Chapter 24, Problem 60P

(a)

To determine

The power received by the city from the space mirror reflecting light.

Answer to Problem 60P

The power received by the city from the space mirror reflecting light is $32.1\text{MW}$.

Explanation of Solution

Given info: The diameter of the space mirror is $200\text{m}$ and the intensity of the sunlight is $1370\text{W}/{\text{m}}^2$. The percentage of light reflected from the space mirror to the city is $74.6\%$.

The formula to calculate the area of space mirror is,

$A=\frac{\pi d^2}{4}$ (1)

Here,

$d$ is the diameter of the space mirror.

Substitute $200\text{m}$ for $d$ in equation (1) to find the value of $A$.

$\begin{array}{c}A=\frac{\pi {\left(200\text{m}\right)}^2}{4}\\ =31415{\text{m}}^2\end{array}$

The formula to calculate the net power received by the mirror station is,

$P=SA$ (2)

Here,

$S$ is the intensity of radiations .

$A$ is the area of space station.

Substitute $31415{\text{m}}^2$ for $A$ and $1370\text{W}/{\text{m}}^2$ for $S$ in the equation (2) to find the value of $P$.

$\begin{array}{c}P=\left(1370\text{W}/{\text{m}}^2\right)\left(31415{\text{m}}^2\right)\\ =43.03\times {10}^6\text{W}\times \frac{{10}^{-6}\text{MW}}{1\text{W}}\\ =43.03\text{MW}\end{array}$

The formula to calculate the power received by the city is,

$P_{\text{city}}=74.6\%\times P$

Substitute $43.03\text{MW}$ for $P$ in the above equation to find the value of $P_{\text{city}}$.

$\begin{array}{c}{P}_{\text{city}}=\frac{74.6\%}{100\%}\times \left(43.03\text{MW}\right)\\ =32.1\text{MW}\end{array}$

Thus, the power received by the city from the space mirror reflecting light is $32.1\text{MW}$.

Conclusion:

Therefore, the power received by the city from the space mirror reflecting light is $32.1\text{MW}$.

(b)

To determine

The intensity of light received by the city.

Answer to Problem 60P

The maximum intensity of light received by the city is $0.63\text{W}/{\text{m}}^2$.

Explanation of Solution

Given info: The diameter of the space mirror is $200\text{m}$ and the intensity of the sunlight is $1370\text{W}/{\text{m}}^2$. The diameter of the city is $8.00\text{km}$.

The formula to calculate the area of space mirror is,

$A=\frac{\pi d^2}{4}$ (1)

Here,

$d$ is the diameter of the space mirror.

Substitute $8.00\text{km}$ for $d$ in equation (1) to find the value of $A$.

$\begin{array}{c}A=\frac{\pi {\left(8.00\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)}^2}{4}\\ =5.03\times {10}^7{\text{m}}^2\end{array}$

The formula to calculate the net power received by the city is,

$P=SA$

Here,

$S$ is the intensity of radiations.

$A$ is the area of the city.

Rewrite the above equation to find the value of $S$.

$S_{\text{city}}=\frac{P_{\text{city}}}{A}$

Substitute $32.06\text{MW}$ for $P_{\text{city}}$ and $5.03\times {10}^7{\text{m}}^2$ for $A$ in the above equation to find the value of $S_{\text{city}}$.

$\begin{array}{c}S=\frac{\left(32.06\text{MW}\times \frac{{10}^6\text{W}}{1\text{MW}}\right)}{\left(5.03\times {10}^7{\text{m}}^2\right)}\\ =0.63\text{W}/{\text{m}}^2\end{array}$

Thus, the maximum intensity of light received by the city is $0.63\text{W}/{\text{m}}^2$.

Conclusion:

Therefore, the maximum intensity of light received by the city is $0.63\text{W}/{\text{m}}^2$.

(c)

To determine

The percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of $7.0°$ above the horizon at noon.

Answer to Problem 60P

The percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of $7.0°$ above the horizon at noon is $0.505\%$.

Explanation of Solution

Given info: The diameter of the space mirror is $200\text{m}$ and the intensity of the sunlight is $1370\text{W}/{\text{m}}^2$. The diameter of the city is $8.00\text{km}$.

The formula to calculate the intensity of light on earth is,

$S^{\prime }=S\times 74.6\%$

Substitute $1370\text{W}/{\text{m}}^2$ for $S$ in the above equation to find the value of $S^{\prime }$.

$\begin{array}{c}{S}^{\prime }=\left(1370\text{W}/{\text{m}}^2\right)\times \frac{74.6\%}{100\%}\\ =1022.02\text{W}/{\text{m}}^2\end{array}$

The sunlight reaches the city St. Petersburg when the sun reaches at an angle of $7.0°$ above the horizon at noon so the vertical component of the intensity of sunlight will be,

${S^{\prime }}_{\text{V}}=S^{\prime }\times \sin \left(7°\right)$ . (3)

Substitute $1022.02\text{W}/{\text{m}}^2$ for $S^{\prime }$ in equation (3) to find ${S^{\prime }}_{\text{V}}$.

$\begin{array}{c}{S^{\prime }}_{\text{V}}=\left(1022.02\text{W}/{\text{m}}^2\right)\times \sin \left(7°\right)\\ =124.55\text{W}/{\text{m}}^2\end{array}$

The formula to calculate the percentage of radiation is,

$S_{\text{percentage}}=\frac{S_{\text{city}}\times 100\%}{{S^{\prime }}_{\text{V}}}$ (4)

Substitute $124.55\text{W}/{\text{m}}^2$ for ${S^{\prime }}_{\text{V}}$ and $0.63\text{W}/{\text{m}}^2$ for $S_{\text{city}}$ in equation (4) to find the value of $S_{\text{percentage}}$.

$\begin{array}{c}{S}_{\text{percentage}}=\frac{0.63\text{W}/{\text{m}}^2\times 100\%}{124.55\text{W}/{\text{m}}^2}\\ =0.505\%\end{array}$

Thus, the percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of $7.0°$ above the horizon at noon is $0.505\%$.

Conclusion:

Therefore, the percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of $7.0°$ above the horizon at noon is $0.505\%$.