1.Distinguish between Cumulatively wound and differentially wound dc machines.
Cumulatively wound and differentially wound are two types of DC machines that differ in the way the field winding is connected to the armature winding.
In a cumulatively wound DC machine, the field winding is connected in series with the armature winding. This means that the current flows through both the armature and field winding in the same direction, resulting in a magnetic field that adds up cumulatively. This type of DC machine is used for applications that require a high starting torque, such as cranes and hoists.
On the other hand, in a differentially wound DC machine, the field winding is connected in parallel with the armature winding. This means that the current flows through the armature and field winding in opposite directions, resulting in a magnetic field that cancels out the effect of the armature field. This type of DC machine is used for applications that require a constant speed, such as fans and pumps
2
A 4-pole generator with 400 armature conductors has a usefu Rdl flux of 0.04 Wb per pole. What is the emf produced if the machine is wave wound and runs at 1200 rpm? What must be the speed at which the machine should be driven to generate the same emf if the machine is lap wound?
Given:
- Number of poles (p) = 4
- Armature conductors (C) = 400
- Useful flux per pole (Φ) = 0.04 Wb
- Speed (N) = 1200 rpm
We can use the formula for calculating the emf (E) generated by a DC generator:
E = Φ * p * C * N / 60
For a wave-wound generator, the number of parallel paths is equal to the number of poles, so the armature conductors per parallel path (C') is given by:
C' = C / p = 400 / 4 = 100
Substituting the given values:
E = 0.04 * 4 * 100 * 1200 / 60 = 320 V
Therefore, the emf produced by the wave-wound generator is 320 V.
For a lap-wound generator, the number of parallel paths is equal to the number of brushes, which is one more than the number of poles. So, for a 4-pole generator, the number of brushes is 5, and the armature conductors per parallel path is:
C' = C / (p+1) = 400 / 5 = 80
To generate the same emf as the wave-wound generator, the speed must be adjusted according to the ratio of armature conductors per parallel path for the two winding types:
N' = N * (C' wave-wound / C' lap-wound) = 1200 * (100/80) = 1500 rpm
Therefore, the speed at which the lap-wound generator should be driven to generate the same emf as the wave-wound generator is 1500 rpm.
3.What is meant by back emf? Explain the principle of torque production in a dc motor.
Back EMF:
When a DC motor is supplied with a voltage, it generates an electromagnetic force that turns the armature. This electromagnetic force, also known as the armature EMF, is directly proportional to the speed of the motor. As the motor rotates faster, the armature EMF increases. When the motor is rotating at a constant speed, the armature EMF is equal and opposite to the applied voltage, which results in zero current flowing through the armature. This phenomenon is known as back EMF or counter EMF.
The principle of torque production in a DC motor:
DC motors operate on the principle of the Lorentz force. When a current-carrying conductor is placed in a magnetic field, a force is exerted on the conductor. The direction of this force is perpendicular to both the direction of current and the direction of magnetic field. In a DC motor, the current flows through the armature winding, which is placed in a magnetic field produced by the field winding. As the armature rotates, the current-carrying conductors in the armature experience a force that is perpendicular to both the direction of current and the direction of magnetic field. This force causes the armature to rotate.
The torque produced in a DC motor is directly proportional to the product of the magnetic field strength, the armature current, and the length of the conductor in the magnetic field. The direction of the torque is given by Fleming's left-hand rule. When a DC voltage is applied to the armature winding, the armature current flows, and the motor starts rotating. As the motor starts rotating, a back EMF is generated, which opposes the applied voltage. The armature current reduces due to the back EMF, which in turn reduces the torque produced by the motor. The motor continues to rotate at a speed at which the back EMF is equal to the applied voltage, and the torque produced is just enough to overcome the load torque.
Therefore, in a DC motor, torque is produced due to the interaction between the magnetic field and the current-carrying conductors in the armature winding, and the motor speed is controlled by the back EMF.
4.Explain about emf equatation
EMF (Electromotive Force) is the force that drives an electric current through a circuit. It is generated when there is a change in magnetic flux through a conductor. The EMF equation relates the magnitude of EMF generated in a conductor to the rate of change of magnetic flux through it.
The EMF equation is given by:
EMF = -N(dΦ/dt)
where, EMF is the electromotive force generated in the conductor, N is the number of turns in the conductor, Φ is the magnetic flux through the conductor, and (dΦ/dt) is the rate of change of magnetic flux.
According to Faraday's law of electromagnetic induction, when there is a change in magnetic flux through a conductor, an EMF is induced in the conductor. The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux through the conductor. The negative sign in the EMF equation indicates the direction of the induced EMF. The induced EMF always opposes the change in magnetic flux that produced it.
The EMF equation is widely used in the design and analysis of electrical machines such as generators, motors, and transformers. The equation helps in calculating the induced EMF in the conductors of these machines and also helps in determining the parameters required for efficient operation of the machine.
5. An 8-pole, 400 V shunt motor has 960 wave connected armature conductors. The Full load armature current is 40 A and the flux per pole is 0.02 Wb. The armature Resistance is 0.1 W and the contact drop is 1 V per brush. Calculate the full load Speed of the motor.
The full load speed of a shunt motor can be calculated using the following formula:
N = (V - Ia Ra - Eb) / (4.44 ΦP)
where N is the full load speed of the motor in rpm, V is the supply voltage, Ia is the armature current, Ra is the armature resistance, Eb is the back EMF, Φ is the flux per pole and P is the number of poles.
To calculate the back EMF, we can use the following formula:
Eb = ΦPN/Z
where N is the speed of the motor in rpm, Z is the number of armature conductors and P is the number of poles.
