Question

In a hot water heating system, there is cylindrical pipe of length $28\mathrm{m}$and diameter $5\mathrm{cm}$. Find the total radiating surface in the system. (Assume$\mathrm{\pi }=\frac{22}{7}$)

Answer 1

Find the curved surface area

Given: Diameter of cylindrical pipe, $\mathrm{d}=5\mathrm{cm}$

Length of cylindrical pipe, $\mathrm{h}=28\mathrm{m}$

So radius of cylindrical pipe is;

$$\begin{gathered} \mathrm{r}=\frac{5}{2}\left[\because \mathrm{r}=\frac{\mathrm{d}}{2}\right] \\ =2.5\mathrm{cm} \\ =\frac{2.5}{100}\left[\because 1\mathrm{m}=100\mathrm{cm}\Rightarrow 1\mathrm{cm}=\frac{1}{100}\mathrm{m}\right] \\ =0.025\mathrm{m} \end{gathered}$$

As, Cylindrical pipe radiates from curved surfaces only, so we have to measure the curved surface area of that pipe.

We know the curved surface area of the cylinder $\mathbf{=}\mathbf{}\mathbf{2}\mathbf{\pi rh}$, where $\mathrm{r}$ is the radius and $\mathrm{h}$is the length of the cylinder.

So, the total radiating surface area of the pipe $=2\mathrm{\pi rh}$

$$\begin{gathered} =2\times \frac{22}{7}\times 0.025\times 28\left[\because \mathrm{\pi }=\frac{22}{7}\right] \\ =\frac{44}{\cancel{7}}\times {\cancel{0.7}}^{0.1} \\ =44\times 0.1 \\ =4.4{\mathrm{m}}^2 \end{gathered}$$

Hence, the total radiating surface area of the pipe is $4.4{\mathrm{m}}^2$.