Chapter 2, Problem 22E

A wind power system with increasing windspeed has the current waveform described by the equation below, delivered to an 80 Ω resistor. Plot the current, power, and energy waveform over a period of 60 s, and calculate the total energy collected over the 60 s time period.

$i(t)=\frac{1}{2}{t}^2\sin \left(\frac{\pi }{8}t\right)\cos \left(\frac{\pi }{4}t\right)A$

To determine

Sketch the current, power, and energy waveform over a period of 60 s to the current waveform of the wind power system. Also calculate the total energy collected over the 60 s time period.

Answer to Problem 22E

The plot for the current, power, and energy waveform is sketched as shown in Figure 1, and the total energy collected over the 60 s time period is $79.024\times {10}^7\text{J}$.

Explanation of Solution

Given data:

The current waveform is,

$i\left(t\right)=\frac{1}{2}{t}^2\sin \left(\frac{\pi }{8}t\right)\cos \left(\frac{\pi }{4}t\right)\text{A}$

The resistor is $R=80\Omega$.

Formula used:

Write the general expression to find the power as follows,

$p=i{\left(t\right)}^{2}R$        (1)

Write the general expression to find the energy as follows,

$w\left(t\right)={\displaystyle \underset{t_0}{\overset{t}{\int }}pdt}$        (2)

Calculation:

Substitute the value of $i\left(t\right)$ and $R$ in equation (1) as follows,

$\begin{array}{c}p={\left[\frac{1}{2}{t}^2\sin \left(\frac{\pi }{8}t\right)\cos \left(\frac{\pi }{4}t\right)\right]}^2\left(80\right)\\ =\left[\frac{1}{4}{t}^4{\sin }^2\left(\frac{\pi }{8}t\right){\cos }^2\left(\frac{\pi }{4}t\right)\right]\left(80\right)\\ =\frac{80}{4}\left[t^4{\sin }^2\left(\frac{\pi }{8}t\right){\cos }^2\left(\frac{\pi }{4}t\right)\right]\end{array}$

$p=20\left[t^4{\sin }^2\left(\frac{\pi }{8}t\right){\cos }^2\left(\frac{\pi }{4}t\right)\right]$        (3)

Applying equation (3) in equation (2) as follows,

$w\left(t\right)=20{\displaystyle \underset{0}{\overset{60}{\int }}{t}^4{\sin }^2\left(\frac{\pi }{8}t\right){\cos }^2\left(\frac{\pi }{4}t\right)dt}$        (4)

Consider the function,

$w\left(t\right)={\displaystyle \int t^4{\sin }^2\left(\frac{\pi }{8}t\right){\cos }^2\left(\frac{\pi }{4}t\right)dt}$        (5)

Consider $u=\frac{t}{8}$ in the above expression as follows,

$\begin{array}{l}u=\frac{t}{8}\\ \frac{du}{dt}=\frac{1}{8}\\ dt=8du\\ t^4=4096u^4\end{array}$

Substituting the values in equation (5) as follows,

$w\left(t\right)=32768{\displaystyle \int u^4{\sin }^2\left(\pi u\right){\cos }^2\left(2\pi u\right)du}$        (6)

Consider,

$x={\displaystyle \int u^4{\sin }^2\left(\pi u\right){\cos }^2\left(2\pi u\right)du}$        (7)

Applying the product to sum formulas in the above expression as follows,

$\begin{array}{c}x={\displaystyle \int u^4\left(\frac{1}{4}-\frac{\cos \left(6\pi u\right)-2\cos \left(4\pi u\right)+3\cos \left(2\pi u\right)}{8}\right)}du\\ ={\displaystyle \int \left(\frac{u^4}{4}-\frac{u^4\cos \left(6\pi u\right)}{8}+\frac{u^4\cos \left(4\pi u\right)}{4}-\frac{3u^4\cos \left(2\pi u\right)}{8}\right)}du\\ =\frac{-1}{8}{\displaystyle \int u^4\cos \left(6\pi u\right)du}+\frac{1}{4}{\displaystyle \int u^4\cos \left(4\pi u\right)du}-\frac{3}{8}{\displaystyle \int u^4\cos \left(2\pi u\right)du}+\frac{1}{4}{\displaystyle \int u^{4}du}\end{array}$

Consider,

$a={\displaystyle \int u^4\cos \left(6\pi u\right)du}$        (8)

