# DC motors, DC generation

1) A 250 V shunt motor has an armature resistance of 0.4 Ω and runs at a speed of 750 rev min–1when taking a full load current of 25 A. Estimate the speed of the motor at no load when the armature current is 3 A, assuming that the flux per pole remains constant. If the field resistance is 110 Ω, calculate the value of the resistance required in the field circuit to increase the speed to 1000 rev min–1if the load torque remains constant.

2) A 20 kW, 500 V d.c. shunt wound motor draws a current of 45 A when running at full load with a speed of 600 rev min–1. On no load, the current drawn from the supply is 5 A. If the armature resistance is 0.3 Ω and the shunt field resistance is 220 Ω, calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full load value, assuming that the flux per pole does not change. (Hint: don’t forget that all armature current is supplying torque. One must decide at some points what is the armature current producing torque and at others, to determine generator emfs, the total armature current).

Answers

1 a) The electric power or the electric energy in the time unit (1 second) is equal to the mechanical power

$P = I^2*R = J*\omega^2$

where $I$ is the current through armature, $R$ is the armature resistance, $J$ is the inertia moment of the armature omega is the angular speed of the motor

$J = I_1^2*R/\omega1^2$

$I_1 =25 A$, $R =0.4 ohm$, $\omega_1 =2*\pi*750/60 sec^{-1}$

$J = 0.04 kg*m^2$

When the flux per pole remains constant we have

$I_2^2*R = J*\omega2^2$

$\omega_2^2 = I_2^2*R/J = 9*0.4/0.04 = 90$

$\omega_2 =9.46 sec^{-1} = 2*\pi*F_2$

$F_2$ is the frequency

$F_2 = 1.5 sec^{-1} =90 min^{-1}$

b)

The force of the magnetic field is $F = k_1*B*I(armature) = k_1*k_2*I(field)*I(armature)$, where B is the magnetic field and $k_1$ and $k_2$ are constants.

If the load torque remains constant the power remains the same

$J*\omega_3^2 = I_2(armature)^2*R$ where $\omega_3 =2*\pi*1000 min^{-1}$

$I_2(armature) = \sqrt{J*\omega_3^2/R} = I_1(armature)*\sqrt{\omega3^2/\omega_1^2} =I_1(armature)*\omega_3/\omega_1 =$

$= 25*1000/750 = 33.3 A$

Now

$I_1(field)*I_1(armature) = I_2(field)*I_2(armature)$

$(U/R_1)*25 = U/(R_1+R_2)*33.3$

$R_1 = 110 ohm$

$R_1/(R_1+R) = 25/33.3$ or $R_1/(R_1+R) =0.75$

$0.75R_1+0.75R = R_1$ or $0.25R_1 =0.75 R$

$R = 36.67 ohms$

2.

$I(armature) = P/U = 20000/500 =40 A$

Therefore 40 A is the current in the armature and 5 A is the current in the shunt field

$40+5 = 45 A$

The torque is proportional to the magnetic force

$T = k_1*B(field)*I(armature)= k_1*k_2*I(field)*I(armature)$

The flux per pole is proportional to the magnetic field

$B = k_2*I(field)$

The flux per pole is constant implies I(field) is constant

$T_1/T_2 = I_1(armature)/I_2(armature)$

$I_2(armature) = 40*0.5 =20 A$

Power is $P_2 = U*I_2(armature) =500 *20 = 10000 W =10 kW$

efficiency is $\eta = P_2/P_1 = 10000/2000 =0.5 =50%$

Shunt field resistance is $220 ohm$ is the same it implies magnetic field B is the same

Load torque is half it means magnetic force is half

$F = k_3*B(field)*I(armature) but F = J*\omega$,

$J$ is the inertia momentum of the armature

$B*I_1(armature) = J*\omega_1$

$B*I_2(armature) = J*\omega_2$

$I_1(armature)/I_2(armature) = \omega_1/\omega_2$

$40/20 = 600/\omega_2$

$omega_2 = 300 rot/min$