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AMAZON multi-meters discounts AMAZON oscilloscope discounts Machine Load No electric motor drive, fixed or variable speed, can be correctly specified without knowing something about the machine that is to be driven, specifically the machine load. For fixed speed drives, it’s often thought to be sufficient to specify only the power requirement in kW at the rated speed. On larger drives, motor manufacturers usually ask for more information about the load, such as the moment of inertia, to ensure that the design of the motor can cope with the acceleration requirements. In the case of AC variable speed drives, more details about the load characteristics are always necessary. The output torque of an AC VSD is considered to be adequate when it:
The selection procedure outlined below applies mainly to single motor AC variable speed drives without special requirements and where the drive is continuous after an initial acceleration period. This means that a standard TEFC squirrel cage induction motor and a standard AC VVVF converter can be used. Multi-motor VSDs and other special applications require further investigation and will be considered in later sections. In general, special applications should be referred to the manufacturers of motors and/or AC converters for special rating calculations. There are a great variety of different types of machine loads that are commonly driven by VSDs, each with different characteristics of torque, inertia, etc. Examples are pumps, fans, crushers, compressors, conveyers, agitators, etc. For VSD applications, most of what needs to be known about the machine load can be covered by the following:
Load Torque The torque required by the driven machine determines the size of the motor because the continuous rated torque of the motor must always be larger than the torque required by the driven machine. The magnitude of the load torque determines the cost of the motor because, as a rule of thumb, the cost of an electric motor is approximately proportional to its rated output torque (not its rated power!). The load torque is not necessarily a fixed value. It can vary with respect to speed, position, angle and time. Another important aspect of the load torque is that the figure should apply at the shaft of the motor. When gearboxes, conveyers or hoists are involved, the actual torque at the machine must be converted to torque at the motor shaft. The conversion formulae are given in ++++ 7.5 to convert the load torque, speed and moment of inertia to motor shaft values. ++++ Torque characteristics of typical types of machine loads as a function of speed angle and time. (Note: k = constant) T = Torque in Nm F = Force in N P = Power in kW v = Velocity in m/sec J = Inertia in kg-m^2 ? = Efficiency in p.u. ++++ Formulae to convert load torque, power and moment of inertia to motor shaft values The load requirements are often given as the mechanical absorbed power (PM kW) at a particular speed (n rev/m). The mechanical load torque may then be calculated from the following formula: T_w = [(kW) 9550 x P_M / n(rev/min)] Nm Where TM = Mechanical torque at the motor shaft in Nm PM = Absorbed load power at the motor shaft in kW n = Actual rotational speed of the motor shaft in rev/min Variable torque machine loads Variable torque machine loads are those which exhibit a variable torque over their entire speed range, such as centrifugal pumps and fans. The torque-speed curve for these loads are shown. ++++ Torque-speed characteristic of a variable torque load The use of variable speed drives for the speed control of pumps and fans are the simplest applications and provide the least number of problems. The reason is that the breakaway starting torque is usually very low and then rises with speed. The following are some of the important factors associated with this type of load:
The manufacturers of modern PWM converters have tried to reduce the cost of VSDs for pump and fan applications by providing reduced performance drives with the following features: • Low over-current capability, typically up to 120% for 30 sec. Constant torque machine loads Constant torque machine loads are those which exhibit a constant torque over their entire speed range, such as conveyors, positive displacement pumps, etc. The torque and power curves for these loads are shown. Torque-speed characteristic of a constant torque load. The following are the potential problems when driving constant torque loads from a converter fed electric motor: • The starting torque is theoretically equal to the full speed load torque but, in practice, the real starting torque can be much higher due to the additional requirements of: - breakaway torque - acceleration torque (dynamic torque) • Running for long periods at low speeds can result in motor thermal over-load, if the load torque is above the motor loadability curve. Separate forced cooling may be necessary in some cases. • Running at speeds above the motor base speed could also be a problem, with increased motor slip and a higher possibility of stalling the motor. The manufacturers of modern PWM converters have tried to overcome some of these problems by providing the following features: • High short time over-current capability, typically up to 150% for 60 sec, which is often required during starting • Voltage boost to compensate for stator volt-drop at low frequencies • Providing adequate motor protection to protect the motor from overload - Motor thermal protection which models low speed cooling reduction - Motor thermistor protection inputs Some of these limitations are illustrated in the loadability curve. Speed Range The selection of the correct size of electric motor for a VSD is affected by the speed range within which it’s expected to run continuously. The important factor is that the motor should be able to drive the load continuously at any speed within the speed range without stalling or overheating the motor, i.e. the torque and thermal capacity of the motor must be adequate for all speeds in the speed range, within the loadability limits. Running the motor at below base speeds (f < 50 Hz) with a standard TEFC cage motor has the following effects on the motor: • Reduces the motor cooling because the cooling fan, which is attached to the motor shaft, runs at reduced speed. Therefore, the temperature rise in the motor will tend to be much higher than expected. ++++ an example of the torque-speed curve for a variable speed pump drive, operating in the range from 10 Hz to 50 Hz. Some comments are: • The load torque is well within the loadability limits at all speeds. • The maximum speed is below the base speed of 50 Hz. The speed range should NOT be increased above 50 Hz because the load torque will exceed the loadability limit of the drive. (Load torque increases as the square of the speed.) • Starting torque is low, so there should be no problems with