By applying integration by parts in the above expression,

$\begin{array}{l}f=u^4{g}^{\prime }=\cos \left(6\pi u\right)\\ f^{\prime }=4u^{3}g=\frac{\sin \left(6\pi u\right)}{6\pi }\end{array}$

Applying the values in equation (8) as follows,

$a=\frac{u^4\sin \left(6\pi u\right)}{6\pi }-{\displaystyle \int \frac{2u^3\sin \left(6\pi u\right)}{3\pi }}du$        (9)

Consider the function,

$b={\displaystyle \int \frac{2u^3\sin \left(6\pi u\right)}{3\pi }}du$

$b=\frac{2}{3\pi }{\displaystyle \int u^3\sin \left(6\pi u\right)}du$        (10)

Solving,

${\displaystyle \int u^3\sin \left(6\pi u\right)}du$

By applying integration by parts in the above expression,

$\begin{array}{l}f=u^3{g}^{\prime }=\sin \left(6\pi u\right)\\ f^{\prime }=3u^{2}g=-\frac{\cos \left(6\pi u\right)}{6\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u^3\sin \left(6\pi u\right)}du=-\frac{u^3\cos \left(6\pi u\right)}{6\pi }-{\displaystyle \int -\frac{u^2\cos \left(6\pi u\right)}{2\pi }}du$        (11)

Consider,

$\frac{-1}{2\pi }{\displaystyle \int u^2\cos \left(6\pi u\right)}du$

Solving,

${\displaystyle \int u^2\cos \left(6\pi u\right)}du$

By applying integration by parts in the above expression,

$\begin{array}{l}f=u^2{g}^{\prime }=\cos \left(6\pi u\right)\\ f^{\prime }=2ug=\frac{\sin \left(6\pi u\right)}{6\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u^2\cos \left(6\pi u\right)}du=\frac{u^2\sin \left(6\pi u\right)}{6\pi }-{\displaystyle \int \frac{u\sin \left(6\pi u\right)du}{3\pi }}$        (12)

Consider,

$\frac{1}{3\pi }{\displaystyle \int u\sin \left(6\pi u\right)du}$

Solving,

${\displaystyle \int u\sin \left(6\pi u\right)du}$

By applying integration by parts in the above expression,

$\begin{array}{l}f=u{g}^{\prime }=\sin \left(6\pi u\right)\\ f^{\prime }=1g=-\frac{\cos \left(6\pi u\right)}{6\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u\sin \left(6\pi u\right)du}=-\frac{u\cos \left(6\pi u\right)}{6\pi }-{\displaystyle \int -}\frac{\cos \left(6\pi u\right)}{6\pi }du$        (13)

Consider,

${\displaystyle \int -}\frac{\cos \left(6\pi u\right)}{6\pi }du$

Let,

$\begin{array}{l}v=6\pi u\\ \frac{dv}{du}=6\pi \\ du=\frac{1}{6\pi }dv\end{array}$

Applying the values in the above expression,

$\begin{array}{c}{\displaystyle \int -}\frac{\cos \left(6\pi u\right)}{6\pi }du=\frac{-1}{36{\pi }^2}{\displaystyle \int \cos \left(v\right)dv}\\ =-\frac{\sin \left(v\right)}{36{\pi }^2}\\ =-\frac{\sin \left(6\pi u\right)}{36{\pi }^2}\left\{\because v=6\pi u\right\}\end{array}$

Substituting the values in equation (13) as follows,

${\displaystyle \int u\sin \left(6\pi u\right)du}=\frac{\sin \left(6\pi u\right)}{36{\pi }^2}-\frac{u\cos \left(6\pi u\right)}{6\pi }$

Substitute the above value in the considered expression $\frac{1}{3\pi }{\displaystyle \int u\sin \left(6\pi u\right)du}$. Therefore,

$\frac{1}{3\pi }{\displaystyle \int u\sin \left(6\pi u\right)du}=\frac{\sin \left(6\pi u\right)}{108{\pi }^3}-\frac{u\cos \left(6\pi u\right)}{18{\pi }^2}$

Substitute the above value in equation (12) as follows,

${\displaystyle \int u^2\cos \left(6\pi u\right)}du=\frac{u^2\sin \left(6\pi u\right)}{6\pi }-\frac{\sin \left(6\pi u\right)}{108{\pi }^3}-\frac{u\cos \left(6\pi u\right)}{18{\pi }^2}$

