Handbook of Small Electric Motors MAZ Part 15 potx

CHAPTER 8 APPLICATION OF MOTORS Chapter Contributors Earl F. Richards William H. Yeadon 8.1 There are many things that must be considered when deciding what type and size of motor is most suitable to one’s application. They can be broken into three general categories—namely, mechanical, electrical, and thermal considerations. The main reference in common use is the standard by the National Electrical Manufacturers Association (NEMA). NEMA Standard MG1 covers standard frame sizing, dimensions, mounting con- figurations, electrical ratings, nominal speeds, and performance requirements of fractional- and integral-horsepower ac and dc motors. Table 8.1 lists some of the main topics. NEMA Standard MG7 covers similar information on servomotors and control systems. Some of the major topics are listed in Table 8.2. 8.1 MOTOR APPLICATION REQUIREMENTS* Any application of a motor must meet some specific output requirement for a given input. Furthermore, the motor must fit in the allotted space and mount securely to the device. NEMA standards provide recommended mounting configurations for almost all applications and define performance ranges for them. Since these configurations are currently available, it behooves one to start the selection process there. There are, * Sections 8.1 to 8.3 contributed by William H. Yeadon,Yeadon Engineering Services, PC. however, many applications where a standard motor just won’t fit. The ever- shrinking computer devices and appliances are a few examples. In these cases, spe- cial shafts,mounting hardware,and cooling provisions may be necessary.Listed here are some characteristics that must be considered. I. Mechanical requirements A. Shaft size and features B. Shaft materials 1. Compatibility with application 2. Materials 3. Finish 4. Hardness C. Bearings 1. Ball or sleeve 2. Side load 3. Axial load 4. Lubrication temperature range 5. Service life D. Mounting system 1. Flange 2. Bolts 3. Studs 4. Assembly method 8.2 CHAPTER EIGHT TABLE 8.1 Main Topics in NEMA Standard MG1 Classification according to size Classification according to application Classification according to electrical type Classification according to environmental protection and methods of cooling Classification according to variability of speed Rating, performance, and testing Complete machines and parts Classification of insulation systems DC motors and generators AC motors and generators AC generators and synchronous motors Single-phase motors Motors for hermetic refrigeration compres- sors Small motors for shaft-mounted fans and blowers Small motors for belted fans and blowers built in frames 56 and smaller Small motors for air conditioning condensers and evaporator fans Application data Small motors and sump pumps Small motors for gasoline-dispensing pumps Small motors for oil burners Small motors for home laundry equipment Motors and jet pumps Small motors for coolant pumps Index TABLE 8.2 Main Topics in NEMA Standard MG7 Referenced standard, definitions, and safety standards Definitions Control definitions Feedback device definitions Motor definitions Safety standards for construction and guide for selection, installation, and operation of motion-control systems Motion-control systems Construction Motors 5. Strength 6. Vibration resistance and damping E. Motor finish 1. Corrosion resistance 2. Material compatibility F. Operating environment 1. Dust 2. Moisture or vapors 3. Heat (required insulation class) II. Electrical requirements A. Power supply type 1. AC a. Voltage b. Frequency 2. DC—voltage B. Electronic drive requirements 1. Voltage 2. Number of phases 3. Control scheme a. Open loop b. Closed loop c. Vector control d. Pulse width modulation e. Phase locked loop f. Linear amplifier g. Pulse frequency modulation h. Silicon controlled rectifier (SCR) controls C. Electromagnetic interference (EMI) or electromagnetic compatibility (EMC) requirements III. Output requirements A. Duty cycle 1. Continuous 2. Intermittent duty—time on, time off, number of cycles B. Rated load torque and speed C. Load acceleration characteristics D. Velocity profile The preceding