II. ENERGY A. FLUID ENERGY SYSTEMS 1. POTENTIAL ENERGY IN FLUID SYSTEMS The same equations used to find potential energy in mechanical systems can be used in fluid systems. When a pump lifts water from a lake to a tank, the work done increases the potential energy of the raised water. E p = m g h II-1 where m is the mass of the water, g is the acceleration of gravity and h is the height to which the water is raised. In English units, where we generally deal with weight instead of mass, it might be preferable to re-write the equation as E p = w h II- where w is the weight of the water (=mg) and h is the height the water is lifted. If the volume is known, the mass (or weight) moved can be found by using the mass density (or weight density). Example II-1: 100 m 3 of water is pumped from a source to a tank 5 meters above the source. Find the potential energy of the water pumped and the work done on the water. h V = 100 m 3 h = 5 m E p =? V = 1000 kg 3 m Solution: Mass: m = ρ ( ) E = mgh = 10 kg 9.8 m p s 5 ( ) ( ) 3 5 100 m = 10 kg 7 5 m =.45 x 10 J Note that the work done on the water is equal to the potential energy the water gained. 1
Example II-: A pump rated at 3 hp pumps 8 ft 3 of water per minute from a lake to a storage tank 75 ft above the lake (use the same picture as in Example II-1). Find the potential energy stored in the tank after ten minutes of pumping. Q V = 8 ft 3 /min. h = 75 ft t = 10 min. E p =? t = 8ft 3 min Solution: Volume pumped: V = Q ( ) V V = 6.4 lb ft 10 min = 80 ft 3 Weight pumped: w = ρ ( ) w 3 Potential Energy: ( ) ( ) 80 ft = 499 lb 5 E = w h = 499 lb 75 ft = 3.74 x 10 ft lb p or we could use the fluid equation, E p =ΔP V: lb 3 5 ΔP V = ρ wh V = 6.4 ( 75 ft)( 80 ft ) = 3.74 x 10 ft lb Ep = 3 ft 3. KINETIC ENERGY OF A FLUID When a fluid is pushed by a piston and its speed increases, it gains kinetic energy, or energy of motion. The kinetic energy of a fluid at any given speed is given by 1 E = mv K (SI units) II-3a and 1 w E = g v K (English units) II-3b In fluid terms, we can also write this as and E E k k 1 = ρv V (SI units) II-4a 1 ρw = v V (English units) II-4b g The following examples show this calculation in each system
Example II-3: Air strikes a set of turbine blades, causing the turbine to rotate. Each second, 3,000 lb of air strikes the blades at 0 ft/sec. Find the kinetic energy of the fluid striking the blades each second. w = 3000 lb v = 0 ft/s E k =? Solution: E k 1 w = v g = 1 0 ft 3 s ( 3000 lb) ft s = 1.88 x 10 4 ft lb Example II-4: Steam under high pressure flows through nozzles and is forced against a turbine. The hot steam (ρ =0.598 kg/m 3 ) flows at 15 m/s. 35 m 3 of steam strikes the turbine blades each second. Find the kinetic energy of the steam hitting the fan each second. STEAM ρ = 0.598 kg/m 3 v = 15 m/s V = 35 m 3 E k =? Solution: 3
E k 1 = ρv V 3 ( 35 m ) 1 kg m = 0.598 15 3 m s = 355 N m = 355 J 3. BERNOULLI'S EQUATION Bernoulli's equation describes the conservation of energy in a fluid system, although it ignores energy losses due to friction. Therefore, it only approximates the actual flow of a fluid through a pipeline. For incompressible fluids (liquids), we can write Bernoulli's Equation in the form P + 1 v + gh = P + 1 v ρ ρ ρ + ρ gh II-5 1 1 1 P is the pressure; the potential energy stored in the fluid per unit volume that forces individual molecules in the fluid to be closer together. In this respect it is similar to the potential energy in a compressed spring. ½ρv is the kinetic energy per unit volume, the energy due to the motion of the fluid. ρgh is the gravitational potential energy per unit volume. This is the potential energy of the fluid due to its position in the gravitational field. The subscripts 1 and represent the system at two different locations. Fluid energy can be stored in each of these forms. If frictional losses are neglected, the total energy of the fluid, that is, the sum of the three energy components, will be unchanged, but the portion of the energy contained in each of the three components can vary. Example II-5: Water flows through the section of pipe shown below. The cross sectional area of the pipe is constant and the pressure measurements are for gauge pressure. a) What is the flow velocity at point (v )? b) What is the pressure at point (P )? 1 v 1 = m/s v =? P 1 = 10 5 N/m P =? h 1 = 3 m Solution: a) Since the cross sectional area of the pipe is constant (A 1 = A ) we can determine from the continuity equation that v =v 1 : Q V = v1 A 1 = v A = constant 4
v = v 1 A 1 A m = v 1 = s Since v =v 1, 1 / ρv 1 = 1 / ρv, so we can drop those terms from Eq. II-5. Furthermore, h =0, so we can drop the ρgh term on the right hand side. We are left with P 1, P and ρgh 1. Recalling that the density of water is 10 3 kg/m 3 and solving for P, we get P = P + ρgh = 10 1 1 5 N m kg + 10 m 9.8 m 3 s 3 ( ) 3 m = 1.9 x 10 5 N m In English units, the weight density is generally used instead of mass density and the acceleration due to gravity. Substitute ρ = ρ w /g in the dynamic pressure expressions and ρ w = ρg in the potential energy expressions to re-write Eq. II-5 as P 1 ρ 1 ρ + II-6 g g w w 1 v1 + ρ wh1 = P + v + ρ wh Example II-6: Water flows through the section of pipe of shown below. The diameter of the pipe at point (d ) is half the diameter of the pipe at point 1 (d = d 1 /). a) What is the velocity at point (v )? b) What is the pressure at point (P )? v 1 = ft/s P 1 = 35 psia 1 v =? P =? h = 5 ft Solution: a) From the continuity equation, or, in terms of the pipe diameter, Q V = v1 A 1 = v A 5
v 1 d1 d π = v π 4 4 d1 d1 v = v 1 = v1 d d / 1 = v = ft 4 = 8 ft 1 s s b) Since h 1 = 0, we can scrap the third term on the left-hand side in Eq. 11-6. The weight density of water (ρ w ) is 6.4 lb/ft3. P 1 must be converted from psia to psfa: P = 35 lb ( 144 in ) = 5040 lb 1 in ft Solving Eq. II-6 for P, we get P = P + 1 ρw g v - 1 ρw g v - h 1 1 ρ w lb = 5040 ft lb = 5040 ft lb 6.4 1 3 + ft ft 3 s lb + 3.9 ft - 6.4 lb ft ft s - 31 lb 6.4 1 3 - ft 8 ft 3 s lb lb = 4670 ft ft ft s lb - 6.4 3 ft ( 5 ft) lb 1 ft lb = 4670 = 3.4 ft 144 in in = 3.4 psia 3.1 Energy Stored in a Hydraulic Accumulator A hydraulic system is any fluid system that uses a liquid as its main working fluid. Some hydraulic systems use a device called an accumulator to store fluid energy. It is basically a sealed tank that holds fluid under pressure. One of its functions is to pressurize and store fluid, making it available, on demand, to supply a brief flow through a hydraulic system. Hydraulic systems can be subjected to sudden increases in pressure, commonly referred to as pressure spikes. Accumulators absorb or dampen these jolts of pressure in a hydraulic system. Accumulators come in a variety of forms; five of the most common types are shown in Fig. II-1. 6
