Finding the Performance Parameters of a Three-Phase Full-Converter Drive The speed of a 20-hp, 300-V, 1800-rpm separately excited dc motor is controlled by a three-phase full-converter drive. The field current is also controlled by a three-phase full converter and is set to the maximum possible

Question

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

The speed of a 20-hp, 300-V, 1800-rpm separately excited dc motor is controlled by a three-phase full-converter drive. The field current is also controlled by a three-phase full converter and is set to the maximum possible value. The ac input is a three-phase, Y-connected, 208-V, 60-Hz supply. The armature resistance is [latex]R_a = 0.25 Ω[/latex], the field resistance is [latex]R_f = 245 Ω[/latex], and the motor voltage constant is [latex]K_v = 1.2 V/A rad/s[/latex]. The armature and field currents can be assumed to be continuous and ripple free. The viscous friction is negligible. Determine (a) the delay angle of the armature converter [latex]α_a[/latex], if the motor supplies the rated power at the rated speed; (b) the no-load speed if the delay angles are the same as in (a) and the armature current at no load is 10% of the rated value; and (c) the speed regulation.

Accepted Answer

[latex]R_a=0.25 \Omega , R_f=245 \Omega , K_\upsilon =1.2 V/A rad/s,V_L= 208 V[/latex], and [latex]\omega =1800 \pi /30 =188.5 rad/s[/latex]. The phase voltage is [latex]V_p=V_L/\sqrt{3}=208/\sqrt{3}=120 V[/latex] and [latex]V_m=120 imes \sqrt{2}=169.7 V[/latex]. Because 1 hp is equal to 746 W, the rated armature current is [latex]I_{rated} = 20 × 746/300 = 49.73 A[/latex]; for maximum possible 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=3\sqrt{3} imes \frac{169.7}{\pi }=280.7 V[/latex]

[latex]I_f=\frac{V_f}{R_f}=\frac{280.7}{245}=1.146 A[/latex]

a. [latex]I_a = I_{rated} = 49.73 A[/latex] and

[latex]E_g = K_vI_f ω= 1.2 × 1.146 × 188.5 = 259.2 V[/latex]

[latex]V_a = 259.2 + I_a R_a = 259.2 + 49.73 × 0.25 = 271.63 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=271.63=\frac{3\sqrt{3}V_m }{\pi }\cos \alpha _a=\frac{3\sqrt{3} imes 169.7 }{\pi } \cos \alpha _a[/latex]

and this gives the delay angle as [latex]α_a = 14.59°[/latex].

b. [latex]I_a = 10\% [/latex] of [latex]49.73 = 4.973 A[/latex] and

[latex]E_{go} = V_a - R_a I_a = 271.63 - 0.25 ×4.973 = 270.39 V[/latex]

From Eq. (14.4), the no-load speed is

[latex]T_d=K_tI_fI_a [/latex] (14.4)

[latex]\omega _0=\frac{E_{go}}{K_\upsilon I_f}=\frac{270.39}{1.2 imes 1.146}=196.62 rad/s[/latex] or [latex]196.62 imes \frac{30}{\pi }= 1877.58 rpm[/latex]

c. The speed regulation is defined as

[latex]\frac{ ext{no-load speed - full-load speed}}{ ext{full-load speed}}=\frac{1877.58 - 1800}{1800}=0.043[/latex] or [latex]4.3\%[/latex]

Question 14.6: Finding the Performance Parameters of a Three-Phase Full-Con...

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