Modern Engineering Thermodynamics

218 CHAPTER 7: Second Law of Thermodynamics and Entropy Transport and Production Mechanisms
CRITICAL THINKING
Notice that, the smaller the difference between T H and T L , the larger the COP defined in Eqs. (7.18) and (7.20) becomes. To
understand this on a physical basis, note that, since the numerator is a fixed value (Q L or Q H ) in both equations and the
denominator represents the net work input (W in ) to the device, it stands to reason that, the smaller is the work input for a
given output, the better the efficiency (i.e., COP) is.
Comparing Eqs. (7.17) and (7.19), we see that
COP heat pump = COP refrig: or
+ 1
air cond:
EXAMPLE 7.3
The temperature outside on a hot summer day is 95°F. You would like your room to
be at 70.°F, so you go out shopping for an air conditioner. If you are going to buy a
Carnot air conditioner, what should be its coefficient of performance?
(COP) = ?
Carnot
AC
Solution
First, draw a sketch of the system (Figure 7.10).
95.°F
Carnot
air
conditioner
70.°F
The unknown is the coefficient of performance of an air conditioner operating between
70.°F and 95°F. Equation (7.20) gives the coefficient of performance for a Carnot air
conditioner as
COP Carnot
=
refrig: or
air cond:
T L
T H − T L
=
70: + 459:67
ð95 + 459:67Þ − ð70: + 459:67Þ = 21
Note that, in the denominator, the temperature difference in this equation can be in
either relative or absolute temperature units because (95 + 459.67) − (70. + 459.67) =
95 − 70. = 25°F or R. The degree size is the same for both the absolute and the relative
temperature scales.
FIGURE 7.10
Example 7.3.
The coefficient of performance of a real air conditioner is sometimes called its energy efficiency rating (EER) and usually ranges
between 3 and 9. The value of 21 calculated in Example 7.3 is unreasonably high, because a Carnot air conditioner is reversible
and has no friction or other internal losses. Therefore, it requires less input work than a real air conditioner.
Exercises
5. The refrigerator in your kitchen maintains a temperature difference of 2.00°C inside when the outside kitchen
temperature is 22.0°C. If it is a Carnot refrigerator, what is its coefficient of performance? Answer: (COP) Carnot ref. = 13.8.
6. If the Carnot air conditioner purchased in Example 7.3 is inserted into the window backward during the winter and
operated as a heat pump, determine its coefficient of performance as a heat pump. Answer: (COP) Carnot HP = 22.
7.7 CLAUSIUS’S DEFINITION OF ENTROPY
Rudolph Clausius extended Thomson’s absolute temperature scale work by rearranging Eq. (7.15) to read
ðQ H Þ rev
= jQ Lj rev
T H T L
and since this applies only to a closed system undergoing a thermodynamic cycle, it can also be written as
∑
cycle
Q
T
= ðQ HÞ rev
T H
− jQ Lj rev
T L
= 0
If we now take an arbitrary thermodynamic closed system cycle and overlay it with an infinite number of
infinitesimal heat engine cycles, as shown in Figure 7.11, then we can extend the finite summation process of
the previous equation into a cyclic integral. Also, since each of these infinite number of heat engines is now