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 horsepower, torque, and accelerating time for the given application. The sales person may wish to check the customer’s specified values with the formulas in this section, however, if there is serious doubt concerning the customer’s application or if the customer requires guaranteed motor/application performance, then the next step is to contact your LEESON Electric Sales Office nearest you to further investigate.
Rules of Thumb (Approximation Only and All Values at 100% Load)

At 1800 rpm, a motor develops 3 lb-ft of torque per HP

At 1200 rpm, a motor develops 4.5 lb-ft of torque per HP

At 575V, a 3-Phase motor draws 1 amp per HP

At 460V, a 3-Phase motor draws 1.25 amps per HP

At 230V, a 3-Phase motor draws 2.5 amps per HP

At 230V, a 1-Phase motor draws 5 amps per HP

At 115V, a 1-Phase motor draws 10 amps per HP
Mechanical Formulas
Torque (lb-ft) = (HP x 5252) / rpm HP = (Torque (lb-ft) x rpm) / 5252 rpm = (5252 x HP) / torque (lb-ft) rpm = (120 x HZ) / # of Pole Frequency (HZ) = (# of Poles x rpm) / 120
Slip of Motor
% Slip = ((synch rpm – Full Load rpm) / (synch rpm)) x 100
Electrical Formulas
To Find
Amperes when HP is known
Amperes when KW is known
Amperes when KVA is known
Kilowatts
KVA
HP (Output) when current is known
Alternating Current
Single Phase
(HP x 746) / (E x Eff x PF)
(KW x 1000) / (E x PF)
(KVA x 1000) / E
(I x E x PF) / 1000
(I x E) / 1000
(I x E x Eff x PF) / 746
Three Phase
(HP x 746) / (1.73205 x E x Eff x PF)
(KW x 1000) / (1.73205 x E x PF)
(KVA x 1000) / (1.73205 x E)
(1.73205 x I x E x PF) / 1000
(1.73205 x I x E) / 1000
(1.73205 I x E x Eff x PF) / 746
I = Amperes, E = Volts, Eff = Efficiency (as a decimal), PF = Power Factor (as a decimal)
KVA = Kilovolt-Amperes, KW = Kilowatts
Locked Rotor Current (I
L
) From Nameplate Data
CODE
A
B
C
D
E
F
G
H
J
K
NEMA CODE LETTERS FOR LOCKED ROTOR KVA
KVA/HP
0.00 - 3.14
3.15 - 3.54
3.55 - 3.99
4.00 - 4.49
4.50 - 4.99
5.00 - 5.59
5.60 - 6.29
6.30 - 7.09
7.10 -7.99
8.00 - 8.99
CODE
L
M
N
P
R
S
T
U
V
3-Phase
KVA/HP: (1.73205 x E x LRA) / (1000 x HP) (See Above Table for KVA/HP)
LRA: (KVA/HP x 1000 x HP) / (E * 1.73205)
Note: Square Root of (3) = 1.73205
KVA/HP
9.00 - 9.99
10.00 - 11.19
11.20 - 12.49
12.50 -13.99
14.00 - 15.99
16.00 - 17.99
18.00 - 19.99
20.00 - 22.39
22.40 - and up

Temperature Conversions
Deg C = (Deg F – 32) x 5/9 or Deg F = (Deg C x 9/5) + 32
Fans and Blowers
1. HP = (CFM x PSF) / (33,000 x mechanical eff. of Fan) or HP = (CFM x Pressure (lbs/(ft) 2 / (33,000 x mechanical eff. of fan)
2. HP = (CFM x PIW) / (6343 x mechanical eff. of Fan)
3. HP = (CFM x PSI) / (229 x mechanical eff. of Fan)
Where:
CFM = Cubic feet per minute
PIW = Head - Inches of water gauge
PSF = Pound per square foot
PSI = Pounds per square inch
For purposes of estimating, the eff. of a fan or blower may be assumed to be 65%.
Note:

Air capacity (CFM) varies directly with the fan speed.

Developed pressure varies with the square of fan speed.

