SME
Kinematics of Mechanisms · Problem type
Torque and power through a train
Use this for “A gearbox takes 50 N·m at 1800 rpm and turns its output at 200 rpm with 95% efficiency. Find the output torque and the power.”
Chart
Pₒᵤₜ = P − PL
9.425 − 0.4712 = 8.954
The flows in add and the flows out take away: Pₒᵤₜ = 8.954.
Input torque Tᵢₙ = 50 N·m
Output torque Tₒᵤₜ = 427.5 N·m
Efficiency η = 95%
This is an example. Type your own values to start a new problem.
Tₒᵤₜ (N·m)
=
η
%
×
Tᵢₙ (N·m)
×
nᵢₙ (rpm)
nₒᵤₜ (rpm)
The bars are the power the shafts carry: P = Tᵢₙ ωᵢₙ goes in, PL = (1 − η)P is lost as heat, and Pₒᵤₜ = Tₒᵤₜ ωₒᵤₜ comes out, ω being each speed in rad/s.
P
Input power
calculated · kW
Pₒᵤₜ
Output power
worked out (not typed) · kW
PL
Power lost
worked out (not typed) · kW
Formulas
P
=
Tᵢₙ
×
2π
×
nᵢₙ
60,000
9.4248
=
50
×
2π
×
1,800
60,000
Tₒᵤₜ
=
(
η
100
)
×
Tᵢₙ
×
(
nᵢₙ
nₒᵤₜ
)
427.5
=
(
95
100
)
×
50
×
(
1,800
200
)
Pₒᵤₜ
=
(
η
100
)
×
P
8.9535
=
(
95
100
)
×
9.4248
PL
=
(
100 −
η
100
)
×
P
0.4712
=
(
100 − 95
100
)
×
9.4248
Assumptions
•
Steady running: no shaft speeds up or slows down.
•
The losses, the share 1 − η of the power, leave as heat.
•
The output speed comes from the train value, nₒᵤₜ = e nᵢₙ.
Step-by-step
WE KNOW
Tᵢₙ = 50 N·m, nᵢₙ = 1,800 rpm, nₒᵤₜ = 200 rpm, η = 95%
FIND
input power (P), output torque (Tₒᵤₜ), output power (Pₒᵤₜ), power lost (PL)
Step 1 · Find input power (P)
Use P = Tᵢₙ × 2π × nᵢₙ ÷ 60,000
Power
is
torque
times
the
shaft
speed
in
rad/s
(2π
×
n
60
);
÷
1000
gives
kW.
P
=
50
×
2π
×
1,800
60,000
P
=
9.42478
kW
Step 2 · Find output torque (Tₒᵤₜ)
Use Tₒᵤₜ = (η ÷ 100) × Tᵢₙ × (nᵢₙ ÷ nₒᵤₜ)
Power
out
is
η
times
power
in,
so
the
torque
grows
as
the
speed
drops,
less
the
losses.
Tₒᵤₜ
=
(
95
100
)
×
50
×
(
1,800
200
)
Tₒᵤₜ
=
95
100
×
50
×
9
Tₒᵤₜ = 0.95 × 50 × 9
Tₒᵤₜ = 47.5 × 9
Tₒᵤₜ = 427.5 N·m
Step 3 · Find output power (Pₒᵤₜ)
Use Pₒᵤₜ = (η ÷ 100) × P
The
share
η
of
the
input
power
reaches
the
output
shaft.
Pₒᵤₜ
=
(
95
100
)
×
9.42478
Pₒᵤₜ = 0.95 × 9.4248
Pₒᵤₜ = 8.9535 kW
Step 4 · Find power lost (PL)
Use PL = ((100 − η) ÷ 100) × P
The rest of the power is lost to friction and churning, as heat.
PL
=
(
100 − 95
100
)
×
9.42478
PL
=
5
100
×
9.4248
PL = 0.05 × 9.4248
PL = 0.4712 kW
Check
Put the numbers back in. Both sides should match.
9.42478
=
50
×
2π
×
1,800
60,000
✓
427.5
=
(
95
100
)
×
50
×
(
1,800
200
)
✓
8.9535
=
(
95
100
)
×
9.42478
✓
0.4712
=
(
100 − 95
100
)
×
9.42478
✓
Topic check
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