Engineering Utilities FIN PDF

Module 1 Building Electrical Systems  Topics to be Discussed: Electricity Theory Building Electrical Materials and Equipment Building Electrical Design Principles  Electrical Theory Electricity – it is a property of matter that results from the presence or movement of el...

Module 1 Building Electrical Systems  Topics to be Discussed: Electricity Theory Building Electrical Materials and Equipment Building Electrical Design Principles  Electrical Theory Electricity – it is a property of matter that results from the presence or movement of electric charge. According to modern theory, matter is electrical in nature.  Structure of Matter Matter – it is anything that occupies space and has mass. Atom – it is the smallest particle of an element that is capable of independent existence. Molecule – it is the smallest particle of matter (an element or a compound) that is capable of free existence. Element – it is a substance that cannot be decomposed any further by chemical action. Compound – it is a combination of two or more elements. Mixture – it is a material composed of two or more substances, each of which retains its own characteristic properties.  Basic Data About the Atom Proton – it is a positively charged particle. Electron – it is negatively charged particle. Neutron – it is a particle with neutral charge (no charge). Atomic Number – it represents the number of protons or electrons of an atom. Atomic Mass – it represents the sum of protons and electrons of an atom. Ion – it is an atom or a group of atoms that has a net positive or negative charge resulting from unequal number of positively charged protons and negatively charged electrons. Shell – it is an energy level or region about the nucleus of an atom in which electrons move. The atom has 5 shells, namely, the K-shell (first orbit), the L-shell (second orbit), the M-shell (third orbit), the N-shell (fourth orbit) and the O-shell (fifth orbit) Valence electrons – is the electrons found in the outermost shell (valence shell) or orbit of an atom.  Basic Data About the Atom Particle Charge Mass 1.602 x 10-21 C, 1.672 x 10-27 Proton positive kg 1.602 x 10-21 C, 9.107 x 10-31 Electron negative kg No charge, 1.672 x 10-27 Neutron neutral kg  Electrical Classification of Materials Electrical Conductors – these are materials that allow the essentially free passage of current when connected to a battery or other source of electrical energy. Electrical Insulators – these are materials that posses low conductivity or offer a relatively high resistance to the flow of electric current. They are also called as dielectrics. Electric Semiconductors – these are materials that have conductivity about midway between good conductors and good insulators.  Electrical Current Electric Current (I) – it is the motion or transfer of charges from one region of a conductor to another. Its unit is the Ampere (A) named after André Marie Ampère.  Sources of Electric Current The following are the methods by which electric current can be produced: Static electricity from friction Thermoelectricity is electricity from heat Piezoelectricity is electricity from pressure Electrochemistry is electricity from a chemical action Photoelectricity is electricity from light Magnetoelectricity is electricity from magnetism  Fundamental Units of Electricity Electric Voltage (V or E) – it is the driving force behind current flow. The unit of voltage is the Volt (V). Voltage level governs the amount of current flow. An increase in voltage causes more current flow and vice versa. Electric Current (I) – it is the motion or transfer of charges from one region of a conductor to another. Its unit is the Ampere (A). Electric Potential Difference – it is the difference in the electric potentials of two charged bodies. Its unit is the volt. Electric Resistance (R) – it is the property of a material that limits the amount of flow of current and converts electric energy to heat energy. Its unit is the Ohm (Ω).  Ohm’s Law According to Ohm’s Law, “At steady- state condition, the voltage across a resistor is directly proportional to the current flowing through it with the temperature remaining constant.” 𝑽𝑽 𝑽 𝑽 𝑽𝑽==𝑰𝑹 𝑰𝑹 𝑹 𝑹== 𝑰𝑰 𝑰=𝑰= 𝑹𝑹 Where: V = applied voltage in volts Georg Simon Ohm I = current drawn in (1787 – 1854) amperes R = resistance in ohms  Conditions for Ohm’s Law Ohm’s law can be applied either to the entire circuit or a part of a circuit When Ohm’s law is applied to a part circuit, part resistance and the potential difference across that part resistance should be used Ohm’s law can be applied to both dc and ac circuits  Limitations of Ohm’s Law Ohm’s law is not applicable to the following: For metals which get heated up due to the flow of current through them For electrolytes where enormous gases are produced on either electrode For vacuum radio valves For arc lamps For semiconductors For gas-filled tubes, in which the ions are generated as a result of current flow For appliances such as metal rectifier and crystal detectors, in which the operation depends on the direction of current  Electrical Power Electrical Power (P) – it is defined as the rate at which electrical energy is expended or used up. It is the rate electric energy is converted into another form, such as light, heat or mechanical energy (or converted from another form into electrical energy). Its unit is the Watt (W). Where: James Watt P = electrical power in watt V = voltage in volt Note: I = current in amperes 1 horsepower (HP) = 746 W R = resistance in ohm  Energy  Energy– it is the capacity to do work. It is measured in Joules (J).  Electrical Energy Consumption – it is the rate at which power is consumed over a specified period in of time. Where: P = electrical power in watt t = time W = electric energy consumed  Electric Circuit Electric Circuit – it is a collection of electrical elements interconnected in some specific way through which electric current flows or it is intended to flow. A circuit consists of a source, conducting parts and a load.  Closed, Open and Short Circuits Closed Circuit – it is an uninterrupted path that allows a continuous flow of current through an electrical circuit. In a building, the circuit is closed when a switch is turned on. Open Circuit. If the path of current flow is interrupted such as if the switch in a circuit is open (turned off), an open circuit results. Short Circuit. If an inadvertent shortcut develops in a circuit that permits current flow through an unintentional path, a short circuit is created. A short circuit occurs when current leaks out of the intended conductor path such as out of a wire with damaged insulation.  