Speed Regulation of DC Motor

---

---

Speed of a DC Motor

The expression for the speed of a DC motor can derived as follows −

The back EMF of a DC motor is given by,

$$\mathrm{E_{b} \: = \: V \: - \: I_{a}R_{a} \:\: \dotso \: (1)}$$

Also,

$$\mathrm{E_{b} \: = \: \frac{NP \: \varphi \: Z}{60 A} \:\: \dotso \: (2)}$$

From eq. (1) & (2), we get,

$$\mathrm{\frac{NP \: \varphi \: Z}{60A} \: = \: V \: - \: I_{a}R_{a}}$$

$$\mathrm{\Rightarrow \: N \: = \: \left(\frac{V \: - \: I_{a}R_{a}}{\varphi}\right) \: \times \: \frac{60A}{PZ}}$$

For a given DC motor, the (60A/PZ) = K (say) is a constant.

$$\mathrm{\therefore \: N \: = \: K \left(\frac{V \: - \: I_{a}R_{a}}{\varphi}\right)}$$

But,

$$\mathrm{(V \: - \: I_{a}R_{a}) \: = \: E_{b}}$$

Therefore,

$$\mathrm{N \: = \: K \left(\frac{E_{b}}{\varphi}\right) \:\: \dotso \: (3)}$$

$$\mathrm{\Rightarrow \: N \: \varpropto \: \frac{E_{b}}{\varphi}\:\: \dotso \:(4)}$$

Hence, the speed of a DC motor is directly proportional to back emf and is inversely proportional to flux per pole.

Speed regulation of a DC Motor

The speed regulation of a motor is defined as the change in the speed from full-load to no-load and is expressed as a percentage of the full-load speed.

$$\mathrm{\% \: \text{ Speed regulation } \: = \: \frac{\text{(No load speed) - (Full load speed)}}{\text{Full load speed}} \: \times \: 100\:\%}$$

$$\mathrm{\Rightarrow \: \% \: \text{ Speed regulation } \: = \: \frac{N_{NL} \: - \: N_{FL}}{N_{FL}} \: \times \: 100\:\%}$$

Numerical Example

A 250 V DC shunt motor takes 6 A at no-load and runs at 1500 RPM. Calculate the speed of the motor when loaded and taking a current of 36 A. The armature and shunt field resistances are 0.3 Ω and 250 Ω respectively. Also, calculate the percentage speed regulation of the motor.

Solution

Let, N₂ is the speed of the motor under loaded condition.

Here,

$$\mathrm{\text{Shunt field current, } \: I_{sh} \: = \: \frac{V}{R_{sh}} \: = \: \frac{250}{250} \: = \: 1\: A}$$

Case 1 - Motor at no-load

$$\mathrm{I_{a1} \: = \: I \: - \: I_{sh} \: = \: 6 \: - \: 1 \: = \: 5 A}$$

$$\mathrm{E_{b1} \: = \: V \: - \: I_{a1}R_{a} \: = \: 250 \: - \: (5 \: \times \: 0.3) \: = \: 248.5 V}$$

$$\mathrm{N_{1} \: = \: 1500 \: RPM \:(given)}$$

Case 2 - Motor on load

$$\mathrm{I_{a2} \: = \: I \: - \: I_{sh} \: = \: 36 \: - \: 1 \: = \: 35 A}$$

$$\mathrm{E_{b2} \: = \: V \: - \: I_{a2}R_{a} \: = \: 250 \: - \: (35 \: \times \: 0.3) \: = \: 239.5 V}$$

In a DC motor, the speed is given by,

$$\mathrm{N \: \propto \: \frac{E_{b}}{\varphi}}$$

Since it is a shunt motor, in which flux is constant, hence,

$$\mathrm{N \: \varpropto \: E_{b}}$$

$$\mathrm{\Rightarrow \: \frac{N_{2}}{N_{1}} \: = \: \frac{E_{b2}}{E_{b1}}}$$

Therefore, speed of the motor under loaded conditions,

$$\mathrm{N_{2} \: = \: N_{1} \: \times \: \frac{E_{b2}}{E_{b1}} \: = \: 1500 \: \times \: \left(\frac{239.5}{248.5}\right) \: = \: 1445.6\:RPM}$$

The percentage speed regulation is −

$$\mathrm{\% \: \text{ speed regulation } \: = \: \frac{N_{NL} \: - \: N_{FL}}{N_{FL}} \: \times \: 100 \: = \: \frac{1500 - 1445.6}{1445.6} \: \times \: 100}$$

$$\mathrm{\% \: \text{ speed regulation } \: = \: 3.76 \: \%}$$