A 100 kW, 200 V, 1000-rpm dc shunt generator has Ra = 0.03 Ω . The generator is driven at rated speed and excitation is such that it gives a rated voltage of 200 V at no-load. The data for the magnetization characteristic of the generator are as follows: If^* (A) 0 1 2.2 3.3 4.5 7.1 EMF(V) 10 32.5

Question

A 100 kW, 200 V, 1000-rpm dc shunt generator has [latex] R_{a} = 0.03 \Omega[/latex]. The generator is driven at rated speed and excitation is such that it gives a rated voltage of 200 V at no-load. The data for the magnetization characteristic of the generator are as follows:

[latex] I_{f}^{*} (A)[/latex]	0	1	2.2	3.3	4.5	7.1
EMF (V)	10	32.5	100	167.5	200	225

Determine the voltage appearing across the generator terminals when [latex] I_{a} = 500 A[/latex]

A series field of 5 turns per pole having a total resistance of [latex] 0.005 \Omega[/latex] is to be added in a long-shunt compound. There are 1200 turns per pole in the shunt field. The generator is to be level-compounded so that the full-load voltage is 200 V when the resistance in the shut field circuit is adjusted to give a no load voltage of 200V. Find the value of the series field diverter resistance to obtain the desired performance.

Assume that the armature reaction AT has been compensated for.

Accepted Answer

The magnetization characteristic is drawn in Fig. 7.58. Drawing the [latex] R_{f} –line[/latex] corresponding to 200 V.

[latex] R_{f}= \frac{200}{4.5}=44.4 \Omega[/latex]

[latex] I_{a}=500A[/latex]

[latex] I_{a}R_{a}=500 imes 0.03=15V[/latex]

Drawing a line parallel to the [latex] R_{f} –line[/latex] at [latex] 15 V (= Q\acute{Q})[/latex]

V(terminal) = 165 V (corresponding to [latex] \acute{Q}[/latex])

With series winding added in the long-shunt compound (cumulative):

Armature circuit resistance [latex] = 0.03 + 0.005 = 0.035 \Omega[/latex]

Armature circuit voltage drop [latex] = 500 imes 0.035 = 17.5 V[/latex]

To compensate for this voltage drop, the operating point on OCC must lie at R, i.e. (200 + 17.5) = 217.5 V. Both full-load and no-load voltage will now be 200 V. From Fig. 7.58 the series field current measured in terms of shunt of field current is

[latex] I_{f,se}=1.75A[/latex]

[latex] AT_{se}= I_{f,se} imes 1200=1.75 imes 1200=2100[/latex]

[latex] I_{se}=\frac{2100}{5} =420A[/latex]

[latex] I_{diverter}=500 – 420 = 80 A[/latex]

[latex] R_{diverter} = 0.005 imes \frac{420}{80}= 0.0265 \Omega[/latex]

Question 7.18: A 100 kW, 200 V, 1000-rpm dc shunt generator has Ra = 0.03 Ω...

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