Question

A 220V,100A dc series motor has armature and field resistances of 0.04 and 0.06 ohm respectively.Running on test, on no load has a seperately excited generator at 1000 rpm gave the following result

Field Current(A)- 25 50 75 100 125 150 175

Terminal Voltage(V)- 66.5 124 158.5 181 198.5 211 221.5

Machine is connected for dynamic braking with braking resistance of 1.5 ohm.Calculate motor current and torque for a speed of 800 rpm.

Answer 1

At 800 rpm with dynamic braking, the motor current is approximately 100A, and the braking torque is maintained at the same level as at 1000 rpm

Back EMF Calculation:

From the provided data, we can see that at 100A field current and 1000 rpm, the terminal voltage is 181V. This tells us the back EMF (
`E_b`) generated by the motor at that operating point is:

`E_b` =
`V - R_a * I_a - R_f * I_f`

`E_b` = 220V - 0.1Ω x 100A - 0.06Ω x 100A = 174V

Speed-Torque Relationship:

For a series motor, torque (T) is proportional to the product of field current (
`I_f`) and armature current (
`I_a`). Additionally, the speed (N) is inversely proportional to the back EMF (
`E_b`):

T ∝
`I_f * I_a`

N ∝
`(1)/(E_b)`

Dynamic Braking Operation:

When connected for dynamic braking, the motor acts as a generator, and its generated current flows through the braking resistor, dissipating energy as heat. This creates a braking torque opposing the motor's rotation.

Applying the Principles:

We need to find Ia at 800 rpm. Since torque needs to be maintained (assuming constant braking force), we can use the speed-torque relationship and adjust If accordingly:

`I_f_(new) = I_f_(old) * ((N_(old) )/(N_(new)))` =
`100A * (1000\: rpm)/(800\: rpm)` = 125A

Now calculate the new back EMF:

`E_b_{new` = 220V - 0.1Ω x Ia (to be determined) - 0.06Ω x 125A

Solve for Ia using the new Eb_new and 1.5Ω braking resistor:

Ia =
`((220V - E_b_(new)) )/((R_a + R_f + R_b))` =
`((220V - 174V))/((0.1 + 0.06 + 1.5) )` ≈ 100A