Home | Articles | Forum | Glossary | Books |

Whenever a machine transforms energy from one form to another, there is always a certain loss. The loss takes place in the machine itself, causing (1) an increase in temperature and (2) a reduction in efficiency. From the standpoint of losses. electrical machines may be divided into two groups: those that have revolving parts (motors. generators, etc.) and those that don’t (transformers, reactors. etc.). Electrical and mechanical losses are produced in rotating machines, while only electrical losses are produced in stationary machines. In this section we analyze the losses in dc machines. but the same losses are also found in most machines operating on alternating current. The study of power losses is important because it gives us a clue as to how they may be reduced. We also cover the important topics of temperature rise and the service life of electrical equipment. We show that both are related to the class of insulation used and that these insulation classes have been standardized.
Mechanical losses are due to bearing friction brush friction, and windage. The friction losses depend upon the speed of the machine and upon the design of the bearings, brushes, commutator, and slip rings. Windage losses depend on the speed and design of the cooling fan and on the turbulence produced by the revolving parts. In the absence of prior information. we usually conduct tests on the machine itself to determine the value of these mechanical losses. Rotating machines are usually cooled by an internal fan mounted on the motor shaft. It draws in cool air from the surroundings. blows it over the windings, and expels it again through suitable vents. In hostile environments, special cooling methods are sometimes used.
-- Totally enclosed, water-cooled, 450 kW, 3600 RPM mo tor for use in a hostile environment. Warm air inside the machine is blown upward and through a water-cooled heat exchanger, situated immediately above the Westinghouse nameplate. After releasing its heat to a set of water-cooled pipes, the cool air reenters the ma chine by way of two rectangular pipes leading into the end bells. The cooling air therefore moves in a closed circuit, and the surrounding contaminated atmosphere never reaches the motor windings. The circular capped pipes located diagonally on the heat exchanger serve as cooling-water inlet and outlet respectively. (Westinghouse) --- Copper losses may be expressed in watts per kilogram. --- Brush contact voltage drop occurs between the brush face and commutator. -- Electrical losses are composed of the following: 1. Conductor I^2 x R losses (sometimes called copper losses) 2. Brush losses 3. Iron losses _____________ 1. Conductor Losses: The losses in a conductor depend upon its resistance and the square of the cur rent it carries. The resistance. in turn. depends upon the length. cross section, resistivity, and temperature of the conductor. The following equations enable us to determine the resistance at any temperature and for any material: … in which R = resistance of conductor (Omega) L = length of conductor m A = cross section of conductor m^2 rho = resistivity of conductor at temperature, t rho = resistivity of conductor at a = temperature coefficient of resistance at 0°C alpha = temperature of conductor 0 C The values of rho and alpha for different materials are listed. In dc motors and generators. copper losses occur in the armature, the series field, the shunt field. the commutating poles. and the compensating winding. These I^2R losses show up as heat, causing the conductor temperatures to rise above ambient temperature. Instead of using the PR equation. we sometimes prefer to express the losses in terms of the number of watts per kilogram of conductor material. The losses are then given... where: P J = current density p = resistivity of the conductor J = density or the conductor 1000 = constant, to take care of units According to this equation. the loss per unit mass is proportional to the square or the current density. For copper conductors. we use densities between 1.5 A/mm^2 and 6 A/mm^2. The corresponding losses vary from 5 W/kg to 90 W/kg. The higher densities require an efficient cooling system to prevent an excessive temperature rise. 2. Brush Losses: The I^2 R losses in the brushes are negligible because the current density is only about 0.1 A/mm^2, which is far less than that used in copper. However, the contact voltage drop between the brushes and commutator may produce significant losses. The drop varies from 0.8v to 1.3v, depending on the type of brush, the applied pressure, and the brush current. -- --- This 150 kW electric oven is used to anneal punched steel laminations. This industrial process, carried out in a controlled atmosphere of 800°C, significantly reduces the iron losses. The laminations are seen as they leave the oven. -- 3. Iron Losses: Iron losses are produced in the armature of a dc machine. They are due to hysteresis and eddy currents. as previously explained. Iron losses depend upon the magnetic flux density, the speed of rotation, the quality of the steel, and the size of the armature. They typically range from 0.5 W/kg to 20 W/kg. The higher values occur in the armature teeth, where the flux density may be as high as 1.7 T. The losses in the armature core are usually much lower. The losses can be minimized by annealing the steel. Some iron losses are also produced in the pole faces. They are due to flux pulsations created as successive armature teeth and slots sweep across the pole face. Strange as it may seem, iron losses impose a mechanical drag on the armature, producing the same effect as mechanical friction. Ex. 1: A dc machine turning at 875 RPM carries an armature winding whose total weight is 40 kg. The cur rent density is 5 A/mm^2 and the operating temperature is 80°c. The total iron losses in the armature amount to 1100 W.
