DC motors. 1. Parallel (shunt) excited DC motor

Transcription

1 DC motors 1. Parallel (shunt) excited DC motor A shunt excited DC motor s terminal voltage is 500 V. The armature resistance is 0,5 Ω, field resistance is 250 Ω. On a certain load it takes 20 A current from the DC supply and it runs at 409 1/min. Calculate the induced voltage, the output power, the efficiency and the torque of the motor if the mechanical and the magnetic losses are 900 W together. Equivalent wiring diagram of the motor: U i I Input power: P in = I = = W The field current: = = = 2 A The armature current (according to KCL): = I = 20 2 = 18 A The losses of the motor: The induced voltage (according to KVL): = 0 = = ,5 = 491 V P b = = = 8838 W The difference between the input and the internal power is due to the electrical losses: P 1 P int = = 1132 W = P av + P f P f = = = 1000 W the field power, P av = I 2 a = ,5 = 162 W the armature losses. 1

2 We can get the output power if we subtract the mechanical and the iron losses from the internal power: P out = P int P iron+mech = = 7938 W The efficiency: η = P out P in = = 0,7938 To calculate the torque of the motor we have to calculate the angular speed first: Ω = 2π n 60 = 2π = 42,83 1/s s = P out Ω = 7938 = 185,33 Nm 42,83 2. Parallel (shunt) excited DC generator Let s see the previous machine operating as generator. Terminal voltage of a shunt excited DC generator is 500 V, the armature resistance is 0,5 Ω, field resistance is 250 Ω. A 25 Ω load is connected to its terminals. Calculate the induced voltage, the output power and the efficiency of the generator if the mechanical and the magnetic losses are 900 W together. What speed should the shaft rotate if the driving torque is 100 Nm? I l The current consumption of the load: I l = U k = 500 = 20 A R t 25 G R l The field current: = = = 2 A The armature current (according to KCL): = I l + = = 22A The output power of the generator: P 2 = I l = = W The induced voltage (according to KVL): 2

3 + = 0 = + = ,5 = 511 V P int = = = W The difference between the internal and the output power is due to the electrical losses: P 1 P int = = 1242 W = P a + P f P f = = = 1000 W is the field power and P a = 2 = ,5 = 242 W is the armature losses. To get the input power we have to add the iron and the mechanical losses to the internal power: P 1 = P b + P vas+súrl = = W The efficiency: η = P 2 P 1 = = 0,8235 To calculate the speed of the motor we have to calculate the angular speed first: Ω = P 1 = = 121,42 1/s h 100 n = Ω 2π 121,42 60 = 60 = 1159,47 1/min 2π 3. Idling Calculate the idling speed of the previous motor (1 st exercise): The basic equations of the DC machines: = k Φ Ω, we can write this equation as: = k Φ 2π 2π n Instead of k we can use c = k as machine factor, so we can write: = c Φ n (sometimes using this equation is more convenient than transform Ω to n or n to Ω) b = k Φ (k is the machine factor, Φ is the flux, Ω is the angular speed and is the armature current) Idling means that the machine rotates unloaded, i.e. the torque of the motor is zero: s = 0 P out = 0 a) First we don t care the losses and calculate the so called ideal idling speed in lossless case. If there are no losses, there are neither iron nor mechanical losses (P iron =0, P mech =0), the internal power of the motor is zero too. (P int =0) P int = int Ω int = 0 = b k Φ = 0, since the flux wasn t change. (The field circuit remains to be connected to the power supply.) 3

4 If there no armature current, there is no voltage drop due to the armature resistance, so the induced voltage is equal to the terminal voltage. So the ideal idling speed can be calculated as: n 0id = c Φ we can get the product of the machine constant and the flux (cφ) from the results of the 1 st exercise: When the speed was 409 1/min (the angular speed is Ω=42,83 1/s), the induced voltage was =491V: c Φ = = 491 = 1,2 Vmin, or n 409 k Φ = = 491 = 11,463 Vs, and Ω 42,82 n 0id = b. Let s take into account the losses: The torque is zero: t =0. c Φ = 500 = 416,67 1/min 1,2 The consumed power of the motor is equal to sum of the filed power and the losses. P int = P irin+mech = 900 W, since the iron and the mechanical losses are independent to the load. The internal torque: int = P int Ω 0 = P iron+mech Ω 0 P iron+mech Ω = 900 = 21 Nm 42,83 To maintain the 21 Nm torque, the motor needs armature current: 0 = int = 21 = 1,83 A k Φ 11,46 So the idling induced voltage: 0 = 0 = 500 1,83 0,5 = 499,083 V The idling speed: n 0 = 0 c Φ = 499,083 = 415,9 1/min 1,2 (It is almost the same than the result of the lossless case. It means that it is more than enough to apply the estimation of the lossless case.) 4. Starting the motor Calculate the required resistance of the starter resistor if the maximum starting armature current is one and a half times greater that the nominal value of the armature current. At the moment of starting there is no flux inside the machine so the induced voltage is zero. It means that the all terminal voltage drops on the low resistance armature: = 0 4

