Construction of Permanent Magnet DC Motor (PMDC Motor)

The design of Permanent Magnet DC motors (PMDC) is same as conventional DC motors. PIMDC motors are smaller and cheaper than other motors. Figure (l) shows the layout of PMDC motors.

Figure 2: Cross-section of PMDC Motor.

From figure (2), it is observed that the stator also provides alternative return path for magnetic material. The rotor consists of various slots for windings, brushes and commutation similar to conventional DC motors

Working Principle of Permanent Magnet DC Motor (PMDC Motor)

When the rotor rotates, the permanent magnets come into alignment with the field magnets and get attracted with the iron (or) steel core of the field magnets. The external magnetic field is produced by this high energy permanent magnets. The attraction of permanent magnets and the iron or steel core produces the torque without consuming power, this is known as ‘attraction phase’. When the rotor rotates and passes the centre of field magnets, a pulse is generated through the field magnet and the permanent field magnet will have same polarity which makes the permanent magnet to repel.

Now, the magnetic field is produced by the field magnets and this magnetic field in combination with the magnetic flux of the permanent magnet produces a torque. For this, power is consumed only for a small (i.e., millisecond) period of time.

This is known as ‘repulsion phase’. As most of the work is done by permanent magnets, so very low power is consumed and hence, the power required to run the motor is very low making the motor highly efficient.

Equivalent circuit of Permanent Magnet DC Motor (PMDC Motor)

Figure 1: Equivalent circuit of PMDC Motor.

Permanent magnet DC motors are similar to normal DC shunt motor, the only difference is that they consist of permanent magnets for producing magnetic flux instead of stationary field winding. By the interaction of flux produced by permanent magnet and armature current, the torque is developed in motor. The equivalent circuit of permanent magnet DC motor is shown in figure. In this circuit, the field winding connection is absent due to permanent magnet. This circuit also helps in analyzing the operation of motor.

Let,

V – DC source voltage

I_a – Armature current

R_a – Armature resistance

E_b – Back e.m.f

K – Constant

N – Speed in r.p.m

n – Speed in r.p.s

P – number of poles

ω – Angular motor speed

T – Torque

A – Number of armature poles

Z – Number of armature conductors

ϕ – Flux per pole.

From the fundamentals, the relation between speed and back e.m.f is given by,

\[{{E}_{b}}=\frac{\phi ZNP}{60A}\]

\[=\frac{\phi ZnP}{A}\text{ }\left[ \begin{align}
& N=60\times n(\text{in r}\text{.p}\text{.s}) \\
& n=\frac{N}{60} \\
\end{align} \right]\]

But we have,

\[\omega =2\pi n\]

\[n=\frac{\omega }{2\pi }\]

\[{{E}_{b}}=\frac{\phi ZP}{A}.\frac{\omega }{2\pi }\]

\[{{E}_{b}}=\frac{\phi ZP}{2\pi A}\omega \]

As flux ‘ϕ’ is constant in case of permanent magnet DC motor, thus we have,

\[K=\frac{\phi ZP}{2\pi A}\]

\[{{E}_{b}}=K\omega …(1)\]

We have a torque equation in terms of current for DC motor as,

\[T=\frac{\phi ZP}{2\pi A}{{I}_{a}}\]

\[=K{{I}_{1}}\]

Also, the supply voltage of motor is given by,

\[V={{E}_{b}}+{{I}_{a}}{{R}_{a}}…(2)\]

Substituting equation (1) in equation (2), we get,

\[V=K\omega +{{I}_{a}}{{R}_{a}}\]

\[K\omega =V-{{I}_{a}}{{R}_{a}}\]

\[\omega =\frac{V-{{I}_{a}}{{R}_{a}}}{K}\]

Hence, by using equivalent circuit and equations, the performance and operation of permanent magnet DC motor can be analyzed.

Advantages of Permanent Magnet DC Motor (PMDC Motor)

1. Manufacturing cost required is less.
2. Size is small.
3. Efficiency is high.
4. Noiseless operation.
5. Speed control is simple.
6. High dynamic response.
7. Reliability is high due to mechanical commutator.
8. Better speed-torque characteristics.

