SME
Engineering Thermodynamics · Problem type
Ideal Diesel cycle
Use this for “An ideal Diesel cycle has a compression ratio of 20 and a cutoff ratio of 2.2, starting at 290 K. Find the heat in, the net work and η.”
Diagram
qₒᵤₜ = qᵢₙ − W = 958 − 605
qₒᵤₜ = 353 kJ/kg
Efficiency = W/qᵢₙ = 605/958 = 63.2%
Carnot limit = 1 − T₁/T₃ = 1 − 300/1,910 = 84.3%
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
T₃ (K)
=
rc
×
T₂ (K)
η
=
1
−
(1
÷
r
0.4
)
×
(
rc
1.4
−
1)
÷
(1.4
×
(
rc
−
1))
T₄
End of expansion temperature
calculated · K
qᵢₙ
Heat in
calculated · kJ/kg
qₒᵤₜ
Heat out
calculated · kJ/kg
wₙₑₜ
Net work
calculated · kJ/kg
Formulas
T₂
=
T₁
×
r
0.4
953.3015
=
300
×
18
0.4
T₃
=
rc
×
T₂
1,906.6029 = 2 × 953.3015
T₄
=
T₃
×
(
rc
r
)
0.4
791.7047
=
1,906.6029
×
(
2
18
)
0.4
qᵢₙ
=
1.005
×
(
T₃
−
T₂
)
958.068 = 1.005 × (1,906.6029 − 953.3015)
qₒᵤₜ
=
0.718
×
(
T₄
−
T₁
)
353.044 = 0.718 × (791.7047 − 300)
wₙₑₜ = qᵢₙ − qₒᵤₜ
605.024 = 958.068 − 353.044
η
=
1
−
(
1
r
0.4
)
×
rc
1.4
− 1
1.4 × (rc − 1)
0.6316
=
1
−
(
1
18
0.4
)
×
2
1.4
− 1
1.4 × (2 − 1)
Assumptions
•
Cold-air standard: cₚ = 1.005 and cᵥ = 0.718 kJ/(kg·K), k = 1.4, all constant.
•
Compression and expansion are isentropic; heat goes in at constant pressure, out at constant volume.
•
r = V₁ ÷ V₂ and the cutoff ratio rc = V₃ ÷ V₂; heats and works are per kilogram.
Step-by-step
WE KNOW
r = 18, rc = 2, T₁ = 300 K
FIND
end of compression temperature (T₂), end of burning temperature (T₃), end of expansion temperature (T₄), heat in (qᵢₙ), 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
×
18
0.4
T₂
=
300
×
3.178
T₂
=
953.30146
K
Step 2 · Find end of burning temperature (T₃)
Use T₃ = rc × T₂
At
constant
pressure
T
grows
with
V,
so
T₃
T₂
=
V₃
V₂
=
rc.
T₃
=
2
×
953.30146
T₃
=
1,906.6029
K
Step 3 · Find end of expansion temperature (T₄)
Use T₄ = T₃ × (rc ÷ r)0.4
Isentropic
expansion
from
V₃
to
V₁:
the
volume
grows
r
rc
times.
T₄
=
1,906.6029
×
(
2
18
)
0.4
T₄
=
1,906.6029
×
0.1111
0.4
T₄
=
1,906.6029
×
0.4152
T₄
=
791.7047
K
Step 4 · Find heat in (qᵢₙ)
Use qᵢₙ = 1.005 × (T₃ − T₂)
Fuel
burns
at
constant
pressure:
cₚ
times
the
rise
from
T₂
to
T₃
.
qᵢₙ = 1.005 × (1,906.6029 − 953.30146)
qᵢₙ = 1.005 × 953.3
qᵢₙ = 958.068 kJ/kg
Step 5 · 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 × (791.7047 − 300)
qₒᵤₜ = 0.718 × 491.7047
qₒᵤₜ = 353.044 kJ/kg
Step 6 · 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 = 958.068 − 353.044
wₙₑₜ = 605.024 kJ/kg
Step 7 · Find thermal efficiency (η)
Use η = 1 − (1 ÷ r0.4) × (rc1.4 − 1) ÷ (1.4 × (rc − 1))
The
Otto
loss,
1
r
to
the
power
k
−
1,
made
larger
by
the
cutoff
factor.
η
=
1
−
(
1
18
0.4
)
×
2
1.4
− 1
1.4 × (2 − 1)
1
18
0.4
=
0.3147
2
1.4
− 1
1.4 × (2 − 1)
=
1.639
1.4
=
1.1707
η
=
1
−
0.3147
×
1.1707
=
0.6316
η
=
0.6316
Check
Put the numbers back in. Both sides should match.
953.30146
=
300
×
18
0.4
✓
1,906.6029 = 2 × 953.30146 ✓
791.7047
=
1,906.6029
×
(
2
18
)
0.4
✓
958.068 = 1.005 × (1,906.6029 − 953.30146) ✓
353.044 = 0.718 × (791.7047 − 300) ✓
605.024 = 958.068 − 353.044 ✓
0.6316
=
1
−
(
1
18
0.4
)
×
2
1.4
− 1
1.4 × (2 − 1)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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