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a triangle similar to pqr, such that pq is proportional to I.R. and qr is proportional to Ir(ARl, and fitting this triangle between the magnetization curve and the field resistance line. EXAMPLE 4.4 The de machine in Example 4.2 is operated as a self-excited generator. (a) The no-load terminal voltage is adjusted to 100 V. Determine the full- load terminal voltage. Neglect armature reaction effects. (b) Repeat (a), assuming that the effect of armature reaction at full load is equivalent to 0.06 field amperes, that is, Ir(AR) = 0.06 A. (c) Determine the maximum value of the armature current that the genera- tor can supply and the corresponding value of the terminal voltage. Assume that Ir(ARJ is proportional to I \u2022. (d) Determine the short-circuit current of the generator. Solution (a) Draw the field resistance line Op such that it intersects the magnetiza- tion curve at 100 V (Fig. E4.4a). IaiFL = 120 A I.R. = 120 X 0.1 = 12 V = pq Fit I.R. = 12 V = a'b' between the magnetization curve and the field resistance line (Fig. E4.4). V, = 80V (b) Construct the triangle pqr with pq = 12 V and qr = 0.06 A, and fit this triangle as abc between the magnetization curve and the field resistance line (Fig. E4.4). v, = 75 v (c) Draw a tangent CR parallel to Op. Extend the triangle pqr to touch the tangent at R. Draw AC parallel to pR and construct the triangle pQR =ABC. Note that pQ = AB represents I.R., QR = BC represents Ir(ARJ, and triangle ABC is the largest triangle that will fit between the 160 chapter 4 DC Machines 120 100 .l!l 80 ~ l:tl"'Go > > U") 0 40 ..... 00 II :::> :::> 20 11 \u2022 amps FIGURE E4.4 magnetization curve and the field resistance line. I.R. = AB = 17 V 17 I.= 0_1 = 170A vt =53 v (d) With the generator terminals short-circuited, V1 = 0 and so Ir = 0 (Fig. E4.3a). The generated voltage is due to residual magnetism and E.= Er = 6 V I.R. = 6 V 6 I.= 0_1 = 60 A Note that because Ir = 0, the machine operates at a low flux level in the linear region of the magnetization curve and so there will be no demagnetiz- ing effect due to armature reaction. \u2022 4.3.3 COMPOUND DC MACHINES Many practical applications require that the terminal voltage remains con- stant when load changes. But when de machines deliver current, the terminal voltage drops because of I.R. voltage drop and a decrease in pole fluxes caused by armature reaction. > DC Generators 161 Series and shunt field wdgs. FIGURE 4.40 Compound de machine. To overcome the effects of I.R. drop and decrease of pole fluxes with armature current, a winding can be mounted on the field poles along with the shunt field winding. This additional winding, known as a series winding, is connected in series with the armature winding and carries the armature current. This series winding may provide additional ampere-turns to in- crease or decrease pole fluxes, as desired. A de machine that has both shunt and series windings is known as a compound de machine. A schematic diagram of the compound machine is shown in Fig. 4.40. Note that in a compound machine the shunt field winding is the main field winding, provid- ing the major portion of the mmf in the machine. It has many turns of smaller cross-sectional area and carries a lower value of current compared to the armature current. The series winding has fewer turns, larger cross- sectional area, and carries the armature current. It provides mmf primarily to compensate the voltage drops caused by I.R. and armature reaction. Figure 4.41 shows the two connections for the compound de machine. In the short-shunt connection the shunt field winding is connected across the armature, whereas in the long-shunt connection the shunt field winding is (a) (b) FIGURE 4.41 Equivalent circuits of compound de machines. (a) Short shunt. (b) Long shunt. + r.: ..\u2022 ... , .