IV. POWER. P = W t. = E t. where W is the work done, t is the time required to do the work, and E is the energy used. 1 horsepower = 1 hp = 550

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1 IV. POWER A. INTRODUCTION Power is the rate at which work is done. Work involves a force or force-like quantity acting on object and causing its displacement in the direction of the force. The time required to do the work does not affect the amount of work done. Lifting a 100 lb weight 10 ft requires the same energy, 1000 ft lb, whether the work is accomplished is 30 seconds or 30 years. Power (P) is a measure of how fast work can be accomplished and allows us to compare machines accordingly. The defining equation for power is: P = W t = E t IV-1 where W is the work done, t is the time required to do the work, and E is the energy used. Power was studied and defined during the Industrial Revolution in England. James Watt, who invented the steam engine, wanted to compare the efficiency of his invention to the most common engine at that time, the donkey. Watt calculated that a typical donkey could perform 550 ft lb of work per second. He defined the unit of power known as the horsepower (hp) as being 550 ft lb/s. 1 horsepower = 1 hp = 550 ft lb s IV-2 In the SI system, power is calculated in units of N m or J, which are also referred to as watts s s (W). When the system of electrical units were established, the units were named for those who had first worked on electricity. Alessandro Volta invented the battery and the volt was named after him. Andre Marie Ampere worked with current and George Simon Ohm studied resistance. The unit of power, the watt, was named after James Watt. Power ratings of machines are given in both watts and horsepower. 1 watt = 1 W = 1 N m s = 1 J s IV-3 1 hp = 746 W = 550 ft lb s IV-4 These values are also listed in the conversion tables in the back of the text. When we measure the power of a machine we can either use the work it does per unit time or measure the energy it uses per unit time. The equations used for calculating power in each system are shown in Table IV

2 ENERGY SYSTEM Mechanical (Translational) Mechanical (Rotational) Fluid Electrical Thermal EQUATION FOR POWER P = W = F d t t = F v P = W τ θ = t t = τ ω Δ P = P V t = ΔP Q V P = V q t = V I P = Q = H κaδt H = t l Table IV-1. Equations for Power. Notice that power is either equal to work or energy divided by time, or power is equal to a force or force-like quantity times a rate. B. POWER IN MECHANICAL SYSTEMS In linear mechanical systems, objects move along straight lines. Work is done when a force pushes or pulls or lifts an object over a distance or height. Power is a convenient way to measure how fast a machine can accomplish its intended work. Fig. IV-1 shows four machines that do work in linear systems. Each will do its work at a certain rate. The maximum rate at which each of these machines can do work is the power rating of that machine. From the equation for linear power, Figure IV-1. Power in linear mechanical systems. P = W t = F d t = F v IV-5 135

3 we can see that it can be determined from the amount of work accomplished per unit time or by multiplying the applied force by the speed or rate of movement (since v = d/ t). In rotational mechanical systems torque causes objects to rotate. Work is done when an applied torque causes rotational movement of an object. Fig. IV-2 shows a few examples of rotational systems. a) windmill b) pipe cutting tool c) another pipe cutting tool d) grinder From the equation for rotational power, Figure IV-2. Power in rotational mechanical systems. P = W t = τ θ t = τ ω IV-6 we can see that power can be determined from the amount of rotational work done per unit time or by multiplying the applied torque by the rotational speed (since ω = θ /t). 136

