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3 years ago

With the aid of a labbled diagram describe the operation of a core type single phase transformer

Engineering

1 answer:

olasank [31]3 years ago

4 0

Answer:

A simple single-phase transformer has each winding being wound cylindrically on a soft iron limb separately to provide a necessary magnetic circuit, which is commonly referred to as “transformer core”. It offers a path for the flow of the magnetic field to induce voltage between two windings.

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Grace [21]

Answer:

THANKS FOR THE POINTS!!!!!!!

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3 years ago

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What do you mean by overflow and underflow error in array?.

Misha Larkins [42]

Answer:

Overflow and underflow are both errors resulting from a shortage of space. On the most basic level, they manifest in data types like integers and floating points. Unlike the physical world, the number stored in a computer exists in a discrete number of digits.

Explanation:

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3 years ago

A typical aircraft fuselage structure would be capable of carrying torsion moment. a)True b)- False

Taya2010 [7]

Answer:

True

Explanation:

An aircraft is subject to 3 primary rotations

1) About longitudinal axis known as rolling

2) About lateral axis known as known as pitching

3) About the vertical axis known as yawing

The rolling of the aircraft induces torsion in the body of the aircraft thus the fuselage structure should be capable of carrying torsion

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3 years ago

Check the answer that best describes the relationship between f(x) and x. (For example if f(x) is Θ(x) check that as your answer

egoroff_w [7]

Answer:

Correct option is A - O(x)

Explanation:

Given f(x) = 10 and g(x) = x

Hence f(x) is O g(x)

As such correct answer is O(x) which is option A

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4 years ago

A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes

ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f = 0.77 for 232 ksi

because for Se 0.5 Su at $10^{6}$ cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = $\frac{(f*Su)^2}{Se}$

a = $\frac{(0.77*1600)^2}{700}$

a = 2168.3 Mpa

coefficient value (b) will be

a = - $\frac{1}{3}$ log $\frac{(f*Su)}{Se}$

b = - $\frac{1}{3}$ log $\frac{(0.77*1600)}{700}$

b = -0.0818

so no of cycle N is

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value

N = $(\frac{ 900}{2168.3})^{1/-0.0818}$

N = 46400

the life (N) of the specimen is 46400 cycles

5 0

3 years ago

With the aid of a labbled diagram describe the operation of a core type single phase transformer​

With the aid of a labbled diagram describe the operation of a core type single phase transformer