Question 16.13: A 200 V, 12 A, 1940 rpm, shunt DC motor has the armature win......

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

V_{t,1}=200 \ V        (1)

I_{t,1}=12\ A        (2)

n_1=1940\ rpm        (3)

R_{a}=0.5\ \Omega        (4)

V_{t,2}=150\ V        (5)

T_{load,2}=T_{load,1}        (6)

φ\propto I_f        (7)

I_f\approx 0        (8)

Since I_f\approx 0, we have:

I_a=I_t        (9)

Figure 16.13 shows the electrical circuit of a shunt DC motor. Applying KVL in the right-hand side mesh of the circuit:

E_{a,1}=V_{t,1}-R_aI_{a,1}=200-(0.5\times 12)=194\ V        (10)

From (6), we can conclude:

T_{e,2}=T_{e,1}        (11)

As we know, the electromagnetic torque (T_e) of an electric machine can be determined by using T_e=k_aφI_a. Therefore, from (11), we can write:

k_aφ_2I_{a,2}=k_aφ_1I_{a,1}\Rightarrow φ_2I_{a,2}=φ_1I_{a,1}\overset{(7)}{\Longrightarrow } I_{f,2}I_{a,2}=I_{f,1}I_{a,1}        (12)

Applying Ohm’s law for the branch of field winding for the first condition:

I_f=\frac{V_t}{R_{fc}  +  R_{fw}}=\frac{V_t}{R_f}        (13)

Solving (9), (12), and (13):

\frac{V_{t,1}}{R_f}I_{a,1}=\frac{V_{t,2}}{R_f}I_{a,2}\Rightarrow I_{a,2}=\frac{V_{t,1}}{V_{t,2}}I_{a,1}=\frac{200}{150}\times 12=16\ A        (14)

Applying Ohm’s law for the branch of field winding for the second condition:

E_{a,2}=V_{t,2}-R_aI_{a,2}=150-(0.5\times 16)=142\ V        (15)

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,1}}{E_{a,2}}=\frac{φ_1ω_1}{φ_2ω_2}=\frac{φ_1n_1}{φ_2n_2}        (16)

Solving (7) and (13):

φ\propto V_t        (17)

Solving (16) and (17):

\frac{E_{a,1}}{E_{a,2}}=\frac{V_{t,1}n_1}{V_{t,2}n_2}        (18)

\Rightarrow \frac{194}{142}=\frac{200}{150}\times \frac{1940}{n_2}\Rightarrow \pmb{n_2\approx 1893\ rpm}

Choice (1) is the answer.