weights vent piston piston a) gravity type b) air-baffle type c) spring type gas pressure relief valve X compressed air diaphragm piston d) bladder type e) piston type Figure II-1. Types of Accumulators Some types of hydraulic systems use energy at a steady rate. The pump and other parts of those systems are selected to meet a steady demand. One steady state hydraulic system is the water supply system in your home or school. You may have noticed that sometimes you can hear the water pipes rattle when a water faucet is turned off too quickly. The rattle, known as water hammer, is caused by a pressure spike that develops when the flow of water is stopped too suddenly. Plumbers sometimes install an accumulator in a water system to prevent water hammer. The accumulator absorbs the pressure pulse, converting it into potential energy, and then gradually releasing the stored energy back into the system. Other types of hydraulic systems use energy intermittently and in short bursts. An example of this type of system is the air operated brake system used on a bus or large truck (Fig. II-). ENGINE BRAKE CONTROL BRAKE CYLINDERS AIR COMPRESSOR ACCUMULATOR Figure II-. Accumulator in air operated brake system 7
In an air brake system, air-driven cylinders control the brake pad movement. An air compressor, powered by the truck engine, supplies pressurized air to the cylinders. However, a large and expensive air compressor would be needed to supply enough air to operate all of the cylinders at the same time. The problem is solved by using a small compressor and an accumulator in the system. The accumulator stores a large volume of pressurized air. The volume is large enough to operate all the brake cylinders at the same time. Many mechanics call the air storage tank a reservoir, but they could just as well call it an accumulator. Some internal combustion engines use turbochargers that boost the engine's efficiency. The engine oil pump lubricates bearings in the turbocharger. When the engine is turned off, the turbocharger turbine continues to turn for a few seconds, but during that time the bearings aren't lubricated because the oil pump is not operating. Some engine manufacturers solve this problem by using an accumulator that sends a flow of pressurized oil to the turbocharger for a few seconds. In this situation, oil in the accumulator is available on demand to lubricate the turbocharger bearings. In Lab 31 ( The Hydraulic Accumulator ) you will see the effects of a hydraulic accumulator on reducing pressure variations in an open system. In the last part of the experiment a shock wave will be produced in the system. You will see a pressure spike travel through the system and investigate how the accumulator reduces the effects of the spike. Fig. II-3 shows the pressure spike in an open fluid system. The large spike on the left indicates a pressure pulse (about 0 psi) traveling through the fluid system without an accumulator in the system. The smaller spike on the right, indicates the longer but smaller pressure pulse (about 3 psi) after it has passed by an accumulator placed in the system. pressure (psig) 1 1 8 4 0 PRESSURE SPIKE BEFORE REACHING ACCUMULATOR PRESSURE SPIKE AFTER PASSING ACCUMULATOR 0 5 1 1 time (seconds) Figure II-3. Effect of accumulator on Pressure Spike 8
3. Air Motors One of the most common types of industrial power hand tools is the pneumatic tool. They are frequently used as nut drivers, drills, air hammers, sanders, grinders and polishers. A compressed gas (usually air) drives pneumatic tools. Work is required to compress the gas from atmospheric pressure to some higher pressure. Most of the work done is retained in the compressed gas in the form of stored energy (remember, pressure is energy per unit volume), which can be distributed through a pipe or hose system to run the pneumatic tools. Compressing a gas causes it to heat up. Many large compressors have radiators through which hot compressed gases travel to be cooled. This process lets a larger volume of gas be stored than is possible when the gas is hot, because cool gas takes up less space than hot gas. 9
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Problem Set 1: Fluid Energy 1. Pressure indicates the energy stored in a unit volume of a fluid due to its pressure (E = P V). Calculate the energy represented by a volume of 8.00 m 3 of gas under an absolute pressure of x 10 5 Pascals. The energy units must be in joules. (Note that 1 Pa = 1 N/m and 1 N m = 1 J).. The kinetic energy per unit volume of a fluid is given by the expression ½ρv. Find the energy represented by a volume of 3.00 m 3 of water flowing at a speed of.5 m/s (note that E K = ½ρv V and ρ water = 1000 kg/m 3 ). 3. The gravitational potential energy per unit volume of a fluid is given by the expression ρgh. Find the total potential energy of 5.00 m 3 of water which is at a height of 15.0 m above the zero E p level. 4. Find the energy, in ft lb, represented by a volume of 8.0 ft 3 of fluid under an absolute pressure of 00 lb/ft. 17
5. Find the energy represented by 8.0 ft 3 of water flowing at a speed of 4.0 ft/s. The weight density of water is 6.4 lb/ft 3. 6. Find the gravitational potential energy of 300 ft 3 of water stored in a tank with respect to the ground 60 ft below. 7. Water is flowing through a level pipeline (h 1 = h ). The water speed is.00 m/s where the pipeline has a cross sectional area of 0.500 m. The absolute pressure is 1.50 x 10 5 Pa. The pipeline narrows to a cross section of 0.00 m. a) Find the speed of the water in the narrow section. Assume that the pipe is full of water and use the continuity equation: Q V = v 1 A 1 = v A = constant b) Find the absolute pressure in the narrow section of the pipe. c) How did the increase in speed affect the pressure? 18
