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Kazeer [188]
3 years ago
11

Freeeeeeee pointsssssss

Engineering
2 answers:
lukranit [14]3 years ago
8 0

Answer: Thanks YSM

Explanation:

sweet-ann [11.9K]3 years ago
4 0

Answer:

yay

Explanation:

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A 15-kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18? Assuming the water t
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Explanation: Reverse card.

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Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 9
dybincka [34]

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The lake can withdraw a maximum of 1.464\times 10^{10} cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

V = V_{in}+V_{p}-V_{e}-V_{out} (Eq. 1)

Where:

V - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

V_{in} - Inflow amount of water, measured in cubic feet per year.

V_{out} - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

V_{p} - Amount of water due to precipitation, measured in cubic feet per year.

V_{e} - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t

V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l} (Eq. 2)

Where:

f_{in} - Average watershed inflow, measured in cubic feet per second.

f_{out} - Average flow to be released, measured in cubic feet per second.

\Delta t - Yearly time, measured in seconds per year.

h_{p} - Change in lake height due to precipitation, measured in feet per year.

h_{e} - Change in lake height due to evaporation, measured in feet per year.

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

If we know that f_{in} = 900\,\frac{ft^{3}}{s}, f_{out} = 300\,\frac{ft^{3}}{s}, \Delta t = 31,536,000\,\frac{second}{yr}, h_{p} = 32\,\frac{in}{yr}, h_{e} = 55\,\frac{in}{yr} and A_{l} = 47,000\,acres, the available amount of water for supply purposes in the Triangle area is:

V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}

The lake can withdraw a maximum of 1.464\times 10^{10} cubic feet per year to provide water supply for the Triangle area.

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