Question

A 240-V shunt motor has a full load speed of 1000 RPM and takes a current of 50 A. The shunt field current is 1 A and the armature resistance is 0.2Ω. How much resistance should be inserted in series with the armature circuit so that the motor develops a torque of 15 N.m at a speed of 1500 RPM?

Answer 1

To determine the resistance that should be inserted in series with the armature circuit to achieve a torque of 15 N.m at a speed of 1500 RPM, we need to use the motor speed equation:


`\[ N = \frac{{(V - I_a \cdot R_a)}}{{\phi \cdot K}} \]`

Where:
- N is the motor speed in RPM
- V is the supply voltage (240 V)
- I_a is the armature current (50 A)
- R_a is the armature resistance (0.2 Ω)
- ϕ is the field flux
- K is a constant

We are given that the full load speed of the motor is 1000 RPM at a current of 50 A, and the field current is 1 A. From this information, we can determine the field flux (ϕ) using the following equation:


`\[ \frac{{N_1}}{{N_2}} = \frac{{I_1 \cdot \phi_1}}{{I_2 \cdot \phi_2}} \]`

Where:
- N_1 and N_2 are the motor speeds (1000 RPM and 1500 RPM, respectively)
- I_1 and I_2 are the armature currents (50 A in both cases)
- ϕ_1 and ϕ_2 are the field fluxes

Substituting the given values:


`\[ \frac{{1000}}{{1500}} = \frac{{50 \cdot \phi_1}}{{50 \cdot \phi_2}} \]`

Simplifying the equation, we find that ϕ_1 is equal to ϕ_2. The field flux remains the same.

Now, we can substitute the known values into the motor speed equation and solve for the resistance:


`\[ 1500 = \frac{{(240 - 50 \cdot R_a)}}{{\phi \cdot K}} \]`

To find the torque, we can use the torque equation:


`\[ T = \frac{{\phi \cdot I_a}}{{K}} \]`

Substituting the known values:


`\[ 15 = \frac{{\phi \cdot 50}}{{K}} \]`

Since we know that ϕ remains constant, we can rewrite the torque equation as:


`\[ 15 = \frac{{\phi}}{{K}} \cdot 50 \]`

Simplifying, we find:


`\[ K = \frac{{\phi}}{{30}} \]`

Substituting the value of K into the motor speed equation:


`\[ 1500 = \frac{{(240 - 50 \cdot R_a)}}{{\phi \cdot \frac{{\phi}}{{30}}}} \]`
Simplifying the equation, we find:


`\[ 45000\phi = 240 - 50 \cdot R_a \]`

Solving for ϕ, we find:


`\[ \phi = \frac{{240 - 50 \cdot R_a}}{{45000}} \]`

Now, we can substitute this value of ϕ into the torque equation to find the resistance:


`\[ 15 = \frac{{\frac{{240 - 50 \cdot R_a}}{{45000}}}}{{30}} \cdot 50 \]`
Simplifying the equation, we find:


`\[ 15 = \frac{{240 - 50 \cdot R_a}}{{9000}} \]`
To solve for R_a, we can cross-multiply:


`\[ 15 \cdot 9000 = 240 - 50 \cdot R_a \]`

Simplifying, we find:


`\[ 135000 = 240 - 50 \cdot R_a \]`
Rearranging the equation, we find:


`\[ R_a = \frac{{240 - 135000}}{{-50}} \]`

Simplifying, we find:

\[ R_a = 2328 \Omega \]

A resistance of 2328 Ω should be inserted in series with the armature circuit to develop a torque of 15 N.m at a speed of 1500 RPM.