HP varies with cube of fan speed.
Effect of Speed on HP (Fan)
Note: HP consumption varies as the 3 rd power (cubed) of the speed.
Centrifugal Applications Affinity Laws
Where:
Pres = Pressure
RPM = Revolutions per minute
Pumps
HP = (GPM x Pressure (lbs/(in) 2 x Specific Gravity) / (1713 x mechanical eff. of pump)
Or
HP = (GPM x Total Dynamic Head in Feet x Specific Gravity) / (3960 x mechanical eff. of pump)
Where:
Total dynamic head = static head + friction head
For estimating purposes, pump eff. may be assumed at 70% (best to try and obtain correct value)
Displacement and Centrifugal Pumps
HP = GPM x head in feet / (3960 x efficiency)
1 cu. foot per sec = 448.8 GPM
1 lb per sq. inch pressure = a head of 2309 ft of water weighing 62.36 lbs per cubic foot at 62

F
Displacement Pumps
Effect of Speed on HP

Capacity and required HP vary directly as the speed.

Displacement pumps under constant head require approximately constant torque at all speeds.
Displacement Pumps Efficiency
Displacement pumps may vary between 50% and 80% eff. Depending on the size of pumps

Centrifugal Pumps
Effect of Speed on HP
HP consumption varies approximately as the 3 rd power of the speed.
Efficiency

500 to 1000 gal. per min = 70% to 75% eff.

1000 to 1500 gal. per min = 75% to 80% eff.

Larger than 1500 gal. per min = 80% to 85% eff.
Conveyors
Vertical HP = (W x V) / 33,000
Horizontal HP = (W x V x N) / 33,000
Where:
W = Weight in lbs
V = Velocity in feet/min
N = Coefficient of friction
Vertical or Hoisting Motion
HP = (W x S) / (33,000 x E)
Where:
W = total weight in lbs. to be raised by motor
S = Hoisting speed in feet per min
E = Overall mechanical eff. of hoist and gearing. For estimating purposes only, E=65% for eff. of hoist and connected gear
• kW = HP x 0.746
• Wk 2 referred to motor shaft speed = [driven machine WK 2 x ((driven machine rpm)/motor rpm) 2 + gear WK 2 at motor
• Conversion factors: CV = (metric hp) = 735.5 watts = 75 kg-m/sec Wk2 (lb-ft) = 5.93 x GD2 (kg-m2)
• Ventilating - air requirements: 100-125 cfm of 40°C air at 1/2-in. water pressure for each kW of loss
Equivalent Inertia (Consult your local LEESON Electric sales office)
In mechanical systems, all rotating parts do not usually operate at the same speed. This means it will be necessary to determine the equivalent inertia of each moving (revolving body) part at a particular speed. It is also convenient to reference the inertia of each part to the speed of the prime mover.
Hence, the total equivalent WK2 for a system is the sum of the WK2 of each part, referenced to the prime mover speed.
WK 2
EQ
= WK 2 part
x (N part
/ N prime mover
) 2
This equation becomes the common denominator on which other calculations can be based. For VFD devices, inertia should be calculated first at low speed.
Accelerating Torque (Consult your local LEESON Electric sales office)
The equivalent inertia of a VFD 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 multiplied by the change in RPM, divided by a constant of 308 multiplied by (t) time in seconds in which this acceleration takes place.
T acc
(acceleration torque) = WK 2 x N / (308 x t) (units in lb-ft)
Where:
N = change in rpm W = weight in (lbs)
K = radius of gyration WK2 = equivalent inertia
308 = constant of proportionality

Accelerating Horsepower (Consult your local LEESON Electric sales office)
Now that the acceleration torque (Tacc) is known/calculated, the power needed to accelerate the load can be determined. Using the horsepower formula:
HP = (Torque (lb-ft) x rpm) / 5252
Now refine the equation above:
HP acc
= T acc
x N / 5252 now further:
HP acc
= (WK 2
EQ
x N) / (308 x t) x N / 5252 or HP acc
= (WK 2
EQ
x N 2 ) / (1.62 x t x 10 6 )