5 = = = 500 = 1000 A It would be an impermissible high current. 0,5 The simplest way to reduce this high current is increasing the resistance of the armature by a serial connected resistor. R st The maximum armature current: i = 1,5 = 1,5 18 = 27 A i I The required starter resistance: =0 U f i = + R st R i = U k R = 500 0,5 = 18,01 A ai Speed control n = c Φ = U k c Φ The speed depends on: a. terminal voltage b. the resistance of the armature circuit c. the flux, which is the dependent upon the field current So the speed can be controlled with one of the three upper parameters a. Let s examine how the terminal voltage influences the speed in cases of the 3 different exciting mode a1. Separately excited DC motor 500 V terminal voltage separately excited DC motor s armature resistance is 0,5 Ω. It takes 20 A current from the DC supply and its speed is /min. Calculate the speed if the terminal voltage is decreased to 400 V while the load is unchanged. I g Since the field current is independent of the armature, the changing of the terminal voltage doesn t influence the flux. U g R g ) 1 = U k1 = ,5 = 490 V U k cφ = 1 n 1 = 2 n 2 = áll. Since is the load (i.e. the torque of the motor) doesn t change, neither the armature current changes. ( b = k Φ 5

6 2 = U k2 = ,5 = 390 V so the speed is: n 2 = 2 n U 1 = = 795,91 1/min i1 490 a2. Serial excited DC motor 500 V terminal voltage serial excited DC motor s armature and field resistance is 5 Ω together. On a certain load it takes 20 A current from the DC supply and its speed is /min. Calculate the speed if the terminal voltage is decreased to 400 V while the load is unchanged. R g The flux of this motor is generated by the armature current. The armature current depends on the load. If the load is constant, the armature current, so the flux stays constant too. ( int = k Φ = k 2 ) = I g U k cφ = 1 n 1 = 2 n 2 = const. 1 = 1 ( + ) = = 400 V 2 = 2 ( + ) = = 300 V n 2 = 2 n U 1 = = 750 1/min. i1 400 a3. Parallel excited DC motor Let s take the motor of the 1 st example. Calculate the speed if the terminal voltage is decreased to 400V while the load is unchanged. We can observe that changing the terminal voltage changes the field current: I g 2 = 2 = 400 = 1,6 A 250 R g U k The flux depends on the field current we can say that the flux changes approximately proportional to the field current. The data of the 1 st example: 1 =500 V, 1 =491 V, 1 =2A, n 1 =409 1/min, cφ 1 =1,2 Vmin. when the field current is 2 =1,6A, cφ 2 = I g2 cφ I 1 = 1,6 1,2 = 0,96 Vmin g1 2 6

7 Because of the decreased flux the motor needs higher armature current to maintain the same torque: int = k φ 1 1 = k φ 2 2 = const. 2 = k φ 1 I k φ a1 = c φ 1 I 2 c φ a1 = 1,2 18 = 22,5 A 2 0,96 The induced voltage: 2 = U k2 2 = ,5 0,5 = 388,75 V The speed: n 2 = 2 = 388,75 = 404,94 1/min c Φ 2 0,96 We can observe that this method has no effect to the shunt excited motor s speed. b. Speed control by increasing the armature resistance Calculate the resistance connected in series with the armature if we want to decrease the speed of the motor of the 1st example to 300 1/min while the load is unchanged. R c I cφ = 1 n 1 = 2 n 2 = const. (since = U f = const. ) 2 = n 2 U n i1 = = 360,15 V = 1 = ,5 = 491 V 2 = 2 ( + R c ) R k = U k2 2 = Calculate the losses of the control resistor: P Rc = 2 R c = ,27 = 2355,48 W ,15 0,5 = 7,27Ω 18 c. Speed control with changing the flux We can control the flux with the field current. The simplest way the control the field current is changing the resistance of the field circuit applying a resistor connected in series with the field winding. I Connect a 150 Ω resistor in series to the field circuit and calculate the speed of the motor of the 1 st example. (The load is constant) R c 7