Disadvantages of Permanent Magnet DC Motor (PMDC Motor)

1. PMDC motor can be demagnetized by armature reaction m.m.f, which in tum makes the motor inoperative.
2. If these motors are used for long periods, then their temperature tends to be higher as compared to wound field motors.
3. Need regular maintenance for brushes and commutator.
4. Density of power is medium.

Applications of Permanent Magnet DC Motor (PMDC Motor)

Some of the  of permanent magnet DC motors are listed as follows.

1. Air-conditioning systems
2. Auto bank machines
3. Bar code readers at supermarkets
4. Ticketing machines
5. Washing machines
6. Vacuum cleaners
7. Blowers
8. Computer disc drives
9. Automobiles staffer
10. Cloth drives
11. Industrial drives
12. Toys and wipers
13. Machine tools
14. Kitchen appliances
15. Optical equipments etc.

Difference between PMDC Motors and DC Motors

The differences between PMDC motors and DC motors are as follows,

The post What is Permanent Magnet DC Motor (PMDC Motor)? Construction, Diagram & Applications appeared first on Electrical and Electronics Blog.

Hopkinson’s Test requires two machines which are identical. The test can be carried out at full-load on the two identical machines. The two machines are mechanically coupled and are electrically adjusted so that, one of them acts as a motor and other as generator. The motor supplies the mechanical power which is utilized to drive the generator. while generator develops electrical power to be utilized in driving the motor. In the absence of losses, the machines will run without any external power supply. But due to the losses, output of generator is not enough to drive the motor and vice-versa. These losses are supplied from supply mains. Hence, in this test the power drawn from the supply mains will be required to overcome the internal losses of two machines. The connections for conducting Hopkinson’s test are as shown in figure 1.

Machine M is started as a motor and supply is given by means of a starter. The main switch of other machine is kept open and the motor is made to run at its normal speed with the help of its shunt regulator. Machine ‘M’ drives machine ‘G’ as a generator. The voltage of ‘G’ is adjusted with the help of its field regulator till voltmeter V₁ reads zero. This shows that, the voltage is same as supply mains both in polarity and magnitude.

Now, the switch ‘S’ is closed. Let, the readings of ammeters A₁, A₂, A₃ and A₄ connected in the circuit measure the current supplied by generator as I₁, line current as I₂, generator field current as I₃ and motor field current as I₄ respectively.

If V is supply voltage then,

Motor input,

P_M = V(I₁ + I₂)

Generator output,

P_G = VI₁

Assuming that both machines have same efficiency (η).

Output of motor = Efficiency × motor Input

= η × V(I₁ + I₂) = Generator input

Generator output = η × Generator input

= η × η × V (I₁ + I₂)…(2)

From equations (1) and (2),

VI₁ = η² V (I₁ + I₂)

\[V{{I}_{1}}={{\eta }^{2}}V({{I}_{1}}+{{I}_{2}})\]

Let,

Resistance of armature of each machine = R_a

Armature copper loss in generator = (I₁ + I₃)² R_a

Armature copper loss in motor = (I₁ + I₂ – I₄)² R_a

Shunt copper loss in generator = VI₃

Shunt copper loss in motor = VI₄

Total copper loss = (I₁ + I₃)² R_a + (I₁ + I₂ – I₄)² R_a + VI₃ + VI₄

Total losses of both machines = Power drawn from supply = VI₂

If the total copper loss is subtracted from total losses, then the stray loss for the motor generator set can be obtained.

Total Stray loss, P_S(T) = VI₂ – [(I₁ + I₃)² R_a + (I₁ + I₂ – I₄)² R_a + VI₃ + VI₄]

And, Stray loss of each machine,

\[{{P}_{S}}=\frac{{{P}_{S}}(T)}{2}\]

Efficiency of Motor

Motor input,

P_M = V(I₁ + I₂)

Total loss,

P_L(M) = Armature loss + Field loss + Stray loss

= (I₁ + I₂ – I₄)² R_a + VI₄ + P_S

Motor output = V(I₁ + I₂) – P_L(M)

\[{{\eta }_{M}}=\frac{V({{I}_{1}}+{{I}_{2}})-{{P}_{L(M)}}}{V({{I}_{1}}+{{I}_{2}})}\]

Efficiency of Generator

Generator output,

P_G = VI₁

Total losses,

P_L(G) = Armature loss + Field loss + Stray loss

= (I₁ + I₃)² R_a + VI₃ + P_S

Generator input = VI₁ + P_L(G)

\[{{\eta }_{G}}=\frac{V{{I}_{1}}}{V{{I}_{1}}+{{P}_{L(G)}}}\times 100\]

Advantages of Hopkinson’s Test :

1. This test is performed on full-load conditions. Therefore, any change in temperature rise and quality of commutations of the two identical machines can be noticed.
2. Due to full load condition, the variation in iron loss because of flux distortion is considered.
3. The power requirement for this test is less in comparison to full load powers of both the machines.