Types of Circuit Connections 1. Series circuit 2. Parallel circuit 3. Combinational Circuit a. Series-parallel circuit – it is a combinational circuit when simplified will result into a series circuit b. Parallel-series circuit – it is a combinational circuit when simplified will result into a parallel circuit.  Series Circuits 1. Series Circuit – the circuit elements are said to be connected in series when they all carry the same current. 2. Properties of a Series Circuit:  The same current flows through all the resistances.  There will be voltage drop across each resistance.  The sum of the voltage drops is equal to the applied voltage. 3. Equivalent Resistance of a Series Circuit: 𝑵𝑵 𝑹𝑹𝑻𝑻==𝑹𝑹𝟏𝟏++𝑹𝑹𝟐𝟐++𝑹𝑹𝟑𝟑++⋯+ ⋯+𝑹𝑹𝑵𝑵==∑ ∑𝑹𝑹𝒏𝒏 𝑰𝑻𝑰= 𝑰 =𝑰 =𝑰 𝑻=𝑰𝟏𝟏=𝑰𝟐𝟐=𝑰𝟑𝟑 𝒏𝒏𝟏𝟏 == 𝑽𝑽𝑻𝑻==𝑽𝑽𝟏𝟏++𝑽𝑽𝟐𝟐++𝑽𝑽𝟑𝟑 Where: RT = total resistance  Sample Problems 1. Four coils having resistances of 3, 5, 10 and 12 ohms are connected in series across 120 V. Determine (a) equivalent resistance of the circuit, (b) current flowing through the circuit and (c) voltage drop across individual coils. Solution: (a) For the equivalent resistance of the circuit 𝑹𝑻=𝟑+5+𝟏0+𝟏𝟐 IT 3 V1 V2 5 𝑹𝑻=𝟑0 𝛀 VT 120 V (b) For the current flowing through the V3 10 circuit 𝑻 𝑰𝑻 = V4 12 𝑹𝑻 (c) For the voltage drop across individual coils.  Sample Problems 2. An electric iron is rated at 1 kW, 250 V. Calculate the current taken by it if it is connected to 220 V supply. Solution: For the electric iron, it is rated at 1 kW, 250 V 𝑽𝟐 (𝟐50 )𝟐 𝑷= 𝑹 𝟏,000= 𝑹 𝑹=6𝟐.5 𝛀 When the electric iron is connected a 220 V supply 𝑰= 𝑽 𝑰= 𝟐𝟐0 𝑹 6𝟐.5 𝑰=𝟑.5𝟐𝑨  Sample Problems 3. Three resistors are connected in series across a 12 V battery. The first resistor has a value of 1 Ω, second has a voltage drop of 4 V and the third has a power dissipation of 12 W. Calculate the value of the circuit current. Solution: IT V1 R1 𝑽𝑻=𝑽 𝟏+𝑽 𝟐+𝑽𝟑 1 𝟏𝟐=𝑽 𝟏+𝟒+𝑽 𝟑 VT V2 R2 12 V 4V 𝑽𝟏+𝑽 𝟑=𝟖 V3 R3 P3 = 12 W For V1, 𝑽𝟏=𝑰 𝑹𝟏 𝑽 𝟏=𝑰 (𝟏) 𝑽𝟏=𝑰 (𝒏umerically) For V3, 𝟑 𝑽 𝟑 =𝟏𝟐 𝑰 𝑷𝟑=𝑽 𝟑𝑰 𝟏𝟐=𝑽 𝑰 Substituting, 𝑽𝟏+𝑽 𝟑=𝟖 𝑰+ 𝟏𝟐= 𝟖 𝑰𝟐−𝟖 𝑰+𝟏𝟐=0 𝑰 𝑰= 6 𝑨 𝑰=𝟐 𝑨  Parallel Circuits 1. Parallel Circuit – circuit elements are connected in parallel when the same voltage is common to all of them. 2. Properties of a Parallel Circuit:  The voltage across each resistance of the parallel combination is the same.  There are as many current paths as the number of branches.  The current in each branch is given by Ohm’s law.  The total current of the circuit is equal to the sum of branch currents. 3. Equivalent Resistance of a Parallel Circuit 𝑽𝑽𝑻𝑻==𝑽𝑽𝟏𝟏==𝑽𝑽𝟐𝟐==𝑽𝑽𝟑𝟑 𝑹𝑹𝑻𝑻== 𝟏𝟏 𝑰𝑻𝑰𝑻==𝑰𝑰𝟏𝟏++𝑰𝟐𝑰𝟐++𝑰𝟑𝑰𝟑 𝑵𝑵 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 ++ ++ +⋯ + = ∑ 𝑹 𝑹𝑹𝟏 𝑹𝑹𝟐 𝑹𝑹𝟑 + ⋯ +𝑹𝑹𝑵 = 𝒏∑ 𝟏 𝟐 𝟑 𝑵 𝒏 =𝟏 𝑹𝒏 =𝟏 𝒏 Where: RT = total resistance  Sample Problems 1. The equivalent resistance of four resistors joined in parallel is 20 ohms. The currents flowing through them are 0.6, 0.3, 0.2 and 0.1 ampere. Find the value of each resistor. Solution: 𝟏 IT VT R1 R2 R3 R4 𝟐 I1 I2 I3 I4 𝟒  Sample Problems 2. Two bulbs of 250 