Efficiency Calculation of DC Shunt Motor by Swinburne’s Test
Question:
A 200-V, 14.92 kW DC shunt motor when tested by the Swinburne method gave the following results:
Running light: armature current was 6.5 A and field current 2.2 A.
With the armature locked, the current was 70 A when a potential difference of 3 V was applied to the brushes.
Estimate the efficiency of the motor when working under full-load conditions.
Given Data:
- Supply Voltage (\(V\)) = \(200\text{ V}\)
- Full-load shaft output (\(P_{\text{out}}\)) = \(14.92\text{ kW} = 14,920\text{ W}\)
- No-load armature current (\(I_{a0}\)) = \(6.5\text{ A}\)
- Shunt field current (\(I_{sh}\)) = \(2.2\text{ A}\)
- Locked rotor voltage across brushes (\(V_{br}\)) = \(3\text{ V}\)
- Locked rotor armature current (\(I_{br}\)) = \(70\text{ A}\)
Step 1: Armature Resistance (\(R_a\))
Locked armature conditionil brush voltage-um current-um upayogichu armature resistance kandethunnu:
$$R_a = \frac{V_{br}}{I_{br}} = \frac{3}{70} \approx 0.04286\ \Omega$$
Step 2: Constant Losses (\(W_c\))
No-load (running light) conditionil ulla losses:
-
No-load armature input:
$$P_{a0} = V \times I_{a0} = 200 \times 6.5 = 1300\text{ W}$$
-
No-load armature copper loss:
$$P_{cu0} = I_{a0}^2 \times R_a = (6.5)^2 \times 0.04286 \approx 1.81\text{ W}$$
-
Iron, friction and windage loss (Stray losses, \(W_m\)):
$$W_m = P_{a0} - P_{cu0} = 1300 - 1.81 = 1298.19\text{ W}$$
-
Shunt field copper loss (\(W_{sh}\)):
$$W_{sh} = V \times I_{sh} = 200 \times 2.2 = 440\text{ W}$$
-
Total constant losses (\(W_c\)):
$$W_c = W_m + W_{sh} = 1298.19 + 440 = 1738.19\text{ W}$$
Step 3: Full-Load Calculations and Efficiency
Method 1: Standard Approximate Method (Common Exam Method)
Full-load input power, output power-inu thulyamaanu ennu karuthiyaal:
$$I_{FL} \approx \frac{P_{\text{out}}}{V} = \frac{14,920}{200} = 74.6\text{ A}$$
Full-load armature current:
$$I_a = I_{FL} - I_{sh} = 74.6 - 2.2 = 72.4\text{ A}$$
Full-load armature copper loss:
$$P_{cu,FL} = I_a^2 \times R_a = (72.4)^2 \times 0.04286 \approx 224.66\text{ W}$$
Total losses at full load:
$$W_{\text{total}} = W_c + P_{cu,FL} = 1738.19 + 224.66 = 1962.85\text{ W}$$
Full-load electrical input power:
$$P_{\text{in}} = P_{\text{out}} + W_{\text{total}} = 14,920 + 1962.85 = 16,882.85\text{ W}$$
Full-load efficiency (\(\eta\)):
$$\eta = \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) \times 100 = \left( \frac{14,920}{16,882.85} \right) \times 100 \approx \mathbf{88.37\%}$$
Method 2: Exact Calculation
Armature-il develop cheytha mechanical power:
$$E_b I_a = P_{\text{out}} + W_m$$
$$(V - I_a R_a) I_a = 14,920 + 1298.19 = 16,218.19$$
$$0.04286 I_a^2 - 200 I_a + 16,218.19 = 0$$
Solving for \(I_a\):
$$I_a \approx 82.55\text{ A}$$
Full-load armature copper loss:
$$P_{cu,FL} = (82.55)^2 \times 0.04286 \approx 292.09\text{ W}$$
Total losses:
$$W_{\text{total}} = 1738.19 + 292.09 = 2030.28\text{ W}$$
Total power input:
$$P_{\text{in}} = 14,920 + 2030.28 = 16,950.28\text{ W}$$
Full-load efficiency (\(\eta\)):
$$\eta = \left( \frac{14,920}{16,950.28} \right) \times 100 \approx \mathbf{88.02\%}$$
Three-Point Starter for DC Shunt Motor
1. Need for a Starter
At starting, the motor armature is stationary, so the back EMF (\(E_b\)) is zero:
$$I_a = \frac{V - E_b}{R_a} = \frac{V - 0}{R_a} = \frac{V}{R_a}$$
Since the armature resistance (\(R_a\)) is very low, a dangerously high starting current flows through the armature. This can blow fuses, damage the commutator, and burn out the windings. A Three-Point Starter inserts variable external resistance into the armature circuit to safely restrict this starting current.
