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Y_Kistochka [10]
3 years ago
5

What are some sources of resistance? (Check all

Engineering
1 answer:
Zarrin [17]3 years ago
7 0

Answer:

All of them.

Explanation:

I take this class rn too and those are answers

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When hermetic refrigerant motor-compressors are designed to operate continuously at currents greater than 156 percent of the rat
ANEK [815]

The nameplate of a hermetic refrigerant motor-compressor that is designed to operate continuously at currents greater than 156% of the rated-load current is marked with branch-circuit selection current.

<h3>What is a hermetic refrigerant motor-compressor?</h3>

A hermetic refrigerant motor-compressor can be defined as a mechanical device that is designed and developed by combining a compressor and motor in a single outer-welded steel shell.

Basically, a hermetic refrigerant motor-compressor is used in the following areas:

  • Household refrigerators
  • Freezers
  • Heat pumps
  • Air coolers
  • Water cooling equipment.
  • Small refrigeration equipment.

According to HSE, the nameplate of a hermetic refrigerant motor-compressor that is designed to operate continuously at currents greater than 156% of the rated-load current is marked with branch-circuit selection current, so as to ensure safety for end users and technicians.

Read more on refrigerants here: brainly.com/question/2928084

6 0
3 years ago
A proposed piping and pumping system has 20-psig static pressure, and the piping discharges to atmosphere 160 ft above the pump.
larisa86 [58]

Answer: (B) 100

Explanation:

Given that;

Pstatic = 20 psig , hz = 160ft, hf = 20ft

Now total head will be;

T.h = hz + hf

T.h= 160 + 20

T.h = 180ft

Minimum pressure = Psatic + egh

we know that specific weight of water is 62.4 (lb/ft3)

so

P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr

P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)

P.min = 20 + 78

P.min = 98 lbf/in²

Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100

7 0
3 years ago
A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?
IceJOKER [234]

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V <em>(Here 208 V is three phase voltage and 120 V is single phase voltage) </em>

<em>Since,</em>

<em />

<em />V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\<em />

V_{1ph} = 120 V

The formula for apparent power in three phase system is given as:

S = \sqrt{3} VI

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A

 

4 0
3 years ago
Yall know what this is called?​
aliya0001 [1]

Answer:

oof no bro

Explanation:

5 0
3 years ago
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
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