Please show all work for points A dc shunt motor drives a centrifugal pump at a

Question

Please show all work for points

A dc shunt motor drives a centrifugal pump at a speed of 1000rpm when the terminal voltage and the line currents are 200V and 50A, respectively. The armature and field resistance are 0.1 ohm and 100 ohm respectively.

A.)Design a starting resistance for a maximum starting current of 120A in the armature circuit.

B.)What resistance should be added to the armature current to reduce the speed to 800 rpm?

C.)If the terminal voltage is reduced by 25%, what is the speed of the motor?

Explanation / Answer

The motor is shunt, it means armature and field resistances are in parallel.

Ia*Ra = If*Rf

Ia/If = Rf/Ra = 100/0.1 =1000

Total current is

If/Ia =10^-3

If =120*10^-3 =0.120 A

Total starting current is I = Ia+If =120+0.120 =120.12 A

Now total internal resistance of the motor is

Req = Rf*Ra/(Rf+Ra) =0.1*100/100.1 =0.0999 ohm

we add an external resistance R to have 120.12 A amximum current

Rtot = Req+R

Rtot = U/I = 200/120.12 =1.665 ohm

R = 1.665-0.0999 =1.5651 ohm

b) The field current resistance and current remains the same.

The power of the armature is P = U^2/Ra (=E/t)

the energy of the armature is E = I*omega^2/2 = K*F^2Â where K is a constant and F the freq.

it means U^2/Ra = K*F^2

(Ra+R)/Ra =(F1/F2)^2 = (1000/800)^2 =1.5625

Ra+R = 1.5625*Ra

R = 0.5625*Ra =0.5625*0.1 =0.056 ohms

c) delivered power P = U^2/R (=E/t)

R is the same before and after reducing voltage.

Again energy E = I*omega^2/2 = K*F^2

it means (U1/U2)^2 = (F1/F2)^2

F2/F1 =U2/U1 = 0.75U1/U1 =0.75

F2 = 0.75*F1 =0.75*1000 =750 rpm

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