Substitute the above value in equation (11) as follows,

${\displaystyle \int u^3\sin \left(6\pi u\right)}du=\frac{u^2\sin \left(6\pi u\right)}{12{\pi }^2}-\frac{\sin \left(6\pi u\right)}{216{\pi }^4}-\frac{u^3\cos \left(6\pi u\right)}{6\pi }+\frac{u\cos \left(6\pi u\right)}{36{\pi }^3}$

Substitute the above value in equation (10) as follows,

$\begin{array}{c}b=\frac{2}{3\pi }\left[\frac{u^2\sin \left(6\pi u\right)}{12{\pi }^2}-\frac{\sin \left(6\pi u\right)}{216{\pi }^4}-\frac{u^3\cos \left(6\pi u\right)}{6\pi }+\frac{u\cos \left(6\pi u\right)}{36{\pi }^3}\right]\\ =\frac{u^2\sin \left(6\pi u\right)}{18{\pi }^3}-\frac{\sin \left(6\pi u\right)}{324{\pi }^5}-\frac{u^3\cos \left(6\pi u\right)}{9{\pi }^2}+\frac{u\cos \left(6\pi u\right)}{54{\pi }^4}\end{array}$

Substitute the above value in equation (9) as follows,

$a=\frac{u^4\sin \left(6\pi u\right)}{6\pi }-\frac{u^2\sin \left(6\pi u\right)}{18{\pi }^3}+\frac{\sin \left(6\pi u\right)}{324{\pi }^5}+\frac{u^3\cos \left(6\pi u\right)}{9{\pi }^2}-\frac{u\cos \left(6\pi u\right)}{54{\pi }^4}$        (14)

Consider the function $m={\displaystyle \int u^4\cos \left(4\pi u\right)du}$ in given in equation x.

$m={\displaystyle \int u^4\cos \left(4\pi u\right)du}$

Applying integration by parts in the above expression,

$\begin{array}{l}f=u^4{g}^{\prime }=\cos \left(4\pi u\right)\\ f^{\prime }=4u^{3}g=\frac{\sin \left(4\pi u\right)}{4\pi }\end{array}$

Applying the values in the above equation as follows,

$m=\frac{u^4\sin \left(4\pi u\right)}{4\pi }-{\displaystyle \int \frac{u^3\sin \left(4\pi u\right)}{\pi }}du$        (15)

Consider,

$\frac{1}{\pi }{\displaystyle \int u^3\sin \left(4\pi u\right)}du$        (16)

Let,

${\displaystyle \int u^3\sin \left(4\pi u\right)}du$

Applying integration by parts in the above expression,

$\begin{array}{l}f=u^3{g}^{\prime }=\sin \left(4\pi u\right)\\ f^{\prime }=3u^{2}g=-\frac{\cos \left(4\pi u\right)}{4\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u^3\sin \left(4\pi u\right)}du=-\frac{u^3\cos \left(4\pi u\right)}{4\pi }-{\displaystyle \int -\frac{3u^2\cos \left(4\pi u\right)}{4\pi }du}$        (17)

Consider,

$-\frac{3}{4\pi }{\displaystyle \int u^2\cos \left(4\pi u\right)du}$        (18)

Let,

${\displaystyle \int u^2\cos \left(4\pi u\right)du}$        (19)

Applying integration by parts in the above expression,

$\begin{array}{l}f=u^2{g}^{\prime }=\cos \left(4\pi u\right)\\ f^{\prime }=2ug=-\frac{\sin \left(4\pi u\right)}{4\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u^2\cos \left(4\pi u\right)du}=\frac{u^2\sin \left(4\pi u\right)}{4\pi }-{\displaystyle \int \frac{u\sin \left(4\pi u\right)}{2\pi }}du$        (20)

Consider,

$\frac{1}{2\pi }{\displaystyle \int u\sin \left(4\pi u\right)}du$        (21)

Let,

${\displaystyle \int u\sin \left(4\pi u\right)}du$

Applying integration by parts in the above expression,

$\begin{array}{l}f=u{g}^{\prime }=\sin \left(4\pi u\right)\\ f^{\prime }=1g=-\frac{\cos \left(4\pi u\right)}{4\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u\sin \left(4\pi u\right)}du=-\frac{u\cos \left(4\pi u\right)}{4\pi }-{\displaystyle \int -\frac{\cos \left(4\pi u\right)}{4\pi }}du$        (22)

