Question 16.15: The magnetization characteristics of a shunt DC motor are sh......
The magnetization characteristics of a shunt DC motor are shown in Fig. 15.3. The field winding resistance of the motor is 80 Ω and assume that the developed electromagnetic torque of the motor is proportional to its speed. When the motor is supplied with 240 V, its armature current and speed are 50 A and 1500 rpm, respectively. Now, with a new load, the field winding resistance is increased to about 120 Ω that results in 77 A for the armature current. Calculate the speed of motor in this new condition.
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1) 885.5 rpm
2) 960 rpm
3) 2100 rpm
4) 2310 rpm
The magnetization characteristics of the shunt DC motor are illustrated in Fig. 16.15.a. In addition, based on the information given in the problem, we have:
T_e\propto n (1)
V_t=240\ V (2)
R_{f,1}=80\ \Omega (3)
I_{a,1}=50\ A (4)
n_1=1500\ rpm (5)
R_{f,2}=120\ \Omega (6)
I_{a,2}=77\ A (7)
Figure 16.15.b shows the electrical circuit of a shunt DC motor. Applying Ohm’s law for the branch of field winding:
I_{f,1}=\frac{V_t}{R_{fc,1} + R_{fw}}=\frac{V_t}{R_{f,1}}=\frac{240}{80}=3\ A (8)
I_{f,2}=\frac{V_t}{R_{fc,2} + R_{fw}}=\frac{V_t}{R_{f,2}}=\frac{240}{120}=2\ A (9)
The equation of the second segment of the magnetization characteristics curve of machine can be formulated as follows:
E_{a}-180=\frac{240-180}{4-1}(I_f-1)\Rightarrow E_a=20I_f+160 (10)
By using the magnetization characteristics curve of machine for I_{f,1}=3\ A and I_{f,2}=2\ A , we have:
I_{f,1}=3\ A \Rightarrow \hat{E}_{a,1}=220\ V (11)
I_{f,2}=2\ A \Rightarrow \hat{E}_{a,2}=200\ V (12)
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,2}}{E_{a,1}}=\frac{φ_2ω_2}{φ_1ω_1}=\frac{φ_2n_2}{φ_1n_1} (13)
Since both induced armature voltages (\hat{E}_{a,1} and \hat{E}_{a,2}) have been extracted from the magnetization characteristics curve, we have n_1=n_2. Hence:
\frac{\hat{E}_{a,2}}{\hat{E}_{a,1}}=\frac{φ_2}{φ_1}=\frac{200}{220} (14)
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,2}}{T_{e,1}}=\frac{φ_2I_{a,2}}{φ_1I_{a,1}} (15)
Solving (14) and (15):
\frac{T_{e,2}}{T_{e,1}}=\frac{200}{220}\times \frac{I_{a,2}}{I_{a,1}} (16)
Solving (1) and (16):
\frac{n_{2}}{n_{1}}=\frac{200}{220}\times \frac{I_{a,2}}{I_{a,1}} (17)
\Rightarrow \frac{n_2}{1500}=\frac{200}{220}\times \frac{77}{50}\Rightarrow \pmb{n_2=2100\ rpm}
Choice (3) is the answer.
