Finding the Performance Parameters of a Single-Phase Semiconverter Drive The speed of a separately excited motor is controlled by a single-phase semiconverter in Figure 14.12a. The field current, which is also controlled by a semiconverter, is set to the maximum possible value. The ac supply

Question

Finding the Performance Parameters of a Single-Phase Semiconverter Drive

The speed of a separately excited motor is controlled by a single-phase semiconverter in
Figure 14.12a. The field current, which is also controlled by a semiconverter, is set to the
maximum possible value. The ac supply voltage to the armature and field converters is one phase, 208 V, 60 Hz. The armature resistance is [latex]R_a = 0.25 Ω[/latex], the field resistance is [latex]R_f = 147 Ω[/latex] , and the motor voltage constant is [latex]K_v = 0.7032 V/A rad/s[/latex]. The load torque is [latex]T_L = 45 N • m[/latex] at 1000 rpm. The viscous friction and no-load losses are negligible. The inductances of the armature and field circuits are sufficient enough to make the armature and field currents continuous and ripple free. Determine (a) the field current [latex]I_f[/latex]; (b) the delay angle of the converter in the armature circuit [latex]α_a[/latex], and (c) the input power factor of the armature circuit converter.

Accepted Answer

[latex]V_s= 208 V, V_m=\sqrt{2} imes 208= 294.16 V , R_a= 0.25 \Omega , R_f = 147 \Omega , T_d =T_L=45 N \cdot m, K_\upsilon= 0.7032 V/A rad/s[/latex],and [latex]\omega =1000 \pi /30=104.72 rad/s[/latex].

a. From Eq. (14.19),

[latex]V_f=\frac{V_m}{\pi }(1+\cos \alpha _f)[/latex] for [latex]0\leq \alpha _f\leq \pi [/latex] (14.19)

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

[latex]V_f=\frac{2V_m}{\pi }=\frac{2 imes 294.16}{\pi }=187.27 V[/latex]

The field current is

[latex]I_f=\frac{V_f}{R_f}=\frac{187.27}{147}=1.274 A[/latex]

b. 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{45}{0.7032 imes 1.274}=50.23 A[/latex]

From Eq. (14.2),

[latex]E_g = K_vωI_f=0.7032 × 104.72 × 1.274 = 93.82 V[/latex]

From Eq. (14.3),

[latex]V_a=R_a I_a + K_vωI_f[/latex] (14.3)

the armature voltage is

[latex]V_a = 93.82 + I_a R_a = 93.82 + 50.23 × 0.25 = 93.82 + 12.56 = 106.38 V[/latex]

From Eq. (14.18),

[latex]V_a=\frac{V_m}{\pi }(1+\cos \alpha _a)[/latex] for [latex]0\leq \alpha _a\leq \pi[/latex] (14.18)

[latex]V_a = 106.38 =(1294.16/\pi) × (1 + \cos \alpha _a )[/latex] and this gives the delay angle as [latex]\alpha _a = 82.2°[/latex].

c. If the armature current is constant and ripple free, the output power is [latex]P_o = V_aI_a = 106.38 × 50.23 = 5343.5 W[/latex]. If the losses in the armature converter are neglected, the power from the supply is [latex]P_a = P_o = 5343.5 W[/latex]. The rms input current of the armature converter, as shown in Figure 14.12, is

[latex]I_{sa}\left(\frac{2}{2\pi }\int_{\alpha _a}^{\pi }{I_a^2d heta } ight)^{1/2}=I_a\left(\frac{\pi -\alpha _a}{\pi } ight)^{1/2}[/latex]

[latex]=50.23\left(\frac{180-82.2}{180} ight)^{1/2}=37.03 A[/latex]

and the input volt–ampere (VA) rating is [latex]VI = V_s I_{sa} = 208 × 37.03 = 7702.24[/latex]. Assuming
negligible harmonics, the input PF is approximately

[latex]PF=\frac{P_o}{VI}=\frac{5343.5}{7702.24}=0.694[/latex] (lagging)

[latex]PF=\frac{\sqrt{2}(1+\cos 82.2°) }{[\pi (\pi -82.2°)]^{1/2}}=0.694[/latex] (lagging)

The input power factor is given by [12]

[latex]PF= \frac{\sqrt{2}(1+\cos \alpha ) }{\sqrt{\pi (\pi +\cos \alpha )} }[/latex]

Question 14.3: Finding the Performance Parameters of a Single-Phase Semicon...

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