Question 16.8: A 5 kW, 200 V, shunt DC generator that has the armature wind......

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

P_g=P_m=5\ kW        (1)

V_{t,g}=V_{t,m}=200\ V        (2)

R_{a}=0.1\ \Omega        (3)

R_{f}=200\ \Omega        (4)

n_g=1000\ rpm        (5)

Figure 16.8.a and Fig. 16.8.b show the electrical circuit of a shunt DC machine in generating and motoring operations, respectively.

Applying Ohm’s law for the branch of field winding in Fig. 16.8.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        (6)

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{5000}{200}=25\ A        (7)

Applying KCL at the top node of the circuit of Fig. 16.8.a:

I_{a,g}=I_{t,g}+I_{f,g}=25+1=26\ A        (8)

Applying KVL in the right-hand side mesh of the circuit of Fig. 16.8.a:

E_{a,g}=V_{t,g}+R_aI_{a,g}=200+(0.1\times 26)=202.6\ V        (9)

Applying KCL at the top node of the circuit of Fig. 16.8.b:

I_{a,m}=I_{t,m}-I_{f,m}=25-1=25\ A        (10)

Applying KVL in the right-hand side mesh of the circuit of Fig. 16.8.b:

E_{a,m}=V_{t,m}-R_{a}I_{a,m}=200-(0.1\times 24)=197.6\ 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,g}}{E_{a,m}}=\frac{φ_{g}ω_{g}}{φ_{m}ω_{m}}=\frac{φ_{g}n_{g}}{φ_{m}n_{m}}        (12)

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

φ_g=φ_m        (13)

Therefore:

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

\Rightarrow \frac{202.6}{197.6}=\frac{1000}{n_m}\Rightarrow \pmb{n_m\approx 975\ rpm}

Choice (2) is the answer.