8. Sea water (ρ w = 64 lb/ft 3 ) flows through a pipe of constant cross section (v = constant). At one location, a gauge pressure is.0 psi. The pipe then bends downward. a) Find the gauge pressure 0 ft below the first location. b) How did the change in height affect the pressure? 9. How high above a valve at the bottom of a large tank (open to the atmosphere at the top) is the water level when 10 m 3 of water flows out of the tank valve per minute and the area of the valve opening is 0.1 m? Assume that the area at the top of the tank (A top ) is much larger than the area of the hole (A hole ). By continuity, therefore v = v top Q V = v top A top = v hole A hole, hole A A hole and we can make the approximation that v top = 0. top v << v top hole 19
10. A horizontal pipe has a cross sectional area of 4.0 in. Downstream the pipe narrows to a cross sectional area of 1.0 in. Gasoline (ρ w = 4 lb/ft 3 ) flows with a speed of 6 ft/s in the larger section and the pressure there is 10 psig. Calculate a) the gasoline speed in the narrow section, and b) the pressure in the narrow section. 0
Worksheet for Problem Set 1 1
Worksheet for Problem Set 1
LEARNING OBJECTIVES FOR LABORATORY EXERCISES 31 AND 3 The learning objectives for Labs 31 and 3 are: 1. Understand the effects of pressure fluctuations in a fluid system.. Be able to describe how an accumulator operates and what purpose it serves in a hydraulic or pneumatic system. 3. Define and be able to calculate fluid potential energy in both SI and English unit systems. 4. Define the effects of fluid energy and be able to calculate fluid kinetic energy problems in both the SI and English unit systems. 5. Understand the purpose of Bernoulli's Equation in the conservation of energy of a fluid system and be able to solve both hydraulic and pneumatic fluid problems for an unknown in both the SI and English unit systems. 6. Understand the concept and effects of viscosity in fluid systems and steps that can be taken to reduce viscosity. 7. Identify workplace applications of the accumulator. LAB 31 OVERVIEW: ENERGY STORED IN A HYDRAULIC ACCUMULATOR A hydraulic system uses liquid under pressure to do work. Pressure spikes and liquid hammers may occur when mechanical changes cause rapid pressure changes in the system. To reduce the effects of such disturbances, an accumulator is sometimes incorporated into the hydraulic system. The accumulator dampens pressure spikes by temporarily storing excessive fluid caused by the pressure increase. By keeping pressure constant in the hydraulic system, the accumulator eliminates some potentially dangerous situations. In the first part of this lab you will build a hydraulic system with two pressure gauges. Then you will observe pressure changes within the system on both gauges. In the second part of this lab you will add an accumulator between the gauges. Again you will observe pressure changes within the system on both gauges. By comparing gauge readings with and without the accumulator, you can see how the accumulator works, and how it affects the system. In the third part of this lab, you will use a shock device to cause pressure spikes. Again, you will observe the operation of the accumulator. 3
LAB 31: ENERGY STORED IN A HYDRAULIC ACCUMULATOR Date OBJECTIVES: SKETCH: DATA TABLE 1 DEGREES FROM FULL CLOSED TO FULL OPEN Valve 1 Valve DATA TABLE Valve Position Pressure Without Accumulator P 1 (psi) Pressure With Accumulator P (psi) Gauge 1 Gauge Gauge 1 Gauge Full Open Max Min Half Open Max Min DATA TABLE 3 Shock Condition No Shock Weight Dropped 3" Weight Dropped 6" Weight Dropped 9" Valve 1 Pressure P 1 Valve Pressure P 4
LAB 31 ANALYSIS: 1. Describe the effect of closing valve halfway.. Describe the changes in the readings for gauge after adding the accumulator. 3. How does the accumulator work? 4. Describe the effect of an accumulator when a sudden pressure spike is experienced in a hydraulic system. 5. Show how the product of pressure and volume units will give the correct units for energy in a) the SI system and b) the British system. USE THE 5-STEP METHOD TO SOLVE THE FOLLOWING: 6. A pressure of 0 psi compresses the air in an accumulator, causing a change in the volume of air of 10 in 3. Compute the amount of energy stored by the compression in units of ft lb. 5
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LAB 3 OVERVIEW: ENERGY IN FLUID SYSTEMS When a fluid is flowing in a system and the cross-sectional area through which it is flowing decreases, its speed increases due to the continuity equation: v A = constant. A change of speed involves an acceleration, which means that a net force is acting. This net force for a fluid can only come from a difference in pressure in different parts of the system. Daniel Bernoulli derived an equation for fluid flow known as Bernoulli's Equation: P 1 + ρgh 1 + ½ ρv = P + ρgh + ½ ρv In the equation, p, h, and v are the pressure, height above some reference level and speed of the fluid at respective points in the system. According to Bernoulli's Equation the sum of the terms on each side of the equation has the same value at all points in an incompressible fluid with negligible viscosity that undergoes laminar flow. The effect of viscosity is to dissipate mechanical energy as heat. If viscosity is not negligible, the sum of the energy terms on one side of the equation will decrease in the direction of flow. Bernoulli applied the Conservation of Energy Principle to derive the equation. It is meant for incompressible fluids but can be used for gases as well as a good approximation. For gases the second term can be neglected since there is very little potential energy difference between two points in a system of low density gases. Thus for gases Bernoulli's Equation reduces to: P 1 + ½ ρv = P + ½ ρv Viscosity of a fluid is a kind of internal friction that prevents neighboring layers of a fluid from sliding freely past one another. In a pipe, fluid next to the wall of the pipe is stationary, while the fluid in the middle of the pipe is traveling at maximum speed. As long as the flow is laminar this speed