Electrical quantities form the foundation for understanding and working safely with refrigeration and air conditioning systems. Mastery of these quantities enables technicians to correctly measure, analyze, and troubleshoot electrical circuits within cooling systems, ensuring optimal performance and safety. This chapter introduces the basic electrical quantities used in the field, focusing on their definitions, units, and practical applications relevant to refrigeration and air conditioning professionals in Kenya.

1.1 Introduction to SI units

The International System of Units (SI) provides the standard framework for measuring electrical quantities in refrigeration and air conditioning work across Kenya and the world. Using SI units ensures consistency, accuracy, and clear communication among technicians, engineers, and suppliers. In this section, we focus on the primary electrical quantities, current, voltage, resistance, power, and energy, and their respective SI units.

1.1.1 Electric Current

Electric current is the flow of electric charge through a conductor, typically measured in amperes (A). It is a fundamental quantity in refrigeration and air conditioning systems, where it represents the rate at which electrons pass through wires powering compressors, fans, and control circuits.

The governing formula for electric current is:

$$I = \frac{Q}{t}$$

where \(I\) is the current in amperes, \(Q\) is the electric charge in coulombs, and \(t\) is the time in seconds.

Worked Examples

Example 1: A refrigeration compressor draws a charge of 300 coulombs over 10 seconds. Calculate the current flowing through the compressor.

Given:
\(Q = 300\,C\), \(t = 10\,s\)

$$I = \frac{Q}{t}$$$$I = \frac{300}{10}$$$$I = 30\,A$$

Answer: 30 amperes

Example 2: An air conditioning fan motor passes a current of 5 A for 2 minutes. Find the total charge that has flowed through the motor.

Given:
\(I = 5\,A\), \(t = 2 \times 60 = 120\,s\)

$$Q = I \times t$$$$Q = 5 \times 120$$$$Q = 600\,C$$

Answer: 600 coulombs

Example 3: A control circuit in a refrigeration unit carries a current of 0.2 A for 15 seconds. Determine the charge transferred.

Given:
\(I = 0.2\,A\), \(t = 15\,s\)

$$Q = I \times t$$$$Q = 0.2 \times 15$$$$Q = 3\,C$$

Answer: 3 coulombs

Example 4: A defrost heater in a refrigeration system draws 8 A for 30 seconds. Calculate the charge flow.

Given:
\(I = 8\,A\), \(t = 30\,s\)

$$Q = I \times t$$$$Q = 8 \times 30$$$$Q = 240\,C$$

Answer: 240 coulombs

Example 5: A refrigeration system’s electronic controller passes a charge of 0.05 C in 0.1 seconds. Find the current.

Given:
\(Q = 0.05\,C\), \(t = 0.1\,s\)

$$I = \frac{Q}{t}$$$$I = \frac{0.05}{0.1}$$$$I = 0.5\,A$$

Answer: 0.5 amperes

1.1.2 Electric Voltage

Voltage is the electric potential difference between two points in a circuit, measured in volts (V). It is the driving force that pushes electric current through components like compressors and fans in cooling systems.

The formula relating voltage, current, and resistance is given by Ohm's law:

$$V = I \times R$$

where \(V\) is voltage in volts, \(I\) is current in amperes, and \(R\) is resistance in ohms.

Worked Examples

Example 1: A refrigeration compressor draws 10 A through a coil with resistance 5 Ω. Calculate the voltage across the coil.

Given:
\(I = 10\,A\), \(R = 5\,\Omega\)

$$V = I \times R$$$$V = 10 \times 5$$$$V = 50\,V$$

Answer: 50 volts

Example 2: An air conditioning unit has a voltage supply of 230 V and a current of 15 A. Find the resistance of the unit.

Given:
\(V = 230\,V\), \(I = 15\,A\)

$$R = \frac{V}{I}$$$$R = \frac{230}{15}$$$$R \approx 15.33\,\Omega$$

Answer: 15.33 ohms

Example 3: A defrost heater operates at 120 V and has a resistance of 60 Ω. Calculate the current passing through it.

Given:
\(V = 120\,V\), \(R = 60\,\Omega\)

$$I = \frac{V}{R}$$$$I = \frac{120}{60}$$$$I = 2\,A$$

Answer: 2 amperes

Example 4: A refrigeration fan coil has a resistance of 8 Ω and draws 3 A. Determine the voltage across it.

Given:
\(R = 8\,\Omega\), \(I = 3\,A\)

$$V = I \times R$$$$V = 3 \times 8$$$$V = 24\,V$$

Answer: 24 volts

Example 5: An electronic controller in an air conditioning system is supplied with 12 V and draws 0.5 A. Calculate the resistance of the controller.

Given:
\(V = 12\,V\), \(I = 0.5\,A\)

$$R = \frac{V}{I}$$$$R = \frac{12}{0.5}$$$$R = 24\,\Omega$$

Answer: 24 ohms

1.1.3 Electrical Resistance

Electrical resistance measures how much a material opposes the flow of electric current, measured in ohms (Ω). Resistance affects the current flow in refrigeration components such as compressor windings and heater elements.