4 1. Efficiency The efficiency (η) of a machine or process can be calculated by dividing the work output by the work input. It also can be calculated by dividing the output power by the input power. Efficiency gives us a measure of how effective a machine or process is. Efficiency = Work Out Work In = Power Out Power In η = W W out in = P P out in IV-7 We often calculate the percent efficiency, which often uses the same variable as the efficiency (η), although we will often write it as %η : W % = W 100% = P out out η 100% IV-8 P in in Example IV-1. Russian weightlifter Vassily Alexeev once lifted 260 kg to a height of 2 meters in 2 seconds. How much horsepower did he generate? Solution: m ( 260 kg) 9.8 ( 2m) 2 2 W F d mgh s kg m J P = = = = = 2550 = 2550 = 2550W 3 t t t 2 s s s 1hp and P = 2550 W = 746W Vassily was a pretty strong guy 3.42hp Example IV-2. The world record for climbing the stairs of the Empire State Building (2004) is 602 seconds (about 10 minutes!). The person who ran the 1000 ft vertical climb weighed 125 lb. What was the average power output during the climb, in horsepower? W P = t F d = t wh = t = ( 125 lb)( 1000 ft) 602 s ft lb = 208 s and ft lb 1hp P = 208 = s ft lb 550 s 0.378hp 137

5 Example IV-3. An automobile travels at a speed of 50 mph when the engine provides a forward thrust of 300 lb. If the engine is 20% efficient, what horsepower does it provide? Solution: P = F v = (300 lb) 50 miles out hr 5280 ft 1 mile 1 hr 3600 s = 2.2 x 10 4 ft lb s = 2.2 x 10 ft lb s 4 1 hp ft lb 550 s = 40 hp From Eq. IV-8, P = in Pout 40 hp 100% = % η 20% ( ) 100% = 200 hp Example IV-4. A winch and a block and tackle system are used to lift a 980 N weight a distance of 2 meters in 2 seconds. The winch pulls the cable 8 meters with a force of 272 N in the same time. Find a) the input power in watts b) the output power in watts c) the efficiency of the block and tackle Solution: a) P = in b) P = out Fin Din t Fout D t out = ( ) ( ) 272 N 8 m = 1088 N m 2 s s ( ) ( ) = 980 N 2 m = 980 W 2 s = 1088 W P c) % = 100% = 980 W out η 100% = 90% P 1088 W in Example IV-5 shows how to find the shaft power produced by a torque in an electric motor. Note that rotational speed, ω, must be in radians/sec Example IV-5. The shaft of an electric motor develops a torque of 0.73 ft lb when it rotates at 1800 rpm. Find the horsepower developed by the shaft. = 0.73 ft lb 1800 rev min Solution: P = ( ) τ ω 1 min 60 s 2 π rad rev = 137 ft lb s 138

6 ft lb 1 hp = 137 s ft lb 550 s = 0.25 hp A flywheel is a device used to store rotational kinetic energy. This stored energy can be used to smooth out the power delivered by the engine cylinders of a car or by a metal press to cut or shape metal forms. Example IV-6 shows a piston in a compressor, being driven by a flywheel, which is powered by an electric motor. Example IV-6. A belt-driven flywheel powers the pistons of a compressor. The flywheel rotates at 1100 rpm (115 rad/s) and delivers a torque of 65 N m. Find the hp of the flywheel. rad = s N m = s 1 hp 746 W Solution: P ( 65 N m) ( 7475 W) = 10 hp = τ ω = 139

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13 Problem Set 7: Mechanical Power 1. A 3500 lb elevator rises 30 ft at constant speed in 10 seconds. Complete parts a-f. a. Find the work done by the cable that lifts the elevator. b. Find the upward speed of the elevator c. Find the elevator power in ft lb/s using P = W/t d. Find the elevator power in ft lb/s using P = F v e. Compare answers c. and d. Are they equal? Should they be? f. Convert the elevator power to horsepower. 2. A force of 25 N is needed to keep a piston moving at 0.55 m/s. a. How much power in N m/s is used to moved the piston? b. How much power in watts is used to move the piston? c. How much power in J/s is used to moved the piston? d. How much work is done on the piston in 2 seconds? 3. A force of 250 lb is applied to push a wagon a distance of 200 ft in 20 minutes. Determine the horsepower expended. 146

