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OverLord2011 [107]
2 years ago
14

Technician a says that personal protective equipment (ppe) does not include clothing. technician b says that the ppe used should

be based on the task you are performing. who is correct?
Engineering
1 answer:
frez [133]2 years ago
4 0

Based on the information provided, the technician who is correct is: B. Technician B.

<h3>What is PPE?</h3>

PPE is an abbreviation for personal protective equipment and it can be defined as a terminology that is used to denote any piece of equipment which offer protection to different parts of the body while working in a potentially hazardous environment.

Additionally, some examples of personal protective equipment (PPE) that are used to protect the different parts of the body are:

  • Respirators
  • Face mask
  • Face shield
  • Gloves
  • Boots
  • Helmet

Based on the information provided, the technician who is correct is Technician B only because PPE should be used based on the task being performed.

Read more on PPE here: brainly.com/question/19131588

#SPJ1

Complete Question:

Technician a says that personal protective equipment (ppe) does not include clothing. technician b says that the ppe used should be based on the task you are performing. who is correct?

Technician A.

Technician B.

Neither Technician A nor Technician B

Both Technician A and Technician B

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(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
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Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

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3 years ago
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