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D.C. MOTOR
➣➣ ➣➣➣^ Motor Principle ➣➣ ➣➣➣ Comparison of Generator and Motor Action ➣➣ ➣➣➣ Significance of the Back e.m.f.—Voltage Equation of a Motor ➣➣ ➣➣➣ Conditions for Maximum Power ➣➣ ➣➣➣ Torque ➣➣ ➣➣➣ Armature Torque of a Motor ➣➣ ➣➣➣^ Shaft Torque ➣➣ ➣➣➣ Speed of a D.C. Motor ➣➣ ➣➣➣ Speed RegulationTorque and Speed of a D.C. Motor ➣➣ ➣➣➣ Motor Characteristics ➣➣ ➣➣➣^ Characteristics^ of^ Series Motors ➣➣ ➣➣➣ Characteristics of Shunt Mo- tors ➣➣ ➣➣➣ Compound Motors ➣➣ ➣➣➣^ Performance Curves ➣➣ ➣➣➣ Comparison of Shunt and Series Motors ➣➣ ➣➣➣ Losses and Efficiency ➣➣ ➣➣➣ Power Stages
C H A P T E R
Learning Objectives
Design for optimum performance and durability in demanding variable speed motor applications. D.C. motors have earned a reputation for dependability in severe operating conditions
Ç
996 Electrical Technology
29.1. Motor Principle
An Electric motor is a machine which converts electric energy into mechanical energy. Its action is based on the principle that when a current-carrying conductor is placed in a magnetic field, it experiences a mechanical force whose direction is given by Fleming’s Left-hand Rule and whose magnitude is given by F = BIl Newton. Constructionally, there is no basic dif- ference between a d.c. generator and a d.c. motor. In fact, the same d.c. ma- chine can be used interchangeably as a generator or as a motor. D.C. motors are also like generators, shunt-wound or series-wound or compound-wound. In Fig. 29.1 a part of multipolar d.c. motor is shown. When its field magnets are excited and its armature conductors
are supplied with current from the supply mains, they experience a force tending to rotate the armature. Armature conductors under N -pole are assumed to carry current downwards (crosses) and those under S -poles, to carry current upwards (dots). By applying Fleming’s Left-hand Rule, the direction of the force on each conductor can be found. It is shown by small arrows placed above each conductor. It will be seen that each conductor can be found. It will be seen that each conductor experiences a force F which tends to rotate the armature in anticlockwise direction. These forces collectively produce a driving torque which sets the armature rotating.
It should be noted that the function of a commutator in the motor is the same as in a generator. By reversing current in each conductor as it passes from one pole to another, it helps to develop a continuous and unidirectional torque.
29.2. Comparison of Generator and Motor Action
As said above, the same d.c. machine can be used, at least theoretically, interchangeably as a generator or as a motor. When operating as a generator, it is driven by a mechanical machine and it develops voltage which in turn produces a current flow in an electric circuit. When operating as a motor, it is supplied by electric current and it develops torque which in turn produces mechanical rotation. Let us first consider its operation as a generator and see how exactly and through which agency, mechanical power is converted into electric power. In Fig. 29.2 part of a generator whose armature is being driven clockwise by its prime mover is shown. Fig. 29.2 ( a ) represents the fields set up independently by the main poles and the armature conductors like A in the figure. The resultant field or magnetic lines on flux are shown in Fig. 29.2 ( b ).
Fig. 29.
Fig. 29.
N N
Armature
Armature (a) (b)
A
+^ +
Principle of Motor^ N S
Motion
Battery
Conductor
N S
998 Electrical Technology
armature flux on the main flux, as shown in Fig. 29.4 ( b ), is two-fold : ( i ) It increases the flux on the left-hand side of the teeth and decreases it on the right-hand side, thus making the distribution of flux density across the tooth section unequal. ( ii ) It inclines the direction of lines of force in the air-gap so that they are not radial but are disposed in a manner shown in Fig. 29.4 ( b ). The pull exerted by the poles on the teeth can now be resolved into two components. One is the tangential component F 1 and the other vertical component F 2. The vertical component F 2 , when considered for all the teeth round the armature, adds up to zero. But the component F 1 is not cancelled and it is this tangential component which, acting on all the teeth, gives rise to the armature torque.
29.3. Significance of the Back e.m.f.
As explained in Art 29.2, when the motor armature ro- tates, the conductors also rotate and hence cut the flux. In ac- cordance with the laws of electromagnetic induction, e.m.f. is induced in them whose direction, as found by Fleming’s Right- hand Rule, is in opposition to the applied voltage (Fig. 29.5). Because of its opposing direction, it is referred to as counter e.m.f. or back e.m.f. E (^) b. The equivalent circuit of a motor is shown in Fig. 29.6. The rotating armature generating the back e.m.f. E (^) b is like a battery of e.m.f. Eb put across a supply mains of V volts. Obviously, V has to drive I (^) a against the opposition of E (^) b. The power required to overcome this opposition is Eb Ia. In the case of a cell, this power over an interval of time is converted into chemical energy, but in the present case, it is converted into mechanical energy.
It will be seen that I (^) a = Net voltage Resistance
b a
V V R
−
where Ra is the resistance of the armature circuit. As pointed out above, E (^) b = Φ ZN × ( P / A ) volt where N is in r.p.s. Back e.m.f. depends, among other factors, upon the armature speed. If speed is high, Eb is large, hence armature current I (^) a , seen from the above equation, is small. If the speed is less, then Eb is less, hence more current flows which develops motor torque (Art 29.7). So, we find that Eb acts like a governor i.e ., it makes a motor self-regulating so that it draws as much current as is just necessary.
29.4. Voltage Equation of a Motor
The voltage V applied across the motor armature has to ( i ) overcome the back e.m.f. E (^) b and ( ii ) supply the armature ohmic drop I (^) a Ra. ∴ V = Eb + I (^) a Ra This is known as voltage equation of a motor. Now, multiplying both sides by I (^) a , we get V Ia = Eb I (^) a + I (^) a^2 R (^) a As shown in Fig. 29.6, V Ia = Eectrical input to the armature E (^) bI (^) a = Electrical equivalent of mechanical power developed in the armature I (^) a^2 Ra = Cu loss in the armature Hence, out of the armature input, some is wasted in I^2 R loss and the rest is converted into me- chanical power within the armature. It may also be noted that motor efficiency is given by the ratio of power developed by the arma-
Fig.29.
Fig. 29.
V
I
Eb
Shunt Field
Ish I
v
Ia
D.C. Motor 999
ture to its input i.e ., Eb I (^) a / V Ia = Eb / V. Obviously, higher the value of E (^) b as compared to V , higher the motor efficiency.
29.5. Condition for Maximum Power
The gross mechanical power developed by a motor is Pm = V I (^) a − I (^) a^2 R (^) a. Differentiating both sides with respect to I (^) a and equating the result to zero, we get d Pm / d Ia = V − 2 I (^) a R (^) a = 0 ∴ I (^) a R (^) a = V / As V = Eb + I (^) a R (^) a and I (^) a Ra = V /2 ∴ E (^) b = V / Thus gross mechanical power developed by a motor is maximum when back e.m.f. is equal to half the applied voltage. This condition is, however, not realized in practice, because in that case current would be much beyond the normal current of the motor. Moreover, half the input would be wasted in the form of heat and taking other losses (mechanical and magnetic) into consideration, the motor efficiency will be well below 50 percent. Example 29.1. A 220-V d.c. machine has an armature resistance of 0.5 Ω. If the full-load armature current is 20 A, find the induced e.m.f. when the machine acts as (i) generator (ii) motor. (Electrical Technology-I, Bombay Univ. 1987)
Fig. 29. Solution. As shown in Fig. 29.7, the d.c. machine is assumed to be shunt-connected. In each case, shunt current is considered negligible because its value is not given. ( a ) As Generator [Fig. 29.7( a )] Eg = V + I (^) a Ra = 220 + 0.5 × 20 = 230 V ( b ) As Motor [Fig 29.7 ( b )] Eb = V − I (^) a R (^) a = 220 − 0.5 × 20 = 210 V Example 29.2. A separately excited D.C. generator has armature circuit resistance of 0.1 ohm and the total brush-drop is 2 V. When running at 1000 r.p.m., it delivers a current of 100 A at 250 V to a load of constant resistance. If the generator speed drop to 700 r.p.m., with field-current unal- tered, find the current delivered to load. (AMIE, Electrical Machines, 2001)
Solution. RL = 250/100 = 2.5 ohms. Eg 1 = 250 + (100 × 0.1) + 2 = 262 V. At 700 r.p.m., E (^) g 2 = 262 × 700/1000 = 183.4 V If I (^) a is the new current, Eg 2 − 2 − ( I (^) a × 0.1) = 2.5 I (^) a This gives I (^) a = 96.77 amp. Extension to the Question : With what load resistance will the current be 100 amp, at 700 r.p.m.? Solution. Eg 2 − 2 − ( I (^) a × 0.1) = R (^) L × I (^) a For I (^) a = 100 amp, and Eg 2 = 183.4 V, RL = 1.714 ohms. Example 29.3. A 440-V, shunt motor has armature resistance of 0.8 Ω and field resistance of 200 Ω. Determine the back e.m.f. when giving an output of 7.46 kW at 85 percent efficiency. Solution. Motor input power = 7.46 × 10 3 /0.85 W
220 V
0.5 W
20 A
Generator
220 V
0.5 W
20 A
Motor (a) (b)
D.C. Motor 1001
φ 60
Z N (^) P a
× = 226
For a Lap-wound armature, P = a ∴ φ = 226 60 1000 32
× ×
= 0.42375 wb
As a motor, I (^) a = 4 amp E (^) b = 200 − 4 × 2 = 192 V = φ ZN /
Giving N = 60 192 0.42375 32
× × = 850 r.p.m.
Tutorial Problems 29.
