25kW Shunt Motor Problem: Rated-Load Speed and Step Resistance Cut Out

[mingming]

Homework Statement

A 25-kW, 230-V shunt motor has an armature resistance of 0.124 Ω and a field-circuit resistance of 95 Ω. The motor delivers rated output power at rated voltage when its armature current is 73.5 A. When the motor is operating at rated voltage, the speed is observed to be 1150 r/min when the machine is loaded such that the armature current is 41.5 A.

a. Calculate the rated-load speed of this motor. In order to protect both the motor and the dc supply under starting conditions, an external resistance will be connected in series with the armature winding (with the field winding remaining directly across the 230-V supply). The resistance will then be automatically adjusted in steps so that the armature current does not exceed 200 percent of rated current. The step size will be determined such that, until all the extemal resistance is switched out, the armature current will not be permitted to drop below rated value. In other words, the machine is to start with 200 percent of rated armature current and as soon as the current falls to rated value, sufficient series resistance is to be cut out to restore the current to 200 percent. This process will be repeated until all of the series resistance has been eliminated.

b. How much resistance should be cut out at each step in the starting operation and at what speed should each step change occur?

Homework Equations

armature voltage is proportional to the speed
Vt= Ea+Ia Ra

The Attempt at a Solution

I have solve for the maximum resistance when the motor is starting Rt= 1.44 ohms
and i have solve for the three resistances to be cut out at each step Ra= 0.78 ohms, Rb= 0.39 ohms and Rc= 0.2 ohms..
MY PROBLEM IS FINDING THE SPEED DURING THE CUT OUT OF RESISTANCE...

 

[KataKoniK]

Thank you for sharing your calculations with us. It seems like you are on the right track with your steps for starting the motor. However, in order to determine the speed at each step, we need to take into account the voltage drop across the armature and field resistances.

First, let's calculate the total resistance of the circuit when the motor is starting, including the external resistance. This can be done by adding the armature resistance (0.124 Ω), field-circuit resistance (95 Ω), and maximum external resistance (1.44 Ω):

Rtotal = 0.124 Ω + 95 Ω + 1.44 Ω = 96.564 Ω

Next, we can use Ohm's law to calculate the voltage drop across the total resistance at each step. This voltage drop will determine the armature voltage and therefore the speed of the motor. We can use the equation Vt = Ea + IaRa to find the armature voltage (Vt) at each step, where Ea is the back emf and Ia is the armature current.

At the first step, the armature current is 200% of rated current (147 A), and the total resistance is 96.564 Ω. Therefore, the voltage drop across the total resistance is:

Vt = IaRtotal = (147 A)(96.564 Ω) = 14,364.108 V

Since the armature voltage is equal to the back emf plus the voltage drop across the armature resistance, we can rearrange the equation to solve for the back emf:

Ea = Vt - IaRa = 14,364.108 V - (147 A)(0.124 Ω) = 14,345.268 V

Now, we can use the equation Vt = Ea + IaRa to solve for the armature voltage at each step, using the back emf calculated above and the armature current at each step (147 A, 105 A, 73.5 A, and 41.5 A). The resulting armature voltages are:

At 200% of rated current (147 A): Vt = 14,364.108 V
At 150% of rated current (105 A): Vt = 11,328.858 V
At 100% of rated current (73.5 A