Disadvantages of Hopkinson’s Test :

1. Availability of two identical machines is difficult.
2. The iron losses of the two machines are different due to different excitation and they cannot be separated.
3. In case of small machines the loads on machines are not equal, this may cause difficulty in analysis.

The post What is Hopkinson’s Test? Circuit Diagram & Theory appeared first on Electrical and Electronics Blog.

Figure 1: Arrangement of Swinburne’s test.

In Swinburne’s test, R and are measured separately so that losses can be calculated easily on various loads. The machine whether motor or generator is run as a shunt motor on no-load, the field rheostat is adjusted to give rated speed and rated voltage when shunt is applied across the terminals of motor.

The circuit diagram for determining the no-load losses of a D.C shunt machine is shown in figure.

Let,

V is the voltage measured by voltmeter

I₀ is the input motor current which is measured by ammeter A₁.

I_sh is the shunt field current measured by ammeter A₂.

We know that,

Input power Output + Losses

Since machine is on no-load,

Input power = Losses                     [Since, Output = 0]

From the above equation, it is clear that no-load input power is used to supply internal losses in the machine that is shunt field copper loss, armature copper loss, stray losses in shunt machine.

Power input = VI₀

Shunt field copper loss = VI_sh

Armature copper loss = I²_a0 = (I₀ – I_sh)²R_a

Here, R_a  is the Armature resistance. Since,

Power input = Total losses

\[V{{I}_{0}}=V{{I}_{sh}}+I_{{{a}_{0}}}^{2}{{R}_{a}}+\text{ Strey losses}\]

\[\text{Strey losses}=V{{I}_{0}}-V{{I}_{sh}}-{{({{I}_{0}}-{{I}_{sh}})}^{2}}{{R}_{a}}\]

If the shunt field copper loss is added to stray loss then the constant loss can be obtained which remains constant irrespective of the load on the machine.

Constant losses = Stray losses + Shunt copper losses

\[=V{{I}_{0}}-{{({{I}_{0}}-{{I}_{sh}})}^{2}}{{R}_{a}}\]

Since constant losses are known, the efficiency can be determined at any load. Let ‘I’ be the load current where the efficiency of machine is to be determined.

Efficiency when Running as Generator

Generator output = VI watts

Total losses = Armature copper loss + Field copper loss + Stray losses

\[=I_{a}^{2}{{R}_{a}}+{{W}_{c}}\]

\[={{(I+{{I}_{sh}})}^{2}}{{R}_{a}}+{{W}_{c}}\]

Input = Output + Losses

\[=VI+{{(I+{{I}_{sh}})}^{2}}{{R}_{a}}+{{W}_{c}}\]

\[\text{Efficiency = }\frac{\text{Output}}{\text{Input}}=\frac{VI}{VI+{{(I+{{I}_{sh}})}^{2}}+{{W}_{C}}}\]

Efficiency when Running as a Motor

Motor input = VI

Total losses = Armature copper loss + W_c

\[=I_{a}^{2}{{R}_{a}}+{{W}_{c}}\]

\[={{(I-{{I}_{sh}})}^{2}}{{R}_{a}}+{{W}_{c}}\]

Motor output = Input – Losses

\[=VI-{{(I-{{I}_{sh}})}^{2}}{{R}_{a}}-{{W}_{c}}\]

\[\text{Efficiency of motor = }\frac{\text{Output}}{\text{Input}}\]

\[=\frac{VI-{{(I-{{I}_{sh}})}^{2}}{{R}_{a}}-{{W}_{c}}}{VI}\]

The post What is Swinburne’s Test of DC Motor? Theory & Diagram appeared first on Electrical and Electronics Blog.

Figure 1: Arrangement of Retardation Test.

