SME
Engineering Thermodynamics
Second law and entropy
Use this for “An engine takes 1200 kJ from a 900 K furnace and does 450 kJ of work, rejecting heat at 320 K. Find its efficiency, the Carnot limit and the entropy generated.”
Diagram
QL = QH − W = 1,000 − 400
QL = 600 kJ
Efficiency = W/QH = 400/1,000 = 40%
Carnot limit = 1 − TL/TH = 1 − 300/800 = 62.5%
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.
QH (kJ)
=
W (kJ)
+
QL (kJ)
ηC
%
=
100%
×
(1
−
TL (K)
÷
TH (K)
)
Sgen (kJ/K)
=
QL (kJ)
÷
TL (K)
−
QH (kJ)
÷
TH (K)
η
Thermal efficiency
calculated · %
Formulas
QH
=
W
+
QL
1,000 = 400 + 600
η
=
100
×
W
QH
40
=
100
×
400
1,000
ηC
=
100
×
(1
−
TL
TH
)
62.5
=
100
×
(1
−
300
800
)
Sgen
=
QL
TL
−
QH
TH
0.75
=
600
300
−
1,000
800
Assumptions
•
The engine runs in a cycle: its own energy and entropy return to the start each cycle.
•
Reservoirs stay at TH and TL in kelvins, however much heat they give or take.
•
Heat in is QH, heat out is QL, both counted as positive amounts.
•
Sgen ≥ 0 for every real engine; Sgen = 0 only for a reversible (Carnot) one.
Step-by-step
WE KNOW
TH = 800 K, TL = 300 K, QH = 1,000 kJ, W = 400 kJ
FIND
heat out to the cold reservoir (QL), thermal efficiency (η), Carnot efficiency (ηC), entropy generated (Sgen)
Step 1 · Find heat out to the cold reservoir (QL)
Use QH = W + QL
First law over a cycle: the heat not turned into work goes to the cold reservoir.
QL
=
QH
−
W
QL = 1,000 − 400
QL = 600 kJ
Step 2 · Find thermal efficiency (η)
Use η = 100 × W ÷ QH
Efficiency is the share of the heat in that comes out as work, as a percent.
η
=
100
×
400
1,000
η
=
40,000
1,000
η
=
40%
Step 3 · Find Carnot efficiency (ηC)
Use ηC = 100 × (1 − TL ÷ TH)
No
engine
between
these
two
reservoirs
can
beat
the
Carnot
efficiency
1
−
TL
TH
.
ηC
=
100
×
(1
−
300
800
)
ηC = 100 × (1 − 0.375)
ηC = 100 × 0.625
ηC = 62.5%
Step 4 · Find entropy generated (Sgen)
Use Sgen = QL ÷ TL − QH ÷ TH
The
cold
reservoir
gains
QL
TL
of
entropy
and
the
hot
one
loses
QH
TH
;
the
engine’s
own
returns
each
cycle.
Sgen
=
600
300
−
1,000
800
Sgen = 2 − 1.25
Sgen = 0.75 kJ/K
Check
Put the numbers back in. Both sides should match.
1,000 = 400 + 600 ✓
40
=
100
×
400
1,000
✓
62.5
=
100
×
(1
−
300
800
)
✓
0.75
=
600
300
−
1,000
800
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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