Losses, Efficiency and Testing of D.C. Machines

 

Chapter 40

Losses, efficiency and testing of D.C. machines (Lesson-40)

40.1 Goals of the lesson

1. To know what are the various power losses which take place in a d.c machine.

2. To understand the factors on which various losses depend upon.

3. To know how to calculate efficiency.

4. To know how to estimate and predict efficiency of a d.c shunt motor by doing simple tests.

40.2 Introduction

In the previous sections we have learnt about the principle of operation of d.c. generators and motors, (starting and speed control of d.c motor). Motors convert electrical power (input power) into mechanical power (output power) while generators convert mechanical power (input power) into electrical power (output power). Whole of the input power can not be converted into the output power in a practical machine due to various losses that take place within the machine. Efficiency η being the ratio of output power to input power, is always less than 1 (or 100 %). Designer of course will try to make η as large as possible. Order of efficiency of rotating d.c machine is about 80 % to 85 %. It is therefore important to identify the losses which make efficiency poor.

In this lesson we shall first identify the losses and then try to estimate them to get an idea of efficiency of a given d.c machine.

40.3 Major losses

Take the case of a loaded d.c motor. There will be copper losses ()22 and aafffIrIRVI= in armature and field circuit. The armature copper loss is variable and depends upon degree of loading of the machine. For a shunt machine, the field copper loss will be constant if field resistance is not varied. Recall that rotor body is made of iron with slots in which armature conductors are placed. Therefore when armature rotates in presence of field produced by stator field coil, eddy current and hysteresis losses are bound to occur on the rotor body made of iron. The sum of eddy current and hysteresis losses is called the core loss or iron loss. To reduce core loss, circular varnished and slotted laminations or stamping are used to fabricate the armature. The value of the core loss will depend on the strength of the field and the armature speed. Apart from these there will be power loss due to friction occurring at the bearing & shaft and air friction (windage loss) due to rotation of the armature. To summarise following major losses occur in a d.c machine.

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1. Field copper loss: It is power loss in the field circuit and equal to2ffIRVI=. During the course of loading if field circuit resistance is not varied, field copper loss remains constant.

 

2. Armature copper loss: It is power loss in the armature circuit and equal to 2aaIR. Since the value of armature current is decided by the load, armature copper loss becomes a function of time.

 

3. Core loss: It is the sum of eddy current and hysteresis loss and occurs mainly in the rotor iron parts of armature. With constant field current and if speed does not vary much with loading, core loss may be assumed to be constant.

 

4. Mechanical loss: It is the sum of bearing friction loss and the windage loss (friction loss due to armature rotation in air). For practically constant speed operation, this loss too, may be assumed to be constant.

 

Apart from the major losses as enumerated above there may be a small amount loss called stray loss occur in a machine. Stray losses are difficult to account. Power flow diagram of a d.c motor is shown in figure 40.1. A portion of the input power is consumed by the field circuit as field copper loss. The remaining power is the power which goes to the armature; a portion of which is lost as core loss in the armature core and armature copper loss. Remaining power is the gross mechanical power developed of which a portion will be lost as friction and remaining power will be the net mechanical power developed. Obviously efficiency of the motor will be given by:

net mechinPPη=

Similar power flow diagram of a d.c generator can be drawn to show various losses and input, output power (figure 40.2).

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It is important to note that the name plate kW (or hp) rating of a d.c machine always corresponds to the net output at rated condition for both generator and motor.

40.4 Swinburne's Test

For a d.c shunt motor change of speed from no load to full load is quite small. Therefore, mechanical loss can be assumed to remain same from no load to full load. Also if field current is held constant during loading, the core loss too can be assumed to remain same.

In this test, the motor is run at rated speed under no load condition at rated voltage. The current drawn from the supply I_L0 and the field current I_f are recorded (figure 40.3). Now we note that:

Input power to the motor, P_in = VI_L0

Cu loss in the field circuit P_fl = VI_f

Power input to the armature, = VI_L0 - VI_f

= V(I_L0 - I_f)

= VI_a0

Cu loss in the armature circuit = 20aaIr

Gross power developed by armature = 200aaVIIr−

= ()00aaaVIrI−

= 00baEI

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Since the motor is operating under no load condition, net mechanical output power is zero. Hence the gross power developed by the armature must supply the core loss and friction & windage losses of the motor. Therefore,

()000corefrictionaaabaPPVIrIEI+=−=

Since, both P_core and P_friction for a shunt motor remains practically constant from no load to full load, the sum of these losses is called constant rotational loss i.e.,

constant rotational loss, P_rot = P_core + P_friction

In the Swinburne's test, the constant rotational loss comprising of core and friction loss is estimated from the above equation.

