10.5 Electric Motors, Engines & Flywheels

10.5 Electric Motors, Engines & Flywheels

Every drive train starts at a prime mover. The machine designer does not design the motor or engine, but must choose one whose torque-speed behavior suits the load, size the shafts and couplings for its torque, check that it can start and accelerate the load, and smooth out torque that fluctuates during each cycle. This section covers the motor and engine relationships that appear in machine design problems.

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1. Three-Phase Induction Motors

The squirrel-cage induction motor is the workhorse of industrial machinery.

• Synchronous speed depends only on supply frequency $f$ and the number of poles $p$:

$$n_s = \frac{120 f}{p} \quad [\text{rpm}]$$

At $60\,\text{Hz}$: 2-pole $= 3{,}600$, 4-pole $= 1{,}800$, 6-pole $= 1{,}200$, 8-pole $= 900\,\text{rpm}$ (at $50\,\text{Hz}$: $3{,}000$, $1{,}500$, $1{,}000$, $750$).

• Slip is the fractional lag of the rotor behind the rotating field: $s = (n_s - n)/n_s$. Full-load slip of general-purpose motors is typically a few percent, which is why a 4-pole nameplate reads about $1{,}750$ to $1{,}780\,\text{rpm}$.
• Torque from power and speed:

$$T\,[\text{lb}\cdot\text{ft}] = \frac{5{,}252\,\text{hp}}{n}, \qquad T\,[\text{N}\cdot\text{m}] = \frac{9{,}549\,P\,[\text{kW}]}{n}$$

• Electrical input for a three-phase motor is $P_{\text{in}} = \sqrt{3}\,V_L I_L\,\text{pf}$, and shaft output is $P_{\text{out}} = \eta P_{\text{in}}$.

NEMA Design Letters (Torque-Speed Character)

Additional selection points:

• Starting current of an across-the-line (direct-on-line) start is commonly about 6 to 7 times full-load current. Soft starters, reduced-voltage starters and variable-frequency drives limit it.
• Service factor (commonly $1.15$ for general-purpose motors) is the continuous overload the motor tolerates at rated voltage and frequency, at the cost of higher winding temperature and shorter insulation life. It is a margin, not a rating to size to.
• Efficiency: NEMA Premium (IE3) efficiency levels are the federal minimum for most general-purpose motors under the US Department of Energy rules in 10 CFR Part 431.
• Variable-frequency drives (VFDs) change speed by changing frequency. Below base speed they hold volts per hertz constant, giving constant torque capability; above base speed the voltage is limited, so the motor runs at constant power with falling torque. Self-cooled motors lose cooling at low speed, which limits continuous low-speed torque.
• Other motor types, such as synchronous, DC, brushless servo and stepper motors, are covered with mechatronics in Section 13.2.

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2. Matching the Motor to the Load

Load Characteristics

• Constant-torque loads (conveyors, positive-displacement pumps, hoists) need the same torque at every speed, so power rises in proportion to speed.
• Variable-torque loads (centrifugal fans and pumps) follow the affinity laws: flow $\propto n$, pressure or torque $\propto n^2$, power $\propto n^3$. Slowing a fan $20\%$ cuts its power to about $0.8^3 = 51\%$.
• Constant-power loads (winders, machine-tool spindles) need high torque at low speed and low torque at high speed.

Reflected Inertia and Acceleration Time

Through a speed reducer with ratio $i = n_{\text{motor}}/n_{\text{load}}$ and efficiency $\eta$, load quantities seen at the motor shaft are

$$J_{\text{reflected}} = \frac{J_{\text{load}}}{i^2}, \qquad T_{\text{load,reflected}} = \frac{T_{\text{load}}}{i\,\eta}$$

The time to accelerate from rest to speed $\omega$ is

$$t = \frac{J_{\text{total}}\,\Delta\omega}{T_{\text{motor,avg}} - T_{\text{load,reflected}}}$$

where $J_{\text{total}}$ includes the motor rotor, couplings and the reflected load inertia. The gear ratio that gives the fastest acceleration of a pure inertia load is $i = \sqrt{J_{\text{load}}/J_{\text{motor}}}$ (inertia matching).

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3. Internal Combustion Engines

• Brake power is measured at the output shaft: $P_b = 2\pi N T$ with $N$ in rev/s. Indicated power is developed on the pistons, and mechanical efficiency is $\eta_m = P_b / P_i$.
• Brake mean effective pressure is a size-independent measure of how hard an engine works:

$$\text{bmep} = \frac{P_b\,n_R}{V_d\,N}$$

where $V_d$ is the total displaced volume, $N$ is in rev/s, and $n_R = 2$ for four-stroke or $1$ for two-stroke engines.

