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fiasKO [112]
2 years ago
7

Adding new equipment or processes may require changes to the PPE requirements for

Engineering
2 answers:
Yuki888 [10]2 years ago
5 0
I think it’s is false I’m not that sure
Vesnalui [34]2 years ago
3 0
True

Explanation- your changing what your using in your lab
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Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
3 years ago
Why were the French and Dutch colonized areas so small compared to the Spanish colonized areas?
Degger [83]
The French and Dutch colonized arenas so small compared to the Spanish colonized areas For farming
4 0
3 years ago
Read 2 more answers
Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equat
PtichkaEL [24]

Answer:

Detailed solution is attached below in three simple steps the problem is solved.

7 0
3 years ago
Water steam enters a turbine at a temperature of 400 o C and a pressure of 3 MPa. Water saturated vapor exhausts from the turbin
yulyashka [42]

Answer:

a) -505.229 kJ/Kg

b) -1.724 kJ/kg

Explanation:

T1 = 400°C

P1 = 3 MPa

P2 = 125 kPa

work output   = 530 kJ/kg

surrounding temperature = 20°C = 293 k

<u>A) Calculate heat transfer from Turbine to surroundings </u>

Q = h2 + w - h1

h ( enthalpy )

h1 = 3231.229 kj/kg

enthalpy at P2

h2 = hg = 2676 kj/kg

back to equation 1

Q = 2676 + 50 - 3231.229  = -505.229 kJ/Kg  ( i.e.  heat is lost )

<u>b) Entropy generation </u>

entropy generation = Δs ( surrounding )  + Δs(system)

                                =  - 505.229 / 293   + 0

                                = -1.724 kJ/kg  

6 0
3 years ago
Determine the magnitude of force P needed to start towing the 40kg crate.Also determine the location of the resultant normal for
erastova [34]

The distance from Point A=500 mm

Explanation:

M := 40kg c := 200mm

μs := 0.3 d := 3

a := 400mm e := 4

b := 800mm

Initial guesses: N_{c} := 200 N P := 50N

Plz refer to the image below

5 0
3 years ago
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