Motor formula
Useful electric motor formula
Motor formula
Useful electric motor formulas.
Mechanical Formulas.
| Torque in lb.ft. = | HP x 5250 rpm |
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HP = | Torque x rpm 5250 |
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rpm = | 120 x Frequency No. of Poles |
| To Find | Alternating Current | |
|---|---|---|
| Single- Phase | Three-Phase | |
| 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 E |
Kva x 1000 1.73 x E |
| Kilowatts | I x E x pf 1000 |
1.73 x I x E x pf 1000 |
| Kva | I x E 1000 |
1.73 x I x E 1000 |
| Horsepower = (Output) | I x E x Eff x pf 746 |
1.73 x I x E x Eff x pff 746 |
Symbols
| 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 |
Basic Horsepower Calculations
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:
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For vertical or hoisting motion:
| HP = | W x S 33,000 x E |
Where:
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For fans and blowers:
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Or
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Or
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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. |
Equivalent Inertia motor formulas
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:
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2This 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 | |||||||
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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:
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Note: reducer RPM = Load RPM
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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:
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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:
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Finally:
| 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.
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