Question

During the month of December, the average river inflow to a reservoir is 10m3/s while the dam bottom outlet releases 22 m3/s. Rainfall is measured as 30 mm uniformly distributed within the month. The average seepage from the reservoir is estimated 0.79 m3/s. The surface areas of the lake are measured from aerial photos as 15 and 11 Km2 and also the staff gauge readings of water surface are recorded as 105m and 102m in the beginning and at the end of the month respectively. (The reservoir volume at the start of the month was 98000 ML.)

a) Calculate average daily evaporation from the lake surface in mm/day.

b) Estimate pan evaporation in mm/day

c) Estimate evaporation from the reservoir surface if average wind velocity is 2.5 m/s and relative humidity is 20%. The mean air and mean water temperatures are the same and equal 20o C. The vapour pressure is recorded as 31 mb.

Answer 1

River Inflow to reservoir, Q_r = 10m³/s

Reservoir Outflow, Q_d = 22m³/s

Rainfall intensity average per month, R = 30mm

Rainfall intensity average per day, R = 1mm

Reservoir Seepage, Q_s = 0.79m³/s

Water Body Area initially = 15km²

Water Body Area at end of month = 11km²

Height of lake initially = 105m

Height of lake at end of month = 102m

Average Evaporation in mm/day

from mass balance of inflow and outflow

Accumulation = Inflow – Outflow – Evaporation Loses

During Start of Month

98000 ML = (10 m³/s + 15*1000*1000*.001m³/s) – (22 m³/s + 0.79 m³/s) – Evaporation (in m³/s)

Evaporation (in m³/s) = 13852.96

Evaporation (in mm/s) = (13852.96/15*1000*1000) = 9.235 * 10^-4 mm/s

Evaporation (in mm/day) = 79.79 mm/day

During End of Month

98000 ML = (10 m³/s + 11*1000*1000*.001m³/s) – (22 m³/s + 0.79 m³/s) – Evaporation (in m³/s)

Evaporation (in m³/s) = 9852.95

Evaporation (in mm/s) = (9852.95/11*1000*1000) = 8.95 * 10^-4 mm/s

Evaporation (in mm/day) = 77.39 mm/day

Average Evaporation in mm/day = 76.59 mm/day

Evaporation from the reservoir surface if average wind velocity is 2.5 m/s

from Lake Mead Equation

E = 0.0331 V (e_s - e) [1 – 0.03 (T – Tw)] 24 h day-1

where E is evaporation rate (mm/day);

V is the wind speed in the lowest 0.5 m above the ground (km/h),

e is vapor pressure (in mm Hg), e_s is saturation vapor pressure,

T is the atmospheric temperature above the water surface in ⁰C, and Tw is the water temperature in ⁰C

from the relationship of relative humidity (RH)

(Actual Vapor Pressure) RH = --------------------------         x100 (Saturation Vapor Pressure)

e = 31mb = 23.25 mm of Hg

RH = 20%

e_{s =} 23.25 * 100 / 20

e_s = 116.25 mm of Hg

V = 2.5 m/s

so from Lake Mead Equation

E = 11.081 mm/day