Given a lapse rate of 3 degrees C/1000 meters and a temp of 20 degrees C at 0 meters:
A.What is the air sitting at 3,000 meters elevation?___degrees C
B.If a sample of air were lifted from 0 meters to 3,000 meters elevation, what would its new temp be?_____C(Assume no condensation is occuring.)
I don't undertsnad how to figure this out at all....
A, 20 - 3 (3000/1000)
=20 - 9
= 11 deg C
B. if it really is "adiabatic",no heat lost or gained by gas during drop in pressure.
for an ideal gas with g =1.4
Po Vo^g = P V^g
To * Vo^(g-1) = T * V^(g-1)
for atmosphere the pressure is approximately
P = Po e^-(M G/R T)y
at 3000 meters using T =293:
M = mol mass of air = 28.8 *10^-3 kg/mol
G = 9.8 m/s^2 gravity
R = 8.315 J/mol deg K gas constant
so
ln (P/1 atm) = -.347
P = .706 atm
so what is the volume ratio V/Vo of a mol for example ?
(V/Vo)^1.4 = =Po/P = 1/.706 = 1.42
1.4 ln (V/Vo) = 1.416
ln V/Vo = 1.01 expands
V/Vo = 2.75 expansion ratio
so what is T actually?
To * Vo^(g-1) = T * V^(g-1)
T = To (Vo/V)^(g-1) with g = 1.4
T = 293 (.364)^.4
ln (T/293) = .4 ln (.364)
ln (T/293) = -.4245
T = 293(.654)
T = 192 K
= -81 C
CHECK MY ARITHMETIC !!!
so what is the volume ratio V/Vo of a mol for example ?
(V/Vo)^1.4 = =Po/P = 1/.706 = 1.42
1.4 ln (V/Vo) = .3507
ln V/Vo = .25046 expands
V/Vo = 1.28 expansion ratio
Vo/V = .778
so what is T actually?
To * Vo^(g-1) = T * V^(g-1)
T = To (Vo/V)^(g-1) with g = 1.4
T = 293 (.778)^.4
ln (T/293) = .4 ln (.778)
ln (T/293) = -.1004
T = 293(.904)
T = 265 K
= -8 C
CHECK MY ARITHMETIC !!!
The way I would do it is as follows:
You gave a lapse rate of 3ºC per 1000m
Air cools the higher you get (to a certain point, but that doesn't interest us for this example)
If the air at 0 is 20ºC and it cools by 3º every 1000m - then the air at 3000m
is 9ºC cooler - namely 11ºC
20 - 9 = 11ºC
________________________________________
Problem B; you didn't give a lapse rate for this example, so I am assuming that you are supposed to use the correct dry adiabatic lapse rate which is:
10ºC for 1000m (or 5.5ºF per 1000')
Thus, assuming that the temp. at 0 is still 20ºC the air at 3000m would be 30ºC cooler (3x10)- namely -10ºC
20 - 30 = -10ºC
At least, that's the way I would do it.
(Actually, the normal lapse rate is 3.5ºF per 1000' or 6.4ºC for 1000m)
The wet adiabatic rate is 3.3ºF per 1000' or 6ºC per 1000m
FALLING air always WARMS at the DRY adiabatic rate
To solve these questions, we need to understand the concept of lapse rate and how it affects temperature with increasing altitude. The lapse rate is the rate at which the temperature decreases with an increase in altitude.
A. To find the temperature at 3,000 meters elevation, we start with the given temperature at 0 meters, which is 20 degrees Celsius. Then we use the lapse rate of 3 degrees Celsius per 1000 meters to calculate the change in temperature per 1000 meters.
The change in temperature for 3000 meters can be calculated as follows:
Change in temperature = (Lapse rate) × (Change in altitude in thousands of meters)
= 3 degrees Celsius/1000 meters × 3 (thousands of meters)
= 9 degrees Celsius
Now, we need to determine the temperature at 3,000 meters elevation, starting from the given temperature of 20 degrees Celsius at sea level (0 meters). To do this, we subtract the change in temperature from the initial temperature.
Temperature at 3,000 meters elevation = Initial temperature - Change in temperature
= 20 degrees Celsius - 9 degrees Celsius
= 11 degrees Celsius
Therefore, the air sitting at 3,000 meters elevation is approximately 11 degrees Celsius.
B. In this case, we are lifting the air sample from 0 meters to 3,000 meters elevation. Since we are raising the air sample from a lower altitude to a higher altitude, we need to follow the same steps as above.
Using the same lapse rate of 3 degrees Celsius per 1000 meters, we calculate the change in temperature for the given change in altitude (3,000 meters).
Change in temperature = (Lapse rate) × (Change in altitude in thousands of meters)
= 3 degrees Celsius/1000 meters × 3 (thousands of meters)
= 9 degrees Celsius
However, this time we need to determine the new temperature after the air sample is lifted. To calculate the new temperature, we add the change in temperature to the initial temperature.
New temperature = Initial temperature + Change in temperature
= 20 degrees Celsius + 9 degrees Celsius
= 29 degrees Celsius
Assuming no condensation is occurring, the new temperature of the air sample lifted from 0 meters to 3,000 meters elevation would be approximately 29 degrees Celsius.