Give two examples where friction is a nuisance

Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

b

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

=99.941%

8 0

3 years ago

Pls ans 10 no. From laws of motion

Vlad1618 [11]

Answer:

40N

Explanation:

Since both weights are connected to one string, you can say that the tensions above each are equal to each other.

If you do the sum of forces for the 4kg mass, then the tension comes out to 40N (if we take gravity to be 10m/s²). But that seemed too good to be true, so I decided to do the work for the 7kg mass as well [which included finding the normal force (N) and plugging it into the sum of forces for the 7kg mass] to find that it also gives 40N as the answer.

If I were to put my process into steps:

Write out the sum of Forces for both masses
Set them equal to each other to find normal force (because this is the only unknown)
Calculate and compare the two tensions to see if they are equal

*This all seems to line up perfectly, but do let me know if my answer doesn't match up with what you might find to he the answer later on.

4 0

3 years ago

An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?

Marizza181 [45]

Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2

6 0

3 years ago

The standard normal curve is not symmetrical. <br> a. True<br> b. False

FrozenT [24]

Answer;

The above statement is false

Explanation;

Symmetrical distribution, commonly known as symmetric distribution or normal distribution, is typically unimodal, meaning it shows only one peak in graph form.

It is a type of distribution where the left side of the distribution mirrors the right side. By definition, a symmetric distribution is never a skewed distribution.

All normal distributions are symmetric and have bell-shaped density curves with a single peak.

7 0

3 years ago

Read 2 more answers

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

$F=ma$

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0

3 years ago

Read 2 more answers