In this method of testing DC machines, the machine runs slightly above its normal speed and supply to the armature is cut-off while the field remains excited. As a result, the armature slows down and its kinetic energy is utilized to overcome rotational losses.

\[\text{K}\text{.E}\text{. of armature = }\frac{1}{2}I{{\omega }^{2}}\]

Where,

I = M.I (i.e., moment of inertia) of armature

ω = Angular speed of armature

P_R = Rate of change of K.E.

\[=\frac{d}{dt}\left( \frac{1}{2}I{{\omega }^{2}} \right)=I\omega \frac{d\omega }{dt}\]

\[=I\left( \frac{2\pi N}{60} \right)\frac{d}{dt}\left( \frac{2\pi N}{60} \right)\]

\[={{\left( \frac{2\pi }{60} \right)}^{2}}IN\frac{dN}{dt}….(1)\]

Hence to determine stray losses, the value of M.I of armature i.e., I and rate of change of speed i.e., \(\frac{dN}{dt}\) must be known. Determination of \(\frac{dN}{dt}\).

The connections for conducting retardation test are shown in figure. The voltmeter V across armature shows the instantaneous back e.m.f. Since E_b ∝ N, the voltmeter can be suitably calibrated to indicate speed. When the supply to armature is cut off the speed of machine decreases with time. The speed or readings of voltmeter are noted at different intervals of time and a curve is drawn between speed and time.

From any point, ‘P’ lying on speed time curve, tangent is drawn meeting the X-axis and Y-axis at A and B respectively. Then,

\[\frac{dN}{dt}=\frac{\text{OB in r}\text{.p}\text{.m}}{\text{OA }\text{in sec}}\]

Determination of M.I (Moment of Inertia) :

First speed time curve is plotted with armature alone. Then a flywheel of known M.I i.e., I₁ is keyed to the shaft and speed time curve is plotted again.

Due to the addition of flywheel, the time taken to lower the speed will be longer. \(\frac{dN}{d{{t}_{1}}}\) and \(\frac{dN}{d{{t}_{2}}}\) will be determined as before, losses will remain the same in both cases since the addition of flywheel will not make much difference.

Rotational losses before the addition of flywheel,

\[{{\left( \frac{2\pi }{60} \right)}^{2}}IN\frac{dN}{d{{t}_{1}}}….(2)\]

Rotational losses after the addition of flywheel,

\[{{\left( \frac{2\pi }{60} \right)}^{2}}(I+{{I}_{1}})N\frac{dN}{d{{t}_{2}}}….(3)\]

Equating equation (2) and (3), we get,

\[{{\left( \frac{2\pi }{60} \right)}^{2}}(I+{{I}_{1}})N\frac{dN}{d{{t}_{2}}}={{\left( \frac{2\pi }{60} \right)}^{2}}IN\frac{dN}{d{{t}_{1}}}\]

\[=I\left( \frac{dN}{d{{t}_{1}}}-\frac{dN}{d{{t}_{2}}} \right)={{I}_{1}}\frac{dN}{d{{t}_{2}}}\]

\[I=\frac{{{I}_{1}}\left( \frac{dN}{d{{t}_{2}}} \right)}{\frac{dN}{d{{t}_{1}}}-\frac{dN}{d{{t}_{2}}}}\]

Substituting the values of I₁,

\(\frac{dN}{d{{t}_{1}}}\) and \(\frac{dN}{d{{t}_{2}}}\) in above equation the value of I can be determined using retardation test.

The post What is Retardation Test (or Running Down Test) of DC Motor? Theory & Circuit Diagram appeared first on Electrical and Electronics Blog.

Figure 1: Arrangement of Brake Test.

Brake test is a direct testing method used to find the efficiency of a DC motor. The arrangement for performing brake test on a DC motor is shown in figure (1). In this test, brakes are applied on a water cooled pulley mounted on the motor shaft depending upon the rating of the machine. The brake rope is fixed with the help of wooden blocks gripping the pulley. One end of the rope is fixed to the earth via spring balance ‘B’ and the other end is connected to a suspended weight ‘W₁‘. Now, the motor is allowed to run under this condition and the load on the motor is so adjusted that it carries full-load current by adding or removing the weights to suspended weight ‘W₁’ Let reading on spring balance ‘B’ be ‘W₂‘.