relationship remains, even though as the viscosity increases the maximum speeds are lower. Viscosity of fluids varies as the temperature changes. The viscosity of liquids decrease as the liquid is heated while gases increase in viscosity as they are heated. In this experiment we will use air stored under pressure in a tank which is supplied by a compressor. Air at high pressure contains potential energy, which can be released as kinetic energy to do work. Air under pressure is directed to the rotor of an air motor, which causes the rotor to turn. The rotor shaft is in turn connected to a shaft pulley and stirring rod, which turns with the motor shaft. Some of the stored energy of the gas is lost to friction in the motor. The remaining potential energy is converted to mechanical energy of stirring various liquids. In this experiment you will observe the effects of stirring several different fluids: air, water and syrup each with a different viscosity. 7
LAB 3: ENERGY OF COMPRESSED AIR Date OBJECTIVES: SKETCH OF LAB SETUP: Fluid of Motor Load Pressure ΔP (psi) Motor Speed ω (RPM) Air Water Syrup Solution Pressure is a measurement of the stored energy per unit volume of a fluid. Show how the units of pressure will result in units of energy per volume, both for the English system and the SI system (Show how N/m can be converted to J/m 3, and how lb/ft can be converted to ft lb/ft 3 ). In Bernoulli's Equation, the value of ρgh (potential energy per unit volume) and ½ ρv (kinetic energy per unit volume) are added to the pressure to give the total energy per unit volume. Show how the units of each expression are the same as pressure units. 8
LAB 3 ANALYSIS 1. As the air passed through the air motor, its pressure dropped to zero gauge pressure, releasing all of its stored energy. What effect did friction of the motor-stirrer moving parts have?. Using what you learned about drag resistance last quarter, how will energy losses due to drag force change (for each rotation) as the speed of the stirrer is increased? 3. Last quarter we also learned that the force of friction is independent of velocity. How will energy losses due to friction change (for each rotation) as the speed of the stirrer is increased? 4. As a liquid is stirred, it acquires kinetic energy. How would the kinetic energy of two liquids of different densities compare if they were stirred at the same speed? Be specific. 5. Using the idea of conservation of energy, explain why the stirring speed varied for the three trials at the same motor pressure. 9
B. ENERGY IN ELECTRICAL SYSTEMS Electrical energy can be stored as potential energy in many ways, but almost all of them require that the energy be converted to another form. Electrical energy is converted into chemical energy when it is stored in a battery. When the battery is used, the chemical energy is converted back to electrical energy. Gravitational potential energy stored as water behind a dam can be converted into electrical energy. Turbine blades in the flow channels at the bottom of the dam are forced to turn by the pressure difference (ΔP) between the upstream (high pressure) and downstream (low pressure) sides of the dam. The rotational mechanical energy of the turbine is connected to an electrical generator, which produces electrical energy. Two other common devices that store electrical energy as potential energy are the capacitor and the inductor. Capacitors and inductors are found in many electrical circuits. 1. CAPACITORS A capacitor is an electrical device used to separate and store electrical charge. It typically consists of two plates of conducting material separated by an insulating material (the dielectric ). or a) common capacitor configuration b) schematic representations Figure II-4. A common capacitor configuration and the schematic representations for capacitors. 30
1.1 How Capacitors Work The circuit in Fig. II-5 contains a battery a switch and a capacitor. In Fig. II-5a the switch is open and there is no net charge on either plate. When the switch is closed, as in Fig. II-5b, electrons flow from the cathode to plate B, giving the plate a negative charge, while at the same time electrons are drawn from plate A to the battery anode, giving that plate a net positive charge. The electrons at plate B would like nothing better than to cross over to plate A, which is low on electrons, but the dielectric material between the two plates prevents this from occurring. Charge is stored on each plate in equal amounts but of opposite charge; the capacitor is said to be charged. When the switch is closed again, the stored charge will be drained from the capacitor, and it is said to be discharged. The rate at which a capacitor charges and discharges is dependent on the size of the capacitor and the amount of resistance in the circuit. Larger capacitors and more resistance mean longer charging and discharging times. Resistor-Capacitor ( RC ) circuits will be discussed next quarter. electrons + PLATE A - PLATE B + - PLATE A +++ +++ --- --- PLATE B a) Switch Open: no charge flows electrons b) Switch Closed: electrons are drawn to plate B and repelled from plate A Figure II-5. A capacitor in a circuit with the switch a) open and b) closed. When placed in electrical circuits, capacitors oppose changes in voltage. (They aren't like resistors. Resistors oppose current flow. A capacitor opposes voltage change.) Capacitors act to smooth out voltage changes in a circuit. The formula for capacitance is: C = q V II-7 where C is the capacitance in farads, q is the charge in coulombs and V is the voltage in volts. Capacitors separate charge; the higher the capacitance (C), the more charge (q) they separate for a given voltage, as we can see from Eq. II-7. Like a resistor, the capacitance will remain constant unless it fails. The capacitance is also dependent upon the area of the plates (A), the distance or gap between the plates (d) and the insulating properties of the material between the plates. The insulating properties of the material are known as the dielectric constant (ε). The greater the area of each plate, the 31
