 2022-01-07

Two sites are being considered for wind power generation. In the first site, the wind blows steadily at $7\frac{m}{s}$ for 3000 hours per year, whereas in the second site the wind blows at $10\frac{m}{s}$ for 1500 hours per year. Assuming the wind velocity is negligible at other times for simplicity, determine which is a better site for wind power generation. Debbie Moore

Step 1
Knows and requirements
Consider the two sites:
First the wind velocity equals ${V}_{1}=7\frac{m}{s}$ for ${t}_{1}=3000$ hours per year.
Second the wind velocity equals ${V}_{2}=10\frac{m}{s}$ for ${t}_{2}=1500$ hours per year.
Neglecting the energy of the wind at other times of the year. Its Ella Williams

Step 1
Givens:
${V}_{1}=7\frac{m}{s}$
${\text{Operation time}}_{1}=3000$ hours per year
${V}_{2}=10\frac{m}{s}$
${\text{Operation time}}_{2}=1500$ hours per year
Solution:
Note: The wind blows streadly. So, $\stackrel{˙}{KE}=\left(\stackrel{˙}{0.5m{V}^{2}}\right)=\stackrel{˙}{m}×0.5{V}^{2}$
The power generation is the time rate of kinetic energy, which can be calculated as follows,
Power $=\mathrm{\Delta }\stackrel{˙}{KE}=\stackrel{˙}{m}\frac{{V}^{2}}{2}$
Regarding that $\stackrel{˙}{m}\propto V$. Then,
Power $\propto {V}^{3}⇒$ Power $=const×{V}^{3}$
Since ${\rho }_{a}$ is constant for both sites and Area is the same as same wind turbine is use turine is used.
For the first site,
${\text{Power}}_{1}=const.×{V}_{1}^{3}=const.×343W$
For the second site,
${\text{Power}}_{2}=cont.×{V}_{2}^{3}=const.×1000W$
Calculating energy generation per year for each of the two sites.
${E}_{year}=Power×\text{Operation time per year}$
For the first site,

For the second site,
${E}_{year,2}=Powe{r}_{2}×{\text{Operation time}}_{2}=const.×1000×1500=const.×1500000\left(W\cdot \frac{hour}{year}\right)$
$\therefore {E}_{year,2}>{E}_{year,1}$ (i.e. Second site is better) karton

Step 1
Explanation:
$\begin{array}{}\\ {V}_{1}=7\frac{m}{s}\\ Tim{e}_{1}=3000\frac{hours}{year}\\ {V}_{2}=10\frac{m}{s}\\ Tim{e}_{2}=1500\frac{hours}{year}\end{array}$
The power generation is the time rate of kinetic energy which can be calculated as:
Powert $=\mathrm{\Delta }KE=m×\frac{{V}^{2}}{2}$
Regarding that $m\propto V$ Then
Power $\propto {V}^{3}⇒Power=constant×{V}^{3}$
Since $\rho a$ is constant for both sides and Area is the same as same wind turbine is used
For First site
$Powe{r}_{1}=const.{V}^{3}$
$Power{t}_{1}=const.\left(7\frac{m}{s}{\right)}^{3}$
$Powe{r}_{1}=const.343Watt$
For second side
$Powe{r}_{2}=const.{V}_{2}^{3}$
$Powe{r}_{2}=conts.\left(10\frac{m}{s}{\right)}^{3}$
$Powe{r}_{2}=const.1000W$
Calculating energy generation per year for each of two sites
${E}_{year}=Power×\text{Operation time per year}$
For First site

${E}_{year1}=const.×343W×3000$
${E}_{year1}=const.10290000\left(W.\frac{hour}{year}\right)$
For Second site

${E}_{year2}=const.×1000W×1500$
${E}_{year2}=const.1500000\left(W.\frac{hour}{year}\right)$
So,
${E}_{year2}>{E}_{year1}$ (Second site is better) xleb123

Let's calculate the energy generated at each site using the formula $E=k·{v}^{3}·t$, where $E$ is the energy generated, $k$ is a constant, $v$ is the wind speed, and $t$ is the time.
For the first site:

For the second site:

To compare the two sites, we can divide the energy generated at the second site by the energy generated at the first site:

Simplifying, we find:

Now we can calculate this ratio to determine which site is better for wind power generation.

$=\frac{{10}^{3}·1500}{{7}^{3}·3000}$
$=\frac{1500000}{343·3000}$
$=\frac{1500000}{1029000}$
Now, let's calculate this ratio:
$\frac{{E}_{2}}{{E}_{1}}\approx 1.459$
Since the ratio $\frac{{E}_{2}}{{E}_{1}}$ is greater than 1, it indicates that the energy generated at the second site is higher than the first site. Therefore, the second site with wind blowing at 10 m/s for 1500 hours per year is a better site for wind power generation compared to the first site with wind blowing at 7 m/s for 3000 hours per year. fudzisako

Step 1: Given:
$E=\frac{1}{2}·\rho ·A·{v}^{3}·t$
where
$E$ is the energy produced,
$\rho$ is the air density (assumed constant),
$A$ is the swept area of the turbine blades,
$v$ is the wind speed, and
$t$ is the time the wind blows.
For the first site, the wind blows at ${v}_{1}=7\phantom{\rule{0.167em}{0ex}}\frac{m}{s}$ for ${t}_{1}=3000$ hours per year.
For the second site, the wind blows at ${v}_{2}=10\phantom{\rule{0.167em}{0ex}}\frac{m}{s}$ for ${t}_{2}=1500$ hours per year.
Let's assume the swept areas of the turbine blades are the same for both sites, so ${A}_{1}={A}_{2}=A$.
Step 2: Now, we can calculate the energy produced at each site:
For the first site:
${E}_{1}=\frac{1}{2}·\rho ·A·\left(7\phantom{\rule{0.167em}{0ex}}\frac{m}{s}{\right)}^{3}·3000\phantom{\rule{0.167em}{0ex}}\text{hours}$
For the second site:
${E}_{2}=\frac{1}{2}·\rho ·A·\left(10\phantom{\rule{0.167em}{0ex}}\frac{m}{s}{\right)}^{3}·1500\phantom{\rule{0.167em}{0ex}}\text{hours}$
To compare the energy production at the two sites, we can calculate the ratio:
$\frac{{E}_{1}}{{E}_{2}}=\frac{\frac{1}{2}·\rho ·A·\left(7\phantom{\rule{0.167em}{0ex}}\frac{m}{s}{\right)}^{3}·3000\phantom{\rule{0.167em}{0ex}}\text{hours}}{\frac{1}{2}·\rho ·A·\left(10\phantom{\rule{0.167em}{0ex}}\frac{m}{s}{\right)}^{3}·1500\phantom{\rule{0.167em}{0ex}}\text{hours}}$
Simplifying the expression, we get:
$\frac{{E}_{1}}{{E}_{2}}=\frac{\left(7\phantom{\rule{0.167em}{0ex}}\frac{m}{s}{\right)}^{3}·3000}{\left(10\phantom{\rule{0.167em}{0ex}}\frac{m}{s}{\right)}^{3}·1500}$
Calculating this ratio will allow us to determine which site is better for wind power generation.

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