1. What do you understand by the term ‘back e.m.f.’? A d.c. motor connected to a 460-V supply has an armature resistance of 0.15 Ω. Calculate ( a ) The value of back e.m.f. when the armature current is 120 A. ( b ) The value of armature current when the back e.m.f. is 447.4 V. [( a ) 442 V ( b ) 84 A ] 2. A d.c. motor connected to a 460-V supply takes an armature current of 120 A on full load. If the armature circuit has a resistance of 0.25 Ω, calculate the value of the back e.m.f. at this load. [ 430 V ] 3. A 4-pole d.c. motor takes an armature current of 150 A at 440 V. If its armature circuit has a resistance of 0.15 Ω, what will be the value of back e.m.f. at this load? [ 417.5 V ]
29.6. Torque
By the term torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius r metre acted upon by a circumferential force of F Newton which causes it to rotate at N r.p.m. (Fig. 29.10).
Then torque T = F × r Newton-metre (N - m) Work done by this force in one revolution = Force × distance = F × 2 π r Joule Power developed = F × 2 π r × N Joule/second or Watt = ( F × r ) × 2 π N Watt Now 2 π N = Angular velocity ω in radian/second and F × r = Torque T ∴ Power developed = T × ω watt or P = T ω Watt Moreover, if N is in r.p.m., then ω = 2 π N /60 rad/ s
∴ P = 2 60
π N × T or P = 2 60
π (^). NT =
NT
29.7. Armature Torque of a Motor
Let Ta be the torque developed by the armature of a motor running at N r.p.s. If Ta is in N / M , then power developed = Ta × 2 π N watt ...( i )
Fig. 29.
1002 Electrical Technology
We also know that electrical power converted into mechanical power in the armature (Art 29.4) = Eb I (^) a watt ... ( ii ) Equating ( i ) and ( ii ) , we get Ta × 2 π N = Eb I (^) a ... ( iii ) Since Eb = Φ ZN × ( P / A ) volt, we have
Ta × 2 π N = Φ ZN (^) ( ) P A.^ I^ a^ or^ Ta^ =^
1 2 π
. Φ ZI 0 ( ) P A N - m
= 0.159 N newton metre ∴ Ta = 0.159 Φ ZIa × ( P / A ) N - m Note. From the above equation for the torque, we find that T (^) a ∝ Φ Ia. ( a ) In the case of a series motor, Φ is directly proportional to I (^) a (before saturation) because field windings carry full armature current ∴ Ta ∝ I (^) a^2
( b ) For shunt motors, Φ is practically constant, hence Ta ∝ I (^) a. As seen from ( iii ) above
Ta = 2
E b Ia π N
N - m - N in r.p.s. If N is in r.p.m., then
Ta = (^60 60) 9. 2 / 60 2 2
E b Ia Eb I (^) a Eb I (^) a E Ib a N N N N
= = = π π π
N-m
29.8. Shaft Torque (Tsh )
The whole of the armature torque, as calculated above, is not available for doing useful work, because a certain percentage of it is required for supplying iron and friction losses in the motor.
The torque which is available for doing useful work is known as shaft torque Tsh. It is so called because it is available at the shaft. The motor output is given by
Output = Tsh × 2 π N Watt provided Tsh is in N-m and N in r.p.s.
∴ Tsh = Output in watts 2 π N
N- m − N in r.p.s
= Output in watts 2 π N / 60
N-m − N in r.p.m.
= 60 output^ 9.55Output 2 N N
= π
N-m.
The difference ( Ta − Tsh ) is known as lost torque and is due to iron and friction losses of the motor. Note. The value of back e.m.f. E (^) b can be found from ( i ) the equation, Eb = V − I (^) a Ra ( ii ) the formula Eb = Φ ZN × ( P / A ) volt Example 29.6. A d.c. motor takes an armature current of 110 A at 480 V. The armature circuit resistance is 0.2 Ω. The machine has 6-poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate (i) , the speed and (ii) the gross torque developed by the armature. (Elect. Machines, A.M.I.E. Sec B, 1989)
Solution. Eb = 480 − 110 × 0.2 = 458 V, Φ = 0.05 W, Z = 864
Now, Eb = (^) ( ) 60
ZN (^) P A
Φ or 458 = (^) ( )
0.05 (^864 ) 60 6
× × N ×
∴ N = 636 r.p.m. Ta = 0.159 × 0.05 × 864 × 110 (6/6) = 756.3 N-m
1004 Electrical Technology
( ii ) Tsh = 9.55 output^ 9.55 37, 300 N 1000
= × = 356 N-m
( iii ) Starting line current = 1.5 × 82.9 = 124.3 A Arm. current at starting = 124.3 − 1.8 = 122.5 A If R is the starter resistance (which is in series with armature), then 122.5 ( R + 0.24) + 2 = 500 ∴ R = 3.825 ΩΩΩΩΩ Example 29.12. A 4-pole, 220-V shunt motor has 540 lap-wound conductor. It takes 32 A from the supply mains and develops output power of 5.595 kW. The field winding takes 1 A. The armature resistance is 0.09 Ω and the flux per pole is 30 mWb. Calculate (i) the speed and (ii) the torque developed in newton-metre. (Electrical Technology, Nagpur Univ. 1992) Solution. I (^) a = 32 − 1 = 31 A ; E (^) b = V − I (^) a Ra = 220 − (0.09 × 31) = 217.2 V
Now, Eb = (^) ( ) 60
ZN (^) P A
Φ ∴ 217.2 = (^) ( ) 30 10 3 540 4 60 4
× − × × N
( i ) ∴ N = 804.4 r.p.m.