a. The copper losses b. The mechanical drag N-m due to the iron losses Solution: a. Referring to Table, the resistivity of copper at 80°C is … Solution: a. Referring to the Table, the resistivity of copper at 80°C is rho = rho_0 (1 + alpha t) = 15.88 (1 + 0.00427 x 80) = 21.3 n Ohm·m The density of copper is 8890 kg/m^3 The specific power loss is: Pc = 1000J^2 p/~ = 1000 X 52 X 21.3/8890 = 60 W/kg Total copper loss is: P = 60 X 40 = 2400 W b. The braking torque due to iron losses can be calculated from p = nT/9.55 1100 = 875 T/9.55 T = 12 N-m or approximately 8.85 ft·lbf
A dc motor running at no-load develops no useful power. However, it must absorb some power from the line to continue to rotate. This no-load power overcomes the friction, windage, and iron losses, and provides for the copper losses in the shunt field. The PR losses in the armature, series field, and commutating field are negligible because the no load current is seldom more than 5% of the nominal full-load current. As we load the machine the current increases in the armature circuit. Consequently, the I^2R losses in the armature circuit (consisting of the armature and all the other windings in series with it) will rise. On the other hand, the no-load losses mentioned above remain essentially constant as the load increases, unless the speed of the machine changes appreciably. It follows that the total losses increase with load. Because they are converted into heat, the temperature of the machine rises progressively as the load increases. However, the temperature must not exceed the maximum allowable temperature of the insulation used in the machine. Consequently, there is a limit to the power that the machine can deliver. This temperature-limited power enables us to establish the nominal or rated power of the machine. A machine loaded beyond its nominal rating will usually over heat. The insulation deteriorates more rapidly, which inevitably shortens the service life of the machine. If a machine runs intermittently. it can carry heavy overloads without overheating, provided that the operating time is short. Thus, a motor having a nominal rating of 10 kW can readily carry a load of 12 kW for short periods. However, for higher loads the capacity is limited by other factors, usually electrical. For in stance, it’s physically impossible for a generator rated at 10 kW to deliver an output of 100 kW, even for one millisecond.
The efficiency of a machine is the ratio of the useful output power P0 to the input power P1. Furthermore, input power is equal to useful power plus the losses p. We can therefore write: The following example shows how to calculate the efficiency of a DC machine.
A dc compound motor having a rating of 10 kW, 1150 RPM, 230 V, 50 A. has the following losses at fill/load: bearing friction loss 40W brush friction loss 50W windage loss 200W (1) total mechanical losses 290W iron losses 420W (3) copper loss in the shunt field 120W copper losses at full load: a. in the armature 500W b. in the series field 25W c. in the commutating winding 70W (4) total copper loss in the armature circuit at full load 595 W Calculate the losses and efficiency at no-load and at 25, 50, 75, 100, and 150 % of the nominal rating of the machine. Draw a graph showing efficiency as a function of mechanical load (neglect the losses due to brush contact drop).
No-load: The copper losses in the armature circuit are negligible at no-load. Consequently. the no load losses are equal to the sum of the mechanical losses (1), the iron losses (2), and the shunt-field losses (3): no-load losses 290 + 420 + 120 = 830W These losses remain essentially constant as the load varies. The efficiency is zero at no-load because no useful power is developed by the motor. 25% load When the motor is loaded to 25% of its nominal rating. the armature current is approximately 25% (or 1/4) of its full load value. Because the copper losses vary as the square of the current. we have the following: copper losses in the armature circuit: = (1/4)^2 x 595 = 37 W no-load losses 830 W total losses = 37 + 830 867W Useful power developed by the motor at 25% load is P0 10 kW X (1/4) 2500 W (= 3.35 hp) power supplied 10 the motor is P1 2500 + 867 3367 W and the efficiency is eta = (P/Pj ) X 100 = (2500/3367) X 100 = 74% In the same way, we find the losses at 50, 75, 100, and 150% of the nominal load: At 50% load the losses are (1/2)^2 X 595 + 830 979 W At 75% load the losses are (3/4)2 X 595 + 830 = 1165W At 100% load the losses are 595 + 830 = 1425 W At 150 % load the losses are (1.5)^2 X 595 + 830 = 2169 W The efficiency calculations for the various loads are listed in Table A and the results are shown graphically.
LOSSES and EFFICIENCY OF A DC MOTOR: The efficiency curve rises sharply as the load in creases. flattens off over a broad range of power, and then slowly begins to fall. This is typical of the efficiency curves of all electric motors. both ac and dc. Electric motor designers usually try to attain the peak efficiency at full-load. In the above calculation of efficiency we could have included the losses due to brush voltage drop. Assuming a constant drop. say. of 0.8 V per brush. the brush loss at full-load amounts to 0.8 V X 50 A x 2 brushes = 80 W. At 50% load, the brush loss would be 40 W. These losses. when added to the other losses. Modify the efficiency curve only slightly. It’s important to remember that at light loads the efficiency of any motor is poor. Consequently. when selecting a motor to do a particular job. we should always choose one having a power rating roughly equal to the load it has to drive. We can prove that the efficiency of any dc ma chine reaches a maximum at that load where the armature circuit copper losses are equal to the no-load losses. In our example this corresponds to a total loss of (830 T 830) 1660 W, an output of 11 811 W (15.8 hp) and an efficiency of 87.68%. You may double-check these results.
The temperature rise of a machine or device is the difference between the temperature of its warmest accessible part and the ambient temperature. It may be measured by simply using two thermometers. However. due to the practical difficulty of placing a thermometer close to the really warmest spot inside the machine. this method is seldom used. We usually rely upon more sophisticated methods. described in the following sections. Temperature rise has a direct bearing on the power rating of a machine or device. It also has a direct bearing on its useful service life. Consequently, temperature rise is a very important quantity.
Apart from accidental electrical and mechanical failures, the life expectancy of electrical apparatus is limited by the temperature of its insulation: The higher the temperature, the shorter its life. Tests made on many insulating materials have shown that the service life of electrical apparatus diminishes approximately by half every time the temperature increases by 10°C. This means that if a motor has a normal life expectancy of eight years at a temperature of 105°c. it will have a service life of only four years at a temperature of 115C, of two years at 125 C, and of only one year at 135°C! The factors that contribute most to the deterioration of insulators are (1) heat, (2) humidity, (3) vibration, (4) acidity, (5) oxidation, and (6) time. Because of these factors. the state of the insulation changes gradually; it slowly begins to crystallize and the transformation takes place more rapidly as the temperature rises. In crystallizing, organic insulators become hard and brittle. Eventually, the slightest shock or mechanical vibration will cause them to break. Under normal conditions of operation, most organic insulators have a life expectancy of eight to ten years provided that their temperature does not exceed 100°C. On the other hand, some synthetic polymers can withstand temperatures as high as 200°C for the same length of time. Low temperatures are just as harmful as high temperatures are, because the insulation tends to freeze and crack. Special synthetic organic insulators have been developed, however. which retain their flexibility at temperatures as low as -60C. --- 6: Factors that may shorten the service life of an insulator. time high temperature humidity chemicals fungus; dust, rodents, ozone noxious gases vibration
Committees and organizations that set standards have grouped insulators into five classes, depending upon their ability to withstand heat. These classes correspond to the maximum temperature levels of: 105°c, 130 °c, 155°c, 180°c, and 220°C {formerly represented by the letters A, B, F, H, and R). This thermal classification (Table B) is a cornerstone in the design and manufacture of electrical apparatus.