Consider,

${\displaystyle \int -\frac{\cos \left(4\pi u\right)}{4\pi }}du$

Let,

$\begin{array}{l}v=4\pi u\\ \frac{dv}{du}=4\pi \\ du=\frac{1}{4\pi }dv\end{array}$

Substituting the values,

$\begin{array}{c}{\displaystyle \int -\frac{\cos \left(4\pi u\right)}{4\pi }}du=\frac{-1}{16{\pi }^2}{\displaystyle \int \cos \left(v\right)}dv\\ =\frac{-\sin \left(v\right)}{16{\pi }^2}\\ =\frac{-\sin \left(4\pi u\right)}{16{\pi }^2}\end{array}$

Substitute the above value in equation (22) as follows,

${\displaystyle \int u\sin \left(4\pi u\right)}du=\frac{\sin \left(4\pi u\right)}{16{\pi }^2}-\frac{u\cos \left(4\pi u\right)}{4\pi }$

Substitute the above value in equation (21) as follows,

$\begin{array}{c}\frac{1}{2\pi }{\displaystyle \int u\sin \left(4\pi u\right)}du=\frac{1}{2\pi }\left[\frac{\sin \left(4\pi u\right)}{16{\pi }^2}-\frac{u\cos \left(4\pi u\right)}{4\pi }\right]\\ =\frac{\sin \left(4\pi u\right)}{32{\pi }^3}-\frac{u\cos \left(4\pi u\right)}{8{\pi }^2}\end{array}$

Substitute the above value in equation (20) as follows,

${\displaystyle \int u^2\cos \left(4\pi u\right)du}=\frac{u^2\sin \left(4\pi u\right)}{4\pi }-\frac{\sin \left(4\pi u\right)}{32{\pi }^3}+\frac{u\cos \left(4\pi u\right)}{8{\pi }^2}$

Substitute the above value in equation (18) as follows

$\begin{array}{c}-\frac{3}{4\pi }{\displaystyle \int u^2\cos \left(4\pi u\right)du}=-\frac{3}{4\pi }\left[\frac{u^2\sin \left(4\pi u\right)}{4\pi }-\frac{\sin \left(4\pi u\right)}{32{\pi }^3}+\frac{u\cos \left(4\pi u\right)}{8{\pi }^2}\right]\\ =-\frac{3u^2\sin \left(4\pi u\right)}{16{\pi }^2}+\frac{3\sin \left(4\pi u\right)}{128{\pi }^4}-\frac{3u\cos \left(4\pi u\right)}{32{\pi }^3}\end{array}$

Substitute the above value in equation (17) as follows

${\displaystyle \int u^3\sin \left(4\pi u\right)}du=\frac{3u^2\sin \left(4\pi u\right)}{16{\pi }^2}-\frac{3\sin \left(4\pi u\right)}{128{\pi }^4}+\frac{3u\cos \left(4\pi u\right)}{32{\pi }^3}-\frac{u^3\cos \left(4\pi u\right)}{4\pi }$

Substitute the above value in equation (16) as follows

$\begin{array}{c}\frac{1}{\pi }{\displaystyle \int u^3\sin \left(4\pi u\right)}du=\frac{1}{\pi }\left[\frac{3u^2\sin \left(4\pi u\right)}{16{\pi }^2}-\frac{3\sin \left(4\pi u\right)}{128{\pi }^4}+\frac{3u\cos \left(4\pi u\right)}{32{\pi }^3}-\frac{u^3\cos \left(4\pi u\right)}{4\pi }\right]\\ =\frac{3u^2\sin \left(4\pi u\right)}{16{\pi }^3}-\frac{3\sin \left(4\pi u\right)}{128{\pi }^5}-\frac{u^3\cos \left(4\pi u\right)}{4{\pi }^2}+\frac{3u\cos \left(4\pi u\right)}{32{\pi }^4}\end{array}$

Substitute the above value in equation (15) as follows

$m=\frac{u^4\sin \left(4\pi u\right)}{4\pi }-\frac{3u^2\sin \left(4\pi u\right)}{16{\pi }^3}+\frac{3\sin \left(4\pi u\right)}{128{\pi }^5}+\frac{u^3\cos \left(4\pi u\right)}{4{\pi }^2}-\frac{3u\cos \left(4\pi u\right)}{32{\pi }^4}$        (23)

Consider the function $n={\displaystyle \int u^4\cos \left(2\pi u\right)}du$ in the expression x.

