Water Surface Elevation of the Lake at the End of the Month

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A lake had a water surface elevation of 103.200 m above datum at the beginning of a certain month.

In that month the lake received an average inflow of 6.0 m3/s from surface runoff sources. In the same period, the outflow from the lake had an average value of 6.5 m3/s.

Further, in that month, the lake received a rainfall of 145 mm and the evaporation from the lake surface was estimated as 6.10 cm.

What will be the water surface elevation of the lake at the end of the month assuming no contribution to or from the groundwater storage?

The average lake surface area may be taken as 5000 hectares.

Please share all steps (if possible).

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3 Answers

  1. There is possibly a set formula to undertake this calculation, but I will use a simple water balance and then the total is either added or subtracted  to the surface elevation height to obtain the answer.   Please note, units of measurement are everything with this problem. 

    Starting with rainfall and evaporation; +0.145 m -0.061 m = +0.084 m; inflow and outflow, +6.0 m3/s -6.5 m3/s = -0.5 m3/s;

    *Assume that month constitutes 31 days, therefore 31 days x 86400 sec/day = 2678400 sec/month x -0.5 m3/s = -1,339,200 m3/month.  (1,339.2 ML)

    *Assume 1 megalitre (1,000,000 L) covers one hectare to 100 mm.  As the surface area of the lake is 5,000 ha, this is 50 ML (5,000 divided by 100 mm) and 1,339.2 ML divided by 50 ML = 26.784 or -2.6784 m

    Our water balance is therefore; +0.084 m in and -2.6784 m out.

    Take one from the other  = -2.67 out.

    The surface datum point at the start of the month was 103.2 less water out of -2.67 equals a height datum at the end of the month of 100.53 metres.

    Regards, Kevin.

     

  2. at the end of the month the lake water surface elevation is 103.2581m.

    the steps are as follows

    Given:

    initial lake water surface elevation (Hi) =103.2m

    inflow to the lake (Qi) = 6.0 m3/s = 6 m3/s * 3600 s/hr *24 hr/day * 30 day/month = 15,552,000 m3 , express in depth by dividing in to the area = 0.31104 m

    outflow from the lake (Qo) = 6.5 m3/s = 6.5 m3/s * 3600 s/hr *24 hr/day * 30 day/month = 16,848,000 m3  = 0.33696 m

    precipitation (P) = 145 mm = 0.145 m

    evaporation from the lake (E) = 6.10 cm = 0.061 m

    lake surface area = 5000 ha = 5 *10^7 m2

    SOLUTION

    Ho at the end of the month = Hi + (all inflows (gain) - all out floss (loss))

    = 103.200 m + (0.31104+0.145 - (0.33696+ 0.061)) m

    =103.200 m + 0.05808 m

    Ho = 103.2581 m

     

  3. The first point is that you have used 'mm' for rainfall and 'cm' for evaporation. It is never a good idea to mix units. Planes have crashed and space missions have failed for this reason.

    You don't say which month- so let us assume 31 days.

    The inflow is 6.0 m3/s and the outflow is 6.5 m3/s, So the net outflow rate is 0.5 m3/s. The volume of outflow is 0.5 (m3/s) * 31 (days) * 86400 (s/day). That is 1,339,200 m3.

    The area of the lake is 50,000 ha, That is 50,000 (ha) * 10,000 (m2/ha) = 500,000,000 m3.

    The drop in lake level is 1,339,200 (m3) / 500,000,000 (m2) = 0.002678 m = 2.678 mm. (which can rounded to 2.7 mm)

    The change in level is +145 mm (rainfall) - 61 mm (evaporation) - 2.7 mm (net outflow) = 81.3 mm

    The final level is 103.20 (m) + 81.3 (mm)/1000 (mm/m) = 103.281 m