发布时间 : 星期日 文章第四章 传热2 - 图文更新完毕开始阅读2050aa2f647d27284b7351e4
解:(a)每米管长的热损失
2?(t1?t4)ql?1r21r31r4ln?ln?ln?1r1?2r2?3r3此处,r1=0.053/2=0.0265m,r2=0.0265+0.0035=0.03m,r3=0.03+0.04=0.07m,r4=0.07+0.02=0.09m2?(500?80)ql??191W/m10.0310.0710.09ln?ln?ln450.02650.070.030.150.07r20.03??2,1rm1?(0.0265?0.03)?0.02825mr10.02652Am1?2?rm1lr?r3?r2m2?0.07?0.03?0.lnr0472m30.rln0720.03Am2?2?rm2lr40.09r?0.07?2,rm3?132(0.07?0.09)Am3?2?rm3l?0.08mQ?t1?t4?1???21Am1???32Am2?3Am3?t1?t4?1?2?r??2??31m1l?22?rm2l?32?rm3lqQt1?t4l?l??1???2??312?rm1?22?rm2?32?rm3?......?191.5W/m(b)保温层界面温度t3
q2?(t1?t3)l?1?lnr2?1lnr31r1?2r2191?2?(500?t3)10.0310.0745ln0.0265?0.07ln0.03解得
t3=132℃