In order to produce unidirectional force (or torque) on the armature conductors of a motor, the conductors under any pole must carry the current in the same direction at all times. This is illustrated in Fig. (4.10). In this case, the current flows away from the observer in the conductors under the N-pole and towards the observer in the conductors under the S-pole. Therefore, when a conductor moves from the influence of N-pole to that of S-pole, the direction of current in the conductor must be reversed. This is termed as commutation. The function of the commutator and the brush gear in a d.c. motor is to cause the reversal of current in a conductor as it moves from one side of a brush to the other. For good commutation, the following points may be noted:

(i) If a motor does not have commutating poles (compoles), the brushes
must be given a negative lead i.e., they must be shifted from G.N.A.
against the direction of rotation of, the motor.
(ii) By using interpoles, a d.c. motor can be operated with fixed brush
positions for all conditions of load. For a d.c. motor, the commutating
poles must have the same polarity as the main poles directly back of
them. This is the opposite of the corresponding relation in a d.c.
generator.
Note. A d.c. machine may be used as a motor or a generator without changing the commutating poles connections. When the operation of a d.c. machine changes from generator to motor, the direction of the armature current reverses. Since commutating poles winding carries armature current, the polarity of commutating pole reverses automatically to the correct polarity.

With no commutating poles used, the brushes are given a forward lead in a d.c. generator and backward lead in a d.c. motor.
(iii) By using commutating poles (compoles), a d.c. machine can be operated with fixed brush positions for all conditions of load. Since commutating poles windings carry the armature current, then, when a machine changes from generator to motor (with consequent reversal of current), the polarities of commutating poles must be of opposite sign.

Therefore, in a d.c. motor, the commutating poles must have the same
polarity as the main poles directly back of them. This is the opposite of
the corresponding relation in a d.c. generator.

N = K (V- I_aR_a)/ Ф = K E_b/ Ф…………………………………………….(i)

T_a α ФI_a…………………………………………………………………………(ii)

If the flux decreases, from Eq.(i), the motor speed increases but from Eq.(ii) the motor torque decreases. This is not possible because the increase in motor speed must be the result of increased torque. Indeed, it is so in this case. When the flux decreases slightly, the armature current increases to a large value. As a result, in spite of the weakened field, the torque is momentarily increased to a high value and will exceed considerably the value corresponding to the load. The surplus torque available causes the motor to accelerate and back e.m.f (E_a=PФZN/60A)to rise. Steady conditions of speed will ultimately be achieved when back e.m.f. has risen to such a value that armature current[I_a = (V- E_a)/ R_a]develops torque just sufficient to drive the load.

Illustration
Let us illustrate the above point with a numerical example. Suppose a 400 V
shunt motor is running at 600 r.p.m., taking an armature current of 50 A. The armature resistance is 0.28 Ω. Let us see the effect of sudden reduction of flux by 5% on the motor.

Initially (prior to weakening of field), we have,

E_a = V-I_aR_a= 400 – 50 × 0.28 = 386 volts

We know that E_b α Ф N. If the flux is reduced suddenly, E_b α Ф because inertia
of heavy armature prevents any rapid change in speed. It follows that when the flux is reduced by 5%, the generated e.m.f. must follow suit. Thus at the instant of reduction of flux, E’_b = 0.95 × 386 = 366.7 volts.

Instantaneous armature current is

I’_a=(V- E’_b)/ R_a =(400-366.7)/0.28=118.9A

Note that a sudden reduction of 5% in the flux has caused the armature current to increase about 2.5 times the initial value. This will result in the production of high value of torque. However, soon the steady conditions will prevail. This will depend on the system inertia; the more rapidly the motor can alter the speed, the sooner the e.m.f. rises and the armature current falls.

But E_b=PФZN/60A

PФZN/60A  = V- I_aR_a

Or  N = (V- I_aR_a)/ Ф ×  60A/ PZ

Or N = K (V- I_aR_a)/ Ф

But         V- I_aR_a = E_a

Therefore N= K E_b/ Ф

Or N α E_b/ Ф

Therefore, in a d.c. motor, speed is directly proportional to back e.m.f. E_b and inversely proportional to flux per pole Ф.

Speed Relations

If a d.c. motor has initial values of speed, flux per pole and back e.m.f. as N₁ ,Ф₁ and E_b1 respectively and the corresponding final values are N₂ ,Ф₂ and E_b2 then,

N₁ α E_b1/ Ф₁ and N₂ α E_b2/ Ф₂

Therefore N₂/ N_{1 =} (E_b2/ E_b1) ×( Ф₁ / Ф₂)

(i) For a shunt motor, flux practically remains constant so that Ф1 = Ф₂.

therefore  N₂/ N_{1 =} E_b2/ E_b1

(ii) For a series motor, Ф α I_a prior to saturation.

therefore N₂/ N_{1 =} (E_b2/ E_b1) × (I_a1/I_a2)

where I_a1 = initial armature current
I_a2 = final armature current

Speed Regulation

The speed regulation of a motor is the change in speed from full-load to no-load and is expressed as a percentage of the speed at full-load i.e.

