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Ivahew [28]
3 years ago
12

Emergency plans are being formulated so that rapid action can be taken in the event of an equipment failure. It is predicted tha

t if a particular pipeline were to rupture, it would release ammonia at a rate of 100 lb/s. It is decided that anyone exposed to potential concentrations exceeding 500 ppm must be evacuated.
1. What recommendation will you make as to the evacuation distance downwind?
Engineering
2 answers:
chubhunter [2.5K]3 years ago
7 0

Answer:

Ammonia gas a hazardous gas to our health, when we are exposed to it for a long time. The gas is lighter than air, that means it's high concentration may not be noticed at the point of leakage, because it flows with the wind direction. Ammonia gas detector are used to determine the concentration of the gas at a particular place. We can use the dispersion modelling software program to know the exact position, where we can place the gas detector, which would be where evacuation is needed.

During evacuation, when the concentration of the gas has increased, a self-contained breathing apparatus should be used for breathing, and an encapsulated suit should be worn to prevent ammonia from reacting with our sweat or any other chemical burn. A mechanic ventilation will also be needed in the place of evacuation, so that the ammonia concentration in that area can be dispersed.

timofeeve [1]3 years ago
6 0

Answer / Explanation:

Ammonia is considered as a high risk to health chemical gas when exposed to the skin and other features of the body such as the eye. Exposure to ammonia in the range of 300 ppm could lead to immediate mortality or life threatening conditions. When ammonia is exposed to fire it has a very high tendency of exploding.

Now, if we refer back to the question where we are being asked on what recommendation we will propose as to the evacuation distance downwind in the event of exposure to ammonia of potential concentrations exceeding 500 ppm.

The first and most important step to take is to move the exposed individuals to a location of fresh air to avoid chocking of the victims and in the case where the victim has inhaled some amount of ammonia, a breathing aid is immediately administered or a self-contained breathing apparatus should be used.

After which we rinse or wash the eyes of the exposed victims with a lukewarm like water.

Most importantly, we make sure we bath all affected areas properly with water. Then we move forward to detecting the source of the leakage.  

Litmus paper or a sulfur stick can be used to detect the source of leaking ammonia. Ammonia can be in gas or liquid form. The liquid form is flammable so only use a burning sulfur stick to detect small leaks of ammonia gas

On doing this, we direct the leak through a flue or vent to properly disperse the leaked or leaking ammonia.

We should also note that during this process, an encapsulated suit should be worn to prevent ammonia from reacting with our sweat or any other chemical burn  

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Explanation:

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To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))     where i=1 to N-1 and N=6

C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))

A area of the polygon can be found by the following formulaA=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i}) where i=1 to N-1

A=\frac{1}{2}[ (x_{1}  y_{2} -x_{2}  y_{1})+ (x_{2}  y_{3} -x_{3}  y_{2})+(x_{3}  y_{4} -x_{4}  y_{3})+(x_{4}  y_{5} -x_{5}  y_{4})+(x_{5}  y_{6} -x_{6}  y_{5})]

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(x_{2}+x_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(x_{3}+x_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(x_{4}+x_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(x_{5}+x_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{x} =15.36

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C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(y_{2}+y_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(y_{3}+y_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(y_{4}+y_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(y_{5}+y_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{y} =22.55

So coordinates for the fire station should be (15.36,22.55)

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