After knowing the value of P_rot from the Swinburne's test, we can fairly estimate the efficiency of the motor at any loading condition. Let the motor be loaded such that new current drawn from the supply is I_L and the new armature current is I_a as shown in figure 40.4. To estimate the efficiency of the loaded motor we proceed as follows:

Input power to the motor, P_in = VI_L

Cu loss in the field circuit P_fl = VI_f

Power input to the armature, = VI_L - VI_f

= V(I_L - I_f)

= VI_a

Cu loss in the armature circuit = 2aaIr

Gross power developed by armature = 2aaVIIr−

= ()aaaVIrI−

= baEI

Net mechanical output power, P_{net mech} = barotEIP−

∴ efficiency of the loaded motor, η = barotLEIPVI−

= netmechinPP

The estimated value of P_rot obtained from Swinburne's test can also be used to estimate the efficiency of the shunt machine operating as a generator. In figure 40.5 is shown to deliver a

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load current I_L to a load resistor R_L. In this case output power being known, it is easier to add the losses to estimate the input mechanical power.

Output power of the generator, P_out = VI_L

Cu loss in the field circuit P_fl = VI_f

Output power of the armature, = VI_L + VI_f

= VI_a

Mechanical input power, P_{in mech} = 2aaaroVIIrP++

∴ Efficiency of the generator, η = LinmechVIP

= 2LaaaroVIVIIrP++

The biggest advantage of Swinburne's test is that the shunt machine is to be run as motor under no load condition requiring little power to be drawn from the supply; based on the no load reading, efficiency can be predicted for any load current. However, this test is not sufficient if we want to know more about its performance (effect of armature reaction, temperature rise, commutation etc.) when it is actually loaded. Obviously the solution is to load the machine by connecting mechanical load directly on the shaft for motor or by connecting loading rheostat across the terminals for generator operation. This although sounds simple but difficult to implement in the laboratory for high rating machines (say above 20 kW), Thus the laboratory must have proper supply to deliver such a large power corresponding to the rating of the machine. Secondly, one should have loads to absorb this power.

40.5 Hopkinson's test

This as an elegant method of testing d.c machines. Here it will be shown that while power drawn from the supply only corresponds to no load losses of the machines, the armature physically carries any amount of current (which can be controlled with ease). Such a scenario can be created using two similar mechanically coupled shunt machines. Electrically these two machines are eventually connected in parallel and controlled in such a way that one machine acts as a generator and the other as motor. In other words two similar machines are required to carry out this testing which is not a bad proposition for manufacturer as large numbers of similar machines are manufactured.

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40.5.1 Procedure

Connect the two similar (same rating) coupled machines as shown in figure 40.6. With switch S opened, the first machine is run as a shunt motor at rated speed. It may be noted that the second machine is operating as a separately excited generator because its field winding is excited and it is driven by the first machine. Now the question is what will be the reading of the voltmeter connected across the opened switch S? The reading may be (i) either close to twice supply voltage or (ii) small voltage. In fact the voltmeter practically reads the difference of the induced voltages in the armature of the machines. The upper armature terminal of the generator may have either + ve or negative polarity. If it happens to be +ve, then voltmeter reading will be small otherwise it will be almost double the supply voltage.

Since the goal is to connect the two machines in parallel, we must first ensure voltmeter reading is small. In case we find voltmeter reading is high, we should switch off the supply, reverse the armature connection of the generator and start afresh. Now voltmeter is found to read small although time is still not ripe enough to close S for paralleling the machines. Any attempt to close the switch may result into large circulating current as the armature resistances are small. Now by adjusting the field current I_fg of the generator the voltmeter reading may be adjusted to zero (E_g ≈ E_b) and S is now closed. Both the machines are now connected in parallel as shown in figure 40.7.

