Electrical M/C’s and Power Utilization 3(2+1)

LESSON 12. EMF and torque equations of DC machines

 

Fig. 7. 1 Flux distribution at no load of a D. C. generator.

The total flux leaving a north pole is given by the area under one of the positive parts of the distribution curve. Similarly, the total flux leaving the armature is  the area of one of the negative parts of the distribution curve. Each positive part and each negative part of the curve may be replaced by a rectangle having the same area. The height of this rectangle will be B maxwells / cm², which is equal to the average value of the flux density under an entire pole pitch.

Now to determine the average electromotive force induced in a single conductor as it passes through the flux of successive poles. Let the total flux leaving a north pole or entering a south pole be φ maxwells. Let A be the pole area in cm², l the active length of the conductor in cm., s the speed of the armature in revolutions per second, and P the number of poles. When the conductor passes through the distance ab, or one pole pitch, the average induced voltage, by equation studied earlier is

e = Blv 10^-8

where,  B is the average flux density, I the active length of the conductor in cm., and v the velocity of the conductor in cm. per second.

Then v = ab/t, where ‘t’ is the time required for the conductor to traverse ab.

Therfore, e = Bl (ab/t) 10^-8 = (φ/t) 10^-8

since Bl (ab)  gives the total flux between  the points a and b as cut by the conductor and is therefore equal to φ.

The time t  = 1/sP

Therefore, the average voltage per conductor is   e =  (φsP) 10^-8

If there are Z such conductors and p paths through the arma­ture, there must be Z / p such conductors in series.  Hence the total voltage generated between brushes is

E = (φsPZ) / (p10⁸)

Example.—A 900 r.p.m., 6-pole generator has a simplex lap winding. There are 300 conductors on the armature. The poles are 10 cm² and the average flux density is 50,000 lines per cm².    What is the voltage induced between brushes?

φ = 10 × 10 × 50,000 = 5,000,000 lines

s =  900/60 = 15 r.p.s ; P = 6; p = 6

E =  5,000,000 ×15 × 6 × 300 / 6 × 10 ⁸ = 225 volts.

The Saturation Curve or No volt characteristic curve:

Fig. 7. 2 Saturation curve

The last equation may be written as

where S = r.p.m.

The quantity within the brackets is constant for a given machine and may be denoted by K. Therefore E = KφS. i.e. the induced emf. in a machine is directly propor­tional to the flux and to the speed. If the speed is kept constant, the induced voltage is directly proportional to the flux, φ.

The flux is produced by the field ampere-turns, and as the turns on the field remain constant, the flux is a function of the field current.  It is not directly proportional to the field current because of the varying permea­bility of the magnetic circuit.

Figure 7.2 shows the relation existing between the field am­pere-turns and the flux per pole.  The flux does not start at zero ordinarily but at some value slightly greater,   owing to the residual magnetism in the machine.   At first the line is prac­tically straight, as most of the reluctance   of    the    magnetic circuit   is   in   the   air-gap.    At the point ‘q’, the iron begins to be saturated arid the curve falls away from the straight line.

The number of field ampere-turns for the air-gap and for the iron can be approximately determined for any point on the curve.

Let it be required to determine the ampere-turns for the gap and for the iron at the point c. From the origin draw ob tan­gent to the saturation curve and also draw the horizontal line ac. The line ob is the magnetization curve of the air gap, if the reluctance of the iron at low saturation be neglected. Therefore, the ampere-turns required by the gap are equal to ab and those required by the iron are equal to bc.

The induced voltage is proportional to the flux, if the speed is maintained constant. Two curves as shown in Fig. 7. 3 below can so be plotted for 1,200 r.p.m. and the other for 900 r.p.m.   The curves are similar, any ordinate of the lower curve being 900/1,200 of the value of the correspond­ing ordinate of the upper curve. Thus, at ordinate ‘ac’

ab/ac = 900/1200 Also at a’b’/a’c’ = 900/1200

If the saturation curve of a generator for one speed is available, saturation curves for other speeds may be readily found by the method.

 

Fig. 7. 3 Saturation   curves different speeds.

Determination of saturation curve.

 

To determine, the saturation curve experimentally, connect the field, in series with an ammeter, across a direct source of power. A voltmeter is connected across the armature terminals under no load.  The readings of voltage for each field current is plotted as the saturation curve. As the voltage drop within the armature due to no load current is negligible, the terminal volts and the induced volts under these conditions are identical.

Fig. 7. 4 Field resistance lines.

Field Resistance Line: If the current in a resistance be plotted against volts, a straight line passing through the origin results. For example, if the resistance of a field circuit be 50 ohms, the current will be 2 amperes when the voltage is 100 volts; 1.5 amperes when the voltage is 75 volts, and 1 ampere when the voltage is 50 volts. This relation is shown in the Fig.7.4 below for some values of resistances. It may be noted that the higher the resistance the greater the slope of the resistance line. The slope of the line is equal to the field resistance in ohms.