There are different methods of **testing of dc machines**.

## Brake Test of DC Machine

Brake test is a direct method. We apply here a brake to a water-cooled pulley mounted on the motor shaft. We fix the brake band with the help of wooden blocks gripping on the pulley. Then we tie one end of the brake band with the ground through a spring balance (S). We hang a weight (W1) to the other end of the brake band. Then we run the motor and increase the weight until the motor takes it rated full load current from the supply mains.

Let us consider, W1 kg is the suspended weight, and W2 kg is the reading of the spring balance while the motor takes its full load current.

Therefore, the net pulling force on the band due to friction is (W1 – W2) kg.

Now, we reconsider R as the radius of the pulley and N is the rpm of the motor. Thus the torque developed at the shaft is,

The mechanical output of the motor will be,

Putting the expression of Tsh we get,

Now if V is the supply voltage and I is the current drawn by the motor at full load condition, we can write the input power to the motor as,

From here we can calculate the overall or commercial efficiency of the dc motor as

The abovementioned brake test for dc motor is only suitable for small sized dc motor. Because, in big dc motor, the braking creates quite a good amount of heat. Hence colling becomes difficult in this case.

## Swinburne’s Test of DC Machine

In this test, we measure the losses separately. From the measured value of the losses, we can determine the efficiency of the machine. Here we can do that without running the machine with any load. Although we can apply this Swinburne’s test only for those machines which have a constant field flux. Such as for shunt dc motor, shunt dc generators, compounded wound dc motors and generators.

### Swinburne’s Test of DC Motor

For illustrating this test, let us take an example of a dc shunt motor. Here, we feed the motor with rated voltage (V). We can get the information of rated voltage and rated speed from its rating plate attached on its body. Now we adjust the speed of the motor to its rated speed by a regulator attached to its shunt field. Then we will take the reading of the ammeter (A_{1}) connected in the supply circuit. This is the no-load current (I) of the motor. At the same time, the ammeter connected to the shunt field circuit gives the shunt field current (I_{sh}) at this condition. Therefore, the no-load current through the armature is (I – I_{sh}).

So, the no-load input power to the motor is VI watts. Among this, the input power to the armature is V(I – I_{sh}) and input power to the shunt field is VI_{sh}.

The input power to the armature at no-load condition consists of iron losses in the core, friction loss in the bearing, windage losses in the rotating surface of the machine, and armature copper losses.

Now, we know the armature current (I-I_{sh}) and we can measure the resistance of the armature circuit. So, we can easily calculate the armature copper loss. For calculating armature copper loss we use ‘hot’ resistance.

### What is hot resistance?

Sometimes we have to measure the resistance of a circuit at room temperature. But when we calculate the voltage drop or copper loss across the resistance, we have to consider the current, flowing to the resistance. But when current flows through the resistance, the temperature fo the resistance rises, which causes an increase of the resistance value. So, if we calculate the voltage drop or copper loss with the measured resistance value it may be erroneous.

So, we have to predict the steady temperature of the resistance at that current. Then we calculate the resistance value of the circuit at that higher temperature for finding out the actual value of voltage drop and copper loss. We refer this calculated resistance at that higher temperature i.e. at the hot condition of the resistor as hot resistance. Without going through the process of calculating hot resistance, let us consider the hot resistance of the armature circuit is R_{a}‘.

Hence actual armature copper loss is

The sum of the constant losses is

### Calculating Efficiency of DC Motor

For that let us see the rated full load current of the machine. Let us consider it as I’. So, the full-load armature current is (I – I_{sh}) = I_{a}‘. The full-load input power is Hence, the output of the motor is

Hence, the efficiency of the dc shunt motor is

### Swinburne’s Test of DC Generator

Here, we run the dc shunt generator at its rated speed with a prime mover.

The shunt field current is Ish as we read from the ammeter A. The rated current and voltage of the generator is I and V respectively. Since the shunt current I_{sh} is constant the shunt current remains the same at the full-load condition of the generator, also. Hence, at the full-load condition, the armature current is I + I_{sh}. Hence the copper loss in the armature is,

The constant loss (Wc) what we have already determined during Swinburne’s test of dc motor is the same when the dc machine runs as a generator. So, we can consider the total loss of the dc generator as,

Therefore, the input power of the generator is

Since the output power is

Because the rated current and voltage of the generator is I and V respectively.

