Electrical quantities form the foundation of instrumentation and control systems in chemical engineering processes. Understanding these quantities enables engineers to design, operate, and troubleshoot instrumentation that monitors and controls process variables such as temperature, flow, pressure, and level. In Kenya’s chemical industries, accurate measurement and control of electrical parameters ensure safety, efficiency, and quality in production.

1.1 Resistance (R), Power (P), Energy (W or E), Frequency (f), Capacitance (C), Inductance (L)

1.1.1 Resistance (R)

Resistance quantifies how much a material opposes the flow of electric current. It is a crucial parameter in designing sensors and actuators in chemical plants, for example, resistance temperature detectors (RTDs). The governing formula for resistance is:

$$ R = \frac{\rho L}{A} $$

where \(R\) is resistance in ohms (\(\Omega\)), \(\rho\) is resistivity (\(\Omega \cdot m\)), \(L\) is length (m), and \(A\) is cross-sectional area (m\(^2\)).

Worked Examples

Example 1:
A copper wire of length 2 m and cross-sectional area \(1 \times 10^{-6}\) m\(^2\) is used in a sensor. Calculate its resistance. Given copper resistivity \(\rho = 1.68 \times 10^{-8}\) \(\Omega \cdot m\).

$$ R = \frac{\rho L}{A} $$

$$ R = \frac{1.68 \times 10^{-8} \times 2}{1 \times 10^{-6}} $$

$$ R = \frac{3.36 \times 10^{-8}}{1 \times 10^{-6}} = 0.0336 \ \Omega $$

Answer: 0.0336 \(\Omega\)

Example 2:
A platinum RTD wire has a resistance of 100 \(\Omega\) at 0 °C. If the wire length is 3 m and diameter is 0.5 mm, calculate the resistivity of platinum.

Cross-sectional area:

$$ A = \pi \times (\frac{d}{2})^2 = 3.1416 \times (\frac{0.0005}{2})^2 = 1.9635 \times 10^{-7} \ m^2 $$

Using

$$ R = \frac{\rho L}{A} \Rightarrow \rho = \frac{R A}{L} $$

$$ \rho = \frac{100 \times 1.9635 \times 10^{-7}}{3} = 6.545 \times 10^{-6} \ \Omega \cdot m $$

Answer: \(6.545 \times 10^{-6}\ \Omega \cdot m\)

Example 3:
In a chemical plant, a sensor circuit requires a 50 \(\Omega\) resistor made from a material with resistivity \(2 \times 10^{-6}\ \Omega \cdot m\). If the wire diameter is 1 mm, calculate the length of wire needed.

Cross-sectional area:

$$ A = \pi \times (\frac{0.001}{2})^2 = 7.854 \times 10^{-7} \ m^2 $$

Using

$$ R = \frac{\rho L}{A} \Rightarrow L = \frac{R A}{\rho} $$

$$ L = \frac{50 \times 7.854 \times 10^{-7}}{2 \times 10^{-6}} = 19.635 \ m $$

Answer: 19.635 m

1.1.2 Power (P)

Power in electrical circuits is the rate at which electrical energy is converted into another form, such as heat or mechanical energy. It is essential in sizing electrical components like motors and heaters in chemical processes. The basic formula for power is:

$$ P = V I $$

where \(P\) is power in watts (W), \(V\) is voltage (V), and \(I\) is current (A). For resistive loads, power can also be expressed as:

$$ P = I^2 R $$

or

$$ P = \frac{V^2}{R} $$

Worked Examples

Example 1:
A heater element in a chemical reactor operates at 220 V and draws 5 A. Calculate the power consumed.

$$ P = V I $$

$$ P = 220 \times 5 = 1100 \ W $$

Answer: 1100 W

Example 2:
A resistive sensor coil has a resistance of 20 \(\Omega\) and carries a current of 3 A. Find the power dissipated.

$$ P = I^2 R $$

$$ P = 3^2 \times 20 = 9 \times 20 = 180 \ W $$

Answer: 180 W

Example 3:
A temperature control unit operates on 110 V and consumes 550 W power. Calculate the resistance of the unit.

$$ P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} $$

$$ R = \frac{110^2}{550} = \frac{12100}{550} = 22 \ \Omega $$

Answer: 22 \(\Omega\)

1.1.3 Energy (W or E)

Electrical energy is the total work done or power consumed over a period. It is vital in chemical plants for cost estimation and energy management. The formula for electrical energy is:

$$ E = P t $$

where \(E\) is energy in joules (J) or kilowatt-hours (kWh), \(P\) is power in watts (W), and \(t\) is time in seconds (s) or hours (h).

Worked Examples

Example 1:
An electric heater consumes 1500 W for 3 hours. Calculate the energy consumed in kilowatt-hours.

$$ E = P t $$

Convert power to kilowatts:

$$ P = \frac{1500}{1000} = 1.5 \ kW $$

$$ E = 1.5 \times 3 = 4.5 \ kWh $$

Answer: 4.5 kWh

Example 2:
A pump motor rated at 2 kW runs for 2.5 hours. Calculate the energy consumed in joules.

