Problem 104-4    Source:   Adapted from Problem 8.22 and Problem 8.23 ​​- page 560 - CHAPMAN, Sthephen J. -    Book: Fundamentals of Electric Machines - 5th Edition - Ed. McGraw-Hill - 2013.

The magnetization curve of an independently excited DC generator is given by the equation E_A = - 0.0000035 F² + 0.043 F + 4. The nominal specifications of the generator are 6 kW, 120 V, 50 A and 1,800 rpm. The Figure 104-04 shows the circuit to consider. The nominal current of the field circuit is 5 A . The following machine data is known:

R_A = 0.18   Ω     R_F = 20   Ω     R_aj = 0   to   40   Ω

V_F = 120 V     N_F = 1,000   turns per pole

Assuming there is no armature reaction, answer the questions below:

a) If this generator is running at no load, what is the range of voltage adjustments that can be obtained by varying R_aj ?

b) If the generator armature current is 50 A, the generator speed is 1,700 rpm and the voltage terminal is 106 V, what will be the field current that must be circulating in the generator?

Solution of the Problem 104-4   

Item a

As the generator is working at no load, that is, without load, then the armature current is equal to zero. Thus, there is no voltage drop across the armature resistance. So we can say that V_g = E_A.

Let's calculate the field current for R_aj = 0, using the equation shown in the circuit of Figure 104-04.

And when R_aj = 40   Ω, the field current will be:

Since we know the number of turns in the field winding and the current flowing through it, we can calculate the magnetomotive force generated by that winding. For the case of I_F = 2 A we have:

And for the case of I_F = 6 A we have:

Since the magnetization curve equation was provided, we can calculate the induced voltage. When I_F = 2 A we have F = 2,000 A.e. Substituting this value in the magnetization equation, we get:

And when I_F = 6 A we have F = 6,000 A.e. Substituting this value in the magnetization equation, we get:

Therefore, the variation of the induced voltage when the rheostat changes from 0 to 40 Ω is between 76 V and 136 V.

Item b

To answer this item, let's calculate the induced voltage, knowing that V_g = 106 V. Using the eq. 104-02 we have:

Note carefully that this value of E_A is for a speed of 1,700 rpm. However, the magnetization equation is for a speed of 1,800 rpm. Therefore, we must find a new induced voltage for 1,800 rpm. This is possible using the eq. 103-21.

Now using this value we can use the magnetization equation and find an easy solution quadratic equation to calculate the magnetomotive force, that is:

The roots of this equation are:

Note that the value of F₂ extrapolates the usual values ​​in electric machines. Then we must disregard it. Thus, we chose the value of F₁. Therefore, the value of I_F is: