Braking Of DC Shunt Motors | Electrical Machines - Electrical Engineering (EE) PDF Download

Braking of DC Shunt Motors

(Refer Slide Time: 00:32)

Speed change by field-weakening - worked example

Consider a DC shunt motor with the following given data.

• Rated terminal voltage, Vt = 240 V
• Armature resistance, Ra = 0.04 Ω
• Field (shunt) resistance (intrinsic), Rf = 100 Ω
• Load current, Il = 200 A
• Initial speed, N1 = 1200 rpm
• Desired speed, N2 = 1500 rpm

We wish to obtain the additional field resistance Rfe that must be inserted in series with the shunt field so that the motor speed increases from 1200 rpm to 1500 rpm while the supply voltage and the load current remain as given.

Analysis and calculation

At Il = 200 A the field current at the starting condition (no extra field resistance) is

If1 = Vt / Rf = 240 / 100 = 2.4 A.

The armature current is

I a1 = Il - If1 = 200 - 2.4 = 197.6 A.

The back EMF (E) at this operating point is

E = Vt - Ia Ra = 240 - 197.6 × 0.04 = 240 - 7.904 = 232.096 V (≈ 232.10 V).

For the same load current Il and same armature current (assuming the same Ia when Rfe is inserted), the back EMF E remains essentially the same. In a linear magnetic circuit the induced emf E is proportional to flux φ and speed N by the relation E ∝ φ N. Thus

E1 = k φ1 N1 and E2 = k φ2 N2, and since E1 = E2 we have φ1 N1 = φ2 N2.

Therefore φ1 / φ2 = N2 / N1 = 1500 / 1200 = 5 / 4.

Assuming linearity, flux is proportional to field current If (for constant field winding geometry), so

If1 / If2 = 5 / 4 ⇒ If2 = (4 / 5) If1.

With constant supply voltage, field current is inversely proportional to the total field circuit resistance, so

Rf_total@1500 = (5 / 4) × Rf@1200 = (5 / 4) × 100 = 125 Ω.

Hence the additional resistance required is

Rfe = Rf_total@1500 - Rf = 125 - 100 = 25 Ω.

Result: Insert a series resistance of 25 Ω in the field circuit to raise speed from 1200 rpm to 1500 rpm under the stated load.

Effect of change in load current with the added field resistance

(Refer Slide Time: 08:07)

With the added Rfe = 25 Ω the total field resistance becomes 125 Ω and the field current becomes If = 240 / 125 = 1.92 A.

Consider the two load cases and compare speeds.

Case A: Il = 200 A (as above)

Ia = Il - If = 200 - 1.92 = 198.08 A (≈ 198.1 A).

E = Vt - Ia Ra = 240 - 198.08 × 0.04 = 240 - 7.9232 = 232.0768 V (≈ 232.08 V).

Case B: Il = 100 A

Ia = Il - If = 100 - 1.92 = 98.08 A.

E100 = Vt - Ia Ra = 240 - 98.08 × 0.04 = 240 - 3.9232 = 236.0768 V (≈ 236.08 V).

The speed ratio is given by N100 / N200 = E100 / E200 ≈ 236.08 / 232.08 ≈ 1.0172.

Thus N100 ≈ 1.0172 × 1500 ≈ 1525.8 rpm (≈ 1530 rpm using rounded values).

Observation: Reducing the load from 200 A to 100 A with the same field weakening causes the speed to rise by ≈ 25-30 rpm. This illustrates that speed regulation with field control (weakening) is relatively poor - load changes produce non-negligible speed variations.

Ward-Leonard method - worked example

(Refer Slide Time: 13:30)

Consider a Ward-Leonard system where two identical DC machines (motor and generator) have the following ratings:

• Terminal voltage, Vt = 230 V
• Power rating, P = 4.5 kW
• Rated motor speed, Nm,rated = 1500 rpm
• Armature resistance (each), Ra = 0.5 Ω

The magnetisation (open-circuit) characteristic of the generator at 1500 rpm gives the following important points (from the data):

• At If = 0.2 A → Eg ≈ 110 V
• At If = 0.3 A → Eg ≈ 148 V
• At If ≈ 0.8 A → Eg ≈ 230 V (this is the motor induced emf at rated speed with Ifm = 0.8 A)
• At If ≈ 1.2 A → Eg ≈ 251 V

The motor field current Ifm is set at 0.8 A. The motor speed must be varied between 300 rpm and 1500 rpm while the motor output power remains Pm = 4.5 kW over the full range. Determine the required range of the generator field current Ifg.

