Jeffrey Annis and Nico Brunner Evaporation Atmospheric Physics Lab Work

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Evaporation

Jeffrey Annis and Nico Brunner

Institute for Atmospheric and Climate Science - IAC ETH

Summer Semester 2007

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I. Introduction

In this experiment, the relationship between wind speed and evaporation is investigated. We measure the evaporation under idealized conditions using a small wind tunnel. We then compare our results from the experiment with calculated values for the evaporation over the sea.

II. Questions to be answered

a. What do you expect for the evaporation for increasing wind velocities?

b. Is the psychrometer - “constant” increasing or decreasing with stronger wind?

c. What should you get for the global evaporation/precipitation?

III. Summary of theory

a. The evaporation is defined as follows:

E =  w (dV/dt) (1/A) Where,

 w = water density [kg / m ³ ]

dV/dt = temporal volume change due to evaporation [m ³ / s]

A = surface area [m ² ]

b. For an infinitesimally small surface the evaporation is described as a function of the wind speed and water vapor gradient:

E =  (e s - e) Where,

E = evaporation [g / cm ² y]

= wind speed [m / s ]

e = water vapor pressure in the atmosphere [hPa]

e s = saturation vapor pressure for temperature T [hPa]

 = constant [s ² / m ² ]

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c. The water vapor pressure is defined as follows:

e = e s -A  p (T-T w ) Where,

A  = psychrometer constant (6.6e ^-4 K ^-1 for an air velocity of 2 m/s) e = water vapor pressure in the atmosphere [hPa]

e s = saturation vapor pressure for temperature T [hPa]

p = air pressure [hPa]

T = temperature of the dry thermometer [°C] (dry-bulb) T w = temperature of the wet bulb thermometer [°C] (wet-bulb) d. Saturation vapor pressure is calculated using the Magnus formulation:

e s = 6.112 exp [(17.67 T) / (T + 243.5)]

The evaporation heat is equal to the heat that is withdrawn from the air surrounding the thermometer. Evaporation increases with higher temperature. At the same relative humidity, the difference between wet-bulb temperature and dry-bulb temperature is larger with rising temperatures. Hence, the error for the measured humidity for the same difference between wet-bulb and drybulb temperature is larger at higher temperatures.

e. Evaporation over the sea is given by the following equation:

E =  C D (q o -q) Where,

C D = drag coefficient (in this experiment treated as constant)

 = air density [kg/m ³ ]

q = specific humidity [kg water/kg air]

q o = specific humidity for saturated air over water at T o

with,

q = 0.622 (e / p)

p = pressure of air [hPa].

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IV. Measurement report

a. The homogeneity of the wind profile was measured by setting the motor speed to position “3“ and recording wind velocities as a function of distance across the wind tunnel channel. The wind channel center is at a distance of 13cm from the left side. The measurements are as follows:

Distance

(cm) 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Wind Speed (m/s)

6.62 6.95 7.1 7.25 7.33 7.35 7.36 7.33 7.28 7.25 7.21 7.17 7.16 7.15 7.1 7.02 6.95 6.79 6.55

b. Then we calibrated the motor positions vs. wind speed at the wind tunnel center. The measurements are as follows:

Wind Speed (m/s)

2 3 4 6 7 8 10 12 14 16

Motor

Speed 0.2 0.7 1.3 2.4 2.7 3.4 4.3 5.1 6.0 6.8

c. The evaporated water volume was then measured for a time span of 10 minutes as a function of the wind velocity. Additionally, one measurement over 30 minutes was conducted in “stagnant air”. For each measurement the dry-bulb and wet-bulb temperature were recorded.

Wind Speed (m/s)

0 (30mins)

2 (10mins)

3 (10mins)

4 (10mins)

6 (10mins)

7 (10mins)

8 (10mins)

10 (10mins)

12 (10mins)

14 (10mins)

16 (10mins) Evaporation

(ml) 0.1 0.3 0.4 0.58 0.6 0.7 1.0 1.0 1.1 1.1 1.2

T (°C) 21 21.2 21.3 21.3 21.5 21.7 21.8 22.0 22.2 22.4 22.7

T

w

(°C) 14.8 14.2 14.1 14.1 14.1 14.1 14.1 14.2 14.2 14.3 14.4

V. Data analysis

a. Homogeneity of wind profile

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b. Evaporation with airflow

First we calculate the evaporation using the equation below:

E =  w (dV/dt) (1/A)

with A = r ² , r = 3.5 cm = 0.035 m, dt = 10min = 600 s,  w = 1000 kg/m ³ dV (ml) dV (m ³ ) Evap (kg m ^-2 s ^-1 )