requirements must be determined before selecting a proper motor that will meet the application requirements. Next, a profile of the load characteris- tics needs to be determined. These characteristics very widely. A given load profile may have a wide requirement range as soon as such things as life, temperature, and humidity vary. For example, if one is to apply a motor to a fan load, the fan charac- teristics must be determined.Typically, a fan is required to move a certain volume of air.That is to say,it must move so many cubic feet per minute (cfm). Figure 8.1 shows a range of cfm values for a fan superimposed on a multispeed motor speed-torque curve. The solid lines show the motor speed-torque curves at high, medium, and low speeds.The dotted lines show fan performance under changing conditions of system pressure.These curves can be obtained from the fan manufacturer.When applying a motor to such a system, the nominal curve is applied to the motor such that it pro- duces the proper volume of airflow at each of the selected speeds. Next, the operat- ing temperature of the motor must be estimated and the performance curves APPLICATION OF MOTORS 8.3 adjusted accordingly. As the temperature of an ac motor increases, motor torque at a given speed generally decreases.The fan curves are again superimposed on the hot speed-torque curves to ensure that the air delivery stays within reasonable limits. The same approach is used with dc motors. If the load is a pump or some device with high starting-torque requirements, this superimposition of load curves must also include the region at and near the stalled- rotor condition. Some ac motors have very weak low-speed curves or high third-harmonic cusp components,such as is shown in Fig. 8.2.In this case it is possible for the fan or pump load to reach the cusp area and remain at that speed because sufficient motor torque does not exist to accelerate the load through the cusp region. 8.2 VELOCITY PROFILES Many devices require that the motor adapt to a changing load, or change speed within a certain time, or stop and reverse direction. During acceleration and decel- eration, additional torque in excess of the normal running torque is required to over- come the motor and system inertia. The torques and currents necessary to accomplish these moves or changes may generally be determined as follows for dc motors having linear speed, torque, and current characteristics. 8.4 CHAPTER EIGHT FIGURE 8.1 Three-speed ac motor with superimposed fan curves. 8.2.1 Acceleration—Motor Subjected to a Step Voltage To find the speed S at a known time t S = S nl ΂ 1 − ΃΄ 1 − exp ΂ − ΃΅ where S nl = no-load speed T L = load torque, oz ⋅ in T st = motor stall torque, oz ⋅ in T m = motor mechanical time constant t = time, s Alternate method: S = ΄ (E t − V br ) − T L ΅΄ 1 − exp ΂΃΅ where E t = terminal voltage V br = brush voltage drop K d = zero source impedance damping coefficient (leads shorted), oz ⋅ in ⋅ s J t = rotor plus load inertia, oz ⋅ in ⋅ s K t = torque constant, (oz ⋅ in)/A (V br is subtracted only in the case of a brush-type dc motor.) K d ᎏ J t K t ᎏ R t 1 ᎏᎏ 0.1047K d t ᎏ T m T L ᎏ T st APPLICATION OF MOTORS 8.5 FIGURE 8.2 Motor speed-torque curve showing excessive third harmonic cusp. To find the rate of change in speed at a known time t: = ΂ 1 − ΃ exp ΂ − ΃ To find the number of revolutions R v at a known time t: R v = ΂ 1 − ΃Ά t − T m ΄ 1 − exp ΂ − ΃΅· 8.2.2 Acceleration—Motor Subjected to a Ramped Voltage To find the speed at a known time t: S = Ά t − T m ΄ 1 − exp ΂ − ΃΅· where K b = back-emf constant,V/krpm E s = slope of ramp voltage,V/s To find the rate of change in speed at a known time t: = ΄ 1 − exp ΂ − ΃΅ To find the number of revolutions R v at a known time t: R v = ΂ − T m Ά t − T m ΄ 1 − exp ΂ − ΃΅·΃ 8.2.3 Deceleration—Leads Open-Circuited or Short-Circuited To find the speed at a known time t: S = where T ft = motor friction plus load friction, oz ⋅ in K x = K f (leads open-circuited) K x = K d + K f (leads short-circuited) K f = infinite source impendence (leads open-circuited) damping coeffi- cient, oz ⋅ in ⋅ s S o = speed, rpm, at time t = o To find the