A reservoir with surface area of250 hectares has saturation vapourpres...
To estimate the average daily evaporation from the lake using the Meysis formula, we need to calculate the wind speed at the water surface level.
First, convert the wind velocity from km/hr to m/s:
16 km/hr = 16,000 m/3600 s = 4.44 m/s
Next, calculate the wind speed at the water surface level using the logarithmic wind profile equation:
u(z) = u(zr) * (ln(z/zo) / ln(zr/zo))
where:
u(z) = wind speed at height z (m/s)
u(zr) = wind speed at reference height zr (m/s)
z = height above the ground surface (m)
zo = roughness length (m)
Assuming a standard roughness length of 0.01 m, and a reference height of 1 m, the equation becomes:
u(z) = 4.44 * (ln(1/0.01) / ln(1/0.01))
u(z) = 4.44 * (ln(100) / ln(100))
u(z) = 4.44 * (4.60517 / 4.60517)
u(z) = 4.44 * 1
u(z) = 4.44 m/s
Now we can calculate the evaporation using the Meysis formula:
E = k * (P - Pw) * u(z) * A
where:
E = evaporation rate (mm/day)
k = empirical coefficient (typically around 0.7)
P = saturation vapor pressure at water surface (mm of Hg)
Pw = actual vapor pressure of air (mm of Hg)
u(z) = wind speed at water surface (m/s)
A = surface area of the reservoir (hectares)
Converting the given values to the appropriate units:
A = 250 hectares = 250,000 m^2
P = 17.54 mm of Hg
Pw = 7.02 mm of Hg
u(z) = 4.44 m/s
Plugging in the values and solving for E:
E = 0.7 * (17.54 - 7.02) * 4.44 * 250,000
E = 0.7 * 10.52 * 4.44 * 250,000
E = 0.7 * 46.5648 * 250,000
E = 8,172,080 mm/day
Therefore, the estimated average daily evaporation from the lake using the Meysis formula is approximately 8,172,080 mm/day.
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