W, 230 V each, are connected across a 200 V supply. Calculate the total power drawn from the supply if the bulbs are connected (a) in parallel and (b) in series. Solution: For each bulb, 𝑽𝟐 (𝟐𝟑0)𝟐 𝑷= 𝟐50= 𝑹𝑏ul𝑏 =𝟐𝟏𝟏.6 𝛀 𝑹 𝑹𝑏ul𝑏  (a) When the bulbs are connected in parallel, VT Rbulb Rbulb 200 V (b) When the bulbs are connected in series, Rbulb Rbulb VT 200 V  Sample 3. Problems Two resistances, one of 30 ohms and another of unknown value are connected in parallel, the total power dissipated in the circuit is 450 watts when the applied voltage is 90 volts. Find the value of the unknown resistance. Solution: 𝑽 𝟐𝑻 (𝟗0)𝟐 𝑷𝑻 = 𝟒50= 𝑹𝑻=𝟏𝟖 𝛀 𝑹𝑻 𝑹𝑻 VT 𝟏 𝟏 𝟏 30 R 𝟏 𝟏 𝟏 90 V = + = + 𝑹𝑻 𝟑0 𝑹 𝟏𝟖 𝟑0 𝑹 PT = 45 W 𝑹=𝟒5 𝛀  Sample Problems 4. Find the equivalent resistance of the circuit given in figure 1 between the following points (a) A and B (b) C and D (c) E and F (d) A and F and (e) A and C. Solution: (a) Between A and B 𝟏 𝑹 𝑨𝐵 = 𝟏 𝟏 + 𝑹 𝑨𝐵 = 5 𝛀 6 𝟏 5 (b) Between C and D 𝟏 𝟑 𝑹 𝐶𝐷 = 𝑹 𝐶𝐷 = 𝛀 𝟏 𝟏 𝟏 𝟐 + + 𝟑 6 6 (c) Between E and F 𝟏 𝑹 𝐸𝐹 = 𝟏 𝟏 𝑹 𝐸𝐹 = 𝟑 𝛀 + 𝟐 6 𝟐 (d) Between A and F By inspection, 𝑹𝑨𝐹= 𝑹 𝑨𝐵 𝑹 𝑨𝐹 = 5 𝛀 6 (e) Between A and C 𝟏 𝑹 𝑨𝐶 = 𝟏 𝟏 𝑹 𝑨𝐶 = 𝟒 𝛀 + 𝟑 𝟐 𝟏+𝟑  Sample Problems 5. What is the value of the unknown resistor R in figure 2 if the voltage drop across the 500-Ωresistor is 2.5 volts? Solution:  Sample Problems 6. A resistance R is connected in series with a parallel combination of two resistances 12 Ω and 8 Ω. Calculate R if the power dissipated in the circuit is 70 W when the applied voltage is 20 V across the circuit. Solution: R VT 20 V 12 18 PT = 70 W 𝑹 = 𝟑𝟐𝛀 = 0. 𝟗𝟏𝛀 𝟑5  Cost of Electrical Energy A utility company will charge its customers for the electric energy consumed. Energy Charge – is the cost of electrical energy consumed. Maximum Demand – is the user’s highest rate at which energy is consumed in kilowatts. Demand Charge – is the billing fee related to the maximum charge.  Sample Problems 1. A 60 W lamp remains lighted for 24 hours per day for 30 days. Determine (a) the electrical energy consumed over this period, (b) the energy charge for the billing period at a rate of Php 5.93/kWh Solution: (a) For the electrical energy consumed over this period  Sample Problems (b) For the energy charge for the billing period at a rate of Php 5.93/kWh  Sample Problems 2. A large residence consumes 1155 kWh of electrical energy over a billing period. Determine the total charge for billing period based on the rate schedule given below: Service Charge Php 261.18 First 1000 kWh of billing period Php 5.92 per kWh Next 2000 kWh of billing period Php 5.52 per kWh Solution: Load Calculation Methods AWG Wire Size Breaker Allowable Intended Unit ( mm² ) Capacity Load (Amps) (Amps) (@ 80 %) 3/0 80 200 160 Service Entrance 1/0 50 150 120 Service Entrance & feeder Wire 3 25 100 80 Service Entrance & Feeder Wire 6 14 60 48 Feeder & Large Appliance wire 8 8 40 32 Feeder & Large Appliance wire 10 5.5 30 24 Dryer Appliances and Air- conditioning 12 3.5 20 16 Laundry Appliances and Bathroom Circuit 14 2.0 15 12 General Lighting & Receptacle Circuit  Mounting Heights of wiring Devices Lighting Switches 1.40 m Convenience Outlets 0.30 m Panel Boards/Cabinets 1.80 m Kilowatt Hour Meter 1.20 m – 1.80 m