rate of change in speed at a known time t: =− ΄΅ exp ΂ − t ΃ K x ᎏ J r T ft + (0.0147 K x S o ) ᎏᎏᎏ 0.1047 J t dS ᎏ dt (T ft + 0.1047 K x S o ) exp ΂ − ᎏ K J t x ᎏ t ΃ − T ft ᎏᎏᎏᎏ 0.1047 K x t ᎏ T m t 2 ᎏ 2 E s ᎏᎏ (0.001)(60)K b t ᎏ T m E s ᎏ 0.001K b dS ᎏ dt t ᎏ T m E s ᎏ 0.001K b t ᎏ T m T L ᎏ T st S nl ᎏ 60 t ᎏ T m T L ᎏ T st S nl ᎏ T m dS ᎏ dt 8.6 CHAPTER EIGHT Time to stop t s ,s: t s = ln ΂ 1 + ΃ To find the number of revolutions R v at a known time t to a maximum time to stop t a : R v = ΂ + S o ΃΄ 1 − exp ΂ − ts ΃΅ − 8.2.4 Deceleration—Plugging Step Voltage (Reversal of Voltage Polarity) S =− ΄ 1 − exp ΂ − ΃΅ + S o exp ΂ − ΃ where E tr = voltage of opposite polarity but not necessarily = E f R t = terminal resistance, Ω Alternate method: S =− ΄ + T ft ΅΄ 1 − exp ΂ − ΃΅ + S o exp ΂ − t ΃ Time to stop t s : t s = T m ln ΂ 1 + ΃ Alternate method: t s = T m ln Ά 1 + ΄΅· A method of linear approximation to determine torque current and heating may be used in many cases.The example that is shown next may be modified to meet any particular problem. Note that it does not take thermal capacitance into account. In some situations thermal capacitance may be significant. See Sec. 8.4 for a method that does account for thermal capacitance. Define a varying load and determine its effect on the heating of the motor. Assume the load and velocity parameters specified here. Refer to the velocity pro- file shown in Fig. 8.3. Using motor parameters from available data: K t = 9.80 (oz ⋅ in)/A J a = 2.32 × 10 −3 oz ⋅ in ⋅ s 2 R t = 2.51 T fi = 0.85 oz ⋅ in R th = 5.50°C/W where R th is the motor thermal resistance. 0.1047 K d S o ᎏᎏ (E tr K t /R t ) + T ft 0.001K b K t S o ᎏᎏ E tr K t + T ft R t K d ᎏ J t t ᎏ T m E tr K t ᎏ R t 1 ᎏᎏ 0.1047K d t ᎏ T m t ᎏ T m E tr K t + T ft R t ᎏᎏ 0.001K t K b T ft t s ᎏᎏ (60)(0.1047)K x K x ᎏ J t T ft ᎏᎏ 0.1047K x J t ᎏ 60K x 0.1047K x S o ᎏᎏ T ft J t ᎏ K x APPLICATION OF MOTORS 8.7 Calculate the required torque for each section of the velocity profile. For section 1, the torque to accelerate must be included. T s1 = T fi + T L + J a α+Jlα α= = = =3142 rad/s 2 T s1 = 0.85 + 10.0 + (2.32 × 10 −3 ) (3142) + (4.32 × 10 −3 ) (3142) = 0.85 + 10.0 + 7.29 + 13.57 = 31.71 oz ⋅ in I s1 == =3.235 A W L,s1 = I 2 s1 R t = (3.235) 2 (2.51) = 26.26 W Find the energy Q s1 . Q s1 = W L,s1 t s1 = (26.26) (0.05) = 1.313 W ⋅ s For section 2, only the friction torque is required. T s2 = T fi + T L = 0.85 + 10.0 = 10.85 oz ⋅ in I s2 == =1.107 A W L,s2 = I 2 s2 R t = (1.107) 2 (2.51) = 3.08 W Q s2 = W L,s2 (t 2 − t 1 ) = (3.08) (0.15 − 0.05) = 0.308 W ⋅ s For section 3,the inertia must be included, but here the frictional torques assist in the deceleration. T s3 = T fi + T L + J a α+J 1 α α= = = =−3142 rad/s 2 −157 ᎏ 0.05 0 − 157 ᎏᎏ 0.2 − 0.15 v 0 − v 1 ᎏ t 3 − t 2 10.85 ᎏ 9.80 T s2 ᎏ K t 31.71 oz ⋅ in ᎏᎏ 9.80 (oz ⋅ in)/A T s1 ᎏ K t 157 rad/s ᎏᎏ 0.05 s 152 − 0 ᎏ 0.05 − 0 v 1 − v 0 ᎏ t 1 − t 0 8.8 CHAPTER EIGHT FIGURE 8.3 Velocity profile. T s3 = 0.85 + 10.0 + (−3142) (2.32 10 −3 ) + (−3142) (4.32 10 −3 ) = 10.85 − 7.29 − 13.57 T s3 =−10.01 oz ⋅ in This just means that the power supply must produce 10.01 oz ⋅ in of torque in the other direction to get the motor to stop in the required time. I s3 == =−1.021 A W s3 = I 2 R t = (−1.021) 2 2.51 = 2.62 W Q s3 = W s3 t s3 = (2.62) (0.05) = 0.131 W ⋅ s For section 4, it is obvious that the torque, losses, and energy are zero. The total energy is Q t = Q s1 + Q s2 + Q s3 + Q s4 = 1.31 + 0.308 + 0.131 + 0.0 Q t = 1.749 W ⋅ s Average power dissipated (lost) P L,av or W L,av is: W L,av == =5.83 W Temperature rise = W L,av R th = (5.83 W) × 5.5°C/W = 32°C average rise Is this a good choice of motor? Assume that the motor has Class B insulation. Then the maximum operating temperature of the copper wire is 130°C.Assume there is a hot spot inside the copper windings that is 10°C hotter than the outside of the windings, just to be safe. Now the maximum temperature becomes 130°C − 10°C = 120°C. If the ambient temperature of the device is 40°C,then you are allowed a max- imum temperature rise of 120°C − 40°C = 80°C. You calculated a 32°C rise under your load profile.You could run the motor 48°C hotter. Therefore, you have chosen more motor than you need. Select a smaller motor. 8.3 CURRENT DENSITY The ultimate goal of motor selection is to find the most cost-effective motor that will meet the performance and service life objectives of the application. This must be done while keeping the motor from exceeding its thermal limits. Motors for many applications are selected by the methods described previously, and while the average losses cause heating that does not exceed the insulation class of the motor, yet fail- ures from apparent overheating occur.Typically, these motors exhibit charred wind- ing insulation with shorted turns. However, the rest of the motor does not appear to have seen excessive heating. 1.749 ᎏ 0.3 Q t ᎏ t 4 −10.01 oz ⋅ in ᎏᎏ 9.80 (oz ⋅ in)/A T s3 ᎏ K t APPLICATION OF MOTORS 8.9 In many cases, the failure occurs when the motor sees high currents for short durations, such as on acceleration, deceleration, plug reversal, or momentary over- loads.When the current densities are calculated under these conditions,they exceed the range of the normal running conditions, which are shown in Table 8.3. Keep in mind that these current densities are guidelines and may not be appropriate in all cases and that motor temperature must be kept within the insulation class of the motor. These densities are commonly exceeded, but the wire insulation grade must be capable of handling the heating. 8.10 CHAPTER EIGHT TABLE 8.3 Current Densities Motor type Density range, A/in 2 Enclosed 2000–4000 Ventilated 6000–8000 These high currents force the electrons to the outer surface of the conductor, causing localized excessive heating at the magnet wire insulation and copper junc- tion. This results in insulation degradation and failure. The insulation in these spots is likely to flake off, resulting in shorted turns. The solution to this problem is to reduce the current density by increasing the diameter of the magnet wire being used.There are times when the motor’s slot fill is already too high to allow for a larger wire. In this case, the higher-temperature magnet-wire insulation may be adjusted to solve the problem. 8.4 THERMAL ANALYSIS FOR A PMDC MOTOR* This section develops an approximate equivalent electrical circuit analog for the thermal characteristics of the permanent-magnet dc (PMDC) motor.Surely it can be understood why the exact equivalent cannot be found with the many variable parameters included; however, the circuit analog will assist in the thermal analysis of the motor and will indicate where problems may exist or improvements may be made. The heat balance in any thermal system states that the heat added to a system must equal the heat transmitted from the system plus the heat stored in the system. In the steady-state mode, the heat transmitted from the system is equal to the heat added to the system, so that the heat stored is a constant; hence, the temperature of the system is a constant. The question then, is whether this is a safe operating tem- perature in terms of the system components. In the case of a motor, is this a safe operating condition—for example, is it within the safe temperature insulation class for the motor windings? A lumped parameter electric circuit analog is used here; however, we must be aware that in order to accurately represent the motor, a distributed parameter model would have to be used. By using lumped parameters, the model parameters are characterized such that materials that offer resistance to heat flow have negligi- * Sections 8.4 to 8.8 contributed by Earl F. Richards. [...]... measurement of radio disturbance characteristics of broadcast receivers and associated equipment EN 55014: 1986 CISPR 14 (1985) ed 2: Limits and methods of measurement of radio interference characteristics of household electrical appliances, portable tools, and similar electrical apparatus EN 55 015: 1986 CISPR 15 (1985) ed 3: Limits and measurement of radio interference characteristics of fluorescent... Dissipation of heat from the surface of a 25-in-diameter rotor with forced axial ventilation through a 0.5-in air gap (From Packer, 1951.) 