Lecture 4. Electrical Power & Energy

Transcription

1 ECE 211 Lectures Page 1 Lecture 4. Electrical Power & Energy Thursday, July 03, :00 PM Textbook Industrial Electricity 8th Edition, by Michael Brumbach, Text Book from Delmar/Cenage Learning Chapter 3 Electrical Power & Energy Reference Book : Electrical Power & Controls, 2nd Edition, 2004, by Timothy L. Skvarenina and William E. Dewitt, from Pearson/Prentice Hall, ISBN Learning Objectives Define the terms relating to electric power and energy Calculate the power and energy necessary to perform various electrical jobs Determine the horsepower necessary to drive various machines Calculate the efficiency of electrical equipment Calculate the cost of operating electrical equipment Work/Work done - may be defined as the overcoming of resistance through a distance Force (F = M* a) newton: N = m kg/s 2 Work = Force * Distance foot-pounds joule: J = N m = m 2 kg/s 2 1 N m = 1 Joule Power - the rate of doing work Power = Work/Time Watt: W = J/s Energy Cannot be destroyed or consumed However, it can be converted from one form to another It is the ability to do work Forms of Energies: Heat, light, mechanical, electrical, chemical Types of Energy Potential energy P = m*g*h Kinetic energy - energy stored in moving mass: Translational: W ke = 1/2 mv 2 m - mass, v - velocity Rotational W ke = 1/2 Iω 2 I - moment of inertia, ω - angular velocity Electric energy storage In Capacitor: W elec = 1/2 C*V 2 In Coil or Inductor: W mag = 1/2 L*I 2 Nuclear energy Energy, work, quantities of heat foot-pounds joule: J = N m = m 2 kg/s 2 1 N m = 1 Joule 1 Joule = foot-pound = kilogram-meter

2 ECE 211 Lectures Page 2 joule: J = N m = m 2 kg/s 2 1 N m = 1 Joule 1 Joule = foot-pound = kilogram-meter Energy Loss and Efficiency Efficiency % eff = Useful energy output / Total energy input x 100 % eff = Power output/power Input x 100 Example of Potential Energy Move the mass up the ramp to an altitude h, by traveling a distance l Friction between mass and the ramp The energy required to move the mass W = m*g*h = mg(l sinθ) = PE (Potential energy) Electrical energy kwh (kilo watt hour) 1 kwh = 3.6 Mega Joules Horsepower (hp) An average horse cloud do work at the rate of 33,000 foot-pounds (44,740 joules) Or, one average horse can move 1000 pounds (453.6 kg) through 33 ft ( m) in 1 minute hp = (ft lb/min)/33,000 1 hp = 746 watts Electric power Measured in watts (W), kw (kilo W), MW (Mega W), GW (Giga W), etc P = E*I P = E* (E/R) = E 2 /R P = (IR)*I = I 2 *R 1 watt per second = 1 joule One watt power is delivered when one "volt" of voltage forces one "ampere" through a circuit Energy and Units Heat energy 1 Btu (British thermal units): the amount of heat required to raise the temperature of 1 pound of water 1 degree 1 Calories (Cal): the amount of heat required to raise temperature of 1 gram (g) of water 1 degree Celsius Electrical energy Power x time kwh (kilo watt hour) Example 1. A machine can place 50,000 lb (22,680 kg) of scrap metal onto a truck 10 ft (3.048 m) high in 5 min. What horsepower is the machine capable of delivering? (3 hp) hp = (ft lb/min)/33,000

3 ECE 211 Lectures Page 3 Example: An elevator is required to lift a load of 500 kg to an altitude of 30 m. 1. How much energy must the motor provide (neglecting any losses)? 2. What horsepower is the machine if it required to lift from ground surface to 30 m height in 2 minute? (1) W = m*g*h = 500 kg* 9.81 N/kg* 30 m = 147,150 N m = 147,150 Joule Since 1 kwh = 3.6 Mega Joule The equivalent electric energy We = 147,150/ = kwh (2) Power = We/time = x 1000 * 60 [watt minute]/2 minute = 1230 watt 1 hp = 746 watt Motor's hp =1230/746 = 1.7 hp chose 2 hp Example 2. How much electrical power is required to operate a water heater that is rated at 240 V, if its resistance (R) is 24Ω? (2400 watt, or 2.4 kw) Example 3: How much power is dissipated by an electric oven if 30A flow through a resistance of 10Ω? (9000 watt, or 9 kw) Example : An extension cord with overload protection is connected to a 120V 15A max outlet. One 900 watt microwave oven is already plug-in to the extension cord Can we connect another 600 watt oven to this extension cord Example. Use the same water heater (P = E 2 /R = 2400W or 2.4 kw )discussed in Example 2. If the water heater is running for one day, how much electrical energy in kwh is consumed? (2.4 kw x 24 hrs = 57.6 kwh) Example 4. The customer, an industrial plant, expends 5000 kwh of electrical energy each working day. The plant is in operation 5 days a week. If the utility company charge $0.03 per kwh for the first 50,000 kwh, and $0.015 per kwh for all energy above 50,000 kwh, how much does the customer pay for a 10-week period? 5000 kwh per day 5 Days a week => 5000 x 5 = kwh 10 Week period => x 10 = kwh Cost First 50,000 kwh ($0.03 dollar per kwh) 50,000 kwh x $0.03/kWh The remaining = kwh ($0.015 kwh per kwh) Mechanical Transmission of Power and Drives The driven machines

