Engineering Synchronous motor coupled to a DC Compound Motor Problem

[Loz30]

Hi All. I'm having trouble working out the solution to this problem.

A Synchronous motor is coupled to a DC Compound Motor.
The DC Machine is rated for 220 V DC. Its Armature Resistance is 1.0 Ω .
The Synchronous Machine is a 4 pole 400 V 50 Hz 3 Phase AC Machine with an efficiency of 80%.

Since Synchronous Motor is not self starting , the DC Machine is run as a Motor first with a Terminal Voltage of 220 V DC. When both the Machines are up to speed AC Power and DC Power (For Field ) are switched on to the AC Machine to Run as Synchronous Motor.

Now the DC Machine starts to run as DC Generator and delivers power at a current of 20 Amps.

1) Determine the input Active power to the Synchronous Motor
2) Determine the Efficiency of DC Generator


I know I can calculate the nominal speed of the AC machine given the frequency and number of poles N=(120f)/p which works out to be 1500RPM but I don't have the torque so I can't calculate the power from that.

I've also tried starting from the DC machine side when acting as a generator, but I'm only able to calculate the armature power from it's resistance and the output current which is P=I²R=20²X1=400W..

I'm not looking for the whole solution, but I would greatly appreciate a little help to get me on the right track so I can work it out on my own.

Thankyou very much!

 

[cnh1995]

Welcome to PF!

What is the field resistance? Is the motor long-shunt or short-shunt?

When you say the generator delivers power at a current of 20A, is it armature current or load current?

 

[Loz30]

Welcome to PF!

What is the field resistance? Is the motor long-shunt or short-shunt?

When you say the generator delivers power at a current of 20A, is it armature current or load current?

Thanks.

The problem doesn't mention what the field resistance is, nor whether the motor is long-shunt or short-shunt..

As for the 20A of current, I would say that the Compound DC motor is separately excited as initially when both the DC motor and AC generator are up to speed it states that the DC Power (field) is switched onto the AC machine. So the armature current would then be equal to the load current I think..I have what I think may be a possible solution as I've been thinking a lot on this problem..

Considering that the armature current = load current = 20A, We have the power across the armature as P_A=I²R=20²x1=400W.
Since the Synchronous AC motor is run up to speed using 220V supply on the DC. The DC machine when acting as a generator would be give out a terminal voltage of 220V? So the output power would be P_OUT=IV=20x 220 = 4400W.

The input power would then be equal to the armature loss plus the output power. So P_IN= P_A+P_OUT=400+4400=4800W.

The efficiency would then equal (P_OUT/P_IN) X100]=(4400/4800) X 100=91.67%
Since the AC motor has an efficiency of 80% the active input power would be equal to the Input power of the DC Generator/.8=4800/.8=6000W

I not exactly 100% on this solution as the output voltage of the DC generator at 220V is giving me doubts..[/SUB]