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Questions Assume: DAR = 10°C per km (10°C km-1 ) MAR = 6°C per km (6°C...

Questions Assume: DAR = 10°C per km (10°C km-1 ) MAR = 6°C per km (6°C km-1 ) DpR = 2°C per km (2°C km-1 ) for “dry” air, DpR = MAR for saturated air

5. Consider a parcel of air that approaches Canada’s west coast from over the Pacific Ocean. The air has a temperature of 12°C and a dew point temperature of 6°C. As the air blows onshore, it is forced to rise from sea level (0 m) over the Coast Mountains of British Columbia (2000 m) and then descends to 400 m (elevation of Kelowna BC). Find:

a) The lifting condensation level (LCL) in metres. (3)

b) The air temperature and dew point temperature of the parcel when it reaches Kelowna. (3)

c) Fully explain any differences between the starting conditions of the air parcel and the final conditions. (3)

Solutions

Expert Solution

1. lifting condensation level is the level at which the temperature of the air parcel is equal to the dew point temperature of the air parcel.

since the air parcel is coming from the Pacific ocean so it is assumed to be a saturated air parcel

so the decrease in the temperature of the air parcel will be according to the MALR ( 6 degrees per km)

let, x be the LCL

therefore,

12-(6/1000) * x = 6- (2/1000) * x

6 = x * (6-2)/1000

6 * 1000 = 4 x

x = 1500 m

therefore the required LCL is 1500 m

2. temperature of the air parcel at the top (2000 m)

= 12 - (6/1000) * 2000 = 12-12 = 0 degree celsius

dew point temperature of the air parcel at the top (2000 m)

= [6 - (2/1000) * 1500 ] - [6 - (6/1000) * 500] = 3 - 3 = 0 degree celsius

the temperature of the air parcel at 400 m at the leeward side is

= 0 + (10/1000) * 400 = 4 degree celsius

the dew point temperature of the air parcel at 400m in the leeward side is

= 0 + (2/1000) * 400 = 0.8 degree celsius

3. relative humidity of the air parcel at the starting point (sea level) is

6 = 12 -[ (100-RH)/5]

6 = (100-RH)/5

RH = 70%

relative humidity of the air parcel at the endpoint (400m in the leeward side ) is

0.8 = 4 - [(100-RH)/5]

3.2 = (100-RH)/5

RH = 84%

so comparing both values we can say that the air parcel in the last point contains more moisture than at the starting point  


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