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40.5.2 Loading the machines

After the machines are successfully connected in parallel, we go for loading the machines i.e., increasing the armature currents. Just after paralleling the ammeter reading A will be close to zero as E_g ≈ E_b. Now if I_fg is increased (by decreasing R_fg), then E_g becomes greater than E_b and both I_ag and I_am increase, Thus by increasing field current of generator (alternatively decreasing field current of motor) one can make E_g > E_b so as to make the second machine act as generator and first machine as motor. In practice, it is also required to control the field current of the motor I_fm to maintain speed constant at rated value. The interesting point to be noted here is that I_ag and I_am do not reflect in the supply side line. Thus current drawn from supply remains small (corresponding to losses of both the machines). The loading is sustained by the output power of the generator running the motor and vice versa. The machines can be loaded to full load current without the need of any loading arrangement.

40.5.3 Calculation of efficiency

Let field currents of the machines be are so adjusted that the second machine is acting as generator with armature current I_ag and the first machine is acting as motor with armature current I_am as shown in figure 40.7. Also let us assume the current drawn from the supply be I₁. Total power drawn from supply is VI₁ which goes to supply all the losses (namely Cu losses in armature & field and rotational losses) of both the machines,

Now:

Power drawn from supply = VI₁

Field Cu loss for motor = VI_fm

Field Cu loss for generator = VI_fg

Armature Cu loss for motor = 2amamIr

Armature Cu loss for generator = 2agagIr

∴Rotational losses of both the machines = ()221fmfgamamagagVIVIVIIrIr−+++

(40.1)

Since speed of both the machines are same, it is reasonable to assume the rotational losses of both the machines are equal; which is strictly not correct as the field current of the generator will be a bit more than the field current of the motor, Thus,

Rotational loss of each machine, ()2212fmfgamamagagrotVIVIVIIrIrP−+++=

Once P_rot is estimated for each machine we can proceed to calculate the efficiency of the machines as follows,

Efficiency of the motor

As pointed out earlier, for efficiency calculation of motor, first calculate the input power and then subtract the losses to get the output mechanical power as shown below,

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Total power input to the motor = power input to its field + power input to its armature

P_inm = fmaVIVI+

Losses of the motor = 2fmamamroVIIrP++

Net mechanical output power P_outm = ()2inmfmamamrotPVIIrP−++

∴ η_m = outminmPP

Efficiency of the generator

For generator start with output power of the generator and then add the losses to get the input mechanical power and hence efficiency as shown below,

Output power of the generator, P_outg = VI_ag

Losses of the generator = 2fgagagroVIIrP++

Input power to the generator, P_ing = ()2outgfgagagrotPVIIrP+++

∴ η_g = outgingPP

 

40.6 Condition for maximum efficiency

 

We have seen that in a transformer, maximum efficiency occurs when copper loss = core loss, where, copper loss is the variable loss and is a function of loading while the core loss is practically constant independent of degree of loading. This condition can be stated in a different way: maximum efficiency occurs when the variable loss is equal to the constant loss of the transformer.

Here we shall see that similar condition also exists for obtaining maximum efficiency in a d.c shunt machine as well.

 

40.6.1 Maximum efficiency for motor mode

 

Let us consider a loaded shunt motor as shown in figure 40.8. The various currents along with their directions are also shown in the figure.

+ Ia IL Figure 40.8: Machine operates as motor

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We assume that field current I_f remains constant during change of loading. Let,

P_rot = constant rotational loss

V I_f = constant field copper loss

Constant loss P_const = P_rot + V I_f

Now, input power drawn from supply = V I_L

Power loss in the armature, = 2aaIr

Net mechanical output power = 2()LaafrotVIIrVIP−−+

= 2LaaconstVIIrP−−

so, efficiency at this load current mη = 2LaaconstLVIIrPVI−−

Now the armature copper loss 2aaIrcan be approximated to 2LaIras aII≈. This is because the order of field current may be 3 to 5% of the rated current. Except for very lightly loaded motor, this assumption is reasonably fair. Therefore replacing I_a by I_f in the above expression for efficiency mη, we get,

mη = 2LLaconstLVIIrPVI−−

= 1LaconstLIrPVVI−−

Thus, we get a simplified expression for motor efficiency mη in terms of the variable current (which depends on degree of loading) I_L, current drawn from the supply. So to find out the condition for maximum efficiency, we have to differentiate mη with respect to I_L and set it to zero as shown below.

mLddIη = 0

or, LaconstLLIrPddIVVI⎛⎞−⎜⎟⎝⎠ = 0

or, 2aconstLrPVVI−+

∴Condition for maximum efficiency is 22LaaaIrIr≈ = P_const

So, the armature current at which efficiency becomes maximum is I_a = constaPr

 

40.6.2 Maximum efficiency for Generation mode

 

Similar derivation is given below for finding the condition for maximum efficiency in generator mode by referring to figure 40.9.

