Principle of Engineering Electrical energy and power

1. Principle of EngineeringElectrical energy and power Electricity Session 3 (2 hours)

2. Energy 能量 • Energy is required to: • Heat up an object (thermal energy) • Move an object (mechanical energy) • Light up a lamp (electrical energy) • Unit: Joule or J • The fundamental unit of energy (as well as heat and work) in the MKS system of units, equal to 1 kg m2 s-2, or 1/4.184 of a calorie. • Energy conversion: • Electrical  Thermal (class example: … ) • Electrical  Mechanical (class example: … ) • Mechanical  Thermal (class example: … )

3. Thermal Mechanical Energy exchange Electrical Sound, light, ….

4. Joule,James(1818-1889) • Joule, James (1818-1889), English physicist who was initially interested in the efficiency of electric motors. He discovered the heat dissipated by a resistor is given by : Heat Energy = I2 R t where I is the current, R the resistance, and t the time Joule’s Law

5. Joule,James(1818-1889) • James Joule discovered heat dissipate in resistor with current flow (Joule’s Law: Heat Energy = I2Rt) • James Joule also discovered the mechanical equivalent of heat in the 1840s in which he found the friction of dropping 772 lbs weight through 1 foot would cause 1 lb water to rise 1oF

6. Law of Energy Conservation • governs all energy-exchange phenomena of nature • Input = Output + Losses • Energy may be “consumed” or “stored” Losses Output Input System

7. Power (電)功率 • Power = rate of energy usage • Energy = how much • Unit: Watt (W) • 100W bulb – uses 100 J/s • 40W bulb – uses 40 J/s Or 1 J = 4.184 calorie for the health consciousYOU!

8. Power Formulae • Power = Rate of Energy Usage = Energy Used / Time • Compare to: • Speed = rate of distance traveled = distance / time Power = Energy / Time or Energy = Power * Time

9. Turning On a 100W Bulb for One Hour • Energy = ? • Energy = 100 W * 3600 s = 360,000 J • Cost = ? • CLP charges by KW-hr: $0.90 per KWhr • The bulb uses 100W * 1 hr = 100 Whr or 0.1 KWhr. • It costs $0.09

10. Electrical Power • Energy is the“Work Done”to movecharges(Q)across apotential difference (V) • Energy (E) = Q V • It takes 1 J to move 1 Coulomb (C) of electrical charge across a potential difference of 1V • Power (P) = Rate of Energy = rate of flow of charges * potential difference = I*V 1 e- = 1.6 x 10-19 Coulomb

11. Electrical Power Dissipation in a Resistor P = VI V = IR I = V/R P = I2 R P = V2 / R Caution: V is the voltage across the resistor in consideration, not the supplied voltage!

12. Power Examples • What is the Power rating of a Hair Dryer for use in Hong Kong that takes 5 A? • P = VI = = 1100 W • What is the Ohmic value of the heating element in the hair dryer? • R = = 44 Ohm • What is Power (heat) consumed in passing a 5A current through a 0.5 Ohm wire? • P = = 12.5 W • Does it matter on the magnitude of voltage so long there is 5A flowing?

13. Switches • Single Pole Single Throw (SPST)單刀單擲 • Single Pole Double Throw (SPDT)單刀雙擲 • Double Pole Double Throw (DPDT)雙刀雙擲 • Single Pole – controls one wire • Single throw – makes one connection (throw) • Double throw – makes two connections • Double Pole – controls two wires

14. SPST Switch • For simple on-off control.

15. SPDT Switch • Single Pole (Connect 1 wire) • Double Throw (to 2 connections) • For 3-way light circuit

16. DPDT Switch • 2 SPDT switches with handles attached • For polarity reversal

17. 3-Way Light Circuit Implement this circuit.

18. Reverse Polarity • Complete the circuit and implement it. motor

19. Relay • Switch is activated electrically by solenoid. • Use a small current to control a large current (IC controls relay and relay controls power consuming devices) • Use DC to control AC • SPST

20. Relay (DPDT) • Use electromagnet to control DPDT switch • Need 6+2 = 8 connectors • Demonstrations/ applications

21. Bread Board • For easy connection of components • Each hole is connected to others in a row via metal strips underneath