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Engineering: environmental fluids
Engineering: environmental fluids

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1.1 The properties of the atmosphere

The atmosphere’s characteristics are important in various aspects of engineering such as the design and operation of aircraft, road and rail vehicles, buildings and other structures, and the all-important weather forecasting. The atmosphere has also provided a source of mechanical power down the ages – for example for windmills and wind pumps, sailing vessels, etc., and now of course a source for electrical power generation with wind-driven turbines and wind farms. The atmosphere’s properties and behaviour of interest in these fields include density, pressure, temperatures, wind speeds, accelerations and turbulence. These are very rarely stable being in a constant state of flux because of the rotation and other motions of the Earth with respect to the Sun and the intermittent heating and cooling cycles which result.

Although vital, the atmosphere height-wise is relatively very thin in relation to the diameter of the Earth. It has been likened to the thickness of the skin on an apple, but there is no real edge or boundary at the upper level. A common rule of thumb is for the upper limit (the Kármán line) to be 100 km from sea level, but in reality the density and pressure continue to diminish with height, with traces of atmosphere being detected at many hundreds of kilometres further up. Even at the height of some of the lower-orbit satellites (say around 150 km), there is a discernible atmosphere which will ultimately slow them down enough for them to fall back to Earth, and all but the largest will burn up before they hit the ground. The largest ones are decommissioned carefully so as to return to Earth in specified safe areas.

The density of the atmosphere at ground/sea level on a still day is taken as 1.225 kg m−3. The mean pressure at this level is stated as 101.325 kPa, which can be read as a mass of air of just over 10 tonnes on each square metre on the Earth’s surface. However, because the pressure reduces with height above ground level, the density decreases in proportion, hence the gradual diminishing with no definite boundary.

An illustrative statistic for the Earth’s atmosphere

As a matter of comparison, if the density of the atmosphere at sea level did remain constant all the way up, what would be the height or thickness of the atmosphere to create the sea-level pressure? Give your answer in km to 3 significant figures.

Solution

From the fundamental law of hydrostatics

cap p sub atm equals rho times g times h

so

equation sequence part 1 h equals part 2 cap p sub atm divided by rho times g equals part 3 101.325 multiplication 10 cubed Pa divided by 1.225 kg m super negative three multiplication 9.81 m s super negative two equals part 4 8431.6 times ellipsis m full stop

Therefore the height of the atmosphere would be 8.43 km (to 3 s.f.).

Activity 4

If the height of the atmosphere was 100.0 km, what would be the atmospheric pressure at sea level if the density was constant at 1.225 kg m−3? Express the answer as a comparison with the standard figure of 101.325 kPa. Give your answer to 3 significant figures.

Answer

From the fundamental law of hydrostatics

cap p sub atm equals rho times g times h

therefore the atmospheric pressure will be

cap p sub atm equals 1.225 kg m super negative three multiplication 9.81 m s super negative two multiplication 100.0 multiplication 10 cubed m equals 1201.725 kPa full stop

Comparing this to the standard figure,

1201.725 kPa divided by 101.325 kPa equals 11.860 times ellipsis

which is 11.9 times (to 3 s.f.) greater than the standard figure.

In reality, the atmosphere’s height and thickness are many times what they would be if the density was constant. The real density at sea level varies from a maximum of approximately 1.4 kg m−3 to a very low density at and above about 6 km height. The reduction of density of the atmosphere is evident at quite low levels; this limits the heights at which aircraft can generate sufficient lift and is why they have to fly so fast to climb high. Figure 1 is based on previously published data from the United Nations International Civil Aviation Organization (ICAO) regarding temperature variation with height. Other properties of interest can be deduced from their relationships with temperature.

Described image
Figure 1 Temperature variation with height in standard atmosphere

Note that from sea level (zero on the vertical axis) the temperature reduces with height in a directly linear manner up to about 11  km altitude. This lower region (0–11  km) is called the troposphere. Above 11  km the temperature stays the same at about –56.5  °C for increasing heights. This region is the stratosphere, and the height at which the constant temperature starts is the isothermal height (sometimes isothermal level), also known as the tropopause. At far greater altitudes there is more variation but because the air is so thin by then the concept of atmospheric temperature does not mean very much.

The values describing the graph in Figure 1 vary a little around the Earth, but typically the sea-level mean (average) temperature is assumed to be 15 °C and the slope or gradient of the graph from zero to 11 km height in degrees per km change in height is −6.5 °C km −1. Above 11 km, the gradient is of course zero as the temperature stays constant.

Calculating the height of the atmosphere at 0 °C

From the above data determine the height  h at which the air temperature reaches 0°C. Give your answer to 2 significant figures.

Solution

In Figure 1, studying the proportions of the slope part of the graph by similar triangles of height (vertical) divided by temperature (horizontal) gives

11000 m divided by 56.5 super degree cap c prefix plus of 15 super degree cap c equals z divided by 15 super degree cap c

so

equation sequence part 1 h equals part 2 15 super degree cap c prefix multiplication of 11 000 m divided by 71.5 super degree cap c equals part 3 2307.69 times ellipsis m full stop

Therefore the height at which the air temperature reaches 0 °C is 2.3 km (to 2 s.f.).

Other properties of air have been deduced or derived from known relationships with temperature and some of these are presented in Figure 2 in non-dimensionalised form so as to fit them all on one graph.