14 4. A 1200 kg elevator lifts a load with a constant velocity of 3.5 m/s when the elevator motor produces 50,000 watts of power. What is the mass of the load? 5. Farmers use many types of agricultural equipment. Many involve a device called a draw bar, which is used to connect attachments, such as plows and wagons, to the tractor. In one case a tractor exerts 6000 lb of pulling force on the draw bar while traveling at 2 mph. Find a) the amount of horsepower delivered to the draw bar at 2 mph, b) the amount of horsepower delivered at 5 mph. 6. An electric motor has a shaft torque of 0.5 ft lb and a rotational speed of 800 rpm (83.7 rad/s). What is the shaft horsepower? 147

15 7. A planer in a mill must have its cutting head rotate at 418 rad/s (4000 rpm). It is powered by a 40 hp electric motor which applies a torque of 70 N m to the cutting head. A drive belt connects the motor to a pulley on the planer. Find a) the power produced by the cutting head in watts, b) the efficiency of the power drive belt assembly. 148

16 Worksheet for Problem Set 7 149

17 Worksheet for Problem Set 7 150

18 LEARNING OBJECTIVES FOR LABORATORY EXERCISES 38 AND 39 The learning objectives for Labs 38 and 39 are: 1. Describe what power means and how it relates to energy in the mechanical, fluid, electrical and thermal systems. 2. Explain how thermal rate and thermal power are the same quantity. 3. Understand the unifying equation for power applies the principle of work or energy divided by time or a force (or force-like quantity) times a rate for mechanical, fluid and electrical systems. 4. Solve mechanical and fluid power problems using both the SI and English unit systems. 5. Define efficiency in terms of power. 6. Measure and determine power for mechanical and fluid systems. 7. Identify workplace applications of mechanical and fluid systems. 151

19 LAB 38 OVERVIEW: LINEAR MECHANICAL POWER Power is the rate of doing work or using energy. It is a measure of the amount of work done in a certain unit of time. The defining equation is P = W/t where W is the work done in joules, and t is the time required to do the work in seconds. The unit of power will then be a watt (W). A watt is one joule per second. Since work is the product of force and distance, the power equation may be mathematically rewritten as: P = F d = F d. t t Since the expression (d/t) is the velocity of the moving force, a convenient form of the power equation is P = F v Mechanical work is done when an object is moved. This motion can be caused by force or torque. When a force is applied, linear motion is produced. When torque is applied, angular motion is produced and the object rotates. When a mechanical system does work or uses energy, some energy is lost to heat and friction. As a result, the total output power that a system produces is actually less than the input power used. The efficiency of the system can be calculated using the equation Pout % η = P in 100% In this experiment, you will build a system that develops linear mechanical power. Then you will put the system to work and find the power it produces. You will also measure the total power input to the system, and find its efficiency. 152

20 LAB 38: LINEAR POWER DATE OBJECTIVES: SKETCH: TABLE 1 DATA Lifting height: D = cm = m Trial Hanging Mass m (grams) Lifting Force (weight) F (N) Time to Raise Load t (sec) Motor Voltage V (volts) Motor Current I (amps) 1 2 TABLE 2 CALCULATIONS Trial Mechanical Power Used P mech (watts) Electrical Power Used P elec (watts) Efficiency η (%)

21 LAB 38 ANALYSIS 1. What were the sources of energy loss which caused the motor and pulley system to have a low efficiency? Be specific. 2. Show how the units of work divided by time are equal to the units of force times velocity for the SI system and the British system: USE THE 5-STEP METHOD TO SOLVE THE FOLLOWING PROBLEMS: 3. A motor-pulley system has an efficiency of 41%. As a load is lifted, it rises 3.20 meters in 6.50 seconds. The input voltage is 110 volts, and the motor current is 3.20 amperes. What is the mass of the load, in kg? 4. A motor has an input voltage of 110 Volts, and draws a current of 2.80 amperes. A) If the efficiency of the system is 25%, what is the output power in horsepower? B) How long will it take for a weight of 50.0 lb to be lifted through a height of 10.0 ft by this system? 154

22 C. POWER IN FLUID SYSTEMS Fluid power is used for numerous applications such as jack hammers, hydraulic cylinders and lifts, to operate aircraft landing gear and to move control surfaces on ships and aircraft. Fluid power is equal to fluid work divided by time or alternatively pressure times fluid volume flow rate: P = W t = ΔP V t = ΔP V = ΔP Q t V IV-9 In Example IV-7 we calculate the fluid power of a pump filling an automobile gas tank. Example IV-7. A gasoline pump moves gasoline from the underground tank to a car's gasoline tank. The pump raises 25 gallons of gasoline to a height of 15 feet in 5 minutes. The weight density of gasoline is 42 lb/ft 3 and 7.48 gallons equals 1 ft 3. a) Find the pressure difference b) Find the volume flow rate c) Find the pump power. h fuel level Solution: h = 42 lb ft a) ΔP = ρ ( ) 15 ft = 630 lb ft w 3 2 b) Q = V V t = 25 gal 5 min = 5 gal 1 ft min 7.48 gal 1 min 60 s 3 3 = ft s c) P = P Q = 630 lb ft 3 Δ V 2 ft s = 6.93 ft lb 1 hp s ft lb 550 s = hp In Example IV-8 we find the fluid power input in a hydraulic cylinder. The mechanical output power is also determined. 155

23 Example IV-8. A hydraulic cylinder has an area of 10 cm 2 and delivers a force of 1000 N while moving a load 10 cm in 1 second. a) Find the mechanical output power. b) Find the fluid power to the cylinder. A Solution: a) P = F v = ( 1000 N ) mech b) P = ΔP Q fluid V 10 cm 1 s 1 m 100 cm = 100 W ΔP = F A = 1000 N 10 cm cm m 4 2 = 10 N 6 m 2 Q V A d = = = t t 2 ( 10 cm )( 10 cm) 1 s 3 1 m 6 10 cm V = m s 3 P = P Q = 10 N 6 fluid Δ V m 2 10 m -4 s 3 = 100 W 156

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30 Problem Set 8: Fluid Power 1. A laser cooling system needs a pump that develops a pressure difference of 35 psi and a flow rate (Q v ) of 1.5 ft 3 /min. a) Find the fluid power developed in ft lb/s. b) Find the fluid power developed in hp. 2. A 1 hp pump produces a pressure difference of 10 5 N/m 2. What is the flow rate (Q v ) in m 3 /s developed by the pump? (Note: 1 hp = 746W) 3. A hydraulic cylinder produces a power of 7500 watts with an output force of 2000 N. What is the speed of the piston in m/s? 4. The same hydraulic cylinder as in problem #3 has an area of 40cm 2. a) What is the pressure difference across the piston face? b) What is the fluid flow rate into the cylinder in m 3 /s? 163

31 5. A pump is used to move gasoline from an underground storage tank to a tank on the surface. The pump raises 230 gallons to a height of 21 feet in 12 minutes. The weight density of gasoline is 42 lb/ft 3. Find the pump power. 6. A hydraulic cylinder has an area of 10 cm 2. It delivers a force of 1000 N and moves a load a distance of 10 cm in 1 second. Find the mechanical power output and the fluid power to the cylinder. 7. An irrigation system is powered by a submersible pump in a well 100 ft deep. The system delivers 50 gallons a minute. The output pressure of the pump is 12,960 lb/ft 2. a) Find the output power of the pump. b) Find the power to raise the water to ground level. c) Find the power left over to distribute water over the ground. 8. A 5 hp (output) water pump has an electrical input power of 5 kw. It is used to fill a 2500 ft 3 tank and must raise the water through a height of 46 ft. Determine the efficiency of the pump and the time required to fill the tank. The weight density of water is 62.4 lb/ft

32 9. Water falling from the height of 30 meters at a rate of 40 m 3 /sec is used to turn a flywheel. If 80 percent of the water s power is delivered to the flywheel and the torque on the flywheel is 20,000 N m, what is the constant angular velocity of the flywheel? 10. How high above a reservoir should a 2000-m 3 tank to be placed if a motor that delivers 20,000 watts is to fill the tank is 40 minutes? 165

