Page 134 - Advanced Thermodynamics for Engineers, Second Edition

6.1 GENERAL CONSIDERATIONS 121

_
It is possible to manipulate these equations to give W in terms of T H , T C , U H A H and the ratio
T 2 /T 1 ¼ s.
From Eqn (6.4),
Q _ C
T 2 ¼ þ T C (6.8)
U C A C
and, from Eqn (6.6),

( )
Q _ H Q _ H Q _ C
T 1 ¼ T 2 ¼ þ T C ;
Q _ C Q _ C U C A C
( )
1 Q _ C
¼ þ T C : (6.9)
s U C A C
Hence,
( _ ) _
_ U H A H Q C U H A H Q H U H A H T C
Q ¼ U H A H T H s U C A C þ T C ¼ U H A H T H U C A C s :
H
Rearranging gives

8 9
T C
> T C 1 > s
< 1
>
>
Q _ H T H s = T H
¼ ¼ ; (6.10)
U H A H T H > U H A H> U H A H
:1 þ > s 1 þ
>
;
U C A C U C A C
and hence,

T C
W _ s T H
ð1 sÞ: (6.11)
¼
U H A H T H U H A H
s 1 þ
U C A C
_
The nondimensional heat addition from the hot reservoir, Q =U H A H T H , and the nondimensional
H
_
power output, W=U H A H T H , can both be calculated as a function of s for given values of T C /T H and
_
_
U H A H /U C A C . The thermal efficiency, h, of the heat engine can also be evaluated as W=Q. Figure 6.2
shows how these parameters vary for T H /T C ¼ 5.0, and two values of U H A H /U C A C ¼ 1.0 and 2.0. It
can be readily seen that the actual power output reaches a maximum value for a particular value of s.
Also shown is that the efficiency of the engine, which is the same for both values of U H A H /U C A C ,
decreases from the Carnot efficiency of 80% at s ¼ T C /T H to 0 at s ¼ 0. The efficiency at maximum
power is 55.3% and this occurs at a value of s ¼ 0.447. This graph confirms that a Carnot cycle has
the highest achievable efficiency but produces zero power output, because when s ¼ T C /T H this
means that the heat transfer is externally reversible, but infinitesimally slow. It would be interesting
to evaluate the efficiency of the engine when it produces its maximum power output, and this will be
done below.