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

Chapter 8 | 453

Thus,

Integrating, we get

The first term on the right-hand side of the final expression above is rec-
ognized as Uand the second term as the exergy destroyed, whose values
were determined earlier. By substituting those values, the total work input to
the heat pump is determined to be 1.0 Btu, proving our claim. Notice that
the system is still supplied with 20.6 Btu of energy; all we did in the latter
case is replace the 19.6 Btu of valuable work by an equal amount of “use-
less” energy captured from the surroundings.
Discussion It is also worth mentioning that the exergy of the system as a
result of 20.6 Btu of paddle-wheel work done on it has increased by 1.0 Btu
only, that is, by the amount of the reversible work. In other words, if the
system were returned to its initial state, it would produce, at most, 1.0 Btu
of work.

 1 20.619.6 2 Btu
1.0 Btu

mcv,avg 1 T 2 T 12 T 0 mcv,avg ln

T 2

T 1

Wnet,in

2

1

a 1 

T 0

T

bmcv dT

dWnet,in

dQH

COPHP

a 1 

T 0

T

bmcv dT

EXAMPLE 8–13 Dropping a Hot Iron Block into Water

A 5-kg block initially at 350°C is quenched in an insulated tank that con-
tains 100 kg of water at 30°C (Fig. 8–40). Assuming the water that vapor-
izes during the process condenses back in the tank and the surroundings are
at 20°C and 100 kPa, determine (a) the final equilibrium temperature,
(b) the exergy of the combined system at the initial and the final states, and
(c) the wasted work potential during this process.

Solution A hot iron block is quenched in an insulated tank by water. The
final equilibrium temperature, the initial and final exergies, and the wasted
work potential are to be determined.
Assumptions 1 Both water and the iron block are incompressible substances.
2 Constant specific heats at room temperature can be used for both the water
and the iron. 3 The system is stationary and thus the kinetic and potential
energy changes are zero, KE
PE
0. 4 There are no electrical, shaft, or
other forms of work involved. 5 The system is well-insulated and thus there is
no heat transfer.
Analysis We take the entire contents of the tank, wateriron block,as the
system. This is a closed systemsince no mass crosses the system boundary
during the process. We note that the volume of a rigid tank is constant, and
thus there is no boundary work.

WATER

Ti = 30°C

100 kg

T 0 = 20°C

Heat P 0 = 100 kPa

IRON

Ti = 350°C

5 kg

FIGURE 8–40

Schematic for Example 8–13.