Chapter 7, Problem 5RE

To determine

To determine that how many kilowatt-hours of electrical energy does the company consume in a typical day.

Answer to Problem 5RE

The company consumes $14400\text{ kilowatt-hours}$ in a typical day.

Explanation of Solution

Given:

The electrical power consumption is modelled by:

  $E\left(t\right)=300\left(2-\cos \left(\frac{\pi t}{12}\right)\right)$

where,

  $\begin{array}{l}t=\text{ time in hours}\\ E\left(t\right)=\text{ Electrical power consumption in kilowatts}\end{array}$

Calculation:

The electrical energy consumption in a typical day is calculated as:

  $\begin{array}{l}\text{Total electrical energy consumption (}{E}_c)={\displaystyle \underset{0}{\overset{24}{\int }}E\left(t\right)\cdot dt}\\ \Rightarrow E_C={\displaystyle \underset{0}{\overset{24}{\int }}300\left(2-\cos \left(\frac{\pi t}{12}\right)\right)\cdot dt}\\ \Rightarrow E_C={\displaystyle \underset{0}{\overset{24}{\int }}600\cdot dt-300{\displaystyle \underset{0}{\overset{24}{\int }}\cos \left(\frac{\pi t}{12}\right)\cdot dt}}\\ \Rightarrow E_C={\left[600t\right]}_0^{24}-300{\left[\frac{12}{\pi }\sin \left(\frac{\pi t}{12}\right)\right]}_0^{24}\\ \Rightarrow E_C=\left(600\times 24-0\right)-300\left[\frac{12}{\pi }\sin \left(2\pi \right)-\frac{12}{\pi }\sin \left(0\right)\right]\\ \Rightarrow E_C=14400-0\\ \Rightarrow E_C=14400\text{ kilowatt-hours}\end{array}$

Hence, $14400\text{ kilowatt-hours}$ of electrical energy, the company consumes in a typical day.