$n={\displaystyle \int u^4\cos \left(2\pi u\right)}du$        (24)

Applying integration by parts in the above expression,

$\begin{array}{l}f=u^4{g}^{\prime }=\cos \left(2\pi u\right)\\ f^{\prime }=4u^{3}g=\frac{\sin \left(2\pi u\right)}{2\pi }\end{array}$

Applying the values in the above equation as follows,

$n=\frac{u^4\sin \left(2\pi u\right)}{2\pi }-{\displaystyle \int \frac{2u^3\sin \left(2\pi u\right)}{\pi }}du$        (25)

Consider,

$\frac{2}{\pi }{\displaystyle \int u^3\sin \left(2\pi u\right)du}$        (26)

Let,

${\displaystyle \int u^3\sin \left(2\pi u\right)du}$        (27)

Applying integration by parts in the above expression,

$\begin{array}{l}f=u^3{g}^{\prime }=\sin \left(2\pi u\right)\\ f^{\prime }=3u^{2}g=-\frac{\cos \left(2\pi u\right)}{2\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u^3\sin \left(2\pi u\right)du}=-\frac{u^3\cos \left(2\pi u\right)}{2\pi }-{\displaystyle \int -\frac{3u^2\cos \left(2\pi u\right)}{2\pi }}$        (28)

Consider,

$\frac{-3}{2\pi }{\displaystyle \int u^2\cos \left(2\pi u\right)du}$        (29)

Let,

${\displaystyle \int u^2\cos \left(2\pi u\right)du}$

Applying integration by parts in the above expression,

$\begin{array}{l}f=u^2{g}^{\prime }=\cos \left(2\pi u\right)\\ f^{\prime }=2ug=\frac{\sin \left(2\pi u\right)}{2\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u^2\cos \left(2\pi u\right)du}=\frac{u^2\sin \left(2\pi u\right)}{2\pi }-{\displaystyle \int \frac{u\sin \left(2\pi u\right)}{\pi }}du$        (30)

Consider,

$\frac{1}{\pi }{\displaystyle \int u\sin \left(2\pi u\right)}du$        (31)

Let,

${\displaystyle \int u\sin \left(2\pi u\right)}du$        (32)

Applying integration by parts in the above expression,

$\begin{array}{l}f=u{g}^{\prime }=\sin \left(2\pi u\right)\\ f^{\prime }=1g=-\frac{\cos \left(2\pi u\right)}{2\pi }\end{array}$

Applying the values in the above equation as follows,

${\displaystyle \int u\sin \left(2\pi u\right)}du=-\frac{u\cos \left(2\pi u\right)}{2\pi }-{\displaystyle \int -\frac{\cos \left(2\pi u\right)}{2\pi }du}$        (33)

Consider,

${\displaystyle \int -\frac{\cos \left(2\pi u\right)}{2\pi }du}$

Let,

$\begin{array}{l}v=2\pi u\\ \frac{dv}{du}=2\pi \\ du=\frac{1}{2\pi }dv\end{array}$

Applying the values in the above equation as follows,

$\begin{array}{c}{\displaystyle \int -\frac{\cos \left(2\pi u\right)}{2\pi }du}=-\frac{1}{4{\pi }^2}{\displaystyle \int \cos \left(v\right)}dv\\ =\frac{-\sin \left(v\right)}{4{\pi }^2}\\ =\frac{-\sin \left(2\pi u\right)}{4{\pi }^2}\end{array}$

Substitute the above value in equation (33) as follows,

${\displaystyle \int u\sin \left(2\pi u\right)}du=\frac{\sin \left(2\pi u\right)}{4{\pi }^2}-\frac{u\cos \left(2\pi u\right)}{2\pi }$

Substitute the above value in equation (31) as follows,

$\frac{1}{\pi }{\displaystyle \int u\sin \left(2\pi u\right)}du=\frac{\sin \left(2\pi u\right)}{4{\pi }^3}-\frac{u\cos \left(2\pi u\right)}{2{\pi }^2}$

Substitute the above value in equation (30) as follows,

${\displaystyle \int u^2\cos \left(2\pi u\right)du}=\frac{u^2\sin \left(2\pi u\right)}{2\pi }-\frac{\sin \left(2\pi u\right)}{4{\pi }^3}+\frac{u\cos \left(2\pi u\right)}{2{\pi }^2}$