### Advantages of Swinburne’s Test

- This is actually a no-load test. Hence, the machine takes only power to compensate for the losses. Thus this test is economical for big size dc machine.
- Swinburne’s test gives the value of constant losses. Hence the test can determine the efficiency of dc motos and efficiency of dc generators at any loading condition.

### Disadvantages of Swinburne’s Test

- In this test, we only use the full load current and voltage taken from the rating plate of the machine. We do not bother about the armature reaction of the machine at its actual loaded condition. Actually, in the loaded condition of the machine, there may be an increase in iron loss due to the distortion of flux patterns because of the armature reaction.
- As it is a no-load test, Swinburne’s test does not predict whether there will be a satisfactory commutation at a loaded condition or not.

## Running Down Test of DC Motor and Generator

We apply this test to shunt motors and shunt generators. By this test, we find out the stray losses of DC machines. Now if we know the shunt copper loss and armature copper loss at full load of the DC machine, we can easily calculate the efficiency of the machine from this **running down test**.

During this test, we run the machine beyond its rated speed. Then suddenly we cut off the supply from the armature. As a result, the armature slows down. At that situation, the kinetic energy of the armature entirely compensates the rotational losses of the machine. The rotational losses are friction losses, windage loses and iron loses.

The kinetic energy of the armature is

Where ‘I’ is the moment of inertia of the armature and ω is the angular velocity.

Again, rational losses (W) is the rate of loss of kinetic energy.

In the above equation, we need to know two quantities

- The moment of inertia (I).
- The rate of change of angular speed with respect to time (dω/dt).

### Method of Finding out dω/dt

For finding out dω/dt, we have to connect one voltmeter across the armature. Then we rotate the machine as a motor with a speed beyond its rated speed. After that suddenly we switch off the supply. Hence, the motor speed falls down. Again the back emf or generated voltage across the armature of the motor is directly proportional to the speed of the motor.

Now the voltmeter connected across the armature will show the fall off voltage across the armature with time. Then we draw a graph showing the voltmeter reading at various time intervals during the falling of speed of the motor. By doing this, we will get a curve as shown in the figure below. From a point P which corresponds to the normal speed of the motor, we draw a tangent AB on the curve.

Then

This is because the speed of the motor (N) is directly proportional to the induced voltage back emf across the armature (E_{b}). Now let us assume that N rpm is rated normal speed of the motor. Again we know that

Therefore from the equation of loss (W) we can write

### Method of Finding out Moment of Inertia

In this method first, we allow to slow down the armature after cutting off the supply mains. This gives a slow down curve as in the case of finding out dω/dt. Now we fit a flywheel of inertia I’. Now after running the motor beyond its rated speed again we cut the supply mains and allow the motor to slow down.

In the second case, because of the moment of inertia of the flywheel along with the moment of inertia of armature, the time taken by the system for slowing down is more. Now let us find out the rate of change of speed with respected time at the normal speed of the motor for both the cases from the curves. Let us consider the value of those rate of the slowdown of the system are dN/dt1 and dN/dt2. But the total loss occurs in the system in both cases would be the same ideally. Hence we can write for the first case,

In the second case,

Therefore finally, we can write

From here we can write

So finally, we can calculate the inertia of the armature (I) as

## Field Test of DC Machines

We perform the **field test** on two similar series motors. The traction series motors are readily available in pair. We mechanically couple these two same dc series motors. Here firstly one dc rotating machine runs as a standard motor. The second one runs as a dc generator with the help of the prime mover coupled with the shaft of the first rotating machine. The variable resistor connected across the second machine (i.e., generator) consumes the power output of the system.

In the **field test of dc machines**, we disconnect the series field of the second machine. Then we reconnect this series field in series with the armature of the first machine (i.e., motor). We do this because we need to make equal copper losses for the field of both identical machines. Since, at this condition, the same current flows through both of the fields.

Again, since both of the mechanically coupled similar dc machines run with equal speed, the friction, windage losses and also iron losses of both machines would be equal.

Let us take, V as supply voltage, I_{1} as input current to the motor, V_{2} as the output voltage of the generator, and I_{2} as the generator load current.

Hence, total intake power to the system isAnd, the output of the system isTherefore,Hence,

So,

Since, both the dc machines are identical

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