Page 122 - Geosystems An Introduction to Physical Geography 4th Canadian Edition

aQuantitativeSOlUTiOn Lapse Rates
This chapter introduced the concept of environmental lapse rates, the measured change in temperature in the atmosphere for a given change in altitude. The global average of environmental lapse rates is the normal lapse rate. We can use lapse rates to calculate temperatures at different altitudes. Being able to do these calculations will benefit your understanding of atmospheric processes described in Chapters 5, 7, and 8.
Expressions of lapse rates are in the form of a change in temperature per unit change in altitude. The normal lapse rate has a value of Δ6.4 C·1000 m−1, which in equivalent terms is Δ6.4 C·km−1 (1000 m = 1 km) or Δ0.64 C·100 m−1. Δ is the symbol for change. The value of 100 m, 1000 m, or 1 km in these expressions represents the size of each step in altitude over which the temperature change is measured.
As seen in Figure 3.3, in the troposphere the temperature typically decreases with an increase in altitude. When you see a lapse rate with a positive value, such as Δ4.0 C · km−1, by conven- tion this represents the usual case of temperature decreasing with increasing altitude. You can interpret Δ4.0 C°·km−1 as meaning, “For every 1 km increase in altitude, the temperature decreases by 4 C°.” It also means: “For every 1 km decrease in altitude, the temperature increases by 4 C°.”
In Figure 3.3b, the temperature at the surface is 15°C. Applying the normal lapse rate, what is the temperature at 10 km altitude?
With 6.4 C°·km−1 there are 10 1–km steps in altitude between the surface (0 km) and 10 km. Because the new altitude is higher than the starting point, this change in altitude is expressed as
a positive value (10 km − 0 km = +10 km). Set up an equation where the answer (T10 km) is calculated as the initial temperature (15°C at 0 km) minus the lapse rate multiplied by the number of steps in altitude:
T10 km = T0 km − (6.4 × 10) = 15 − (6.4 × 10)
= 15 − 64 = −49°C
Check Figure 3.3b to confirm that the temperature for
10 km was calculated correctly. Note that if the normal lapse rate were expressed as C°·km−1, there would be 100 steps in altitude instead of 10, but with the corresponding temperature change for each step being 1/10th the magnitude (0.64 C° instead of 6.4 C°), the final answer would be the same as above.
How would we set up the calculation if we started with the temperature of −49°C at 10 km and decreased altitude to 0 km? The altitude difference is 0 km − 10 km = −10 km. Compare this setup with the equation above:
T0 km = T10 km − (6.4 × −10) = −49 − (6.4 × −10) = −49 − (−64)
= −49 + 64
= 15°C
Notice that we set up the right side of the equation as before, initial temperature minus lapse rate multiplied by the number
of steps in altitude. The multiplication of the two negative signs results in a positive temperature change for this decrease in altitude.
Lapse rate calculations do not always involve the normal lapse rate or “round” temperatures and differences in altitude. Two examples with these types of variations are now presented.
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First, with an environmental lapse rate of ∆7.6 C° · km−1, what is the change in temperature between a surface at 2 km and a point in the atmosphere 3 km above it? The altitude of the point for which we want to calculate the temperature is 5 km (3 km above a surface at 2 km), but the change in altitude is only 3 km.
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= T2 km − (22.8)
If the temperature at 2 km is, for example, 28.9°C, then the
temperature 3 km higher is 28.9 − 22.8 = 6.1°C.
Second, if the temperature at Winnipeg (238 m above
sea level) is 9.7°C and the environmental lapse rate is
∆5.9 C° · km−1, what is the temperature outside an airplane flying above the city at 12 000 m? The difference in altitude is 12000 – 238 = 11762 m or 11.762 km.
T12 km = T238 m − (5.9 × 11.762) = 9.7 − (5.9 × 11.762) = 9.7 − 69.4
= −59.7°C
T
5 km
= T − (7.6 × 3) 2 km
Note that sometimes the lapse rate is inverted; temperature increases with an increase in altitude. In these cases we need to show the lapse rate as a negative, or state explicitly that the rate is inverted. Moving between the altitudes of 3500 m and 5500 m (a 2 km difference) with an inverted lapse rate of ∆4.5 C° · km−1, what is the temperature change?
T5.5 km = T3.5 km − (−4.5 × 2) = T3.5 km − (−9)
= T3.5 km + 9
The temperature is 9 C° greater at 5.5 km than at 3.5 km altitude. With an inversion, a decrease in altitude would result in a decrease in temperature. Finally, an isothermal condition occurs when there is no change in temperature with a change in altitude. Here the lapse rate would be ∆0 C°·km−1.