Described image
Figure 2 Properties of the standard atmosphere. Note that kinematic viscosity eta increases with altitude, so the inverse ratio eta sub zero divided by eta is shown.

The constants used to non-dimensionalise the properties are depicted on the graph with the subscript ‘0’ and correspond to the values at sea level. Note that the height is now on the horizontal axis with the 11 km height marked as a vertical line. The properties shown are the local speed of sound (sonic velocity), c , the dynamic viscosity, eta , the kinematic viscosity, nu , the density, rho , and the pressure, cap p . For instance, it can be seen from the graph that the speed of sound c sub zero  = 340m s−1 at sea level and decreases linearly through the troposphere; above the tropopause it remains constant, given approximately by

equation sequence part 1 c equals part 2 c sub zero multiplication 0.865 equals part 3 294 m s super negative one full stop

In the troposphere two properties, pressure and density, can be modelled by simple expressions as follows.

For pressure:

cap p divided by cap p sub zero equals left parenthesis one minus z divided by z sub cap c right parenthesis super 5.256
Equation label: (Equation 1)

where cap p is the absolute pressure, cap p sub zero (=  cap p sub atm ) is the standard sea-level value of 101.3kPa, z is the altitude under consideration and z sub cap c is a constant value of 44 300 m.

For density:

rho divided by rho sub zero equals left parenthesis one minus z divided by z sub cap c right parenthesis super 4.256
Equation label: (Equation 2)

where rho is the required density, rho sub zero is the standard sea-level value of 1.225kg m−3, z is the altitude under consideration and z sub cap c is a constant value of 44 300 m.

Above the isothermal level of the tropopause, the pressure and density are based on the values at this isothermal level, as:

equation sequence part 1 cap p divided by cap p sub one equals part 2 rho divided by rho sub one equals part 3 exp left parenthesis negative z minus z sub one divided by z sub d right parenthesis
Equation label: (Equation 3)

where z sub d is a constant value of 6377 m and cap p sub one , rho sub one and z sub one are the values of pressure, density and altitude at the isothermal level, and which will be, respectively, 22.6 kPa, 0.364 kg m−3 and 11000 m.

Reductions in atmospheric pressure with height

What is the percentage reduction in atmospheric pressure at the height when the air temperature drops to 0 °C? Give your answer to 3 significant figures.

Solution

Rearranging Equation 1 to find pressure  cap p gives

cap p equals left parenthesis one minus z divided by z sub cap c right parenthesis super 5.256 multiplication cap p sub zero full stop

From Example 2, the height of the atmosphere at which the air temperature reaches 0 °C was found to be z equals 2307.69 m , so

equation sequence part 1 cap p equals part 2 left parenthesis one minus 2307.69 m divided by 44 300.0 m right parenthesis super 5.256 multiplication left parenthesis 101.3 multiplication 10 cubed Pa right parenthesis equals part 3 76.47 times ellipsis multiplication 10 cubed Pa full stop

Therefore the percentage reduction in pressure is

left parenthesis 101.3 minus 76.47 times ellipsis right parenthesis kPa divided by 101.3 kPa multiplication 100 percent equals 24.5 percent left parenthesis to three s full stop f full stop right parenthesis full stop

Activity 5

What is the percentage reduction in atmospheric density at the height when the air temperature reaches 0 °C, compared with the value at sea level? Give your answer to 3 significant figures.

Answer

Equation 2,

rho divided by rho sub zero equals left parenthesis one minus z divided by z sub cap c right parenthesis super 4.256

can be rearranged to find density

equation sequence part 1 rho equals part 2 1.225 kg m super negative three multiplication left parenthesis one minus 2307.69 m divided by 44 300.0 m right parenthesis super 4.256 equals part 3 0.9755 times ellipsis kg m super negative three full stop

Therefore the percentage reduction in density is

left parenthesis 1.225 minus 0.9755 times ellipsis right parenthesis kg m super negative three divided by 1.225 kg m super negative three equals 20.4 percent left parenthesis to three s full stop f full stop right parenthesis full stop

Atmospheric model equations

The troposphere denotes the part of Earth’s atmosphere from an altitude of zero to 11 000 m. In this region the local atmospheric pressure can be evaluated from the expression in equation (4) as

cap p divided by cap p sub zero equals left parenthesis one minus z divided by z sub cap c right parenthesis super 5.256

where cap p sub zero is the standard sea-level value of 101.3 kPa, z is the altitude in metres under consideration and z sub cap c is a constant value of 44 300 m.

Also in the troposphere, the local air density can be evaluated from the expression in equation (5) as

rho divided by rho sub zero equals left parenthesis one minus z divided by z sub cap c right parenthesis super 4.256

where rho is the required density, rho sub zero is the standard sea-level value of 1.225kg m−3, z is the altitude under consideration and z sub cap c is a constant value of 44300 m.

The top level of the troposphere is known as the tropopause, and the height of the lower boundary of the tropopause, 11 000 m, is known as the isothermal height. Above this height, the pressure and density are both related by the expression in equation (6) as

equation sequence part 1 cap p divided by cap p sub one equals part 2 rho divided by rho sub one equals part 3 exp left parenthesis negative z minus z sub one divided by z sub d right parenthesis

where z sub d is a constant value of 6377 m and cap p sub one , rho sub one and z sub one are the values of the pressure, density and altitude at the isothermal level, and hence will be respectively 22.6 kPa, 0.364kg m−3 and 11000 m.