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21 January, 12:17

7. The water storage in a reach of river is ~ 8.7 x107 m3. Streamflow into the river reach is ~ 95m3/sec; streamflow out of that reach is ~ 87 m3/sec. Groundwater flows into this reach of river at 3 m3/min, and daily evaporation is ~ 8 mm/day. Assuming these flows stay constant over 24 hours, what is the change in total storage in the reach after one full day?

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  1. 21 January, 12:27
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    dV = (0.002175 - 0.008*A)

    Explanation:

    Given:

    - Volume of water storage initially V_o = 8.7*10^7 m^3

    - Stream Flow into river Q_s, in = 95 m^3 / s

    - Stream Flow into river Q_s, out = 87 m^3 / s

    - Ground Water flow in Q_g, in = 3 m^3 / min

    - Evaporation from pond Q_e = 8 mm/day

    - Time taken into consideration t = 1 day

    Find:

    change in total storage in the reach after one full day

    Solution:

    We will use the flow balance to determine the final volume at the end of the day.

    (Q_in - Q_out) * t = dV

    (Q_s, in + Q_g, in - Q_s, out - Q_e) * 1 day = dV

    (95/86400 + 3/1440 - 87/86400 - 0.008*Area of pond) = dV

    dV = (0.002175 - 0.008*A)
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