% Speed regulation = [( N.L. speed - F.L.speed)/F.L.speed ] × 100

=[(N_o -N)/N] × 100

where N_o = No – load .speed
N = Full – load speed

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]]> http://electricalandelectronics.org/2009/09/14/speed-of-a-d-c-motor/feed/ 0 http://electricalandelectronics.org/2009/09/14/shaft-torque-tsh/ http://electricalandelectronics.org/2009/09/14/shaft-torque-tsh/#comments Mon, 14 Sep 2009 14:02:48 +0000 arjun http://electricalandelectronics.org/?p=2001 The torque which is available at the motor shaft for doing useful work is known as shaft torque. It is represented by Tsh. Fig. (4.9) illustrates the concept of shaft torque. The total or gross torque T_a developed in the armature of a motor is not available at the shaft because a part of it is lost in overcoming the iron and frictional losses in the motor. Therefore, shaft torque Tsh is somewhat less than the armature torque T_a. The difference T_a – T_sh is called lost torque.

T_a - T_sh =9.55 × iron and frictional losses/N

For example, if the iron and frictional losses in a motor are 1600 W and the
motor runs at 800 r.p.m., then,

T_a - T_sh =9.55 × 1600 /800 =19.1 N-m

As stated above, it is the shaft torque T_sh that produces the useful output. If the speed of the motor is N r.p.m., then,

Output in watts= 2πN T_sh/60

or T_{sh =}Output in watts /(2πN /60 ) N-m

or T_{sh =} 9.55 ×Output in watts /N     N-m

Brake Horse Power (B.H.P.)

W.D./revolution = force x distance moved in 1 revolution

F × 2π r = 2π×T_sh J

W.D./minute = 2π N T_sh J

W.D./ sec= 2πNT_sh J/60 jS^-1 or watt=2πNT_sh J/(60 ×746) H.P.

Useful output power =2πNT_sh J/(60 ×746) H.P.

or B.H.P. =2πNT_sh J/(60 ×746)

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]]> http://electricalandelectronics.org/2009/09/14/shaft-torque-tsh/feed/ 0 http://electricalandelectronics.org/2009/09/13/armature-torque-of-d-c-motor/ http://electricalandelectronics.org/2009/09/13/armature-torque-of-d-c-motor/#comments Sun, 13 Sep 2009 19:07:48 +0000 arjun http://electricalandelectronics.org/?p=1996 Torque is the turning moment of a force about an axis and is measured by the
product of force (F) and radius (r) at right angle to which the force acts i.e.
D.C. Motors torque
T = F × r

In a d.c. motor, each conductor is acted upon by a circumferential force F at a distance r, the radius of the armature (Fig. 4.8). Therefore, each conductor exerts a torque, tending to rotate the armature. The sum of the torques due to all armature conductors is known as gross or armature torque (T_a).

Let in a d.c. motor
r = average radius of armature in m
l = effective length of each conductor in m

Z = total number of armature conductors

A = number of parallel paths

i = current in each conductor = I_a/A
B = average flux density in Wb/m²
Φ = flux per pole in Wb
P = number of poles

Force on each conductor, F = B i l newtons

Torque due to one conductor = F × r newton- metre
Total armature torque, T_a = Z F r newton-metre
= Z B i l r

Now i = I_a/A, B = Φ/a where a is the x-sectional area of flux path per pole at
radius r. Clearly, a = 2πr l /P.

T_a = Z × (Ф/a)×( I_a/A)×l×r

T_a = Z × (ФP/2πr l)×( I_a/A)×l×r = Z Ф I_aP/(2πA) N-m

or T_a = 0.159Z Ф I_a(P/A) N-m……………………………………(i)

so Ta

T_a α Ф I_a

Hence torque in a d.c. motor is directly proportional to flux per pole and
armature current.

(i) For a shunt motor, flux Φ is practically constant.

T_a α  I_a

(ii) For a series motor, flux Φ is directly proportional to armature current I_a
provided magnetic saturation does not take place.

T_a α I_a²

up to magnetic saturation

Alternative expression for Ta

E_b = PФZN/60A

(60×E_b) /N= PФZ/A

From Eq.(i), we get the expression of Ta as:

T_a =0.159×(60× E_b/N)× I_a

T_a =9.55×( E_bI_a/N)

(ii) Series-wound motor in which the field winding is connected in series with the armature [See Fig. 4.5]. Therefore, series field winding carries the armature current. Since the current passing through a series field winding is the same as the armature current, series field windings must be designed with much fewer turns than shunt field windings for the same m.m.f. Therefore, a series field winding has a relatively small number of turns of thick wire and, therefore, will possess a low resistance.

