The angle of incidence of the sun's rays formula. Solar radiation

To be maximum is very important collector orientation and angle. To absorb the maximum amount, the plane of the solar collector must always be perpendicular to the sun's rays. However, the sun shines on the Earth's surface depending on the time of day and year. always at a different angle. Therefore, for the installation of solar collectors, it is necessary to know the optimal orientation in space. To assess the optimal orientation of the collectors, the rotation of the Earth around the Sun and around its axis, as well as changes in the distance from the Sun, are taken into account. To determine the position or, it is necessary to take into account basic angular parameters:

Latitude of the installation site φ;

Hourly angle ω;

Solar declination angle δ;

The angle of inclination to the horizon β;

Azimuth α;

Installation site latitude(φ) shows how much the place is north or south of the equator, and makes an angle from 0 ° to 90 °, counted from the plane of the equator to one of the poles - north or south.

hour angle(ω) converts local solar time to the number of degrees the sun travels across the sky. By definition, the hour angle is zero at noon. The earth rotates 15° in one hour. In the morning the angle of the sun is negative, in the evening it is positive.

Sun declination angle(δ) depends on the rotation of the Earth around the Sun, since the orbit of rotation has an elliptical shape and the axis of rotation itself is also tilted, the angle changes during the year from 23.45° to -23.45°. The declination angle becomes equal to zero twice a year on the days of the spring and autumn equinoxes.

The declination of the sun for a particular day is determined by the formula:

Tilt to the horizon(β) is formed between the horizontal plane and the solar panel. For example, when mounting on a sloping roof, the angle of inclination of the collector is determined by the slope of the roof slope.

Azimuth(α) characterizes the deviation of the absorbing plane of the collector from the south direction, when the solar collector is oriented exactly to the south, azimuth = 0°.

The angle of incidence of the sun's rays on an arbitrarily oriented surface, having a certain value of azimuth α and angle of inclination β, is determined by the formula:

If in this formula we replace the value of the angle β by 0, then we get an expression for determining the angle of incidence of the sun's rays on a horizontal surface:

The intensity of the solar radiation flux for a certain position of the absorbing panel in space is calculated by the formula:

Where J s and J d are the intensity of the fluxes of direct and diffuse solar radiation incident on a horizontal surface, respectively.

Solar collector position coefficients for direct and diffuse solar radiation.

To ensure that the absorber receives the maximum (for the estimated period) amount of solar energy, the collector is mounted in an inclined position with the optimal angle of inclination to the horizon β, which is determined by the calculation method and depends on the period of use of the solar system. With a southern orientation of the collector for year-round solar systems β = φ, for seasonal solar systems β = φ–15°. Then the formula will take the form, for seasonal solar systems:

For year-round:

Solar collectors oriented to the south and mounted at an angle of 30° to 65° relative to the horizon, allow you to achieve the maximum absorption value. But even under certain deviations from these conditions, it can generate a sufficient amount of energy. A low-angle installation is more efficient if the solar collectors or solar arrays cannot be oriented to the south.

For example, if the solar panels are oriented to the southwest, with an azimuth of 45° and an inclination angle of 30°, then such a system will be able to absorb up to 95% of the maximum amount of solar radiation. Or, when oriented east or west, up to 85% of energy can be delivered to the collector when the panels are installed at an angle of 25-35°. If the angle of inclination of the collector is larger, then the amount of energy entering the surface of the collector will be more uniform, for supporting heating, this installation option is more efficient.

Often the orientation of the solar collector depends on, the installation of the collector is carried out on the roof of the building, so it is very important at the design stage to take into account the possibility of optimal installation of the collectors.

In the same geographical point at different times of the day, the sun's rays fall on the earth at different angles. By calculating this angle and knowing the geographic coordinates, astronomical time can be accurately calculated. The opposite effect is also possible. With the help of a chronometer showing the exact astronomical time, you can georeference a point.

You will need

- gnomon;
- ruler;
- horizontal surface;
- liquid level to establish a horizontal surface;
- calculator;
- tables of tangents and cotangents.

