Question

ercise 8.1:

1. assume that at sea level a parcel of air has a temperature of 40° c, and a dew point temperature of

0° c. calculate the temperature found along the slopes of the mountain using the appropriate
adiabatic rate. put your answers in the blanks and show your work.
rising air
3,500 m
2,000 m
40°c
sea level
sea level
figure 1

Explanation:

Step1: Calculate the Lifting Condensation Level (LCL)

The formula to find the LCL is \(z=\frac{T - T_d}{0.0098}\) (where \(T\) is the air temperature, \(T_d\) is the dew - point temperature). Given \(T = 40^{\circ}C\) and \(T_d=0^{\circ}C\), then \(z=\frac{40 - 0}{0.0098}\approx4082\ m\). But since our mountain is \(3500\ m\), we will use the dry adiabatic rate (\(\Gamma_d = 10^{\circ}C/1000\ m\)) for the rising air on the windward side.

Step2: Calculate temperature at \(2000\ m\) on windward side

Using the dry adiabatic formula \(T = T_0-\Gamma_d\times h\). Here \(T_0 = 40^{\circ}C\), \(\Gamma_d = 10^{\circ}C/1000\ m\), and \(h = 2000\ m\). So \(T=40-10\times\frac{2000}{1000}=40 - 20=20^{\circ}C\)

Step3: Calculate temperature at \(3500\ m\) on windward side

Again, using \(T = T_0-\Gamma_d\times h\). Here \(h = 3500\ m\), so \(T=40-10\times\frac{3500}{1000}=40 - 35 = 5^{\circ}C\)

On the leeward side, we use the dry adiabatic rate (assuming no moisture condenses on the way down).

Step4: Calculate temperature at sea - level on leeward side

Using \(T = T_{top}+\Gamma_d\times h\). The temperature at the top (\(h = 3500\ m\)) is \(T_{top}=5^{\circ}C\), and \(h = 3500\ m\). So \(T=5 + 10\times\frac{3500}{1000}=5+35 = 40^{\circ}C\)

Answer:

• At \(2000\ m\) on the windward side: \(20^{\circ}C\)
• At \(3500\ m\) on the windward side: \(5^{\circ}C\)
• At sea - level on the leeward side: \(40^{\circ}C\)