A 230 V dc shunt motor having armature resistance of 2 Ω draws an armature current of 5 A to drive a constant torque load at 1250 rpm. At no load it draws a current of 1 A. (a) A resistance of 15 Ω is added in series to the armature. Find the motor speed with load torque as above. Also determine

Question

A 230 V dc shunt motor having armature resistance of [latex] 2 \Omega [/latex] draws an armature current of 5 A to drive a constant torque load at 1250 rpm. At no load it draws a current of 1 A.

(a) A resistance of [latex] 15 \Omega [/latex] is added in series to the armature. Find the motor speed with load torque as above. Also determine the speed regulation.

(b) A resistance of [latex] 15 \Omega [/latex] is shunted across the armature and [latex] 10 \Omega [/latex] in series with the supply line (as in Fig. 7.96(a). Calculate the load speed and speed regulation.

(c) Compare the power wasted in external resistance (s) in parts (a) and (b).

Rotational loss torque is negligible. The armature reaction effect is to be ignored.

Accepted Answer

[latex] E_{a} =230-2 imes 5=220 V [/latex]

[latex] n =1250 rpm , \omega=130.9 rad / s [/latex]

[latex] E_{a}=K_{a} \Phi \omega ; \Phi [/latex] is constant; constant shunt field current, no armature reaction effect

[latex] 220 =K_{a} \Phi imes 130.9 [/latex]

Or [latex] K_{a} \Phi =1.68 [/latex]

(a) [latex] R_{e}=15 \Omega [/latex] in series

No load speed

[latex] I_{a o}=1 A [/latex]

[latex] E_{a}=230-(15+2) imes 1=213 V [/latex]

[latex] \omega_{0}=\frac{213}{1.68}=126.8 rad / s [/latex]

Load torque constant

As [latex] \Phi [/latex] is constant [latex] I_{a}=5 A [/latex]

[latex] E_{a}=230-(15+2) imes 5=145 V [/latex]

[latex] \omega=\frac{145}{1.68}=86.3 rad / s [/latex]

Speed regulation [latex] =\frac{126.8-86.3}{86.3} imes 100=46.9 \% [/latex]

(b) From the Thevenin equivalent of Fig. 7.96(c)

[latex] R_{1}=10 \Omega, R_{2}=15 [/latex]

[latex] \beta=\frac{R_{2}}{R_{1}+R_{2}}=\frac{15}{10+15}=0.6 [/latex]

[latex] V_{ TH }=230 imes 0.6=230 V [/latex]

[latex] R_{ TH }=\beta R_{1}=0.6 imes 10=6 \Omega [/latex]

No load speed

[latex] I_{a o} =1 A [/latex]

[latex] E_{a} =V_{ TH }-\left(R_{ TH }+R_{a} ight) I_{ ao } [/latex]

[latex] =230-(6+2) imes 1=130 V [/latex]

[latex] \omega_{0} =\frac{130}{1.68}=77.38 rad / s [/latex]

On load

[latex] I_{a} =5 A [/latex]

[latex] E_{a} =138-(6+2) imes 5=98 V [/latex]

[latex] \omega =\frac{98}{1.68}=58.33 rad / s [/latex]

Speed regulation [latex] =\frac{77.38-58.33}{58.33} imes 100=32.6 \% [/latex]

Observation Speed regulation is much better (less) in shunted armature control than in rheostatic control.

(c) Power loss

(i) Rheostatic control

[latex] P_{e}=(5)^{2} imes 15=375 W [/latex]

(ii) Shunted armature control

[latex] V_{a} ext { (across armature) }=98+2 imes 5=108 V [/latex]

[latex] P(15 \Omega) =\frac{(108)^{2}}{15}=777.6 W [/latex]

[latex] I(15 \Omega) =\frac{108}{15}=7.24 A [/latex]

[latex] I(10 \Omega) =7.2+5=12.2 A [/latex]

[latex] P(10 \Omega) =(12.2)^{2} imes 10=1488.4 W [/latex]

Then [latex] P_{e} =777.6+1488.4=2266 W [/latex]

Observation External power loss is far larger in shunted armature control than in rheostatic control. In fact it is much larger than power of the motor [latex] (230 imes 5 = 1150 W)[/latex] being controlled.

Remark Shunted armature control is employed for very small motors where speed regulation requirement is stringent.

Question 7.50: A 230 V dc shunt motor having armature resistance of 2 Ω dra...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions