HELP WITH ASSIGNMENT 1

Question

                                             

            Question 1

1.1        Calculate the speed of a four-pole series generator having a wave-wound armature with 315 conductors and resistance of 0.6 ohm supplying a load of 50 kW at 1000 V. The resistance of the field-winding brush contact resistance is 0.4 ohm. The field sets up a flux per pole of 0.1 Wb.                                                                                                                        

 

1.2        What type of winding would be used for the following:

1.2.1     A high-voltage low-current.

1.2.2     A high-current low-voltage DC machine.                                                             

1.3        The open-circuit characteristics of a shunt-excited DC machine are as follows:

Terminal Voltage (V)	80	160	240	320	340	360
Field Current (A)	4	8	12	20	24	30

 

Plot a graph and determine the open-circuit voltage if the field circuit resistance is 12 ohms.          

1.4        A short-shunt compound generator supplies a load of 100 A. It has a shunt-field resistance of 20 ohms, an armature resistance of 0.3 ohm and a field resistance of 0.2 ohm.

Determine the armature EMF if the terminal voltage is 180 V.                                           

Question 2

2.1        A sinusoidal alternating current supply has a maximum value of 396.04 V and a periodic time of 50 milliseconds.

Determine the following:

2.1.1     The RMS value of the voltage.

2.1.2     The average value of the voltage.

2.1.3     The frequency.

2.1.4     The instantaneous value two milliseconds after the commencement of the cycle.  

2.2        A 48 kVA 4 800/48 V, 50 Hz single-phase transformer has 3000 turns on the primary winding.

Determine the following:

2.2.1     The turn ratio.

2.2.2     The number of secondary turns.

2.2.3     The secondary full-load current.

2.2.4     The maximum value of the core flux.                                                                  

2.3        In a certain circuit having TWO parallel branched the instantaneous branch circuits are represented by:

 

 

2.3.1     Determine the total current and write it in this form:

                                                                                     

 

2.3.2     Represent these currents by drawing a phasor diagram.                                       

 

Answer 1

(2.1)

Step 1:

The maximum/peak voltage is .

Time period of sinusoidal function is ms.

(2.1.1)

Find RMS value of the voltage.

The RMS voltage is .

Where is peak voltage.

RMS voltage is 280.042 V.

Step 2:

(2.1.2)

Find the average value of the voltage.

The average voltage is .

Where is peak voltage.

Average voltage is 252.126 V

Step 3:

(2.1.3)

Find the frequency.

Frequency is resiprocal of time period.

Where is time period.

Frequency is 20 Hz.

Comments

Contd....

Step 4:

(2.1.4)

Find the instantaneous value after two milliseconds.

The instantaneous voltage V_i value after a time of 2ms is given as:

Where is angular frequency,

             t  is time.

The instantaneous voltage V_i  value after a time of 2ms

The instantaneous voltage V_i value after a time of 2ms is 98.49 V.

Solution :

(2.1.1) RMS voltage is 280.042 V.

(2.1.2) Average voltage is 252.126 V.

(2.1.3) Frequency is 20 Hz.

(2.1.4) The instantaneous voltage V_i value after a time of 2ms is 98.49 V.

Answer 2

(2.2)

Step 1:

Transformer rating 48 kVA.

Frequency is .

Number of turns in primary winding is .

Primary voltage is .

Secondary voltage is .

(2.2.1)

Find the turn ratio.

Turn ratio is given as .

Turns ratio is 1:100.

Step 2:

(2.2.2)

Find the number of secondary winding turns.

The number turns can be find using the formula .

Substitute , and .

The number of turns in secondary winding is 30.

Step 3:

(2.2.3)

Find the secondary full-load current.

The formula for full-load current is .

Secondary voltage is .

Transformer rating 48 kVA.

Secondary full-load current is

The secondary full-load current is 577.35 A.

Comments

Step 4:

(2.2.4)

 Find the maximum value of the core flux.

Formula for the maximum core flux is .

Substitute , and .

