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Efficiency of D.C. Machines

9.3. Efficiency of D.C. Machines:

Generator: 6

Thus the efficiency increases with increase in load current, reaches a maximum value when load current equals the value given by eqn. (27) and then starts decreasing,

Efficiency curve. The efficiency of a machine is different at different values of power output. As the output increases, the efficiency increases till it reaches a maximum value. As the output is further increased, the efficiency starts decreasing. A graph of efficiency vs. output is called efficiency curve. A typical efficiency curve is shown in Fig. 96. The machines are so designed as to give maximum efficiency at or near the rated output of the machine. Since the generators operate at a constant terminal
voltage V, the efficiency curve of a generator can be drawn between efficiency and load current I:7

Fig. 96. Efficiency curve.

 

Example 43. The armature and shunt field resistances of 110 V shunt generator running at 800 r.p.m. are 0.12 Ω, 55 n respectively. It supplies 220 lamps each rated at 60 W, 110 V, The friction, windage and core loss of the machine is equal to 410 W. Calculate:

(i)                           Armature copper loss on [ull-load and shunt field copper loss .

(ii)                        Torque in Nm exerted by the prime-mover,

(iii)                      Mechanical, electrical and overall efficiencies.

Solution. Terminal voltage, V = 110 V

Speed of the generator, N = 800 r.p.m.

Armature resistance, Ra = 0.12 Ω

Shunt field resistance, Rsh = 55 n

Friction, windage and core loss = 410 W8

Armature copper loss   = 1786 W. (Ans.)

 

Shunt field copper loss             = VIsh = 110 × 2 = 220 W. (Ans.)

 

[Shunt field copper loss = Ish2Rsh = 22 × 55 = 220 W]

 

(ii) Total copper loss = 1786 + 220 = 2006 W

 

Core, friction and windage loss = 410 W

 

Total loss = 2006 + 410 = 2416 W

 

Output = 220 × 60 = 13200 W

 

Input = output + losses = 13200 + 2416 = 15616 W

 

If T is the torque in Nm exerted by the prime mover 9=186.4 NM

Torque exerted by the prime mover

= 186.4 Nm. (Ans.)

(iii) Electrical power generated in armature

=EbIa

= mechanical input – stray losses

= 15616 – 410 = 15206 W

Mechanical power input = 15616 W.

10

Example 44. Find the efficiency at full-load and at half-load for a 1200 kW, 600 V shunt generator using the following data:

Armature resistance = 0.005 Ω,

Shunt field resistance = 60 Ω,

Brush contact drop = 1 V per brush,

Mechanical and iron losses at rated load = 20 kW,

Stray load loss = 1.2% of output.

Solution. Terminal voltage, V = 600 V

Full-load output = 1200 kW

Armature resistance, Ra = 0.005 Ω

Shunt field resistance, Rsh = 60 Ω

Brush contact drop = 1 V per brush

Mechanical losses at rated load = 20 kW

Stray load loss = 1.2% of output

Efficiency at full-load and at half-load:

(i) At full-load11

Armature current,

 

Ia = I + Ish

 

= 2000 + 10 = 2010 A

 

Armature copper loss

 

= Ia2Ra = 20102 × 0.005 = 20200 W

 

Shunt field loss = VIsh = 600 × 10 = 6000 W

 

Brush contact loss = 2 × 2010 = 4020 W

 

Mechanical and iron losses = 20 × 1000 = 20000 W

 

Stray load losses = 1.2% of full-load output 12

Armature current, Ia = 1000 + 10 = 1010 A

 

Armature copper loss, Ia2Ra = 10102 × 0.005 = 5100 W

 

Shunt field loss = VIsh = 600 × 10 = 6000 W

 

Brush contact loss = 2 × 1010 = 2020 W

 

Mechanical and iron losses = 20000 W

 

Stray load losses = 1.2% of 600000 = 7200 W13

Hence, efficiency at half-load = 93.7%. (Ans.)

 

Example 45. The following data pertains to a short-shunt compound generator:

 

Armature resistance = 0.04 Ω,

 

Series field resistance = 0.025 Ω,

 

Shunt field resistance = 45 Ω,

 

Contact drop / brush = 1 V,

 

Magnetic losses = 2.5 k W,

 

Mechanical losses = 1 k W.

