Question

Derive the relation between the commercial unit of energy and SI unit of energy.

Answer 1

1. kWh is a commercial unit of energy. $1kWh$is the amount of energy consumed at the rate of $1$ kilowatt for $1$ hours.
2. SI unit of energy is Joule.

Calculating the number of joules in 1 kWh,

$$\begin{gathered} \mathbf{1}\mathit{k}\mathit{i}\mathit{l}\mathit{o}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{=}\mathbf{1}\mathit{k}\mathit{i}\mathit{l}\mathit{o}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{\times }\mathbf{1}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{} \\ \mathbf{1}\mathit{k}\mathit{i}\mathit{l}\mathit{o}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{=}\mathbf{1000}\mathbf{}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{\times }\mathbf{1}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{} \end{gathered}$$

But $1watt=\frac{1Joule}{1second}$………(i)

Substituting in equation (i)

$\mathbf{1}\mathit{k}\mathit{i}\mathit{l}\mathit{o}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{=}\mathbf{1000}\frac{\mathbf{J}\mathbf{o}\mathbf{u}\mathbf{l}\mathbf{e}\mathbf{s}}{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{d}\mathbf{s}}\mathbf{\times }\mathbf{1}\mathit{h}\mathit{o}\mathit{u}\mathit{r}$

As $\mathbf{1}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{=}\mathbf{60}\mathbf{\times }\mathbf{60}$seconds

Thus, $\mathbf{1}\mathit{k}\mathit{i}\mathit{l}\mathit{o}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{=}\mathbf{1000}\mathbf{}\frac{\mathbf{J}\mathbf{o}\mathbf{u}\mathbf{l}\mathbf{e}\mathbf{s}}{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{d}\mathbf{s}}\mathbf{\times }\mathbf{60}\mathbf{\times }\mathbf{60}\mathbf{}\mathit{s}\mathit{e}\mathit{c}\mathit{o}\mathit{n}\mathit{d}\mathit{s}$

or $\mathbf{1}\mathbf{}\mathit{k}\mathit{i}\mathit{l}\mathit{o}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{=}\mathbf{3600000}\mathbf{}\mathit{J}\mathit{o}\mathit{u}\mathit{l}\mathit{e}\mathit{s}$ or $\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{J}\mathbf{)}$

Hence, the relation between the commercial unit of energy and the SI unit of energy is $\mathbf{1}\mathbf{}\mathit{k}\mathit{i}\mathit{l}\mathit{o}\mathit{w}\mathit{a}\mathit{t}\mathit{t}\mathbf{}\mathit{h}\mathit{o}\mathit{u}\mathit{r}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{J}\mathit{o}\mathit{u}\mathit{l}\mathit{e}\mathit{s}$.