8 The field current: 2 = 500 = = 1,25 A R g + R c We suppose that the flux is proportional to the field current: I g1 = 2 A cφ 1 = 1,2 Vmin and when I g2 = 1,25 A cφ 2 = I g2 cφ I 1 = 1,25 1,2 = 0,75 Vmin g1 2 Since the flux decreases, the motor need to take more current to maintain the same torque: b = k φ 1 1 = k φ 2 2 = const. 2 = k φ 1 I k φ a1 = c φ 1 I 2 c φ a1 = 1,2 18 = 28,8 A 2 0,75 The induced voltage: 2 = 2 2 = ,8 0,5 = 485,6 V The speed: n 2 = 2 = 485,6 = 647,47 1/min c Φ 2 0,75 We can see that this method causes significant change of the speed. Calculate the losses of the control resistance: P Rc = I g 2 R sz = 1, = 234,375 W this is ten times smaller than the losses of the previous method. 6. Nominal power of a 550 V serial excited DC motor is 50 kw. The armature resistance is 0,2 Ω, the field resistance is 0,1Ω. It takes 100 A current from the DC supply and its speed is 650 1/min. Calculate the efficiency, the losses, the induced voltage and the torque of the motor! The nominal power always means the maximum output power given in the motor catalogue. R g The nominal torque of the motor: = I g U k s = P out Ω = P 2 2π n The input power: 60 = π = 734,56Nm P in = = = W The efficiency: 8

9 η = P out P in = = 0,909 The induced voltage: = ( + ) = (0,1 + 0,2) = 520 V P b = = = W The losses: Armature losses: P a = I 2 a = ,2 = 2000 W The field power: P f = I 2 f = ,1 = 1000 W The sum of the iron and the mechanical power: P iron+mech = P b P 2 = = 2000 W How does the speed change if the load decreases in half? The load means the torque of the motor. The relationship between the torque and the current: b = kφ The serial excited motor s flux depends on the armature current. We suppose that the flux is proportional to the armature current, so the torque is proportional to the square of the current: Calculating the current consumption: b = k 2 the torque decreases to t2 = t1 = 734,56 = 367,28 Nm, so 2 2 b1 2 = b2 2 2 = 1 b2 = 100 0,5 = A b1 1 2 when the speed is n 1 =650 1/min, the induced voltage is 1 =520 V volt, so: cφ 1 = 1 = 520 = 0,8 Vmin n The flux is generated by the armature current so when the current consumption decreases to 2 =70,71A, the flux decreases too: the induced voltage: cφ 2 = 2 cφ I 1 = 70,71 0,8 = 0,5657 Vmin a = 2 ( + R g ) = ,71 (0,1 + 0,2) = 528,787 V the speed: 9

10 n 2 = 2 = 528,787 = 934,75 1/min cφ 2 0,5657 We can observe that the load influences significantly the speed of the motor! 7. Serial excited generator Terminal voltage of a serial excited generator at a 25 Ω load is 500 V. The resistance of the armature and the field circuit is 1 Ω together. Calculate the efficiency of the generator if sum of the iron and the mechanical losses is 1,5 kw. Calculate the speed of the generator if the driving torque is 240 Nm. R g The current consumed by the load, the armature current and the field current is the same. G =I g =I t U k R t = = I l = = 500 = 50 A R l 10 The output power: P out = I l = = W The induced voltage: = + ( + ) = = 550 V P int = = = W The input power: P in = P int + P iron+mech = = W The efficiency: η = P 2 P 1 = = 0,862 The speed: n = 60 2π Ω = 60 2π P 1 = 60 h 2π = 1153,87 1/min Terminal voltage of a 400 V separately excited DC motor is 400 V. It takes 22 A, its speed is 955 1/min. The resistance of the armature is 2 Ω, the field resistance is 200Ω. Calculate the induced voltage, the losses, the efficiency and the torque of the motor if the sum of R mechanical and the iron losses is the 4 % of the a output power. The input power: U f 10

11 P in = = = 8800 W The field power: P f = U f = U f 2 = = 800 W The induced voltage: = = = 356 V P int = = = 7832 W The difference between the input and the internal power is due to the armature losses: P 1 P b = P a = 2 = = 968 W We get the output power if we subtract the mechanical and the iron losses from the internal power: P out = P int P iron+mech = P int 0,04 P out P 2 = P b 1,04 = 7832 = 7530,77 W 1,04 P vas+súrl = ,77 = 301,23 W When we calculate the efficiency, we have to take into account the power of the field circuit which is independent to the armature circuit. The power of the field circuit have to add to the input power: The torque: η = s = P out Ω P out = 7530,77 P in + P f = 0,784 = P out = 60 = 75,29Nm 2π n 2π Other questions: a) Calculate the speed of the motor if the terminal voltage of the armature is decreased to 300 V? b) Calculate the speed of the motor if the filed voltage is decreased to 300 V. c) Calculate the speed of the motor if a 45 Ω resistor is connected in series to the armature. d) Calculate the resistance of the starter resistor if the maximum allowed armature current is 1,5 times greater than the nominal armature current. 11