• Brake thermal efficiency is $\eta_{th} = P_b / (\dot{m}_f\,Q_{HV})$, and brake specific fuel consumption is $\text{bsfc} = \dot{m}_f / P_b$.
• The air-standard Otto cycle sets an ideal limit: $\eta = 1 - 1/r^{\,k-1}$. With compression ratio $r = 10$ and $k = 1.4$, $\eta = 1 - 10^{-0.4} = 0.60$; real spark-ignition engines achieve far less.
• Engine output torque pulses with each firing stroke, so engine-driven machinery needs flywheels, torsionally soft couplings or torsional vibration dampers, and drivetrain shafts must be checked for fluctuating torque ($T_a \ne 0$).

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4. Flywheels

A flywheel stores kinetic energy $E = \tfrac{1}{2} I \omega^2$ so that a machine with fluctuating demand (punch press, crusher, reciprocating compressor) or fluctuating supply (engine) runs at nearly constant speed. Define the coefficient of speed fluctuation

$$C_s = \frac{\omega_{\max} - \omega_{\min}}{\omega_{\text{avg}}}, \qquad \omega_{\text{avg}} \approx \frac{\omega_{\max} + \omega_{\min}}{2}$$

The energy that must be absorbed or released between the speed extremes is

$$\Delta E = \tfrac{1}{2} I (\omega_{\max}^2 - \omega_{\min}^2) = I\,\omega_{\text{avg}}^2\,C_s \quad \Longrightarrow \quad I = \frac{\Delta E}{C_s\,\omega_{\text{avg}}^2}$$

Allowable $C_s$ ranges from about $0.2$ for crushers and hammering machinery down to well below $0.01$ for electrical generators. A rim-type flywheel stores most of its inertia in the rim ($I \approx m r^2$), while a solid disk has $I = \tfrac{1}{2} m r^2$. The rim's hoop stress, roughly $\sigma \approx \rho v^2$ for a thin rotating ring, limits the safe rim speed.

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5. Worked Example: Motor, Drive and Flywheel Checks

Part A: Induction Motor Data

A 4-pole, $60\,\text{Hz}$, $460\,\text{V}$ motor is rated $15\,\text{hp}$ at $1{,}760\,\text{rpm}$, with full-load efficiency $92.4\%$ and power factor $0.86$.

1. Synchronous speed and slip: $n_s = 120(60)/4 = 1{,}800\,\text{rpm}$; $s = (1{,}800 - 1{,}760)/1{,}800 = 2.22\%$.
2. Full-load torque: $T = 5{,}252(15)/1{,}760 = 44.8\,\text{lb}\cdot\text{ft}$ ($60.7\,\text{N}\cdot\text{m}$).
3. Electrical input: $P_{\text{in}} = 15(0.7457)/0.924 = 12.11\,\text{kW}$.
4. Line current: $I = 12{,}110/(\sqrt{3} \times 460 \times 0.86) = 17.7\,\text{A}$.

Part B: Reflected Inertia and Acceleration

The motor drives a conveyor pulley through a $10{:}1$ reducer. The pulley and belt have $J_{\text{load}} = 12\,\text{kg}\cdot\text{m}^2$, and the rotor has $J_{\text{motor}} = 0.08\,\text{kg}\cdot\text{m}^2$.

• Reflected load inertia: $12/10^2 = 0.12\,\text{kg}\cdot\text{m}^2$, so $J_{\text{total}} = 0.20\,\text{kg}\cdot\text{m}^2$.
• If the motor supplies an average of $60\,\text{N}\cdot\text{m}$ more than the reflected load torque during starting, reaching $1{,}760\,\text{rpm}$ ($184.3\,\text{rad/s}$) takes $t = 0.20(184.3)/60 = 0.61\,\text{s}$.

Part C: Punch-Press Flywheel

A press needs $\Delta E = 3{,}000\,\text{J}$ per stroke from its flywheel, which averages $300\,\text{rpm}$ ($\omega_{\text{avg}} = 31.42\,\text{rad/s}$) with $C_s = 0.10$.

$$I = \frac{\Delta E}{C_s\,\omega_{\text{avg}}^2} = \frac{3{,}000}{0.10 (31.42)^2} = \frac{3{,}000}{98.70} = 30.4\,\text{kg}\cdot\text{m}^2$$

A thin steel rim at $r = 0.50\,\text{m}$ needs about $m = I/r^2 = 122\,\text{kg}$. The speed swings about $\pm 5\%$ (roughly $285$ to $315\,\text{rpm}$), the rim speed is $v = 31.42(0.50) = 15.7\,\text{m/s}$, and the rim hoop stress is only about $\rho v^2 = 7{,}850(15.7)^2 = 1.9\,\text{MPa}$.