Standards organizations have also established a max. ambient temperature, which is usually 40 C. This standardized temperature was established for the following reasons: 1. It enables electrical manufacturers to foresee the worst ambient temperature conditions that their machines are likely to encounter. 2. It enables them to standardize the size of their machines and to give performance guarantees. ---- CLASSES OF INSULATION SYSTEMS Class: - 105°C A
- 130°C B
- 155°C F
- 180°C H
- 200°C N
- 220°C R
- 240°C S
- above 240°C C
Materials or combinations of materials such as cotton, silk. and paper when suitably impregnated or coated or when immersed in a dielectric liquid such as oil. Other materials or combinations of materials may be included in this class if by experience or accepted tests they can be shown to have comparable thermal life at 105 C. Materials or combinations of materials such as mica. glass fiber. Asbestos, etc. with suitable bonding substances. Other materials or combinations of materials may be included in this class if by experience or accepted tests they can be shown to have comparable thermal life at 130°C. Materials or combinations of materials such as mica, glass fiber, asbestos, etc., with suitable bonding substances. Other materials or combinations of materials may be included in this class if by experience or accepted tests they can be shown to have comparable life at 155 C. Materials or combinations of materials such as silicone elastomer, mica, fiber, asbestos, etc. with suitable bonding substance, such as appropriate silicone resins. Other materials or combinations of materials may be included in this class if by experience or accepted tests they can be shown to have comparable life at 150°C. Materials or combinations of materials which by experience or accepted tests can be shown to have the required thermal life at 200°C. Materials or combinations of materials which by experience or accepted tests can be shown to have the required thermal life at 220°C. Materials or combinations of materials which by experience or accepted tests can be shown to have the required thermal life at 240°C. Materials consisting entirely of mica, porcelain, glass, quartz. and similar inorganic materials. Other materials or combinations of materials may be included in this class if by experience or accepted tests they can be shown to have the required thermal life at temperatures above 240 C. ----- --- Typical limits of some dc and ac industrial machines, according to the insulation class: (1) Shows the maximum permissible temperature of the insulation to obtain a reasonable service life (2) Shows the maximum permissible temperature using the resistance method (3) Shows the limiting ambient temperature --- The temperature of a machine varies from point to point. but there are places where the temperature is warmer than anywhere else. This hottest-spot temperature must not exceed the maximum allowable temperature of the pm1icular class of insulation used. Hot-spot temperature limits for class A, B, E and H insulation. They are the temperature limits previously mentioned. The maximum ambient temperature of 40 C is also shown (curve 3). The temperature difference between curve 1 and curve 3 gives the max. temperature rise for each insulation class. This limiting temperature rise enables the manufacturer to establish the physical size of the motor. relay, and so forth. he intends to put on the market. Thus. for class B insulation. the maximum allowable temperature rise is (130 minus 40) = 90°C, To show how the temperature rise affects the size of a machine. suppose a manufacturer has designed and built a 10 kW motor using class B insulation. To test the motor he places it in a constant ambient temperature of 40°C and loads it up until it delivers 10 kW of mechanical power. Special temperature detectors, located at strategic points inside the ma chine, record the temperature of the windings. After the temperatures have stabilized (which may take several hours). the hottest temperature is noted. and this is called the hot-spot temp. If the hotspot temperature so recorded is, say 147°C the manufacturer would not be permitted to sell his product. The reason is that the temperature rise (147°C - 40°C) = 107°C exceeds the maximum permissible rise of 90°C for class B insulation. On the other hand, if the hottest-spot temperature is only 100°C the temperature rise is (100° - 40) = 60 C. The manufacturer immediately perceives that he can make a more economical design and still remain within the permissible temperature rise limits. For instance, he can reduce the conductor size until