4 ECE 211 Lectures Page 4 Mechanical Transmission of Power and Drives The driven machines Presses, lathes, elevators, pumps, saws The driving machines Engines, turbines, electric motor The connections between driving machines and driven machines Belts on pulleys Chain of sprockets Gear assemblies Direct drives Drives & Speed Requirements Belts, pulleys The speed of the driven machine differs from the speed of motor N1/N2 = D2/D1 N motor /N machine = D motor /D machine Chains on sprockets Gear assemblies & drive N1/N2 = T2/T1 N1 = speed of the driving gear (rpm) N2 = speed of driven gear T1 = number of teeth in the driving gear T2 = number of teeth in the driven gear Direct Drives Machine & Motor may be mounted on the same shaft Example: M-G set (Motor-Generator) Shaft coupling Example 5: The nameplate of a motor indicates that it rotates at a speed of 1700 rpm. The diameter of the pulley on the motor shaft is 6 inch (15.24 cm). If the machine must be driven at a speed of 850 rpm, what size pulley must be used on the machine? N1 = 1700 rpm, N2 = 850 rpm, D1 = 6 inch, find D2? D2 = 12 inches

5 ECE 211 Lectures Page 5 Example 6: A motor operates at a speed of 3250 rpm. The machine it is driving requires a speed of 650 rpm. If the machine pulley is 20 inch (50.8 cm), what size pulley must be installed on the motor? N1 = 3250 rpm, N2 = 650 rpm, D2 = 20 inch, find D1? D1 = 4 inches Example 7: A motor operates at a speed of 500 rpm. The pulley attached to its shaft has a diameter of 20 inch (50.8 cm). The machine that the motor is driving has a pulley that is 8 inch (20.32 cm) in diameter. At what speed will the machine operate? N1 = 500 rpm, D1 = 20 in, D2 = 8 in, find N2? N2 = 1250 rpm Example 8: The driving gear on a motor has 60 teeth and the driven gear has 90 teeth. If the driven gear revolves at 100 rpm, what is the speed of the driving gear? N1 =?, N2 = 100 rpm, T1 = 60, T2 = 90. N1 = 150 rpm Example 9: The driving gear on a motor revolves at a speed of 1150 rpm and has 25 teeth. The speed of the machine must be 115 rpm. How many teeth must there be on the driven gear? N1 = 1150 rpm, N2 = 115 rpm, T1 = 25 teeth, T2 =? T2 = 250 teeth Example 10: Drive gear A on a motor has 20 teeth and revolves at a speed of 2250 rpm. Gear A drives a second gear, gear B, which as 30 teeth and meshes with gear C, which has 40 teeth. What is the speed of gear C? Find N C 1125 rpm, from N A /N C = T C /T A Find N B = 1500 rpm, from N A /N B = T B /T A Find N C = 1125 rpm, N B /N C = T C /T B Direction of Rotation

6 ECE 211 Lectures Page 6 Torque & Horsepower hp = T*N/5252 T = torque, in pound-feet (lb ft) N =speed in rpm Example 11: How much toque is required to drive a machine rated at 10 hp? The speed of the machine is 50 rpm. ( lb ft) Motor: Starting Torque - the amount of torque developed at the instant the motor is energized Type of load Max torque required to produce motion Starting Current The value of current required by the motor from the instant it is energized until it reaches its rated current Other Factors in Selecting Drives Safety requirements Space requirements Size and type of shafts Horizontal or vertical shaft Distance from the motor to the machine Sizing Motors hp = (W*h)/(33000 * Eff) W = weight lifted in pounds (lb) h = height lifted, in feet per minute (ft/min) Eff = efficiency Example 12: What size motor is required to drive a hoist if it must lift 100 lb (45.6 kg) to a height of 30 ft (9.144 m) in 1 minute? The hoist is 90% efficient. hp = (W*h)/(33000 * Eff) = (1000 * 30)/(33000*0.9) = 30,000/29,700 1 hp