Question

An air mass has a temperature of 18°C and a relative humidity of 42%. a) Determine the vapor pressure and saturation vapor pressure of the air. You can use tables from the Gupta text, or the equations presented in class (which are alsc in the evapotranspiration references on the camino site). Or you may use both to be more confident of your work. b) Assume the air mass rises and cools at a rate of -10 C/km. At what elevation would condensation begin to occur?

Answer 1

Temperature (T) = 18⁰C

Relative Humidity (RH) = 42% = 0.42

1. CALCULATION OF SATURATION VAPOR PRESSURE AND VAPOR PRESSURE

CALCULATION USING FORMULA

Saturation Vapor Pressure =

e_s = 6.11 x 10^{7.5T/(237.3+T)}

      = 6.11 x 10^{7.5*18/(237.3+18)}

   =20.64 mb                      (mb = millibar)

RH = e / e_s                           (e = vapor pressure)

e = e_s x RH = 20.64 x 0.42 = 8.67 mb

CALCULATION FROM TABLE OF SATURATION VAPOR PRESSURE

From table,

At T=17.8⁰C, e_s=20.3 mb

At T=18.3⁰C, e_s=21.1 mb

Interpolating above values to obtain e_s at T = 18⁰C

e_s = 20.3 + { (21.1-20.3)/(18.3-17.8)}x(17-17.8)

    = 20.62 mb

e = 20.62 x 0.42 = 8.66 mb

Hence, both values from formula and table are approximately equal as can be seen above.

Saturation Vapor Pressure = 20.64 mb

Vapor Pressure = 8.67 mb

2. CALCULATION OF ELEVATION AT WHICH CONDENSATION WILL BEGIN

Condensation begins when the atmosphere can no longer hold any more water vapor. That is when Relative Humidity is 1 (100%).

Thus RH =1 is achieved, when vapor pressure (e) equals to saturation vapor pressure (e_s)

As the air mass rises, its temperature decreases. We need to find the temperature at which saturation vapor pressure is equal to the vapor pressure

That is, the temperature at which, e_s = e = 8.67 mb

e_s = 6.11 x 10^{7.5T/(237.3+T)}

8.67= 6.11 x 10^{7.5T/(237.3+T)}

Or,

10^{7.5T/(237.3+T)} = 8.67 / 6.11 = 1.419

Taking Log Base 10 on both sides,

7.5T/(237.3+T) = log 1.419 = 0.152

7.5T = 36.07 +0.152T

7.348T = 36.07

T = 36.07/7.348 = 4.909⁰C

Also from Saturation Vapor Pressure Table, the saturation vapor pressure is 8.7 mb at 5⁰C and our required e_s is 8.67 mb. Thus 4.909⁰C found from formula tallies with table value.

Hence, condensation will occur at 4.909⁰C.

Present Temperature = 18⁰C

Air Mass cools at a rate of -10⁰C / km, that is

10⁰C cooling is achieved by a rise in 1 km

1⁰C cooling is achieved by a rise in 1/10 km

(18-4.909)⁰C cooling is achieved by a rise in (1/10)*(18-4.909) km

                                                                               =1.3091 km

Hence, at an elevation of 1.3091 km above present position, the air mass will start to condense.