Duty cycles and motor heating

A motor rarely fails because it lacks torque. It fails because it runs too hot for too long and its winding insulation ages. So the question "is the motor big enough?" is mostly a question about heat.

Heat, insulation and time

Losses, mostly copper losses $I^2R$ , heat the windings. The insulation sets the limit:

Insulation class	Max. winding temperature	Allowed rise at 40 °C ambient (by resistance)
B	130 °C	80 K
F	155 °C	105 K
H	180 °C	125 K

A common design practice is class F insulation used with a class B rise: the extra 25 K is margin. A rough rule of thumb: every 10 K hotter halves insulation life.

A motor is a big lump of iron and copper, so it heats slowly. A first-order model captures this well:

\theta(t) = \theta_\infty\left(1 - e^{-t/\tau}\right), \qquad \theta_\infty \propto \text{losses}

The thermal time constant τ ranges from a few minutes for small motors to an hour or more for large ones. A load that lasts much less than τ barely warms the motor; what counts is the average heating over many cycles.

FoundationStart here if this is new to you

A motor heats like a pot of water on a stove. Turn the flame up for ten seconds and the water hardly changes. Leave it high for an hour and the water boils. What decides whether the water boils is the average flame over a long time, not one short blast.

Duty types

IEC 60034-1 names the standard patterns of use. A motor's rating only holds for the duty written on its nameplate.

Code	Duty	Typical load
S1	continuous, constant load	pumps, fans, compressors
S2	short time, then cools fully (S2 30 min)	sluice gates, valves
S3	intermittent periodic: load, rest, load…	hoists, cranes
S4, S5	as S3 with starting (S4) and electric braking (S5)	frequently started conveyors (S4), crane travel drives (S5)
S6	continuous running, intermittent load	presses idling between strokes, mixers, saws
S7, S8	continuous with braking (S7), or with speed changes (S8)	reversing drives, rolling mills, winders
S9, S10	non-periodic (S9), or discrete load levels (S10)	VFD applications

For S3 and S6, the cyclic duration factor is the loaded share of each cycle, $\text{CDF} = t_\text{on} / (t_\text{on} + t_\text{off})$ , with standard values 15, 25, 40 and 60 %.

The RMS torque method

If copper losses dominate, heating goes as torque squared. A cycle of torques $T_i$ lasting $t_i$ heats the motor like a constant torque of

T_\text{rms} = \sqrt{\frac{\sum T_i^2\, t_i}{\sum t_i}}

This is the calculation of Lab I. The Lab I cycle, 80 N·m for 10 s then 0 for 15 s, gives $T_\text{rms} = 80\sqrt{10/25} = 50.6$ N·m and a CDF of 40 %.

The same method works on power when the speed is constant. In Tutorial 1.2, the parts hoist needs 2.62 kW for 10 s in every 60 s: $P_\text{rms} = 2.62\sqrt{10/60} = 1.07$ kW. The heating would allow a tiny motor, but the hoist must still deliver 2.62 kW and start its load: the booklet sizes it on the peak and on starting torque.

Predict first

The Lab I cycle peaks at 80 N·m with T_rms = 50.6 N·m. Which rated torque should the motor have?

80 N·m, for the peakAt least 50.6 N·m, then check the peak separately32 N·m, since it is loaded only 40 % of the time

Try it: duty cycle and motor heating

Step 1 torque80 N·m

Step 1 duration10 s

Step 2 torque0 N·m

Step 2 duration15 s

Motor rated torque50 N·m

Try the motor whose rated torque equals T_rms, then one size smaller.

Load torque
T_rms
Motor rated torque
Winding rise

The motor overheats: its rated torque is below T_rms. Pick a bigger motor.

T_rms: 50.6 N·m
Peak torque: 80 N·m
Cyclic duration factor: 40 %
Highest rise: 81 K

T_rms = √( Σ Tᵢ² tᵢ / Σ tᵢ )

Try the three cycles. With the hoist, set the rated torque to 50 N·m, just below $T_\text{rms}$ , and watch the rise creep above 80 K. With the press, the motor stays cool but its 120 N·m peak exceeds 80 % of breakdown: the peak decides, not the heat.

ExplorerGo deeper: derivations and open questions

Short-time rating. A motor loaded for $t_\text{on}$ and then allowed to cool fully (S2) only reaches $\theta_\infty(1 - e^{-t_\text{on}/\tau})$ . It may therefore carry more than its continuous rating: power can be scaled by $\sqrt{1/(1 - e^{-t_\text{on}/\tau})}$ . For $t_\text{on} = 10$ min and $\tau = 20$ min that is 1.59. Where does the square root come from?

Cooling at rest. A fan-cooled motor cools more slowly when it stands still, because its fan stops. Some methods weight the rest time by a factor $k \approx 0.25$ –0.5 in the denominator of $T_\text{rms}$ . Does that make the motor bigger or smaller? And what happens at low speed on a VFD?

When RMS fails. $T_\text{rms}$ assumes losses $\propto T^2$ . Iron and friction losses do not follow torque, and at low speed the cooling changes. For S4 and S5, the starting and braking currents (5 to 8 times rated) add heat that no load torque shows. How would you extend the thermal model of the widget to count starts per hour?