Rev Up Your Skills: Electric Motor Design Exercises for Trinidad

1. Convert speed to angular velocity — First convert the rotational speed from revolutions per minute to radians per second. Since one revolution equals 2π radians and there are 60 seconds in a minute, multiply by 2π and divide by 60.
   $$\omega =n\times \frac{2\pi }{60}$$
2. Apply power formula — Use the mechanical power formula where power equals torque multiplied by angular velocity.
   $$P=T\times \omega$$
3. Calculate final power — Substitute the values and compute the mechanical power output.
   $$P=12\times 157.08$$

$$1885\text{ W}$$

→ The motor produces approximately 1885 watts of mechanical power.

1. Calculate efficiency ratio — Divide the output power by the input power to get the efficiency as a decimal.
   $$\eta =\frac{P_{out}}{P_{in}}$$
2. Convert to percentage — Multiply the decimal efficiency by 100 to express it as a percentage.
   \eta_{%} = \eta \times 100 ParseError: Expected '}', got 'EOF' at end of input: …\eta \times 100

$$84\text{ \%}$$

→ The motor has an efficiency of 84%.

1. Calculate initial power — First calculate the power using the initial torque and speed values.
   $$P=T_1\times {\omega }_1=T_1\times \frac{2\pi n_1}{60}$$
2. Set up equation for new conditions — With constant power, set the new power equal to the initial power and solve for the new speed.
   $$P=T_2\times {\omega }_2=T_2\times \frac{2\pi n_2}{60}$$
3. Solve for new speed — Rearrange the equation to find the new speed $n_2$.
   $$n_2=n_1\times \frac{T_1}{T_2}$$

$$800\text{ rpm}$$

→ The new speed would be 800 rpm.

1. Calculate maximum flux — The maximum flux Φ_max equals the flux density multiplied by the pole face area.
   $${\mathrm{\Phi }}_{max}=B\times A$$
2. Use induced voltage formula — For a sinusoidal supply, the RMS voltage relates to the number of turns, frequency, and maximum flux.
   $$V=4.44\times f\times N\times {\mathrm{\Phi }}_{max}$$
3. Solve for turns per coil — Rearrange the formula to find the number of turns required per coil.
   $$N=\frac{V}{4.44\times f\times {\mathrm{\Phi }}_{max}}$$

$$312\text{ turns}$$

→ Each coil requires approximately 312 turns.

1. Convert power to watts — Convert the mechanical power output from horsepower to watts.
   $$P_{mech}=5\text{ hp}\times 745.7=3728.5\text{ W}$$
2. Calculate input power — The input electrical power equals the mechanical power divided by efficiency.
   $$P_{in}=\frac{P_{mech}}{\eta }$$
3. Calculate input current — For single-phase AC, input current equals input power divided by (voltage × power factor).
   $$I=\frac{P_{in}}{V\times PF}$$
4. Determine wire area — The wire cross-sectional area equals current divided by maximum current density.
   $$A_{wire}=\frac{I}{J_{max}}$$

$$4.9{\text{ mm}}^2$$

→ The minimum wire cross-sectional area should be at least 4.9 square millimetres.

1. Apply flux continuity — The magnetic flux remains constant through the magnetic circuit, so Φ_stator = Φ_rotor = Φ_airgap.
   $${\mathrm{\Phi }}_s={\mathrm{\Phi }}_r=\mathrm{\Phi }=0.008\text{ Wb}$$
2. Calculate stator flux density — Flux density in the stator equals flux divided by stator area.
   $$B_s=\frac{\mathrm{\Phi }}{A_s}$$
3. Calculate rotor flux density — Flux density in the rotor equals flux divided by rotor area.
   $$B_r=\frac{\mathrm{\Phi }}{A_r}$$

$$B_s=0.8\text{ T},B_r=0.89\text{ T}$$

→ The stator flux density is 0.8 tesla and the rotor flux density is approximately 0.89 tesla.

1. Calculate total direct cost — Total direct cost equals material cost plus labor cost for all motors.
   $$C_{direct}=N\times (C_{mat}+C_{lab})$$
2. Calculate overhead costs — Overhead costs are 15% of the total direct costs.
   $$C_{overhead}=O_{rate}\times C_{direct}$$
3. Calculate total production cost — Total cost equals direct costs plus overhead costs.
   $$C_{total}=C_{direct}+C_{overhead}$$

$$142600\text{ TTD}$$

→ The total production cost for 100 motors is 142,600 Trinidad and Tobago dollars.

1. Use synchronous speed formula — The synchronous speed $n_s$ equals 120 times frequency divided by the number of poles.
   $$n_s=\frac{120f}{N_p}$$
2. Rearrange to solve for frequency — Rearrange the synchronous speed formula to isolate frequency.
   $$f=\frac{n_s\times N_p}{120}$$
3. Substitute values — Plug in the synchronous speed (assumed equal to motor speed) and number of poles.
   $$f=\frac{1000\times 6}{120}$$

$$50\text{ Hz}$$

→ The supply frequency should be 50 hertz.

1. Convert output power to watts — Convert the mechanical power output from horsepower to watts.
   $$P_{out}=7.5\text{ hp}\times 745.7=5592.75\text{ W}$$
2. Calculate input power — Input power equals output power divided by efficiency.
   $$P_{in}=\frac{P_{out}}{\eta }$$
3. Calculate total losses — Total losses equal input power minus output power.
   $$P_{loss}=P_{in}-P_{out}$$
4. Calculate copper losses — Copper losses equal total losses minus iron losses.
   $$P_{cu}=P_{loss}-P_{iron}$$
5. Calculate percentage copper losses — Percentage copper losses equals copper losses divided by total losses times 100%.
   $$Pc{t}_{cu}=\frac{P_{cu}}{P_{loss}}\times 100$$

$$P_{loss}=1007\text{ W},Pc{t}_{cu}=80\text{ \%}$$

→ Total losses are 1007 watts, with copper losses accounting for approximately 80% of the total losses.

1. Calculate required torque — Use the power formula to calculate the torque required for the specified power and speed.
   $$P=T\times \omega =T\times \frac{2\pi n}{60}$$
2. Rearrange for torque — Rearrange the power equation to solve for torque.
   $$T=\frac{P\times 60}{2\pi n}$$
3. Compare motor types — 4-pole motors typically operate around 1500 rpm (synchronous speed at 50 Hz), while 6-pole motors operate around 1000 rpm. At 720 rpm, a 6-pole motor would be more suitable.

$$T=398\text{ N·m},\text{6-pole motor recommended}$$

→ The crane requires approximately 398 newton-metres of torque. A 6-pole motor would be more suitable as it can provide higher torque at lower speeds while meeting the 90% efficiency requirement.