more charge they can hold. The greater the dielectric constant, the harder it is for the charges to cross from one plate to the other, and the more charge they can hold. The smaller the distance between the plates, the stronger the electric field is across the plates and the more charge the plates can hold. The capacitance is directly proportional to the area and the dielectric constant, and inversely proportional to the distance between the plates: ε C = A d II-8. By varying these quantities, capacitors with a wide range of values can be manufactured. Fixed capacitors are built with one value. Variable capacitors have a range of values obtained by varying the areas of the plates. Capacitors come in different forms, some of which can be seen in Fig. II-6, but in each case charges of opposite sign build up on each plate and are prevented from crossing by the insulation. Fig. II-6a shows a common parallel plate capacitor. Fig. II-6b shows a tubular capacitor. Here the plates are made of metal foil and the insulator is made of wax paper or plastic. Because the metal and insulation are flexible materials, the layers can be rolled into a tube giving a small package for a large plate area. The capacitors shown in Fig. s II-6a and II-6b are called fixed capacitors, because their capacitance cannot be adjusted. Fig. II-6c shows a variable capacitor. Here two sets of metal plates mesh without touching. The more the plates mesh, the higher the capacitance. Controlling the amount of mesh between the plates varies the capacitance. insulation metal foil a) parallel plate capacitor b) tubular capacitor c) variable capacitor Figure II-6. Types of Capacitors 1. Units of Capacitance Capacitance is a quantity which gives the ratio of net charge stored on either plate to the voltage difference across the plates (C = q/v). Capacitance is said to equal one farad (F) when one volt (V) is applied and a charge of one coulomb (C) moves from one capacitor plate to the other (1 F = 1 C/V). 3
The farad is a very large unit of capacitance. As a result, most capacitors found in electrical circuits are a small fraction of a farad. Values for many capacitors are given in microfarads (written μf) or picofarads (written pf). The prefix micro (μ) is equal to one millionth. Therefore, a capacitance of 1 μf is equal to one millionth of a farad. A picofarad is one millionth of a microfarad. 1.3 Energy Stored in the Capacitor Work must be done to separate the positive and negative charges on the capacitor plates. This work is stored in the capacitor as potential energy. As a battery or some other voltage source charges the capacitor, the voltage difference between the plates builds up (see Fig. II-7). As the charge builds up, so does the voltage difference. + - + - - electron BATTERY + missing electron BATTERY VOLTMETER - - - - - - - - - - - - CAPACITOR PLATES - - - - - - - - - - - - VOLTMETER + - + - - - - - - + - + CAPACITOR PLATES - - - - - - - - - - - - - - - - a) uncharged capacitor b) charged capacitor Figure II-7. Work or energy to charge a capacitor Energy stored in a capacitor depends on its capacitance rating and how much voltage is used to charge the capacitors. The formula for electrical potential energy stored in a capacitor is: E = p 1 CV II-8 where E p is the stored energy in joules, C is the capacitance in farads and V is the voltage used to charge the capacitor (in volts). It might be helpful to compare the energy stored in a capacitor with the energy stored in a stretched spring. Recall that the potential energy stored in a spring is E p = ½ kd, where d is the distance the spring is stretched from its equilibrium position and k is the spring constant. A spring with a large spring constant (a stiff spring) that is stretched a large distance will have a large amount of potential energy. Now consider a capacitor with a large capacitance (C). Energy is stored by creating a large voltage difference (ΔV) across the capacitor. In much the same fashion as with the spring, the potential energy (E p = ½ CV ) stored in the capacitor will be large. 33
Electrical energy units are always metric (SI). Thus, capacitance always is measured in farads (1 farad = 1 F = 1 C /J) or fractions of a farad. Voltages are measured in volts, charge is measured in coulombs, and electrical energy is measured in joules, so in an equation like E p = ½ CV, there is only one set of units to use. Example II-7. A single-phased, 40 V motor has a capacitor in its starting circuit. The capacitor is rated at 50 μf (50 x 10-6 farads). Find the charge stored on the capacitor plates and the potential energy stored in the capacitor. V = 40 V C = 50 x 10-6 F q =? Solution: Charge stored on the plates: Since C = q, V -6 ( ) ( ) q = C V = 50 x 10 F 40 V = 0.01 C Note: Since 1 coulomb of charge equals 6.5 x 10 18 electrons, 0.01 coulombs of charge equals 7.5 x 10 16 electrons. Therefore, 7.5x10 16 electrons have been added to one capacitor plate and removed from the other. Energy stored in the capacitor: Ep = 1 CV = 1 = 1.44 F V = 1.44 J ( 50 x 10 F ) ( 40 V) -6 Note: ( coulomb) C J 1 volt = 1 joule/coulomb and 1farad = 1, so 1 F V = 1 joule J C = 1 J. INDUCTORS An inductor is an electrical device used in electrical circuits to oppose changes in current. It isn't like a resistor, which opposes current. An inductor opposes a change in current. A typical inductor is shown in Fig. II-8. 34
current core Figure II-8. Basic inductor configuration..1 How Inductors Work An inductor is essentially a coil of wire. Recall from our discussion of solenoids that a current through a wire generates a magnetic field that is concentric about the wire, and if we coil the wire up we generate a magnetic field in the same shape as that formed by a bar magnet. Flowing current produces the magnetic field. It is proportional to the current, the number of coils of wire and the magnetic permeability of the material in the core, be it air, iron or some other material. We can write this as NI B = μ II-9 l where B is the intensity of the magnetic field, μ is the magnetic permeability of the material, N is the number of coils of wire, I is the current and l is the length of the solenoid. Note that a timevarying current will produce a time-varying magnetic field. Thus when the current changes, so does the magnetic field. It turns out that a changing magnetic field produces