Answer:
Heat transfer rate = 20.08 W
Step-by-step explanation:
Given that
D= 1 cm
K= 380 W/m.°C
Wall temperature = 200°C
Surrounding temperature = 30°C
![h=15\ W/m^2C](https://img.qammunity.org/2020/formulas/engineering/college/556zvmwgvbykqnjqv8o40hv12nek4c5a6j.png)
Rod is long so this is the case of long fin .
Heat transfer in long fin given as
![Q=√(hPKA)\ \Delta T](https://img.qammunity.org/2020/formulas/engineering/college/bc61xhjz07hz9681le0fbmlbsfp3ymropr.png)
Here
P= π D
P = 3.14 x 0.01 m
P= 0.0314 m
![A=(\pi)/(4)* 0.01^2\ m^2](https://img.qammunity.org/2020/formulas/engineering/college/crfthbz2yyx3dwnkwzpq5lj3a5egvl2zed.png)
![A=7.8* 10^(-5)\ m^2](https://img.qammunity.org/2020/formulas/engineering/college/k5gwpnyb16we29rjebuaanyvui8uc2vnh5.png)
![Q=√(hPKA)\ \Delta T](https://img.qammunity.org/2020/formulas/engineering/college/bc61xhjz07hz9681le0fbmlbsfp3ymropr.png)
![Q=\sqrt{15* 0.0314* 380* 7.8* 10^(-5)}\ (200-30)](https://img.qammunity.org/2020/formulas/engineering/college/v8b208lsoa90k2j90anlpqupfel8iowv09.png)
Q=20.08 W
So heat transfer rate = 20.08 W