We assume that field current I_f remains constant during change of loading. Let,

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P_rot = constant rotational loss

V I_f = constant field copper loss

Constant loss P_const = P_rot + V I_f

Net output power to load = V I_L

Power loss in the armature, = 2aaIr

Mechanical input power = 2()LaafrotVIIrVIP+++

= 2LaaconstVIIrP++

so, efficiency at this load current gη = 2LLaaconstVIVIIrP++

As we did in case of motor, the armature copper loss 2aaIrcan be approximated to 2LaIras aII≈. So expression for gηbecomes,

gη = 2LLLaconstVIVIIrP++

Thus, we get a simplified expression for motor efficiency gηin terms of the variable current (which depends on degree of loading) I_L, current delivered to the load. So to find out the condition for maximum efficiency, we have to differentiate gηwith respect to I_L and set it to zero as shown below.

gLddIη = 0

or, 2LLLLaconstVIddIVIIrP⎛⎞⎜⎟++⎝⎠ = 0

∴Simplifying we get the condition as 22LaaaIrIr≈ = P_const

So, the armature current at which efficiency becomes maximum is I_a = constaPr

Thus maximum efficiency both for motoring and generating are same in case of shunt machines. To state we can say at that armature current maximum efficiency will occur which will make variable loss = constant loss. Eventually this leads to the expression for armature current for maximum efficiency as aconstIPr=.

 

40.7 Tick the correct answer

 

 

1. A 5 kW, 230 V, d.c shunt motor is operating at rated condition with 82% efficiency. Power drawn form the supply and the total loss are respectively

 

(A) 5 kW and 900 W (B) 6.1 kW and 1100 W

(C) 6.1 kW and 900 W (D) 5 kW and 1100 W

 

2. 10 kW, 230 V, d.c shunt generator is operating at rated condition with 80% efficiency. Input power and the total loss are respectively

 

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(A) 5 kW and 2.5 kW (B) 12.5 kW and 1 kW

(C) 12.5 kW and 2.5 kW (D) 5 kW and 1 kW

 

 

3. A 10 kW, 220 V, d.c shunt motor is drawing 53.78 A from supply mains at rated condition. The machine has r_a = 0.3 Ω and R_f = 220 Ω. The core and friction loss together is

 

(A) 220 W (B) 1831.6 W (C) 867.7 W (D) 743.9 W

 

4. A 10 kW, 220 V, d.c shunt generator is supplying a load at rated condition. Input mechanical power is found to be 12.5 kW. The machine has r_a = 0.3 Ω and R_f = 220 Ω. The core and friction loss together is

 

(A) 1660.3 W (B) 619.7 W (C) 839.7 W (D) 2500 W

 

5. During no load test on a 220 V, shunt motor having r_a = 0.7 Ω and R_f = 220 Ω, the armature current is recorded to be 4 A. The rotational loss of the motor is

(A) 880 W (B) 11.2 W (C) 1088.8 W (D) 868.8 W

 

 

6. A d.c shunt motor with r_a = 0.8 Ω, has 200 W field circuit loss and 250 W rotational loss. Motor will operate maximum efficiency when loaded to carry an armature current of about

(A) 17.68 A (B) 23.72 A (C) 15.81 A (D) 53.76 A

 

 

40.8 Solve the following

 

 

1. A 220 V d.c shunt motor has armature and field resistance as 0.8 Ω and 200 Ω. During Swinburne's test, current drawn from the supply is found to be 2.5 A. Estimate the efficiency of the machine,

 

 

(i) When it is running as a motor drawing a line current of 40 A from the 220 V supply.

 

(ii) When it is running as a generator delivering a load current of 40 A at 220 V.

 

 

2. Two similar coupled machines of same rating, each having an armature resistance of 0.5 Ω are connected for Hopkinson's test. Test data recorded as follows:

 

Supply voltage = 230 Volts

Total line current drawn from the supply = 8 A

Field current of the machine running as generator = 3 A

Field current of the machine running as motor = 2 A

Generator armature current = 17 A

 

(i) Estimate the rotational loss of each machine.

 

(ii) Estimate the efficiency of the generator.

 

(iii) Estimate the efficiency of the motor.

 

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