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Dafna11 [192]
2 years ago
6

Which pipe for water is best for construction?

Engineering
1 answer:
MAVERICK [17]2 years ago
4 0

Likely, CPVC. Most recommended.

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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
6) Describe the differences between the troposphere and stratosphere.
vodka [1.7K]

Answer:

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Explanation:

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3 years ago
A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa
Fed [463]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
4 0
3 years ago
Read 2 more answers
Air enters a tank through an area of 0.2 ft2 with a velocity of 15 ft/s and a density of 0.03 slug/ft3. Air leaves with a veloci
Mademuasel [1]

Answer:

please find attached.

Explanation:

4 0
3 years ago
Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m
lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

Q = A \times V

where A is Area

A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2

Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s

minimum flow rate provided by pump is 0.02513 m^3/s

5 0
3 years ago
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