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Hitman42 [59]
3 years ago
8

Once Joe Martin reports his concerns to senior management at corporate headquarters and requests that the Ethicana plant operati

ons be suspended until the faulty equipment, safety and operational issues are corrected it would be unethical for Joe to file a report with the Ethicana Worker Safety and Environmental Protection Agency if the plant continues in operation without repairs.
Engineering
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

Answer: It would be unethical to file a report on safety without proper repairs.

Explanation: As a top staff in the company Joe, complained to his senior management at their headquarters about the faulty parts and unsafe working conditions at the Ethicana plant. After filing his complaint nothing is done to correct it, therefore for Joe to give a safety report for an unsafe working environment is highly unethical and is against what engineering stands for in the area of occupational health safety and environment. It is best for him to stand his ground that the faulty equipments be fixed and that the unsafe working conditions at the Ethicana operations plants be corrected for the sake of the working population there and also for the benefit of the environment.

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(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length
EastWind [94]

This question is incomplete, the complete question is;

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2l. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

D₃ = _____D.

{ the tolerance is +/-3% }

Answer:

the diameter of the second pipe D₃ is 1.13D

Explanation:

Given the data in the question;

Length = 2l

pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.

Now, we know that for Laminar Flow;

V' = πD⁴ΔP / 128μL

where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃

Hence,

V'₁ = πD⁴ΔP₁₋₂ / 128μL  = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL

so

D₃ = D( ΔP₁₋₂ / ΔP₂₋₃ )^{\frac{1}{4}

we substitute

D₃ = D( 1.657 )^{\frac{1}{4}

D₃ = D( 1.134568 )

D₃ = 1.13D

Therefore, the diameter of the second pipe D₃ is 1.13D

8 0
3 years ago
A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame
denis23 [38]

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature T_\infity = 15 degree celcius

T_S =  220 + 273 = 473 K

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

Q = hA(T_S - T_{\infty}) \eta

= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8

Q = 125.538 W

5 0
3 years ago
The temperature in a pressure cooker is 130 degree C while the water is boiling. Determine the pressure inside the cooker.
emmasim [6.3K]

Answer:

The pressure inside the cooker is 1.0804 atm.

Explanation:

Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.

<u>For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.</u>

So, Standard conditions:

T₁ = 373.15 K

P₁ = 1 atm

Given ,

The water boils at Temperature =  130 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So, the temperature, T₂ = (130 + 273.15) K = 403.15 K

To find pressure inside the cooker (P₂) :

<u>Applying Amontons's Law as:</u>

\frac {P_1}{T_1}=\frac {P_2}{T_2}

So,

P_2=\frac {P_1}{T_1} \times {T_2}

P_2=\frac {1 atm}{373.15 K} \times {403.15 K}

P_2=1.0804 atm

<u>Thus, The pressure inside the cooker is 1.0804 atm.</u>

6 0
3 years ago
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Kipish [7]
My best guess is b but I honestly don’t know
8 0
3 years ago
Read 2 more answers
The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i
koban [17]

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            b=25m/sin(60)\\=28.867

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N

Noting from the Bernoulli  equation that

Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0
3 years ago
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