33 Worksheet for Problem Set 8 166

34 Worksheet for Problem Set 8 167

35 LAB 39 OVERVIEW: FLUID POWER IN HYDRAULIC SYSTEMS The rate at which work is done is called power (P). In a fluid system, a pressure difference (ΔP) causes a certain volume of fluid (V) to move from one point to another. If you measure the time (t) it takes to do this, you can find fluid power using the following equation. Δ P = P V t This lab has two parts. In the first part you will measure the power a pump uses to move water from one place to another. The water stays at the same level, but is pumped at different flow rates. The pump is required to work harder to move the water faster. This means that a pump pumping at 12 gal/min requires more power than a similar pump pumping at 6 gal/min. You will show this by observing the current and voltage the pump needs at two different pumping rates. In the second part you will measure the power the pump uses to lift water from one level to another level. You will see that the pump has to work harder to lift water a greater distance. 168

36 LAB 39: FLUID POWER DATE OBJECTIVES: SKETCH: DATA TABLE 1 Trial # Voltage V elec (volts) Current I (amps) Electrical Power P elec (Watts) DATA TABLE 2 Trial # inches of Hg Pressure Difference ΔP P 1 P 2 psi psi ΔP psi ΔP lb/ft 2 Elapsed Time t (sec) Fluid Power P fluid (ft lb/s) Volume of Fluid Moved: V fluid = ft 3 169

37 LAB 39 ANALYSIS 1. What effect on the fluid rate did the increase in electrical power input have? (Trials 1 & 2) 2. Was there any effect on the fluid flow rate when the water was pumped to a higher level between trials 1 and 3? 3. What relationship did you find between increased fluid flow rate and fluid power? USE THE 5-STEP METHOD TO SOLVE THE FOLLOWING PROBLEMS: 4. Water is pumped to a height 40 ft above a reservoir to fill a 3500 ft 3 storage tank. A) How much output power must be delivered if the tank is to be filled in 25 minutes? The weight density of water is 62.4 lb/ft 3. B) If the pump system is 40% efficient, how much pressure must the pump provide? 5. Water is pumped through a pipe system that has a fluid resistance of 0.82 psi/gpm. What is the fluid power when the system delivers 200 gallons in a time of 5.0 minutes? 170

38 D. POWER IN ELECTRICAL SYSTEMS Electric power is integral to almost every part of our day-to-day lives. Without electrical power in our homes we lose light, heat, air conditioning, telephones, electrical appliances, etc. It is so essential that in many places, such as hospitals, they provide for emergency power using gas or diesel powered electrical generators. Power in electrical systems follow the unifying equations for power. Power = Work time or Power = Force - like Quantity x Rate Δ P = V q t IV-10 where ΔV is the voltage difference, q is the charge moved and t is the time in which the electrical work is done. Since the current, I = q/t, P = ΔV I IV-11 This latter equation, P = ΔV I (sometimes given as just V I), is most frequently used in power calculations. Remembering Ohm's Law: ΔV = IR, we can derive two additional equations for electrical power: P = I 2 R IV-12 P = ( Δ V) R 2 IV-13 Both of these forms are useful when the resistance is known and either the current or the voltage is also known. 1. Units of Electric Power The basic unit of electrical power is the watt. 1 watt = 1 volt ampere: 1 W = V A 1 watt = 1 ampere squared ohm: 1 W = 1 A 2 Ω 1 watt = 1 volt squared/ohm: 1 W = 1 V 2 Ω 171

39 Table IV-2 shows the relationships between power units. This table can be used to convert from one type of power unit to any other type of power unit. Power watt or J/s ft lb/s hp Btu/hr cal/s 1 watt = x ft lb/s = x hp = Btu/hr = x cal/s x Table IV-2. Power conversion table. Example IV-9. An electric motor produces an output shaft power of 0.25 hp. It operates at 24 V and 9.47 A. Find a) the input power and b) the efficiency. Solution: a) P = V I = ( 24 V ) ( 9.47 A ) in Δ = 227 W P = 0.25 hp 746 W out hp b) ( ) = W c) % η = P P out in 100% = W 100% = 82% 227 W Example IV-10. A 110 V electric motor has an output power of 1 hp and it is 88% efficient. Find the input current. Solution: P = in P out η 100% = 1 hp 746 W hp ( ) 88% 100% = 848 W P = ΔV I I = in PV = 848 W in Δ 110 V = 7.71 A Remember that W V = J s J C = C s = A. 172