Instruction

Find a strictly horizontal surface. Check it with a level. Both bubble and electronic devices can be used. If you are using a liquid level, the bubble should be exactly in the center. For the convenience of further work, fix a sheet of paper on the surface. It is best to use graph paper in this case. As a horizontal surface, you can take a sheet of thick, durable plywood. It should not have depressions and bumps.
Draw a dot or a cross on graph paper. Install the gnomon vertically so that its axis coincides with your mark. A gnomon is a rod or pole installed strictly vertically. Its top has the shape of a sharp cone.
Place a second dot at the end point of the gnomon's shadow. Designate it as point A, and the first - as point C. You should know the height of the gnomon with sufficient accuracy. The larger the gnomon, the more accurate the result will be.
Measure the distance from point A to point C in any way you can. Please note that the units of measurement are the same as the height of the gnomon. If necessary, convert to the most convenient units.
On a separate sheet of paper, draw a drawing using the data obtained. In the drawing, you should get a right-angled triangle, in which the right angle C is the location of the gnomon, the leg CA is the length of the shadow, and the leg CB is the height of the gnomon.
Calculate angle A using tangent or cotangent using the formula tgA=BC/AC. Knowing the tangent, determine the actual angle.
The resulting angle is the angle between the horizontal surface and the sunbeam. The angle of incidence is the angle between the perpendicular dropped onto the surface and the beam. That is, it is equal to 90º-A.

Memo for solving problems on the topic "Earth as a planet of the solar system"

To perform tasks to determine the height of the Sun above the horizon at various points located on the same parallel, it is necessary to determine the noon meridian using the data on the time of the Greenwich meridian. The midday meridian is determined by the formula:

(12 hours - Greenwich meridian time) * 15º - if the meridian is in the Eastern Hemisphere;

(Greenwich meridian time - 12 noon) * 15º - if the meridian is in the Western Hemisphere.

The closer the meridians proposed in the task are to the noon meridian, the higher the Sun will be in them, the farther - the lower.

Example1. .

Determine in which of the points indicated by letters on the map of Australia, on March 21, the sun will beuppermost above the horizon at 5 am GMT solar time. Write down the rationale for your answer.

Answer. At point A

Point A is closer than other points to the midday meridian (12 - 5) * 15º \u003d 120º East.

Example2. Determine in which of the points indicated by letters on the map of North America the Sun will be located below everything above the horizon at 18:00 GMT. Write down your reasoning.

Answer. At point A (18-12)*15º =90 º

2. To perform tasks to determine the height of the Sun above the horizon at various points that are not on the same parallel, and when there is an indication of the day of the winter (December 22) or summer (June 22) solstice, you need

remember that the Earth moves counterclockwise and the more east the point is, the earlier the Sun will rise above the horizon .;

to analyze the position of the points indicated in the assignment relative to the polar circles and tropics. For example, if the question contains an indication of the day - December 20, this means a day close to the day of the winter solstice, when polar night is observed in the territory north of the Arctic Circle. This means that the further north the point is located, the later the Sun will rise above the horizon, the further south, the earlier.

Determine in which of the points indicated by letters on the map of North America, on December 20, the Sun first of all on the Greenwich meridian time will rise above the horizon. Write down your reasoning.

Answer. At point C.

Point A is located to the east of point C, and point C is to the north (December 20, the shorter the day, the closer to the north pole).

Determine which of the parallels: 20° N, 10° N, at the equator, 10° S, or 20° S. - will there be a maximum length of day on the day when the Earth is in orbit in the position shown in the figure by number 3? Justify your answer.

Answer.The maximum duration will be at latitude 20 S.

At point 3, the Earth is on the day of the winter solstice - December 22, in conditions of longer daylight - the Southern Hemisphere. Point A occupies the southernmost position.

On which of the parallels indicated in the figure by letters, on December 22, the daylight hours are the shortest?

4. To determine the geographical latitude of the area, the dependence of the angle of incidence of the sun's rays on the latitude of the area is taken into account. On the days of the equinox(March 21 and September 23), when the rays of the Sun fall vertically on the equator, the formula is used to determine the geographical latitude:

90 º - angle of incidence of the sun's rays = latitude of the area (north or south is determined by the shadows cast by objects).

