Finding the Performance of a Three-Phase Full-Converter Drive with Field Control The speed of a 20-hp, 300-V, 900-rpm separately excited dc motor is controlled by a three-phase full converter. The field circuit is also controlled by a three-phase full converter. The ac input to the armature and

Question

Finding the Performance of a Three-Phase Full-Converter Drive with Field Control

The speed of a 20-hp, 300-V, 900-rpm separately excited dc motor is controlled by a three-phase full converter. The field circuit is also controlled by a three-phase full converter. The ac input to the armature and field converters is three-phase, Y-connected, 208 V, 60 Hz. The armature resistance is [latex]R_a = 0.25 Ω[/latex], the field circuit resistance is [latex]R_f = 145 Ω[/latex] , and the motor voltage constant is [latex]K_v = 1.2 V/A rad/s[/latex]. The viscous friction and no-load losses can be considered negligible. The armature and field currents are continuous and ripple free. (a) If the field converter is operated at the maximum field current and the developed torque is [latex]T_d = 116 N · m[/latex] at 900 rpm, determine the delay angle of the armature converter [latex]α_a[/latex]. (b) If the field circuit converter is set for the maximum field current, the developed torque is [latex]T_d = 116 N · m[/latex], and the delay angle of the armature converter is [latex]α_a = 0[/latex], determine the speed of the motor. (c) For the same load demand as in (b), determine the delay angle of the field converter if the speed has to be increased to 1800 rpm.

Accepted Answer

[latex]R_a = 0.25 Ω, R_f = 145 Ω, K_v = 1.2 V/A rad/s[/latex], and [latex]V_L = 208 V[/latex]. The phase voltage is [latex]V_p=208/\sqrt{3}=120 V[/latex] and [latex]V_m=\sqrt{2} imes 120=169.7 V[/latex].

a. [latex]T_d = 116 N[/latex] · m and [latex]ω= 900 π/30 = 94.25 rad/s[/latex]. For maximum field current, [latex]α_f = 0=[/latex]. From Eq. (14.28),

[latex]V_f=\frac{3\sqrt{3}V_m }{\pi }\cos \alpha _f[/latex] for [latex]0\leq \alpha _f\leq \pi[/latex] (14.28)

[latex]V_f=\frac{3 imes \sqrt{3} imes 169.7}{\pi }=280.7 V[/latex]

[latex]I_f=\frac{280.7}{145}=1.936 A[/latex]

From Eq. (14.4),

[latex]T_d = K_t I_f I_a[/latex] (14.4)

[latex]I_a=\frac{T_d}{K_\upsilon I_f}=\frac{116}{1.2 imes 1.936}=49.93 A[/latex]

[latex]E_g = K_vI_f ω= 1.2 × 1.936 × 94.25 = 218.96 V[/latex]

[latex]V_a = E_g + I_a R_a = 218.96 + 49.93 × 0.25 = 231.44 V[/latex]

From Eq. (14.27),

[latex]V_a=\frac{3\sqrt{3}V_m }{\pi }\cos \alpha _a[/latex] for [latex]0\leq \alpha _a\leq \pi[/latex] (14.27)

[latex]V_a=231.44=\frac{3 imes \sqrt{3} imes 169.7 }{\pi }\cos \alpha _a[/latex]

which gives the delay angle as [latex]α_a = 34.46°[/latex].

b. [latex]α_a = 0[/latex] and

[latex]V_a=\frac{3 imes \sqrt{3} imes 169.7 }{\pi }=280.7 V[/latex]

[latex]E_g = 280.7 - 49.93 × 0.25 = 268.22 V[/latex]

and the speed

[latex]\omega =\frac{E_g}{K_\upsilon I_f}=\frac{268.22}{1.2 imes 1.936}=115.45 rad/s[/latex] or 1102.5 rpm

c. [latex]ω= 1800 π/30 = 188.5 rad/s[/latex]

[latex]E_g = 268.22 V = 1.2 × 188.5 × I_f[/latex] or [latex]I_f = 1.186 A[/latex]

[latex]V_f = 1.186 × 145 = 171.97 V[/latex]

From Eq. (14.28),

[latex]V_f=171.97=\frac{3 imes \sqrt{3} imes 169.7 }{\pi }\cos \alpha _f[/latex]

which gives the delay angle as [latex]α_f = 52.2°[/latex].

Question 14.7: Finding the Performance of a Three-Phase Full-Converter Driv...

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