breakaway. • The acceleration torque is high, so the drive can be expected to quickly reach its maximum speed, if fast acceleration is required. However, with pumps, a long acceleration time is normally desirable to prevent water hammer. ++++ Example of speed range and torque curve of a variable speed pump drive when controlled by a PWM-type VVVF converter. Running the motor at above base speeds (f > 50 Hz) with a standard TEFC cage motor has the following effect on the motor: • The air-gap flux is reduced because the V/f ratio is reduced. Consequently, there is a reduction in the output torque capability of the motor. The torque is reduced in proportion to the frequency. The load torque is not permitted to exceed the pullout torque of the motor, even for a short period, otherwise the motor will stall. The maximum torque allowed at above-synchronous speeds depends on the motor characteristics and frequency as follows: … where Tp = Pull-out torque (maximum torque) of the motor in Nm. f = Actual frequency in the above-synchronous range in Hz 0.6 = Factor of safety ++++ an example of the torque-speed curve for a variable speed conveyor drive, operating in a similar range from 10 Hz to 50 Hz. ++++ Example of speed range and torque curve of a variable speed conveyor drive when controlled by a PWM-type VVVF converter. Some comments on this application are: • The load torque falls outside the loadability limits at low speeds below 28 Hz. There could be problems running the motor continuously at speeds below 28 Hz. • Although the maximum speed is below the base speed of 50 Hz, but the speed range could be increased above 50 Hz to take advantage of the loadability characteristic above 50 Hz. (Load torque remains constant with increases in speed.) • Starting torque is high, with a high breakaway, so there may be some problems with breakaway. • Acceleration torque is small, so the drive ramp-up time may have to take place over a long period to avoid exceeding the VSD current limit. Machine load inertia During acceleration and deceleration, the moment of inertia of the load imposes an additional dynamic acceleration torque on the motor. The dynamic acceleration torque is the extra torque that is required to change the kinetic state of the load as it accelerates from one speed to another. The moment of inertia and the required acceleration time together affect the motor torque and consequently the size and cost of the motor. The dynamic acceleration torque TA is calculated as follows: … where dn = Change in speed during acceleration in rev/sec dt = The time it takes to effect the speed change sec J = Moment of inertia of the drive system in kg-m^2 … this can be rewritten as follows, with the speed in rev/min: When running at a constant speed, a motor must deliver a torque corresponding to the machine load torque TL. During acceleration, there is the added requirement for the acceleration torque TA. So the total torque required from the motor must be greater than the sum of the load torque TL and the dynamic torque TA. The motor must be selected to provide this total torque without exceeding its load capacity. Example: A conveyor drive is to be accelerated from zero to a speed of 1500 rev/min in 10 secs. The moment of inertia of the load JL = 4.0 kg-m^2 . The torque of the conveyor load, referred to the motor shaft, is a constant at 520 Nm. The motor being considered is a 110 kW, 1480 rev/m motor with a JM = 1.3 kg-m^2 . Is this motor adequate for this duty? The total moment of inertia of the drive system is ... During acceleration, the dynamic torque required is ... The machine load is a constant torque type with a value given above as ... During acceleration, the motor must supply a total torque T_Tot of ... The rated motor torque may be obtained from the manufacturer's tables or calculated from the rated power as follows: Because TN = T_Tot, the motor is evidently suited for the drive requirements. When the motor drives the mechanical load through a gearbox or pulleys, the inertia of the load must be 'referred' to the motor shaft using the formula given. Where JM = Inertia at the motor shaft JL = Inertia at the load shaft Example: A 5.5 kW motor of rated speed 1430 rev/min and rotor inertia of 0.03 kg-m^2 drives a machine at 715 rev/min via a 2:1 pulley and belt drive. The inertia of the mechanical load is 5.4 kg-m^2, running at 715 rev/min at full rated speed. If ...the load is a constant torque load with an absorbed power of 4.5 kW at 715 rev/min, what is the acceleration time for this drive system from standstill to full load speed of 715 rev/min? Assume that the full motor torque is 150% of rated torque and is constant over the acceleration period. The rated output torque of the motor is given by: The maximum output torque is 150% during the acceleration period ... The absorbed power of the load is 4.5 kW at 715 rev/m, which gives a load torque of ... This needs to be converted to the motor shaft by the pulley ratio ... The acceleration torque is the difference between maximum motor torque and load torque referred to the motor shaft ... The Inertia of the mechanical load referred to motor shaft is .... If a gearbox is used, its inertia of the gearbox itself, referred to the motor, should also be taken into consideration. Therefore, to calculate the overall acceleration time of the drive system, the simple formula above may be applied, provided that the acceleration torque remains constant and the drive accelerates linearly in a uniform time. Where t = Total acceleration time in sec. J-Tot = Moment of inertia of the (motor + load) in kg-m^2 n = Final speed of the drive in rev/min TA = Acceleration torque of the drive system in Nm Assuming that the acceleration torque remains constant over the acceleration period, the minimum acceleration time of the conveyor drive system is: This formula can be used as a rough estimate for acceleration time, but it’s only an approximation because the acceleration torque is seldom a constant value, it’s the difference in two changing values being the motor torque and the load torque. There are two alternative methods of achieving a more accurate result: • Use a computer program to accurately calculate the result. This technique is used by large engineering companies, motor manufacturers and vendors. • Use a manual graphical system to calculate the acceleration time. The first step in calculating the acceleration torque is to clearly define the motor and the load torque-speed characteristics. The motor torque-speed curve is usually available from the manufacturer. If this is not available, important points on the curve are usually given in the motor manufacturer's catalogue in the form of the 4 points given below. • Starting breakaway torque • Pull-up torque and speed • Pull-out torque (or breakdown torque) and speed • Rated full-load torque and speed. |
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