( ii ) Tsh = 9.55 × output in watts^ 9.55 5, 595 N 804.
= × =^ 66.5 N-m
Example 29.13 ( a ). Find the load and full-load speeds for a four-pole, 220-V, and 20-kW, shunt motor having the following data : Field–current = 5 amp, armature resistance = 0.04 ohm, Flux per pole = 0.04 Wb, number of armature-conductors = 160, Two-circuit wave-connection, full load current = 95 amp, No load current =9 A. Neglect armature reaction. (Bharathithasan Univ. April 1997) Solution. The machine draws a supply current of 9 amp at no load. Out of this, 5 amps are required for the field circuit, hence the armature carries a no-load current of 4 amp. At load, armature-current is 90 amp. The armature-resistance-drop increases and the back e.m.f. decreases, resulting into decrease in speed under load compared to that at No-Load. At No Load : Eao = 220 − 4 × 0.04 = 219.84 volts Substituting this, 0.04 × 160 × ( N /60) × (4/2) = 219. No-Load speed, N 0 = 1030.5 r.p.m. At Full Load : Armature current = 90 A , E (^) a = 200 − 90 × 0.04 = 216.4 V N = (216.4/219.84) × 1030.5 = 1014.4 r.p.m. Example 29.13 ( b ). Armature of a 6-pole, 6-circuit D.C. shunt motor takes 400 A at a speed of 350 r.p.m. The flux per pole is 80 milli-webers, the number of armature turns is 600, and 3% of the torque is lost in windage, friction and iron-loss. Calculate the brake-horse-power.
(Manonmaniam Sundaranar Univ. Nov. 1998) Solution. Number of armature turns = 600 Therefore, Z = Number of armature conductors = 1200 If electromagnetic torque developed is T Nw − m, Armature power = T ω = T × 2 π 350/ = 36.67 T watts To calculate armature power in terms of Electrical parameters, E must be known. E = φ Z ( N /60) ( P / A )
D.C. Motor 1005
= 80 × 10 −^3 × 1200 × (350/60) × (6/6) = 560 volts With the armature current of 400 A , Armature power = 560 × 400 watts Equating the two, T = 560 × 400/36.67 = 6108.5 Nw − m. Since 3 % of this torque is required for overcoming different loss-terms,
Net torque = 0.97 × 6180.5 = 5925 Nw - m For Brake-Horse-Power, net output in kW should be computed first. Then “kW” is to be con- verted to “BHP”, with 1 HP = 0.746 kW.
Net output in kW = 5925 × 36.67 × 10 −^3 = 217.27 kW Converting this to BHP, the output = 291.25 HP Example 29.13 ( c ). Determine the torque established by the armature of a four-pole D.C. motor having 774 conductors, two paths in parallel, 24 milli-webers of pole-flux and the armature current is 50 Amps. (Bharathiar Univ. April 1998)
Solution. Expression for torque in terms of the parameters concerned in this problem is as follows :
T = 0.159 φ Z I (^) a p / a Nw - m Two paths in parallel for a 4-pole case means a wave winding. T = 0.159 × (24 × 10 −^3 ) × 774 × 50 × 4/ = 295.36 Nw-m Example 29.13 ( d ). A 500-V D.C. shunt motor draws a line-current of 5 A on light-load. If armature resistance is 0.15 ohm and field resistance is 200 ohms, determine the efficiency of the machine running as a generator delivering a load current of 40 Amps.
(Bharathiar Univ. April 1998) Solution. ( i ) No Load, running as a motor : Input Power = 500 × 5 = 2500 watts Field copper-loss = 500 × 2.5 = 1250 watts Neglecting armature copper-loss at no load (since it comes out to be 2.5^2 × 0.15 = 1 watt), the balance of 1250 watts of power goes towards no load losses of the machine running at rated speed. These losses are mainly the no load mechanical losses and the core-loss.
( ii ) As a Generator, delivering 40 A to load : Output delivered = 500 × 40 × 10 −^3 = 20 kW Losses : ( a ) Field copper-loss = 1250 watts ( b ) Armature copper-loss = 42.5 2 × 0.15 = 271 watts ( c ) No load losses = 1250 watts Total losses = 2.771 kW Generator Efficiency = (20/22.771) × 100 % = 87.83 % Extension to the Question : At what speed should the Generator be run, if the shunt-field is not changed, in the above case? Assume that the motor was running at 600 r.p.m. Neglect armature reaction.
Solution. As a motor on no-load, E (^) b 0 = 500 − Ia ra = 500 − 0.15 × 2.5 = 499.625 V As a Generator with an armature current of 42.5 A,
D.C. Motor 1007
Solution. Eb = V − I (^) a Ra = 200 − 45 (0.5 + 0.3) = 164 V
Now Eb = (^) .( ) 60
ZN (^) P A
Φ (^) volt
∴ 164 = 18 10 3 280 4 4 60 4
× − × × × N × ∴^ N^ = 488 r.p.m.
