Question 16.9: In a 10 kW, 200 V, shunt DC motor, the armature winding resi......

Based on the information given in the problem, we have:

P_g=P_m=10\ kW        (1)

V_{t,g}=V_{t,m}=200\ V        (2)

R_{a}=0.1\ \Omega        (3)

R_{f}=200\ \Omega        (4)

n_g=713\ rpm        (5)

Figure 16.9.a and Fig. 16.9.b show the electrical circuit of a shunt DC machine in generating and motoring operations, respectively.

The rated terminal current of the machine can be calculated as follows:

I_{t,g}=I_{t,m}=\frac{P_g}{V_{t,g}}=\frac{10000}{200}=50\ A        (6)

Applying Ohm’s law for the branch of field winding in Fig. 16.9.a:

I_{f,g}=I_{f,m}=\frac{V_{t,g}}{R_{f,c}  +  R_{fw}}=\frac{V_{t,g}}{R_f}=\frac{200}{200}=1\ A        (7)

Applying KCL at the top node of the circuit of Fig. 16.9.a:

I_{a,g}=I_{t,g}+I_{f,g}=50+1=51\ A        (8)

Applying KVL in the right-hand side mesh of the circuit of Fig. 16.9.a:

E_{a,g}=V_{t,g}+R_{a}I_{a,g}=200+(0.1\times 51)\simeq 205\ V        (9)

Applying KCL at the top node of the circuit of Fig. 16.9.b:

I_{a,m}=I_{t,m}-I_{f,m}=50-1=49\ A        (10)

Applying KVL in the right-hand side mesh of the circuit of Fig. 16.9.b:

E_{a,m}=V_{t,m}-R_{a}I_{a,m}=200-(0.1\times 49)\simeq 195\ V        (11)

As we know, the induced armature voltage of an electric machine can be determined by using E_a = k_aφω. Therefore, we can write:

\frac{E_{a,m}}{E_{a,g}}=\frac{φ_{m}ω_{m}}{φ_{g}ω_{g}}=\frac{φ_{m}n_{m}}{φ_{g}n_{g}}        (12)

Since I_{f,g}=I_{f,m}, we have:

φ_g=φ_m        (13)

Therefore:

\frac{E_{a,m}}{E_{a,g}}=\frac{n_m}{n_g}        (14)

\Rightarrow \frac{195}{205}=\frac{713}{n_m}\Rightarrow \pmb{n_m= 750\ rpm}

Choice (4) is the answer.