SME
Engineering Thermodynamics · Problem type
Ideal Otto cycle
Use this for “An ideal Otto cycle has a compression ratio of 8, starts at 290 K and takes in 750 kJ/kg of heat. Find T₂, the net work and η.”
Diagram
qₒᵤₜ = qᵢₙ − W = 800 − 468
qₒᵤₜ = 332 kJ/kg
Efficiency = W/qᵢₙ = 468/800 = 58.5%
Carnot limit = 1 − T₁/T₃ = 1 − 300/1,840 = 83.7%
Some heat always goes to the cold reservoir: no engine turns all its heat into work.
This is an example. Type your own values to start a new problem.
T₂ (K)
=
T₁ (K)
×
r
0.4
qᵢₙ (kJ/kg)
=
0.718
×
(
T₃ (K)
−
T₂ (K)
)
η
=
1
−
1
÷
r
0.4
T₄
End of expansion temperature
calculated · K
qₒᵤₜ
Heat out
calculated · kJ/kg
wₙₑₜ
Net work
calculated · kJ/kg
Formulas
T₂
=
T₁
×
r
0.4
722.4674
=
300
×
9
0.4
qᵢₙ
=
0.718
×
(
T₃
−
T₂
)
800 = 0.718 × (1,836.6735 − 722.4674)
T₄
=
T₃
r
0.4
762.667
=
1,836.6735
9
0.4
qₒᵤₜ
=
0.718
×
(
T₄
−
T₁
)
332.1949 = 0.718 × (762.667 − 300)
wₙₑₜ = qᵢₙ − qₒᵤₜ
467.8051 = 800 − 332.1949
η
=
1
−
1
r
0.4
0.5848
=
1
−
1
9
0.4
Assumptions
•
Cold-air standard: cᵥ = 0.718 kJ/(kg·K) and k = 1.4, both constant.
•
Compression and expansion are isentropic; heat goes in and out at constant volume.
•
The compression ratio is r = V₁ ÷ V₂; heats and works are per kilogram of air.
Step-by-step
WE KNOW
r = 9, T₁ = 300 K, qᵢₙ = 800 kJ/kg
FIND
end of compression temperature (T₂), peak temperature (T₃), end of expansion temperature (T₄), heat out (qₒᵤₜ), net work (wₙₑₜ), thermal efficiency (η)
Step 1 · Find end of compression temperature (T₂)
Use T₂ = T₁ × r0.4
Isentropic
compression:
T
rises
by
the
compression
ratio
r
to
the
power
k
−
1.
T₂
=
300
×
9
0.4
T₂
=
300
×
2.408
T₂
=
722.4674
K
Step 2 · Find peak temperature (T₃)
Use qᵢₙ = 0.718 × (T₃ − T₂)
Divide
qᵢₙ
by
cᵥ,
then
add
T₂
.
T₃
=
T₂
+
qᵢₙ
0.718
T₃
=
722.4674
+
800
0.718
T₃
=
722.4674
+
1,114
T₃
=
1,836.674
K
Step 3 · Find end of expansion temperature (T₄)
Use T₄ = T₃ ÷ r0.4
Isentropic
expansion
through
the
same
ratio:
divide
T₃
by
r
to
the
power
k
−
1.
T₄
=
1,836.674
9
0.4
T₄
=
1,836.674
2.408
T₄
=
762.667
K
Step 4 · Find heat out (qₒᵤₜ)
Use qₒᵤₜ = 0.718 × (T₄ − T₁)
Heat
leaves
at
constant
volume:
cᵥ
times
the
drop
from
T₄
back
to
T₁
.
qₒᵤₜ = 0.718 × (762.667 − 300)
qₒᵤₜ = 0.718 × 462.667
qₒᵤₜ = 332.1949 kJ/kg
Step 5 · Find net work (wₙₑₜ)
Use wₙₑₜ = qᵢₙ − qₒᵤₜ
Over a whole cycle ΔU = 0, so the net work is the heat in less the heat out.
wₙₑₜ = 800 − 332.1949
wₙₑₜ = 467.8051 kJ/kg
Step 6 · Find thermal efficiency (η)
Use η = 1 − 1 ÷ r0.4
The ideal Otto efficiency depends on the compression ratio alone.
η
=
1
−
1
9
0.4
η
=
1
−
1
2.408
η
=
1
−
0.4153
η
=
0.5848
Check
Put the numbers back in. Both sides should match.
722.4674
=
300
×
9
0.4
✓
800 = 0.718 × (1,836.674 − 722.4674) ✓
762.667
=
1,836.674
9
0.4
✓
332.1949 = 0.718 × (762.667 − 300) ✓
467.8051 = 800 − 332.1949 ✓
0.5848
=
1
−
1
9
0.4
✓
Topic check
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