Problem 103-8    Source:   Example 4.1 - page 135 - SEN, P.C. -    Book: Principles of Electric Machines and Power Electronics - 3rd Edition - Ed. Wiley - 2014.

A 4 pole DC machine has an armature with a radius equal to 125 mm and an effective length of 250 mm. The poles cover 75% of the periphery of the armature. The armature winding has 33 coils and each coil has 7 turns . Coils are accommodated in 33 slots. The magnetic flux density, B, at each pole is 0.75 T. Considering the armature winding as overlapping, determine:

a) the constant, K_a of the machine.

b) the induced armature voltage, E_A, when the rotor turns at 1,000 rpm.

c) the power, P_m, developed by the machine when the armature current is 400 A .

d) the current in the coils and the electromagnetic torque developed.

Solution of the Problem 103-8   

Item a

To find the value of K_a of the machine we need to calculate the number of conductors, Z, and the number of paths, a. For a lap winding we must use the eq. 102-04 and eq. 102-08, as below:

Replacing the variables by their respective values, we find:

Replacing the variables by their respective values ​​and considering a simplex machine, where m = 1, we have:

With these calculations we can calculate the machine constant K_a using eq. 102-14 shown below.

So, replacing the variables by their respective values, we get:

Item b

To calculate the induced voltage in the armature, E_A, we will use eq. 102-13, shown below.

However, we do not know the value of the magnetic flux. For its calculation we must determine the value of the pole area. With the given data, we get:

So the flow is:

Therefore, replacing the variables by their respective values ​​in eq. 102-13, we get:

Item c

The power in the machine shaft is given by eq. 103-17, repeated below.

Therefore, replacing the variables by their respective values ​​in eq. 103-17, we get:

Item d

The current in each coil is given by:

And to calculate the torque developed by the machine we use the eq. 103-16, shown below.

Using basic algebra in this equation, we find:

Where do we use the fact that ω = 2 π x n /60 = 104.72   rad/s.