PLEASE HELP ITS DUE TODAY ILL POST THE OTHER HALF

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

2 years ago

I need a science fair project that doesn't need anything on the pic and it can be about any subject thanks.

tigry1 [53]

you could make a self propelled car all you need is cardboard, wheels, and a balloons or rubber bands

7 0

3 years ago

Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro

luda_lava [24]

Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$

8 0

3 years ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

2 years ago

A stationary 90 kg baseball pitcher throws a 0.15 kg baseball forward

Marina CMI [18]

Answer: momentum = 6kgm/s

Explanation:

given that the baseball pitcher is at stationary position, his velocity will be equal to zero. If velocity is zero, his linear momentum will therefore equal to zero.

Linear momentum is the product of mass and velocity. Given that the baseball has

Mass M = 0.15 kg

Velocity V = 40 m/s

Momentum = MV

Momentum = 0.15 × 40 = 6 kgm/s

4 0

2 years ago