Question

ENGINEERING HYDROLOGY CIVL431 Assignment1 Problem 11 The following data were observed at Demirkopru Dam site on Gedir River in June 1971. Find the monthly evaporation in mm from the reservoir surface of this dam. If there were a Class A pan installed nearby the reservoir, what would be the monthly evaporation observed from this pan? Storage at the beginning of the month 4XY.X-10 m Storage at the end of the month 47XY -10 m Average surface area of reservoir in the month 13X.Y km Mean inflow IX.X m's Amount of water used for power generation 4YZ *10m Depth of rainfall for the month : NONE Average seepage rate in the month :ZYZ.ZXIus Problem ? For the same reservoir in Problem the area of lake surface decreased by 3% in the next month, July 1971, while the seepage, evaporation, inflow and water used for power are the same. There was no rain in this month either Determine the change in storage for this month Problem A certain basin is expected to receive XYZ mm of rainfall in June, where evaporation from soil surface and transpiration are estimated together as YZ mm. Total infiltrated water depth in the same month is assumed to be iz mm. If the area of the basin is XYZ km and there are no other losses, determine the surface runoff from this basin in mm depth, m'volume and Its average rate Problem + Assume that during a storm it rained 12 mm over a basin of 7X km area. There exists a lake with ZX km area in the basin and it does not contribute to the surface flow. If the total infiltration depth was 9 mm for this storm and there were no other losses, determine the volume of surface runoff from the basin Problems Assume that a lake with an area of 7 hectares has an average inflow of QXYZm's and outflow of ZYX lus during a certain month. Total precipitation in this month is 3X.Z mm and the decrease in storage is sXYZ00 ml Determine the evaporation from this lake in m' and in mm for this month assuming negligible seepage from the lake. Note that X Y Z are the last 3 digits of your student number. Just insert them as they are Eg Mohamed Saced: 17352395 Y Z

Answer 1

Problem 1

Storage at beginning of month S1= 439.5*10⁶ m³

Storage at the end of month S2= 453.9*10⁶ m³

Change in storage $\Delta$ S = S2 - S1 = 453.9*10⁶ -439.5*10⁶ = 14.4*10⁶ m³

Mean inflow = 13.3 m³/s

Inflow volume for the month of June,1971 , I = 13.3*30*24*60*60 = 34.47*10⁶ m³

Amount of water used for power production = Op = 49.5*10⁶ m³

Average seepage rate in the month = 595.53 l/s

Seepage volume for the month Os= 595.53*10^-3*30*24*60*60 = 1.54*10⁶ m³

Precipitation P = 0

Using water budget equation ,

P + I = Op + Os + $\Delta$ S +E_{​​​​​​L}

0 + 34.47*10⁶ = 49.5*10⁶ + 1.54*10⁶ + 14.4 * 10⁶ + E_L

E_L = -30.97*10⁶ ​​​​ m³

Depth of evaporation = E_L/ Surface area = -30.97*10⁶ /33.9*10⁶ = -0.91 m

Evaporation cannot be negative. So evaporation is taken as zero.

Pan evaporation = Lake evaporation/ pan coefficient

Pan coefficient taken as 0.7 usually

Problem 2

Surface area = 0.97x Actual Surface area = 0.97*33.9 = 32.883 km²

Mean inflow = 13.3 m³/s

Inflow volume I = 13.3*31*24*60*60 = 35.62*10⁶ m³

Op = 49.5 m³

Os = 595.53 *10^-3*31*24*60*60 = 1.595 *10⁶ m³

E_L = 0

P + I = Op + Os + $\Delta$ S + E_L

35.62*10⁶ = 49.5*10⁶ + 1.595*10⁶ + $\Delta$ S

$\Delta$S = -15.475*10⁶ m³

Decline in water level = 15.475*10⁶/32.883*10⁶ = 0.47 m =47 cm

Problem 3

Precipitation P= 395 mm

Evapotranspiration E_T = 95 mm

Infiltration I = 15 mm

Area A = 395 km²

Assuming no other losses,

Runoff R = Precipitation (P) - Evapotranspiration (E_T) - Infiltration (I)

= P - E_T - I = 395 - 95 - 15 = 285 mm

Runoff depth = 285 mm = 0.285 m

Runoff volume = Runoff depth *Area = 0.285*395*10⁶ = 112.575*10⁶ m³

There are 30 days in the month of June. So,

Rate of runoff = Runoff volume/ Duration = 112.575*10⁶ / (30*24*60*60) = 43.432 m³/s = 43432 lt/s

Problem 4

Precipitation = 15 mm

Area of catchment = 73 km²

Area of lake = 5.3 km²

Area contributing to surface runoff = 73 - 5.3 = 67.7 km²

Total Infiltration depth = 9 mm

Runoff depth = Precipitation- Losses = Precipitation- Infiltration = 15-9 = 6 mm

Volume of surface runoff = Runoff depth * Area contributing to surface runoff = 6*10^-3 * 67.7 *10⁶ = 406200 m³

Problem 5

Surface area of lake = 79 ha

Inflow rate = 0.395 m³/s

Inflow volume = 0.395*30*24*60*60 = 1.024*10⁶ m³

Outflow rate = 593 lt/s = 0.593 m³/s

Outflow volume = 0.593*30*24*60*60 = 1.537*10⁶ m³

Precipitation = 33.5 mm

Precipitation volume = 33.5*10^-3*790000 = 0.0265 * 10⁶ m³

Change in storage $\Delta$ s = -539500 m³

P + I = O + $\Delta$ S + E

E = P + I - O - $\Delta$ S = 0.0265*10⁶ + 1.024*10⁶ - 1.537*10⁶ + 0.5395*10⁶

E = 0.053*10⁶ m³

Evaporation depth = Evaporation/ Surface area = 0.053*10⁶ / 790000 = 0.067 m = 67 mm