Chapter 8 | 455(c) The wasted work potential is equivalent to the exergy destroyed, which can
be determined from XdestroyedT 0 Sgenor by performing an exergy balance on
the system. The second approach is more convenient in this case since the
initial and final exergies of the system are already evaluated.
Net exergy transfer Exergy Change
by heat, work, and mass destruction in exergyDiscussion Note that 219.4 kJ of work could have been produced as the
iron was cooled from 350 to 31.7°C and water was heated from 30 to
31.7°C, but was not.
XdestroyedX 1 X 2 315 95.6219.4 kJ0 XdestroyedX 2 X 1XinXout¬ Xdestroyed¢Xsystem
EXAMPLE 8–14 Exergy Destruction
during Heat Transfer to a Gas
A frictionless piston–cylinder device, shown in Fig. 8–41, initially contains
0.01 m^3 of argon gas at 400 K and 350 kPa. Heat is now transferred to the
argon from a furnace at 1200 K, and the argon expands isothermally until
its volume is doubled. No heat transfer takes place between the argon and
the surrounding atmospheric air, which is at T 0 300 K and P 0 100 kPa.
Determine (a) the useful work output, (b) the exergy destroyed, and (c) the
reversible work for this process.
Solution Argon gas in a piston–cylinder device expands isothermally as a
result of heat transfer from a furnace. The useful work output, the exergy
destroyed, and the reversible work are to be determined.
Assumptions 1 Argon at specified conditions can be treated as an ideal gas
since it is well above its critical temperature of 151 K. 2 The kinetic and
potential energies are negligible.
Analysis We take the argon gascontained within the piston–cylinder device
as the system. This is a closed systemsince no mass crosses the system
boundary during the process. We note that heat is transferred to the system
from a source at 1200 K, but there is no heat exchange with the environment
at 300 K. Also, the temperature of the system remains constant during the
expansion process, and its volume doubles, that is, T 2 T 1 and V 2 2 V 1.
(a) The only work interaction involved during this isothermal process is the
quasi-equilibrium boundary work, which is determined from
This is the total boundary work done by the argon gas. Part of this work is
done against the atmospheric pressure P 0 to push the air out of the way, and
it cannot be used for any useful purpose. It is determined from Eq. 8–3:
WsurrP 01 V 2 V 12 1 100 kPa 231 0.020.01 2 m^3 4a1 kJ
1 kPa#m^3b1 kJ2.43 kPa#m^3 2.43 kJ
WWb
21P dVP 1 V 1 lnV 2
V 1 1 350 kPa 21 0.01 m^32 ln0.02 m^3
0.01 m^3400 K
350 kPaArgonP 0 = 100 kPaT 0 = 300 KFurnace
TR = 1200 K
QRFIGURE 8–41
Schematic for Example 8–14.⎭⎪⎬⎪⎫ ⎭⎪⎬⎪⎫ ⎭⎪⎬⎪⎫