Three Phase Full Converters MCQ Quiz - Objective Question with Answer for Three Phase Full Converters - Download Free PDF

Calculation of the Required Firing Angle:

The given problem involves a 750 Hp, 250 V, 1200 r/min DC motor connected to a 208 V, 3-phase, 60 Hz line through a 3-phase bridge converter. We aim to calculate the required firing angle under rated full load conditions.

Given Data:

• Motor Power (P) = 750 Hp (1 Hp = 746 W, so P = 750 × 746 W = 559500 W)
• Armature Voltage (V_a) = 250 V
• Armature Current (I_a) = 2500 A
• Armature Resistance (R_a) = 4 mΩ = 0.004 Ω
• Line Voltage (V_L) = 208 V
• Frequency (f) = 60 Hz
• Motor Speed = 1200 r/min

Step-by-Step Solution:

1. Calculate the Back EMF (E):

The back EMF (E) can be calculated using the voltage equation of the motor:

V_a = E + I_a × R_a

Rearranging for E:

E = V_a - I_a × R_a

Substitute the values:

E = 250 - (2500 × 0.004) = 250 - 10 = 240 V

2. RMS Line Voltage of the Converter:

The RMS line voltage of the converter is given as V_L = 208 V. The peak line voltage (V_m) can be calculated as:

V_m = √2 × V_L

Substitute the value of V_L:

V_m = √2 × 208 = 294.96 V

3. Average Output Voltage of the Converter:

The average output voltage (V_dc) of a 3-phase bridge converter is given by:

V_dc = (3√2 / π) × V_m × cos(α)

Here, α is the firing angle. Rearranging to solve for cos(α):

cos(α) = (V_dc × π) / (3√2 × V_m)

Since V_dc is the back EMF (E), we use E = 240 V:

cos(α) = (240 × π) / (3√2 × 294.96)

Substitute the values:

cos(α) = (240 × 3.1416) / (3 × 1.414 × 294.96)

cos(α) = 753.984 / 1250.89 = 0.6026

4. Calculate the Firing Angle (α):

Take the inverse cosine of 0.6026 to find α:

α = cos^-1(0.6026)

α ≈ 52.75°

5. Adjust the Angle to the Closest Option:

The closest option to the calculated firing angle is 45°. Therefore, the correct option is:

Option 1: 45°

Additional Information

Analysis of Other Options:

Option 2 (208°): This option is far beyond the practical firing angle range for a 3-phase bridge converter. Firing angles typically range between 0° and 90°, where higher angles reduce the average output voltage. Thus, 208° is not valid.

Option 3 (27°): While 27° is within the acceptable range for a firing angle, it does not match the calculated value of 45°. This option is incorrect.

Option 4 (32°): Similarly, 32° is a feasible firing angle but does not correspond to the calculated value. This option is also incorrect.

Conclusion:

The correct firing angle under rated full load conditions is 45°, as calculated. This value ensures that the converter provides the required average output voltage to the DC motor under the given operating conditions. Understanding the relationship between firing angle, line voltage, and output voltage is critical for analyzing and designing converter-fed motor systems.

Explanation:

Calculation of DC Voltage Generated by the Inverter

To calculate the DC voltage generated by the inverter, let us analyze the given data step-by-step:

• DC source voltage, V_source = 16 kV = 16000 V
• Internal resistance of the DC source, R_internal = 1 Ω
• DC current supplied by the source, I_DC = 900 A
• Inverter output voltage, V_inverter = 12 kV = 12000 V

Step 1: Voltage Drop Across the Internal Resistance

The internal resistance of the DC source causes a voltage drop, which can be calculated using Ohm's Law:

Voltage Drop (V_drop) = I_DC × R_internal

Substitute the given values:

V_drop = 900 A × 1 Ω = 900 V

Step 2: DC Voltage at the Terminals of the Source

The DC voltage at the terminals of the source is the source voltage minus the voltage drop across the internal resistance:

V_terminals = V_source - V_drop

Substitute the values:

V_terminals = 16000 V - 900 V = 15100 V

Step 3: DC Voltage Generated by the Inverter

The inverter generates a DC voltage that corresponds to the terminal voltage of the source because it is directly connected to the source.

Thus, the DC voltage generated by the inverter is:

V_inverter = V_terminals = 15100 V

Final Answer: The DC voltage generated by the inverter is 15100 V.

Analysis of Other Options:

Let us now analyze why the other options are incorrect:

Option 1: 27000 V

This option is incorrect because the DC source voltage is only 16000 V, and the internal resistance causes a voltage drop. The DC voltage generated by the inverter cannot exceed the source voltage. The value of 27000 V is unrealistic and does not match the calculations.

Option 2: 15100 V

This is the correct answer as derived above. The terminal voltage of the DC source, after accounting for the voltage drop across the internal resistance, is 15100 V, which is the voltage generated by the inverter.

