EVAPORATION PANS

3.3 EVAPORATION PANS

(i) Floating pans (made of GI) of 90 cm square and 45 cm deep are mounted on a raft floating in water. The volume of water lost due to evaporation in the pan is determined by knowing the volume of water required to bring the level of water up to the original mark daily and after making allowance for rainfall, if there has been any.

(ii) Land pan. Evaporation pans are installed in the vicinity of the reservoir or lake to determine the lake evaporation. The IMD Land pan shown in Fig. 3.2 is 122 cm diameter and

122 cm dia. 122 cm dia.

Point gauge

stilling well, 10.2 cm dia 0.9-mm thick

2 cm

25.5 cm, land pan 25.5 cm, land pan copper sheet

10 cm, wooden cribs

Fig. 3.2 IMD land pan

25.5 cm deep, made of unpainted GI; and set on wood grillage 10 cm above ground to permit circulation of air under the pan. The pan has a stilling well, vernier point gauge, a thermometer with clip and may be covered with a wire screen. The amount of water lost by evaporation from the pan can be directly measured by the point gauge. Readings are taken twice daily at 08.30 and 17.30 hours I.S.T. The air temperature is determined by reading a dry bulb thermometer kept in the Stevenson’s screen erected in the same enclosure of the pan. A totalising anemometer is normally mounted at the level of the instrument to provide the wind speed information

WATER LOSSES

required. Allowance has to be made for rainfall, if there has been any. Water is added to the pan from a graduated cylinder to bring the water level to the original mark, i.e., 5 cm below the

top of the pan. Experiments have shown that the unscreened pan evaporation is 1.144 times that of the screened one.

(iii) Colarado sunken pan. This is 92 cm square and 42-92 cm deep and is sunk in the ground such that only 5-15 cm depth projects above the ground surface and thus the water level is maintained almost at the ground level. The evaporation is measured by a point gauge.

Pan coefficient —Evaporation pan data cannot be applied to free water surfaces di- rectly but must be adjusted for the differences in physical and climatological factors. For ex- ample, a lake is larger and deeper and may be exposed to different wind speed, as compared to

a pan. The small volume of water in the metallic pan is greatly affected by temperature fluc- tuations in the air or by solar raditions in contrast with large bodies of water (in the reservoir) with little temperature fluctuations. Thus the pan evaporation data have to be corrected to obtain the actual evaporation from water surfaces of lakes and reservoirs, i.e., by multiplying by a coefficient called pan coefficient and is defined as

Lake evaporation

Pan coefficient =

Pan evaporation

and the experimental values for pan coefficients range from 0.67 to 0.82 with an average of 0.7. Example 3.1 The following are the monthly pan evaporation data (Jan.-Dec.) at

Krishnarajasagara in a certain year in cm.

The water spread area in a lake nearby in the beginning of January in that year was

2.80 km 2 and at the end of December it was measured as 2.55 km 2 . Calculate the loss of water due to evaporation in that year. Assume a pan coefficient of 0.7 . Solution Mean water spread area of lake

cone formula

Annual loss of water due to evaporation (adding up the monthly values)

= 228.7 cm

Annual volume of water lost due to evaporation

= 4.29 × 10 6 m 3 or 4.29 Mm 3

Example 3.1 (a) Compute the daily evaporation from a Class A pan if the amounts of water added to bring the level to the fixed point are as follows:

Day: 1 2 3 4 5 6 7 Rainfall (mm):

14 6 12 8 0 5 6 Water

3 0 0 7 4 3 added (mm):

(removed)

HYDROLOGY

What is the evaporation loss of water in this week from a lake (surface area = 640 ha) in the vicinity, assuming a pan coefficient of 0.75?

+ water added

Solution Pan evaporation, E p , mm = Rainfall

or − water removed

Pan evaporation in the week = ∑ E p = 63 mm

Pan coefficient

0.75 = E

∴ Lake evaporation during the week E L

= 63 × 0.75 = 47.25 mm

Water lost from the lake

=A.E L = 640 ×

= 30.24 ha.m ~

0.3 Mm

uniform intensity of 15 mm /hr is 21.6 Mm 3 . If the area of the basin is 300 km 2 , find the average infiltration rate and the runoff coefficient for the basin.

Solution (i) Infiltration loss F p = Rainfall (P) – Runoff (R)

= 3 mm/hr

6 hr

(ii) Yield = C A P

21.6 × 10 6 m 3 = C(300 × 10 6 m 2

C = 0.8

Piche Evaporimeter

It is usually kept suspended in a Stevenson screen. It consists of a disc of filter paper kept constantly saturated with water from a graduated glass tube Fig. 3.3. The loss of water from the tube over a known period gives the average rate of evaporation. Though it is a simple instrument, the readings obtained are often more erratic than those from standard pans.

Measures to Reduce Lake Evaporation

The following are some of the recommended measures to reduce evaporation from water sur- faces :

(i) Storage reservoirs of more depth and less surface area, i.e., by choosing a cross section of the reservoir like a deep gorge Fig. 3.4 ; while the surface water is exposed to tem- perature gradients the deeper waters are cool; from this standpoint a large reservoir is prefer- able to a number of small reservoirs (while it is the reverse from the point of flood control).

WATER LOSSES

Suspension in Stevenson screen

1 Water level 2 2 Graduated

3 glass tube with water 4

Water

Porous cup

with water

Fig. 3.3. Piche evaporimeter Minimum surface area

to reduce evaporation

Cool deep water

Fig. 3.4 Reservoir in a deep gorge

(ii) By growing tall trees like Causerina on the windward side of the reservoirs to act as wind breakers.

(iii) By spraying certain chemicals or fatty acids and formation of films. By spreading a manomolecular layer of acetyl alcohol (hexadecanol) C 16 H 33 OH over the reservoir surface (from boats)—a film is formed on the surface which is only 0.015 micron (approx.) in thickness. It is

a polar compound and it has great affinity for water on one side (hydrophylic) and repels water on the other side (hydrophobic). The film will only allow precipitation from the top into it but

will not allow water molecules to escape from it. This method is readly effective when the wind velocities are less. If the wind velocity is more, it will sweep the film off the water surface and

deposit it on the bank. However the film is pervious to O 2 and CO 2 . About 2.2 kg (22 N) of acetyl alcohol is required to cover an area of 1 ha of reservoir surface. It is best suited for small and medium size reservoirs.

(iv) By allowing flow of water, temperature is reduced and evaporation is reduced; i.e., by designing the outlet works so that the warmer surface water can be released. (v) By removing the water loving weeds and plants like Phreatophytes from the periph- ery of the reservoir.

HYDROLOGY

(vi) By straightening the stream-channels the exposed area of the water surface (along the length) is reduced and hence evaporation is reduced.

(vii) By providing mechanical coverings like thin polythene sheets to small agricultural ponds and lakes.

(viii) By developing undergound reservoirs, since the evaporation from a ground water table is very much less than the evaporation from a water surface. (ix) If the reservoir is surrounded by huge trees and forest, the evaporation loss will be less due to cooller environment.