 

Find the efficiency of the generator when it delivers 400 A at 440 V.

 

Solution. Terminal voltage, V =440 V

 

Load current, I = 400 A

 

Armature resistance, Ra = 0.04 Ω

 

Series field resistance, Rse = 0.025 Ω

 

Shunt field resistance, Rsh = 45 Ω

 

Copper drop/brush = 1 V

 

Magnetic losses = 2.5 kW = 2500 W

 

Mechanical losses = 1 kW = 1000 W

 

Refer Fig. 97

 

Output of the generator = VI = 440 × 400 = 176000 W

 

Voltage across shunt field winding = V + IRse

 

= 440 + 400 × 0.025 = 450 V14

Fig. 97

 

Series field copper loss = I2Rse = 4002 × 0.025 = 4000 W

 

Shunt field copper loss = VIsh = 450 × 10 = 4500 W

 

Magnetic losses =2500 W

 

Mechanical losses =1000 W

 

Total losses = 6724 + 820 + 4000 + 4500 + 2500 + 1000 = 19544 W 15

Hence, efficiency of the generator’” 90%. (Ans.)

 

HIGHLIGHTS

 

  1. 1.   Basic type of D.C. machine is that of commutator type. This is actually an alternating current (A.C.)machine, hut furnished with a special device, a commutator, which under certain conditions converts alternating current into direct current.
  2. 2.   An electrical generator is a machine which converts mechanical energy (or power) into electrical energy (or power). It works on the following principle: “Whenever a conductor cuts magnetic flux, dynamically induced e.m.f. is produced in it according to Faraday’s'Laws of Electromagnetic induction”.
  3. 3.   A D.C. machine consists of two main parts :

 

-        Stationary Part: designed mainly for producing magnetic flux.

 

-        Rotating Part: called armature, where ‘mechanical energy is converted into electrical (electric generator) or, conversely, electrical energy into mechanical (electric motor).

 

The various parts of a D.C. machine are enumerated below:

 

  1. Frame
  2. Field poles
  3. Commutating poles
  4. Armature
  5. Commutator
  6. Brush gear
  7. Armature shaft bearings
  8. Armature windings.
  9. 4.   According to the degree of closure produced by winding, armature windings are of following two type« :

 

-        Open coil winding

 

-        Closed coil winding

 

The closed armature windings are of two types:

 

-        Ring winding

 

-        Drum winding

 

In general there are two types of drum armature windings:

 

-        Lap winding

 

-        Wave winding.

 

  1. 5.   ‘Lap winding’ is suitable for comparatively low voltage but high current generators whereas ‘wave winding’ is used for high voltage, low current machines.
  2. 6.   In the ‘lap winding’ the finish of each coil is connected to the start of the next coil so that winding or commulator pitch is unity.

 

In ‘wave winding’ the finish of coil is connected to the start of another coil well away from the first coil.

 

  1. 7.   The rated output of an electrical machine is its duty in the conditions for which it has been designed by the manufacturer.The rated duty is characterised by values given on the machine’s rating plate, and termed rated value, as, for instance: rated output, rated voltage, rated current, rated speed, etc.
  2. 8.   E.m.f equation of a generator is given as follows: 15= number of slots × number of conductors/slot

    N = rotational speed of armature, r.p.m.

    a = number of parallel paths in armature

    For a wave wound generator: a = 2 ]

    For a lap wound generator: a = p

    1. 1.   According to method of excitation D.C. generators are classified as follows:

    -        Separately excited generators

    -        Self excited generators.

    Self excited generators can be divided in accordance with how the field winding is connected into generators as follows:

    -        Shunt wound generators

    -        Series wound generators

    -        Compound wound generators :

    1. Short shunt
    2. Long shunt.
    3. 2.   The electric motor is a machine which converts electrical energy into mechanical energy.
    4. 3.   Voltage equation of a motor is V= Eb + laRa.
    5. 4.   Condition for maximum power developed in armature is Eb =V/2.
    6. 5.   Armature torque of a motor is given by Ta = 0.159 Zɸp × Ia/a Nm. 16In a D.C. motor reversal is accomplished by changing the polarity of either the armature or the field; but not by changing both.