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

Chapter 9 | 529

EXAMPLE 9–10 Second-Law Analysis of an Otto Cycle

Determine the exergy destruction associated with the Otto cycle (all four
processes as well as the cycle) discussed in Example 9–2, assuming that
heat is transferred to the working fluid from a source at 1700 K and heat is
rejected to the surroundings at 290 K. Also, determine the exergy of the
exhaust gases when they are purged.

Solution The Otto cycle analyzed in Example 9–2 is reconsidered. For spec-
ified source and sink temperatures, the exergy destruction associated with the
cycle and the exergy purged with the exhaust gases are to be determined.
Analysis In Example 9–2, various quantities of interest were given or deter-
mined to be

Processes 1-2 and 3-4 are isentropic (s 1
s 2 , s 3
s 4 ) and therefore do
not involve any internal or external irreversibilities; that is, Xdest,12
0 and
Xdest,34
0.
Processes 2-3 and 4-1 are constant-volume heat-addition and heat-rejection
processes, respectively, and are internally reversible. However, the heat trans-
fer between the working fluid and the source or the sink takes place through a
finite temperature difference, rendering both processes irreversible. The
exergy destruction associated with each process is determined from Eq. 9–32.
However, first we need to determine the entropy change of air during these
processes:

Also,

Thus,

For process 4-1, s 1 s 4
s 2 s 3
0.7540 kJ/kg · K, qR,41
qout

381.83 kJ/kg, and Tsink
290 K. Thus,

xdest,41
T 0 c1s 1 s 42 sys

qout

Tsink

d

82.2 kJ>kg

1 290 K2c0.7540 kJ>kg#K

800 kJ>kg

1700 K

d

xdest,23
T 0 c1s 3 s 22 sys

qin

Tsource

d

qin
800 kJ>kg¬and¬Tsource
1700 K

0.7540 kJ>kg#K

1 3.50452.4975 2 kJ>kg#K 1 0.287 kJ>kg#K 2 ln

4.345 MPa

1.7997 MPa

s 3 s 2 
s° 3 s° 2 R ln

P 3

P 2

T 3
1575.1 K¬¬ wnet
418.17 kJ>kg

T 2
652.4 K¬¬ qout
381.83 kJ>kg

T 1
290 K¬¬ qin
800 kJ>kg

T 0
290 K¬¬ P 3
4.345 MPa

r
8 ¬¬ P 2
1.7997 MPa