Option 3: 16000 V

This option assumes that there is no voltage drop across the internal resistance of the DC source. However, the problem explicitly states that the internal resistance is 1 Ω and the current is 900 A, leading to a voltage drop of 900 V. Thus, the terminal voltage is less than 16000 V, making this option incorrect.

Option 4: 16400 V

This option is incorrect as it exceeds the source voltage of 16000 V. Additionally, it does not account for the voltage drop caused by the internal resistance of the source. Such a value is not physically possible in this scenario.

Conclusion:

The correct answer is Option 2: 15100 V. This value accurately accounts for the voltage drop across the internal resistance of the DC source and represents the DC voltage generated by the inverter.

Additional Information

Important Considerations:

• The internal resistance of the DC source plays a significant role in determining the terminal voltage. For high currents, even small resistances can cause substantial voltage drops.
• In practical applications, minimizing the internal resistance of the source is crucial to maximize the efficiency of the system.
• Six-pulse inverters are commonly used in industrial applications due to their simplicity and cost-effectiveness. However, they may introduce harmonics into the system, which need to be mitigated using filters.

Multiquadrant Operation of Electric Drives:

A quadrant diagram is drawn by plotting the speed-torque characteristics of the load/motor for all four quadrant operations.

There are basically two modes of operation:

(i) Motoring Mode: The electrical energy is converted into mechanical energy and supports the motion.

(ii) Braking mode: The mechanical energy is converted into electrical energy and the motor works as a generator and opposes the motion. The fig. below shows the torque and speed coordinates for both forward and reverse motions.

We know that motor can provide motoring and braking operation for both forward and reverse directions.

The power is given as Power = Speed × Torque.

Now, if the power developed is positive then the operation is motoring. If the power developed is negative, the operation is braking.

Quadrant I - Forward Motoring:

In this region, the power and torque are both positive, so the power developed is positive and the machine works as a motor supplying mechanical energy.

Quadrant II - Forward Braking:

In this region, the speed is positive, but torque is negative, so power developed is negative and the machine works under braking opposing the motion.

Quadrant III - Reverse Motoring:

In this region, the speed is negative and the torque is also negative, so power developed is positive and so the machine works under reverse motoring mode.

Quadrant IV - Reverse Braking:

In this region, the torque is positive and the speed is negative, so the power developed is negative, so the machine works as a braking mode in the reverse direction.

Three-Phase Full Converters:

If all diode is replaced by thyristor, a three-phase full converter bridge formed as shown below,

Explanation:

To operate the motor in reverse motoring the current should be negative. So for that the firing angle α should be greater than π/2.

Additional Information

Four-Quadrant Operation of the Hoist:

As the load is resistive the peak instantaneous output voltage is 

Vm = √2 Vrms

\({V_m} = 400√ 2 \;V\)

Ripple frequency of three-phase full-wave converter:

Figure: output voltage waveform of a three-phase full-wave rectifier.

From the above output voltage waveform, we can observe that for a complete one cycle of input supply, we got 6 pulses in the output.

So, the three-phase full-wave ac to dc converter is a 6-pulse converter.

Then the ripple frequency of the output f0 = m × f

Where,

m = number of pulses in the output per one complete cycle of the input

f = supply voltage frequency

For three-phase ac to dc  full wave converter f0 = 6 × f

Concept:

Ripple frequency at the output = m × supply frequency

fo = m × fs

Where m = types of the pulse converter

Calculation:

A three-phase full-wave AC to DC converter is a 6-pulse converter

Number of pulses (m) = 6

fo = 6 × supply voltage frequency

∴ f₀ = 6 x 50

f₀ = 300 Hz

In a six-pulse thyristor bridge rectifier, the harmonics present are = 6 k ± 1

So, the harmonics are = 5, 7, 11, 13, ...

lowest harmonic component = 5^th harmonic supply frequency = 50 Hz

Concept:

The output voltage of a three-phase full converter is given by

\({V_0} = \frac{{3\sqrt 3 {V_{mp}}}}{\pi }\cos \alpha\)

V_mp is maximum supply phase voltage

α is firing angle

Calculation:

Given that, rms supply line voltage (V_rms) = 440 V

Maximum supply phase voltage \({V_{mp}} = \frac{{440\sqrt 2 }}{{\sqrt 3 }}\)

Firing angle (α) = 45°

\({V_0} = \frac{{3\sqrt 3 }}{\pi } \times \frac{{440\sqrt 2 }}{{\sqrt 3 }}\cos 45 = 420.16\;V\)

Concept:

For a fully controlled three phase bridge rectifier,

\({I_{rms}} = \sqrt {\frac{2}{3}} \times {I_0}\)

Calculation:

