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# What is Speed Control of DC Shunt Motors?

The speed of a DC shunt is given by,

$$\mathrm{𝑁 \varpropto\frac{𝐸_{𝑏}}{\varphi}}$$

$$\mathrm{⇒ 𝑁 = 𝐾 (\frac{𝑉 − 𝐼_{𝑎}𝑅_{𝑎}}{\varphi})\: … (1)}$$

It is clear from the equation (1) that the speed of a DC shunt motor can be changed by two methods −

- Flux Control Method
- Armature Resistance Control Method

## Flux Control Method

The *flux control method* is based on the principle that by varying the field flux ϕ, the speed of DC shunt motor can be changed.

$$\mathrm{𝑁 \varpropto\frac{1}{\varphi}}$$

In this method, a variable resistance (called *field rheostat*) is connected in series with the shunt field winding. By increasing the resistance of the field rheostat, the shunt field current Ish can be reduced and hence the field flux. Thus, by the flux control method, the speed of a DC shunt can only be increased above the normal speed.

The flux control method is frequently used for the speed control of DC shunt motors because it is simple and inexpensive method.

## Advantages

The flux control method for the speed control of DC shunt motor has following advantages −

- It is a simple and convenient method.
- It is an inexpensive method as the small power loss in the field rheostat due to the small value of I
_{sh}. - The speed control using flux control method is independent of the load on the machine.

## Disadvantages

Following are the disadvantages of the flux control method −

- By this method, only speeds higher than the normal speed can be obtained because the total resistance of the field circuit cannot be decreased below shunt field winding resistance (R
_{sh}). - In flux control method, there is a limit to the maximum speed obtainable, because if the field flux is too much weakened, the commutation becomes poorer.

## Armature Resistance Control Method

The *armature resistance control method* is based on the principle that by varying the voltage available across the armature, the back EMF of the motor can be changed, which in turn changes the speed of the shunt motor.

In this method, a variable resistance RC (called *controller resistance*) is inserted in series with the armature.

The speed of DC shunt motor is given by,

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

Also,

$$\mathrm{𝐸_{𝑏} = 𝑉 − 𝐼_{𝑎}(R_{a} + 𝑅_{𝐶})}$$

Thus, due to the voltage drop in the controller resistance, the back EMF is decreased and hence the speed of the motor. The maximum speed that can be obtained using armature resistance control method is the speed corresponding to R_{C} = 0, i.e., the normal speed. Therefore, by this method only speed below the normal speed can be obtained.

## Disadvantages

The armature resistance control method has following disadvantages −

- A large amount of power being wasted in the controller resistance since it carries full armature current.
- The output and efficiency of the motor being decreased.
- This method of speed control results in the poor speed regulation.
- The speed changes with the variation in the load because the speed depends upon the voltage drop across the controller resistance and hence on the armature current demanded by the load.

## Numerical Example

A 250 V DC shunt motor having an armature resistance of 0.3 Ω carries an armature current of 60 A and runs at 800 RPM. If the flux is decrease by 15 % by the field rheostat. Find the speed of the motor assuming the load torque remains the same.

## Solution

**Case 1 – Normal conditions −**

$$\mathrm{𝑁_{1} = 800 RPM ; \:𝐼_{𝑎1} = 60 A}$$

$$\mathrm{\therefore 𝐸_{𝑏1} = 𝑉 − 𝐼_{𝑎1}R_{a} = 250 − (50 × 0.3) = 235 V}$$

**Case 2 – Final conditions after 15 % reduction in the flux −**

$$\mathrm{\varphi_{2} = 0.85\varphi_{2}}$$

$$\mathrm{⇒\frac{\varphi_2}{\varphi_1}= 0.85}$$

Since the load torque is given to be constant, hence,

$$\mathrm{\varphi_{1}𝐼_{𝑎1} = \varphi_{2}𝐼_{𝑎2}}$$

$$\mathrm{⇒ 𝐼_{𝑎2} =\frac{\varphi_1}{\varphi_2} \times 𝐼_{𝑎1} =\frac{1}{0.85} \times 60 = 70.58 A}$$

$$\mathrm{\therefore 𝐸_{𝑏2} = 𝑉 − 𝐼_{𝑎2}R_{a} = 250 − (70.58 × 0.3) = 228.83 V}$$

Therefore,

$$\mathrm{\frac{𝑁_{2}}{𝑁_{1}}=\frac{𝐸_{𝑏2}}{𝐸_{𝑏1}}\times\frac{\varphi_{1}}{\varphi_2}}$$

$$\mathrm{⇒𝑁_{2}=\frac{228.83}{235}\times\frac{1}{0.85}× 800}$$

$$\mathrm{\therefore\:𝑁_{2}= 916.46 RPM}$$

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