The annual storage in Broad River watershed is 0 cm/y. Annual precipitation is 100 cm/y and evapotranspiration is 50 cm/y. The s
andy loam soil in the watershed has an infiltration rate of 5 × 10-7 cm/s. The area of the watershed is 40 km2. Calculate the annual flow at the outlet of Broad River. What is the runoff coefficient for this watershed?
2 answers:
Answer:
The annual flow at the outlet of Broad River is 13688480 m³
The runoff coefficient for the watershed is 0.342212
Explanation:
Here, we have
The annual storage in the watershed = 0 cm/y
Annual precipitation = 100 cm/y
Annual evapotranspiration = 50 cm/y
infiltration rate = 5 × 10-7 cm/s
Watershed area = 40 km²
From the question, annual infiltration rate is given by
Annual infiltration rate = 5 × 10-7 cm/s × 31557600 s/year = 15.7788 cm/y
Therefore, annual flow =
Annual precipitation - Annual evapotranspiration - Annual infiltration rate - Annual storage in the watershed
Annual flow = 100 cm/y - 50 cm/y - 15.7788 cm/y - 0 cm/y = 34.2212 cm/y
The area of the watershed = 40 km²
Therefore, the volume of the annual flow is given by;
Height of annual flow × area of the watershed
Volume of annual flow = 34.2212 cm × 40 km² = 1368.848 cm·km²
= 13688480 m³
The runoff coefficient = =
The amount of precipitation received is given by
100 cm × 40 km² = 4000 cm·km² = 40000000 m³
Runoff coefficient = = 0.342212.
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Answer:
Part A
The mass plow rate, is approximately 97.0 lbm/s
Part B
The power used to overcome friction, is approximately 1.9 hp
Explanation:
The efficiency of the pump, η = 80%
The power input to the pump, P = 20 hp
The speed of the water through the pipe, v = 10 ft./s
The diameter of the pipe, d = 5.2 inches = 13/30 ft.
The free surface of the pool above the lake, h = 80 ft.
The density of the water, ρ = 62.4 lbm/ft.³
Part A
The mass plow rate, = Q × ρ
Where;
ρ = 62.4 lbm/ft³
Q = A × v
A = The cross-sectional area of the pipe
∴ Q = π·d²/4 × v = π × ((13/30 ft.)²)/4 × 10 ft.s ≈ 1.4748 ft.³/s
∴ The mass plow rate, ≈ 1.4748 ft.³/s × 62.4 lbm/ft.³ = 97.02752 lbm/s
The mass plow rate, ≈ 97.0 lbm/s
Part B
The power to pump the water at the given rate, =
·g·h
∴ = 97.02752 lbm/s × 32.1740 ft./s² × 80 ft. ≈ 14.1130725 Hp
≈ 14.1130725 Hp
The power output of the pump, = 0.8 × 20 hp = 16 hp
Therefore, the power used to overcome friction, =
-
∴ ≈ 16 hp - 14.1130725 Hp ≈ 1.8869275 hp
The power used to overcome friction, ≈ 1.9 hp