While increasing the load, the load is increased in step by step manner until full-load condition is achieved and for each step various values of voltmeter, ammeter, the suspension weights i.e., W₁ and W₂ and speed of the motor are noted and the efficiency of the rotor can be calculated as follows,

The net pull on the rope due to friction at the pulley is given as,

\[=({{W}_{1}}-{{W}_{2}})\text{ kg wt}\]

\[\text{= 9}\text{.81 (}{{W}_{1}}-{{W}_{2}}\text{) Newton}\]

The shaft torque T_sh developed by the motor is given by,

\[{{\text{T}}_{sh}}=({{W}_{1}}-{{W}_{2}})\text{ R kg-mt}\]

\[\text{= 9}\text{.81 }({{W}_{1}}-{{W}_{2}})\text{ R N-m}\]

Where,

R = Radius of pulley (mts)

Motor power output is given by,

\[{{P}_{out}}={{T}_{sh}}\times \omega \text{ watts}\]

\[\text{=}\frac{2\pi \times 9.81}{60}({{W}_{1}}-{{W}_{2}})\text{ RN watts}\]

\[\text{= 1}\text{.027 N }({{W}_{1}}-{{W}_{2}})\text{ R watts}\]

Where,

Let,

V = Supply voltage

I = Full-load current taken by the motor.

Input power is given is,

\[{{\text{P}}_{in}}=VI\text{ watts}\]

The efficiency is given by,

\[%\eta =\frac{\text{Output power}}{\text{Input power}}\times 100\]

\[=\frac{\frac{2\pi \times 9.81}{60}({{W}_{1}}-{{W}_{2}})\text{ RN}}{VI}\times 100\]

\[=\frac{1.027({{W}_{1}}-{{W}_{2}})\text{ RN}}{VI}\times 100\]

Advantages of Brake Test

1. Test requires no other machines. Hence, it reduces the cost and energy.
2. This method is very simple.
3. Very much convenient for small motors.

Disadvantages of Brake Test

1. This test is performed with only small motors. In case of large motors it is difficult to dissipate the large amount of heat generated at the brake.
2. The output of the motor cannot be determined accurately because of the unavoidable errors occurring in spring balance which leads to incorrect determination of the internal losses or efficiency of a DC machine.
3. In case of series wound machines care should be taken so that the brake applied is tight otherwise the motor will attain dangerously high speed, which leads to severe damage.

The post What is Brake Test of DC Machine? Theory, Circuit Diagram & Advantages appeared first on Electrical and Electronics Blog.

1. Armature control method
2. Field control method
3. Ward-Leonard method.

Armature Control Method

In this method of speed control, an external variable resistance is connected between the armature terminals and the line.

When the resistance added is maximum due to higher resistance, the current flowing through the armature would be reduced proportionally. The field flux remains unchanged thus, with the reduction in the armature current, the torque is also reduced.

\[T={{k}_{a}}\oint{{{I}_{a}}}\]

As the torque of the machine is reduced, the speed also reduces. If the armature resistance is kept minimum, the rated current flows through the armature keeping the speed at rated value. Thus by analysis, it could be noted that the speed variation is possible only below the rated speed but not above the rated speed. The major advantage is that due to higher resistance in the armature the power losses considered would be large. So, it is used for intermittent periods only. The circuit and the characteristics are shown in figure (1).

Advantages of Armature control method

1. This method of speed control is simple.
2. Speed can be easily varied.
3. Speeds below base speed or normal speed can be achieved easily.
4. Initial cost required is low.
5. Economical for short time speed control applications.

Disadvantages of Armature control method

1. The operating cost increases for low speeds.
2. There occurs power loss in controller resistance.
3. Output and efficiency is reduced due to power loss.
4. Speed regulation is poor.

Field Control Method

In field control method, a varying external resistance is connected in the field circuit. When the resistance added to the field is zero then the current through the armature would be the rated value. As the resistance increases the current through the field reduces thereby, reducing the flux of the machine. But this reduction of current in the field increases the current through the armature above the rated value and thus, the operating torque also increases which in turn increases the speed above the rated value. While starting, the field resistance should be kept minimum else huge current flows though the armature which damages the armature coils. So, from this it can be said that the variation of speed above the rated speed can also be obtained by increasing the field resistance.

