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ddd [48]
3 years ago
15

50POINTS

Engineering
2 answers:
maxonik [38]3 years ago
7 0

Answer:

Ensure that all material and energy inputs and outputs are as inherently safe and benign as possible. Minimize the depletion of natural resources. Prevent waste. Develop and apply engineering solutions while being cognizant of local geography, aspirations, and cultures.Green engineering is the design, commercialization, and use of processes and products that minimize pollution, promote sustainability, and protect human health without sacrificing economic viability and efficiency.The goal of environmental engineering is to ensure that societal development and the use of water, land and air resources are sustainable. This goal is achieved by managing these resources so that environmental pollution and degradation is minimized.

Explanation:i helped

VikaD [51]3 years ago
3 0

Explanation:

One possible reason for this query could be the amount of trees per say would have to be cut down to construct the green engineering facilities. Another thing could be distance from other facilities. For example you wouldn't want a green engineering facility next to a fossil fuel burning facility. (That would be pointless.) Before comensing the green engineering process, you would also like to ask local officers for any enviornmental information that could be used during the making of this, since there may be endangered animals or things that could be affected by the construction. You would also not like a location near a big town or city since the noise or the very mild pollution may bother the locals.

#learnwithbrainly

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Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

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T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

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Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

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Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

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