Electric Power and Machines: free practice, theory and problems

1. The transformer

What is it about?

Electricity from the power station is sent out at high voltage and transformed down in several steps before it reaches the socket. The transformer does this with two coils around an iron core, with no moving parts and very small losses.

Concepts and formulas

• The turns ratio uses the numbers of turns $N_1$ (primary) and $N_2$ (secondary). For an ideal transformer:

• The power is the same on both sides (ideally): $U_1 I_1 = U_2 I_2$. Down in voltage means up in current.
• Transformers are rated in apparent power in kVA. For three-phase: $I = \dfrac{S}{\sqrt3\,U}$.
• Efficiency: $\eta = \dfrac{P_{out}}{P_{out} + P_{loss}}$. Large transformers often exceed 98 %.
• The transformer only works on alternating current, because it needs a changing magnetic field.
• Power is transmitted at high voltage because the current is then low, and the line losses ($I^2 R$) become small.

How to solve the problems

1. Write down the ratio $N_1/N_2$.
2. Voltage follows the turns, current goes the opposite way.
3. Three-phase: use $\sqrt3$ in the current formula.

Example

A transformer has 1000 turns on the primary side and 50 on the secondary side. The primary voltage is 230 V.

1. $U_2 = U_1\cdot N_2/N_1 = 230\cdot 50/1000 = 11.5$ V.
2. If the secondary delivers 10 A, the primary draws $10\cdot 50/1000 = 0.5$ A.

Common mistakes

• Flipping the ratio for current: fewer turns give lower voltage but higher current.
• Forgetting $\sqrt3$ for three-phase.
• Thinking a transformer works on direct current.

Concepts in this part

Practise the transformer in the app →

2. Electric motors

What is it about?

Electric motors account for almost half of the electricity used in industry: pumps, fans, conveyors, compressors. The most common is the induction motor, which is robust, cheap and runs directly on three-phase alternating current.

Concepts and formulas

• The synchronous speed is the speed of the rotating magnetic field. With frequency $f$ and $p$ poles:

A 4-pole motor at 50 Hz has $n_s = 1500$ rpm.

• Slip: the rotor of an induction motor turns slightly slower than the field.

• Torque from power and speed:

• Efficiency: $\eta = P_{mech}/P_{el}$. The rating plate gives the mechanical shaft power.
• A variable frequency drive changes the frequency and thus the speed, which saves a lot of energy in pumps and fans.

How to solve the problems

1. Find $n_s$ from frequency and number of poles.
2. Slip or rotor speed: $n = n_s(1 - s)$.
3. Torque: $9550\cdot P/n$.

Example

A 4-pole motor at 50 Hz runs at 1440 rpm and delivers 11 kW.

1. $n_s = 120\cdot 50/4 = 1500$ rpm.
2. $s = (1500 - 1440)/1500 = 0.04 = 4$ %.
3. $T = 9550\cdot 11/1440 \approx 73$ Nm.

Common mistakes

• Mixing up number of poles and number of pole pairs. $n_s = 60f/\text{pole pairs}$ is the same as $120f/\text{poles}$.
• Mixing rad/s and rpm when computing torque.
• Thinking an induction motor runs exactly synchronously. Then no current is induced in the rotor, and there is no torque.

Concepts in this part

Practise electric motors in the app →

3. The power system and energy

What is it about?

Norway gets almost all of its electricity from hydropower, which is sent through a grid of lines and transformers to every home and business. Here you learn to calculate energy and cost, line losses, voltage drop and how much power a hydropower plant delivers.

Concepts and formulas

• Power $P$ (W, kW) is how fast energy is used. Energy $E = P\cdot t$ is measured in kWh. A 2 kW heater for 3 hours uses 6 kWh.
• Electricity cost: energy times price per kWh.
• Loss in a line: $P_{loss} = I^2 R$. Double voltage for the same power gives half the current and a quarter of the loss.
• Voltage drop across a line with resistance $R$: $\Delta U = I R$. Often given as a percentage of the nominal voltage.
• Hydropower: the power from the head $H$ and the flow $Q$:

• The grid frequency (50 Hz) is kept stable by production and consumption being in balance at every moment.

How to solve the problems

1. Distinguish between power (kW) and energy (kWh).
2. Losses and voltage drop: find the current first.
3. Hydropower: $\rho = 1000$ kg/m³ and $g = 9.81$ m/s².

Example

A power plant has a head of 100 m, a flow of 10 m³/s and an efficiency of 0.9.

1. $P = 1000\cdot 9.81\cdot 10\cdot 100\cdot 0.9$.
2. $P \approx 8\,829\,000$ W $\approx 8.83$ MW.

Common mistakes

• Mixing kW and kWh.
• Computing losses with $I R$ instead of $I^2 R$.
• Forgetting the efficiency in the hydropower formula.

Concepts in this part

Practise the power system and energy in the app →