How Do Motor Losses Determine Peak Efficiency Load?

The calculation yields a direct design rule: peak efficiency occurs where the constant loss equals the load-varying loss. For the model L=A+BX^2, the peak-load fraction is X_peak=sqrt(A/B). A peak at rated load gives A/B=1; a peak at 80% rated load gives A/B=0.64.

1. Loss-model prerequisites

Before anything else, confirm what each term represents. The derivation depends on these definitions remaining consistent from the loss data through the final efficiency check.

Symbol	Meaning	Required interpretation
A	Loss independent of load	No-load or constant-loss component in watts
B	Load-varying loss at rated load	Variable-loss component in watts when X=1
X	Per-unit load	Output load divided by rated output load
P100	Rated output power	Output power at X=1
L	Total loss	L=A+BX^2

The model also requires output power to follow Pout=X P100. Treat X as load fraction, not automatically as per-unit current. Current may contain a no-load component and may not track output load linearly. Substitute measured current for X only after establishing the required relationship for the motor and operating range.

1. Express each load point as rated-output fraction. For example, 80% load becomes X=0.80.
2. Keep A and B in the same power unit.
3. Confirm that B is the variable portion alone, not total rated-load loss.

Do not move on until evaluating L at rated load produces L(1)=A+B.

2. Efficiency expression

Calculate motor efficiency from output power and total loss:

efficiency=Pout/(Pout+L)

Substituting the model terms gives:

efficiency=(X P100)/(X P100+A+BX^2)

Finding the maximum directly is possible, but minimizing inverse efficiency isolates the load-dependent loss terms more cleanly:

1/efficiency=(Pout+L)/Pout=1+L/Pout

The constant 1 does not change the location of the minimum. Therefore, maximize efficiency by minimizing:

L/Pout=(A+BX^2)/(X P100)=(A/X+BX)/P100

Because P100 is constant with respect to X, it also does not affect the minimizing value. The remaining objective is:

f(X)=A/X+BX

1. Insert a positive trial load fraction into both the original efficiency equation and f(X).
2. Check that reducing f(X) raises the calculated efficiency.
3. Retain the operating domain X>0; inverse efficiency is undefined at zero output.

The setup is verified when the same X that minimizes f(X) maximizes the full efficiency expression.

3. Peak-efficiency condition

Differentiate the reduced objective with respect to load fraction:

df/dX=-A/X^2+B

At an interior efficiency maximum, set the derivative to zero:

-A/X^2+B=0

Rearranging gives the governing condition:

A=BX^2

The right-hand side is the load-varying loss at the selected load. Peak efficiency therefore occurs at the load where variable loss equals constant loss. Solving for load gives:

X_peak=sqrt(A/B)

For positive A, B, and X, the second derivative is 2A/X^3, which is positive. The stationary point minimizes inverse efficiency and therefore maximizes efficiency.

Peak-efficiency location	Calculation	Required ratio
Rated load	A/B=(1.00)^2	1.00
80% rated load	A/B=(0.80)^2	0.64

Confirm the result by checking A=BX_peak^2 with the selected ratio before applying motor data.

4. Ratio-to-load decision path

The ratio A/B determines the location of the modeled peak. It does not by itself give the efficiency value; that calculation also needs rated output power and the absolute losses.

Observed or calculated condition	Interpretation within the model	Next check
A/B=1	Constant and rated-load variable losses are equal	Peak occurs at X=1
A/B<1	Rated-load variable loss exceeds constant loss	Peak occurs below rated load
A/B>1	Constant loss exceeds rated-load variable loss	Calculated peak lies above rated load
Measured curve disagrees with sqrt(A/B)	The two-term loss model or extracted loss values do not describe the tested range	Recheck loss separation and compare additional load points

1. For a known peak location, calculate A/B=X_peak^2.
2. For known loss components, calculate X_peak=sqrt(A/B).
3. Compare X_peak with the intended operating range. A mathematical peak above rated load is not authorization to operate above the motor rating.
4. Calculate actual efficiency with the full expression rather than using the loss ratio alone.

Do not move on until the ratio and peak load reproduce one another through both inverse calculations.

5. Loss extraction from test data

Separate total loss from its components before calculating the peak. At zero mechanical output, the model gives L(0)=A. At rated load, it gives L(1)=A+B.

1. At the no-load point, calculate input real power using the topology and measured quantities. For the stated three-phase example, A=sqrt(3) V I PF, producing 1891 W from 460 V, 32.2 A, and PF=0.0737.
2. At rated load, calculate total loss as Pelec-Pout.
3. Subtract the already identified constant loss: B=Pelec-Pout-A.
4. Using the stated rated values 460 V, 114 A, PF=0.873, and 100 hp at 745.7 W/hp, obtain B=2832 W.
5. Calculate X_peak=sqrt(1891/2832)=0.817, which rounds to about 82% rated load.

The critical correction is the subtraction of A. Assigning Pelec-Pout directly to B treats all rated-load loss as load-varying even though A is still present. That incorrect assignment gives B=4723.5 W and predicts a misleading peak near 63% load.

Verify the extraction by adding the components back together: rated total loss must equal 1891 W+2832 W, apart from rounding.

6. Symptom and cause checks

Symptom	Likely calculation cause	Correction
Predicted peak is far below the plotted efficiency maximum	Rated total loss was assigned to B	Use B=Pelec-Pout-A
Result states A=B at every load	The peak condition was treated as an identity for all X	Apply A=BX^2 only at X_peak
80% peak produces A/B=0.80	The square-law term was omitted	Calculate 0.80^2=0.64
Current fraction was used as load fraction	X was redefined during the calculation	Return to rated-output fraction or establish a current-to-load mapping
One ratio cannot fit the measured efficiency curve	Actual losses do not follow one constant term plus one squared-load term over the full range	Extract losses at several load points and inspect the residual from A+BX^2

A real motor can have loss components whose load dependence is not captured by this two-term expression. Use the formula as a model check: if measured total loss minus A+BX^2 changes systematically with load, refine the loss representation before relying on its predicted peak.

Do not move on until the corrected calculation explains both the loss data and the observed neighborhood of the efficiency maximum.

7. End-to-end verification

1. Record rated output power as P100 and convert every test point to X=Pout/P100.
2. Determine A from the no-load loss under the stated model.
3. At rated load, calculate total loss from input power minus output power.
4. Calculate B=(rated total loss)-A.
5. Predict the peak with X_peak=sqrt(A/B).
6. At that load, verify the balance A=BX_peak^2.
7. Calculate efficiency at points below, at, and above the predicted peak using (X P100)/(X P100+A+BX^2).
8. Confirm that the center point has the highest calculated efficiency and compare it with measured efficiency data at nearby loads.

If the modeled and measured peaks differ materially, repeat the loss extraction from measured input and output power at several loads. The final acceptance check is a local maximum in measured efficiency near the calculated X_peak, with the extracted losses satisfying L(0)=A and L(1)=A+B.

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