Speed and 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,

$$\mathrm{E_{b} = V − I_{a}R_{a} … (1)}$$

Also,

$$\mathrm{𝐸_{b} =\frac{NP\varphi𝑍}{60𝐴}\:… (2)}$$

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

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

$$\mathrm{⇒ N = (\frac{V − I_{a}R_{a}}{\varphi}) \times\frac{60A}{PZ}}$$

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

$$\mathrm{\therefore N = K (\frac{V − I_{a}R_{a}}{\varphi})}$$

But,

$$\mathrm{(V − I_{a}R_{a}) = E_{b}}$$

Therefore,

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

$$\mathrm{⇒ N \varpropto\frac{E_{b}}{\varphi}\:......(4)}$$

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.

$$\mathrm{\% \:Speed regulation =\frac{(No\:load \:speed) − (Full\: load \:speed)}{Full\: load \:speed} × 100\:\%}$$

$$\mathrm{⇒ \% \:Speed\: regulation =\frac{N_{NL} − N_{FL}}{N_{FL}}× 100\:\%}$$

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,

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

Case 1 – Motor at no-load −

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

$$\mathrm{E_{b1} = V − I_{a1}R_{a} = 250 − (5 × 0.3) = 248.5 V}$$

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

Case 2 – Motor on load −

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

$$\mathrm{E_{b2} = V − I_{a2}R_{a} = 250 − (35 × 0.3) = 239.5 V}$$

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,

$$\mathrm{𝑁 \varpropto 𝐸_{𝑏}}$$

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

Therefore, speed of the motor under loaded conditions,

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

The percentage speed regulation is −

$$\mathrm{\% speed\:regulation =\frac{N_{NL}− N_{FL}}{N_{FL}}× 100 =\frac{1500 − 1445.6}{1445.6}× 100}$$

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