Substitute the above value in equation (29) as follows,

$\begin{array}{c}\frac{-3}{2\pi }{\displaystyle \int u^2\cos \left(2\pi u\right)du}=\frac{-3}{2\pi }\left[\frac{u^2\sin \left(2\pi u\right)}{2\pi }-\frac{\sin \left(2\pi u\right)}{4{\pi }^3}+\frac{u\cos \left(2\pi u\right)}{2{\pi }^2}\right]\\ =\frac{-3u^2\sin \left(2\pi u\right)}{4{\pi }^2}+\frac{3\sin \left(2\pi u\right)}{8{\pi }^4}-\frac{3u\cos \left(2\pi u\right)}{4{\pi }^3}\end{array}$

Substitute the above value in equation (28) as follows,

${\displaystyle \int u^3\sin \left(2\pi u\right)du}=\frac{3u^2\sin \left(2\pi u\right)}{4{\pi }^2}-\frac{3\sin \left(2\pi u\right)}{8{\pi }^4}-\frac{u^3\cos \left(2\pi u\right)}{2\pi }+\frac{3u\cos \left(2\pi u\right)}{4{\pi }^3}$

Substitute the above value in equation (26) as follows,

$\begin{array}{c}\frac{2}{\pi }{\displaystyle \int u^3\sin \left(2\pi u\right)du}=\frac{2}{\pi }\left[\frac{3u^2\sin \left(2\pi u\right)}{4{\pi }^2}-\frac{3\sin \left(2\pi u\right)}{8{\pi }^4}-\frac{u^3\cos \left(2\pi u\right)}{2\pi }+\frac{3u\cos \left(2\pi u\right)}{4{\pi }^3}\right]\\ =\frac{3u^2\sin \left(2\pi u\right)}{2{\pi }^3}-\frac{3\sin \left(2\pi u\right)}{4{\pi }^5}-\frac{u^3\cos \left(2\pi u\right)}{{\pi }^2}+\frac{3u\cos \left(2\pi u\right)}{2{\pi }^4}\end{array}$

Substitute the above value in equation (25) as follows,

$n=\frac{u^4\sin \left(2\pi u\right)}{2\pi }-\frac{3u^2\sin \left(2\pi u\right)}{2{\pi }^3}+\frac{3\sin \left(2\pi u\right)}{4{\pi }^5}+\frac{u^3\cos \left(2\pi u\right)}{{\pi }^2}-\frac{3u\cos \left(2\pi u\right)}{2{\pi }^4}$        (34)

Consider the function ${\displaystyle \int u^{4}du}$ in the expression x as follows,

${\displaystyle \int u^{4}du}=\frac{u^5}{5}$        (35)

Substitute the calculated values of equation (34), (35), (23), and (14) in the expression x as follows,

$\begin{array}{l}x=-\frac{u^4\sin \left(6\pi u\right)}{48\pi }+\frac{u^2\sin \left(6\pi u\right)}{144{\pi }^3}-\frac{\sin \left(6\pi u\right)}{2592{\pi }^5}-\frac{u^3\cos \left(6\pi u\right)}{72{\pi }^2}+\frac{u\cos \left(6\pi u\right)}{432{\pi }^4}\\ +\frac{u^4\sin \left(4\pi u\right)}{16\pi }-\frac{3u^2\sin \left(4\pi u\right)}{64{\pi }^3}+\frac{3\sin \left(4\pi u\right)}{512{\pi }^5}+\frac{u^3\cos \left(4\pi u\right)}{16{\pi }^2}-\frac{3u\cos \left(4\pi u\right)}{128{\pi }^4}\\ -\frac{3u^4\sin \left(2\pi u\right)}{16\pi }+\frac{9u^2\sin \left(2\pi u\right)}{16{\pi }^3}-\frac{9\sin \left(2\pi u\right)}{32{\pi }^5}-\frac{3u^3\cos \left(2\pi u\right)}{8{\pi }^2}+\frac{9u\cos \left(2\pi u\right)}{16{\pi }^4}+\frac{u^5}{20}\end{array}$

