ELEC201 Electrical Technology By Dr Munaf Salim Najim

ELEC-201 Electrical Technology By Dr. Munaf Salim Najim Al-Din Department of Electrical & Computer Engineering University of Nizwa 1

Course Contents: • • • Review of current, voltage and power, Magnetically coupled circuits, Transformers: Single phase and Three phase, DC machines, AC machines, Single phase induction motor, Three phase induction motor, Elements of power system, Power generation: Conventional-Thermal, Hydro, Nuclear, Gas power plants; Nonconventional-Solar, Wind. 2

Text Book: D. V. Kems Jr. and J. D. Erwin, “Essentials Electrical and Computer Engineering”, Pearson-Prentice Hall, 2004. REFERENCES: 1. A. R. Hambley, “Electrical Engineering Principle and Applications” Pearson-Prentice Hall, 2005. 2. B. L. Theraja and A. K. Theraja, “Electrical Technology” S. Chand (India), 2000. 3

UNIT 1 Review of Current, Voltage and Power Figure 1 4

Figure 2 5

Electrical Circuits Fluid Flow Battery Pump Charge Fluid Conductors (no resistance) Pipes (frictionless) Current Flow rate Voltage Pressure differential Switches Valves Resistance Constriction in pipe (turbulence, heat) 6

Basic Electric Circuits Elements and Terminology Figure 3 7

Figure 4 8

• • Example 1: How many number of electrons pass through a cross section of a wire carrying 1 A in 1 s time? dq = i. dt = 1 A. 1 s = 1 Coulomb. The charge of an electron is -1. 6 x 10 -19 Coulomb, Therefore the number of electrons in one Coulomb Example 2 9

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• Example 3: A lightning bolt carries 10000 A of current and lasts for 40 microseconds. What is the charge deposited on an object hit by the lightning strike? Since I = Δq/Δt, Therefore the charge =Ix Δt =10000 x 40 x 10 -6 C = 0. 4 Coulombs 11

• Voltage or Potential and Potential Difference: Energy is required for the movement of charge from one point to another. The work per unit charge is called voltage or potential. • The potential difference V between two points is the energy required to move one coulomb of charge from one point to the other point. • If W = Energy required to move the charge from one point to other, Q = Charge transferred between two points, V = Potential difference between two points, Then 12

Example-1: What is the work done (or energy required) by a flash light battery of 1. 5 V to transfer 0. 5 Coulomb of charge through the bulb? W = V. Q = 1. 5 x 0. 5 J = 0. 75 J Example-2: A 20 V delivers 0. 1 J of energy (a) How much charge was delivered by the Battery? (b) If the process in (a) is occurred in 1µs what was the current? (a) Since V = W/Q, therefore Q = W/V = 0. 1/20 = 0. 005 Coulomb (b) I = Δq/Δt = 0. 005/1 x 10 -6 = 5 x 103 A 13

• Power is defined as the time rate at which energy (w) is produced or consumed depending on whether the element is a source of power or an user of power, respectively. which can be rewritten as Thus the energy where T= t 2 -t 1 = Time duration. 14

• Power p = V. I = (IR). I , p = I 2 R, (As V = IR) Also as I = V/R, p = V 2/R Thus the power p = V. I p = I 2 R, p = V 2/R and UNITS: Voltage Current Power Energy Volt, Ampere, Watt. Hour 15

Example-1: The voltage across a lamp is 12 V and the current through the lamp is 60 m. A. Determine the resistance of the lamp and the power consumed by the lamp. Ans. R = 200 Ohm, P = 720 m. W. Example-2: A speaker which converts electrical energy into sound energy is of 8 Ohm and the maximum power rating is of 200 W. Determine the maximum current that can be delivered to the speaker. Ans. I = 5 A. 16

Example-3: An industrial load is served by a power company generator through a 100 Km long single phase transmission line of resistance 0. 1 Ohm/Km. The load absorbs 1. 2 MW of power. Determine the amount of power that must be supplied by the generator if the line voltage at the load is: (a) 12 KV, (b) 120 KV. (a) VL = 12 KV, PL = 1. 2 MW, as P = Vx. I, therefore, IL = PL/VL = 1. 2 x 106/ 12 x 103 =100 A, The line losses = PL = IL 2 x Rline = 1002 x 2 x 100 x 0. 1 =0. 2 MW Therefore the power that must be supplied by the generator PG = Pline + PLoad = 0. 2 MW + 1. 2 MW =1. 4 MW (b) VL = 120 KV, PL = 1. 2 MW, as P = Vx. I, therefore, IL = PL/VL = 1. 2 x 106/ 120 x 103 =10 A, The line losses = PL = IL 2 x Rline = 102 x 2 x 100 x 0. 1 =2 KW Therefore the power that must be supplied by the generator PG = Pline + PLoad = 2 KW + 1. 2 MW = 0. 002 MW + 1. 2 MW =1. 202 MW 17

Example-4 18

Electrical Circuits • Electrical Circuit – Various types of circuit elements connected in closed paths by conductors • Circuit elements – Resistances, inductances, capacitances, voltage sources, etc. • Voltage Sources – Induce flow of electrons (charge) through conductors and other circuit elements – Energy is transferred between circuit elements, resulting in a useful 19 function

Circuit Elements 1 - Resistor (R) (Ohm ) 2 - Inductor (L) (Henry) 3 - Capacitor (Farads) 4 - Voltage Source 5 - Current Source 20

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Conductors and Insulators Conductors Semiconductors Insulators Gold Silicon Glass Silver Germanium Plastic Copper Gallium Arsenide Ceramics Aluminum Rubber 22

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Electrical Circuit Analysis • Steps – Model physical components mathematically • Voltage or current sources • Resistors, inductors, capacitors, • Diodes, transistors, transformers, electric motors – Determine system of equations for unknown element characteristics • Solve using linear algebra methods 30

Kirchhoff’s Current Law (KCL) • KCL: The net current entering a node is zero. • Also, the net current leaving a node is zero. • In other words, the total amount of current ENTERING a node must equal the total amount of current LEAVING a node. 31

Understanding KCL 32

KCL Examples 33

Series Circuit Elements • When elements are connected end to end, they are connected in SERIES • All circuit elements have IDENTICAL Currents 34

• Identify the groups of circuit elements that are connected in series 35

Kirchhoff’s Voltage Law (KVL) • KVL: The algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit. • To add voltages algebraically, we must be consistent: 36

Understanding KVL 37

• Identify the groups of circuit elements that are connected in parallel 38

Parallel Circuit Elements • When both ends of one element are connected to corresponding ends of another, they are connected in PARALLEL • All circuit elements have IDENTICAL Voltages 39

KVL: Conservation of Energy 40

DC and AC Electricity • DC-when the current is constant with time it is called direct current (DC). i(t) DC t i(t) = Idc = Constant, Similarly V(t) = Vdc = Constant 41

• AC-when the current varies with time, reversing direction periodically, it is called alternating current (AC). AC i(t) T Imm V t i(t) = Im Sin(ωt + θ), Similarly V(t) = Vm Sin(ωt + θ), where ω = 2πf = 2π / T, Where f = frequecy, T (= 1/f) = Time period and θ = Phase angle 42

DC superimposed on AC AC DC v(t) Vm VA t V(t) = DC + AC V(t) = VA + Vm Sin(ωt ) 43

B 120 0 1200 Three Phase supply: v(t) A B 0 20 A 1 C C t Phase A, B, and C are 1200 apart respectively from each other as shown above. VA =Vm. Sin(ωt), VB = Vm Sin(ωt + 1200+ ) and VC = Vm Sin(ωt - 1200 ) 44

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