0.3 3.0e ^-7 1.299e ^-4

0.4 4.0e ^-7 1.732e ^-4

0.58 5.8e ^-7 2.512e ^-4

0.6 6.0e ^-7 2.598e ^-4

0.7 7.0e ^-7 3.032e ^-4

1 10.0e ^-7 4.331e ^-4

1.1 11.0e ^-7 4.764e ^-4

1.2 12.0e ^-7 5.197e ^-4 Second, we use the equation below to determine :

E =  (e s - e)

E (kg m

^-2

s

^-1

) u (m/s) e

s

(hPa) e (hPa)  (s

²

m

^-2

)

0.000130 2 25.16505102 20.54505102 44.34205637

0.000173 3 25.32000225 20.56800225 38.32029563

0.000251 4 25.32000225 20.56800225 41.67332149

0.000260 6 25.63241664 20.74841664 27.96345897

0.000303 7 25.94820584 20.93220584 27.22757847

0.000433 8 26.10737565 21.02537565 33.59246694

0.000433 10 26.42828512 21.28028512 26.52943543

0.000476 12 26.75264714 21.47264714 23.71068292

0.000476 14 27.08049320 21.73449320 20.0725358

0.000520 16 27.57886433 22.10086433 18.6984575

If we use linear regression to find alpha, we get y = 15.621x + 0.0001 R ² = 0.886, so  ~ = 16 s ² m ^-2

c. Evaporation without airflow

For the situation without airflow, the evaporation was considerably

less. This is because there is no airflow to disturb the boundary layer and

cause the water vapor gradient to induce evaporation. This shows how

important airflow is in addition to the humidity gradient in causing

evaporation.

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d. Evaporation over the sea E =  air C D (q o -q) Where, C D = 10 ^-3

u = 4 ms ^-1

 air = 1.0 kg/m ³

e s = 6.112 exp [(17.67 T) / (T + 243.5)]

e = e s -A  p (T-T w )

e so = 6.112 exp [(17.67 T o ) / (T o + 243.5)]

e o = e so -A  p (T o -T w ) q = 0.622 (e / p) p = 1000 hPa e so = 21.96, e o = 18.0 , q o = 1.120e ^-2 e s = 18.17, e = 16.19 , q = 1.007e ^-2

E = 4.52e ^-6 kg m ^-2 s ^-1 for over the sea, using u = 4 ms ^-1 VI. Results

Our calculations of evaporation did not correspond well to the value we derived for evaporation over the sea. We calculated E = 4.52e ^-6 kg m ^-2 s ^-1 for over the sea, but found E 4m/s = 2.51e ^-6 kg m ^-2 s ^-1 . An error in the units is probably responsible for the large difference.

VII. Error analysis and discussion

VIII. Final discussion of results

Jeffrey Annis and Nico Brunner Evaporation Atmospheric Physics Lab Work

Evaporation

Jeffrey Annis and Nico Brunner

Institute for Atmospheric and Climate Science - IAC ETH

Summer Semester 2007

I. Introduction

In this experiment, the relationship between wind speed and evaporation is investigated. We measure the evaporation under idealized conditions using a small wind tunnel. We then compare our results from the experiment with calculated values for the evaporation over the sea.

II. Questions to be answered

a. What do you expect for the evaporation for increasing wind velocities?

b. Is the psychrometer - “constant” increasing or decreasing with stronger wind?

c. What should you get for the global evaporation/precipitation?

III. Summary of theory

a. The evaporation is defined as follows:

E =  w (dV/dt) (1/A) Where,

 w = water density [kg / m 3 ]

dV/dt = temporal volume change due to evaporation [m 3 / s]

A = surface area [m 2 ]

b. For an infinitesimally small surface the evaporation is described as a function of the wind speed and water vapor gradient:

E =  <u> (e s - e) Where,

E = evaporation [g / cm 2 y]

< u > = wind speed [m / s ]

e = water vapor pressure in the atmosphere [hPa]

e s = saturation vapor pressure for temperature T [hPa]

 = constant [s 2 / m 2 ]

c. The water vapor pressure is defined as follows:

e = e s -A  p (T-T w ) Where,

A  = psychrometer constant (6.6e -4 K -1 for an air velocity of 2 m/s) e = water vapor pressure in the atmosphere [hPa]

e s = saturation vapor pressure for temperature T [hPa]

p = air pressure [hPa]

T = temperature of the dry thermometer [°C] (dry-bulb) T w = temperature of the wet bulb thermometer [°C] (wet-bulb) d. Saturation vapor pressure is calculated using the Magnus formulation:

e s = 6.112 exp [(17.67 T) / (T + 243.5)]

e. Evaporation over the sea is given by the following equation:

E =  C D <u> (q o -q) Where,

C D = drag coefficient (in this experiment treated as constant)

 = air density [kg/m 3 ]

q = specific humidity [kg water/kg air]

q o = specific humidity for saturated air over water at T o

with,

q = 0.622 (e / p)

p = pressure of air [hPa].

