Respuesta :
Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s
The rate of withdrawal is 10.11 cubic meter per second
Given-
Total reservoir is 1375 hectare. which is equal to [tex]1375\times 10^4[/tex] meter square.
The average seepage loss from the reservoir is 2.85 cm or 0.0285 m.
Total precipitation on the reservoir is 18.5 cm or 0.185 m.
Total evaporation is 9.5 cm or 0.085 m.
The average inflow into the reservoir is [tex]0.5\times10^6[/tex] cubic meter per day.
The total inflow in a month can be calculate is
[tex]=0.5\times 30=1500\times10^4[/tex]
Net inflow is equal to the total precipitation on the reservoir subtract by all the losses.It can be represent as,
[tex]Q_{net}=0.185 - 0.095 - 0.025[/tex]
[tex]Q_{net}=0.065[/tex]
Total volume inflow is equal to the product of net inflow and total reservoir,
[tex]V_{net} =1375\times 10^4\times Q_{net}[/tex]
[tex]V_{net} =1375\times 10^4\times 0.065[/tex]
[tex]V_{net} =89.375[/tex]
The constant rate of withdrawal in cubic meter can be calculated by adding the net inflow in a month, total volume inflow and the reservoir.
[tex]Q=1375\times 10^4\times 0.75+89.375\times 10^4+1500\times10^4[/tex]
[tex]Q=2620\times10^4[/tex]
For per second withdrawal,
[tex]Q=\dfrac{2620\times10^4}{30\times24\times60\times60}[/tex]
[tex]Q=10.11[/tex]
Hence, the rate of withdrawal is 10.11 cubic meter per second.
For more about the flow, follow the link below,
https://brainly.com/question/1410288
