Question 11.3: A 500-V, 100-hp, 2500 r/min, separately excited dc motor has......

From Eq. 11.4

T_{\mathrm{mech}}=\frac{E_aI_a}{ω_m}=K_{\mathrm{f}}I_{\mathrm{f}}I_a \quad \quad \quad(11.4)

I_{\mathrm{a}}=\frac{T_{\mathrm{mech}}}{K_{\mathrm{f}} I_{\mathrm{f}}}

and from Eq. 11.5

I_{\mathrm{a}}=\frac{(V_{\mathrm{a}}-E_{\mathrm{a}})}{R_{\mathrm{a}}}\quad \quad \quad (11.5)

I_{\mathrm{a}}=\frac{V_{\mathrm{a}}-E_{\mathrm{a}}}{R_{\mathrm{a}}}=\frac{V_{\mathrm{a}}-K_{\mathrm{f}} I_{\mathrm{f}} \omega_{\mathrm{m}}}{R_{\mathrm{a}}}

Hence we can solve for ω_m

\omega_{\mathrm{m}}=\frac{V_{\mathrm{a}}-\left(\frac{T_{\text {mech}} R_{\mathrm{a}}}{K_{\mathrm{f}} I_{\mathrm{f}}}\right)}{K_{\mathrm{f}} I_{\mathrm{f}}}

and the speed in r/min as

n=\left(\frac{30}{\pi}\right) \omega_{\mathrm{m}}

Finally, the field current is

I_{\mathrm{f}}=\frac{V_{\mathrm{f}}}{R_{\mathrm{f}}}=\frac{300}{109}=2.75 \mathrm{~A}

and the rated full-load torque is given by

T_{\text {rated }}=\frac{P_{\text {rated }}}{\left(\omega_{\mathrm{m}}\right)_{\text {rated }}}=\frac{100 \times 746}{2500 \times\left(\frac{\pi}{30}\right)}=285 \mathrm{~N} \cdot \mathrm{m}

Figure 11.5 is the desired plot. Notice that the speed drops approximately 63 r/min as the torque is increased from zero to full-load, independent of the armature voltage and machine speed.

Here is the MATLAB script: