Question 16.16: A shunt DC motor that has been connected to a variable mecha......

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

V_t=200\ V        (1)

\omega_A=157\ rad/sec        (2)

I_{t,A}=20\ A        (3)

\omega_B=167.5\ rad/sec        (4)

I_{t,B}=18\ A        (5)

I_f\approx 0\ A        (6)

From (6), we have:

I_a=I_t        (7)

Method 1: 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,A}}{E_{a,B}}=\frac{φ_Aω_A}{φ_Bω_B}        (8)

Since V_{t,B}=V_{t,A}, we have:

I_{f,B}=I_{f,A}\Rightarrow φ_B=φ_A        (9)

Therefore:

\frac{E_{a,A}}{E_{a,B}}=\frac{\omega_A}{\omega_B}        (10)

Figure 16.16 shows the electrical circuit of a shunt DC motor. Applying KVL in the right-hand side mesh of the circuit:

E_a=V_t-R_aI_a        (11)

Solving (10) and (11):

\frac{V_t-R_aI_{a,A}}{V_t-R_aI_{a,B}}=\frac{\omega_A}{\omega_B}        (12)

\Rightarrow \frac{200-20R_a}{200-18R_a}=\frac{157}{167.5}\Rightarrow R_a\simeq 4\ \Omega        (13)

The air gap power, when the motor is under load A, can be calculated as follows:

P_{ag,A}=P_{in,A}-P_{cu,A}=V_tI_{t,A}-R_aI_{a,A^2}        (14)

\Rightarrow P_{ag,A}=(200\times 20)-(4\times 20^2)=2400\ W        (15)

The developed electromagnetic torque, when the motor is under load A, can be calculated as follows:

\pmb{T_{e,A}}=\frac{P_{ag,A}}{\omega_A}=\frac{2400}{157}\approx \pmb{15.2\ N.m}        (16)

The air gap power, when the motor is under load B, can be calculated as follows:

P_{ag,B}=P_{in,B}-P_{cu,B}=V_tI_{t,B}-R_aI_{a,B^2}        (17)

\Rightarrow P_{ag,B}=(200\times 18)-(4\times 18^2)=2304\ W        (18)

The developed electromagnetic torque, when the motor is under load B, can be calculated as follows:

\pmb{T_{e,B}}=\frac{P_{ag,B}}{\omega_B}=\frac{2304}{167.5}\approx \pmb{13.7\ N.m}        (19)

Method 2: As we know, the induced armature voltage of an electric machine can be determined by using E_a=k_aφω=k^\prime_aω. Applying KVL in the right-hand side mesh of the circuit of Fig. 16.16:

V_t=E_a+R_aI_a=k^\prime_a\omega+R_aI_a        (20)

\Rightarrow \begin{cases} V_t=k^\prime_a\times \omega_A+R_aI_{a,A} \\ V_t=k^\prime_a\times \omega_B+R_aI_{a,B} \end{cases} \Rightarrow \begin{cases} 200=k^\prime_a\times 157+20R_a \\ 200=k^\prime_a\times 167.5+18R_a \end{cases}         (21)

\Rightarrow R_a=4\ \Omega,\ k^\prime_a=0.7625        (22)

As we know, the electromagnetic torque of an electric machine can be determined by using T_e=k_aφI_a=k^\prime_aI_a. Therefore, we can write:

\pmb{T_{ag,A}}=k^\prime_aI_{a,A}=0.7625\times 20\approx \pmb{15.2\ N.m}\\[0.3cm] \pmb{T_{ag,B}}=k^\prime_aI_{a,B}=0.7625\times 18\approx \pmb{13.7\ N.m}

Choice (3) is the answer.