\u2022 ·~ 162 chapter 4 DC Machines connected across the series combination of armature and series winding. The equations that govern the steady-state performance are as follows. Short Shunt V, = Ea - laRa - J,Rsr I,= Ia- Ir where Rr is the resistance of the series field windings. Long Shunt V, = E. - IaCRa + Rsr) I,= Ia- Ir v, Ir = ---'-- Rrw + Rrc (4.26) (4.27) (4.28) (4.29) (4.30) For either connection, assuming magnetic linearity, the generated voltage is (4.31) where <l>,h is the flux per pole produced by the mmf of the shunt field winding <I>sr is the flux per pole produced by the mmf of the series field winding When these two fluxes aid each other the machine is called a cumulative compound machine, and when they oppose each other the machine is called a differential compound machine. Note that both shunt field mmf and series field mmf act on the same magnetic circuit. Therefore, the total effective mmf per pole is (4.32) (4.33) where Nr is the number of turns per pole of the shunt field winding Nsr is the number of turns per pole of the series field winding FAR is the mmf of the armature reaction From Eq. 4.33, Nsr Ir(eff) = Ir :!::: Nr lsr - Ir(AR) (4.34) The voltage-current characteristics of the compound de generator are shown in Fig. 4.42. With increasing armature current the terminal voltage may rise (overcompounding), decrease (undercompounding), or remain es- sentially flat (flat compounding). This depends on the degree of compound- p DC Generators 163 Flat compound Undercompound Differential compound L---L---------L---------------------~~ O la(rated) FIGURE 4.42 V-I characteristics of compound de gen- erators. ing, that is, the number of turns of the series field winding. For differential compounding (i.e., mmf of the series field winding opposed to that of the shunt field winding) the terminal voltage drops very quickly with increasing armature current. In fact, the armature current remains essentially constant. This current-limiting feature of the differentially compounded de generator makes it useful as a welding generator. EXAMPLE 4.5 The de machine in Example 4.2 is provided with a series winding so that it can operate as a compound de machine. The machine is required to provide a terminal voltage of 100 Vat no load as well as at full load (i.e., zero voltage regulation) by cumulatively compounding the generator. If the shunt field winding has 1200 turns per pole, how many series turns per pole are required to obtain zero voltage regulation. Assume a short-shunt connection and that the series winding has a resistance R,r = 0.01 D. Solution VtiNL = 100 v From Example 4.3(b), Rr = 100 D. Now, from Fig. 4.41a, I.= Ir +It 120 = Ir +It (E.4.5a) r.: ~'"" r· .,,, ;::. 164 chapter 4 DC Machines Also, from Fig. 4.41a IrRr = ItRsr + Vt Vt + ItRsr I r = ___:___:___:.:_ Rr 100 X It X 0.01 Ir = 100 From Eqs. E4.5a and E4.5b From Fig. 4.41a, I 1 = LOlA It= 118.99 A = 100 + 118.99 X 0.01 + 120 X 0.1 = 113.2V (E.4.5b) From the magnetization curve (Example 4.2 and Fig. 4.24) the shunt field current required to generate Ea = 113.2 Vis 1.45 A(= Ir(eff)). From Eq. 4.34 Nsr 1.45 = 1.01 + 1200 X 118.99- 0.06 Nsr = 5.04 turns per pole \u2022 4.3.4 SERIES GENERA TOR The circuit diagram of a series generator is shown in Fig. 4.43. The series field winding provides the flux in the machine when the armature current flows through it. Note that the field circuit is not complete unless a load is + FIGURE 4.43 Equivalent circuit of a de series generator. F I DC Generators 165 FIGURE 4.44 Magnetization curve (E. versus I.) and I.(R. + R,) versus I \u2022. connected to the machine. The equations governing the steady-state opera- tion are E. = V, + I.(R. + R,r) /,=I. (4.35) (4.36) The magnetization curve E. versus I. (Fig. 4.44) for the series machine can be obtained by separately exciting the series field. To obtain the terminal voltage-current characteristic (i.e., V1 versus /,), draw

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