Question

A river basin has an average annual precipitation of 2500 mm/year and a drainage area of 10000 km2. The average discharge at the outlet of the basin is 200 m3/s. What is the average evapotranspitation (mm/year) in this region? What is the runoff coefficient? Assuming that the basin has average precipitation and a constant runoff coefficient, estimate the average discharge in a subbasin of 2000 km2.

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Answer to a math question A river basin has an average annual precipitation of 2500 mm/year and a drainage area of 10000 km2. The average discharge at the outlet of the basin is 200 m3/s. What is the average evapotranspitation (mm/year) in this region? What is the runoff coefficient? Assuming that the basin has average precipitation and a constant runoff coefficient, estimate the average discharge in a subbasin of 2000 km2.

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Adonis

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106 Answers

\text{PQ} = \text{Precipitação} \times \text{Área de drenagem}

= 2500 \frac{\text{mm}}{\text{ano}} \times 10000 \text{km}^2

= \left(2500 \times 10^{-3} \frac{\text{m}}{\text{ano}} \right) \times \left(10000 \times 10^6 \text{m}^2\right)

= 25000 \times 10^3 \text{m}^3/\text{ano}

= 2.5 \times 10^{10} \text{m}^3/\text{ano}

Q = \text{Vazão} \times \text{Tempo}

= 200\frac{\text{m}^3}{\text{s}} \times (365 \times 24 \times 60 \times 60)\frac{\text{s}}{\text{ano}}

= 200 \frac{\text{m}^3}{\text{s}} \times 31536000 \frac{\text{s}}{\text{ano}}

= 6.3072 \times 10^9 \text{m}^3/\text{ano}

ET = \text{What we need to calculate} = PQ - Q

= 2.5 \times 10^{10} - 6.3072 \times 10^9 \text{m}^3/\text{ano}

= 1.86928 \times 10^{10} \text{m}^3/\text{ano}

To transform the result to mm (per year), divide it by the area again and then express in mm/year.

ET = \frac{1.86928 \times 10^{10} \text{m}^3/\text{ano}}{10^7 \text{m}^2}

ET = 1869.28 \text{mm/ano}

Coeficiente de escoamento = \frac{\text{Q}}{PQ}

= \frac{6.3072 \times 10^9 \text{m}^3/\text{ano}}{2.5 \times 10^{10} \text{m}^3/\text{ano}}

= 0.2523

Now calculating average flow for a sub-basin of 2000 km^2:

Q_{sub} = Coeficiente de escoamento \times PQ_{sub}

= 0.2523 \times (2500 \times 10^{-3} \frac{\text{m}}{\text{ano}} \times 2000 \times 10^6 \text{m}^2)

= 0.2523 \times (5 \times 10^9 \text{m}^3/\text{ano})

= 1.2615\times 10^9 \text{m}^3/\text{ano}

Convert this to m^3/s:

Q_{sub}= \frac{1.2615\times 10^9 \frac{\text{m}^3}{\text{ano}}}{365 \times 24 \times 60 \times 60 \text{s}}

= \frac{1.2615\times 10^9}{31536000} \frac{\text{m}^3}{\text{s}}

≈ 40 \text{m}^3/\text{s}

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