the hot-spot temperature rise is very dose to 90°C. Obviously, this reduces the weight and cost of the windings. But the manufacturer also realizes that the reduced conductor size now enables him to reduce the size of the slots. This, in turn, reduces the amount of iron. By thus redesigning the motor. the manufacturer ultimately ends up with a machine that operates within the permissible temperature rise limits and has the smallest possible physical size, as well as lowest cost. In practice, it’s not convenient to carry out performance tests in a controlled ambient temperature of 40 C. The motor is usually loaded to its rated capacity in much lower (and more comfortable) ambient temperatures. Toward this end, it has been established by standards-setting bodies that, for testing purposes, the ambient temperature may lie anywhere between 10°C and 40 C. The hottest-spot temperature is recorded as before. If the temperature rise under these conditions is equal to or less than 9 0° C (for class B insulation), the manufacturer is allowed to sell his product.
Soln: The hot-spot temperature rise is 125°-32°=93°C --- the permissible hot-spot temperature rise for class F insulation is (155° - 40°) = 115°C The motor easily meets the temperature standards. The manufacturer could reduce the size of the motor and thereby market a competitive product.
The hot-spot temperature rise is rather difficult to measure because it has to be taken at the very inside of a winding. This can be done by embedding a small temperature detector, such as a thermocouple or thermistor. However, this direct method of measuring hot-spot temperature is costly, and is only justified for larger machines. To simplify matters. accepted standards permit a second method of determining temperature rise. It’s based upon the average winding temperature, measured by resistance, rather than the hot--spot temperature. The maximum allowable average winding temperatures for the various insulation classes are shown in curve 2. For example, in the case of class B insulation, an average winding temperature of 120 C is assumed to correspond to a hot-spot temperature of 130°C. Consequently, an average temperature rise of (120 - 40°) = 80 C is assumed to correspond to a hot-spot temperature rise of (130° - 40) = 90°C. The average temperature of a winding is found by the resistance method. It consists of measuring the winding resistance at a known winding temperature. and measuring it again when the ma chine is hot. For example. if the winding is made of copper, we can use the following equation to determine its average temperature … Where: t_2 = average temperature of the winding when hot 234 = a constant equal to 1/alpha: = 1/0.00427 R2= hot resistance of the winding R1 = cold resistance of the winding t_1=temperature of the winding when cold Knowing the hot winding temperature by the resistance method, we can immediately calculate the corresponding temperature rise by subtracting the ambient temperature. If this temperature rise falls within the permissible limit (80°C for class B insulation). the product is acceptable from a standards point of view. Note that when performance tests are carried out using the resistance method, the ambient temperature must again lie between 10°C and 40°C. If the winding happens to be made of aluminum wire but the number 234 must be changed to 228.
a. The average temperature of the winding, at full load b. The full-load temperature rise by the resistance method c. Whether the motor meets the temperature standards Soln: a. The average temperature of the shunt field at full-load is t-, (R2/R1) (234 + t) - 234 (30/22) (234 + 19) 234 = 111° C. b. The average temperature rise at full-load is 111°-24° = 87°c. c. The maximum allowable temperature rise by resistance for class B insulation is (120°-40°) = 80°c. Consequently, the motor does not meet the standards. Either its rating will have to be reduced, or the cooling system improved, before it can be put on the market. Alternatively, it may be rewound using class F insulation. As a very last resort, its size may have to be increased. A final word of caution: temperature rise standards depend not only on the class of insulation, but also on the type of apparatus (motor, transformer, relay, etc.), the type of construction (drip-proof, to tally enclosed, etc.), and the field of application of the apparatus (commercial, industrial, naval. etc.). Consequently, the pertinent standards must always be consulted before conducting a heat-run test on a specific machine or device. ---8 100 kW, 2000 RPM motor; mass: 300 kg. ---9 100 kW, 1000 RPM motor; mass: 500 kg.