Advantages of Field control method

The advantages of speed control of D.C motor using flux control method are as follows.

1. Speed can be increased upto twice the rated speed.
2. It is a very simple.
3. Less amount of power is wasted as the field rheostat is connected across the field.
4. Independent of load.
5. It is the most efficient and economical method.

Disadvantages of Field control method

1. Armature reaction is greater and results in instability at high speeds.
2. Speeds below the normal speed cannot be controlled.
3. Poor commutation.

Ward-Leonard Method

Ward Leonard system is used to control the speed of DC shunt motors. It comes under the category of voltage control method. In this method, the basic adjustable voltage to the armature is achieved by means of an adjustable voltage generator. The arrangement of Ward Leonard system is shown in figure (3).

As shown in figure, Ward-Leonard system consists of two motors M₁ and M₂. M₁ is the motor whose speed is to be controlled. A motor-generator set consists of either DC or an A.C motor and is directly coupled to the generator G.

The input to the generator ‘G’ is fed from motor ‘M₂‘ and the output of generator ‘G’ is applied to motor ‘M₁ Motor ‘M2’ operates at constant speed. The voltage applied to motor ‘M₁‘ is varied by connecting the field of generator to a regulator and hence variable voltage is applied to the armature of the main motor ‘M₁‘ . Thus, speed control is achieved. The switch RS is used to reverse the direction of field current of generator ‘G’ i.e., when switch RS is in connected position, reverse voltage is generated in generator ‘G’ and is applied to motor ‘M₁‘ due to which the direction of rotation of motor ‘M₁‘ reverses.

Advantages of Ward Leonard method

1. Speed can be controlled over the range from zero to normal speed in both the directions.
2. Speed regulation is smooth.
3. For large motors when a D.C supply is unavailable, this method is very much suitable.

Disadvantages of Ward Leonard method

1. This method is expensive as it requires two extra machines.
2. The overall efficiency of the system for light loads is low.
3. There is a possibility for the armature winding to get damage when the applied voltage across the armature increases more than the rated voltage i.e., for a very long duration.

Comparison between different Methods of speed control of DC Shunt Motor

The post Speed Control for DC Shunt Motor – Methods & Diagram appeared first on Electrical and Electronics Blog.

A four point starter is a protective device used in shunt and compound DC motors to limit the high starting current in the absence of back e.m.f. The construction of four point starter is similar to three point starter except the connection of NVC (No-volt coil). The N-V coil connected in series with field winding in three point starter is replaced in four point starter by connecting it across the supply through an additional terminal, N as shown in figure. In this type of arrangement, when the armature touches stud number 1, the line current divides into three parts. In which, first part passes through starting resistance, series field and armature. The second part passes through shunt field winding and the third part passes through N-V coil and protective resistance.

In four-point starter since N-V coil current is independent of shunt field circuit current, any change of current in shunt field does not affect the current passing through the N-V coil. Thus, the electromagnetic pull exerted by the N-V coil will always be sufficient to hold the handle in RUN position under all the operating conditions and prevents the spiral spring from resisting the starting arm to OFF position. The function of No-volt release and overload release in four point starter is similar to three point starter.

Difference between 3 -point and 4-point starters.

The post What is a 4 Point Starter? Working Principle, Construction & Diagram appeared first on Electrical and Electronics Blog.

The speed-torque characteristics of DC shunt motor are also known as mechanical characteristics. They can be obtained from torque-current and speed-current characteristics of DC shunt motor.

The expression for back e.m.f in a DC motor is given by,

\[{{E}_{b}}={{k}_{a}}\phi N\text{     }\left[ {{k}_{a}}=\frac{ZP}{60A} \right]…(1)\]

Also, we have,

\[{{E}_{b}}=V-{{I}_{a}}{{R}_{a}}…(2)\]

On equating equations (1) and (2), we get,

\[{{k}_{a}}\phi N=V-{{I}_{a}}{{R}_{a}}\]

\[N=\frac{1}{{{k}_{a}}\phi }\left[ V-{{I}_{a}}{{R}_{a}} \right]…(3)\]

The expression for torque of a DC motor is given by,

\[T={{k}_{a}}\phi {{I}_{a}}\]

\[{{I}_{a}}=\frac{T}{{{k}_{a}}\phi }\]