Case 1 - Motor speed Nm = 300 rpm

Induced emf of the motor scales with speed, so

Em(300) = (300 / 1500) × Em(1500) = (1 / 5) × 230 = 46 V.

Output power Pm = 4.5 kW = 4500 W. The armature current at 300 rpm is

Ia(300) = Pm / Em(300) = 4500 / 46 ≈ 97.8 A.

The generator must supply the armature current and must overcome the armature drops in both machines, therefore the required generator terminal (open-circuit) voltage is

Eg = Em(300) + 2 Ra Ia(300) = 46 + 2 × 0.5 × 97.8 = 46 + 97.8 = 143.8 V.

From the magnetisation data, Eg ≈ 143.8 V lies between the values at If = 0.2 A (110 V) and If = 0.3 A (148 V). Interpolating linearly gives Ifg ≈ 0.29 A.

Case 2 - Motor speed Nm = 1500 rpm

Em(1500) = 230 V (from the OCC at Ifm = 0.8 A).

Ia(1500) = Pm / Em(1500) = 4500 / 230 ≈ 19.57 A (≈ 19.6 A).

Eg = Em(1500) + 2 Ra Ia(1500) = 230 + 2 × 0.5 × 19.57 = 230 + 19.57 = 249.57 V (≈ 249.6 V).

From the magnetisation data Eg ≈ 249.6 V corresponds to approximately Ifg ≈ 1.18 A (interpolating near the point If ≈ 1.2 A gives ≈ 251 V).

Result: The generator field current must be varied approximately over the range Ifg ≈ 0.29 A to 1.18 A to cover the motor speed range from 300 rpm to 1500 rpm while delivering 4.5 kW.

Braking and stopping of DC shunt motors

(Refer Slide Time: 25:35)

Stopping (braking) of a motor is as important as starting and speed control. Consider an elevator (lift) drive which has an elevator cage weight Wcg and a counterweight Wcc. The motor must be capable of rotating in either direction depending on upward or downward motion. Different operating situations require different braking strategies. The principal electrical braking methods for a DC shunt motor are:

• Regenerative braking
• Dynamic (rheostatic) braking
• Plugging (reverse supply) braking

Regenerative braking

(Refer Slide Time: 29:24)

When the load torque tends to drive the motor above its no-load speed (for example, when lowering a lightly loaded elevator so that the machine is driven by the load), the motor acts as a generator and feeds power back to the supply. The machine remains motoring for torques up to the point where the operating point crosses into the generating region; in the generating region the armature current reverses sign and an opposing torque is produced which tends to slow the machine. This is the regenerative braking mode.

Important points:

• Regenerative braking requires the supply (or converter/inverter) to accept returned energy.
• In regenerative mode the machine often runs at a speed higher than the rated no-load speed; it cannot by itself bring the machine to rest unless the applied armature voltage is reduced to zero (or energy is diverted), so regenerative alone may not stop the machine completely.

Dynamic (rheostatic) braking

(Refer Slide Time: 36:18)

In dynamic braking the motor is disconnected from the supply and its armature is connected to an external braking resistor Rb. The motor therefore acts as a generator: the kinetic energy of the rotating system is converted into electrical energy and dissipated in Rb as heat.

With terminal voltage set to zero and the braking resistance Rb inserted the torque-speed characteristic becomes a straight line through the origin whose slope depends on Rb. The torque in braking mode is given by the generator equation:

T = (k φ) / (Ra + Rb) × (k φ) ω = (k φ)² / (Ra + Rb) × ω (braking torque proportional to speed)

Thus when dynamic braking is applied the operating point moves smoothly from the motoring characteristic to the braking characteristic and the speed decelerates until stop. An external braking resistor is necessary to limit the generated current; without it the regenerated current can be large (comparable with starting currents) and undesirable.

