Question

A lake with a surface area of 525 acres was monitored over a period of time. During onemonth period the inflow was 30 cfs (ie. ft3 /sec), the outflow was 27 cfs, and a 1.5 in seepage loss was measured. During the same month, the total precipitation was 4.25 inches. Evaporation loss was estimated as 6 inches. Estimate the storage change for this lake during the month.

Answer 1

Answer:

The storage of the lake has increased in $4.58\times 10^{6}$ cubic feet during the month.

Explanation:

We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:

$\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}$

Where $\Delta V_{storage}$ is the monthly storage change of the lake, measured in cubic feet.

Monthly inflow

$V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)$

$V_{inflow} = 77.76\times 10^{6}\,ft^{3}$

Monthly outflow

$V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)$

$V_{outflow} = 66.98\times 10^{6}\,ft^{3}$

Seepage losses

$V_{seepage} = s_{seepage}\cdot A_{lake}$

Where:

$s_{seepage}$ - Seepage length loss, measured in feet.

$A_{lake}$ - Surface area of the lake, measured in square feet.

If we know that $s_{seepage} = 1.5\,in$ and $A_{lake} = 525\,acres$, then:

$V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)$

$V_{seepage} = 2.86\times 10^{6}\,ft^{3}$

Evaporation losses

$V_{evaporation} = s_{evaporation}\cdot A_{lake}$

Where:

$s_{evaporation}$ - Evaporation length loss, measured in feet.

$A_{lake}$ - Surface area of the lake, measured in square feet.

If we know that $s_{evaporation} = 6\,in$ and $A_{lake} = 525\,acres$, then:

$V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)$

$V_{evaporation} = 11.44\times 10^{6}\,ft^{3}$

Precipitation

$V_{precipitation} = s_{precipitation}\cdot A_{lake}$

Where:

$s_{precipitation}$ - Precipitation length gain, measured in feet.

$A_{lake}$ - Surface area of the lake, measured in square feet.

If we know that $s_{precipitation} = 4.25\,in$ and $A_{lake} = 525\,acres$, then:

$V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)$

$V_{precipitation} = 8.10\times 10^{6}\,ft^{3}$

Finally, we estimate the storage change of the lake during the month:

$\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}$

$\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}$

The storage of the lake has increased in $4.58\times 10^{6}$ cubic feet during the month.