Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 8 | 451

(d) Noting that the decrease in the exergy of the steam is the exergy sup-
plied and the useful work output is the exergy recovered, the second-law effi-
ciency for this process can be determined from

That is, 44.8 percent of the work potential of the steam is wasted during
this process.

hII

Exergy recovered

Exergy supplied

Wu

X 1 X 2

5.3

35.025.4

0.552 or 55.2%

EXAMPLE 8–12 Exergy Destroyed during Stirring of a Gas

An insulated rigid tank contains 2 lbm of air at 20 psia and 70°F. A paddle
wheel inside the tank is now rotated by an external power source until the
temperature in the tank rises to 130°F (Fig. 8–38). If the surrounding air is
at T 0
70°F, determine (a) the exergy destroyed and (b) the reversible work
for this process.

Solution The air in an adiabatic rigid tank is heated by stirring it by a pad-
dle wheel. The exergy destroyed and the reversible work for this process are to
be determined.
Assumptions 1 Air at about atmospheric conditions can be treated as an
ideal gas with constant specific heats at room temperature. 2 The kinetic
and potential energies are negligible. 3 The volume of a rigid tank is con-
stant, and thus there is no boundary work. 4 The tank is well insulated and
thus there is no heat transfer.
Analysis We take the aircontained within the tank as the system. This is a
closed system since no mass crosses the system boundary during the
process. We note that shaft work is done on the system.
(a) The exergy destroyed during a process can be determined from an exergy
balance, or directly from Xdestroyed
T 0 Sgen. We will use the second approach
since it is usually easier. But first we determine the entropy generated from
an entropy balance,

Net entropy transfer Entropy Change

by heat and mass generation in entropy

Taking cv
0.172 Btu/lbm · °F and substituting, the exergy destroyed
becomes

19.6 Btu

1 530 R 21 2 lbm 21 0.172 Btu>lbm#°F 2 ln

590 R

530 R

Xdestroyed
T 0 Sgen
T 0 mcv ln

T 2

T 1

Sgen
mcv ln

T 2

T 1

0 Sgen
¢Ssystem
m°cv ln

T 2

T 1

R ln

V 2

V 1

¢

SinSout¬ Sgen
¢Ssystem

AIR

m = 2 lbm

T 1 = 70°F

P 1 = 20 psia

T 0 = 70°F

Wpw

FIGURE 8–38

Schematic for Example 8–12.

0

→

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