Question 16.4: A 50-hp shunt-connected dc motor has the magnetization curve...

The equivalent circuit is shown in Figure 16.20.The field current is given by
I_F = \frac{V_T}{ R_F + R_{adj}} = \frac{240} {10 + 14} = 10 A
Next, we use the magnetization curve to find the machine constant K \phi for this value of field current. From the curve shown in Figure 16.19, we see that the induced armature voltage is E_A = 280 V at I_F = 10 A and n_m = 1200 rpm. Thus, rearranging Equation 16.15 and substituting values, we find that the machine constant is

E_A = K \phi ω_m       (16.15)
K \phi = \frac{E_A} {ω_m} = \frac{280} {1200(2π/60)} = 2.228

We assume that the rotational power loss is proportional to speed. This is equivalent to assuming constant torque for the rotational loss. The rotational-loss torque is
T_{rot} = \frac{P_{rot}}{ ω_m} = \frac{1450}{ 1200(2π/60)} = 11.54 Nm
Thus, the developed torque is
T_{dev} = T_{out} + T_{rot} = 250 + 11.54 = 261.5 Nm
Now, we use Equation 16.16 to find the armature current:

T_{dev} = K \phi I_A         (16.16)
I_A = \frac{T_{dev}} {K \phi }= \frac{261.5}{ 2.228} = 117.4 A
Then, applying Kirchhoff’s voltage law to the armature circuit, we have
E_A = V_T − R_AI_A = 240 − 0.065(117.4) = 232.4 V
Solving Equation 16.15 for speed and substituting values, we get
ω_m = \frac{E_{A}} {K \phi } = \frac{232.4 }{2.228} = 104.3 rad/s
or
n_m = ω_m\left(\frac{ 60} {2π}\right) = 996.0 rpm
To find efficiency, we first compute the output power and the input power,
given by
P_{out} = T_{out}ω_m = 250(104.3) = 26.08 kW
P_{in} =V_TI_L = V_T(I_F + I_A) = 240(10 + 117.4) = 30.58 kW
η= \frac{P_{out}} {P_{in}} × 100\% = \frac{26.08} {30.58} × 100% = 85.3%