DC motor efficiency which formula to use?

Question

I'm doing a lab report for a shunt DC motor evaluation. I seem to be stuck on the efficiency vs torque of the motor.

I have plotted a graph of measured \$\frac{P_{out}}{P_{in}}\cdot 100\%\$, which shows that the motor becomes less effcient with more load. (Increments from 0.1Nm to 1.1Nm then stalls.)

This is fine as it shows me that the motor is running aroud 50% efficiency at 0.1Nm and 18% at 1Nm.

My issue is when I try to use some more involved formulas the graph seem to flip around using the recorded values and shows that the motor is most efficient just before stall condition up to 95% so I'm unsure which to use.

The formulas I have been given are

$$ P_{out}= 2\pi T \frac{N}{60} \\P_{in} = V_f I_f + V_a I_a$$

Calculate these then put into the previous formula \$\frac{P_{out}}{P_{in}}\cdot 100\%\$.

Any ideas why the results are flipping?

To add here are my results from the Lab I have calculated the Pin and Pout using Ohms law V.I from the armature values Pin and Field values Pout i have also added the equation 5,6,7 calculations on the right hand side

Here are my calculations using the Eq 5,6,7 from above excell doing the calculations

Hopefully this all makes sense thanks

Answer 1

Eq5, Eq6 and Eq7 are correct for a motor with a separately excited field. If the motor has a shunt field, the field and armature voltages would be equal. Since they are not quite equal, the motor is treated like it has a separately excited field.

The Excel spreadsheet must be set up for some other conditions. Since nothing you have posted indicates that anything else is applicable, Eq5, Eq6 and Eq7 should be used.

The RPM increasing with load would seem to indicate that something is going on that is not explained. However I don't see a reason that should have much effect the efficiency very much.

Answer 2

\$ V_f I_f\$ is a pure loss, it is needed for the excitation. If the excitation winding would be made of superconductor, then you could fill the current only once and close the windings contacts. The other solution is to use permanent magnets instead.
Then you have a copper loss \$I_a^2 R_a\$ and a iron loss in the rotor \$P_{fe}=k_{fe}\Omega\$, as the rotor field is AC, but let's say that you can omit this for simpler calculation.
Post the formulas you did before, so we can take a look and discuss.

EDIT:
You could rewrite the efficiency formula from loss. \$P_{in}=P_{EM} + P_{loss} = P_{EM}+ P_{Cu}+P_{Fe}\$, where \$P_{EM}\$ is the electromagnetic energy and it is converted into a mechanical energy, so let's make a new formula: $$\eta=\dfrac{P_{mech}}{P_{mech}+P_{loss}}=\dfrac{P_{mech}}{P_{mech}+V_fI_f+I_a^2R_a}$$ The formula is omits the iron losses. At least you won't get over 100% efficiency, additionally you need a resistance of the armature \$R_a\$

Answer 3

What I think is going on here is that you are operating on the lighter than peak efficiency part to the curve. A shunt DC motor has two places where the efficiency goes to zero. One is when speed=0. The other is where torque=0. The maximum efficiency of a DC motor is at higher speeds when lightly loaded. The measurements seem to have some error because you can't be more than 100% efficient. Putting that aside, the way you get what you call inverted is by going from very light loads up to the peak. After the peak you start to go down. At this site; Electric Equipment You'll find this graph;

Notice how starting at very low loads the efficency is going up as you increase the load. Effiency is the green curve.