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nalin [4]
3 years ago
7

A three-phase transformer bank consists of 3 single-phase transformers to handle 400 kVA witha 34.5kV/13.8kV voltage ratio. Find

the rating of each individual single-phase transformer in the bank (high voltage, low voltage, turns ratio, and apparent power) for the following transformer connections:
Engineering
1 answer:
Contact [7]3 years ago
6 0

Answer:

(a) high voltage = 19.9 kV, low voltage = 7.97 kV, turns ratio = 2.50:1, apparent power = 133 kVA

(b) high voltage = 19.9 kV, low voltage = 13.8 kV, turns ratio = 1.44:1, apparent power = 133 kVA

(c) high voltage = 34.5 kV, low voltage = 7.97 kV, turns ratio = 4.33:1, apparent power = 133 kVA

(d) high voltage = 34.5 kV, low voltage = 13.8 kV, turns ratio = 2.50:1, apparent power = 133 kVA

Explanation:

The turn ratio can be estimated by taking the ratio of the values of the high voltage and the low voltage. For example, the turn ratio for (a) is 19.9 kV/7.97 kV = 2.50:1. Similarly, the apparent power for each transformer is calculated as 400 kVA/3 = 133 kVA. Furthermore, the high and low voltages for the Δ are the same for the three-phase transformer i.e. 34.5 kv/13.8 kV. The high and high voltages for Y connection is calculated as 34.5 kV/1.73 = 19.9 kV and 13.8 kV/1.73 = 7.97 kV.

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See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

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A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after
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<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

Explanation:

<h2>Given:</h2>

m = 1,250 kg

v_i = 30 km/hr

F = 1,700 N

t = 10 s

<h2>Required:</h2>

Final velocity

<h2>Equation:</h2><h3>Force</h3>

F = ma

where: F - force

m - mass

a - acceleration

<h3>Acceleration</h3>

a = \frac{v_f \:-\:v_i}{t}

where: a - acceleration

v_i - initial velocity

v_f - final velocity

t - time elapsed

<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>

F = ma

Substitute the value of F and m

(1700 N) = (1250 kg)(a)

a = \frac{1700\:N}{1250\:N}

a = 1.36 m/s²

<h3>Solve for final velocity using the formula for acceleration</h3>
  • Convert 30 km/hr to m/s

= \frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}

= 8.33 m/s

  • Substitute the value of a, v_i and t

a = \frac{v_f \:-\:v_i}{t}

1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}

(10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s

v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s

v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s

v_f\: =\:  21.93\: m/s

  • Convert to km/hr

= \frac{21.93\:m}{s}\:×\:\frac{1\:km}{1000\:m}\:×\:\frac{3600\:s}{1/:hr}

= 78.95\: km/hr

<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
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