On the days of the solstices (June 22 and December 22), it must be taken into account that the rays of the Sun fall vertically (at an angle of 90º) on the tropic (23.5 º N and 23.5º S). Therefore, to determine the latitude of the area in the illuminated hemisphere (for example, June 22 in the Northern Hemisphere), the formula is used:

90º- (angle of incidence of the sun's rays - 23.5º) = latitude of the area

To determine the latitude of the area in the unlit hemisphere (for example, December 22 in the Northern Hemisphere), the formula is used:

90º - (angle of incidence of the sun's rays + 23.5º) = latitude of the area

Example1.

Determine the geographical coordinates of the point if it is known that on the days of the equinox the noon Sun is above the horizon at a height of 40º (the shadow of the object falls to the north), and local time is ahead of the Greenwich meridian time by 3 hours. Write down your calculations and reasoning

Answer. 50 º N, 60 º E

90 º - 40 º = 50 º ( NL , because the shadow of objects falls to the north in the northern hemisphere)

(12-9)x15 =60º ( o.d. , because local time is ahead of Greenwich Mean Time, so the point is located to the east)

Example2.

Determine the geographical coordinates of a point located in the United States, if it is known that on March 21 at 17 o'clock in the solar time of the Greenwich meridian at this point it is noon and the Sun is at an altitude of 50 ° above the horizon. Write down your reasoning.

Answer. 40ºN, 75ºW

90 º -50 º =40 º ( NL -because USA is in the northern hemisphere

(17h -12h)*15 = 75º (h.d., because it is located from the Greenwich meridian to the west for 3 time zones)

Example3.

Determine the geographical latitude of the place if it is known that on June 22 the noon Sun is above the horizon at an altitude of 35º NL Record your calculations.

Answer.78,5 º NL

90 º -(35 º -23.5 º ) = 78.5 s.l.

5. To determine the meridian (geographical longitude of the area), on which the point is located, according to the time of the Greenwich meridian and local solar time, it is necessary to determine the time difference between them. For example, if it is noon (12 o'clock) on the Greenwich meridian, and the local solar time at the specified point is 8 o'clock, the difference (12-8) is 4 hours. The length of one time zone is 15º. To determine the desired meridian, the calculation is 4 x 15º = 60º. To determine the hemisphere in which a given meridian is located, you need to remember that the Earth rotates from west to east (counterclockwise). So, if the time of the Greenwich meridian is greater than at a given point, the point is in the Western Hemisphere (as in the proposed example). If the time of the Greenwich meridian is less than the given point, the point is in the Eastern Hemisphere.

Example.

On what meridian is the point located, if it is known that at noon, according to the Greenwich meridian time, the local solar time is 16 hours in it? Write down your reasoning.

Answer. The point is on meridian 60º o.d.

16h. -12h. = 4 hours (time difference)

4x15 º = 60 º

Eastern longitude, because at the point 16.00, when it is still 12.00 at Greenwich (i.e. the point is located to the east)

The most important source from which the surface of the Earth and the atmosphere receive thermal energy is the Sun. It sends a colossal amount of radiant energy into the world space: thermal, light, ultraviolet. Electromagnetic waves emitted by the Sun propagate at a speed of 300,000 km/s.

The heating of the earth's surface depends on the angle of incidence of the sun's rays. All the sun's rays hit the Earth's surface parallel to each other, but since the Earth has a spherical shape, the sun's rays fall on different parts of its surface at different angles. When the Sun is at its zenith, its rays fall vertically and the Earth heats up more.

The totality of radiant energy sent by the Sun is called solar radiation, it is usually expressed in calories per surface area per year.

Solar radiation determines the temperature regime of the Earth's air troposphere.

It should be noted that the total amount of solar radiation is more than two billion times the amount of energy received by the Earth.

Radiation reaching the earth's surface consists of direct and diffuse.

Radiation that comes to Earth directly from the Sun in the form of direct sunlight in a cloudless sky is called straight. It carries the greatest amount of heat and light. If our planet had no atmosphere, the earth's surface would receive only direct radiation.

However, passing through the atmosphere, about a quarter of the solar radiation is scattered by gas molecules and impurities, deviates from the direct path. Some of them reach the Earth's surface, forming scattered solar radiation. Thanks to scattered radiation, light also penetrates into places where direct sunlight (direct radiation) does not penetrate. This radiation creates daylight and gives color to the sky.