Plugging (reverse supply) braking

(Refer Slide Time: 42:06)

Plugging is a method in which the armature supply polarity is reversed while the field polarity is usually left unchanged. The effect is to produce a large opposing torque immediately. In practice the machine is disconnected from the normal supply and the polarity of the applied armature voltage is reversed; an additional resistance Rp is inserted to limit the current.

(Refer Slide Time: 44:15)

When plugging is applied the torque-speed characteristic shifts so that an opposing torque acts on the motor at the existing speed. The operating point jumps from the motoring point to the plugging characteristic and the speed reduces. Because the plugging torque is large, plugging gives fast deceleration.

Important caution: If the supply polarity remains reversed long enough the motor will eventually accelerate in the reverse direction. Therefore the supply must be disconnected and the motor isolated near the instant that the speed reaches (or approaches) zero.

Equivalent circuit and power during plugging

(Refer Slide Time: 47:37)

While the rotor is still turning in the original direction (positive speed) and the supply polarity is reversed, the equivalent circuit shows the reversed supply voltage, the induced back EMF E (due to rotation), the armature resistance Ra and the plugging resistance Rp. The armature current magnitude is given by

Ia = (V + E) / (Ra + Rp)

Here V and E are algebraic magnitudes taking the reversed polarity into account. During plugging both the internal generated emf and the external source supply energy which is dissipated in Ra and Rp; thus plugging is inefficient and wastes energy in resistances.

Worked example - plugging resistance, current and torques

(Refer Slide Time: 49:40)

Given: a DC shunt motor with the following data:

• Terminal voltage, V = 220 V
• Rated speed, Nrated = 970 rpm
• Rated armature current, Iarated = 100 A
• Armature resistance, Ra = 0.05 Ω

We wish to apply plugging from an initial speed N = 1000 rpm and limit the inrush plugging current to at most twice the rated armature current (i.e. Ia,max ≤ 200 A). Find the required plugging resistance Rp, the braking torque at the initial speed, and the torque at zero speed.

Step-by-step calculation

At rated armature current (100 A) and rated speed 970 rpm the back EMF is

Erated = V - Iarated Ra = 220 - 100 × 0.05 = 220 - 5 = 215 V.

At N = 1000 rpm the induced emf scales with speed, so

E1000 = (1000 / 970) × 215 = 1.0309278 × 215 = 221.65 V (approx).

To ensure Ia ≤ 200 A when plugging is applied (supply polarity reversed), the total resistance (Ra + Rp) must satisfy

Ra + Rp ≥ (V + E1000) / Ia,max = (220 + 221.65) / 200 = 441.65 / 200 = 2.20825 Ω.

Therefore the plugging resistance required is

Rp = 2.20825 - Ra = 2.20825 - 0.05 = 2.15825 Ω ≈ 2.16 Ω.

Braking torque at initial speed (N = 1000 rpm) with Ia = 200 A is obtained from electrical power balance: electrical power developed = E × Ia = mechanical power developed = T × ω. Thus

ω1000 = 2π × (1000 / 60) = 104.7198 rad/s (approx).

Tinitial = E1000 × Ia / ω1000 = 221.65 × 200 / 104.7198 ≈ 423.5 N·m.

At zero speed (E = 0) the armature current is

Ia0 = V / (Ra + Rp) = 220 / 2.20825 ≈ 99.66 A.

The torque at zero speed (proportional to Ia) is

T0 = Tinitial × (Ia0 / 200) = 423.5 × (99.66 / 200) ≈ 211.1 N·m.

Summary of results for the example:

• Required plugging resistance Rp ≈ 2.16 Ω
• Initial braking torque at 1000 rpm ≈ 423.5 N·m
• Torque at zero speed (with the resistance connected) ≈ 211.1 N·m

Remarks:

• Plugging produces a high braking torque and rapid deceleration, but if the reversed supply is left connected the motor will restart in the opposite direction. Therefore the supply must be opened before the motor reverses.
• Plugging is inefficient: both the source and the generated emf contribute to losses in Ra and Rp during braking.
• Dynamic (rheostatic) braking is generally used to dissipate kinetic energy in a resistor and is simpler; regenerative braking is energy-efficient if the supply (or converter) can accept returned energy but cannot usually bring the machine to rest without additional means.

Thank you.