Substituting the above value of I_a in equation (3), we get,

\[N=\frac{1}{{{k}_{a}}\phi }\left[ V-\left( \frac{T}{{{k}_{a}}\phi } \right){{R}_{a}} \right]\]

\[N=\frac{V}{{{k}_{a}}\phi }-\frac{T{{R}_{a}}}{{{({{k}_{a}}\phi )}^{2}}}…(4)\]

As the torque in a DC motor increases, flux (ϕ) decreases. This is because when torque is increased, armature current increases resulting in reduction of air gap flux ϕ i.e., due to armature reaction and saturation. Therefore, for increased torque there is an increase in the value \(\frac{T}{{{\phi }^{2}}}\) of equation (4) causing a drop in the speed of the motor. However, when the effect of armature reaction is neglected, the value of flux (ϕ) remains constant due to which for increase in torque there will not be much drop in speed of the DC shunt motor. The speed-torque characteristics of a DC shunt motor are drawn as shown below. Dotted line in above figure represents the ideal speed- torque characteristics of a DC shunt motor. But, due to the reduction in flux caused by the armature reaction, the curve is obtained as shown by the thick line.

Speed-Torque Characteristics of DC Series Motor

The speed-torque characteristics of DC series motor can be obtained or derived from the torque-current and speed-current characteristics of DC series motor. Therefore, armature torque and speed has inverse relationship as shown in figure (3).

We know that,

\[N\propto \frac{{{E}_{b}}}{\phi }\propto \frac{{{E}_{b}}}{{{I}_{a}}}\]

Since on no-load the speed is dangerously high, the series motors are never started on no-load. When the motor is connected across supply mains without load, the current drawn is small and hence ϕ is small and the speed tends to increase \(\left[ N\propto \frac{{{E}_{b}}}{\phi } \right]\). With increase in speed, E_b increases \(\left[ {{E}_{b}}=\frac{\phi ZN}{60}\times \frac{P}{A} \right]\) thus the field current decreases \(\left[ {{I}_{a}}=\frac{V-{{E}_{b}}}{{{R}_{a}}} \right]\) which in tum leads to decrease in flux and hence speed increases gradually. This process continues until the armature gets damaged. Hence, series motors are not suitable for the services where the load may be entirely removed.

The post Speed Torque Characteristics of DC Motor (Shunt & Series) appeared first on Electrical and Electronics Blog.

Consider a DC motor with stator and rotor. When the stator coils are energized, a stator magnetic flux is set up and follows the path as shown in figure (a) with no current in the rotor conductor.

When rotor conductor carries current. magnetic flux is set up and flows in the direction as shown in figure (b) with no current in the stator coil.

Now when both the stator coil and the rotor conductor carries current, then an interaction between the stator flux and the flux produced by the rotor takes place and the resultant magnetic field is set up as shown in figure (c). The rotor conductors experience a force in the upward direction and develop a torque. Thus, the torque is produced in a DC motor when the stator flux and flux produced by the rotor conductor interacts with each other.

Derivation for Torque Equation of a DC Motor

Let,

F — Force in Newton

r — Radius of armature in meter

T_a — Armature torque in N-m

S — Circumferential distance

f — Flux/pole in Wb

P — Number of poles

Z — Number of armature conductors

A — Number of armature paths

I_a — Armature current

N — Speed of armature in r.p.m. Since, torque is the twisting movement produced across the armature.

\(\frac{N}{60}\) = Speed in r.p.m

dt = \(\frac{60}{N}\) = Time taken for one revolution.

Mechanical work done per second is given as,

\[\text{Mechanical W}\text{.D / sec = }\frac{F\times \text{ Circumferential distance}}{\text{Time}}\]

\[=\frac{F\times S}{{}^{60}/{}_{N}}\]

\[=F\times 2\pi r\times \frac{N}{60}\text{       }\left[ S=2\pi r \right]\]

\[=F\times r\times \frac{2\pi N}{60}\]

\[=\frac{{{T}_{a}}\times 2\pi N}{60}\text{ N-m / sec  }\left[ {{T}_{a}}=F\times r \right]\]

The electrical work done per second,

\[\text{Electrical W}\text{.D / sec = }{{E}_{b}}\times {{I}_{a}}\]

\[=\frac{\phi PN}{60}\times \frac{Z}{A}{{I}_{a}}\text{     }\left[ {{E}_{b}}=\frac{\phi PN}{60}\times \frac{Z}{A} \right]\]

