Question 16.5: A series-connected dc motor runs at nm1 = 1200 rpm while dri...
A series-connected dc motor runs at n_{m1} = 1200 rpm while driving a load that demands a torque of 12 Nm. Neglect the resistances, rotational loss, and saturation effects. Find the power output. Then, find the new speed and output power if the load torque increases to 24 Nm.
Since we are neglecting losses, the output torque and power are equal to the developed torque and power, respectively. First, the angular speed is
ω_{m1} = n_{m1} × \frac{2π}{ 60} = 125.7 rad/s
and the output power is
P_{dev1} = P_{out1} = ω_{m1}T_{out1} = 1508 W
Setting R_A = R_F = 0 in Equation 16.34 gives
T_{dev} = \frac{KK_FV^2_T} {(R_A + R_F + KK_Fω_m)^2} (16.34)
T_{dev} = \frac{KK_FV^2_T} {(R_A + R_F + KK_Fω_m)^2} = \frac{V^2_T} {KK_Fω^2_m}
Thus, for a fixed supply voltage V_T, torque is inversely proportional to speed squared, and we can write
\frac{T_{dev1}} {T_{dev2}} = \frac{ω^2_{m2}} {ω^2_{m1}}
Solving for ω_{m2} and substituting values, we have
ω_{m2 }= ω_{m1} \sqrt{\frac{T_{dev1}} {T_{dev2}}} = 125.7\sqrt {\frac{12} {24}} = 88.88 rad/s
which corresponds to
n_{m2} = 848.5 rpm
Finally, the output power with the heavier load is
P_{out2} = T_{dev2}ω_{m2} = 2133 W