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

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; Non-conventional-Solar, Wind.

Text Book: D. V. Kems Jr. and J. D. Erwin, A. R. Hambley, “Essentials Electrical and Computer Engineering”, Pearson-Prentice Hall, 2004. REFERENCES: 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.

UNIT 1 Review of Current, Voltage and Power Figure 1

Figure 2

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)

Basic Electric Circuits Elements and Terminology Figure 3

Figure 4

Example1: How many number of electrons pass through a cross section of a wire carrying 1A in 1s time? dq = i.dt = 1A.1s = 1 Coulomb. The charge of an electron is -1.6x10-19Coulomb, Therefore the number of electrons in one Coulomb Example 2

Example3: A lightning bolt carries 10000A 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 =10000x40x10-6C = 0.4Coulombs

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

Example-1: What is the work done (or energy required) by a flash light battery of 1.5V to transfer 0.5 Coulomb of charge through the bulb? W = V.Q = 1.5x0.5 J = 0.75J Example-2: A 20V delivers 0.1J 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 = 0.005Coulomb (b) I = Δq/Δt = 0.005/1x10-6 = 5x103A

Power 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= t2-t1 = Time duration.

Power p = V.I = (IR).I , (As V = IR) p = I2R, Also as I = V/R, p = V2/R Thus the power p = V.I and p = V2/R UNITS: Voltage Volt, Current Ampere, Power Watt Energy Watt.Hour

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

(a) VL = 12KV, PL = 1.2MW, as P = VxI, 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 1.2MW of power. Determine the amount of power that must be supplied by the generator if the line voltage at the load is: (a) 12KV, (b) 120KV. (a) VL = 12KV, PL = 1.2MW, as P = VxI, therefore, IL = PL/VL = 1.2x106/ 12x103 =100A, The line losses = PL = IL2x Rline = 1002x2x100x0.1 =0.2MW Therefore the power that must be supplied by the generator PG = Pline + PLoad = 0.2MW + 1.2MW =1.4MW (b) VL = 120KV, PL = 1.2MW, as P = VxI, therefore, IL = PL/VL = 1.2x106/ 120x103 =10A, The line losses = PL = IL2x Rline = 102x2x100x0.1 =2KW PG = Pline + PLoad = 2KW + 1.2MW = 0.002MW + 1.2MW =1.202MW

Example-4

Electrical Circuits Electrical Circuit Circuit elements 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 function

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

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

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

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.

Understanding KCL

KCL Examples

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

Identify the groups of circuit elements that are connected in series

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:

Understanding KVL

Identify the groups of circuit elements that are connected in parallel

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

KVL: Conservation of Energy

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

AC-when the current varies with time, reversing direction periodically, it is called alternating current (AC). i(t) t AC T i(t) t AC Vm Im T T T 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

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

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