圆筒壁导热计算举例

圆筒壁导热计算举例【例2-1】在一60mmx3.5mm里层为40mm的氧化镁粉，平均导热系数九=0.07W-m-i-K-1,外层为20mm的石棉层，其平均导热系数九=0.15W-m-i-K-1。现用热电偶测得管内壁的温度为500°C,最外层面温度为80°C,管壁的导热系数九=4呵•m-1•K-1。试求每米管长的热损失及保温层界面的温度。解:(a)每米管长的热损失2兀G-1)TOC\o"1-5"\h\zq=14—i1r1r1rIn亠+lnr+In亠九r九r九r112233此处，r=0.053=0.0265m,r=0.0265+0.0035=0.03i22r=0.03+0.04=0.07m,r=0.07+0.02=0.09m34(b)2x3.14x(500一80)19丄加亞+丄in007+j450.02650.070.030.150.070.030.09=191W-m-i保温层界面温度t2n(t-1)q=13姣11r1rin―2+in—3入r入r11222x3.14x(500-13)1.0.031「0.07in+in450.02650.070.03解得：t=132°C3