Given that, output load current (I₀) = 100 A

In a fully controlled three-phase bridge converter,

RMS value of the A.C. line current

\({I_{rms}} = \sqrt {\frac{2}{3}} \times {I_0} = \sqrt {\frac{2}{3}} \times 100 = 81.64\;A\)A

RMS current \({I_a} = \sqrt {\frac{2}{3}} \times 10 = 8.16\ A\)

Fundamental current \({I_{{a_1}}} = \frac{{\sqrt 6 }}{\pi }{I_o} = 0.78 \times 10 = 7.8\;A\)

\(\begin{array}{l} THD = \sqrt {\frac{1}{{D{F^2}}} - 1} \\ DF = \frac{{{I_{{a_1}}}}}{{{I_a}}} = \frac{{0.78 \times 10}}{{0.816 \times 10}} = 0.955\\ THD = \sqrt {{{\left( {\frac{1}{{0.955}}} \right)}^2} - 1} = 31\% \end{array}\)

Three-Phase Full Converters:

If all diode is replaced by thyristor, a three-phase full converter bridge formed as shown below,

• A three-phase input supply is connected to the terminal A, B, C, and RLE load is connected across the output terminal.
• It worked as three-phase AC to DC converter for firing angle delay 0 <α ≤ 90.
• The positive group of SCRs is fired at an interval of 120° and a similarly negative group of SCRs fired at an interval of 120°, but SCR of both groups are fired at an internal of 60° or commutation occurs at every 60°.
• At any time tow SCRs, one from the positive group and one from the negative group must be conducted together for the source to energize the load.
   

Waveform:

Voltage and Current Equation:

The average output voltage (V₀) is given by,

\(V_0=\frac{3V_m}{\pi}cosα\ \longrightarrow 0≤ α≤ \pi\)

Where V_m is the maximum value of line voltage.

Voltage and current waveform for α = 30° and for constant current can be drawn as,

From waveform RMS value of Thyristor current (I_T = i_T1) will be,

\(I_T=\sqrt{I_0^2\times \frac{2\pi}{3}\times \frac{1}{2\pi}}=I_0\times \sqrt\frac{1}{3}\)

Where I₀ is constant DC load current.

Application:

Given,

I₀ = 150 A

From the above concept,

The rms value of the current flowing through each thyristor of converter is \({I_T} = {I_o}\sqrt {\frac{1}{3}} = \frac{{150}}{{\sqrt 3 }}\)

Concept:

The average output voltage of the three-phase full converter is given by:

\(V_o(avg) = {V_{ml} \over {2\pi\over N}}cosα\)

where, V_ml = Maximum value of line voltage 

N = No. of pulses produced

α = Firing angle

Calculation:

Given, N = 6

Also, \(V_{ml} = \sqrt{3}V_m\)

where, V_m = Maximum value of phase voltage

\(V_o(avg) = {3\sqrt{3}V_{m} \over \pi}cosα\)

\({I_{0\left( {max} \right)}} = \frac{{{V_{0max}}}}{R} = \frac{{400}}{{10}} = 40A\)

Input transformer kVA rating = \(\sqrt 3 \times {V_l}{I_l}\)

 \({I_l}\) = Rms value of line current on ac diode \(= {I_0} \times \sqrt {\frac{2}{3}}\)

kVA rating of transformer \(= \sqrt 3 \times 400 \times 40 \times \sqrt {\frac{2}{3}}\)

= 22.6kVA

The Correct option is 3

Concept:

The output voltage of 3 phase 6 pulse bridge rectifiers is:

VO = \(\boldsymbol{\mathbf{}\frac{3V_{ML}}{\Pi }}\) = \(\frac{3\sqrt{3} V_{MP}}{\Pi }\)

Rms Line voltage = 415 V

Out voltage = V_O

Calculation:

Pick inverse voltage 3 phase 6 pulse bridge rectifiers if the safety factor is 2 

V _peak =   \(\ {\sqrt{2}\times saftey factor\times V_{rms} = \sqrt{2}\times 2.1\times 415}\) = 1232.49

Peak inverse voltage =V_peak

Additional Information Thyristor:

• A thyristor is a four-layer semiconductor device, consisting of alternating P-type and N-type materials (PNPN). A thyristor usually has three electrodes: an anode, a cathode, and a gate, also known as a control electrode.

Concept:

Ripple frequency of three-phase full-wave converter:

Figure: output voltage waveform of three-phase full-wave rectifier.

From the above output voltage waveform, we can observe that for a complete one cycle of input supply we got 6 pulses in the output.

So, the three-phase full-wave ac to dc converter is a 6-pulse converter.

Then the ripple frequency of the output f₀ = m × f

Where,

m = number of pulses in the output per one complete cycle of the input

f = supply voltage frequency

Solution:

For single-phase full-wave bridge circuit f0 = 2 × f

For a three-phase full-wave converter f₀ = 6 × f

Hence, the ratio output ripple-frequency to the supply-voltage frequency = f0 / f = 6