Finding the Performance Parameters of a Single-Phase Full-Converter Drive The speed of a separately excited dc motor is controlled by a single-phase full-wave converter in Figure 14.13a. The field circuit is also controlled by a full converter and the field current is set to the maximum possible

Question

Finding the Performance Parameters of a Single-Phase Full-Converter Drive

The speed of a separately excited dc motor is controlled by a single-phase full-wave converter in Figure 14.13a. The field circuit is also controlled by a full converter and the field current is set to the maximum possible value. The ac supply voltage to the armature and field converters is one phase, 440 V, 60 Hz. The armature resistance is [latex]R_a = 0.25 Ω[/latex], the field circuit resistance is [latex]R_f = 175 Ω[/latex], and the motor voltage constant is [latex]K_v = 1.4 V/A rad/s[/latex]. The armature current corresponding to the load demand is [latex]I_a = 45 A[/latex]. The viscous friction and no-load losses are negligible. The inductances of the armature and field circuits are sufficient to make the armature and field currents continuous and ripple free. If the delay angle of the armature converter is [latex]\alpha _a=60°[/latex] and the armature current is [latex]I_a = 45 A[/latex], determine (a) the torque developed by the motor [latex]T_d[/latex], (b) the speed ω, and (c) the input PF of the drive.

Accepted Answer

[latex]V_s=440 V, V_m=\sqrt{2} imes 440 =622.25 V ,R_a=0.25 \Omega ,\alpha _a=60°[/latex] ,and [latex]K_v = 1.4 V/A rad/s[/latex].

a. From Eq. (14.21),

[latex]V_f=\frac{2V_m}{\pi }\cos \alpha _f[/latex] for [latex]0\leq \alpha _f\leq \pi[/latex] (14.21)

the maximum field voltage (and current) would be obtained for a delay angle of [latex]α_f = 0[/latex] and

[latex]V_f=\frac{2V_m}{\pi }=\frac{2 imes 622.25}{\pi }=396.14 V[/latex]

The field current is

[latex]I_f=\frac{V_f}{R_f}=\frac{396.14}{175}=2.26 A[/latex]

From Eq. (14.4), the developed torque is

[latex]T_d = T_L = K_vI_f I_a = 1.4 × 2.26 ×45 = 142.4 N ·m[/latex]

From Eq. (14.20),

[latex]V_a=\frac{2V_m}{\pi }\cos \alpha _a[/latex] for [latex]0\leq \alpha _a\leq \pi [/latex] (14.20)

the armature voltage is

[latex]V_a=\frac{2V_m}{\pi }\cos 60=\frac{2 imes 622.25}{\pi }\cos 60=198.07 V[/latex]

The back emf is

[latex]E_g = V_a - I_a R_a = 198.07 - 45 × 0.25 = 186.82 V[/latex]

b. From Eq. (14.2),

[latex]E_g = K_vωI_f[/latex] (14.2)

the speed is

[latex]\omega =\frac{E_g}{K_\upsilon I_f}=\frac{186.82}{1.4 imes 2.26}=59.05[/latex] rad/s or 564 rpm

c. Assuming lossless converters, the total input power from the supply is

[latex]P_i = V_a I_a + V_f I_f = 198.07 × 45 + 396.14 × 2.26 = 9808.4 W[/latex]

The input current of the armature converter for a highly inductive load is shown in Figure 14.13b and its rms value is [latex]I_{sa} = I_a = 45 A[/latex]. The rms value of the input current of field converter is [latex]I_{sf }= I_f = 2.26 A[/latex]. The effective rms supply current can be found from

[latex]I_s = (I_{sa}^2 + I_{sf}^2 )^{1/2}[/latex]

[latex]= (45^2 + 2.26^2)^{1/2} = 45.06 A[/latex]

and the input VA rating, [latex]VI = V_s I_s = 440 × 45.06 = 19,826.4[/latex]. Neglecting the ripples, the input power factor is approximately

[latex]PF=\frac{P_i}{VI}=\frac{9808.4} {19,826.4}=0.495[/latex] ( lagging)

From Eq. (10.7),

[latex]PF=\frac{I_{s1}}{I_s}\cos(-\alpha )=\frac{2\sqrt{2} }{\pi }\cos \alpha[/latex] (10.7)

[latex]PF=\left(\frac{2\sqrt{2} }{\pi } ight)\cos \alpha _a=\left(\frac{2\sqrt{2} }{\pi } ight)\cos 60°=0.45[/latex] (lagging)

Question 14.4: Finding the Performance Parameters of a Single-Phase Full-Co...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions