Drives and their loads

Choosing a motor starts with the machine it has to turn, not with the motor. This chapter works from the load back to the motor: what the load asks for, how that demand reaches the motor shaft, which motor can meet it, and whether the motor stays cool while doing so.

What an electric drive is

An electric drive is the whole chain that turns electrical energy into controlled motion. Power flows from the supply to the load; measurements flow back to a controller that decides what the converter does next.

The parts of an electric drive

The converter can be as simple as a contactor (on or off), or a soft starter, or a variable-frequency drive (VFD) that sets speed and torque continuously.
The transmission (gearbox, belt, rope drum, rack) matches the fast, low-torque motor to the slow, high-torque machine.
The load is the machine itself: a pump, a hoist, a conveyor belt, a fan.

FoundationStart here if this is new to you

Think of a bicycle. Your legs are the motor, the chain and gears are the transmission, the road and the hill are the load. Your brain is the controller and your eyes and legs are the sensors. You change gear so that your legs can turn at a comfortable speed whatever the hill does. A drive does the same thing, only with electricity.

The three load families

What matters about a load is how its torque changes with speed. Almost every industrial load belongs to one of three families. In per unit of the rated values:

Family	Torque	Power	Examples
Constant torque	$T = T_n$	$P \propto n$	conveyors, hoists, extruders, piston pumps
Variable (quadratic) torque	$T \propto n^2$	$P \propto n^3$	centrifugal pumps and fans
Constant power	$T \propto 1/n$	$P = P_n$	winders, machine-tool spindles

Try it: the three load families

Speed50 %

Constant torque
Variable torque
Constant power

Torque needed at this speed: 25 %
Power needed at this speed: 13 %

Centrifugal pumps and fans. Torque grows as speed squared and power as speed cubed: at half speed the fan needs only 25 % torque and 12.5 % power. This is why a VFD saves so much energy on fans.

Predict first

A fan runs at 80 % of its rated speed. What share of its rated power does it need?

80 %64 %About 51 %

Real loads add a breakaway torque at standstill (static friction, a conveyor full of wet sand) that can be larger than the running torque. The motor's starting torque must beat it.

Carrying the load to the motor shaft

The motor never sees the load directly: it sees it through the transmission. For a gearbox with ratio $i = \omega_m / \omega_L$ (motor faster than load) and efficiency $\eta$ , when the motor drives the load:

\omega_m = i\,\omega_L, \qquad T_m = \frac{T_L}{i\,\eta}

Inertia is referred through the square of the ratio, and a mass $m$ moving in a straight line at speed $v$ adds its own equivalent inertia:

J_\text{eq} = J_m + \frac{J_L}{i^2} + m\left(\frac{v}{\omega_m}\right)^2

Worked example: hoist

A hoist lifts $m = 800$ kg at $v = 0.5$ m/s with a drum of radius $r = 0.25$ m, through a gearbox with $i = 75$ and $\eta = 0.9$ .

Drum speed: $\omega_L = v / r = 2$ rad/s, so the motor turns at $\omega_m = 75 \times 2 = 150$ rad/s, about 1432 rpm.
Drum torque: $T_L = m g r = 800 \times 9.81 \times 0.25 = 1962$ N·m. At the motor: $T_m = 1962 / (75 \times 0.9) = 29.1$ N·m.
Check with power: $T_m \omega_m = 29.1 \times 150 = 4.36$ kW $= m g v / \eta$ .
The 800 kg adds only $800 \times (0.5 / 150)^2 = 0.0089$ kg·m² at the motor shaft. A drum of 12 kg·m² adds $12 / 75^2 = 0.0021$ kg·m². Through a large ratio, the motor's own rotor often dominates the inertia.

Where motor and load meet

In steady state the motor settles where its torque curve crosses the load curve: $T_m(\omega) = T_L(\omega)$ . Any surplus torque accelerates the inertia, $J_\text{eq}\,d\omega/dt = T_m - T_L$ (lesson A3).

The crossing must also be stable: if the speed rises a little, the load torque must exceed the motor torque so that the drive slows back down. On an induction motor this holds everywhere between breakdown and synchronous speed, which is why that is the working region (next lesson).

ExplorerGo deeper: derivations and open questions

Stability condition. Linearise $J\,d\Delta\omega/dt = (\partial T_m/\partial\omega - \partial T_L/\partial\omega)\,\Delta\omega$ . A disturbance dies out when

\frac{\partial T_L}{\partial \omega} > \frac{\partial T_m}{\partial \omega}

Check it on the constant-torque load and the induction motor curve: which crossings are stable when the load line cuts the curve twice? What changes for a fan load?

Optimal gear ratio. For a pure inertia load driven to maximum acceleration, the ratio that maximises $\dot\omega_L$ for a given motor torque is $i = \sqrt{J_L / J_m}$ . Derive it from $\dot\omega_L = i T_m / (i^2 J_m + J_L)$ .