2. Circuit Diagram
3. Three Main Terminals
- L (Line Terminal): Connected to the incoming positive supply via the Overload Release (OLR) coil.
- A (Armature Terminal): Connected directly to the armature winding of the DC shunt motor.
- F (Field Terminal): Connected to the shunt field winding through the No-Volt Coil (NVC).
4. Construction & Working Operation
-
Starting Position (Stud 1):
To start the motor, the handle is gently pulled clockwise to touch Stud 1. At this point:
- The entire starting resistance is connected in series with the armature winding, effectively limiting the inrush starting current.
- The brass arc establishes a circuit directly feeding full supply voltage to the shunt field winding through the NVC, securing maximum flux (\(\Phi\)) and maximum starting torque.
- Gradual Acceleration (Studs 2 to 4): As the motor picks up speed, back EMF (\(E_b\)) builds up. The starting handle is moved progressively across studs 2, 3, and 4, gradually cutting out the external starting resistance.
- Normal Running Position (Stud 5 / RUN): In the final 'RUN' position, all starting resistance is completely cut out of the armature circuit, and the motor runs at rated speed.
5. Protective Devices in 3-Point Starter
- No-Volt Coil (NVC) / Under-Voltage Protection: The NVC is an electromagnet connected in series with the shunt field. In the 'RUN' position, it magnetically attracts the soft-iron keeper on the starting handle, holding it firmly against the tension of the spiral return spring. If the power supply fails or voltage drops below a safe limit, the NVC loses its magnetism, and the return spring pulls the handle back to the 'OFF' position.
- Overload Release (OLR) / Overload Protection: The OLR is an electromagnet connected in series with the main line terminal (\(L\)). If the motor draws excessive current due to overloading, the magnetic pull of the OLR lifts its movable iron plunger. This bridges the tripping contacts, short-circuiting the NVC coil. As a result, the NVC demagnetizes immediately, releasing the handle back to the 'OFF' position and cutting off the motor.
Define torque in dc motor and compare armature torque and shaft
torque
Torque in DC Motor: Armature Torque vs. Shaft Torque
1. Definition of Torque
Torque is defined as the turning or twisting moment of a force about an axis of rotation. In a DC motor, it is the rotational force produced on the armature conductors when current-carrying conductors interact with the magnetic field of the stator poles.
Mathematically, torque (\(T\)) is expressed as:
$$T = F \times r \quad (\text{N}\cdot\text{m})$$
where \(F\) is the tangential force exerted on the armature conductors (in Newtons), and \(r\) is the radius of the armature core (in meters).
2. Types of Torque in a DC Motor
A. Armature Torque (\(T_a\))
Armature Torque (Gross Torque): The total gross electromagnetic torque developed inside the armature winding due to electromechanical energy conversion.
The standard expression for armature torque is:
$$T_a = \frac{1}{2\pi} \left(\frac{P Z}{A}\right) \Phi I_a = 0.159 \times \Phi I_a \times \left(\frac{P Z}{A}\right) \quad (\text{N}\cdot\text{m})$$
Also, in terms of mechanical power developed:
$$E_b I_a = \omega T_a = \left(\frac{2\pi N}{60}\right) T_a \implies T_a = \frac{E_b I_a}{\left(\frac{2\pi N}{60}\right)} = 9.55 \times \frac{E_b I_a}{N} \quad (\text{N}\cdot\text{m})$$
B. Shaft Torque (\(T_{sh}\))
Shaft Torque (Net Output Torque): The useful net torque available at the motor shaft for driving the mechanical load. It is always less than the armature torque because a part of the gross torque is consumed to overcome internal iron losses (hysteresis and eddy current) and mechanical losses (friction and windage).
$$T_{sh} = T_a - T_f$$
where \(T_f\) is the lost torque due to mechanical and core friction losses.
