Ask question

Ask question

All categories

3 years ago

7

A 60-Hz single-phase transformer with capacity of 150 kVA has the following parameters: RP = 0.35 Ω RS = 0.002 Ω Rc = 5.2 kΩ XP

= 0.5 Ω XS = 0.008 Ω Xm = 1.1 kΩ The primary transformer voltage is 2.8 kV and the secondary is 230 V. The transformer is connected to a variable load (0 to 300 kW) with a lagging power factor of 0.83 and a load voltage equal to the rated transformer secondary. Determine: (a) the total input impedance of the transformer when the secondary is shorted; and (b) the input current, voltage, power and power factor at full load (150 kW). (c) Plot the voltage regulation versus load, and determine the load that produces 5% regulation.

1 answer:

worty [1.4K]3 years ago

8 0

Answer:

Explanation:

Total Resistance, Rt = (0.35 + 0.002 + 5200 + 0.5 + 0.008 + 1100) = 6300 ohms

a) Capacity of transformer, Pt = 150KVA = 150,000 W

Input Voltage, Vp = 2.8 KV = 2800 V

Current, Ip = 150000/2800 = 53.57 A

Input impedance, Zp = Vp/Ip = 2800/5357 = 52.27 ohms

b) i) Input current = 53.57 A

ii) Voltage, V = Ip * Rt = 53.57 X 6300 = 337.5 KV

iii) Power, P = I² * Rt = (53.57)² X 6300 = 18.08 MW

iv) Power factor = 0.83

You might be interested in

Determine the kinetic energy of the ball immediately after it is hit. (You must provide an answer before moving to the next part

earnstyle [38]

The question is incomplete. Here is the complete question.

A baseball palyer hits a 5.1 oz baseball with an initial velocity of 140ft/sat an angle of 40° with the horizontal as shown. Determine

a) The kinetic energy of the ball immediately after it is hit

b) The kinetic energy of the ball when it reaches its maximum height

c) The maximum height above the ground reached by the ball.

Answer: a) KE = 131.64 J

b) KE = 0

c) h = 126 ft

Explanation: <u>Kinetic</u> <u>energy</u> is the energy an object posses due to its motion. It can be calculated as $KE=\frac{1}{2}mv^{2}$

a) Kinetic energy's unit is Joule. So, we have to transform ounce in kg and ft/s in m/s for the units to correspond:

m = 5.1(0.02835)

m = 0.1445 kg

v = 140 ft = 42.67 m/s

Then, kinetic energy is

$KE=\frac{1}{2}(0.1445)(42.67)^{2}$

KE = 131.64 J

Kinetic energy immediately after the ball is hit is 131.64 J.

b) At its maximum height, the ball has its highest potential energy. Because of the law of conservation of energy, when potential energy is maximum, kinetic energy is minimum and vice-versa. So, at the maximum height, kinetic energy is 0.

c) This type of motion is <u>projectile</u> <u>motion</u>. The maximum height on a projectile motion can be determined by

$v_{y}^{2}=v_{0y}^{2}-2g\Delta y$

When h is maximum, $v_{y}=0$

Velocity of the ball has an angle with the horizontal, so initial velocity at the y-axis is

$v_{0y}=v_{0}sin(\theta)$

Substituting and solving

$v_{y}^{2}=v_{0}^{2}sin^{2}(\theta)-2gh$

$0=(42.67)^{2}sin^{2}(40)-2(9.8)h$

$19.6h=(42.67)^{2}(0.643)^{2}$

$h=\frac{(1820.73)(0.4132)}{19.6}$

h = 38.4 m

Transforming into ft: h = 126 ft

The maximum height above the ground reached by hte ball is 126 feet.

6 0

3 years ago

On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A

Anvisha [2.4K]

From the law of atmosphere

$N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}$

Where

$n_0$ = constant and is number density where the height y = 0cm

$n_V$ = Number density at height y=3cm

Kb = Boltzmann constant $= 1.38*10^{-23}J/K$

$T=20K$

$m = 10^{-19}kg$

Re-arranging the equation to have the value of the gravity,

$\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}$

$ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}$

Since it is 30% of value above surface, therefore $N_v = 0.3n_0$

$ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}$

$g = -\frac{KbT ln(0.3)}{my}$

$g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}$

$g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}$

$g = 1.104*10^{-1}m/s^2$

$g = 0.1m/s^2$

Therefore the correct answer is C.

4 0

4 years ago

An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign

dimaraw [331]

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

7 0

3 years ago

Which federal agency helps foreign governments with international conservation efforts?

TEA [102]

Answer:

US Fish and Wildlife Service

Explanation:

The federal agency that helps foreign governments with international conservation efforts are the US Fish and Wildlife Service. This organization focus includes enforcing federal wildlife laws, protecting endangered species, manage migratory birds, restoring nationally significant fisheries; conserving and restoring wildlife habitat, such as wetlands, and even helping foreign governments with their international conservation efforts.

5 0

4 years ago

Read 2 more answers

An astronomer would most likely use parallax to

12345 [234]

The correct answer is<span> measure the distance to a star.

It is</span> the semi-angle of inclination between two sight-lines to the star, as observed when the Earth is on opposite sides of the Sun in its orbit. This is used to measure the distance to closer stars.

4 0

3 years ago