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MOTOR FORMULAS AND CALCULATIONS
|The formulas and calculations which appear below should be used for estimating purposes only. It is the responsibility of the customer to specify the required motor Hp, Torque and accelerating time for his application. The salesman may wish to check the customers specified values with the formulas in this section, however, if there is serious doubt concerning the customers application or if the customer requires guaranteed motor/application performance, the customer should engage an electrical engineer to do the exact calculations. |
For a detailed explanation of each formula, Click on the links below to go right to it.
Deg C = (Deg F - 32) x 5/9
Temperature Conversion Formula
|t =||WK2 x rpm|
308 x T av.
|-----||WK2 = inertia in lb.ft.2|
t = accelerating time in sec.
T = Av. accelerating torque lb.ft.
|T =||WK2 x rpm|
308 x t
|inertia reflected to motor = Load Inertia||Load rpm|
|ns =||120 x f|
|-----||f =||P x ns|
|-----||P =||120 x f|
|HP =||T x n|
|-----||T =||5250 HP|
|-----||n =||5250 HP|
|% Slip =||ns - n|
|D||4.0 -4.49||I||7.1 -7.99||P||12.5-13.99||V||22.4 & Up|
|E||4.5 -4.99||K||8.0 -8.99||R||14.0-15.99|
|I||=||current in amperes|
|E||=||voltage in volts|
|kW||=||power in kilowatts|
|kVA||=||apparent power in kilo-volt-amperes|
|HP||=||output power in horsepower|
|n||=||motor speed in revolutions per minute (RPM)|
|ns||=||synchronous speed in revolutions per minute (RPM)|
|P||=||number of poles|
|f||=||frequency in cycles per second (CPS)|
|T||=||torque in pound-feet|
|EFF||=||efficiency as a decimal|
|PF||=||power factor as a decimal|
In mechanical systems, all rotating parts do not usually operate at the same speed. Thus, we need to determine the "equivalent inertia" of each moving part at a particular speed of the prime mover.
The total equivalent WK2 for a system is the sum of the WK2 of each part, referenced to prime mover speed.
The equation says:
This equation becomes a common denominator on which other calculations can be based. For variable-speed devices, inertia should be calculated first at low speed.
Let's look at a simple system which has a prime mover (PM), a reducer and a load.
|WK2 = 100 lb.ft.2||WK2 = 900 lb.ft.2|
(as seen at output shaft)
|WK2 = 27,000 lb.ft.2|
The formula states that the system WK2 equivalent is equal to the sum of WK2parts at the prime mover's RPM, or in this case:
Note: reducer RPM = Load RPM
The WK2 equivalent is equal to the WK2 of the prime mover, plus the WK2 of the load. This is equal to the WK2 of the prime mover, plus the WK2 of the reducer times (1/3)2, plus the WK2 of the load times (1/3)2.
This relationship of the reducer to the driven load is expressed by the formula given earlier:
In other words, when a part is rotating at a speed (N) different from the prime mover, the WK2EQ is equal to the WK2 of the part's speed ratio squared.
In the example, the result can be obtained as follows:
The WK2 equivalent is equal to:
|WK2EQ = lb.ft.2pm + 100 lb.ft.2Red + 3,000 lb.ft2Load |
WK2EQ = 3200 lb.ft.2
The total WK2 equivalent is that WK2 seen by the prime mover at its speed.
|To Find||Alternating Current|
|Amperes when horsepower is known||HP x 746|
E x Eff x pf
|HP x 746|
1.73 x E x Eff x pf
|Amperes when kilowatts are known||kW x 1000|
E x pf
|kW x 1000|
1.73 x E x pf
|Amperes when kVA are known||kVA x 1000|
|kVA x 1000|
1.73 x E
|Kilowatts||I x E x pf|
|1.73 x I x E x pf|
|kVA||I x E|
|1.73 x I x E|
|Horsepower = (Output)||I x E x Eff x pf|
|1.73 x I x E x Eff x pff|
I = Amperes; E = Volts; Eff = Efficiency; pf = Power Factor; kVA = Kilovolt-amperes; kW = Kilowatts
|Three Phase: IL =||577 x HP x kVA/HP|
|See: kVA/HP Chart|
|Single Phase: IL =||1000 x HP x kVA/HP|
|EXAMPLE:||Motor nameplate indicates 10 HP, 3 Phase, 460 Volts, Code F. |
|IL @ ELINE = IL @ EN/P x||ELINE|
|EXAMPLE:||Motor has a locked rotor current (inrush of 100 Amperes (IL) at the rated nameplate voltage (EN/P) of 230 volts. |
What is IL with 245 volts (ELINE) applied to this motor?
IL @ 245 V. = 100 x 254V/230V
IL @ 245V. = 107 Amperes
Horsepower is work done per unit of time. One HP equals 33,000 ft-lb of work per minute. When work is done by a source of torque (T) to produce (M) rotations about an axis, the work done is:
|radius x 2 x rpm x lb. or 2 TM|
When rotation is at the rate N rpm, the HP delivered is:
For vertical or hoisting motion:
|HP =||W x S|
33,000 x E
For fans and blowers:
For purpose of estimating, the eff. of a fan or blower may be assumed to be 0.65.
|Note:||Air Capacity (cfm) varies directly with fan speed. Developed Pressure varies with square of fan speed. Hp varies with cube of fan speed.|
For estimating, pump efficiency may be assumed at 0.70.
The equivalent inertia of an adjustable speed drive indicates the energy required to keep the system running. However, starting or accelerating the system requires extra energy.
The torque required to accelerate a body is equal to the WK2 of the body, times the change in RPM, divided by 308 times the interval (in seconds) in which this acceleration takes place:
The constant (308) is derived by transferring linear motion to angular motion, and considering acceleration due to gravity. If, for example, we have simply a prime mover and a load with no speed adjustment:
|WK2 = 200 lb.ft.2||WK2 = 800 lb.ft.2|
The WK2EQ is determined as before:
|WK2EQ = WK2pm + WK2Load|
|WK2EQ = 200 + 800|
|WK2EQ = 1000 ft.lb.2|
If we want to accelerate this load to 1800 RPM in 1 minute, enough information is available to find the amount of torque necessary to accelerate the load.
The formula states:
In other words, 97.4 lb.ft. of torque must be applied to get this load turning at 1800 RPM, in 60 seconds.
Note that TAcc is an average value of accelerating torque during the speed change under consideration. If a more accurate calculation is desired, the following example may be helpful.
The time that it takes to accelerate an induction motor from one speed to another may be found from the following equation:
The Application of the above formula will now be considered by means of an example. Figure A shows the speed-torque curves of a squirrel-cage induction motor and a blower which it drives. At any speed of the blower, the difference between the torque which the motor can deliver at its shaft and the torque required by the blower is the torque available for acceleration. Reference to Figure A shows that the accelerating torque may vary greatly with speed. When the speed-torque curves for the motor and blower intersect there is no torque available for acceleration. The motor then drives the blower at constant speed and just delivers the torque required by the load.
In order to find the total time required to accelerate the motor and blower, the area between the motor speed-torque curve and the blower speed-torque curve is divided into strips, the ends of which approximate straight lines. Each strip corresponds to a speed increment which takes place within a definite time interval. The solid horizontal lines in Figure A represent the boundaries of strips; the lengths of the broken lines the average accelerating torques for the selected speed intervals. In order to calculate the total acceleration time for the motor and the direct-coupled blower it is necessary to find the time required to accelerate the motor from the beginning of one speed interval to the beginning of the next interval and add up the incremental times for all intervals to arrive at the total acceleration time. If the WR2 of the motor whose speed-torque curve is given in Figure A is 3.26 ft.lb.2 and the WR2 of the blower referred to the motor shaft is 15 ft.lb.2, the total WR2 is:
|15 + 3.26 = 18.26 ft.lb.2,|
And the total time of acceleration is:
|t = 2.75 sec.|
Curves used to determine time required to accelerate induction motor and blower
|T1 = 46 lb.ft.||T4 = 43.8 lb.ft.||T7 = 32.8 lb.ft.|
|T2 = 48 lb.ft.||T5 = 39.8 lb.ft.||T8 = 29.6 lb.ft.|
|T3 = 47 lb.ft.||T6 = 36.4 lb.ft.||T9 = 11 lb.ft.|
Sales Orders are often entered with a note under special features such as:
-----"Suitable for 10 starts per hour"
----"Suitable for 3 reverses per minute"
-----"Motor to be capable of accelerating 350 lb.ft.2"
-----"Suitable for 5 starts and stops per hour"
Orders with notes such as these can not be processed for two reasons.
Obtaining this information and checking with the product group before the order is entered can save much time, expense and correspondence.
Duty cycle refers to the detailed description of a work cycle that repeats in a specific time period. This cycle may include frequent starts, plugging stops, reversals or stalls. These characteristics are usually involved in batch-type processes and may include tumbling barrels, certain cranes, shovels and draglines, dampers, gate- or plow-positioning drives, drawbridges, freight and personnel elevators, press-type extractors, some feeders, presses of certain types, hoists, indexers, boring machines, cinder block machines, key-seating, kneading, car-pulling, shakers (foundry or car), swaging and washing machines, and certain freight and passenger vehicles. The list is not all-inclusive. The drives for these loads must be capable of absorbing the heat generated during the duty cycles. Adequate thermal capacity would be required in slip couplings, clutches or motors to accelerate or plug-stop these drives or to withstand stalls. It is the product of the slip speed and the torque absorbed by the load per unit of time which generates heat in these drive components. All the events which occur during the duty cycle generate heat which the drive components must dissipate.
Because of the complexity of the Duty Cycle Calculations and the extensive engineering data per specific motor design and rating required for the calculations, it is necessary for customer to refer to an electrical engineer for motor sizing with a duty cycle application.
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