an induced voltage that opposes the voltage that produced the current in the first place! This is known as Faraday s Law of Induction. The induced voltage, in turn, produces an induced current in the direction opposing the change in current that produced it. Any increase in the current through an inductor will produce an increase in its magnetic field, which will produce an induced current in the direction opposite to the current increase, thus retarding the rate at which the current increase occurs. Any decrease in current will be retarded in the same manner. An inductor acts something like a huge flywheel. Flywheels are sometimes used with motors to keep them running smoothly, at constant speed. The energy associated with changes in speed are stored in the flywheel for later use. In a similar manner, an inductor in a circuit keeps the current flowing at a smooth, constant speed. It resists the increases or decreases in current and stores the energy produced by the current change. Inductors are important in circuits where filtering is needed. In such instances, they're called chokes or filter chokes because they smooth out - or filter - current flow. 35
The rate at which the current changes in through an inductor (ΔI/Δt) is proportional to the potential difference (V) and inversely proportional to the inductance (L). A larger potential difference allows the current to change faster, but a larger inductor slows the rate of change down. We can write this as ΔI Δt = V L II-10 The formula for inductance is then L = V ΔI Δt II-11. Basic Inductors Inductor configurations vary, but the principle remains the same. Sometimes the wire is wrapped around a hollow tube and the inductor is called an air-core inductor. In other cases, the coil of wire is wound around a cylindrical core of material, usually iron, which acts to increase the inductance. Some inductors are variable; they have a movable core, which can be moved into or out of the core area. The core has a higher magnetic permeability than air, so the more core material inside the inductor, the greater the magnetic field generated. We can use a movable core to increase or decrease the inductance as needed. Some common examples of inductors are illustrated in Fig. II- 9. 36
Figure II-9. Basic Inductors The value of inductance for a given coil depends on the material in the core area, the number of loops in the wire coil and the length of the wrapped core. These factors are all considered when making an inductor. By changing some or all of them, inductors can be made to have any value of inductance desired..3 Units of Inductance An inductor is rated in terms of how much voltage it can develop to oppose a certain rate of change in current. Inductance is measured in units called henries. An inductor rated at one henry (1 H) will develop a voltage of one volt whenever the current is changing at the rate of one ampere per second. L = V I t 1 H = 1 V A s which means we can write 1 henry in more than one form: 1 henry = 1 volt ampere / second volt second = 1 ampere = 1 ohm second = 1 J s C.4 Energy Stored in an Inductor Energy is stored in an inductor when a current increases from zero to some value as it flows through the coil. If the current in the circuit drops to zero, the energy stored in the inductor is released again as current. Compare the buildup and release of energy in an inductor with the mechanical act of pushing a wagon or cart up a hill. Suppose you stop at some point along the hill. At this point, a certain amount of work has been done on the cart and stored as potential energy in the cart. As long as you hold the cart at that position, the energy remains stored in the cart. When you remove your hands, the cart rolls back down the hill, releasing its stored energy. Energy stored in an inductor acts much the same way. The stored energy is released and used when the source that causes current to flow in the inductor is disconnected. The energy stored in an inductor is given by E = p 1 LI II-1 37
where E p is the stored (potential) energy in joules, L is the inductance in henries and I is the current in amperes. Note the similarities between the equations for energy stored in an inductor, the energy stored in a capacitor and the translational kinetic energy of a body in motion. Example II-8. An inductor with an inductance of 8 henries and a resistance of ohms is connected to a 30V DC power source. The current is steady. Find the energy stored by the inductor. + - ΔV = 30 V L = 8 H R = Ω Solution: The current is found from Ohm s Law: I = V R The energy stored in the inductor is = 30 V Ω = 15 A E = p 1 LI = 1 (8 H) (15 A) = 900 H A = 900 J Note: 1 H = 1 J s C, and 1 A = 1 C s, so 1 H A = 1 J s C C s = 1 J 38
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Problem Set : Capacitance and Inductance 1. Describe the function of an electrical capacitor.. The quantity of charge that builds up on either plate of a one farad capacitor with a voltage of one volt is. 3. Electrical (E p, E K ) may be stored in a capacitor or an inductor. 4. Capacitance can be calculated from the stored charge (q) and the voltage (V) across the plates. Write the equation for capacitance. 5. Capacitance in measured in the units called: a) ohms b) henries c) farads d). coulombs 6. The potential energy of the charge stored in a capacitor can be found from the equation: a) E p = ½mv b) E p =½CV c) E p = ½Iω d) E p = ½LI 7. Find the charge stored on the capacitor plates if a capacitor rated at 0 μf when the voltage difference is 100 V. 8. Find the potential energy, in joules, that is stored in the capacitor in problem 7. 9. An inductor opposes changes in ( current, resistance). 46
10. Inductance is measured in a) ohms b) henries c) farads d) coulombs 11. The value of inductance ( increases, decreases) when an iron-core inductor is substituted for an otherwise similar air-core inductor. 1. Inductance for a given coil depends on a) material of the core b) number of loops in the wire coil c) length and cross sectional area of the coil d. all of the above 13. The equation for E p stored in an inductor is a) E p = ½ mv b) E p = ½C V c) E p = ½ I ω d) E p = ½ L I 14. An inductor has an inductance of 8 henries and I = 15 A. Find E p in joules stored in the inductor. 15. A 7.0 μf capacitor has 6.0 volts applied across its plates. a) How much charge will be stored on each plate? b) How much energy is stored in the charged capacitor? 16. A 40.0 μf capacitor is charged to 30 V. It discharges through a 00 Ω resistor. Find the electrical energy released by the capacitor. 47
