A PMDC motor has an armature resistance of 4.2 Ω . When 6 V supply is connected to the motor it runs at a speed 12,125 rpm drawing a current of 14.5 mA on no-load (a) Calculate its torque constant (b) What is the value of rotational loss? With an applied voltage of 6 V, (c) calculate the stalled

Question

A PMDC motor has an armature resistance of [latex] 4.2 \Omega [/latex]. When 6 V supply is connected to the motor it runs at a speed 12,125 rpm drawing a current of 14.5 mA on no-load

(a) Calculate its torque constant

(b) What is the value of rotational loss?

With an applied voltage of 6 V,

(c) calculate the stalled torque and stalled current of the motor (motor shaft held stationary)

(d) at a gross output of 1.6 W, calculate the armature current and efficiency. Assume that the rotational loss varies as square of speed.

(e) calculate the motor output at a speed of 10,250 rpm and the efficiency.

Accepted Answer

No load [latex] V_{t}=6 V , I_{ mo }=14.5 mA , n=12125 rpm [/latex] or [latex] \omega=1269.7 rad / s [/latex]

(a) [latex] E_{a}=6-14.5 imes 10^{3} imes 4.2= [/latex] 5.939 V

[latex] 5.939=K_{m} \omega=K_{m} imes 1269.7 [/latex]

Or [latex] K_{m}=4.677 imes 10^{-3} [/latex]

(b) Rotational loss, [latex] P_{r o t}=E_{a} I_{a} [/latex]; there is no load

[latex] =5.939 imes 14.5 imes 10^{-3}=0.0861 W [/latex]

(c) Stalled current

[latex] \omega=0 [/latex] so [latex] E_{a}=0 [/latex]

[latex] I_{a}( stall )=\frac{6}{4.2}=1.4285 A [/latex]

Torque (stall) [latex] =K_{m} I_{a} ext { (stall) }=4.677 imes 10^{-3} imes 1.428=6.67 m Nm [/latex]

(d) [latex] P_{ ext {out }}( ext { gross })=1.6 W =E_{a} I_{a} [/latex]

[latex] \left(6-4.2 I_{a} ight) I_{a}=1.6 [/latex]

[latex] 4.2 I_{a}^{2}-6 I_{a}+1.6=0 [/latex]

Solving we find [latex] I_{a}=0.354 A , 1.074 A [/latex]

Thus

[latex] I_{a}=0.354 A [/latex]; higher value rejected

[latex] E_{a}=6-0.854 imes 4.2=4.513 V =K_{m} \omega [/latex]

[latex] \omega=\frac{4.513 imes 10^{3}}{4.677}=965 rad / s [/latex]

Rotational loss (proportional to square of speed)

[latex] P_{r o t}=0.0861 imes\left(\frac{965}{1269.7} ight)^{2}=[/latex] 0.05 W

[latex] P_{ ext {out }}( ext { net })=P_{ ext {out }}( ext { gross })-P_{ ext {rot }}=1.6-0.05=[/latex] 1.55 W

Power input, [latex] P_{i}=V_{t} I_{a}=6 imes 0.354= [/latex] 2121 W

[latex] \eta=\frac{1.55}{2.124} imes 100=73 \% [/latex]

(e) Motor speed, n = 10250 rpm or [latex] \omega=1073.4 [/latex] rad / s

[latex] E_{a}=K_{m}^{\omega}=4.513 imes 10^{-3} imes 1073.4= [/latex] 4.844 V

[latex] I_{a}=\frac{6-4.844}{4.2}=[/latex] 0.275 A

[latex] P_{ ext {out }}( ext { gross })=P_{e}=E_{a} I_{a}=4.844 imes 0.275= [/latex] 1.332 W

[latex] P_{r o t}=0.0861 imes\left(\frac{1073.4}{1269.7} ight)^{2}=[/latex] 0.0615 W

[latex] P_{o u t}( ext { net })=1.332-0.0615=[/latex] 1.27 W

[latex] P_{i n}=6 imes 0.275=[/latex] 1.65 W

[latex] \eta=\frac{1.27}{1.65} imes 100=77 \% [/latex]

Question 7.63: A PMDC motor has an armature resistance of 4.2 Ω . When 6 V ...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions