1. So, we now know that the efficiency of Carnot Engine is:
Let’s take some realistic numbers:
~ 600 K is the temperature of the
so-called “superheated steam”
typically used for driving steam
turbines in power plants. Ambient
air or river water are typically
used as “sinks”.
Hot source:
600 K
Heat sink:
300 K

2. 50% of the thermal energy is converted to work.
The other 50% is wasted….
One can say: well, 50% is not spectacular, but still
not so bad!
Unfortunately – the truth is even sadder. The efficiency
of the Carnot Cycle is the highest possible theoretical
efficiency that a thermal engine can attain – because
it is an ideally reversible cycle.
The operation of all practical thermal engines
necessarily involves irreversible processes, which even
further lowers the efficiency.
Let’s demonstrate that, using as an example something
we can call a “PRACTICAL CARNOT ENGINE”

3. In the Carnot Cycle, heat periodically
is transferred to the gas, and then
out of the gas (here, through the
cylinder “front plate”).
Heat can flow only if the temperatures
at both sides of the plate are different.
The heat transferred through the plate
over a time period t is:
Where σ if the thermal conduction
coefficient.
Then, the thermal power transferred
across the plate is:

4. If we want the temperatures of the isothermal processes to be the same
as these of the “hot source” and the “heat sink”, as illustrated below:
In such situation, T = 0, and no thermal power is transferred to the
gas and out of the gas. The cycle will take an infinitely long time! –
meaning that NO POWER is delivered!

5. So, in order to enable the heat to be transferred, let’s lower the
temperature in the isothermal expansion by, say, 50 K, and let’s
increase the temperature in the isothermal compression by the
same value:
Now heat power is trans-
ferred and converted to
mechanical power, but
note that the the conver-
sion efficiency is no longer
50%, but:

6. From the preceding page:
Perhaps T = 100 K is a better
option?

7. The T = 100 K option seems to provide a slightly higher power
output than the T = 50 K one.
But perhaps there is even a better option? We have to investigate
the function:
to find the T value for which it has a maximum.
One can use the standard analytical procedure, but it is far more
Instructive to use a graphical method (next slide).

8. The blue curve is the product of the “red” and the “green” functions
(out of scale). It has a maximum at about 81 K.
Efficiency
(left scale)
Thermal power
input

9. So, always keep in mind that the famous Carnot’s
However, the “heat-to-work” conversion efficiency is then only:
Nearly two times less than the maximum theoretical efficiency!
It means that 73.4%, nearly ¾ of the input energy, goes
to the sink, and only slightly more than ¼ is converted
to usable work.
So, always keep in mind that the famous Carnot’s
formula gives only the maximum effi-
ciency allowed by the Second Law of Thermodynamics,
but not that attainable in practical engines!

10. A thorough analysis of the “practical Carnot Engine” based on the
assumption that one can use two different ΔT values, yields a result
for the maximum attainable efficiency of a Carnot engine designed to
deliver maximum power. It’s a remarkably simple formula, look!
Instead of the Tc/Th ratio, you just take its square root, that’s all.
The text to the
left is copied
from a well-
known thermal
physics text by
Herbert Callen.
In the table,
efficiencies cal-
culated using
the (4.29) equa-
tion are com-
pared with data
from real power
plants.