Substitute the above value in equation (6) as follows,

$\begin{array}{l}w\left(t\right)=-\frac{2048u^4\sin \left(6\pi u\right)}{3\pi }+\frac{2048u^2\sin \left(6\pi u\right)}{9{\pi }^3}-\frac{1024\sin \left(6\pi u\right)}{81{\pi }^5}-\frac{4096u^3\cos \left(6\pi u\right)}{9{\pi }^2}+\frac{2048u\cos \left(6\pi u\right)}{27{\pi }^4}\\ +\frac{2048u^4\sin \left(4\pi u\right)}{\pi }-\frac{1536u^2\sin \left(4\pi u\right)}{{\pi }^3}+\frac{192\sin \left(4\pi u\right)}{{\pi }^5}+\frac{2048u^3\cos \left(4\pi u\right)}{{\pi }^2}\\ -\frac{768u\cos \left(4\pi u\right)}{{\pi }^4}-\frac{6144u^4\sin \left(2\pi u\right)}{\pi }+\frac{18432u^2\sin \left(2\pi u\right)}{{\pi }^3}-\frac{9216\sin \left(2\pi u\right)}{{\pi }^5}\\ -\frac{12288u^3\cos \left(2\pi u\right)}{{\pi }^2}+\frac{18432u\cos \left(2\pi u\right)}{{\pi }^4}+\frac{8192u^5}{5}\end{array}$

Substitute $u=\frac{t}{8}$ in the above expression as follows,

$\begin{array}{l}w\left(t\right)=-\frac{t^4\sin \left(\frac{3\pi t}{4}\right)}{6\pi }+\frac{32t^2\sin \left(\frac{3\pi t}{4}\right)}{9{\pi }^3}-\frac{1024\sin \left(\frac{3\pi t}{4}\right)}{81{\pi }^5}-\frac{8t^3\cos \left(\frac{3\pi t}{4}\right)}{9{\pi }^2}\\ +\frac{256t\cos \left(\frac{3\pi t}{4}\right)}{27{\pi }^4}+\frac{t^4\sin \left(\frac{\pi t}{2}\right)}{2\pi }-\frac{24t^2\sin \left(\frac{\pi t}{2}\right)}{{\pi }^3}+\frac{192\sin \left(\frac{\pi t}{2}\right)}{{\pi }^5}+\frac{4t^3\cos \left(\frac{\pi t}{2}\right)}{{\pi }^2}\\ -\frac{96t\cos \left(\frac{\pi t}{2}\right)}{{\pi }^4}-\frac{3t^4\sin \left(\frac{\pi t}{4}\right)}{2\pi }+\frac{288t^2\sin \left(\frac{\pi t}{2}\right)}{{\pi }^3}-\frac{9216\sin \left(\frac{\pi t}{4}\right)}{{\pi }^5}\\ -\frac{24t^3\cos \left(\frac{\pi t}{4}\right)}{{\pi }^2}+\frac{2304t\cos \left(\frac{\pi t}{4}\right)}{{\pi }^4}+\frac{x^5}{20}\end{array}$

Applying the limits to the above expression over a period of 60 s,

$\begin{array}{c}w\left(t\right)=20{\displaystyle \underset{0}{\overset{60}{\int }}w\left(t\right)dt}\\ =20\left[\frac{349920000{\pi }^4+56160000{\pi }^2-1301120}{9{\pi }^4}\right]\\ =20\left[\frac{6240000}{{\pi }^2}+\frac{1301120}{9{\pi }^4}+38880000\right]\\ =20\left[3.9512\times {10}^7\right]\end{array}$

$w\left(t\right)=79.024\times {10}^7\text{J}$

Matlab code to plot for wind power waveform:

t_end = 60;   % End time in seconds

t_pts = 600; % Number of points for time vector

t=linspace(0,t_end,t_pts);  % Define time vector

dt=t_end/t_pts; % Separation between time points

R=80; % Resistance in ohms

for i=1:t_pts; % Iterate for each point in time

    current(i)=0.5*t(i)^2*sin(pi/8*t(i))*cos(pi/4*t(i));

    p(i)=current(i)^2*R;

end

w=cumsum(p)*dt; % Energy from cumulative sum times time separation

figure(1)

subplot(3,1,1); plot(t,current,'r'); % Plot voltage

ylabel('Current (A)');

subplot(3,1,2); plot(t,p,'r') % Plot power

ylabel('Power (W)')

subplot(3,1,3); plot(t,w,'r') % Plot energy

xlabel('Time (seconds)')

ylabel('Energy (J)')

Matlab output:

Figure 1 shows the plot for the current, power, and energy waveform.

Conclusion:

Thus, the plot for the current, power, and energy waveform is sketched as shown in Figure 1, and the total energy collected over the 60 s time period is $79.024\times {10}^7\text{J}$.