Question

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?

Answer 1

Answer:

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)