# Excercises Electric Machinery Fundamentals

## Problem 2-18



In [1]:

%pylab inline




Populating the interactive namespace from numpy and matplotlib



### Description

A 10-kVA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 480-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V.



In [2]:

Sw = 10e3  # [VA]
Vp = 600   # [V]

Vh = 480   # [V] which is also the load voltage
Vl = 120   # [V]



#### (a)

• Sketch the transformer connection that will do the required job.

#### (b)

• Find the kilovoltampere rating of the transformer in the configuration.

#### (c)

• Find the maximum primary and secondary currents under these conditions.

#### (d)

The transformer in Problem 2-18 is identical to the transformer in Problem 2-17, but there is a significant difference in the apparent power capability of the transformer in the two situations.

• Why?
• What does that say about the best circumstances in which to use an autotransformer?

### SOLUTION

#### (a)

For this configuration, the common winding must be the larger of the two windings. The transformer connection is shown below:

#### (b)

The kVA rating of the autotransformer can be found from the equation:

$$S_{IO} = \frac{N_{SE} + N_C}{N_{SE}}S_W = \frac{N_{SE} + 4N_{SE}}{N_{SE}}S_W$$


In [3]:

n = Vh/Vl  # = Nc/Nse
n




Out[3]:

4.0




In [4]:

Sio = (1 + n)/1 * Sw
print('''
Sio = {:.1f} kVA
==============
'''.format(Sio/1000))




Sio = 50.0 kVA
==============



#### (c)

The maximum primary current for this configuration will be:

$$I_P = \frac{S}{V_P}$$


In [5]:

Ip = Sio/Vp
print('''
Ip = {:.1f} A
===========
'''.format(Ip))




Ip = 83.3 A
===========



and the maximum secondary current is:

$$I_S = \frac{S}{V_S}$$


In [6]:

Is = Sio/Vh
print('''
s = {:.0f} A
=========
'''.format(Is))




s = 104 A
=========



#### (d)

Note that the apparent power handling capability of the autotransformer is much higher when there is only a small difference between primary and secondary voltages. Autotransformers are normally only used when there is a small difference between the two voltage levels.