SME
Manufacturing Processes · Problem type
Cutting power and motor power
Use this for “Turning steel at 150 m/min, 0.25 mm/rev and 2 mm deep needs 2.8 J/mm³. Find the cutting power, and the motor power at 85 % efficiency.”
Diagram
MRR = vfd = 75 cm³/min.
Depth and feed are enlarged in the detail.
Specific energy u = 2.8 J/mm³.
Cutting power Pc = 3.5 kW.
Motor power Pₘₒₜₒᵣ = 4.12 kW.
This is an example. Type your own values to start a new problem.
MRR (cm³/min)
=
v (m/min)
×
f (mm/rev)
×
d (mm)
Pc (kW)
=
u (J/mm³)
×
MRR (cm³/min)
Pₘₒₜₒᵣ (kW)
=
Pc (kW)
η
Formulas
MRR
=
v
×
f
×
d
75 = 150 × 0.25 × 2
Pc
=
u
×
MRR
60
3.5
=
2.8
×
75
60
Pₘₒₜₒᵣ
=
Pc
η
4.1176
=
3.5
0.85
Assumptions
•
The specific energy u is the work to remove 1 mm³ of this material, read from a table.
•
Pc = u × MRR; with u in J/mm³ and MRR in cm³/min, divide by 60 for kW.
•
The motor loses power in the drive: Pₘₒₜₒᵣ = Pc ÷ η.
Step-by-step
WE KNOW
v = 150 m/min, f = 0.25 mm/rev, d = 2 mm, u = 2.8 J/mm³, η = 0.85
FIND
removal rate (MRR), cutting power (Pc), motor power (Pₘₒₜₒᵣ)
Step 1 · Find removal rate (MRR)
Use MRR = v × f × d
Speed × feed × depth: m/min × mm × mm is 1000 mm³/min, which is 1 cm³/min.
MRR = 150 × 0.25 × 2
MRR = 37.5 × 2
MRR = 75 cm³/min
Step 2 · Find cutting power (Pc)
Use Pc = u × MRR ÷ 60
Energy
for
each
mm³
times
mm³
each
second
is
power:
J/mm³
×
cm³/m
in
60
is
kW.
Pc
=
2.8
×
75
60
Pc
=
210
60
Pc = 3.5 kW
Step 3 · Find motor power (Pₘₒₜₒᵣ)
Use Pₘₒₜₒᵣ = Pc ÷ η
The motor supplies the cutting power plus the drive’s losses.
Pₘₒₜₒᵣ
=
3.5
0.85
Pₘₒₜₒᵣ = 4.1176 kW
Check
Put the numbers back in. Both sides should match.
75 = 150 × 0.25 × 2 ✓
3.5
=
2.8
×
75
60
✓
4.1176
=
3.5
0.85
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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