8.23 FIGURE 8 .15 1951.) Dissipation of heat from rotor surface (From Packer, FIGURE 8.16 Dissipation of heat from the surface of a 25-in-diameter 36in-long rotor (From Packer, 1951.) 8.24 APPLICATION OF MOTORS FIGURE 8.17 Dissipation of heat from the surface of a 25-in-diameter... thickness = 0.4 0.4 R3 = ᎏᎏ = 0.89 (0. 015) (29.7) Thermal conductance of end caps R4: L R4 = ᎏ kA where L = thickness of end caps = 0.25 A = area of end caps = 2 × π(3.875)2/4 × 0.6 = 14.14 in2 0.25 R4 = ᎏᎏ = 0.003 (5.41)(14.14) (8.12) APPLICATION OF MOTORS 8.27 Conductance of shell not covered by magnets R5: L R5 = ᎏ kA (8.13) where L = 0.212 in A = π (diameter of shell) (length) − (area covered by... ft/min Length of channel = 20 in 0.02–0.08 W/ (in2 ⋅ °C) 1200–5000 ft/min TABLE 8.8 Typical Values of Cooling Coefficient Part Cylindrical surface of dc armature c 0. 015 to 0.035 ᎏᎏ 1 + 0 1u u Remarks Armature peripheral speed Smaller values for large, open machines Lower values for forced cooling Cylindrical surfaces of stator and rotor 0.03 to 0.05 ᎏᎏ 1 + 0 1u Relative peripheral speed Back of stator... low-voltage electrical installations in the frequency range 3–148.5 kHZ, Part 1: General requirements, frequency bands, and electromagnetic disturbances EN 55011: 1989 CISPR 11 (1990) ed 2: Limits and methods of measurement of radio disturbance characteristics of industrial, scientific, and medical (ISM) radiofrequency equipment EN 55013: 1988 CISPR 13 (1975) ed 1 + Amdt 1 (1983): Limits and methods of measurement... Convection of air gap to end cap R15: 1 R15 = ᎏ hA (8.18) where A = 14.4 in2 (from R4) 1 R15 = ᎏᎏ = 1.98 0.035(14.4) Convection of air gap to shell not covered by magnet R16: 1 R16 = ᎏ hA (8.19) where A = 26.3 1 R16 = ᎏᎏ = 0.51 0.075(26.3) convection of shell covered by magnets to ambient air R17: 1 R17 = ᎏ hA (8.20) where A = (3.8) (8.8) (0.75) = 78.8 in2 1 R17 = ᎏᎏ = 1.58 0.008(7877) Convection of end... R= ᎏ a where l = length of path (conductor) a = cross-sectional area (area of conductor) p = electrical resistivity constant In most cases, motors will be composed of many different materials whose thermal resistances will be different Also, in most cases there will be multiple thermal paths, and series and parallel arrangements will be required Thermal bonding between surfaces of different materials... Ventilating ducts in core 0. 015 to 0.025 ᎏᎏ 1 + 0.1u 0.08 to 0.2 ᎏᎏ u Based on total coil surface Based on total coil surface Based on exposed coil surface only Commutator peripheral speed Surface includes a proportion for risers Air velocity in ducts Taken as about 10 percent of peripheral speed of core 8.19 APPLICATION OF MOTORS TABLE 8.9 Weights and Thermal Properties of Materials Pe lb/in3 Material... constant in seconds This will give a form of Fig 8.22 5 RTH CTH = time constant, s time constant ∴ CTH = ᎏᎏ RTH J/C° or (W ⋅ s)/C° 8.5 SUMMARY OF MOTOR CHARACTERISTICS AND TYPICAL APPLICATIONS The most common motors, their characteristics, and their typical applications are listed in Tables 8.11 and 8.12 Keep in mind that there are thousands of variations of motors and applications, and only general... current I flowing in a wire sets up an electric field as well as a magnetic field The electric field E (Fig 8.23) is perpendicular to the magnetic field and is along the length of the wire If a pair of conductors carrying time-varying current of the same TABLE 8.13 European Harmonized Emission Standards Standard Description EN 50081-1: 1991 EMC generic emission standard, Part 1: Residential, commercial, . fans Application data Small motors and sump pumps Small motors for gasoline-dispensing pumps Small motors for oil burners Small motors for home laundry equipment Motors and jet pumps Small motors for coolant. hermetic refrigeration compres- sors Small motors for shaft-mounted fans and blowers Small motors for belted fans and blowers built in frames 56 and smaller Small motors for air conditioning condensers and. testing Complete machines and parts Classification of insulation systems DC motors and generators AC motors and generators AC generators and synchronous motors Single-phase motors Motors for hermetic refrigeration