Total solar radiation

All the rays of the sun that hit the earth are total solar radiation i.e., the totality of direct and diffuse radiation (Fig. 1).

Rice. 1. Total solar radiation per year

Distribution of solar radiation over the earth's surface

Solar radiation is distributed unevenly over the earth. It depends:

1. on the density and humidity of the air - the higher they are, the less radiation the earth's surface receives;

2. from the geographical latitude of the area - the amount of radiation increases from the poles to the equator. The amount of direct solar radiation depends on the length of the path that the sun's rays travel through the atmosphere. When the Sun is at its zenith (the angle of incidence of the rays is 90 °), its rays hit the Earth in the shortest way and intensively give off their energy to a small area. On Earth, this occurs in the band between 23° N. sh. and 23°S sh., i.e. between the tropics. As you move away from this zone to the south or north, the length of the path of the sun's rays increases, i.e., the angle of their incidence on the earth's surface decreases. The rays begin to fall on the Earth at a smaller angle, as if gliding, approaching the tangent line in the region of the poles. As a result, the same energy flow is distributed over a larger area, so the amount of reflected energy increases. Thus, in the region of the equator, where the sun's rays fall on the earth's surface at an angle of 90 °, the amount of direct solar radiation received by the earth's surface is higher, and as you move towards the poles, this amount is sharply reduced. In addition, the length of the day at different times of the year also depends on the latitude of the area, which also determines the amount of solar radiation entering the earth's surface;

3. from the annual and daily movement of the Earth - in the middle and high latitudes, the influx of solar radiation varies greatly according to the seasons, which is associated with a change in the noon height of the Sun and the length of the day;

4. on the nature of the earth's surface - the brighter the surface, the more sunlight it reflects. The ability of a surface to reflect radiation is called albedo(from lat. whiteness). Snow reflects radiation especially strongly (90%), sand is weaker (35%), chernozem is even weaker (4%).

Earth's surface, absorbing solar radiation (absorbed radiation), heats up and radiates heat into the atmosphere (reflected radiation). The lower layers of the atmosphere largely delay terrestrial radiation. The radiation absorbed by the earth's surface is spent on heating the soil, air, and water.

That part of the total radiation that remains after reflection and thermal radiation of the earth's surface is called radiation balance. The radiation balance of the earth's surface varies during the day and seasons of the year, but on average for the year it has a positive value everywhere, with the exception of the icy deserts of Greenland and Antarctica. Radiation balance reaches its maximum values at low latitudes (between 20°N and 20°S) - over 42*10 2 J/m 2 , at a latitude of about 60° in both hemispheres it decreases to 8*10 2 - 13 * 10 2 J / m 2.

The sun's rays give up to 20% of their energy to the atmosphere, which is distributed throughout the entire thickness of the air, and therefore the heating of the air caused by them is relatively small. The sun heats the earth's surface, which transfers heat to the atmospheric air due to convection(from lat. convection- delivery), i.e., the vertical movement of air heated at the earth's surface, in place of which colder air descends. This is how the atmosphere receives most of its heat, on average three times more than directly from the sun.

The presence of carbon dioxide and water vapor does not allow the heat reflected from the earth's surface to freely escape into outer space. They create Greenhouse effect, due to which the temperature drop on Earth during the day does not exceed 15 ° C. In the absence of carbon dioxide in the atmosphere, the earth's surface would cool down by 40-50 °C overnight.

As a result of the growth in the scale of human economic activity - the burning of coal and oil at thermal power plants, emissions from industrial enterprises, an increase in car emissions - the content of carbon dioxide in the atmosphere increases, which leads to an increase in the greenhouse effect and threatens global climate change.

The sun's rays, having passed through the atmosphere, fall on the surface of the Earth and heat it, and that, in turn, gives off heat to the atmosphere. This explains the characteristic feature of the troposphere: a decrease in air temperature with height. But there are times when the upper layers of the atmosphere are warmer than the lower ones. Such a phenomenon is called temperature inversion(from lat. inversio - turning over).