Question 16.14: The no-load current of a 220 V shunt DC motor at the speed o......
The no-load current of a 220 V shunt DC motor at the speed of 1000 rpm is 2 A. If the full-load current of the machine is 35 A and the armature winding resistance is 0.2 Ω, what is the full-load electromagnetic torque developed? Assume that the magnetic flux remains constant and ignore the current of field circuit.
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1) 71 N. m
2) 65 N. m
3) 73 N. m
4) 77 N. m
Based on the information given in the problem, we have:
V_t=220\ V (1)
n_{NL}=1000\ rpm (2)
I_{t,NL}=2\ A (3)
I_{t,FL}=35\ A (4)
R_a=0.2\ \Omega (5)
φ_{FL}=φ_{NL} (6)
I_f\approx 0\ A (7)
Since I_f\approx 0\ A, we have I_a\approx I_t.
Figure 16.14 shows the electrical circuit of a shunt DC motor. Applying KVL in the right-hand side mesh of the circuit:
E_{a,NL}=V_t-R_aI_{a,NL}=220-(0.2\times 2)=219.6\ V (8)
The developed electromagnetic torque in the no-load condition can be calculated as follows:
T_{e,NL}=\frac{P_{e,NL}}{\omega_{NL}}=\frac{E_{a,NL}I_{a,NL}}{n_{NL}\frac{2\pi}{60}}=\frac{219.6 \times 2}{1000 \times \frac{2\pi}{60}}=4.2\ N.m (9)
As we know, the electromagnetic torque of an electric machine can be determined by using T_e=k_aφI_a. Therefore, we can write:
\frac{T_{e,FL}}{T_{e,NL}}=\frac{φ_{FL}I_{a,FL}}{φ_{NL}I_{a,NL}} (10)
Since the terminal voltage is constant, we have:
I_{f,FL}=I_{f,NL}\Rightarrow φ_{FL}=φ_{NL} (11)
Therefore:
\frac{T_{e,FL}}{T_{e,NL}}=\frac{I_{a,FL}}{I_{a,NL}} (12)
\Rightarrow \frac{T_{e,FL}}{4.2}=\frac{35}{2}\Rightarrow \pmb{T_{e,FL}\approx 73\ N.m}
Choice (3) is the answer.
