DC Shunt Motor: calculation of the efficiency

[Master1022]

Homework Statement

A 230 V DC shunt motor has an armature resistance of $0.2 \Omega$ and a field resistance of $216 \Omega$. At full rated load, the output power is $18642$ W and the armature current is 90 A. Find the rotational losses at this full load condition and therefore find the motor efficiency in this case.

Relevant Equations

$V = E + I_a R_a$
$P = I V$

Hi,

My main question is: should we be subtracting the rotational power losses from the motor output power in our efficiency calculation?

My approach:
Will skip extra work as I have done the above parts of the question correctly.
I used power conservation to state that:
$$ P_{input} = P_{motor} + P_{field} + P_{armature} + P_{losses} $$
We end up with: $P_{motor} = 18642$ W, $P_{field} = \frac{230^2}{216} = 244.907$ W, $P_{input} = 230(90 + \frac{230}{216}) = 20944.907$ W, and therefore $P_{losses} = 437.5$ W.

When it comes to calculating the efficiency, I was always under the impression that we did:
$$ \eta = \frac{want}{pay for} = \frac{P_{motor}}{P_{input}} = \frac{18642}{20944.907} = 0.89 $$
However, in a previous question the solution takes into account the 'frictional losses (rotational losses)' such that we do:
$$ \eta = \frac{P_{motor} - P_{losses}}{P_{input}} $$
in that solution we subtract the 'windage losses'... and $P_{motor} = I_a \cdot (V - I_a R_a)$

Potential reasons for the calculation method:
1. Negligible effect by excluding the $P_{losses}$: perhaps, but the inclusion of the value does change the efficiency by 3% which doesn't seem that insignificant to me

I am struggling to see why the solution methods differ and would really appreciate any help.

 

[Baluncore]

I am struggling to see why the solution methods differ and would really appreciate any help.

Electrical losses include I²R heating of field and armature. They can be calculated based on measurements.

Rotating losses include windage, bearing and commutator friction. They can either be estimated, or calculated by subtraction of measured shaft power from input power. But that is not always necessary.

DC shunt motor; 230 V; Armature resistance; 0.2 Ω; Field resistance; 216 Ω;
At full rated load, the shaft power; 18642 W; and the armature current; 90 A.

Armature heat; I²R = 90 * 90 * 0.2 = 1620 W.
Field heat; V·I = 230 V * ( 230 V / 216 R) = 244.9 W.
Total resistive heat loss = 1864.9 W.

The total input power can be computed fully from the data.
Input armature power = 230 V * 90 A = 20,700. W; (Supplies rotating and I²R losses).
Input field power = 244.9 W
Total input power = 20,700. + 244.9 = 20,944.9 W

Input power 20,944.9 W
Output power 18,642. W
Efficiency = 18,642. / 20,944.9 =0.8900

The rotating windage and friction losses were not ignored in the efficiency computation.
They simply did not need to be computed.

Energy audit; input = output shaft + armature heat + field heat + rotating.
Rotating = input - output shaft - armature heat - field heat;
Total rotating losses; 20,944.9 - 18,642. - 1,864.9 = 438.0 W

 

[Master1022]

@Baluncore : thank you very much for taking the time to respond to me!

Electrical losses include I²R heating of field and armature. They can be calculated based on measurements.

Rotating losses include windage, bearing and commutator friction. They can either be estimated, or calculated by subtraction of measured shaft power from input power. But that is not always necessary.

DC shunt motor; 230 V; Armature resistance; 0.2 Ω; Field resistance; 216 Ω;
At full rated load, the shaft power; 18642 W; and the armature current; 90 A.

Armature heat; I²R = 90 * 90 * 0.2 = 1620 W.
Field heat; V·I = 230 V * ( 230 V / 216 R) = 244.9 W.
Total resistive heat loss = 1864.9 W.

The total input power can be computed fully from the data.
Input armature power = 230 V * 90 A = 20,700. W; (Supplies rotating and I²R losses).
Input field power = 244.9 W
Total input power = 20,700. + 244.9 = 20,944.9 W

Agreed

Input power 20,944.9 W
Output power 18,642. W
Efficiency = 18,642. / 20,944.9 =0.8900

The rotating windage and friction losses were not ignored in the efficiency computation.
They simply did not need to be computed.

Okay, thank you for confirming that

I now have two quick follow up questions (which perhaps suggests that I don't actually understand this):
1) What is the rotating loss part of that does the output power equal mathematically? Is it $iE$ and $T \omega$?
2) Is the core loss inherently included in any of the other terms (eg. output power)?

This probably doesn't make sense, but this is what I was thinking:

If we do KVL around the loop including the supply and the armature:
$$ V = E + I_a R_a $$
and then multiply by $I_a$ to get:
$$ V I_a = E I_a + I_a ^ 2 R_a $$

We could then include $V I_f$ on both sides to get:
$$ V (I_a + I_f) = E I_a + I_a ^ 2 R_a + V I_f \rightarrow VI = E I_a + I_a R^2 + V I_f $$
and this doesn't, on the face of it, look like it agrees with the expression that we wrote down from conservation of energy/power...
I was told by my tutor that we don't include these rotational loss terms in the voltage equation, so does one of the above terms include the windage losses in it? If so, it seems that $I_a E$ is the most likely contender. Therefore, perhaps that is why, in the previous solution, they had to subtract the loss term from their calculation as the loss term is included in the $I_a E$ term?

Just to clarify, they did use $I_a \cdot (V - I_a R_a)$ to calculate the power of the motor and there was a separate value listed for the windage loss.

I understand that working from total input power and subtracting the losses will yield the correct output motor power, but was just wondering whether $iE$ yields the same answer (which it looks like it does if the windage losses are 0)

 

[Baluncore]

The motor sits in a box, between two bulkheads. The power supply enters through one wall. The shaft comes out through the other. Losses can be seen as heat radiated from the box. The field winding may be separated, but trying to introduce more partitions of the motor will make identification and analysis of efficiency more difficult.

Core losses in the armature, due to commutation of the armature field, may be significant.
If the field current remains steady, core losses in the field will not be significant.

2) Is the core loss inherently included in any of the other terms (eg. output power)?

Electrical input energy supplies all losses. Electrical energy flows into the motor through the power supply. Only mechanical energy departs the motor through the output shaft. Energy flow to satisfy all other losses is modeled as heating of the motor, with equilibrium cooling to the local environment sink.

The speed-independent losses will include the heating of conductors due to I²R.

Speed-dependent losses will include armature core and windage, plus brush and bearing frictional losses. I cannot see how the speed-dependent parameters can be separated from each other, they must be estimated.

A DC motor will have a load dependent speed. Speed dependent back emf regulates armature current and therefore torque. Maximum power will be at about half the no-load RPM.

It gets complicated. KVL around the armature loop may need to include the speed dependent back EMF.

 

[Master1022]

Thank you for replying once again. However, I am still unsure of a few things. Would you be able to explain in more simple terms?

Electrical input energy supplies all losses. Electrical energy flows into the motor through the power supply. Only mechanical energy departs the motor through the output shaft. Energy flow to satisfy all other losses is modeled as heating of the motor, with equilibrium cooling to the local environment sink.

The speed-independent losses will include the heating of conductors due to I²R.

Speed-dependent losses will include armature core and windage, plus brush and bearing frictional losses. I cannot see how the speed-dependent parameters can be separated from each other, they must be estimated.

Apologies there may have been some laziness on my part. When I said "windage losses" or "rotational losses", I was basically just referring to the umbrella of "other" losses, which I will denote as $C$.

Does that mean the most reliable way to get the output power is to start with the input and subtract losses as follows?
$$ P_{motor} = IV - I_a^2 R_a - C - I_f V $$
When I was doing $I_a E$ to get the output motor power in other questions, it just so happened that there were no losses ($C = 0$) and thus the equality held up...

I suppose the calculation could be sped up by excluding the field losses and just considering the armature loop:
$$ P_{motor} = I_a V - I_a^2 R_a - C $$

It gets complicated. KVL around the armature loop may need to include the speed dependent back EMF.

Is that not what the $E$ represents in my equation $V = E + I_a R_a$? E is the back-emf