Given data:
Number of poles (P) = 8
Supply voltage (V) = 400 V
Armature current (Ia) = 40 A
Armature resistance (Ra) = 0.1 Ω
Flux per pole (Φ) = 0.02 Wb
Number of armature conductors (Z) = 960
Contact drop per brush = 1V (so total contact drop = 2V)
Now, let's calculate the back EMF:
Eb = ΦPN/Z = (0.02 x 8 x 960) / 2 = 76.8 V
Next, let's calculate the voltage drop due to armature resistance:
Ia Ra = 40 x 0.1 = 4 V
Total voltage drop due to contact drop and armature resistance = 4 V + 2 V = 6 V
Now, let's substitute the values in the formula for full load speed:
N = (V - Ia Ra - Eb) / (4.44 ΦP) = (400 - 4 - 76.8) / (4.44 x 0.02 x 8) = 750 rpm
Therefore, the full load speed of the motor is 750 rpm.
Unit 2
2. A single phase 230V/15V, 50 Hz transformer has the secondary full load current of 8A. It has 45 turns on the secondary. Calculate (a) the voltage per tum (b) the number of primary turns (c) the full loadprimary current and (d) the KVA output of the transformer.
Given data:
Secondary voltage (Vs) = 15 V
Secondary full load current (Is) = 8 A
Frequency (f) = 50 Hz
Number of secondary turns (Ns) = 45
(a) The voltage per turn can be calculated using the formula:
Vs / Ns = Vp / Np
where Vp is the primary voltage and Np is the number of primary turns.
Vs / Ns = Vp / Np
15 / 45 = Vp / Np
Vp / Np = 15/45
Vp / Np = 1/3 V
So, the voltage per turn is 1/3 V.
(b) The number of primary turns can be calculated using the formula:
Vp / Vs = Np / Ns
Vp / 15 = Np / 45
Np = 45 x Vp / 15
We know that Vp / Np = 1/3 V
Substituting this value in the above equation, we get:
Np = 45 x 3 = 135
So, the number of primary turns is 135.
(c) The full load primary current can be calculated using the formula:
Ip = Is x Ns / Np
Ip = 8 x 45 / 135
Ip = 2.67 A
So, the full load primary current is 2.67 A.
(d) The KVA output of the transformer can be calculated using the formula:
KVA = Vp x Ip / 1000
We know that Vp = 230 V (given)
Substituting the values, we get:
KVA = 230 x 2.67 / 1000
KVA = 0.614 KVA
So, the KVA output of the transformer is 0.614 KVA.
3.Distinguish between core type and shell type transformers.
Core-type and shell-type transformers are two common types of transformers that differ in their construction and performance characteristics. Here are the main differences between them:
1. Construction:
Core-type transformers have a core made of laminated steel sheets with the windings wrapped around it. The primary winding is usually wound first and then the secondary winding is wound around it. In contrast, shell-type transformers have a core with two sets of windings that are placed on opposite legs of the core. The primary and secondary windings are wound side by side around the central core.
2. Size:
Core-type transformers are usually smaller in size and weight than shell-type transformers of similar power ratings. This is because core-type transformers have a shorter magnetic path length and require less core material.
3. Efficiency:
Shell-type transformers are generally more efficient than core-type transformers. This is because the magnetic flux in shell-type transformers flows around the core, which reduces the amount of flux leakage and minimizes core losses. In contrast, core-type transformers have a longer magnetic path and are more prone to flux leakage and core losses.
4. Leakage Reactance:
Core-type transformers have a higher leakage reactance compared to shell-type transformers. Leakage reactance is the leakage of magnetic flux from the windings and is higher in core-type transformers because the windings are wrapped around the core. In contrast, shell-type transformers have windings on opposite legs of the core, which reduces the leakage reactance.
5. Cost:
Core-type transformers are generally less expensive than shell-type transformers because they require less core material and are easier to manufacture. However, this cost advantage is offset by their lower efficiency, higher leakage reactance, and limited power handling capability.
In summary, core-type transformers are smaller, less expensive, and have higher leakage reactance than shell-type transformers. In contrast, shell-type transformers are more efficient, have lower leakage reactance, and can handle higher power levels.
Draw and explain the equivalent circuit of a Single-phase transformer.
The equivalent circuit of a single-phase transformer is a simplified representation of the transformer's electrical characteristics. It is commonly used to analyze the transformer's performance and to design the transformer for a specific application. The equivalent circuit includes two ideal components, the primary and secondary winding, and three loss components, the core losses, copper losses in the primary winding, and copper losses in the secondary winding.
The equivalent circuit of a single-phase transformer is shown below:
In the above circuit diagram, the primary and secondary windings are represented as ideal inductors L1 and L2, respectively. The mutual inductance between the two windings is represented by M. The core losses are represented by the resistor Rc, while the copper losses in the primary and secondary windings are represented by the resistors R1 and R2, respectively.
The primary and secondary windings are assumed to have negligible resistance and are represented as ideal inductors. The mutual inductance M is responsible for the transfer of power between the primary and secondary windings. The core losses Rc are due to eddy currents and hysteresis losses in the core material. The copper losses R1 and R2 are due to the resistive losses in the primary and secondary windings, respectively, and are proportional to the current flowing through them.
The equivalent circuit of a single-phase transformer can be used to determine the transformer's voltage regulation, efficiency, and power transfer characteristics. By analyzing the equivalent circuit, one can determine the transformer's operating parameters under different load conditions and select an appropriate transformer for a given application.