Question 7.9: A permanent-magnet dc motor is known to have an armature res......
A permanent-magnet dc motor is known to have an armature resistance of 1.03 Ω. When operated at no load from a dc source of 50 V, it is observed to operate at a speed of 2100 r/min and to draw a current of 1.25 A. Find (a) the torque constant K_m, (b) the no-load rotational losses of the motor and (c) the power output of the motor when it is operating at 1700 r/min from a 48-V source.
a. From the equivalent circuit of Fig. 7.20, the generated voltage E_a can be found as
\begin{aligned} E_a &= V_t – I_aR_a \\& = 50 – 1.25 × 1.03 = 48.7 V\end{aligned}
At a speed of 2100 r/min,
\begin{aligned} ω_m &= \left( \frac{2100 ~r}{min} \right) × \left( \frac{2π ~rad}{r} \right) × \left( \frac{1 ~min}{60 ~s} \right) \\&= 220 ~rad/sec \end{aligned}
Therefore, from Eq. 7.26,
E_a=K_m ω_m \quad \quad \quad (7.26)
K_m = \frac{E_a}{ω_m} = \frac{48.7}{220}=0.22 ~V/(rad/sec)
b. At no load, all the power supplied to the generated voltage E_a is used to supply rotational losses. Therefore
\text{Rotational losses}= E_aI_a = 48.7 × 1.25 = 61 ~W
c. At 1700 r/min,
ω_m = 1700 \left( \frac{2π}{60} \right)= 178 ~rad/sec
and
E_a = K_m ω_m = 0.22 × 178 = 39.2 ~V
The input current can now be found as
I_a=\frac{V_t – E_a}{R_a}=\frac{48 – 39.2}{1.03}= 8.54 ~A
The electromagnetic power can be calculated as
P_{mech} =E_aI_a = 39.2 × 8.54 = 335 ~W
Assuming the rotational losses to be constant at their no-load value (certainly an approximation), the output shaft power can be calculated:
P_{shaft} =P_{mech} – \text{rotational losses} = 274 ~W