LEARNING OBJECTIVES FOR LABORATORY EXERCISES 33 AND 34 The learning objectives for Labs 33 and 34 are: 1. Describe what potential energy means in an electrical system.. Describe what a capacitor is, the basics of its construction and how it works. 3. Define capacitance and be able to work problems concerning electrical energy stored in capacitor. 4. Describe what an inductor is, the basics of its construction and how it works. 5. Define inductance and be able to work problems concerning electrical energy stored in an inductor. 6. Identify workplace applications of capacitors and inductors. 7. Understand and be able to use the units of capacitance and inductance. 8. Understand the relationship between thermal energy and work. Be able to use the mechanical equivalent of heat. 9. Describe the three ways, heat energy is transferred. Be able to work problems involving changes of state. 10. Describe the role of heat energy in the Law of Conservation of Energy. 11. Understand the processes and losses involved in converting fluid energy to electrical energy and heat energy. Give workplace applications of these energy conversions. 48
LAB 33: ENERGY STORED IN A CAPACITOR - TAKE HOME EXERCISE LAB OBJECTIVES: Upon completion of this lab, you should be able to: 1. Interpret a strip chart recorder of capacitor energy discharged through a resistor.. Calculate the energy stored in a capacitor. Equipment Required: Ruler and pencil OVERVIEW A capacitor is an electrical device that stores electrical energy. In a sense it is similar to a battery, except a battery generates its own electrical energy as a result of chemical action. A capacitor does not generate anything. It stores electrical energy from another source on its plates. This energy can be released when needed. A capacitor consists of two metal plates separated by an in insulator. The amount of charge a capacitor can hold for a given potential difference is known as its capacitance. Capacitance is measured in farads (F). The capacitance is equal to one farad (1 F) when an applied potential difference across the capacitor plates produces a charge of one coulomb on one of its plates. The formula for capacitance is C = q V where C is the capacitance in farads, q is the charge in coulombs and V is the voltage in volts. The farad is a relatively large unit of capacitance. Because of this, most capacitance values are usually given in microfarads (μf) or picofarads (pf). As previously described, a capacitor stores electrical energy on its plates. One plate fills up with electrons from a power source. The other plate loses electrons and winds up with an excess of positive charge. The result is a temporary battery, as shown below. - + electron flow (arrows electronflow) - - Capacitor Plates - + + - + - + + electron flow 49
The energy stored in a capacitor depends on two factors: its capacitance rating and the amount of voltage used to charge the plates. The formula for electrical potential energy (E p ) stored in a capacitor is: E = p 1 C V where C is the capacitance in farads and V is the voltage used to charge the capacitor, in volts. E p is measured in joules. Capacitance is set by three factors. The first is the area (size) of the plates. The second factor is the distance between the plates. The third factor is the type of insulation material (the dielectric ) between the plates. By varying any or all of these factors you can build capacitors with a wide range of values. Most capacitors are built with a fixed value and are called fixed capacitors. Others are made with movable plates to vary the plate area or distance between plates and are called variable capacitors. Once you know the capacitance and charging voltage you can find the potential energy stored in a capacitor. You can find the energy stored in a capacitor using a strip chart recorder, which plots voltage change over the period of time that the capacitor is charging or discharging. The graph below shows the energy released by discharging a 50 μf capacitor through a 400 kω resistor after being charged to 15 V DC. As we have discussed, capacitors store electrical energy on their plates. One plate fills up with electrons from a power source. The other plate loses the same amount of electrons and winds up with an excess of positive charge. The result is a temporary battery. See the figure below: 15 1 A 9 15 V DC B R 1 = 400 kω C 1 = 50 μf 6 3 0 0 0 40 60 80 100 50
The strip chart is a graph of voltage versus time. The horizontal lines on the grid mark the voltage reading. See the figure below. 15 Perforated Edge of Sheet 5 lines 1 0 lines Chart Span 9 6 15 lines 10 lines 3 5 lines 0 0 lines 0 0 40 60 80 100 Zero Volt Line Perforated Edge of Sheet The voltage reading on the strip chart is equal to the number of grid lines multiplied by the voltage increment. The maximum vertical range is called the span. The span is set on the chart recorder by the user. The recorder may be set for full scale (span control setting). An input of 15 volts will start the pen marking at the top of the strip chart if the span is set for 15 volts. Divide the span control setting by the number of horizontal lines to find the voltage increment. Voltage Increment = Voltage Span Setting # of Horizontal Lines If the voltage span setting is 0 volts and the # of horizontal lines is 5, then the voltage 0 V increment is = 0.8 V 5 lines line. 51
The vertical lines on the grid mark the time increments. See the figure below. 15 Chart Length Measures Time 0 0 0 40 60 80 100 The speed of the recorder may be preset or adjustable. The actual time of recording is the time increment multiplied by the number of vertical lines. You will find the time increment using the following equation: Time Increments = 1 (# of Lines per Inch ) ( Chart Speed) If the number of vertical lines per inch is 5 lines/inch and the chart speed is 10 inches per minute, then the time increment is 0.0 minutes per line. To change the time increment from minutes per line to seconds per line: Time Increment = 0.0 min 60 sec line min = 1. sec line In this Laboratory Homework Exercise you will use the time increment and voltage increment to interpret a strip chart reading. 5
