Answer:
Step-by-step explanation:
given data:
innner radius r_i = = 5cm
outer radius r_0 = 7 cm
temperature at outer surface = 80 degree celcius
temperature at inner surface = 100 degree celcius
thermal resistance per unit length is given as
![R _(th) = (1)/( 2\pi kl) ln ( ro)/(ri)](https://img.qammunity.org/2020/formulas/engineering/college/edmy6a0sfww77b3u9zyqh4zp2qowq6x5ig.png)
![= (1)/( 2\pi k*1) ln ( 7)/(5)](https://img.qammunity.org/2020/formulas/engineering/college/ctvcoq8jddphj7l56a5ud382n3cts6br1c.png)
![= (0.053)/(k)](https://img.qammunity.org/2020/formulas/engineering/college/qqufichungqi7lsu1g0or1qakh6in84ibo.png)
heat loss rate per unit length
![q = (Ti-To)/(R_(th))](https://img.qammunity.org/2020/formulas/engineering/college/w46wxa8l7n6g2n08wxi4xay791fgxdvzps.png)
![q = (100-80)/((0.053)/(k))](https://img.qammunity.org/2020/formulas/engineering/college/gznzr7fuljw6sw24q1og02guf2o6ewgh7p.png)
q = 373.474 K
1) for pure COPPER
k = 387 W/m degree celcius
q = 373.474 * 387 = 144534.438 W
2) for pure ALUMINIUM
k = 200 W/m degree celcius
q = 373.474 * 200 = 74694.84 W
3)
1) for pure IRON
k = 62W/m degree celcius
q = 373.474 * 62 = 23155.416 W