40 Example IV-11. A 5 Ω resistor is in a circuit with a 10 V battery. Find the power delivered to the resistor. Solution: P = ( ) ( ) Δ V R 2 2 = 10 V 5 Ω = 20 W 2. The Kilowatt-Hour Meter The kilowatt-hour meter on the side of your house is used by Puget Sound Energy to measure the electrical energy consumption in your house. The bill you pay is for kilowatt-hours (kwh) consumed. One kilowatt-hour is the amount of energy used by a 1000 watt device operated for one hour. In units of joules, 1000 W kw J 1 s 1 W 3600 s hr 6 ( 1kWh) = 3.6 x 10 J = 3.6 million joules A watt is a unit of power. A watt-hour is a unit of power times a unit of time and thus represents energy (E = P t). The kilowatt-hour meter measures energy consumed. The meter is a little electric motor whose speed of rotation depends on the voltage and current in the circuit that it is measuring. As you use electrical energy, the motor spins, turning a horizontal disc and a system of gears. These gears cause the pointers to rotate and point to the correct number on the meter register. In the kilowatt-hour meter lab you will learn how to read and use the meter. 173

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47 Problem Set 9: Electrical Power 1. An electric toaster operates at voltage of 110 V and current of 9.09 amperes. How much power does it use? 2. A 100 watt light bulb operates at 110 V. How much current flows through the bulb? 3. What is the maximum current that should flow through a resistor rated at 50 ohms and 100 watts? 4. What is the maximum voltage drop that should be applied across a resistor rated at 100 ohms and 2 watts? 5. If a 5 hp motor operates for 20 hours, how much electrical energy (in kwh) is consumed? 180

48 6. A winch and a block are used to lift a crate weighing 980 N through a distance of 2 meters in 2 seconds. The winch pulls a cable through a distance of 8 meters with a force of 272 N in the same time. a) Find the input power in watts supplied to the block and tackle by the winch. b) Find the output power in watts of the block and tackle used in lifting the crate. c) Find the efficiency of the system. 7. An electric motor develops a shaft torque of 0.73 ft lb when the shafts rotates at 1800 rpm. Find the horsepower developed by the shaft. 8. An electric motor produces an output shaft power of 0.25 hp while operating at 24 V and 9.47 A. a) Find the input power b) Find the efficiency 181

49 9. A 110 V electric motor has an output power of 1hp and is 88% efficient. Find the current required to run the motor. 182

50 Worksheet for Problem Set 9 183

51 Worksheet for Problem Set 9 184

52 LEARNING OBJECTIVES FOR LABORATORY EXERCISES 41 AND 42 The learning objectives for Labs are: 1. Show that power in electrical systems obeys the unifying principle of work divided by time or a force-like-quantity times a rate. 2. Describe and be able to use units for electrical power in both the SI and English unit systems. 3. Describe a watt-hour meter and how it operates. 4. Understand the difference between a series shunt motor and shunt wound motor and be able to explain when you would use each type of motor. 5. Be able to understand and work problems of power in electrical systems. 185

53 LAB 41 OVERVIEW: THE WATT-HOUR METER All electrical measurements are made in SI units. Energy is measured in Joules, and power is measured in Watts. The relationship between voltage, current, and power is shown by the equation: P = V I where P is the electrical power in watts, V is the voltage difference in volts and I is the current in amperes. Since power is defined as the rate at which energy is used, the energy used in a system can be determined using the following equation: Energy Used = Power x Time One watt of power acting for one second uses one joule of energy. This is too small a unit for many practical applications. The amount of energy used by an appliance can be measured in watts times hours, or watt-hours. Your electric bill is figured by how much electrical energy in watt-hours you use each month. Watt-hour meters often measure energy in units of kilowatt-hours. These units are convenient for everyday use. Remember that a kilowatt-hour is a unit of energy. It is equal to 1000 watt-hours or 3,600,000 joules. 186

