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USPshnik [31]
3 years ago
5

What are the four causes of electrical faults?

Engineering
1 answer:
Arada [10]3 years ago
6 0

Answer:

Electrical faults are also caused due to human errors such as selecting improper rating of equipment or devices, forgetting metallic or electrical conducting parts after servicing or maintenance, switching the circuit while it is under servicing, etc.

Explanation:

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A clutch is chattering. Technician A says that uneven wear on the flywheel could be the cause. Technician B says that worn or br
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Both Technician A and B are correct

Explanation:

Due to a contamination of the clutch disc friction surface, the clutch chatters

The contamination that causes clutch chattering includes

1) Worn or broken motor mounts

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3) Clutch link damage

4) Warped flywheel, due to overheating, the flywheel can become warped such that the non uniform surface interfaces the clutch resulting in clutch chattering

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; How do engineers make a difference in the world and with proof ?​
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Engineers are the reason healthcare has improved so dramatically throughout the years. The advancements in medical technology are down to the work conducted by engineers as well as the creation of devices that help to save lives and improve the quality of life for others. New developments are being created by engineers constantly, for example, the Crossrail project in London is creating a new transport system throughout the South East to create shorter journey times and easier connections. The existing transport system wouldn’t be where it is today without engineers.
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3 years ago
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A short framing member that fills the space<br> between the rough sill and the soleplate is a
Jet001 [13]

A short framing member that fills the space between the rough sill and the soleplate is a cripple stud.

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3 years ago
A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness
Bingel [31]

Answer:

a) heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

Explanation:

Assumptions:

  1. Constant properties
  2. Steady state conditions
  3. Negligible effect of radiation
  4. Negligible constant resistance between tube and insulation
  5. one dimensional radial conduction

a) What is the heat gain per unit tube length

R_{conv,i}'=\frac{1}{2\pi r_1h_i}

d_1=36mm Therefore r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}

r_2=2mm=2*10^{-3}m

k_{st}=14.2W/m.k

h_o=6W/m^2

h_i=400W/m^2

R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W

R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W

R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W

R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W

heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

r_3=r_1+r_2+10mm=30mm=0.03m

R_{conv,i}' and R_{cond,st}' are the same, but R_{conv,o}' changes.

Therefore:

R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W

R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W

The total resistance R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W

heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

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Answer:

It is used for solving optimization problems in which,given some functional,one seeks the function minimizing or minimizing it.

Explanation:

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