DC machines - Mandatory work/practical work PDF

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Mandatory work/practical work...

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DC Machine Theory. When the input to an electrical machine is electrical energy and the output is mechanical energy, the machine is called an electric motor. Hence, an electric motor converts electrical energy into mechanical energy. An electric generator behaves in the opposite manner , i.e. it converts mechanical energy into electrical energy. Therefore a motor and generator are the same type of system, with the outputs reversed. Efficiency , η = (output power/input power)*100% In an electric motor, conductors rotate in a uniform magnetic field. Consider the single loop conductor mounted between permanent magnets in the following fig. A voltage is applied between points A and B.

A force ‘F’ acts on the loop because of the interaction of the magnetic field of the permanent magnets and the magnetic field created by the current flowing in the loop. F = BIL where, B = magnetic flux density of the field I = current flowing in the conductor L = effective length of the conductor (Aside: use the left-hand rule to show the direction of the two components of the force acting on the sides of the conductor)

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If the loop is free to rotate , then after 1800 , the conductors will appear as:

For rotation to continue in the same direction , it is necessary for the current to flow in the direction shown above, i.e. from D to C and F to E.

This apparent reversal in the direction of current is achieved by a process called commutation. As outlined in the following fig, when a direct voltage is applied at terminals A and B, current flow will always be away from the commutator for the part of the conductor adjacent to the N-pole and always towards the commutator for the part adjacent to the Spole. Therefore the forces act to give continuous rotation in an anti-clockwise direction.

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In practice there are many conductors on the rotating part of a dc machine and these are attached to many commutator segments. A schematic diagram of a basic dc machine is shown below:

The basic parts are: (a) a stationary part called the stator, having 

A steel ring called a yoke ,to which are attached

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

The magnetic poles ,around which are the



Field windings(current passing through these windings creates the electromagnetic field)

(b) a rotating called the armature mounted in bearings housed in the stator and having: 

A laminated cylinder of ironor steel called the core on which slots are cut to house the



Armature winding ,i.e. a single or multi-loop conductor system and



The commutator.

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The average emf induced in a single conductor on the armature of a dc machine is given by: fluxcut / rev 2 p  timeof 1rev 1 / n

where: p = number of pairs of poles Φ = is the flux (Webers) , entering or leaving a pole n= speed of rotation (rev/s)

Therefore the average emf per conductor is : 2pΦ n volts

If there are Z conductors connected in series, then the average emf is: 2pΦ nZ volts Note: for a given machine, the number of pairs of poles ‘p’ and the number of conductors connected in series ‘Z’ are constant; hence the generated emf, E , is proportional to Φ n. ω = 2πn = angular velocity(rad/s) Hence the generated emf , E , is proportional to Φ ω.

(i)

The power on the shaft of dc machine = Tω (Watts) where, T = torque (N/m) and ω = angular velocity (rad/s)

(ii)

The power developed by the armature current =EIa ,where: E=generated emf (volts) and Ia=armature current (amps). 5

Therefore, neglecting losses, Tω = EIa But E is proportional to Φ ω. Hence, Tω = Φ ω Ia (i.e. T is proportional to Φ Ia )

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Shunt-wound and Series-wound Machines When the field winding of a dc machine is connected in parallel with the armature, the machine is said to be shunt-wound. When the field winding of a dc machine is connected in series with the armature, the machine is said to be series-wound.

Note: depending on whether a machine is series-wound or shunt-wound, it behaves differently when a a load is applied.

Shunt-wound motor characteristics Torque is proportional to Φ Ia .For the shunt-wound motor, the field winding is connected in parallel with the armature circuit and hence, the applied voltage gives a constant field current (i.e. a shunt-wound motor is a constant flux machine).Since Φ is constant, it follows that torque is proportional to Ia. The corresponding characteristic is shown below.

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The armature circuit of a dc motor has resistance (due to the armature winding and brushes).Let this resistance = Ra.

Applying KVL, V = E + IaRa =>

E = V - IaRa

But E is proportional to Φn. =>

n is proportional to E/Φ

=>speed of rotation, n is proportional to (V - IaRa)/ Φ The corresponding speed/armature current characteristic may be derived from this last equation and is given below.

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As the load on the shaft of the motor increases, Ia increases and the speed drops slightly (this would represent a fall of 10% between no-load and full-load in practice) Since torque is proportional to armature current , the theoretical speed/torque characteristic is:

The corresponding characteristics for the series-wound motor are shown in the following figs.

Example: The shaft torque of a diesel motor driving a 100V d.c. shunt-wound generator is 25 Nm. The armature current of the generator is 16A at this value of torque. If the shunt field regulator is adjusted so that the flux is reduced by 15%, the torque increases to 35 Nm. Determine the armature current at this new value of torque.

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Shaft torque  Ia , T = kIa T1 = k1Ia1 and T2 = k2Ia2 T1 / T2 = (k1Ia1)/ (k2Ia2) = (1Ia1)/ (2Ia2) 25/35 = (1x 16)/ (0.851x Ia2)  Ia2 = 26.35A Example: A 150V d.c. motor develops a shaft torque of 25Nm at 1400 rev/min. If the efficiency is 80% , determine the current supplied to the motor. efficiency

= [T(2n)]/VI

80= [25(2)(1400/60)]/(150)I I = Example: A 200V d.c. shunt-wound motor has an armature resistance of 0.4 and at a certain load , has an armature current of 30A at a speed of 1350 rev/min. If the load on the shaft of the motor is increased so that the armature current increases to 45A , determine the speed of the motor (assuming the flux remains constant).

En E = V-IaRa 

E1 = 200 – (30 x 0.4) = 188V ,and E2 = 200 – (45 x 0.4) = 182V E1/E2 = 1n1/2n2 Since the flux is constant, 1=2

 

188/182 = 1(1350/60)/1n2 n2 = 21.78 revs/s or 1307 rev/min

Exercise: A dc shunt-wound generator running at constant speed, generates a voltage of 150V at a certain value of field current. Determine the change in the generated voltage when the field current is reduced by 20%,assuming the flux is proportional to the field current. (generated voltage reduced to 120V).

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Exercise: A 100 V dc generator supplies a current of 15A when running at 1500 rev/min.If the torque on the shaft driving the generator is 12 N/m , determine: 

the efficiency of the generator.

(79.6%)



The power loss in the generator.

(385 watts)

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