Total input = 200 × 45 = 9,000 W ; Cu loss = I (^) a^2 Ra = 45^2 × 0.8 = 1,620 W Iron + Friction losses = 800 W ; Total losses = 1,620 + 800 = 2,420 W Output = 9,000 − 2,420 = 6,580 W
∴ Tsh = 9 × 55 × 6580 488
= 128 N-m
Let F be the pull in newtons at the rim of the pulley. Then F × 0.205 = 128.8 ∴ F = 128.8/0.205 N = 634 N Example 29.16. A 4-pole, 240 V, wave connected shunt motor gives 1119 kW when running at 1000 r.p.m. and drawing armature and field currents of 50 A and 1.0 A respectively. It has 540 conductors. Its resistance is 0.1 Ω. Assuming a drop of 1 volt per brush, find (a) total torque (b) useful torque (c) useful flux / pole (d) rotational losses and (e) efficiency. Solution. Eb = V − I (^) a Ra − brush drop = 240 − (50 × 0.1) − 2 = 233 V Also I (^) a = 50 A
( a ) Armature torque Ta = 9.55 b^ a
E I N
N-m = 9.55 × 233 50 1000
× = 111 N-m
( b ) Tsh = 9.55 output^ 9.55 11, N 1000
= × =^ 106.9 N-m
( c ) Eb = (^) ( ) 60
ZN P A
Φ (^) × volt
∴ 233 = (^) ( ) (^540 1000 ) 60 2
Φ × × × ∴ ΦΦΦΦΦ = 12.9 mWb ( d ) Armature input = V Ia = 240 × 50 = 12,000 W Armature Cu loss = I (^) a^2 Ra = 50^2 × 0.1 = 250 W ; Brush contact loss = 50 × 2 = 100 W ∴ Power developed = 12,000 − 350 = 11,650 W ; Output = 11.19 kW = 11,190 W ∴ Rotational losses = 11,650 − 11,190 = 460 W ( e ) Total motor input = VI = 240 × 51 = 12,340 W ; Motor output = 11,190 W ∴ Efficiency = 11,190^100 12, 240
× = 91.4 %
Example 29.17. A 460-V series motor runs at 500 r.p.m. taking a current of 40 A. Calculate the speed and percentage change in torque if the load is reduced so that the motor is taking 30 A. Total resistance of the armature and field circuits is 0.8 Ω. Assume flux is proportional to the field current. (Elect. Engg.-II, Kerala Univ. 1988) Solution. Since Φ ∝ Ia , hence T ∝ Ia^2
∴ T 1 ∝ 402 and T 2 ∝ 30 2 ∴ 2 1
9 16
T T
=
∴ Percentage change in torque is = 1 2 1
100 7 100 16
T T T
− × = × = 43.75 %
Now E (^) b 1 = 460 − (40 × 0.8) = 428 V ; E (^) b 2 = 460 − (30 × 0.8) = 436 V 2 2 1 1 1 2
b a b a
N E I N E I
= × ∴^2 436 40 500 428 30
N = × ∴^ N 2 =^ 679 r.p.m.
1008 Electrical Technology
Example 29.18. A 460-V, 55.95 kW, 750 r.p.m. shunt motor drives a load having a moment of inertia of 252.8 kg-m^2_. Find approximate time to attain full speed when starting from rest against full-load torque if starting current varies between 1.4 and 1.8 times full-load current._
Solution. Let us suppose that the starting current has a steady value of (1.4 + 1.8)/2 = 1.6 times full-load value.
Full-load output = 55.95 kW = 55,950 W ; Speed = 750 r.p.m. = 12.5 r.p.s. F.L. shaft torque T = power/ω = power/2π N = 55,950 π × (750/60) = 712.4 N-m During starting period, average available torque = 1.6 T − T = 0.6 T = 0.6 × 712.4 = 427.34 N-m This torque acts on the moment of inertial I = 252.8 km-m^2.
∴ 427.4 = 252.8 × 2 12. d 252. dt dt
ω^ π × = × ,^ ∴^ dt^ =^ 46.4 s
Example 29.19. A 14.92 kW, 400 V, 400 -r.p.m. d.c. shunt motor draws a current of 40 A when running at full-load. The moment of inertia of the rotating system is 7.5 kg-m 2. If the starting current is 1.2 times full-load current, calculate (a) full-load torque (b) the time required for the motor to attain the rated speed against full-load. (Electrical Technology, Gujarat Univ. 1988) Solution. ( a ) F.L. output 14.92 kW = 14,920 W ; Speed = 400 r.p.m. = 20/3 r.p.s Now, T ω = output ∴ T = 14,920/2π × (20/3) = 356 N-m ( b ) During the starting period, the torque available for accelerating the motor armature is = 1.2 T − T = 0.2 T = 0.2 × 356 = 71.2 N-m
Now, torque = I
d dt
ω ∴ 71.2 = 7.5 ×
2 (20 / 3) dt
π × ∴ dt = 4.41 second
29.9. Speed of a D.C. Motor
From the voltage equation of a motor (Art. 27.4), we get
Eb = V − I (^) a Ra or (^) ( ) 60
ZN P A
Φ = V − I (^) a Ra
∴ N = a^ a ( 60 ) V I R (^) A ZP
− × Φ
r.p.m.