Convert time to seconds:

$$ t = 2.5 \times 3600 = 9000 \ s $$

Calculate energy:

$$ E = P t = 2000 \times 9000 = 18,000,000 \ J $$

Answer: 18,000,000 J

Example 3:
A chemical process uses a 500 W motor for 45 minutes daily. Calculate monthly energy consumption in kWh (30 days).

Convert time to hours:

$$ t = \frac{45}{60} = 0.75 \ h $$

Daily energy:

$$ E_{daily} = \frac{500}{1000} \times 0.75 = 0.375 \ kWh $$

Monthly energy:

$$ E_{monthly} = 0.375 \times 30 = 11.25 \ kWh $$

Answer: 11.25 kWh

1.1.4 Frequency (f)

Frequency is the number of cycles per second of an alternating current (AC) signal. It affects the operation of AC motors and instrumentation in chemical plants. The unit of frequency is hertz (Hz). Frequency relates to time period \(T\) by:

$$ f = \frac{1}{T} $$

where \(T\) is the period in seconds.

Worked Examples

Example 1:
An AC power supply completes one cycle every 0.02 seconds. Calculate its frequency.

$$ f = \frac{1}{T} = \frac{1}{0.02} = 50 \ Hz $$

Answer: 50 Hz

Example 2:
A vibrating sensor produces 120 cycles in 2 seconds. Determine the frequency.

$$ f = \frac{\text{Number of cycles}}{\text{Time}} = \frac{120}{2} = 60 \ Hz $$

Answer: 60 Hz

Example 3:
A signal has frequency 400 Hz. Calculate its period.

$$ T = \frac{1}{f} = \frac{1}{400} = 0.0025 \ s $$

Answer: 0.0025 s

1.1.5 Capacitance (C)

Capacitance is the ability of a component to store electrical charge. Capacitors are used in chemical process control for filtering and timing circuits. The basic relation for capacitance is:

$$ C = \frac{Q}{V} $$

where \(C\) is capacitance in farads (F), \(Q\) is charge in coulombs (C), and \(V\) is voltage in volts (V). The capacitance of a parallel plate capacitor is:

$$ C = \frac{\varepsilon A}{d} $$

where \(\varepsilon\) is permittivity of the dielectric (F/m), \(A\) is plate area (m\(^2\)), and \(d\) is plate separation (m).

Worked Examples

Example 1:
A capacitor stores 0.005 C of charge at 10 V. Find its capacitance.

$$ C = \frac{Q}{V} = \frac{0.005}{10} = 0.0005 \ F = 500 \ \mu F $$

Answer: 500 \(\mu F\)

Example 2:
A parallel plate capacitor has plates of area 0.02 m\(^2\), separated by 1 mm, with air as dielectric \(\varepsilon = 8.85 \times 10^{-12} F/m\). Calculate the capacitance.

$$ C = \frac{\varepsilon A}{d} = \frac{8.85 \times 10^{-12} \times 0.02}{0.001} = 1.77 \times 10^{-10} \ F $$

Answer: 177 pF

Example 3:
A capacitor of 10 \(\mu F\) is charged to 100 V. Calculate the stored charge.

$$ Q = C V = 10 \times 10^{-6} \times 100 = 0.001 \ C $$

Answer: 0.001 C

1.1.6 Inductance (L)

Inductance is the property of a conductor by which a change in current induces an electromotive force (EMF) opposing the change. Inductors are used in chemical engineering control systems to smooth signals and filter noise. The voltage induced in an inductor is:

$$ V = L \frac{dI}{dt} $$

where \(L\) is inductance in henrys (H), and \(\frac{dI}{dt}\) is the rate of change of current (A/s).

Worked Examples

Example 1:
An inductor of 0.5 H carries a current increasing at 3 A/s. Calculate the induced voltage.

$$ V = L \frac{dI}{dt} = 0.5 \times 3 = 1.5 \ V $$

Answer: 1.5 V

Example 2:
A coil of inductance 2 H has voltage 10 V induced when current changes. Find the rate of change of current.

$$ V = L \frac{dI}{dt} \Rightarrow \frac{dI}{dt} = \frac{V}{L} = \frac{10}{2} = 5 \ A/s $$

Answer: 5 A/s

Example 3:
In a chemical plant control circuit, an inductor generates 6 V when the current changes at 4 A/s. Calculate the inductance.

$$ L = \frac{V}{\frac{dI}{dt}} = \frac{6}{4} = 1.5 \ H $$

Answer: 1.5 H

Chapter Summary

This chapter introduced the fundamental electrical quantities essential for control and instrumentation, starting with resistance, which measures how much a material opposes electric current. Power was explored as the rate of energy transfer in an electrical circuit, followed by energy itself, defined as the total work done or consumed over time. Frequency was covered as the number of cycles per second in alternating current systems, while capacitance and inductance were explained as properties of components that store energy in electric and magnetic fields respectively. The chapter also distinguished between conductors, insulators, and semiconductors, highlighting their roles in electrical circuits. Factors affecting conductivity such as temperature and material type were discussed to understand their impact on circuit performance. Finally, practical aspects of conductors and insulators in circuits were examined alongside methods for measuring work and power, providing a comprehensive foundation for further study in electrical control systems.