Ohm’s law relates resistance, voltage, and current:

$$R = \frac{V}{I}$$

Worked Examples

Example 1: A refrigeration motor is supplied with 240 V and draws 20 A. Calculate the resistance of the motor windings.

Given:
\(V = 240\,V\), \(I = 20\,A\)

$$R = \frac{V}{I}$$$$R = \frac{240}{20}$$$$R = 12\,\Omega$$

Answer: 12 ohms

Example 2: A defrost heater has a resistance of 40 Ω and operates on 200 V. Find the current it draws.

Given:
\(R = 40\,\Omega\), \(V = 200\,V\)

$$I = \frac{V}{R}$$$$I = \frac{200}{40}$$$$I = 5\,A$$

Answer: 5 amperes

Example 3: A refrigeration fan coil draws 4 A with a voltage drop of 48 V. Calculate the resistance.

Given:
\(I = 4\,A\), \(V = 48\,V\)

$$R = \frac{V}{I}$$$$R = \frac{48}{4}$$$$R = 12\,\Omega$$

Answer: 12 ohms

Example 4: An air conditioning compressor coil has a resistance of 8 Ω and draws 15 A. Find the voltage across the coil.

Given:
\(R = 8\,\Omega\), \(I = 15\,A\)

$$V = I \times R$$$$V = 15 \times 8$$$$V = 120\,V$$

Answer: 120 volts

Example 5: A control circuit has resistance 50 Ω and is connected to a 230 V supply. Calculate the current.

Given:
\(R = 50\,\Omega\), \(V = 230\,V\)

$$I = \frac{V}{R}$$$$I = \frac{230}{50}$$$$I = 4.6\,A$$

Answer: 4.6 amperes

1.1.4 Electrical Power and Energy

Electrical power is the rate at which electrical energy is consumed or converted, measured in watts (W). Power is crucial in refrigeration systems to determine compressor motor ratings and energy consumption.

The formula for electric power is:

$$P = V \times I$$

Electrical energy consumed is power multiplied by time, measured in watt-hours (Wh) or kilowatt-hours (kWh):

$$E = P \times t$$

where \(E\) is energy in watt-hours, \(P\) is power in watts, and \(t\) is time in hours.

Worked Examples

Example 1: A refrigeration compressor operates at 230 V and draws 10 A. Calculate its power consumption.

Given:
\(V = 230\,V\), \(I = 10\,A\)

$$P = V \times I$$$$P = 230 \times 10$$$$P = 2300\,W$$

Answer: 2300 watts

Example 2: An air conditioning fan runs at 240 V and 5 A for 3 hours. Calculate the energy consumed in kilowatt-hours.

Given:
\(V = 240\,V\), \(I = 5\,A\), \(t = 3\,h\)

$$P = V \times I$$$$P = 240 \times 5$$$$P = 1200\,W = 1.2\,kW$$

$$E = P \times t$$$$E = 1.2 \times 3$$$$E = 3.6\,kWh$$

Answer: 3.6 kilowatt-hours

Example 3: A defrost heater with power rating 1500 W operates for 2 hours. Calculate the energy used.

Given:
\(P = 1500\,W = 1.5\,kW\), \(t = 2\,h\)

$$E = P \times t$$$$E = 1.5 \times 2$$$$E = 3\,kWh$$

Answer: 3 kilowatt-hours

Example 4: A refrigeration system compressor consumes 3 kW running for 4.5 hours. Find the total energy consumed.

Given:
\(P = 3\,kW\), \(t = 4.5\,h\)

$$E = P \times t$$$$E = 3 \times 4.5$$$$E = 13.5\,kWh$$

Answer: 13.5 kilowatt-hours

Example 5: An air conditioning system operates at 220 V and 8 A for 6 hours. Calculate power and energy consumption.

Given:
\(V = 220\,V\), \(I = 8\,A\), \(t = 6\,h\)

$$P = V \times I$$$$P = 220 \times 8$$$$P = 1760\,W = 1.76\,kW$$

$$E = P \times t$$$$E = 1.76 \times 6$$$$E = 10.56\,kWh$$

Answer: Power = 1760 watts; Energy = 10.56 kilowatt-hours

Chapter Summary

This chapter began by introducing the International System of Units (SI) as the foundation for measuring electrical quantities. It then detailed the specific SI units used for various electrical parameters such as current, voltage, resistance, and power. The fundamental relationship between voltage, current, and resistance was explored through Ohm's law, providing a basis for solving electrical circuit problems. Practical calculations involving power, current, voltage, and resistance were demonstrated to reinforce understanding. The chapter also covered the common instruments used to measure these electrical parameters accurately. Further, it explained calculations related to capacitors, emphasizing their role in storing electrical energy. Finally, the concept of electric charge and the electrostatic field was introduced to highlight the forces and interactions at the microscopic level in electrical phenomena.