In terms of useful output power (\(P_{\text{out}}\)):
$$P_{\text{out}} = \omega T_{sh} = \left(\frac{2\pi N}{60}\right) T_{sh} \implies T_{sh} = \frac{P_{\text{out}}}{\left(\frac{2\pi N}{60}\right)} = 9.55 \times \frac{P_{\text{out}}}{N} \quad (\text{N}\cdot\text{m})$$
3. Comparison Between Armature Torque and Shaft Torque
| Feature | Armature Torque (\(T_a\)) | Shaft Torque (\(T_{sh}\)) |
|---|---|---|
| Definition | Gross torque developed internally in the armature. | Net useful torque delivered at the output shaft. |
| Magnitude | Higher (\(T_a > T_{sh}\)). | Lower due to rotational losses. |
| Power Association | Corresponds to gross electrical power converted (\(E_b I_a\)). | Corresponds to net mechanical shaft output power (\(P_{\text{out}}\)). |
| Loss Consideration | Does not account for iron, friction, and windage losses. | Derived after subtracting mechanical & iron losses (\(T_a - T_f\)). |
| Formula | $$T_a = 9.55 \times \frac{E_b I_a}{N}$$ | $$T_{sh} = 9.55 \times \frac{P_{\text{out}}}{N}$$ |
| Practical Significance | Indicates electromagnetic performance and torque generation capability. | Determines the actual load-driving capacity of the motor. |
$$\text{Lost Torque } (T_f) = T_a - T_{sh} = 9.55 \times \frac{\text{Mechanical \& Iron Losses}}{N}$$
Three-Point Starter for DC Shunt Motor
1. Need for a Starter
When a DC motor is at rest, the armature is stationary, meaning the back EMF (\(E_b\)) is zero. The armature current is governed by Ohm's law:
$$I_a = \frac{V - E_b}{R_a} = \frac{V - 0}{R_a} = \frac{V}{R_a}$$
Because the armature winding resistance (\(R_a\)) is extremely small, connecting the motor directly across the full supply voltage would cause an enormous inrush current (typically 10 to 20 times the rated full-load current). This excessive current can:
- Blow out fuses or trip line breakers.
- Damage the commutator surface and produce severe brush sparking.
- Burn out the armature winding insulation due to excessive \(I^2 R\) heating.
To prevent this, a Three-Point Starter inserts an external variable starting resistance into the armature circuit at standstill and gradually cuts it out as the motor gains speed and establishes back EMF.
2. Schematic Diagram
3. Main Terminals
- L (Line Terminal): Connected to the positive DC power supply line through the Overload Release (OLR) coil.
- A (Armature Terminal): Connected directly to the motor's armature terminal.
- F (Field Terminal): Connected to the motor's shunt field winding in series with the No-Volt Coil (NVC).
4. Construction and Operation
-
Starting Step (Stud 1):
The operator pulls the handle clockwise from the 'OFF' position until it touches Stud 1. At this point:
- The full bank of starting resistance is placed directly in series with the armature winding, suppressing the heavy starting inrush current.
- The handle simultaneously makes contact with the continuous brass arc, which feeds the full line voltage directly to the shunt field winding via the NVC coil. This establishes maximum field flux (\(\Phi\)) right away, producing high starting torque.
- Intermediate Steps (Studs 2 to 4): As the rotor begins turning, it produces a proportional back EMF (\(E_b\)). This opposing voltage naturally curbs the armature current. The operator advances the lever smoothly across studs 2, 3, and 4, cutting out sections of the external starting resistance in steps.
-
Normal Running Position (Stud 5 / 'RUN'):
Once the handle reaches Stud 5:
- All external starting resistance is completely removed from the armature circuit, allowing the motor to run at full rated speed.
- The soft-iron piece attached to the handle is held firmly in place by the magnetic pull of the energised No-Volt Coil (NVC), working against the tension of the spiral return spring.
5. Built-in Protective Mechanisms
- No-Volt Release (NVC) — Low/Zero Voltage Protection: If the incoming power line fails or line voltage drops below a minimum threshold, the current through the shunt circuit drops. The NVC demagnetizes and loses its grip on the handle. The spiral spring promptly snaps the handle back to the 'OFF' position, preventing sudden, uncontrolled restarts when supply voltage is restored.
- Overload Release (OLR) — Over-Current Protection: If the motor experiences an excessive mechanical overload, the resulting heavy line current flows through the series OLR coil. This creates a strong electromagnetic field that pulls up the movable iron plunger. The plunger bridges two fixed tripping contacts beneath it, short-circuiting the NVC coil. Deprived of current, the NVC loses its magnetic hold, allowing the return spring to snap the handle back to 'OFF' and disconnect the motor.