Problem 104-5    Source:  Problem prepared by the website author.

A DC generator in the independent configuration of 80 kw and 6 poles, has 56 slots with 14 conductors in each slot. The armature winding is connected in 6 parallel paths. Since the flux per pole equal to 0.03 Wb, R_A = 0.1 Ω and the machine running at 800 rpm at full load, determine:

a) the induced voltage at full load;;

b) the armature current;

c) the terminal voltage of the generator.

Solution of the Problem 104-5   

Item a

We will first calculate the number of conductors, Z.

Knowing that a = 6, P = 6 and Z = 784, we can calculate the constant K_a through eq. 102-14, or:

Now we can calculate the value of E_A using eq. 102-13.

Item b

The power of the generator is given by P_g = V_g I_A. From this relationship we have V_g = P_g / I_A. On the other hand, the induced voltage is given by eq. 104-02. Replacing this relation in eq. 104-02, we get:

By working this relation algebraically, we will obtain an equation of the second degree with an easy solution. Thus, we get:

Substituting for numerical values, we find:

The roots of this equation are:

Note that the value of I_A = 2,841 A   extrapolates the power generated by the machine. Then we must disregard it. So, we choose the value below.

Item c

To find the terminal voltage of the generator we will use the power ratio, that is: