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Q43 PE

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Found in: Page 262

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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# (a) Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be $$4.00 \times {10^{26}}{\rm{ W}}$$.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of $$1.30{\rm{ kW}}/{{\rm{m}}^2}$$ reaches Earth’s surface. Calculate the area in $${\rm{k}}{{\rm{m}}^2}$$ of solar energy collectors needed to replace an electric power plant that generates $$750{\rm{ MW}}$$ if the collectors convert an average of $$2.00\%$$ of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs $$\left( {1.05 \times {{10}^{20}}{\rm{ J}}} \right)$$? Australia’s energy needs $$\left( {5.4 \times {{10}^{18}}{\rm{ J}}} \right)$$? China’s energy needs $$\left( {6.3 \times {{10}^{19}}{\rm{ J}}} \right)$$? (These energy consumption values are from 2006.)

(a) The power per square meter reaching the Earth’s upper atmosphere is $$1.42{\rm{ kW}}/{{\rm{m}}^2}$$.

(b) The area of the solar collector needed to replace an electric power plant is $$28.8{\rm{ k}}{{\rm{m}}^2}$$, the area of solar collector to meet the energy demand of United States, Australia and China are $$128076{\rm{ k}}{{\rm{m}}^2}$$, $$6576.9{\rm{ k}}{{\rm{m}}^2}$$ and $$76538.5{\rm{ k}}{{\rm{m}}^2}$$, respectively.

See the step by step solution

## Power

Power is a scalar quantity which is defined how fast the energy is being used.

Mathematically,

$$P = \frac{E}{T}$$

Here, E is the amount of energy consumed and T is the time.

## Power per square meter reaching Earth’s upper atmosphere

(a)

The power reaching the earth upper atmosphere is,

$$P = \frac{{{P_{Sun}}}}{{4\pi {r^2}}}$$

Here, $${P_{Sun}}$$ is the power output of the Sun $$\left( {{P_{Sun}} = 4.00 \times {{10}^{26}}{\rm{ W}}} \right)$$, and r is the distance between the Sun and the Earth $$\left( {r = 149.6 \times {{10}^6}{\rm{ m}}} \right)$$.

Putting all known values,

\begin{aligned}P &= \frac{{\left( {4.00 \times {{10}^{26}}{\rm{ W}}} \right)}}{{4\pi \times {{\left( {149.6 \times {{10}^9}{\rm{ m}}} \right)}^2}}}\\ &= \left( {1422.28{\rm{ W}}/{{\rm{m}}^2}} \right) \times \left( {\frac{{1{\rm{ kW}}}}{{1000{\rm{ W}}}}} \right)\\ &= 1.42{\rm{ kW}}/{{\rm{m}}^2}\end{aligned}

Therefore, the power per square meter reaching the Earth’s upper atmosphere is $$1.42{\rm{ kW}}/{{\rm{m}}^2}$$.

## Area of the solar collector needed

(b)

Since the efficiency of the solar collector is 2% . Therefore, the solar power being converted into electric power is,

$$P' = {P_s} \times 0.02$$

Here, $${P_s}$$ is the energy reaches Earth’s surface $$\left( {{P_s} = 1.30{\rm{ kW}}/{{\rm{m}}^2}} \right)$$.

Putting all known values,

\begin{aligned}P' &= \left( {1.30{\rm{ kW}}/{{\rm{m}}^2}} \right) \times 0.02\\ &= 26{\rm{ W}}/{{\rm{m}}^2}\end{aligned}

The electric power plant generates a power of $${P_P} = 750{\rm{ MW}}$$. Therefore, the area of solar energy collector to replace this electric power plant is,

$$A = \frac{{{P_P}}}{{P'}}$$

Putting all known values,

\begin{aligned}A& = \frac{{750{\rm{ MW}}}}{{26{\rm{ W}}/{{\rm{m}}^2}}}\\ &= \frac{{\left( {750{\rm{ MW}}} \right) \times \left( {\frac{{10{\rm{ W}}}}{{1{\rm{ MW}}}}} \right)}}{{26{\rm{ W}}/{{\rm{m}}^2}}}\\ &= 2.88 \times {10^7}{\rm{ }}{{\rm{m}}^2} \times \left( {\frac{{1{\rm{ k}}{{\rm{m}}^2}}}{{{{10}^6}{\rm{ }}{{\rm{m}}^2}}}} \right)\\ &= 28.8{\rm{ k}}{{\rm{m}}^2}\end{aligned}

Therefore, the area of the solar collector needed to replace an electric power plant is $$28.8{\rm{ k}}{{\rm{m}}^2}$$.

