Question 16.11: The no-load characteristics of a shunt DC machine at the spe......

Based on the information given in the problem, we know that:

V_{t,1}=240\ V        (1)

n_{NL,1}=2000\ rpm        (2)

V_{t,2}=180\ V        (3)

Moreover, the no-load characteristics of the machine at the speed of 1000 rpm are as follows:

(Table 1)

The no-load characteristics of the machine at the speed of 2000 rpm can be updated as follows:

(Table 2)

Now, we can extract the field winding current from the no-load characteristics of the machine at the speed of 2000 rpm as follows:

\hat{E}_{a,1}|_{2000\ rpm}\approx V_{t,1}=240\ V\Rightarrow I_{f,1}=4\ A        (4)

Figure 16.11 shows the electrical circuit of a shunt DC motor. Applying Ohm’s law for the branch of field winding (the first condition):

R_f=\frac{V_{t,1}}{I_{f,1}}=\frac{240}{4}=60\ \Omega        (5)

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

I_{f,2}=\frac{V_{t,2}}{R_f}=\frac{180}{60}=3\ A        (6)

Now, we can extract the emf of the motor from its no-load characteristics at the speed of 2000 rpm as follows:

I_{f,2}=3  A\ \Rightarrow \hat{E}_{a,2}|_{2000\ rpm}=220\ V        (7)

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_{1}ω_{1}}{φ_{2}ω_{2}}=\frac{φ_{1}n_{1}}{φ_{2}n_{2}}        (8)

\Rightarrow \frac{\hat{E}_{a,1}}{\hat{E}_{a,2}}|_{2000\ rpm}=\frac{φ_1}{φ_2}\Rightarrow \frac{φ_1}{φ_2}=\frac{240}{220}        (9)

Now, for both no-load conditions, we have the relation below:

\frac{V_{t,1}}{V_{t,2}}\approx\frac{E_{a,1}}{E_{a,2}}=\frac{φ_1}{φ_2}\frac{n_{NL,1}}{n_{NL,2}}        (10)

\Rightarrow \frac{180}{240}=\frac{240}{220}\times \frac{2000}{n_{NL,2}}\Rightarrow \pmb{n_{NL,2}\approx 1636\ rpm}

Choice (3) is the answer.