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Vanyuwa [196]
3 years ago
5

An industrial system is rated at 600 v the system includes a partially exposed terminal block thats mounted on a bulkhead a new

electrical apprentice who isnt considered to be qualified according to OSHA's standards approches the energized block the closest the apprentice should come to the block is?
Engineering
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

10 ft or 3.048 m (for unqualified person)

1.09 ft or 0.33 m (for qualified person)

Explanation:

OSHA has prescribed "Minimum Approach Distance - MAD" to ensure safety of personnel working with electrical power lines.

For a unqualified person:

The minimum approach distance is 10 ft or 3.048 m

For a qualified person:

For a voltage in the range of 300V to 750V (Phase to Ground or Phase to Phase) the minimum approach distance is 1.09 ft or 0.33 m

Since 600V lies in the above range hence it is applicable on 600V.

OSHA has developed a full list of minimum approach distance according to the level of voltages (50V to 72000V)

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Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm.
bija089 [108]

Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

<u>Given the following data:</u>

  • Emf = 520 Volts
  • Speed = 660 r.p.m
  • Number of armature conductors = 144 slots
  • Number of poles = 4 poles
  • Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

E = \frac{\theta ZN}{60} × \frac{P}{A}

<u>Where:</u>

  • E is the electromotive force in the DC generator.
  • Z is the total number of armature conductors.
  • N is the speed or armature rotation in r.p.m.
  • P is the number of poles.
  • A is the number of parallel paths in armature.
  • Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

Z = 144 × 2 × 3

Z = 864

Substituting the given parameters into the formula, we have;

520 = \frac{\theta (864)(660)}{60} × \frac{4}{2}

520 = \theta (864)(11) × 2

520 = 19008 \theta \\\\\Theta = \frac{520}{19008}

<em>Magnetic flux </em><em>=</em><em> 0.0274 Weber.</em>

Therefore, the magnetic flux per pole is 0.0274 Weber.

Read more: brainly.com/question/15449812?referrer=searchResults

5 0
3 years ago
un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti
777dan777 [17]
Ecfñnokg pinogdf gabn Etta r
7 0
3 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
3 years ago
How many hours should it take an articulated wheel loader equipped with a 4-yd^3 bucket to load 3000 yd^3 of gravel (average mat
densk [106]

Answer:

17 hours 15 minutes

Explanation:

See attached picture.

4 0
3 years ago
During his military campaign in what is now Germany, Julius Caesar lead his army of 40,000 soldiers to the western bank of the R
CaHeK987 [17]

Answer:

identifying a problem

Explanation:

its right

5 0
3 years ago
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