Now V − I (^) a Ra = Eb ∴ N = Eb^ ( 60 A ) ZP × Φ
r.p.m. or N = K b
E Φ It shows that speed is directly proportional to back e.m.f. E (^) b and inversely to the flux Φ on N ∝ Eb /Φ.
For Series Motor Let N 1 = Speed in the 1st case ; I (^) a 1 = armature current in the 1st case Φ 1 = flux/pole in the first case N 2 , I (^) a 2 , Φ 2 = corresponding quantities in the 2nd case. Then, using the above relation, we get
N 1 ∝ 1 1
E b Φ where E (^) b 1 = V − I (^) a 1 Ra ; N 2 ∝ 2 2
E b Φ
where Eb 2 = V − I (^) a 2 Ra
∴ 2 1
N N
= 2 1 1 2
b b
E E
Φ × Φ
Prior to saturation of magnetic poles ; Φ ∝ I (^) a ∴ 2 2 1 1 1 2
b a b a
N E^ I N E I
= ×
1010 Electrical Technology
Now, speed may be found either by using the relation for Eb or Ta as given in Art. Eb = Φ ZN × ( P/A ) or 439.7 = 34.6 × 10 −^3 × 944 × N × 2 ∴ N = 6.73 r.p.s. or 382.2 r.p.m.
Example 29.21. A 250-V shunt motor runs at 1000 r.p.m. at no-load and takes 8A. The total armature and shunt field resistances are respectively 0.2 Ω and 250 Ω. Calculate the speed when loaded and taking 50 A. Assume the flux to be constant. (Elect. Engg. A.M.Ae. S.I. June 1991)
Solution. Formula used : 0
N N
= (^0 ) 0 0 0
b (^) ; Since (given); b b b
E (^) N E E N E
Φ × Φ = Φ = Φ I (^) sh = 250/250 = 1 A Eb 0 = V − I (^) a 0 Ra = 250 − (7 × 0.2) = 248.6 V; Eb = V − I (^) a Ra = 250 − (49 × 0.2) = 240.2 V ∴ 1000
N (^) = 240. ;
N = 9666.1 r.p.m.
Example 29.22. A d.c. series motor operates at 800 r.p.m. with a line current of 100 A from 230-V mains. Its armature circuit resistance is 0.15 Ω and its field resistance 0.1 Ω. Find the speed at which the motor runs at a line current of 25 A, assuming that the flux at this current is 45 per cent of the flux at 100 A. (Electrical Machinery - I, Banglore Univ. 1986)
Solution.^2 1
N N = 2 1 2 1 1 1 2 2
; 0.45 or 1
b b
E E
Φ Φ × Φ = Φ = Φ Φ E (^) b 1 = 230 − (0.15 + 0.1) × 100 = 205 V; Eb 2 = 230 − 25 × 0.25 = 223.75 V 2 800
N = 223.75^1 ; 2 205 0.
× N = (^) 1940 r.p.m.
Example 29.23. A 230-V d.c. shunt motor has an armature resistance of 0.5 Ω and field resistance of 115 Ω. At no load, the speed is 1,200 r.p.m. and the armature current 2.5 A. On application of rated load, the speed drops to 1,120 r.p.m. Determine the line current and power input when the motor delivers rated load. (Elect. Technology, Kerala Univ. 1988)
Solution. N 1 = 1200 r.p.m., Eb 1 = 230 − (0.5 × 2.5) = 228.75 V N 2 = 1120 r.p.m., E (^) b 2 = 230 − 0.5 I (^) a 2
Now, 2 1
N N
= (^2 2 ) 1
1120 230 0. ; 1200 228.
b a a b
E I I E
− ∴ = = 33 A
Line current drawn by motor = I (^) a 2 + I (^) sh = 33 + (230/115) = 35 A Power input at rated load = 230 × 35 = 8,050 W
Example 29.24. A belt-driven 100-kW, shunt generator running at 300 r.p.m. on 220-V bus- bars continues to run as a motor when the belt breaks, then taking 10 kW. What will be its speed? Given armature resistance = 0.025 Ω, field resistance = 60 Ω and contact drop under each brush = 1 V, Ignore armature reaction. (Elect. Machines (E-3) AMIE Sec.C Winter 1991)
Solution. As Generator [Fig. 29.12 ( a )] Load current, I = 100,000/220 = 454.55 A ; I (^) sh = 220/60 = 3.67 A I (^) a = I + I (^) sh = 458.2 A ; I (^) a Ra = 458.2 × 0.025 = 11. Eb = 220 + 11.45 + 2 × 1 = 233.45 V ; N 1 = 300 r.p.m.
D.C. Motor 1011
3.67 A
454.55 A
Ia I
Bus Bars
Ish
220 V
3.67A (^) Ia 45.45A
Bus Bars
Ish
220V
I
A
(a) (b)
- 025 W 60 W
Fig. 29. As Motor [Fig. 29.12 ( b )] Input line current = 100,000/220 = 45.45 A; I (^) sh = 220/60 = 3.67 A Ia = 45.45 − 3.67 = 41.78 A; I (^) a Ra = 41.78 × 0.025 = 1.04 V; Eb 2 = 220 − 1.04 − 2 × 1 = 216.96 V 2 1
N N =^
2 1 1 2 1 2
b (^) ; since because (^) sh is constant b
E I E
Φ × Φ = Φ Φ
∴ 2 300
N = (^2) 216.96 ; 279 r.p.m.