Although maximum allowable temperature rise establishes the nominal power rating of a machine, its basic physical size depends upon the power and speed of rotation. Consider the 100 kW, 250 V, 2000 RPM generator shown. Suppose we have to build another generator having the same power and voltage, but running at half the speed. To generate the same voltage at half the speed, we either have to double the number of conductors on the armature or double the flux from the poles. Consequently, we must either increase the size of the armature, or increase the size of the poles. In practice, we increase both. We conclude that for a given power output, a low-speed machine is always bigger than a high-speed machine. This is true for both ac and dc machines. Basically, the size of a machine depends uniquely upon its torque. Thus, a 100 kW, 2000 RPM motor has about the same physical size as a 10k W motor running at 200 RPM because they develop the same torque. Low-speed motors are therefore much more costly than high-speed motors of equal power. Consequently, for low-speed drives, it’s often cheaper to use a small high-speed motor with a gear box than to use a large low-speed motor directly coupled to its load.
++1 Name the losses in a dc motor. ++2 What causes iron losses and how can they be reduced? ++3 Explain why the temperature of a machine increases as the load increases. ++4 What determines the power rating of a ma chine? ++5 If we cover up the vents in a motor. its output power must be reduced. Explain. ++6 If a motor operates in a cold environment may we load it above its rated power? Why? ++7 Name some of the factors that contribute to the deterioration of organic insulators. ++8 A motor is built with class H insulation. What maximum hot-spot temperature can it withstand?
++9 A dc motor connected to a 240 V line produces a mechanical output of 160 hp. Knowing that the losses are 12 k W. calculate the input power and the line current. ++10 A 115 V dc generator delivers 120 A to a load. If the generator has an efficiency or 81%, calculate the mechanical power needed to drive it, hp. ++11 Calculate the full-load current of a 250 hp. 230 V dc motor having an efficiency of 92 %. ++12 A machine having class B insulation attains a temperature of 208C (by resistance) in a torrid ambient temperature of 180 C. a. What is the temperature rise? b. Is the machine running too hot and, if so, by how much? ++13 The efficiency of a motor is always low when it operates at 10% of its nominal power rating. Explain. ++14 Calculate the efficiency of the motor in Example 2 when it delivers an output of 1 hp. ++15 An electric motor driving a skip hoist with draws 1.5 metric tons of minerals from a trench 20 m deep every 30 seconds. If the hoist has an overall efficiency of 94%, calculate the power output of the motor in horsepower and in kilowatts. ++16 Thermocouples are used to measure the internal hot-spot winding temperature of a 1200 kW ac motor, insulated class F. If the motor runs at full-load, what is the maximum temperature these detectors should indicate in an ambient temperature of 40 C? 30T'? 14°C! ++17 A 60 hp ac motor with class F insulation has a cold winding resistance of 12 n at 23°c. When it runs at rated load in an ambient temperature of 310c. the hot winding resistance is found to be 17.4 n. a. Calculate the hot winding temperature. b. Calculate the temperature rise of the motor. c. Could the manufacturer increase the name plate rating of the motor? Explain. ++18 An electric motor has a normal life of eight years when the ambient temperature is 3()°C. If it’s installed in a location where the ambient temperature is 60°C. what is the new probable service life of the motor? ++19 A No. 10 round copper wire 210m long carries a current of 12A. Knowing that the temperature of the conductor is 105°C. calculate the following: a. The current density A/mm^2 b. The specific copper losses [W/kg]
++20 An aluminum conductor operates at a current density of 2 A/mm^2 a. If the conductor temperature is 120C. calculate the specific losses [W / kg]. b. Express the current density in circular mils per ampere. ++21 The temperature rise of a motor is roughly proportional to its losses. On the other hand, its efficiency is reasonably constant in the range between 50 % and 150% of its nominal rating. Based on these facts. if a 20 kW motor has a full load temperature rise of 80°C. what power can it deliver at a temperature rise of 105°C? ++22 An electromagnet (insulated class A) situated in a particularly hot location has a service life of two years. What is its expected life span if it’s rewound using class F insulation? ++23 An 11 kW ac motor having class B insulation would normally have a service life of 20,000 h, provided the winding temperature by resistance does not exceed 120°C. By how many hours is the service life reduced if the motor runs for 3h at a temperature (by resistance) of 200°C?
