Finding the Speed and Torque Response of a Converter-Fed Drive A 50-kW, 240-V, 1700-rpm separately excited dc motor is controlled by a converter, as shown in the block diagram Figure 14.29. The field current is maintained constant at If = 1.4 A and the machine back emf constant at If = 1.4 A and

Question

Finding the Speed and Torque Response of a Converter-Fed Drive

A 50-kW, 240-V, 1700-rpm separately excited dc motor is controlled by a converter, as shown in the block diagram Figure 14.29. The field current is maintained constant at [latex]I_f = 1.4 A[/latex] and the machine back emf constant is [latex]K_v = 0.91 V/A rad/s[/latex]. The armature resistance is [latex]R_m = 0.1 Ω[/latex] and the viscous friction constant is [latex]B = 0.3 N · m/rad/s[/latex]. The amplification of the speed sensor is [latex]K_1 = 95 mV/rad/s[/latex] and the gain of the power controller is [latex]K_2 = 100[/latex]. (a) Determine the rated torque of the motor. (b) Determine the reference voltage [latex]V_r[/latex] to drive the motor at the rated speed. (c) If the reference voltage is kept unchanged, determine the speed at which the motor develops the rated torque. (d) If the load torque is increased by 10% of the rated value, determine the motor speed. (e) If the reference voltage is reduced by 10%, determine the motor speed. (f) If the load torque is increased by 10% of the rated value and the reference voltage is reduced by 10%, determine the motor speed. (g) If there was no feedback in an open-loop control, determine the speed regulation for a reference voltage of [latex]V_r = 2.31 V[/latex]. (h) Determine the speed regulation with a closed-loop control.

Accepted Answer

[latex]I_f = 1.4 A, K_v = 0.91 V/A rad/s, K_1 = 95 mV/rad/s, K_2 = 100, R_m = 0.1 Ω, B = 0.3 N · m/rad/s[/latex], and [latex]ω_{rated} = 1700 π/30 = 178.02 rad/s[/latex].

a. The rated torque is [latex]T_L = 50,000/178.02 = 280.87 N ·m[/latex].

b. Because [latex]V_a = K_2 V_r[/latex], for open-loop control Eq. (14.70)

[latex]\Delta \omega =\frac{K_2K_\upsilon I_f}{R_mB+(K_\upsilon I_f)}\Delta V_r[/latex] (14.70)

gives

[latex]\frac{\omega }{V_a}=\frac{\omega }{K_2V_r}=\frac{K_\upsilon I_f}{R_mB+(K_\upsilon I_f)^2}=\frac{0.91 imes 1.4}{0.1 imes 0.3+(0.91 imes 1.4)^2}=0.7707[/latex]

At rated speed,

[latex]V_a=\frac{\omega }{0.7707}=\frac{178.02}{0.7707}=230.98 V[/latex]

and feedback voltage,

[latex]V_b = K_1 ω= 95 × 10^{-3 }× 178.02 = 16.912 V[/latex]

With closed-loop control, [latex](V_r - V_b)K_2 = V_a[/latex] or [latex](V_r - 16.912)× 100 = 230.98[/latex], which gives the reference voltage, [latex]V_r = 19.222 V[/latex].

c. For [latex]V_r = 19.222 V[/latex] and [latex]ΔT_L = 280.87 N · m[/latex], Eq. (14.95)

[latex]\Delta \omega =-\frac{R_m}{R_mB+(K_\upsilon I_f)^2+K_1K_2K_\upsilon I_f}\Delta T_L[/latex] (14.95)

gives

[latex]\Delta \omega =-\frac{0.1 imes 280.86}{0.1 imes 0.3 + (0.91 imes 1.4)^2 + 95 imes 10^{-3} imes 100 imes 0.91 imes 1.4}[/latex]

= -2.04 rad/s

The speed at rated torque,

[latex]ω= 178.02 - 2.04 = 175.98 rad/s [/latex] or 1680.5 rpm

d. [latex]ΔT_L = 1.1 × 280.87 = 308.96 N · m[/latex] and Eq. (14.95) gives

[latex]\Delta \omega =-\frac{0.1 imes 308.96}{0.1 imes 0.3 + (0.91 imes 1.4)^2 + 95 imes 10^{-3} imes 100 imes 0.91 imes 1.4}[/latex]

= -2.246 rad/s

The motor speed

[latex]ω= 178.02 - 2.246 = 175.774 rad/s[/latex] or 1678.5 rpm

e. [latex]ΔV_r = -0.1 × 19.222 = -1.9222 V[/latex] and Eq. (14.94)

[latex]\Delta \omega =-\frac{K_2K_\upsilon I_f}{R_mB+(K_\upsilon I_f)^2+K_1K_2K_\upsilon I_f}\Delta V_r[/latex] (14.94)

gives the change in speed,

[latex]\Delta \omega =-\frac{100 imes 0.91 imes 1.4 imes 1.9222}{0.1 imes 0.3+(0.91 imes 1.4)^2+95 imes 10^{-3} imes 100 imes 0.91 imes 1.4}[/latex]

= -17.8 rad/s

The motor speed is

[latex]ω= 178.02 - 17.8 = 160.22 rad/s [/latex] or 1530 rpm

f. The motor speed can be obtained by using superposition:

[latex]ω= 178.02 - 2.246 - 17.8 = 158 rad/s[/latex] or 1508.5 rpm

g. [latex]ΔV_r = 2.31 V[/latex] and Eq. (14.70)

[latex]\Delta \omega =\frac{K_2K_\upsilon I_f}{R_mB+(K_\upsilon I_f)^2}\Delta V_r[/latex] (14.70)

gives

[latex]\Delta \omega =\frac{100 imes 0.91 imes 1.4 imes 2.31}{0.1 imes 0.3+(0.91 imes 1.4)^2}=178.02 rad/s[/latex] or 1700 rpm

and the no-load speed is [latex]ω= 178.02 rad/s[/latex] or 1700 rpm. For full load, [latex]ΔT_L = 280.87 N · m[/latex], Eq. (14.71)

[latex]\Delta \omega =-\frac{R_m}{R_mB+(K_\upsilon I_f)}\Delta T_L[/latex] (14.71)

gives

[latex]\Delta \omega =-\frac{0.1 imes 280.87}{0.1 imes 0.3 + (0.91 imes 1.4)^2}=-16.99 rad/s[/latex]

and the full-load speed

[latex]ω= 178.02 - 16.99 = 161.03 rad/s[/latex] or 1537.7 rpm

The speed regulation with open-loop control is

[latex]\frac{1700-1537.7}{1537.7}=10.55\%[/latex]

h. Using the speed from (c), the speed regulation with closed-loop control is

[latex]\frac{1700-1680.5}{1680.5}=1.16\%[/latex]

Note: By closed-loop control, the speed regulation is reduced by a factor of approximately 10, from 10.55% to 1.16%.

Question 14.13: Finding the Speed and Torque Response of a Converter-Fed Dri...

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