N =
Example 29.25. A d.c. shunt machine generates 250-V on open circuit at 1000 r.p.m. Effective armature resistance is 0.5 Ω, field resistance is 250 Ω , input to machine running as a motor on no- load is 4 A at 250 V. Calculate speed of machine as a motor taking 40 A at 250 V. Armature reaction weakens field by 4%. (Electrical Machines-I, Gujarat Univ. 1987) Solution. Consider the case when the machine runs as a motor on no-load. Now, I (^) sh = 250/250 = 1 A; Hence, I (^) a 0 = 4 − 1 = 3A; Eb 0 = 250 − 0.5 × 3 = 248.5 V It is given that when armature runs at 1000 r.p.m., it generates 250 V. When it generates 248.5 V, it must be running at a speed = 1000 × 248.5/250 = 994 r.p.m. Hence, N 0 = 994 r.p.m. When Loaded I (^) a = 40 − 1 = 39 A ; Eb = 250 − 39 × 0.5 = 230.5 V Also, Φ 0 /Φ = 1/0. N E = 0
230.5 1 994 248.5 0.
b b
E (^) N E
∴ = × (^) N = 960 r.p.m.
Example 29.26. A 250-V shunt motor giving 14.92 kW at 1000 r.p.m. takes an armature current of 75 A. The armature resistance is 0.25 ohm and the load torque remains constant. If the flux is reduced by 20 percent of its normal value before the speed changes, find the instantaneous value of the armature current and the torque. Determine the final value of the armature current and speed. (Elect. Engg. AMIETE (New Scheme) 1990) Solution. Eb 1 = 250 − 75 × 0.25 = 231.25 V, as in Fig. 29.13. When flux is reduced by 20%, the back e.m.f. is also reduced instantly by 20% because speed remains constant due to inertia of the heavy armature (Art. 29.11).
∴ Instantaneous value of back e.m.f. ( Eb ) (^) inst = 231.25 × 0. = 185 V ( I (^) a ) (^) inst = [ V − ( Eb ) (^) inst]/ R (^) a = (250 − 185)/0.25 = 260 A
Fig. 29.
D.C. Motor 1013
( a ) Ta =
196 48
1031
× × (^) = 87.1 N-m
Example 29.29. The armature circuit resistance of a 18.65 kW 250-V series motor is 0.1 Ω , the brush voltage drop is 3V, and the series field resistance is 0.05. When the motor takes 80 A, speed is 600 r.p.m. Calculate the speed when the current is 100 A.
(Elect. Machines, A.M.I.E. Sec. B, 1993) Solution. E (^) b 1 = 250 − 80 (0.1 + 0.05) − 3 = 235 V. E (^) b 2 = 250 − 100 (0.1 + 0.05) − 3 = 232 V Since Φ ∝ I (^) a , hence, Φ 1 ∝ 80, Φ 2 ∝ 100, Φ 1 /Φ 2 = 80/
Now 2 1
N N = 2 1 2 1 2
or 232 80 600 235 100
b b
E N E
Φ × = × Φ
; N 2 = 474 r.p.m.
Example 29.30. A 220-volt d.c. series motor is running at a speed of 800 r.p.m. and draws 100 A. Calculate at what speed the motor will run when developing half the torque. Total resistance of the armature and field is 0.1 ohm. Assume that the magnetic circuit is unsaturated. (Elect. Machines ; A.M.I.E. Sec. B, 1991)
Solution.^2 1
N N = 2 1 2 1 1 2 1 2
b b a b b a
E E I E E I
Φ × = × Φ (∴ Φ ∝^ I^ a ) Since field is unsaturated, Ta ∝ Φ I (^) a ∝ I (^) a^2. (∴ T 1 ∝ I (^) a 12 and T 2 ∝ I (^) a 22 )
or T 2 / T 1 = ( I (^) a 2 / I (^) a 1 ) 2 or 1/2 = ( I (^) a 2 / I (^) a 1 ) 2 ; I (^) a 1 = I (^) a 1 / 2 = 70.7 A
E (^) b 1 = 220 − 100 × 0.1 = 210 V ; E (^) b 2 = 220 − 0.1 × 70.7 = 212.9 V
∴ 2 800
N
212.9 100 210 70.
× (^) ; N 2 = 1147 r.p.m.
Example 29.31. A 4-pole d.c. motor runs at 600 r.p.m. on full load taking 25 A at 450 V. The armature is lap-wound with 500 conductors and flux per pole is expressed by the relation.
Φ = (1.7 × 10 −^2 × I 0.5 ) weber where 1 is the motor current. If supply voltage and torque are both halved, calculate the speed at which the motor will run. Ignore stray losses. (Elect. Machines, Nagpur Univ. 1993) Solution. Let us first find R (^) a.
