)；苯蒸气在80℃的冷凝潜热为395×103J/kg。问此换热器是否合用?
一列管式换热器，管径为φ25mm×2.5mm，传热面积为10m2(按管外径计)。今拟用于使80℃的饱和苯蒸气冷凝、冷却到50℃。苯走管外，流量为1.25kg/s；冷却水走管内与苯逆流，流量为6kg/s，进口温度为10℃。现已估算出苯冷凝、冷却时的对流传热系数分别为1600W/(m2·K)、850W/(m2·K)；水的对流传热系数为2500W/(m2·K)。忽略管两侧污垢热阻和管壁热阻。已知水、苯(液体)的比热容分别为4.18×103J/(kg·K)、1.76×103J/(kg·K)；苯蒸气在80℃的冷凝潜热为395×103J/kg。问此换热器是否合用?
题目解答
答案
解析
步骤 1：计算冷凝段和冷却段的传热量
冷凝段传热量 ${Q}_{1}={m}_{n}{r}_{n}={m}_{c}{c}_{pc}({t}_{12}-{t}_{12})$
冷却段传热量 ${Q}_{2}={m}_{1}{C}_{Ph}({t}_{5}-{t}_{12})={m}_{c}{C}_{Rx}({t}_{12}-{t}_{c})$
将已知数据代入有
${Q}_{1}=1.25\times 395\times {10}^{3}=6\times 4.18\times {10}^{3}({t}_{1}-{t}_{12})$
${Q}_{i}=1.25\times 1.76\times {10}^{3}(80-50)=6\times 4.18\times {10}^{3}({t}_{2}-10)$
解上两式得到
${Q}_{1}=494kW,{Q}_{2}=66.0kW$
${t}_{12}={12.6}^{\circ }C$ ,${t}_{{c}_{2}}={32.3}^{\circ }C$
步骤 2：计算冷凝段的传热面积
对冷凝段:
$\Delta {t}_{m}=\dfrac {(80-32.3)-(80-12.6)}{\ln [ (80-32.3)!(80-12.6)] }={56.9}^{\circ }C$
$\dfrac {1}{{K}_{1}}=\dfrac {{d}_{0}}{{a}_{1}{d}_{1}}+\dfrac {1}{{a}_{0}}=\dfrac {25}{20}\times \dfrac {1}{2500}+\dfrac {1}{1600}$
${K}_{1}=889W/({m}^{2}\cdot K)$
${A}_{1}=\dfrac {{Q}_{1}}{{K}_{1}\Delta {L}_{m}}=\dfrac {4.94\times {10}^{3}}{889\times 56.9}=9.77{m}^{2}$
步骤 3：计算冷却段的传热面积
对冷却段:
$\Delta {C}_{m}=\dfrac {(80-12.6)-(50-10)}{\ln [ (80-12.6)/(50-10)] }={52.5}^{\circ }C$
$\dfrac {1}{{K}_{2}}=\dfrac {{d}_{0}}{{a}_{1}{d}_{1}}+\dfrac {1}{{a}_{0}}=\dfrac {25}{20}\times \dfrac {1}{2500}+\dfrac {1}{850}$
${K}_{2}=596W/({m}^{2}\cdot K)$
${A}_{2}=\dfrac {{Q}_{2}}{{K}_{2}\Delta {C}_{m}}=\dfrac {66.0\times {10}^{3}}{596\times 52.5}=2.11{m}^{2}$
步骤 4：计算总传热面积
$A={A}_{1}+{A}_{2}=9.77+2.11=11.88{m}^{2}\gt 10{m}^{2}$
冷凝段传热量 ${Q}_{1}={m}_{n}{r}_{n}={m}_{c}{c}_{pc}({t}_{12}-{t}_{12})$
冷却段传热量 ${Q}_{2}={m}_{1}{C}_{Ph}({t}_{5}-{t}_{12})={m}_{c}{C}_{Rx}({t}_{12}-{t}_{c})$
将已知数据代入有
${Q}_{1}=1.25\times 395\times {10}^{3}=6\times 4.18\times {10}^{3}({t}_{1}-{t}_{12})$
${Q}_{i}=1.25\times 1.76\times {10}^{3}(80-50)=6\times 4.18\times {10}^{3}({t}_{2}-10)$
解上两式得到
${Q}_{1}=494kW,{Q}_{2}=66.0kW$
${t}_{12}={12.6}^{\circ }C$ ,${t}_{{c}_{2}}={32.3}^{\circ }C$
步骤 2：计算冷凝段的传热面积
对冷凝段:
$\Delta {t}_{m}=\dfrac {(80-32.3)-(80-12.6)}{\ln [ (80-32.3)!(80-12.6)] }={56.9}^{\circ }C$
$\dfrac {1}{{K}_{1}}=\dfrac {{d}_{0}}{{a}_{1}{d}_{1}}+\dfrac {1}{{a}_{0}}=\dfrac {25}{20}\times \dfrac {1}{2500}+\dfrac {1}{1600}$
${K}_{1}=889W/({m}^{2}\cdot K)$
${A}_{1}=\dfrac {{Q}_{1}}{{K}_{1}\Delta {L}_{m}}=\dfrac {4.94\times {10}^{3}}{889\times 56.9}=9.77{m}^{2}$
步骤 3：计算冷却段的传热面积
对冷却段:
$\Delta {C}_{m}=\dfrac {(80-12.6)-(50-10)}{\ln [ (80-12.6)/(50-10)] }={52.5}^{\circ }C$
$\dfrac {1}{{K}_{2}}=\dfrac {{d}_{0}}{{a}_{1}{d}_{1}}+\dfrac {1}{{a}_{0}}=\dfrac {25}{20}\times \dfrac {1}{2500}+\dfrac {1}{850}$
${K}_{2}=596W/({m}^{2}\cdot K)$
${A}_{2}=\dfrac {{Q}_{2}}{{K}_{2}\Delta {C}_{m}}=\dfrac {66.0\times {10}^{3}}{596\times 52.5}=2.11{m}^{2}$
步骤 4：计算总传热面积
$A={A}_{1}+{A}_{2}=9.77+2.11=11.88{m}^{2}\gt 10{m}^{2}$