Question 7.16: A 100 kW, 200 V, long-shunt commutative compound generator h...

A 100 kW, 200 V, long-shunt commutative compound generator has equivalent armature resistance of 0.03 \Omega and a series field resistance of 0.004 \Omega . There are 1200 shunt turns per pole and 5 series turn per pole. The data of its magnetization characteristic at 1000 rpm is given below:

I_{f}(A)

0 1 2.2 3.3 4.2 5.3 7.1
E_{a}(V) 11 33 105 150 175 200

225

 

(a) Calculate the terminal voltage and rated output current for shunt field current of 5 A and a speed of 950 rpm. Neglect armature reaction effect.

(b) Calculate the number of series turns per pole for the generator to level compound with same shunt field current, series field resistance and generation speed to be the same as in part (a). The demagnetization due to armature reaction in equivalent shunt field current is 0.001875 I_{a}  where I_{a}  is the armature current.

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The schematic diagram of long-shunt compound generator is drawn in Fig. 7.56(a)

Line current, I_{L} = \frac{100\times 10^{3}}{200} =500A

Shunt field current, I_{f} = 5 A  (given)

From Fig. 7.56(a)

I_{a}=500+5=505A

 

I_{f,eq}=I_{f}+\frac{N_{se}}{N_{f}} I_{se}

 

= 5 + \frac{5}{1200} \times 505 = 7.1 A

From the magnetization characteristic of Fig. 7.56(b) (as per data given)

(a)                          E_{a} = 225 V at 1000 rpm

E_{a} (950 rpm) = 225 \times \frac{950}{1000} = 213.75 V

From Fig. 7.56(a)

V_{t}=E_{a}-I_{a} (R_{a}+R_{se})= 213.75 – 505 (0.03 + 0.004)

The generator is under – compounded (armature reaction ignored)

(b)                           I_{fd}=0.001875 \times 505 = 0.95 A (demag)

For level compound

V_{t}=200V

 

E_{a}= 200 + 505 (0.03 + 0.04) = 217.17 (950 rpm)

From the magnetization characteristic

I_{f}(net) =7.5A

Excitation balance equation

I_{f}+\frac{N_{se}}{N_{f}} I_{a}-I_{fd}=I_{f}(net)

 

5 + 505 \left(\frac{N_{se}}{1000} \right) – 0.95 = 7.5

which gives                        N_{se} = 6.87 or 7 turns

7 16 1
7 16 2

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