Respuesta :
Answer:
The storage of the lake has increased in [tex]4.58\times 10^{6}[/tex] 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:
[tex]\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}[/tex]
Where [tex]\Delta V_{storage}[/tex] is the monthly storage change of the lake, measured in cubic feet.
Monthly inflow
[tex]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)[/tex]
[tex]V_{inflow} = 77.76\times 10^{6}\,ft^{3}[/tex]
Monthly outflow
[tex]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)[/tex]
[tex]V_{outflow} = 66.98\times 10^{6}\,ft^{3}[/tex]
Seepage losses
[tex]V_{seepage} = s_{seepage}\cdot A_{lake}[/tex]
Where:
[tex]s_{seepage}[/tex] - Seepage length loss, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{seepage} = 1.5\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]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)[/tex]
[tex]V_{seepage} = 2.86\times 10^{6}\,ft^{3}[/tex]
Evaporation losses
[tex]V_{evaporation} = s_{evaporation}\cdot A_{lake}[/tex]
Where:
[tex]s_{evaporation}[/tex] - Evaporation length loss, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{evaporation} = 6\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex]
[tex]V_{evaporation} = 11.44\times 10^{6}\,ft^{3}[/tex]
Precipitation
[tex]V_{precipitation} = s_{precipitation}\cdot A_{lake}[/tex]
Where:
[tex]s_{precipitation}[/tex] - Precipitation length gain, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{precipitation} = 4.25\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]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)[/tex]
[tex]V_{precipitation} = 8.10\times 10^{6}\,ft^{3}[/tex]
Finally, we estimate the storage change of the lake during the month:
[tex]\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}[/tex]
[tex]\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}[/tex]
The storage of the lake has increased in [tex]4.58\times 10^{6}[/tex] cubic feet during the month.
The volume of water gained and the loss of water through flow,
seepage, precipitation and evaporation gives the storage change.
Response:
- The storage change for the lake in a month is 1,582,823.123 ft.³
How can the given information be used to calculate the storage change?
Given parameters:
Area of the lake = 525 acres
Inflow = 30 ft.³/s
Outflow = 27 ft.³/s
Seepage loss = 1.5 in. = 0.125 ft.
Total precipitation = 4.25 inches
Evaporator loss = 6 inches
Number of seconds in a month is found as follows;
[tex]30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds[/tex]
Number of seconds in a month = 2592000 s.
Volume change due to flow, [tex]V_{fl}[/tex] = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³
1 acre = 43560 ft.²
Therefore;
525 acres = 525 × 43560 ft.² = 2.2869 × 10⁷ ft.²
Volume of water in seepage loss, [tex]V_s[/tex] = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³
Volume gained due to precipitation, [tex]V_p[/tex] = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³
Volume evaporation loss, [tex]V_e[/tex] = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³
- [tex]Storage \, change \, \Delta V = \mathbf{V_{fl} - V_s + V_p - V_e}[/tex]
Which gives;
ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123
- The storage change, ΔV = 1,582,823.123 ft.³
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