54 LAB 41: THE WATT-HOUR METER OBJECTIVES: SKETCH: DATE DATA TABLE "kh"number (watt-hours per revolution) Volt Meter Reading V = Divisions between numbers on disc Device Number of Revolutions Voltmeter Reading Energy Used (kwh) Hot Plate Television 187

55 ANALYSIS: THE WATT-HOUR METER 1. Did the line voltage remain the same for each test you ran? If not, describe how it varied and why you think it changed. 2. Explain why "kilowatt-hours" are energy units, even though kilowatt is a power unit. 3. How did the energy used by the appliances that produce noticeable heat compare with the other appliances for the same one-minute interval? 4. Using the information above, list the common home appliances that have the greatest energy consumption. 5. If you were gone on a two-week vacation, how could you save energy on an electric water heater without turning it off? Use the 5-step method to solve the following problem. 6. A 100 watt light bulb is left on overnight for a time of 8 hours. How much energy does it use in kilowatt-hours? in joules? 188

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57 LAB 42 OVERVIEW: MOTOR GENERATOR Electric Motors and generators play an important part in our industrial society. We find many types of motors and generators used in homes, schools, factories, and offices. Generators are most often used in power plants. Motors are more common, being used in many electrical tools and appliances. The main parts of generators and motors are very similar to one another. They differ only in how they are used. Electric motors and generators are devices that convert power from one form to another. Generators convert rotating mechanical power to electrical power. Motors convert electrical power to rotating mechanical power. The power input (P IN ) to a motor or generator is the product of the voltage and current input. The power output (P OUT ) of a motor or generator is the product of the voltage and current output. This is shown in the equation P = V I. Recall that linear power can be expressed as the product of force and velocity. Rotational power may also be expressed in the convenient form P = τ ω. The efficiency of a motor or generator is equal to the ratio of output power to input power. This is given in the equation below: Efficiency = = P out η x 100% P in In this experiment you will study the hookup and operation of DC motors and generators. For information and not part of this experiment there are two basic types of DC motors and generators, series-wound and shunt-wound. Each of these motors has a special purpose or use. The series-wound motor has its field windings connected in series with the armature windings. The field windings in a series-wound motor are a few turns of large diameter wire. This causes the motor to have high torque output. Speed regulation is not good, but serieswound motors can handle heavy loads. The shunt-wound motor has its field windings connected in parallel with the armature windings. The field windings in a shunt-wound motor are many turns of small diameter wire. This motor has low torque output, but very good speed regulation. It is used where constant speeds are important. It runs at a higher speed than a series-wound motor for a given source voltage. 190

58 LAB 42: EFFICIENCY OF MOTORS AND GENERATORS DATE OBJECTIVES: SKETCH: DATA TABLE 1 MOTOR Input Voltage (V) MOTOR Input Current (A) GENERATOR Output Voltage (V) GENERATOR Output Current (A) 3 VDC 4 VDC 6 VDC DATA TABLE 2 Motor Voltage V in (VDC) Input Power P in (Watts) Output Power P out (Watts) Efficiency % SAMPLE CALCULATIONS: 191

59 LAB 42 ANALYSIS 1. How are motors and generators the same? How are they different? 2. Explain the function of the motor in this experiment. 3. Explain the function of the generator in the experiment. 4. In a hydroelectric dam system, what provides the same function as the motor did in this experiment? Be specific. 5. What caused the low efficiency for this experiment? 6. How would you design a more efficient system? Use the 5-step method to solve the following problem. 7. A DC electric motor has an input of 12 A at 24 V. It rotates at 1200 rpm and produces a force of 1.5 lb on a de Prony brake with a lever arm of 9.0 inches. Find the efficiency of the motor. 192

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