The maximum value of the core flux is 7.202 m Wb.

Solution :

(2.2.1) Turns ratio is 1:100.

(2.2.2) The number of turns in secondary winding is 30.

(2.2.3) The secondary full-load current is 577.35 A.

(2.2.4) The maximum value of the core flux is 7.202 m Wb.

Answer 3

(2.3.1)

Step 1:

The two branch currents are and .

If the polar form of equation is then complex form of .

Complex form of equation is

Complex form of equation is

For parallel connection : .

Hence the resultant current is .

in polar form can be written as .

Hence .

Step 2:

(2.3.2)

Phasor diagram :

Plot the phasor diagram .

Solution :

.

 

Answer 4

(1.2)

Step 1:

Winding is a twisting movement or course of wire.

Types of winding are

                Coil winding machine

                Film winding machine

                Rope winding machine

                wave winding

                Paper winding machine

                Foil winding machine

(1.2.1)

For high-voltage low-current applications wave winding is used.

Wave winding is one type of armature winding. In this winding the end of one coil is connected to the starting of another coil of the same polarity as that of the first coil.

In this winding only two brushes are required but more parallel brushes can be added to make it equal to the number of poles. If one or more brushes set poor contacts with the commutator, satisfactory operation is still possible.

For a given number of poles and armature conductors it gives more emf than that of lap winding. Hence wave winding is used in high voltage and low  current  machines. This winding is suitable for small generators circuit with voltage rating 500-600V.

(1.2.2)

For high-current low-voltage applications foil winding is preferred.

Solution :

(1.2.1) High-voltage low-current applications wave winding is used.

(1.2.2) High-current low-voltage applications foil winding is preferred.

Answer 5

(1.1)

Step 1:

For a four pole generator :

The number of conductors Z  is 315 conductors.

Armature resistance is 0.6 .

Brush resistance is 0.4 .

The total armature resistance R_a is 0.6 + 0.4 = 1 .

Flux per pole is 0.1 Wb.

DC generator supplies load of 50 kW at 1000 v.

Armature current is .

The EMF of a DC generator : ,

is the emf of the DC Generator.

.

where Z  is the number of conductors.

N  is the speed of the DC generator.

is the magnetic flux.

P  is the number of magnetic poles.

a is the number of parallel circuits.

Step 2:

Calculate the emf of the DC Generator.

Emf of the DC Generator is 950 v.

EMF of a DC generator : .

For a four pole generator : P = 4

For wave wound winding : a = 2

Speed of the DC generator is 905 rpm.

Solution :

Speed of the DC generator is 905 rpm.

Answer 6

(1.3)

Step 1:

The open-circuit characteristics of a shunt-excited DC machine are given in a table.

Plot the open-circuit characteristics with given values.

Graph:

Draw the coordinate plane.

Plot the points from the table.

Connect the points to a smooth curve.

The graph shows the open-circuit characteristics of a shunt-excited DC machine.

Step 2:

Find the open-circuit voltage if the field circuit resistance is 12 ohms.

Draw a 12 ohms resistance line on open-circuit characteristics curve.

Consider the points (0, 0) and (4, 48) with a slope of 12.

Now draw a line connecting (0, 0) and (4, 48) on open-circuit characteristics curve.

Now observe the graph:

The 12 ohm resistance line intersects the open-circuit characteristics curve at (30, 360).

Therefore the open-circuit voltage when the field circuit resistance 12 ohms is 360 V.

Solution :

The open-circuit voltage when the field circuit resistance 12 ohms is 360 V.

Answer 7

(1.4)

Step 1:

The short-shunt compound generator supplies a load current of 100 A.

Shunt field resistance is 20 .

Armature resistance is 0.3 .

Field resistance is 0.2 .

Terminal voltage is 180 v.

DC short shunt compound generator :

EMF generated by the DC generator is , where .

Step 2:

Calculate .

Armature current .

EMF generated by the DC generator is .

.

EMF generated by the DC generator is .

Solution:

EMF generated by the DC generator is .