(iii) Compound-wound motor which has two field windings; one connected in parallel with the armature and the other in series with it. There are two types of compound motor connections (like generators). When the shunt field winding is directly connected across the armature terminals [See Fig. 4.6], it is called short-shunt connection. When the shunt winding is so connected that it shunts the series combination of armature and series field [See Fig. 4.7], it is called long-shunt connection

The compound machines (generators or motors) are always designed so that the flux produced by shunt field winding is considerably larger than the flux produced by the series field winding. Therefore, shunt field in compound machines is the basic dominant factor in the production of the magnetic field in the machine.

V = applied voltage
E_b = back e.m.f.
R_a = armature resistance
I_a = armature current

Since back e.m.f. Eb acts in opposition to the applied voltage V, the net voltage across the armature circuit is V- E_b. The
armature current I_a is given by;

I_a = (V – E_b)/ R_a

or V = E_{b +} I_aR_{a ……………………………..}(i)

This is known as voltage equation of the d.c. motor.

Power Equation

If Eq.(i) above is multiplied by I_a throughout, we get,

VI_a = E_bI_{a +}I²_aR_a

VI_a= electric power supplied to armature (armature input)
E_bI_a = power developed by armature (armature output)
I²_aR_a = electric power wasted in armature (armature Cu loss)

Thus out of the armature input, a small portion (about 5%) is wasted as a I²_aRa and the remaining portion E_bI_a is converted into mechanical power within the armature.

Condition For Maximum Power

The mechanical power developed by the motor is P_m= E_bI_a

Now P_m=VI_{a -}I²_aR_a

Since, V and R_a are fixed, power developed by the motor depends upon armature current. For maximum power, dP_m/dI_a should be zero.

dP_m/dI_{a =} V – 2I_aR_a

or I_aR_{a =} V/2

Now, V = E_{b +} I_aRa =E_{b +} V/2

therefore Eb=  V/2

Hence mechanical power developed by the motor is maximum when back e.m.f. is equal to half the applied voltage.

Limitations
In practice, we never aim at achieving maximum power due to the following reasons:
(i) The armature current under this condition is very large—much excess of rated current of the machine.
(ii) Half of the input power is wasted in the armature circuit. In fact, if we take into account other losses (iron and mechanical), the efficiency will be well below 50%.

Consider a shunt wound motor shown in Fig. (4.2). When d.c. voltage V is applied across the motor terminals, the field magnets are excited and armature conductors are supplied with current. Therefore, driving torque acts on the armature which begins to rotate. As the armature rotates, back e.m.f. E_b is induced which opposes the applied voltage V. The applied voltage V has to force current through the armature against the back e.m.f.  E_b . The electric work done in overcoming and causing the current to flow against Eb is converted into mechanical energy developed in the armature. It follows, therefore, that energy conversion in a d.c. motor is only possible due to the production of back e.m.f. E_b.

Net voltage across armature circuit = V – E_b

If  R_a is the armature circuit resistance, then, I_a = (V – E_b)/ R_a

Since V and R_a are usually fixed, the value of E_b will determine the current drawn by the motor. If the speed of the motor is high, then back e.m.f. E_b (= P Φ ZN/60 A) is large and hence the motor will draw less armature current and viceversa.

Significance of Back E.M.F.

The presence of back e.m.f. makes the d.c. motor a self-regulating machine i.e., it makes the motor to draw as much armature current as is just sufficient to develop the torque required by the load.

Armature current,I_a = (V – E_b)/ R_a

(i) When the motor is running on no load, small torque is required to overcome the friction and windage losses.  Therefore, the armature current I_a is small and the back e.m.f. is nearly equal to the applied voltage.
(ii) If the motor is suddenly loaded, the first effect is to cause the armature to slow down. Therefore, the speed at which the armature conductors move through the field is reduced and hence the back e.m.f. E_b falls. The decreased back e.m.f. allows a larger current to flow through the armature and larger current means increased driving torque. Thus, the driving torque increases as the motor slows down. The motor will stop slowing down when the armature current is just sufficient to produce the increased torque required by the load.
(iii) If the load on the motor is decreased, the driving torque is momentarily in excess of the requirement so that armature is accelerated. As the armature speed increases, the back e.m.f. E_b also increases and causes the armature current I_a to decrease. The motor will stop accelerating when the armature current is just sufficient to produce the reduced torque required by the load.

It follows, therefore, that back e.m.f. in a d.c. motor regulates the flow of armature current i.e., it automatically changes the armature current to meet the load requirement