SME
Power Systems
Transformers and machines
Use this for “A 4-pole, 60 Hz induction motor runs at 1746 rpm. Find the slip and rotor frequency.”
Chart
This is an example. Type your own values to start a new problem.
nₛ (rpm)
=
120
×
f (Hz)
÷
p
s
=
(
nₛ (rpm)
−
n (rpm)
)
÷
nₛ (rpm)
fᵣ (Hz)
=
s
×
f (Hz)
Pag
Air-gap power
example
Pcu
Rotor copper loss
calculated
Pconv
Converted power
calculated
T
Torque
calculated · N·m
Formulas
nₛ
=
120
×
f
p
1,800
=
120
×
60
4
s
=
nₛ −
n
nₛ
0.03
=
1,800 − 1,746
1,800
fᵣ
=
s
×
f
1.8 = 0.03 × 60
Pcu
=
s
×
Pag
300 = 0.03 × 10,000
Pconv
=
(1
−
s
)
×
Pag
9,700 = (1 − 0.03) × 10,000
T
=
Pconv
2π ×
n
÷ 60
53.0516
=
9,700
2π × 1,746 ÷ 60
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
Steady state, motoring: the rotor turns slower than the field, so the slip is between 0 and 1.
•
p counts poles, not pole pairs; the field turns at nₛ = 120f ÷ p in rpm.
•
Of the air-gap power, the share s is lost in the rotor copper and the rest is converted.
•
Friction and windage are left out of T, so the shaft torque is a little less.
Step-by-step
WE KNOW
f = 60 Hz, p = 4, n = 1,746 rpm, Pag = 10 kW
FIND
synchronous speed (nₛ), slip (s), rotor frequency (fᵣ), rotor copper loss (Pcu), converted power (Pconv), torque (T)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to W).
Pag = 10 kW = 10,000 W (1 kW = 1000 W)
Step 2 · Find synchronous speed (nₛ)
Use nₛ = 120 × f ÷ p
The field turns once per pole pair each cycle; 60 s a minute and 2 poles a pair make 120.
nₛ
=
120
×
60
4
nₛ
=
7,200
4
nₛ = 1,800 rpm
Step 3 · Find slip (s)
Use s = (nₛ − n) ÷ nₛ
Slip
is
how
far
the
rotor
falls
behind
the
field,
as
a
share
of
the
field’
s
speed.
s
=
1,800 − 1,746
1,800
s
=
54
1,800
s
=
0.03
Step 4 · Find rotor frequency (fᵣ)
Use fᵣ = s × f
The
rotor
slips
past
the
field
at
s
×
nₛ,
so
its
currents
run
at
s
×
f
.
fᵣ = 0.03 × 60
fᵣ = 1.8 Hz
Step 5 · Find rotor copper loss (Pcu)
Use Pcu = s × Pag
The
rotor
copper
takes
the
share
s
of
the
power
crossing
the
gap.
Pcu = 0.03 × 10,000
Pcu = 300 W
Step 6 · Find converted power (Pconv)
Use Pconv = (1 − s) × Pag
What the rotor copper doesn’t take is turned into mechanical power.
Pconv = (1 − 0.03) × 10,000
Pconv = 0.97 × 10,000
Pconv = 9,700 W
Step 7 · Find torque (T)
Use T = Pconv ÷ (2π × n ÷ 60)
Power
is
torque
times
angular
speed,
and
2π
n
60
turns
rpm
into
rad/
s
.
T
=
9,700
2π × 1,746 ÷ 60
T
=
9,700
182.841
T
=
53.0516
N·m
Step 8 · Convert the answers
Change each answer back to the unit you chose.
Pcu = 300 W = 0.3 kW (1 kW = 1000 W)
Pconv = 9,700 W = 9.7 kW (1 kW = 1000 W)
Check
Put the numbers back in. Both sides should match.
1,800
=
120
×
60
4
✓
0.03
=
1,800 − 1,746
1,800
✓
1.8 = 0.03 × 60 ✓
300 = 0.03 × 10,000 ✓
9,700 = (1 − 0.03) × 10,000 ✓
53.0516
=
9,700
2π × 1,746 ÷ 60
✓
Topic check
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