IV. Measurement report

a. The homogeneity of the wind profile was measured by setting the motor speed to position “3“ and recording wind velocities as a function of distance across the wind tunnel channel. The wind channel center is at a distance of 13cm from the left side. The measurements are as follows:

Distance

(cm) 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Wind Speed (m/s)

6.62 6.95 7.1 7.25 7.33 7.35 7.36 7.33 7.28 7.25 7.21 7.17 7.16 7.15 7.1 7.02 6.95 6.79 6.55

b. Then we calibrated the motor positions vs. wind speed at the wind tunnel center. The measurements are as follows:

Wind Speed (m/s)

2 3 4 6 7 8 10 12 14 16

Motor

Speed 0.2 0.7 1.3 2.4 2.7 3.4 4.3 5.1 6.0 6.8

c. The evaporated water volume was then measured for a time span of 10 minutes as a function of the wind velocity. Additionally, one measurement over 30 minutes was conducted in “stagnant air”. For each measurement the dry-bulb and wet-bulb temperature were recorded.

Wind Speed (m/s)

0 (30mins)

2 (10mins)

3 (10mins)

4 (10mins)

6 (10mins)

7 (10mins)

8 (10mins)

10 (10mins)

12 (10mins)

14 (10mins)

16 (10mins) Evaporation

(ml) 0.1 0.3 0.4 0.58 0.6 0.7 1.0 1.0 1.1 1.1 1.2

T (°C) 21 21.2 21.3 21.3 21.5 21.7 21.8 22.0 22.2 22.4 22.7

T

(°C) 14.8 14.2 14.1 14.1 14.1 14.1 14.1 14.2 14.2 14.3 14.4

V. Data analysis

a. Homogeneity of wind profile

b. Evaporation with airflow

First we calculate the evaporation using the equation below:

E =  w (dV/dt) (1/A)

with A = r 2 , r = 3.5 cm = 0.035 m, dt = 10min = 600 s,  w = 1000 kg/m 3 dV (ml) dV (m 3 ) Evap (kg m -2 s -1 )

0.3 3.0e -7 1.299e -4

0.4 4.0e -7 1.732e -4

0.58 5.8e -7 2.512e -4

0.6 6.0e -7 2.598e -4

0.7 7.0e -7 3.032e -4

1 10.0e -7 4.331e -4

 w = water density [kg / m ³ ]

dV/dt = temporal volume change due to evaporation [m ³ / s]

A = surface area [m ² ]

E = evaporation [g / cm ² y]

 = constant [s ² / m ² ]

A  = psychrometer constant (6.6e ^-4 K ^-1 for an air velocity of 2 m/s) e = water vapor pressure in the atmosphere [hPa]

 = air density [kg/m ³ ]

with A = r ² , r = 3.5 cm = 0.035 m, dt = 10min = 600 s,  w = 1000 kg/m ³ dV (ml) dV (m ³ ) Evap (kg m ^-2 s ^-1 )

0.3 3.0e ^-7 1.299e ^-4

0.4 4.0e ^-7 1.732e ^-4

0.58 5.8e ^-7 2.512e ^-4

0.6 6.0e ^-7 2.598e ^-4

0.7 7.0e ^-7 3.032e ^-4

1 10.0e ^-7 4.331e ^-4

1 10.0e ^-7 4.331e ^-4

1.1 11.0e ^-7 4.764e ^-4

1.1 11.0e ^-7 4.764e ^-4

1.2 12.0e ^-7 5.197e ^-4 Second, we use the equation below to determine :

If we use linear regression to find alpha, we get y = 15.621x + 0.0001 R ² = 0.886, so  ~ = 16 s ² m ^-2

d. Evaporation over the sea E =  air C D <u> (q o -q) Where, C D = 10 ^-3

u = 4 ms ^-1

 air = 1.0 kg/m ³

e o = e so -A  p (T o -T w ) q = 0.622 (e / p) p = 1000 hPa e so = 21.96, e o = 18.0 , q o = 1.120e ^-2 e s = 18.17, e = 16.19 , q = 1.007e ^-2

E = 4.52e ^-6 kg m ^-2 s ^-1 for over the sea, using u = 4 ms ^-1 VI. Results

Our calculations of evaporation did not correspond well to the value we derived for evaporation over the sea. We calculated E = 4.52e ^-6 kg m ^-2 s ^-1 for over the sea, but found E 4m/s = 2.51e ^-6 kg m ^-2 s ^-1 . An error in the units is probably responsible for the large difference.