L LAB PROCEDURE A. 1. Count the number of horizontal lines in the entire voltage span. # lines in the measuring span = Line Drawn By Recorder Pen as Capacitor Voltage Changed 15 1 9 6 3 15 V DC 50 μf 400 kω Capacitor Energy Released Through Resistor -Chart Grid - 5 Lines Per Inch Across the Span and Along the Length -Span Control Setting = 15 V -Chart Speed Setting = 3 in./min. 0 0 0 40 60 80 100. Use the data above to find the voltage increment: Voltage Increment = Span Control Setting # Horizontal Lines Voltage increment = volts line 3. Count the number of vertical lines in any of the one inch sections measuring time in the above figure. Vertical Lines / One Inch Section = Time: 5 lines/inch Chart Speed: 3 in./min. 4. Compute the time increment from the data above: Time Increment sec line 1 60 sec = in # Lines min Chart Speed min Inch sec = line 53
5. Find the voltage across the capacitor for each of the marked segments on the chart in the figure below. Record the values in the spaces on the graph. Capacitor Voltage (volts) = (# Lines ) 0.6 volts line 15 1 9 Voltage Axis 6 3 0 0 0 40 60 80 100 A B C D E Time Axis There are five primary horizontal and vertical grid lines on the actual chart. You will find the average voltage for each segment using the equation below: (# volts at Start) (# volts at End) + V avg = Use the values for voltage increment for each of the five segments and record them in DATA TABLE 1. 54
DATA TABLE Segment Time Interval t (sec) Average Voltage V avg (volts) Average Current I avg (μamps) Average Energy per Segment E avg (joules) A B C D E E experimental = J E theoretical = J 6. Find the time interval in seconds for each segment. The time is equal to the number of vertical lines in the segment multiplied by the time increment. Record the values in DATA TABLE 1. 7. Find the current flowing through the resistor in each of the time increments. Do this by dividing the average voltage of each segment by the resistor value, 400,000 Ω. Record these currents in the data table. 8. Find the average energy for each segment, using the equation below. Record these values. E avg (J) = V avg (V) I avg (A) t (sec) Note: Remember to convert your values for current to amps before calculating the energy. 9. Add the energy stored in the capacitor for each of the segments. This result is the experimental value of the total energy stored in the capacitor. Record this value for E exp in DATA TABLE 1. 10. Compute the theoretical value of the total energy stored in the capacitor using the equation: E = 1 CV theoretical and record it in DATA TABLE 1. Remember that the capacitance is in farads, not microfarads. 55
QUESTIONS 1. Find the percentage difference between the theoretical energy released by the capacitor and the experimental value. Use the equation: % Difference = % = E - E theoretical experimental η E theoretical 100%. Which segment shows the largest amount of energy released? 3. What is the physical construction of a capacitor? 4. What is the unit of capacitance? 5. Show by canceling units that the product of the equation below is a unit of energy: E = p 1 CV 6. If the resistor were replaced with one of a higher value, how would the discharge current be affected? 7. Considering your answer to #6, how would the time to discharge the capacitor energy be affected? 56
3. MULTIPLE ENERGY CONVERSIONS: GENERATING ELECTRICITY In the U.S., electrical energy is typically generated by either reciprocating engines (either gasoline or diesel powered) or turbines driven by fluid energy. Reciprocating engines are normally used for temporary electric power in remote areas or as emergency electric power when normal sources are interrupted, although the generation of electricity from wind and solar energy is becoming increasingly common. Different fluids are used to drive turbine generators. Air, in the form of wind power, is used in areas where a prevailing wind is available. High temperature gas is used to power gas turbine generators. Water is used at hydroelectric dams in many locations. Steam is one of the most common fluids used to drive turbine generators. The heat required for heating liquid water into steam can be obtained from burning fossil fuels (coal, oil and natural gas), from the fission process used in nuclear power plants and from geothermal and solar energy sources. The (kinetic) fluid energy delivered to the turbine is given by E K = ΔP Q V t II-1 where E k is the fluid kinetic energy, ΔP is the pressure difference, Q V is the volume flow rate and t is the time duration of the flow. The mechanical energy in the rotating parts of the turbine and generator can be determined from: E K = ½ I ω II-13 where E K is the rotational mechanical energy, I is the moment of inertia and ω is the angular speed. Electrical energy produced in the generator can be determined from E E = V I t II-14 where E E is the electrical energy, V is the voltage across the load, I is the current through the load and t is the time duration of interest. At each step in the process of producing electrical energy - heat to fluid, fluid to mechanical, mechanical to electrical - energy is lost. The electrical energy generated (energy out) will always be less than the initial energy used (energy in). Losses include heat losses incurred while transferring energy to the fluid, drag resistance of the fluid flow, losses across the turbine blades or vanes due to clearances, friction in the bearings of rotating parts, heat lost in heating mechanical components (steam systems) and electrical resistance in electrical components and wiring of the generator. All of these losses show up as heat energy. In Lab 34 we will use the potential energy of compressed air to turn an air motor. The air motor is directly coupled to the shaft of a DC permanent magnet motor, which serves as a generator. The electric current output from the generator will flow through a load (a resistor in the lab). This 57