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Modern Engineering Thermodynamics

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10.11 Energy Efficiency Based on the Second Law 347<br />

and, if T 0 = T H , the second law efficiency of a refrigerator or air conditioner that rejects heat to the local environment<br />

reduces to<br />

Second law efficiency of a refrigerator or air conditioner that rejects heat to the local environment<br />

When T 0 = T H , ε R/AC<br />

=<br />

COP actual<br />

ref or air cond<br />

COP Carnot<br />

ref or air cond<br />

(10.35)<br />

where the coefficient of performance of a Carnot refrigerator or air conditioner given in Eq. (7.20) is used. The following<br />

example illustrates the use of this material.<br />

EXAMPLE 10.14<br />

A common window air conditioner has an actual COP of 8.92. If the inside temperature is T 0 = T L = 20.0°C and the outside<br />

temperature is T H = 35.0°C, determine the second law availability efficiency of this air conditioner.<br />

Solution<br />

First draw a sketch of the system (Figure 10.19).<br />

The unknown is the second law availability efficiency of this air conditioner.<br />

Equation (7.20) gives the coefficient of performance of a Carnot refrigerator or<br />

air conditioner as<br />

COP = 8.92<br />

COP Carnot =<br />

ref: or air cond:<br />

T L<br />

T H − T L<br />

=<br />

20:0 + 273:15 K<br />

ð35:0 + 273:15 KÞ − ð20:0 + 273:15 KÞ = 19:6<br />

then, Eq. (10.35) can be used to determine the second law efficiency as<br />

T inside = 20.0°C<br />

Window<br />

air<br />

conditioner<br />

T outside = 35.0°C<br />

ε R/AC<br />

=<br />

COP actual<br />

ref: or air cond:<br />

COP Carnot<br />

ref: or air cond:<br />

= 8:92 = 0:455 = 45:5%<br />

19:6<br />

Exercises<br />

41. Determine the second law thermal efficiency of the air conditioner<br />

in Example 10.14 if the outside air temperature is increased from<br />

35.0°C to40.0°C. Assume all the other variables remain unchanged.<br />

Answer: ε R/AC<br />

= 60.9%.<br />

FIGURE 10.19<br />

Example 10.14.<br />

42. If the inside air temperature in Example 10.14 increases from 20.0°C to 22.0°C, determine the new second<br />

law thermal efficiency of the window air conditioner. Assume all the other variables remain unchanged.<br />

Answer: ε R/AC<br />

= 39.3%.<br />

43. When the outside temperature in Example 10.14 increases from 35.0°C to 40.0°C, the actual COP of the air<br />

conditioner decreases from 8.92 to 7.5. Determine the new second law thermal efficiency of the air conditioner.<br />

Answer: ε R/AC<br />

= 51.2%.<br />

When we deal with heat exchangers in which the fluids do not mix, the desirable result is the increase in<br />

temperature of the cold stream (we could choose the desired result to be the decrease in temperature of the hot<br />

stream if we wished). Thus, the corresponding _A desired result is the increase in the available energy rate of the cold<br />

stream, or from Figure 10.20a,<br />

_A desirable result = _m C ða f 4 − a f 3 Þ<br />

and the source availability is the decrease in the available energy rate of the hot stream:<br />

_A initial or = _m H ða f 1 − a f 2 Þ<br />

net input<br />

Then, the second law availability efficiency for a nonmixing heat exchanger is the ratio of these two terms:<br />

Second law efficiency of a nonmixing heat exchanger<br />

ε nonmixing HX = _m Cða f 4 − a f 3 Þ<br />

_m H ða f 1 − a f 2 Þ<br />

(10.36)

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