## Area of the solar collector to meet the energy demand of United States

The energy consumption of United States in a year is $${E_{US}} = 1.05 \times {10^{20}}{\rm{ J}}$$. Therefore, power consumption of United States in a year is,

$${P_{US}} = \frac{{{E_{US}}}}{{1{\rm{ year}}}}$$

Putting all known values,

\begin{aligned}{P_{US}} &= \frac{{1.05 \times {{10}^{20}}{\rm{ J}}}}{{\left( {1{\rm{ year}}} \right) \times \left( {\frac{{365{\rm{ day}}}}{{1{\rm{ year}}}}} \right) \times \left( {\frac{{24{\rm{ hr}}}}{{1{\rm{ day}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ hr}}}}} \right)}}\\ &= 3.33 \times {10^{12}}\;{\rm{W}}\end{aligned}

The area of solar energy collector to meet the energy demand in United States is,

$${A_{US}} = \frac{{{P_{US}}}}{{P'}}$$

Putting all known values,

\begin{aligned}{A_{US}} &= \frac{{3.33 \times {{10}^{12}}{\rm{ W}}}}{{26{\rm{ W}}/{{\rm{m}}^2}}}\\ &= 1.28076 \times {10^{11}}{\rm{ }}{{\rm{m}}^2} \times \left( {\frac{{1{\rm{ k}}{{\rm{m}}^2}}}{{{{10}^6}{\rm{ }}{{\rm{m}}^2}}}} \right)\\ &= 128076{\rm{ k}}{{\rm{m}}^2}\end{aligned}

Therefore, the area of solar energy collector to meet the energy demand of United States is $$128076{\rm{ k}}{{\rm{m}}^2}$$.

## Area of the solar collector to meet the energy demand of Australia

The energy consumption of Australia in a year is $${E_A} = 5.4 \times {10^{18}}{\rm{ J}}$$. Therefore, power consumption of Australia in a year is,

$${P_A} = \frac{{{E_A}}}{{1{\rm{ year}}}}$$

Putting all known values,

\begin{aligned}{P_A} &= \frac{{5.4 \times {{10}^{18}}{\rm{ J}}}}{{\left( {1{\rm{ year}}} \right) \times \left( {\frac{{365{\rm{ day}}}}{{1{\rm{ year}}}}} \right) \times \left( {\frac{{24{\rm{ hr}}}}{{1{\rm{ day}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ hr}}}}} \right)}}\\ &= 1.71 \times {10^{11}}\;{\rm{W}}\end{aligned}

The area of solar energy collector to meet the energy demand in Australia is,

$${A_A} = \frac{{{P_A}}}{{P'}}$$

Putting all known values,

\begin{aligned}{A_A} &= \frac{{1.71 \times {{10}^{11}}{\rm{ W}}}}{{26{\rm{ W}}/{{\rm{m}}^2}}}\\ &= 6.5769 \times {10^9}{\rm{ }}{{\rm{m}}^2} \times \left( {\frac{{1{\rm{ k}}{{\rm{m}}^2}}}{{{{10}^6}{\rm{ }}{{\rm{m}}^2}}}} \right)\\ &= 6576.9{\rm{ k}}{{\rm{m}}^2}\end{aligned}

Therefore, the area of solar energy collector to meet the energy demand of Australia is $$6576.9{\rm{ k}}{{\rm{m}}^2}$$.

## Area of the solar collector to meet the energy demand of China

The energy consumption of China in a year is $${E_C} = 6.3 \times {10^{19}}{\rm{ J}}$$. Therefore, power consumption of Australia in a year is,

$${P_C} = \frac{{{E_C}}}{{1{\rm{ year}}}}$$

Putting all known values,

\begin{aligned}{P_C} &= \frac{{6.3 \times {{10}^{19}}{\rm{ J}}}}{{\left( {1{\rm{ year}}} \right) \times \left( {\frac{{365{\rm{ day}}}}{{1{\rm{ year}}}}} \right) \times \left( {\frac{{24{\rm{ hr}}}}{{1{\rm{ day}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ hr}}}}} \right)}}\\ &= 1.99 \times {10^{12}}\;{\rm{W}}\end{aligned}

The area of solar energy collector to meet the energy demand in China is,

$${A_C} = \frac{{{P_C}}}{{P'}}$$

Putting all known values,

\begin{aligned}{A_C} &= \frac{{1.99 \times {{10}^{12}}{\rm{ W}}}}{{26{\rm{ W}}/{{\rm{m}}^2}}}\\ &= 7.65385 \times {10^{10}}{\rm{ }}{{\rm{m}}^2} \times \left( {\frac{{1{\rm{ k}}{{\rm{m}}^2}}}{{{{10}^6}{\rm{ }}{{\rm{m}}^2}}}} \right)\\ &= 76538.5{\rm{ k}}{{\rm{m}}^2}\end{aligned}

Therefore, the area of solar energy collector to meet the energy demand of China is $$76538.5{\rm{ k}}{{\rm{m}}^2}$$.

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