Topic: Heat Transfer
If the inner and outer surfaces of a hollow cylinder (having radii $r_{1}$ and $r_{2}$ and length L) are at temperatures $t_{1}$ and $t_{2}$ then rate of radial heat flow will be
- A. $\frac{k}{2\pi L}\frac{t_{1}-t_{2}}{\log\frac{r_{2}}{r_{1}}}$
- B. $\frac{k}{2\pi L k}\frac{t_{1}-t_{2}}{\log\frac{r_{2}}{r_{1}}}$
- C. $\frac{2\pi L}{k}\frac{t_{1}-t_{2}}{\log\frac{r_{2}}{r_{1}}}$
- D. $2\pi Lk\frac{t_{1}-t_{2}}{\log\frac{r_{2}}{r_{1}}}$
Answer: D. $2\pi Lk\frac{t_{1}-t_{2}}{\log\frac{r_{2}}{r_{1}}}$
Explanation: For a hollow cylinder with inner and outer radii $r_{1}$ and $r_{2}$ and length L at temperatures $t_{1}$ and $t_{2}$, the thermal resistance is $R_{th} = \frac{\log(r_{2}/r_{1})}{2\pi kL}$. The heat transfer is $Q = \frac{\Delta T}{R_{th}} = \frac{(t_{1}-t_{2})2\pi kL}{\log(r_{2}/r_{1})} = 2\pi Lk\frac{t_{1}-t_{2}}{\log(r_{2}/r_{1})}$.