As 1 N-m/sec= 1 watt, equating mechanical and electrical power, we get,

\[\frac{{{T}_{a}}\times 2\pi N}{60}=\frac{\phi PN}{60}\times \frac{Z}{A}{{I}_{a}}\]

\[{{T}_{a}}\times 2\pi =\frac{\phi PZ{{I}_{a}}}{A}\]

\[{{T}_{a}}=\frac{1}{2\pi }\times \frac{\phi PZ{{I}_{a}}}{A}\]

\[{{T}_{a}}=\frac{0.159}{9.81}\frac{\phi PZ{{I}_{a}}}{A}\text{ kg-m}\]

\[{{T}_{a}}=\frac{0.0162\phi PZ{{I}_{a}}}{A}\text{ kg-m (MKS}\text{ unit)}\]

Shaft Torque

Shaft torque is the torque which is available for doing useful work. It is usually denoted by T and is always available at the shaft.

\[\text{Motor output = }\omega {{\text{T}}_{sh}}\text{ watts = }\frac{2\pi N}{60}.{{T}_{sh}}\text{ watts}\]

Shaft torque,

\[{{\text{T}}_{sh}}=\frac{\text{Output in watts}}{\frac{2\pi N}{60}}\]

\[=\frac{60}{2\pi }\times \frac{\text{Output}}{N}\text{ N-m = 9}\text{.55 }\times \text{ }\frac{\text{Output}}{N}\text{ N-m}\]

Lost Torque

When the mechanical power developed is transmitted to the shaft of the motor, there occurs a power loss due to friction, windage and iron loss. Therefore the torque required to overcome these losses is known as lost torque or lost torque. The total armature torque is the combination of shaft torque and lost torque.

\[{{T}_{a}}={{T}_{sh}}+{{T}_{f}}\]

Lost torque,

\[{{T}_{f}}={{T}_{a}}-{{T}_{sh}}\]

The post DC Motor Torque Equation – Theory, Diagram & Derivation appeared first on Electrical and Electronics Blog.

Figure 1: Interpoles.

Interpoles are small poles placed in between the main poles of a D.C generator as shown in figure (1). Interpoles are also known as commutating poles or com poles. In generators, the polarity of the inter pole is same as the main pole which is ahead of it in the direction of rotation. The following are the two functions performed by interpoles in a D.C generator.

1. Neutralization of reactance voltage
2. Neutralization of cross magnetizing effect of armature reaction.

Neutralization of Reactance Voltage :

When the position of brushes is fixed at GNA, then due to armature reaction, air gap flux which exists along GNA is also along brush axis or interpolar region. Reactance voltage is induced in the coil undergoing commutation due to the presence of air gap flux. This reactance voltage leads to delayed commutation and sparking at brushes. The function of interpoles or commutating poles is to generate e.m.f exactly in opposition to the reactance voltage, so that the effect of reactance voltage is nullified and the commutation process is improved. Also, as armature current increases, the air gap flux and there by reactance voltage increases. Therefore. the interpole winding is connected in series with armature so that commutating e.m.f also increases whenever there is a rise in armature current. This method of attaining sparkless commutation by using interpoles is known as voltage commutation.

Neutralization of Cross Magnetizing Effect of Armature Reaction :

Cross magnetizing effect of armature reaction in inter-polar region is minimized by interpoles. As interpoles are connected in series with armature windings the m.m.f produced by them opposes the m.m.f produced by armature conductor in interpolar region. If the armature current increases, the armature reaction also increases. Due to the series connection of armature and interpoles, the interpole m.m.f also increases proportionately with armature m.m.f but in opposite direction. The process of neutralization of cross magnetizing effect of armature reaction by using interpoles is as shown in figure (2).

In the above figure 2,

Of represents m.m.f due to main poles

Oa represents cross magnetizing m.m.f due to armature

bc represents m.m.f due to interpoles.

As bc is in opposite direction to Oa, the m.m.f due to interpoles cancels out cross magnetizing m.m.f.

The post What are Interpoles in DC Machines? Theory, Connection Diagram & Uses appeared first on Electrical and Electronics Blog.