++24 A reel of No. 2/0 single copper conductor has a resistance of 0.135 ohms at a temperature of 25°C. Calculate the approximate weight of the conductor in pounds. ++25 The Table lists the properties of commercially available copper conductors. In an electrical installation. it’s proposed to use a No.4 A WG conductor in an area where the operating temperature of the conductor may be as high as 70°C. Calculate the resistance under these conditions of a 2-conductor cable No.4 AWG that is 27 meters long. ++26 The shunt field of a 4-pole dc motor has a total resistance of 56 ohms at 25"C. By scraping off the insulation. it’s found that the bare copper wire has a diameter of 0.04 inches. Determine the AWG wire size, and calculate the weight of the wire per pole, in kilograms. ++27 The National Electrical Code allows a maximum current of 65 A in a No.6 gauge cop per conductor, type RW 75. A 420 ft cable is being used on a 240 V dc circuit to carry a current of 48 A. Assuming a maximum operating temperature of 70"C. calculate the following: a. The power loss, in watts, in the 2-conductor cable b. The approximate voltage at the load end if the voltage at the service panel is 243 V. ++28 In Problem 27. if the voltage drop in the cable must not exceed J 0 V when it’s carrying a current of 60 A. what minimum conductor size would you recommend? Assume a maximum operating temperature of 70°C. ++29 A dc busbar 4 inches wide, ¼” thick, and 30 ft long carries a current of 2500 A. Calculate the voltage drop if the temperature of the busbar is 105°C. What is the power loss per meter? ++30 Equation 3 gives the resistance/temperature relationship of copper conductors, namely, t2 = R2/R1 (234 + t1) - 234 Using the information given, deduce a similar equation for aluminum conductors. ++31 The commutator of a 1.5 hp, 2-pole, 3000 RPM dc motor has a diameter of 63 mm. Calculate the peripheral speed in feet per minute and in miles per hour. ++32 The following information is given on the brushes used in the motor of Problem 31: number of brushes: 2 current per brush: 15 A brush dimensions: 5/8 in. wide, 5/16 in thick, 3/4 in. long. (The 5/16 in. x 5/8 in. area is in contact with the commutator) resistivity of brush: 0.0016 Ohm in. brush pressure: 1.5 lbf brush contact drop: 1.2 V coefficient of friction: 0.2 Calculate the following: a. The resistance of the brush body in ohms b. The voltage drop in the brush body c. The total voltage drop in one brush, including the contact voltage drop d. The total electrical power loss (in watts) due to the two brushes e. The frictional force of one brush rubbing against the commutator surface (in lbf and in newtons) f. The frictional energy expended by the two brushes when the commutator makes one revolution (in joules) g. The power loss due to friction, given the speed of 3000 RPM h. The total brush loss as a percent of the 1.5 hp motor rating --- A VARMECA induction motor, including its variable-speed controller and gear reducer, is housed under a trans parent plastic cover. The entire unit is being water-sprayed to demonstrate its ability to operate continuously under harsh environmental conditions. (Emerson Electric). Also see: Generating Electrical Power |

Top of Page | PREV: | NEXT: | Basics of Industrial Motor Control: part 2: ELECTRIC DRIVES | HOME |