Now N = 60 b r.p.m. E A Z P
Φ ^
∴ 600 = (^2) 0. 60 4 1.7 10 25 500 4
E b −
× × × × × ∴ Eb = 10 × 1.7 × 10 −^2 × 5 × 500 = 425 V I (^) a Ra = 450 − 425 = 25 V ; Ra = 25/25 = 1.0 Ω Now in the Ist Case T 1 ∝ Φ 1 I 1 ∴ T 1 ∝ 1.7 × 10 −^2 × 25 × 25
Similarly T 2 ∝ 1.7 × 10 −^2 × 1 × I ; Now T 1 = 2 T 2 ∴ 1.7 × 10 −^2 × 125 = 1.7 × 10 −^2 × I 3/2^ × 2 ∴ I = (125/2)2/3^ = 15.75 A E (^) b 1 = 425 V ; E (^) b 2 = 225 − (15.75 × 1) = 209.3 V
Using the relation 2 1
N N = 2 1 1 2
b b
E E
Φ × Φ ; we have
1014 Electrical Technology
2 600
N
2 2
209.3 1.7^10 (^425) 1.7 10 15.
− −
× × × × ×
; N 2 = 372 r.p.m.
Tutorial Problems 29.
1. Calculate the torque in newton-metre developed by a 440-V d.c. motor having an armature resistance of 0.25 Ω and running at 750 r.p.m. when taking a current of 60 A. [325 N-m] 2. A 4-pole, lap-connected d.c. motor has 576 conductors and draws an armature current of 10 A. If the flux per pole is 0.02 Wb, calculate the armature torque developed. **[18.3 N-m]
- (** a ) A d.c. shunt machine has armature and field resistances of 0.025 Ω and 80 Ω respectively. When connected to constant 400-V bus-bars and driven as a generator at 450 r.p.m., it delivers 120 kW. Calculate its speed when running as a motor and absorbing 120 kW from the same bus-bars. ( b ) Deduce the direction of rotation of this machine when it is working as a motor assuming a clockwise rotation as a generator. [( a ) 435 r.p.m. ( b ) Clockwise] 4. The armature current of a series motor is 60 A when on full-load. If the load is adjusted to that this current decreases to 40-A, find the new torque expressed as a percentage of the full-load torque. The flux for a current of 40 A is 70% of that when current is 60 A. [46%] 5. A 4-pole, d.c. shunt motor has a flux per pole of 0.04 Wb and the armature is lap-wound with 720 conductors. The shunt field resistance is 240 Ω and the armature resistance is 0.2 Ω. Brush contact drop is 1V per brush. Determine the speed of the machine when running ( a ) as a motor taking 60 A and ( b ) as a generator supplying 120 A. The terminal voltage in each case is 480 V. [972 r.p.m. ; 1055 r.p.m.] 6. A 25-kW shunt generator is delivering full output to 400-V bus-bars and is driven at 950 r.p.m. by belt drive. The belt breaks suddenly but the machine continues to run as a motor taking 25 kW from the bus-bars. At what speed does it run? Take armature resistance including brush contact resistance as 0.5 Ω and field resistance as 160 Ω. [812.7 r.p.m.] ( Elect. Technology, Andhra Univ. Apr. 1977 ) 7. A 4-pole, d.c. shunt motor has a wave-wound armature with 65 slots each containing 6 conductors. The flux per pole is 20 mWb and the armature has a resistance of 0.15 Ω. Calculate the motor speed when the machine is operating from a 250-V supply and taking a current of 60 A. [927 r.p.m.] 8. A 500-V, d.c. shunt motor has armature and field resistances of 0.5 Ω and 200 Ω respectively. When loaded and taking a total input of 25 kW, it runs at 400 r.p.m. Find the speed at which it must be driven as a shunt generator to supply a power output of 25 kW at a terminal voltage of 500 V. [442 r.p.m.] 9. A d.c. shunt motor runs at 900 r.p.m. from a 400 V supply when taking an armature current of 25 A. Calculate the speed at which it will run from a 230 V supply when taking an armature current of 15 A. The resistance of the armature circuit is 0.8 Ω. Assume the flux per pole at 230 V to have decreased to 75% of its value at 400 V. [595 r.p.m.] 10. A shunt machine connected to 250-A mains has an armature resistance of 0.12 Ω and field resistance of 100 Ω. Find the ratio of the speed of the machine as a generator to the speed as a motor, if line current is 80 A in both cases. [1.08] ( Electrical Engineering-II, Bombay Univ. April. 1977, Madras Univ. Nov. 1978 ) 11. A 20-kW d.c. shunt generator delivering rated output at 1000 r.p.m. has a terminal voltage of 500 V. The armature resistance is 0.1 Ω, voltage drop per brush is 1 volt and the field resistance is 500 Ω. Calculate the speed at which the machine will run as a motor taking an input of 20 kW from a 500 V d.c. supply. [976.1 r.p.m.] ( Elect. Engg-I Bombay Univ. 1975 ) 12. A 4-pole, 250-V, d.c. shunt motor has a lap-connected armature with 960 conductors. The flux per pole is 2 × 10 −^2 Wb. Calculate the torque developed by the armature and the useful torque in newton-metre when the current taken by the motor is 30A. The armature resistance is 0.12 ohm and the field resistance is 125 Ω. The rotational losses amount to 825 W. [85.5 N-m ; 75.3 N-m] ( Electric Machinery-I, Madras Univ. Nov. 1979 )
