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3 years ago

As an engineer which types of ethical issues or problem you can face in industrial environment.

Engineering

1 answer:

Anarel [89]3 years ago

8 0

Explanation:

Answer ⬇️

Social and ethical issues in engineering, ethical principles of engineering, professional code of ethics, some specific social problems in engineering practice: privacy and data protection, corruption, user orientation, digital divide, human rights, access to basic services.

➡️dhruv73143(⌐■-■)

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3 years ago

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

dybincka [34]

Answer:

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The maximum amount of water that can be withdrawn from the lake is represented by the following formula: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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3 years ago

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Answer:

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2 years ago

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